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Existence and blow-up of solutions in Hénon-type heat equation with exponential nonlinearity

Existence and blow-up of solutions in Hénon-type heat equation with exponential nonlinearity 1IntroductionIn this article, we study the Hénon-type heat equation with exponential growth: (1.1)vt=Δv+∣x∣βev,x∈RN,t>0,v(x,0)=v0(x),x∈RN,\left\{\phantom{\rule[-1.25em]{}{0ex}}\begin{array}{ll}{v}_{t}=\Delta v+| x{| }^{\beta }{e}^{v},\hspace{1.0em}& x\in {{\mathbb{R}}}^{N},\hspace{1em}t\gt 0,\\ v\left(x,0)={v}_{0}\left(x),\hspace{1.0em}& x\in {{\mathbb{R}}}^{N},\end{array}\right.where N≥3N\ge 3, 0<β<20\lt \beta \lt 2, and v0∈C(RN){v}_{0}\in C\left({{\mathbb{R}}}^{N}). In this whole article, we make the assumption that there exist constants σ∈(0,2)\sigma \in \left(0,2)and C>0C\gt 0, such that (1.2)−Ce∣x∣2−σ≤v0(x)≤CinRN.-C{e}^{| x{| }^{2-\sigma }}\le {v}_{0}\left(x)\le C\hspace{1.0em}{\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N}.We denote T=T(v0)>0T=T\left({v}_{0})\gt 0by the maximal existence time of the unique classical solution of problem (1.1). It is obvious that under the constraint (1.2), the solution vvof (1.1) on RN{{\mathbb{R}}}^{N}is always bounded below as long as vvexists. If T(v0)<∞T\left({v}_{0})\lt \infty , then limt→T(v0)supx∈RNv(t,x)=+∞{\mathrm{lim}}_{t\to T\left({v}_{0})}{\sup }_{x\in {{\mathbb{R}}}^{N}}v\left(t,x)=+\infty . We say that such solution vvblows up in finite time and its blow-up time is T(v0)T\left({v}_{0}). Otherwise, we call that vvis the global solution if T=∞T=\infty .To make a clear statement of results for (1.1) with exponential nonlinearity, we begin by reviewing some established findings with regard to the global existence and blow-up of the following heat equation: (1.3)ut−Δu=∣x∣β∣u∣p−1u,x∈RN,t>0,u(x,0)=u0(x),x∈RN,\left\{\phantom{\rule[-1.25em]{}{0ex}}\begin{array}{ll}{u}_{t}-\Delta u=| x{| }^{\beta }| u{| }^{p-1}u,\hspace{1.0em}& x\in {{\mathbb{R}}}^{N},\hspace{1em}t\gt 0,\\ u\left(x,0)={u}_{0}\left(x),\hspace{1.0em}& x\in {{\mathbb{R}}}^{N},\end{array}\right.where p>1p\gt 1. It can be proved that for β≠0\beta \ne 0, N≥3N\ge 3, p>N+βN−2p\gt \frac{N+\beta }{N-2}, the function ϕ∗(x)=Bp,N,β1p−1∣x∣−2+βp−1,Bp,N,β=2+βp−1N−2−2+βp−1.{\phi }_{\ast }\left(x)={B}_{p,N,\beta }^{\frac{1}{p-1}}| x{| }^{-\tfrac{2+\beta }{p-1}},\hspace{1em}{B}_{p,N,\beta }=\frac{2+\beta }{p-1}\left(N-2-\frac{2+\beta }{p-1}\right).is a positive singular stationary solution of (1.3). We will find that whether the solutions of the problem (1.3) exist globally in time or blow up in finite time strongly depended on the decay rate of u0{u}_{0}. Precisely, the critical decay rate of the u0{u}_{0}is given by ∣x∣−2+βα−1| x{| }^{-\tfrac{2+\beta }{\alpha -1}}.We start by taking the case β=0\beta =0in (1.3) into account. Let pF=1+2/N{p}_{F}=1+2\hspace{0.1em}\text{/}\hspace{0.1em}Nbe the Fujita exponent (see [9,10]). It has been proved that all solutions of (1.3) with u0≥0{u}_{0}\ge 0, u0≢0{u}_{0}\not\equiv 0, blow up in finite time if and only if p≤pFp\le {p}_{F}(see [10,11]). On the other hand, for sufficient small initial data, there exist global in time solutions of (1.3) when p>pFp\gt {p}_{F}. Lee and Ni [20] proved that if p>pFp\gt {p}_{F}and u0(x)=λψ(x)(λ>0){u}_{0}\left(x)=\lambda \psi \left(x)\left(\lambda \gt 0), then the solution uuof (1.3) exists globally in time (T(u0)=∞)\left(T\left({u}_{0})=\infty )for λ>0\lambda \gt 0small enough, where ψ≢0\psi \not\equiv 0is a nonnegative function satisfying limsup∣x∣→∞∣x∣2p−1ψ(x)<∞{\mathrm{limsup}}_{| x| \to \infty }| x{| }^{\tfrac{2}{p-1}}\psi \left(x)\lt \infty . On the contrary, if liminf∣x∣→∞∣x∣2p−1ψ(x){\mathrm{liminf}}_{| x| \to \infty }| x{| }^{\tfrac{2}{p-1}}\psi \left(x)large enough, then the solution uuof (1.3) blows up in finite time (T(u0)<∞)\left(T\left({u}_{0})\lt \infty ). Equation (1.3) has one parameter family {Vτ}τ>0{\left\{{V}_{\tau }\right\}}_{\tau \gt 0}of radially symmetric regular stationary solutions for the supercritical case p≥N+2N−2p\ge \frac{N+2}{N-2}with N≥3N\ge 3, where each Vτ{V}_{\tau }satisfies lim∣x∣→∞∣x∣2p−1Vτ(∣x∣)=Bp,N,0{\mathrm{lim}}_{| x| \to \infty }| x{| }^{\tfrac{2}{p-1}}{V}_{\tau }\left(| x| )={B}_{p,N,0}. The articles [14,15] showed the stability results of Vτ{V}_{\tau }.Recently, Quittner and Souplet [28] considered the model (1.4)ut−Δu=λu+∣u∣p−1u,x∈RN,t>0,u(x,0)=u0(x),x∈RN,\left\{\phantom{\rule[-1.25em]{}{0ex}}\begin{array}{ll}{u}_{t}-\Delta u=\lambda u+| u{| }^{p-1}u,\hspace{1.0em}& x\in {{\mathbb{R}}}^{N},\hspace{1em}t\gt 0,\\ u\left(x,0)={u}_{0}\left(x),\hspace{1.0em}& x\in {{\mathbb{R}}}^{N},\end{array}\right.with p>1p\gt 1and λ∈R\lambda \in {\mathbb{R}}. They derived that T(u0)<∞T\left({u}_{0})\lt \infty under the assumptions 0≤u0∈L∞(RN),∫RNu0(x)π−n2e−∣x∣2dx>c≔(max(0,2n−λ))1/(p−1),0\le {u}_{0}\in {L}^{\infty }\left({{\mathbb{R}}}^{N}),\hspace{1.0em}\mathop{\int }\limits_{{{\mathbb{R}}}^{N}}{u}_{0}\left(x){\pi }^{-\tfrac{n}{2}}{e}^{-| x{| }^{2}}{\rm{d}}x\gt c:= {({\rm{\max }}\left(0,2n-\lambda ))}^{1\text{/}\left(p-1)},or 0≤u0∈L∞(RN),liminf∣x∣→∞∣x∣2/(p−1)u0(x)>μ1/(p−1),0\le {u}_{0}\in {L}^{\infty }\left({{\mathbb{R}}}^{N}),\hspace{1.0em}\mathop{\mathrm{liminf}}\limits_{| x| \to \infty }| x{| }^{2\text{/}\left(p-1)}{u}_{0}\left(x)\gt {\mu }^{1\text{/}\left(p-1)},where μ>0\mu \gt 0be the first eigenvalue of the Dirichlet Laplacian in the unit ball of RN{{\mathbb{R}}}^{N}. Wang [33] obtained the optimal bound of u0{u}_{0}, which classifies the global existence and blow-up of solutions for large pp. The Joseph-Lundgren exponent pJL≔1+4N−4−2N−1ifN≥11,∞ifN≤10,{p}_{{\rm{JL}}}:= \left\{\begin{array}{ll}1+\frac{4}{N-4-2\sqrt{N-1}}\hspace{1.0em}& {\rm{if}}\hspace{1.0em}N\ge 11,\\ \infty \hspace{1.0em}& {\rm{if}}\hspace{1.0em}N\le 10,\end{array}\right.plays a key role in the study of the stability of stationary solutions. The previous article [33] showed that if p≥pJLp\ge {p}_{{\rm{JL}}}, and u0{u}_{0}satisfies liminf∣x∣→∞∣x∣2p−1u0(x)>Bp,N,0{\mathrm{liminf}}_{| x| \to \infty }| x{| }^{\tfrac{2}{p-1}}{u}_{0}\left(x)\gt {B}_{p,N,0}, then solutions of (1.3) blow up in finite time. Otherwise, if p≥pJLp\ge {p}_{{\rm{JL}}}and liminf∣x∣→∞∣x∣2p−1u0(x)=Bp,N,0{\mathrm{liminf}}_{| x| \to \infty }| x{| }^{\tfrac{2}{p-1}}{u}_{0}\left(x)={B}_{p,N,0}, then there exists a global in time solution of (1.3). Later, the previous article [23] generalized the results of [33] to the case p≥pFp\ge {p}_{F}. We refer to previous articles [1–3,17,19,34] and the references therein for more results.For the case β≠0\beta \ne 0, the Joseph-Lundgren exponent is defined by p˜JL≔(N−2)2−2(β+2)(β+N)+2(β+2)(N+β)2−(N−2)2(N−2)(N−10−4β),ifN≥11+4β,+∞,ifN≤10+4β.{\tilde{p}}_{{\rm{JL}}}:= \left\{\begin{array}{ll}\frac{{\left(N-2)}^{2}-2\left(\beta +2)\left(\beta +N)+2\left(\beta +2)\sqrt{{\left(N+\beta )}^{2}-{\left(N-2)}^{2}}}{\left(N-2)\left(N-10-4\beta )},\hspace{1.0em}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{1em}N\ge 11+4\beta ,\\ +\infty ,\hspace{1.0em}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{1em}N\le 10+4\beta .\end{array}\right.The previous article [33] proved that when N≥3N\ge 3, N+βN−2<p<p˜JL\frac{N+\beta }{N-2}\lt p\lt {\tilde{p}}_{{\rm{JL}}}and β>−2\beta \gt -2, if the initial value u0(x){u}_{0}\left(x)satisfies, 0≤u0(x)≤ϕ∗(x),0\le {u}_{0}\left(x)\le {\phi }_{\ast }\left(x),then (1.3) with the nonnegative initial value u0(x)∈C(RN){u}_{0}\left(x)\in C\left({{\mathbb{R}}}^{N})has a global solution u(x)u\left(x)such that 0≤u(x)≤ϕ∗(x)and∣∣u(⋅,t)∣∣L∞(RN)→0ast→∞.0\le u\left(x)\le {\phi }_{\ast }\left(x)\hspace{1.0em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1.0em}| | u\left(\cdot ,\hspace{0.33em}t)| {| }_{{L}^{\infty }\left({{\mathbb{R}}}^{N})}\to 0\hspace{1em}\hspace{0.1em}\text{as}\hspace{0.1em}\hspace{1em}t\to \infty .Moreover, the aforementioned result holds with 0≤u0(x)≤η˜ϕ∗(x)0\le {u}_{0}\left(x)\le \tilde{\eta }{\phi }_{\ast }\left(x)if p>N+βN−2p\gt \frac{N+\beta }{N-2}for some η˜∈(0,1)\tilde{\eta }\in \left(0,1). Later, the previous article [27] proved that the Cauchy problem for (1.3) possesses the Fujita exponent N+2+βN\frac{N+2+\beta }{N}with β>−1\beta \gt -1. Furthermore, the solutions of (1.3) blow up in finite time for all nontrivial nonnegative initial value u0{u}_{0}if 1<p≤N+2+βN1\lt p\le \frac{N+2+\beta }{N}, and there are both global and blowing-up solutions if p>N+2+βNp\gt \frac{N+2+\beta }{N}. Recently, the previous articles [12,13] proved the existence of sign-changing solutions of (1.3) with singular initial value u0{u}_{0}. For more general results on this direction, one can see the previous articles [4,5,7,18,21,25,26,29,30,35,36] and the references therein.The previous article [32] investigated the existence, uniqueness, and blow-up solutions of the reaction-diffusion equation with the exponential growth: (1.5)ut=Δu+λeu,x∈RN,t>0,λ>0.{u}_{t}=\Delta u+\lambda {e}^{u},\hspace{1.0em}x\in {{\mathbb{R}}}^{N},\hspace{1em}t\gt 0,\hspace{0.33em}\lambda \gt 0.The author proved that if u0(x)=U3(x)+c{u}_{0}\left(x)={U}_{3}\left(x)+cis a continuous function away from 0 and assume that lim∣x∣→0(u0(x)−U3(x))<c∗{\mathrm{lim}}_{| x| \to 0}\left({u}_{0}\left(x)-{U}_{3}\left(x))\lt {c}^{\ast }, then the Cauchy problem (1.5) admits a weak solution, at least locally in time, where c∈Rc\in {\mathbb{R}}, U3(x){U}_{3}\left(x)is the singular solution of (1.5): U3(x)≔−2log∣x∣+log(2N−4)λ.{U}_{3}\left(x):= -2\log | x| +\log \frac{\left(2N-4)}{\lambda }.On the contrary, if lim∣x∣→0(u0(x)−U3(x))>c∗{\mathrm{lim}}_{| x| \to 0}\left({u}_{0}\left(x)-{U}_{3}\left(x))\gt {c}^{\ast }, then there is no solution even locally in time, and instantaneous blow-up occurs. Recently, Fujishima [8] considered the problem (1.1) with β=0\beta =0. Precisely, the author proved that if N≥3,u0∈C(RN),liminf∣x∣→∞(2log∣x∣+u0(x))>L∗N\ge 3,\hspace{1.0em}{u}_{0}\in C\left({{\mathbb{R}}}^{N}),\hspace{1.0em}\mathop{\mathrm{liminf}}\limits_{| x| \to \infty }\left(2\log | x| +{u}_{0}\left(x))\gt {L}^{\ast }and −Ce∣x∣2−ε≤u0(x)≤CinRN-C{e}^{| x{| }^{2-\varepsilon }}\le {u}_{0}\left(x)\le C\hspace{1.0em}{\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N}for some constants ε\varepsilon in (0,2)\left(0,2)and C>0C\gt 0, then T(u0)<∞T\left({u}_{0})\lt \infty , where L∗{L}^{\ast }can be described in two cases: L∗>log(2N−4){L}^{\ast }\gt \log \left(2N-4)with 3≤N≤93\le N\le 9and L∗=log(2N−4){L}^{\ast }=\log \left(2N-4)with N≥10N\ge 10. The previous article [8] obtained the optimal bound of u0{u}_{0}, which classifies the global existence and blow-up of solutions of (1.1). For more results on this direction, one can also refer to [6,24] and the references therein.Inspired by previous works [8,33], the purpose of this article is to consider the existence and blow-up of solutions of the Hénon-type heat equation with exponential nonlinearity. As in [31], it is obvious that for any initial value v0{v}_{0}satisfying (1.2), there exists T=T(v0)>0T=T\left({v}_{0})\gt 0such that (1.1) has a unique classical solution in C2,1(RN×(0,T))∩C(RN×[0,T)){C}^{2,1}\left({{\mathbb{R}}}^{N}\times \left(0,T))\cap C\left({{\mathbb{R}}}^{N}\times \left[0,T)). Moreover, it is clear that the function V∗(x)=−(2+β)log∣x∣+log[(2+β)(N−2)]{V}_{\ast }\left(x)=-\left(2+\beta )\log | x| +\log {[}\left(2+\beta )\left(N-2)]is a singular stationary solution of (1.1). Hence, we guess that the critical decay rate at the space infinity for the existence of global in time solutions of (1.1) is given by −(2+β)log∣x∣-\left(2+\beta )\log | x| . More specifically, we investigate the case where v0{v}_{0}decays to −∞-\infty at space infinity with the rate −(2+β)log∣x∣-\left(2+\beta )\log | x| and obtain the optimal bound of lim∣x∣→∞[(2+β)log∣x∣+v0(x)]{\mathrm{lim}}_{| x| \to \infty }{[}\left(2+\beta )\log | x| +{v}_{0}\left(x)], which is the crucial indicator to classify the global in time solutions and blow-up solutions of the problem (1.1). To accomplish this, we study the properties of forward self-similar solutions of (1.1), which is defined by (1.6)v(x,t)=−1+β2logt+ωxt.v\left(x,t)=-\left(1+\frac{\beta }{2}\right)\log t+\omega \left(\frac{x}{\sqrt{t}}\right).A direct computation shows that if vvis a solution of (1.1), then ω\omega satisfies the elliptic equation: (1.7)Δω+12y⋅∇ω+∣y∣βeω+1+β2=0inRN.\Delta \omega +\frac{1}{2}y\cdot \nabla \omega +| y{| }^{\beta }{e}^{\omega }+1+\frac{\beta }{2}=0\hspace{1.0em}{\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N}.In particular, if ω=ω(r)\omega =\omega \left(r)(r=∣y∣r=| y| ) is radially symmetric about the origin, then ω\omega satisfies ω′(0)=0\omega ^{\prime} \left(0)=0and (1.8)ω″+N−1r+r2ω′+rβeω+1+β2=0,r>0.{\omega }^{^{\prime\prime} }+\left(\frac{N-1}{r}+\frac{r}{2}\right)\omega ^{\prime} +{r}^{\beta }{e}^{\omega }+1+\frac{\beta }{2}=0,\hspace{1.0em}r\gt 0.We will look into the specific behavior of the function ω\omega . Comparing to the previous work [8], we encounter a new difficulty here to obtain the precise behavior of the function ω\omega . That is, since the energy functional is not monotone decreasing, we cannot obtain the global and decay estimates of the function ω\omega directly. Motivated by the previous works [12,13,16], we can overcome the difficulty by deducing the new inequality of the energy functional.One refers to ω\omega as a solution of (1.7) if ω∈C2([0,∞))\omega \in {C}^{2}\left(\left[0,\infty ))is a classical solution of (1.7). We are interested in solutions ω\omega of (1.7) satisfying (1.9)limr→∞[(2+β)logr+ω(r)]=D,D∈R.\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\omega \left(r)]=D,\hspace{1em}D\in {\mathbb{R}}.Set SD≔{ω∈C2([0,∞)):ωis a solution of (1.8) satisfying (1.9)},{S}_{D}:= \left\{\omega \in {C}^{2}\left(\left[0,\infty )):\omega \hspace{0.33em}\hspace{0.1em}\text{is a solution of (1.8) satisfying (1.9)}\hspace{0.1em}\right\},and D∗≔sup{D∈R:SD≠∅}.{D}^{\ast }:= \sup \left\{D\in R:{S}_{D}\ne \varnothing \right\}.Then we have the following main results.Theorem 1.1Let N≥3N\ge 3, 0<β<20\lt \beta \lt 2. Then the constant D∗{D}^{\ast }defined byD∗≔sup{D∈R:SD≠∅}{D}^{\ast }:= \sup \left\{D\in R:{S}_{D}\ne \varnothing \right\}is finite. Furthermore, D∗>log[(2+β)(N−2)]andSD∗≠∅,if3≤N<10+4β,D∗=log[(2+β)(N−2)]andSD∗=∅,ifN≥10+4β.\left\{\phantom{\rule[-1.25em]{}{0ex}}\begin{array}{ll}{D}^{\ast }\gt \log {[}\left(2+\beta )\left(N-2)]\hspace{0.33em}and\hspace{0.33em}{S}_{{D}^{\ast }}\ne \varnothing ,\hspace{1.0em}& if\hspace{0.33em}3\le N\lt 10+4\beta ,\\ {D}^{\ast }=\log {[}\left(2+\beta )\left(N-2)]\hspace{0.33em}and\hspace{0.33em}{S}_{{D}^{\ast }}=\varnothing ,\hspace{1.0em}& if\hspace{0.33em}N\ge 10+4\beta .\end{array}\right.Next we state the sufficient condition for the solutions of (1.1) to blow up in finite time.Theorem 1.2Let N≥3N\ge 3, 0<β<20\lt \beta \lt 2, and v0∈C(RN){v}_{0}\in C\left({{\mathbb{R}}}^{N})be an initial function satisfying (1.2). If v0{v}_{0}satisfiesliminf∣x∣→∞[(2+β)log∣x∣+v0(x)]>D∗,\mathop{\mathrm{liminf}}\limits_{| x| \to \infty }{[}\left(2+\beta )\log | x| +{v}_{0}\left(x)]\gt {D}^{\ast },then T(v0)<∞T\left({v}_{0})\lt \infty .Remark 1.3Consider solutions of the following equation: (1.10)v″+N−1rv′+rβev=0,{v}^{^{\prime\prime} }+\frac{N-1}{r}v^{\prime} +{r}^{\beta }{e}^{v}=0,with v(0)=αv\left(0)=\alpha , v′(0)=0v^{\prime} \left(0)=0, α∈R\alpha \in {\mathbb{R}}, that is, the radially symmetric stationary solution vvof (1.1). According to Lemma 4.1, we show that limr→∞[(2+β)logr+v(r)]=log[(2+β)(N−2)].\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+v\left(r)]=\log {[}\left(2+\beta )\left(N-2)].Thus, every solution of (1.10) has the same decay bound with V∗(x)=−(2+β)log∣x∣+log[(2+β)(N−2)]{V}_{\ast }\left(x)=-\left(2+\beta )\log | x| +\log {[}\left(2+\beta )\left(N-2)]. Then by Theorem 1.1, we know that for any D≤D∗D\le {D}^{\ast }, there exists a global in time solution of (1.1) with lim∣x∣→∞[(2+β)log∣x∣+v0(x)]=D.\mathop{\mathrm{lim}}\limits_{| x| \to \infty }{[}\left(2+\beta )\log | x| +{v}_{0}\left(x)]=D.Therefore, one can observe that the optimal decay bound is −(2+β)log∣x∣+D∗-\left(2+\beta )\log | x| +{D}^{\ast }by Theorem 1.2.Remark 1.4To prove Theorem 1.2, we define the function ψ\psi as follows: (1.11)ψ(y,s)≔1+β2log(1+t)+v(x,t),y=x1+t,s=log(1+t).\psi (y,s):= \left(1+\frac{\beta }{2}\right)\log \left(1+t)+v\left(x,t),\hspace{1.0em}y=\frac{x}{\sqrt{1+t}},\hspace{1.0em}s=\log \left(1+t).Then ψ\psi satisfies (1.12)ψs=Δψ+12y⋅∇ψ+∣y∣βeψ+1+β2inRN×(0,∞),ψ(y,0)=ψ0(y)inRN,\left\{\begin{array}{ll}{\psi }_{s}=\Delta \psi +\frac{1}{2}y\cdot \nabla \psi +| y{| }^{\beta }{e}^{\psi }+1+\frac{\beta }{2}\hspace{1.0em}& {\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N}\times \left(0,\infty ),\\ \psi (y,0)={\psi }_{0}(y)\hspace{1.0em}& {\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N},\end{array}\right.where ψ0=v0{\psi }_{0}={v}_{0}.The structure of this article is as follows: In Section 2, we present some preliminary results. In Section 3, we obtain the existence of radially symmetric forward self-similar solutions for (1.1) as well as its asymptotic behavior at space infinity. In Section 4, we investigate the multiplicity of the forward self-similar solutions. In Section 5, we prove Theorem 1.2 by using the previous results.2Preliminaries2.1Super- and sub-solution methodsIn this section, we review some definitions of continuous weak super-solution and sub-solution. We introduce the comparison principle for problems (1.1), (1.7), and (1.12).Definition 1(i)Function vvis a continuous weak super-solution of (1.1) in RN×[0,T]{{\mathbb{R}}}^{N}\times \left[0,T]if v∈C(RN×[0,T])v\in C\left({{\mathbb{R}}}^{N}\times \left[0,T])satisfies v(x,0)≥v0(x)v\left(x,0)\ge {v}_{0}\left(x)in RN{{\mathbb{R}}}^{N}and ∫RNv(x,t)ξ(x,t)t=0t=T′≥∬RN×[0,T′][v(x,t)(∂tξ(x,t)+Δξ(x,t))+∣x∣βev(x,t)ξ(x,t)]dxdt{\left.\underset{{\mathbb{R}}}{\overset{N}{\int }}v\left(x,t)\xi \left(x,t)\right|}_{t=0}^{t=T^{\prime} }\ge \mathop{\iint }\limits_{{{\mathbb{R}}}^{N}\times \left[0,T^{\prime} ]}{[}v\left(x,t)\left({\partial }_{t}\xi \left(x,t)+\Delta \xi \left(x,t))+| x{| }^{\beta }{e}^{v\left(x,t)}\xi \left(x,t)]{\rm{d}}x{\rm{d}}tfor all T′∈[0,T]T^{\prime} \in \left[0,T]and all nonnegative test functions ξ∈C2,1(RN×[0,T])\xi \in {C}^{2,1}\left({{\mathbb{R}}}^{N}\times \left[0,T])such that supp ξ(⋅,t)\xi \left(\cdot ,t)is compact in RN{{\mathbb{R}}}^{N}for all t∈[0,T]t\in \left[0,T]. If the reverse inequalities of the aforementioned inequalities hold, then function vvis referred to as a continuous weak sub-solution of (1.1) with the initial value v0{v}_{0}.(ii)Function ω\omega is a continuous weak super-solution of (1.7) if ω∈C(RN)\omega \in C\left({{\mathbb{R}}}^{N})satisfies ∫RNωΔη−12y⋅∇η−N2η+∣y∣βeω+1+β2ηdy≤0\mathop{\int }\limits_{{{\mathbb{R}}}^{N}}\left[\omega \left(\Delta \eta -\frac{1}{2}y\cdot \nabla \eta -\frac{N}{2}\eta \right)+\left(| y{| }^{\beta }{e}^{\omega }+1+\frac{\beta }{2}\right)\eta \right]{\rm{d}}y\le 0for all nonnegative test functions η∈C2(RN)\eta \in {C}^{2}\left({{\mathbb{R}}}^{N})with compact support in RN{{\mathbb{R}}}^{N}. If the reverse inequalities of the aforementioned inequalities hold, then function ω\omega is referred to as a continuous weak sub-solution of (1.7).(iii)Function ψ\psi is a continuous weak super-solution of (1.12) in RN×[0,S]{{\mathbb{R}}}^{N}\times \left[0,S]if ψ∈C(RN×[0,S])\psi \in C\left({{\mathbb{R}}}^{N}\times \left[0,S])satisfies ψ(y,0)≥ψ0(y)\psi (y,0)\ge {\psi }_{0}(y)in RN{{\mathbb{R}}}^{N}and ∫RNψ(y,s)η(y,s)s=0s=S′≥∬RN×[0,S′]ψ∂sη+Δη−12y⋅∇η−N2η+∣y∣βeψ+1+β2ηdyds{\left.\underset{{\mathbb{R}}}{\overset{N}{\int }}\psi (y,s)\eta (y,s)\right|}_{s=0}^{s=S^{\prime} }\ge \mathop{\iint }\limits_{{{\mathbb{R}}}^{N}\times \left[0,S^{\prime} ]}\left[\psi \left({\partial }_{s}\eta +\Delta \eta -\frac{1}{2}y\cdot \nabla \eta -\frac{N}{2}\eta \right)+\left(| y{| }^{\beta }{e}^{\psi }+1+\frac{\beta }{2}\right)\eta \right]{\rm{d}}y{\rm{d}}sfor all S′∈[0,S]S^{\prime} \in \left[0,S]and all nonnegative test functions η∈C2,1(RN×[0,S])\eta \in {C}^{2,1}\left({{\mathbb{R}}}^{N}\times \left[0,S]), which satisfies supp η(⋅,s)\eta \left(\cdot ,s)is compact in RN{{\mathbb{R}}}^{N}for all s∈[0,S]s\in \left[0,S]. If the reverse inequalities of the aforementioned inequalities hold, then function ψ\psi is referred to as a continuous weak sub-solution of (1.12).On account of [23, Lemmas 2.1, 2.3] and [33, Lemma 1.3], we can obtain the comparison principle as follows.Lemma 2.1(i)Suppose that v¯\overline{v}and v̲\underline{v}are bounded above and satisfy v¯−v̲≥−AeB∣x∣2\overline{v}-\underline{v}\ge -A{e}^{B| x{| }^{2}}in RN×[0,T]{{\mathbb{R}}}^{N}\times \left[0,T]for some constants A,B>0A,B\gt 0. Then one has that v̲(x,t)≤v¯(x,t)\underline{v}\left(x,t)\le \overline{v}\left(x,t)in RN×[0,T]{{\mathbb{R}}}^{N}\times \left[0,T], and there exists a classical solution vvof (1.1) satisfying v̲≤v≤v¯\underline{v}\le v\le \overline{v}in RN×[0,T]{{\mathbb{R}}}^{N}\times \left[0,T], where v¯\overline{v}and v̲\underline{v}are continuous weak super-solution and sub-solution of (1.1) in RN×[0,T]{{\mathbb{R}}}^{N}\times \left[0,T], respectively.(ii)If ψ¯\overline{\psi }and ψ̲\underline{\psi }are bounded above and satisfy ψ¯(y,s)−ψ̲(y,s)≥−AeB∣y∣2\overline{\psi }(y,s)-\underline{\psi }(y,s)\ge -A{e}^{B| y{| }^{2}}in RN×[0,S]{{\mathbb{R}}}^{N}\times \left[0,S]for some constants A,B>0A,B\gt 0, then one gets that ψ̲≤ψ¯\underline{\psi }\le \overline{\psi }in RN×[0,S]{{\mathbb{R}}}^{N}\times \left[0,S]and there exists a classical solution ψ\psi of (1.12) satisfying ψ̲≤ψ≤ψ¯\underline{\psi }\le \psi \le \overline{\psi }in RN×[0,S]{{\mathbb{R}}}^{N}\times \left[0,S], where ψ¯\overline{\psi }and ψ̲\underline{\psi }be continuous weak super-solution and sub-solution of (1.12) in RN×[0,S]{{\mathbb{R}}}^{N}\times \left[0,S], respectively.By using the method of [23, Lemmas 2.4, 2.5], with the help of Lemma 2.1(ii), we obtain some results on the monotonicity.Lemma 2.2If ψ̲0(ψ¯0){\underline{\psi }}_{0}\left({\overline{\psi }}_{0})is a continuous weak sub-solution (super-solution) of (1.7) and ψ\psi is a global in time solution of (1.12) with ψ0(y)≥ψ̲0(y)(≤ψ¯0(y)){\psi }_{0}(y)\ge {\underline{\psi }}_{0}(y)\left(\le {\overline{\psi }}_{0}(y))for y∈RNy\in {{\mathbb{R}}}^{N}, then one obtains ψ(y,s)≥ψ̲0(y)(≤ψ¯0(y))\psi (y,s)\ge {\underline{\psi }}_{0}(y)\left(\le {\overline{\psi }}_{0}(y))for y∈RN,s≥0y\in {{\mathbb{R}}}^{N},s\ge 0.Lemma 2.3If ψ0{\psi }_{0}is a continuous weak sub-solution (super-solution) of (1.7) and ψ\psi is a global in time solution of (1.12), then one deduces that ψ\psi is nondecreasing (nonincreasing) in ss.Next we prepare a lemma about the construction of comparison functions. For a rigorous proof of this lemma, the readers can refer to [23, Lemma 2.6].Lemma 2.4(i)Assume that n∈Nn\in {\mathbb{N}}(N={1,2,3,…}{\mathbb{N}}=\left\{1,2,3,\ldots \right\}) with n≥3n\ge 3and ωi=ωi(∣y∣)(i=1,2,…,n){\omega }_{i}={\omega }_{i}\left(| y| )\left(i=1,2,\ldots ,n)is regular radially symmetric sub-solutions of (1.7). If there exist constants R1<R2<⋯<Rn−1{R}_{1}\lt {R}_{2}\hspace{0.33em}\lt \cdots \lt {R}_{n-1}such that ωi(Ri)=ωi+1(Ri){\omega }_{i}\left({R}_{i})={\omega }_{i+1}\left({R}_{i})and ωi′(Ri)≤ωi+1′(Ri){\omega }_{i}^{^{\prime} }\left({R}_{i})\le {\omega }_{i+1}^{^{\prime} }\left({R}_{i})for i=1,2,…,n−1i=1,2,\ldots ,n-1, then the function ω̲\underline{\omega }defined byω̲(r)≔ω1(r)forr∈[0,R1],ωi+1(r)forr∈[Ri,Ri+1],(i=1,2,…,n−2)ωn(r)forr∈[Rn−1,∞),\underline{\omega }\left(r):= \left\{\begin{array}{ll}{\omega }_{1}\left(r)\hspace{1.0em}& for\hspace{1.0em}r\in \left[0,{R}_{1}],\\ {\omega }_{i+1}\left(r)\hspace{1.0em}& for\hspace{1.0em}r\in \left[{R}_{i},{R}_{i+1}],\left(i=1,2,\ldots ,n-2)\\ {\omega }_{n}\left(r)\hspace{1.0em}& for\hspace{1.0em}r\in \left[{R}_{n-1},\infty ),\end{array}\right.is a continuous weak sub-solution of (1.7).(ii)Suppose that n∈Nn\in {\mathbb{N}}with n≥3n\ge 3and ωi=ωi(∣y∣)(i=1,2,…,n){\omega }_{i}={\omega }_{i}\left(| y| )\left(i=1,2,\ldots ,n)is regular radially symmetric super-solutions of (1.7). If there exist constants R1<R2<⋯<Rn−1{R}_{1}\lt {R}_{2}\hspace{0.33em}\lt \cdots \lt {R}_{n-1}such that ωi(Ri)=ωi+1(Ri){\omega }_{i}\left({R}_{i})={\omega }_{i+1}\left({R}_{i})and ωi′(Ri)≥ωi+1′(Ri){\omega }_{i}^{^{\prime} }\left({R}_{i})\ge {\omega }_{i+1}^{^{\prime} }\left({R}_{i})for i=1,2,…,n−1i=1,2,\ldots ,n-1, then one obtains that the function ω¯\overline{\omega }defined byω¯(r)≔ω1(r)forr∈[0,R1],ωi+1(r)forr∈[Ri,Ri+1],(i=1,2,…,n−2)ωn(r)forr∈[Rn−1,∞),\overline{\omega }\left(r):= \left\{\begin{array}{ll}{\omega }_{1}\left(r)\hspace{1.0em}& for\hspace{1.0em}r\in \left[0,{R}_{1}],\\ {\omega }_{i+1}\left(r)\hspace{1.0em}& for\hspace{1.0em}r\in \left[{R}_{i},{R}_{i+1}],\left(i=1,2,\ldots ,n-2)\\ {\omega }_{n}\left(r)\hspace{1.0em}& for\hspace{1.0em}r\in \left[{R}_{n-1},\infty ),\end{array}\right.is a continuous weak super-solution of (1.7).2.2Convergence properties of solutionsIn this subsection, we study the asymptotic behavior of solution for (1.12) and obtain that the solution of (1.12) converges to a solution of (1.8). We first prove the following results for the convergence.Proposition 2.5Let ψ\psi be a global solution of (1.12) such that ψ=ψ(r,s)\psi =\psi \left(r,s)is spatially radially symmetric about the origin where r=∣y∣r=| y| . Assume that ψ(r,s)\psi \left(r,s)is either nonincreasing or nondecreasing in s≥0s\ge 0for each fixed r≥0r\ge 0, and put ω(r)≔lims→∞ψ(r,s)\omega \left(r):= {\mathrm{lim}}_{s\to \infty }\psi \left(r,s)for r≥0r\ge 0. Then we have the following results: (i)If ω∈L∞([0,∞))\omega \in {L}^{\infty }\left(\left[0,\infty )), then ω∈C2([0,∞))\omega \in {C}^{2}\left(\left[0,\infty ))and satisfies (1.8) with ω′(0)=0\omega ^{\prime} \left(0)=0.(ii)If ψ(r,s)\psi \left(r,s)is nondecreasing in s≥0s\ge 0and nonincreasing in r≥0r\ge 0, ω∉L∞([0,∞))\omega \notin {L}^{\infty }\left(\left[0,\infty )), then ω∈C2((0,∞))\omega \in {C}^{2}\left(\left(0,\infty ))and satisfies (1.8) with limr→0ω(r)=∞{\mathrm{lim}}_{r\to 0}\omega \left(r)=\infty .We first show the following lemma.Lemma 2.6Suppose that ψ\psi is a global solution of (1.12), and ψ=ψ(r,s)\psi =\psi \left(r,s), where r=∣y∣r=| y| is radially symmetric and ψ(r,s)\psi \left(r,s)is either nonincreasing or nondecreasing in s≥0s\ge 0for each fixed r≥0r\ge 0, and put ω(r)≔lims→∞ψ(r,s)\omega \left(r):= {\mathrm{lim}}_{s\to \infty }\psi \left(r,s)for r≥0r\ge 0. Then ω∈Lloc1(RN)\omega \in {L}_{{\rm{loc}}}^{1}\left({{\mathbb{R}}}^{N})is a solution of (1.8) under the mean of the distribution. Furthermore, (i)If ω∈L∞(RN)\omega \in {L}^{\infty }\left({{\mathbb{R}}}^{N}), then ω∈C2(RN)\omega \in {C}^{2}\left({{\mathbb{R}}}^{N})and is a classical solution of (1.8).(ii)If ω∉L∞(RN)\omega \notin {L}^{\infty }\left({{\mathbb{R}}}^{N})and nonincreasing in r>0r\gt 0, then ω∈C2(0,∞)\omega \in {C}^{2}\left(0,\infty )and is a solution of (1.8) satisfying limr→0ω(r)=∞{\mathrm{lim}}_{r\to 0}\omega \left(r)=\infty .ProofOur first goal is to show that ω∈Lloc1(RN)\omega \in {L}_{{\rm{loc}}}^{1}\left({{\mathbb{R}}}^{N}). It is easy to verify the case where ψ(y,s)\psi (y,s)is nonincreasing in s≥0s\ge 0, then we need to only consider that ψ(y,s)\psi (y,s)is nondecreasing in s≥0s\ge 0.Let R>0R\gt 0, and set BR≔{x∈RN:∣x∣<R}{B}_{R}:= \left\{x\in {{\mathbb{R}}}^{N}:| x| \lt R\right\}. Consider the eigenvalue problem (2.1)∇⋅(ρ∇ϕ)+λϕρ=0inBR,ϕ=0on∂BR,\left\{\phantom{\rule[-1.25em]{}{0ex}}\begin{array}{ll}\nabla \cdot \left(\rho \nabla \phi )+\lambda \phi \rho =0\hspace{1.0em}& {\rm{in}}\hspace{1.0em}{B}_{R},\\ \phi =0\hspace{1.0em}& {\rm{on}}\hspace{1.0em}\partial {B}_{R},\end{array}\right.where ρ(x)≔exp∣x∣24\rho \left(x):= \exp \left(\frac{| x{| }^{2}}{4}\right). Let λ0∈R{\lambda }_{0}\in {\mathbb{R}}denote the first eigenvalue of (2.1) and ϕ0{\phi }_{0}be the corresponding eigenfunction with ∫BRϕ0ρdx=1{\int }_{{B}_{R}}{\phi }_{0}\rho {\rm{d}}x=1. By [23, Lemma A.1], we know λ0>0{\lambda }_{0}\gt 0. Let us first define Ψ(s)≔∫BRψ(∣y∣,s)ϕ0(y)ρ(y)∣y∣βdy.\Psi \left(s):= \mathop{\int }\limits_{{B}_{R}}\psi \left(| y| ,s){\phi }_{0}(y)\rho (y)| y{| }^{\beta }{\rm{d}}y.We rewrite (1.12) in the form (2.2)ψsρ=∇⋅(ρ∇ψ)+eψρ∣y∣β+1+β2ρinRN×(0,∞).{\psi }_{s}\rho =\nabla \cdot \left(\rho \nabla \psi )+{e}^{\psi }\rho | y{| }^{\beta }+\left(1+\frac{\beta }{2}\right)\rho \hspace{1.0em}{\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N}\times \left(0,\infty ).By multiplying (2.2) with ϕ0∣y∣β{\phi }_{0}| y{| }^{\beta }and integrating on BR{B}_{R}, we obtain Ψ′(s)=∫BRψs(∣y∣,s)ϕ0(y)ρ∣y∣βdy=−∫BR∇ψ⋅∇(ϕ0∣y∣β)ρdy+∫BReψϕ0ρ∣y∣2βdy+∫BR1+β2ϕ0ρ∣y∣βdy.\begin{array}{rcl}\Psi ^{\prime} \left(s)& =& \mathop{\displaystyle \int }\limits_{{B}_{R}}{\psi }_{s}\left(| y| ,s){\phi }_{0}(y)\rho | y{| }^{\beta }{\rm{d}}y\\ & =& -\mathop{\displaystyle \int }\limits_{{B}_{R}}\nabla \psi \cdot \nabla ({\phi }_{0}| y{| }^{\beta })\rho {\rm{d}}y+\mathop{\displaystyle \int }\limits_{{B}_{R}}{e}^{\psi }{\phi }_{0}\rho | y{| }^{2\beta }{\rm{d}}y+\mathop{\displaystyle \int }\limits_{{B}_{R}}\left(1+\frac{\beta }{2}\right){\phi }_{0}\rho | y{| }^{\beta }{\rm{d}}y.\end{array}By multiplying (2.1) with ψ∣y∣β\psi | y{| }^{\beta }and integrating by parts, we obtain ∫∂BRψ∂ϕ0∂νρ∣y∣βdS−∫BR∇(ψ∣y∣β)⋅∇ϕ0ρdy+λ0∫BRψϕ0ρ∣y∣βdy=0.\mathop{\int }\limits_{\partial {B}_{R}}\psi \frac{\partial {\phi }_{0}}{\partial \nu }\rho | y{| }^{\beta }{\rm{d}}S-\mathop{\int }\limits_{{B}_{R}}\nabla (\psi | y{| }^{\beta })\cdot \nabla {\phi }_{0}\rho {\rm{d}}y+{\lambda }_{0}\mathop{\int }\limits_{{B}_{R}}\psi {\phi }_{0}\rho | y{| }^{\beta }{\rm{d}}y=0.It follows from maximum principle that ∂ϕ0∂ν<0\frac{\partial {\phi }_{0}}{\partial \nu }\lt 0on ∂BR\partial {B}_{R}.From the aforementioned analysis, we obtain Ψ′(s)=−∫∂BRψ∂ϕ0∂νρ∣y∣βdS−λ0∫BRψϕ0ρ∣y∣βdy+∫BReψϕ0ρ∣y∣2βdy+∫BR1+β2ϕ0ρ∣y∣βdy.\Psi ^{\prime} \left(s)=-\mathop{\int }\limits_{\partial {B}_{R}}\psi \frac{\partial {\phi }_{0}}{\partial \nu }\rho | y{| }^{\beta }{\rm{d}}S-{\lambda }_{0}\mathop{\int }\limits_{{B}_{R}}\psi {\phi }_{0}\rho | y{| }^{\beta }{\rm{d}}y+\mathop{\int }\limits_{{B}_{R}}{e}^{\psi }{\phi }_{0}\rho | y{| }^{2\beta }{\rm{d}}y+\mathop{\int }\limits_{{B}_{R}}\left(1+\frac{\beta }{2}\right){\phi }_{0}\rho | y{| }^{\beta }{\rm{d}}y.Using Jensen’s inequality yields (2.3)Ψ′(s)≥−∫∂BR∣R∣βψ∂ϕ0∂νρdS−λ0Ψ+∫BReψϕ0ρ∣y∣βdy≥−∫∂BR∣R∣βψ∂ϕ0∂νρdS−λ0Ψ+eΨ\Psi ^{\prime} \left(s)\ge -\mathop{\int }\limits_{\partial {B}_{R}}| R{| }^{\beta }\psi \frac{\partial {\phi }_{0}}{\partial \nu }\rho {\rm{d}}S-{\lambda }_{0}\Psi +\mathop{\int }\limits_{{B}_{R}}{e}^{\psi }{\phi }_{0}\rho | y{| }^{\beta }{\rm{d}}y\ge -\mathop{\int }\limits_{\partial {B}_{R}}| R{| }^{\beta }\psi \frac{\partial {\phi }_{0}}{\partial \nu }\rho {\rm{d}}S-{\lambda }_{0}\Psi +{e}^{\Psi }for all s≥0s\ge 0. Next we consider two cases: ω(r0)=∞\omega \left({r}_{0})=\infty for some r0>0{r}_{0}\gt 0or ω(r)<∞\omega \left(r)\lt \infty for any r>0r\gt 0.Let us assume initially that there exists some r0>0{r}_{0}\gt 0satisfying ω(r0)=∞\omega \left({r}_{0})=\infty . Then we could choose some s1>0{s}_{1}\gt 0such that ψ(r0,s)>0\psi \left({r}_{0},s)\gt 0for all s≥s1s\ge {s}_{1}. Thus, by (2.3), we have (2.4)Ψ′(s)≥eΨ−λ0Ψ,s≥s1.\Psi ^{\prime} \left(s)\ge {e}^{\Psi }-{\lambda }_{0}\Psi ,\hspace{1.0em}s\ge {s}_{1}.If lims→∞Ψ(s)=∞{\mathrm{lim}}_{s\to \infty }\Psi \left(s)=\infty , we may choose some s2≥s1{s}_{2}\ge {s}_{1}such that eΨ−λ0Ψ≥eΨ2for anys≥s2.{e}^{\Psi }-{\lambda }_{0}\Psi \ge {e}^{\tfrac{\Psi }{2}}\hspace{1.0em}\hspace{0.1em}\text{for any}\hspace{0.1em}\hspace{0.33em}s\ge {s}_{2}.In view of (2.4), we see Ψ′(s)≥eΨ2for alls≥s2,\Psi ^{\prime} \left(s)\ge {e}^{\tfrac{\Psi }{2}}\hspace{1.0em}\hspace{0.1em}\text{for all}\hspace{0.1em}\hspace{0.33em}s\ge {s}_{2},and we have Ψ(s)≥−2loge−Ψ(s2)2−12(s−s2),s≥s2,\Psi \left(s)\ge -2\log \left({e}^{-\tfrac{\Psi \left({s}_{2})}{2}}-\frac{1}{2}\left(s-{s}_{2})\right),\hspace{1.0em}s\ge {s}_{2},which leads to a contradiction; thus, Ψ\Psi exists for all s≥0s\ge 0.Next we assume that ω(r)<∞\omega \left(r)\lt \infty for any r>0r\gt 0. From (2.3), there exist some C≥0C\ge 0such that Ψ′(s)≥eΨ−λ0Ψ−C\Psi ^{\prime} \left(s)\ge {e}^{\Psi }-{\lambda }_{0}\Psi -Cfor all s≥0s\ge 0. If lims→∞Ψ(s)=∞Ψ(s)=∞{\mathrm{lim}}_{s\to \infty }\Psi \left(s)=\infty \Psi \left(s)=\infty , we may choose some s3≥0{s}_{3}\ge 0such that eΨ−λ0Ψ−C≥eΨ2for anys≥s3.{e}^{\Psi }-{\lambda }_{0}\Psi -C\ge {e}^{\tfrac{\Psi }{2}}\hspace{1.0em}\hspace{0.1em}\text{for any}\hspace{0.1em}\hspace{0.33em}s\ge {s}_{3}.Then we yield a contradiction in the same way as mentioned earlier.Thus, we infer the boundedness of the function Ψ\Psi in [0,∞)\left[0,\infty ). Since ψ\psi is nondecreasing in ss, so is Ψ\Psi . According to the monotone convergence theorem, we have ∫BRω(∣y∣,s)ϕ0(y)ρ(y)∣y∣βdy<∞.\mathop{\int }\limits_{{B}_{R}}\omega \left(| y| ,s){\phi }_{0}(y)\rho (y)| y{| }^{\beta }{\rm{d}}y\lt \infty .Since R>0R\gt 0is arbitrarily, we obtain ω∈Lloc1(RN)\omega \in {L}_{{\rm{loc}}}^{1}\left({{\mathbb{R}}}^{N}).It remains to show that ω\omega is a solution of (1.7) in the sense of distribution. For ς>0\varsigma \gt 0, we have ∫RNψ(∣y∣,s+ς)η(y)dy∣s=0s=1=∫01∫RNψ(∣y∣,s+ς)Δη−12y⋅∇η−N2η+∣y∣βeψ+1+β2ηdyds,\underset{{\mathbb{R}}}{\overset{N}{\int }}\psi \left(| y| ,s+\varsigma )\eta (y){\rm{d}}y{| }_{s=0}^{s=1}=\underset{0}{\overset{1}{\int }}\mathop{\int }\limits_{{{\mathbb{R}}}^{N}}\psi \left(| y| ,s+\varsigma )\left(\Delta \eta -\frac{1}{2}y\cdot \nabla \eta -\frac{N}{2}\eta \right)+\left(| y{| }^{\beta }{e}^{\psi }+1+\frac{\beta }{2}\right)\eta {\rm{d}}y{\rm{d}}s,where η∈C2(RN)\eta \in {C}^{2}\left({{\mathbb{R}}}^{N})with compact support. Letting ς→∞\varsigma \to \infty in the aforementioned equation, we obtain 0=∫RNω(y)Δη−12y⋅∇η−N2η+∣y∣βeψ+1+β2ηdy.0=\mathop{\int }\limits_{{{\mathbb{R}}}^{N}}\omega (y)\left(\Delta \eta -\frac{1}{2}y\cdot \nabla \eta -\frac{N}{2}\eta \right)+\left(| y{| }^{\beta }{e}^{\psi }+1+\frac{\beta }{2}\right)\eta {\rm{d}}y.It turns out that ω\omega is a solution of (1.7) in the sense of distribution.If ω∈L∞(RN)\omega \in {L}^{\infty }\left({{\mathbb{R}}}^{N}), we have ω∈C2(RN)\omega \in {C}^{2}\left({{\mathbb{R}}}^{N})by applying the regularity theory of elliptic equations; therefore, ω\omega is a classical solution of (1.8). Thus, (i)\left(i)follows.If ω∉L∞(RN)\omega \notin {L}^{\infty }\left({{\mathbb{R}}}^{N})and nonincreasing in r>0r\gt 0for any s≥0s\ge 0and by ω∈Lloc1(RN)\omega \in {L}_{{\rm{loc}}}^{1}\left({{\mathbb{R}}}^{N}), we have limr→0ω(r)=∞{\mathrm{lim}}_{r\to 0}\omega \left(r)=\infty and ω∈L∞(RN\Br)\omega \in {L}^{\infty }\left({{\mathbb{R}}}^{N}\backslash {B}_{r})for any r>0r\gt 0. Therefore, ω∈C2(RN\0)\omega \in {C}^{2}\left({{\mathbb{R}}}^{N}\backslash 0)by the regularity theory and ω\omega satisfies (1.8) in (RN\0)\left({{\mathbb{R}}}^{N}\backslash 0). Thus, (ii)\left(ii)follows.□Proof of Proposition 2.5(i)Owing to Lemma 2.6(i), we have that ω=ω(r)\omega =\omega \left(r)is a classical radial solution of (1.8). Thus, ω∈C2([0,∞))\omega \in {C}^{2}\left(\left[0,\infty ))and satisfies (1.8) with ω′(0)=0\omega ^{\prime} \left(0)=0.(ii)Since ψ(r,s)\psi \left(r,s)is nonincreasing in r≥0r\ge 0, ω(r)\omega \left(r)is nonincreasing in r≥0r\ge 0. Then Lemma 2.6(ii) implies that ω∈C2((0,∞))\omega \in {C}^{2}\left(\left(0,\infty ))and satisfies (1.8) with limr→0ω(r)=∞{\mathrm{lim}}_{r\to 0}\omega \left(r)=\infty .□3Existence and asymptotic of forward self-similar solutionsIn this section, we prove the existence of forward self-similar solutions for (1.1), that is, the solutions for (1.7). We also obtain the asymptotic behavior of solutions at space infinity.3.1Existence of forward self-similar solutionsIn this subsection, we prove the existence of the solution for (1.7) by using the contraction mapping argument. Let ω\omega be a solution of (1.7) and define function τ(y)≔eω(y)\tau (y):= {e}^{\omega (y)}, then τ\tau satisfies (3.1)Δτ+12y⋅∇τ+1+β2τ+∣y∣βτ2−∣∇τ2∣τ=0inRN,τ>0inRN.\Delta \tau +\frac{1}{2}y\cdot \nabla \tau +\left(1+\frac{\beta }{2}\right)\tau +| y{| }^{\beta }{\tau }^{2}-\frac{| {\nabla \tau }^{2}| }{\tau }=0\hspace{1.0em}{\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N},\hspace{1.0em}\tau \gt 0\hspace{0.33em}{\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N}.Conversely, if τ\tau is a positive solution of (3.1), then the function defined by ω(y)≔logτ(y)\omega (y):= \log \tau (y)is a solution of (1.7). Studying solutions of (1.7) is therefore equivalent to analyzing the solutions of (3.1). In particular, if τ=τ(r)\tau =\tau \left(r)is radially symmetric about the origin, where r=∣y∣r=| y| , then τ\tau satisfies (3.2)τ″+N−1r+r2τ′+1+β2τ+rβτ2−(τ′)2τ=0in(0,∞),τ′(0)=0.{\tau }^{^{\prime\prime} }+\left(\frac{N-1}{r}+\frac{r}{2}\right)\tau ^{\prime} +\left(1+\frac{\beta }{2}\right)\tau +{r}^{\beta }{\tau }^{2}-\frac{{\left(\tau ^{\prime} )}^{2}}{\tau }=0\hspace{1.0em}{\rm{in}}\hspace{0.33em}\left(0,\infty ),\hspace{1.0em}\tau ^{\prime} \left(0)=0.Now consider the following integral equation: (3.3)τ(r)=αexp−∫0rσN(s)−1∫0stβτ(t)+1+β2σN(t)dtds,\tau \left(r)=\alpha \exp \left\{-\underset{0}{\overset{r}{\int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}\left[{t}^{\beta }\tau \left(t)+1+\frac{\beta }{2}\right]{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\},where α>0\alpha \gt 0. One can easily to check that (3.3) is equivalent to problem (3.2) with τ(0)=α\tau \left(0)=\alpha . We first introduce two functions σN=σN(r){\sigma }_{N}={\sigma }_{N}\left(r)and γN=γN(r){\gamma }_{N}={\gamma }_{N}\left(r)defined by (3.4)σN(r)≔rN−1er24,γN(r)≔∫0rσN(s)−1∫0sσN(t)dtds,r>0.{\sigma }_{N}\left(r):= {r}^{N-1}{e}^{\tfrac{{r}^{2}}{4}},\hspace{1.0em}{\gamma }_{N}\left(r):= \underset{0}{\overset{r}{\int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s,\hspace{1.0em}r\gt 0.Functions σN=σN(r){\sigma }_{N}={\sigma }_{N}\left(r)and γN=γN(r){\gamma }_{N}={\gamma }_{N}\left(r)are useful to study the radially symmetric solutions ω=ω(∣y∣)\omega =\omega \left(| y| )of (1.7) as ∣y∣→∞| y| \to \infty . We state following lemmas, which describe the asymptotic behavior of functions σN=σN(r){\sigma }_{N}={\sigma }_{N}\left(r)and γN=γN(r){\gamma }_{N}={\gamma }_{N}\left(r). The proof of the following lemmas can be found in [8, Lemmas 2.1 and 2.2].Lemma 3.1For N≥1N\ge 1, the limit(3.5)limr→∞(γN(r)−2logr)\mathop{\mathrm{lim}}\limits_{r\to \infty }\left({\gamma }_{N}\left(r)-2\log r)exists in R{\mathbb{R}}. In particular, it holds(3.6)supr≥0∣γN(r)−2log(1+r)∣<∞.\mathop{\sup }\limits_{r\ge 0}| {\gamma }_{N}\left(r)-2\log \left(1+r)| \lt \infty .Lemma 3.2For N≥1N\ge 1, inequality∫0s(1+t)−2σN(t)dt≤CsN−4es24\left|\underset{0}{\overset{s}{\int }}{\left(1+t)}^{-2}{\sigma }_{N}\left(t){\rm{d}}t\right|\le C{s}^{N-4}{e}^{\tfrac{{s}^{2}}{4}}holds for all sufficiently large s>0s\gt 0and C>0C\gt 0is a constant.We are now turning to the existence of solutions for (3.2).Proposition 3.3Let N≥1N\ge 1, 0<β<20\lt \beta \lt 2. For any α>0\alpha \gt 0, there exists a unique positive solution τ∈C2([0,∞))\tau \in {C}^{2}\left(\left[0,\infty ))of (3.2) satisfying τ(0)=α\tau \left(0)=\alpha .ProofFirst, we will prove the local existence and uniqueness. Let g(r)≔rβeω+1+β2,g\left(r):= {r}^{\beta }{e}^{\omega }+1+\frac{\beta }{2},then (1.8) is equivalent to (3.7)rN−1er24ω′(r)′=−g(r)rN−1er24.\left({r}^{N-1}{e}^{\tfrac{{r}^{2}}{4}}\omega ^{\prime} \left(r)\right)^{\prime} =-g\left(r){r}^{N-1}{e}^{\tfrac{{r}^{2}}{4}}.Since ω′(0)=0\omega ^{\prime} \left(0)=0, integrating of (3.7) with respect of rrgives (3.8)ω′(r)=−r1−Ne−r24∫0rg(t)tN−1et24dt=−σN(r)−1∫0rg(t)σN(t)dt.\omega ^{\prime} \left(r)=-{r}^{1-N}{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{r}{\int }}g\left(t){t}^{N-1}{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t=-{\sigma }_{N}{\left(r)}^{-1}\underset{0}{\overset{r}{\int }}g\left(t){\sigma }_{N}\left(t){\rm{d}}t.Integrating once more, we obtain (3.9)ω(r)=logα−∫0rσN(s)−1∫0sg(t)σN(t)dtds.\omega \left(r)=\log \alpha -\underset{0}{\overset{r}{\int }}{\sigma }_{N}{\left(s)}^{-1}\underset{0}{\overset{s}{\int }}g\left(t){\sigma }_{N}\left(t){\rm{d}}t{\rm{d}}s.Therefore, τ\tau satisfies the following integral equation: (3.10)τ(r)=αexp−∫0rσN(s)−1∫0stβτ(t)+1+β2σN(t)dtds.\tau \left(r)=\alpha \exp \left\{-\underset{0}{\overset{r}{\int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}\left[{t}^{\beta }\tau \left(t)+1+\frac{\beta }{2}\right]{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\}.Next we use a contraction mapping argument to prove existence and uniqueness of solution τ∈C([0,δ])\tau \in C\left(\left[0,\delta ])of (3.10) for δ>0\delta \gt 0small enough. To accomplish that, we choose δ>0\delta \gt 0and define Jα,δ≔{τ∈C([0,δ]):τ(0)=α}.{J}_{\alpha ,\delta }:= \left\{\tau \in C\left(\left[0,\delta ]):\tau \left(0)=\alpha \right\}.For τ∈Jα,δ\tau \in {J}_{\alpha ,\delta }, H[τ](r)≔αexp−∫0rσN(s)−1∫0stβτ(t)+1+β2σN(t)dtds,H\left[\tau ]\left(r):= \alpha \exp \left\{-\underset{0}{\overset{r}{\int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}\left[{t}^{\beta }\tau \left(t)+1+\frac{\beta }{2}\right]{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\},Then we obtain 0<H[τ](r)≤α0\lt H\left[\tau ]\left(r)\le \alpha , H[τ](0)=αH\left[\tau ]\left(0)=\alpha , and obvious H[τ](r)∈Jα,δH\left[\tau ]\left(r)\in {J}_{\alpha ,\delta }. Furthermore, by the elementary inequality (3.11)∣e−A−e−B∣≤∣A−B∣forA,B>0,| {e}^{-A}-{e}^{-B}| \le | A-B| \hspace{1.0em}{\rm{for}}\hspace{0.33em}A,B\gt 0,we have sup0<r<δ∣H[τ1](r)−H[τ2](r)∣≤αsup0<r<δ∫0rσN(s)−1∫0stβ∣τ1(t)−τ2(t)∣σN(t)dtds≤α∫0δσN(s)−1sβ⋅σN(s)⋅sds⋅sup0<r<δ∣τ1(r)−τ2(r)∣=α∫0δs1+βds⋅sup0<r<δ∣τ1(r)−τ2(r)∣=α2+βδ2+βsup0<r<δ∣τ1(r)−τ2(r)∣\begin{array}{rcl}\mathop{\sup }\limits_{0\lt r\lt \delta }| H\left[{\tau }_{1}]\left(r)-H\left[{\tau }_{2}]\left(r)| & \le & \alpha \mathop{\sup }\limits_{0\lt r\lt \delta }\underset{0}{\overset{r}{\displaystyle \int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\displaystyle \int }}{t}^{\beta }| {\tau }_{1}\left(t)-{\tau }_{2}\left(t)| {\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\\ & \le & \alpha \underset{0}{\overset{\delta }{\displaystyle \int }}{\sigma }_{N}{\left(s)}^{-1}{s}^{\beta }\cdot {\sigma }_{N}\left(s)\cdot s{\rm{d}}s\cdot \mathop{\sup }\limits_{0\lt r\lt \delta }| {\tau }_{1}\left(r)-{\tau }_{2}\left(r)| \\ & =& \alpha \underset{0}{\overset{\delta }{\displaystyle \int }}{s}^{1+\beta }{\rm{d}}s\cdot \mathop{\sup }\limits_{0\lt r\lt \delta }| {\tau }_{1}\left(r)-{\tau }_{2}\left(r)| \\ & =& \frac{\alpha }{2+\beta }{\delta }^{2+\beta }\mathop{\sup }\limits_{0\lt r\lt \delta }| {\tau }_{1}\left(r)-{\tau }_{2}\left(r)| \end{array}for all τ1,τ2∈Jα,δ{\tau }_{1},{\tau }_{2}\in {J}_{\alpha ,\delta }. Thus, by selecting δ=α−12+β\delta ={\alpha }^{-\tfrac{1}{2+\beta }}, we obtain the unique positive solution τ∈C([0,δ])\tau \in C\left(\left[0,\delta ])of (3.3). And τ\tau satisfies 0<τ(r)≤α0\lt \tau \left(r)\le \alpha in [0,δ]\left[0,\delta ]. It is easy to see that τ∈C1([0,δ])\tau \in {C}^{1}\left(\left[0,\delta ]), τ′(0)=0\tau ^{\prime} \left(0)=0, and (3.12)τ′(r)=−τ(r)σN(r)−1∫0rtβτ(t)+1+β2σN(t)dt,\tau ^{\prime} \left(r)=-\tau \left(r){\sigma }_{N}{\left(r)}^{-1}\underset{0}{\overset{r}{\int }}\left[{t}^{\beta }\tau \left(t)+1+\frac{\beta }{2}\right]{\sigma }_{N}\left(t){\rm{d}}t,by (3.10). Therefore, τ∈C2((0,δ])\tau \in {C}^{2}\left(\left(0,\delta ])and satisfies (3.2).Finally, we infer from (3.12) that τ′(r)−τ′(0)r=τ′(r)r=−τ(r)r−Ne−r24∫0rtβτ(t)+1+β2tN−1et24dt=−τ(r)r−Ne−r24tNNtβτ(t)+1+β2et24t=0t=r−∫0rtNN⋅tβτ′(t)+βtβ−1τ(t)+t2tβτ(t)+1+β2et24dt=−τ(r)r−Ne−r24rNNrβτ(r)+1+β2er24+O(rN+1)\begin{array}{rcl}\frac{\tau ^{\prime} \left(r)-\tau ^{\prime} \left(0)}{r}& =& \frac{\tau ^{\prime} \left(r)}{r}=-\tau \left(r){r}^{-N}{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{r}{\displaystyle \int }}\left[{t}^{\beta }\tau \left(t)+1+\frac{\beta }{2}\right]{t}^{N-1}{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\\ & =& -\tau \left(r){r}^{-N}{e}^{-\tfrac{{r}^{2}}{4}}\left\{{\left[\frac{{t}^{N}}{N}\left({t}^{\beta }\tau \left(t)+1+\frac{\beta }{2}\right){e}^{\tfrac{{t}^{2}}{4}}\right]}_{t=0}^{t=r}-\underset{0}{\overset{r}{\displaystyle \int }}\frac{{t}^{N}}{N}\cdot \left[{t}^{\beta }\tau ^{\prime} \left(t)+\beta {t}^{\beta -1}\tau \left(t)+\frac{t}{2}\left({t}^{\beta }\tau \left(t)+1+\frac{\beta }{2}\right)\right]{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\right\}\\ & =& -\tau \left(r){r}^{-N}{e}^{-\tfrac{{r}^{2}}{4}}\left\{\phantom{\rule[-1.25em]{}{0ex}},\frac{{r}^{N}}{N}\left[{r}^{\beta }\tau \left(r)+1+\frac{\beta }{2}\right]{e}^{\tfrac{{r}^{2}}{4}}+O\left({r}^{N+1})\right\}\end{array}for r>0r\gt 0small enough, which indicates that τ″(0){\tau }^{^{\prime\prime} }\left(0)exists and equals to −αN1+β2-\frac{\alpha }{N}\left(1+\frac{\beta }{2}\right).On the other hand, since τ\tau satisfies (3.3) in [0,δ]\left[0,\delta ], we observe that τ\tau also satisfies (3.2) for r>0r\gt 0small enough. Moreover, we infer from (3.2) that limr→0τ″(r)=−(N−1)limr→0τ′(r)r−α1+β2=−(N−1)τ″(0)+Nτ″(0)=τ″(0).\mathop{\mathrm{lim}}\limits_{r\to 0}{\tau }^{^{\prime\prime} }\left(r)=-\left(N-1)\mathop{\mathrm{lim}}\limits_{r\to 0}\frac{\tau ^{\prime} \left(r)}{r}-\alpha \left(1+\frac{\beta }{2}\right)=-\left(N-1){\tau }^{^{\prime\prime} }\left(0)+N{\tau }^{^{\prime\prime} }\left(0)={\tau }^{^{\prime\prime} }\left(0).Thus, we obtain τ∈C2([0,δ])\tau \in {C}^{2}\left(\left[0,\delta ]), which shows the local existence and uniqueness.For the purpose of proving the global existence, we introduce the energy of the solution τ\tau (3.13)I[τ](r)≔12∣τ′(r)∣2+12+β4∣τ(r)∣2+13rβ∣τ(r)∣3.I\left[\tau ]\left(r):= \frac{1}{2}| \tau ^{\prime} \left(r){| }^{2}+\left(\frac{1}{2}+\frac{\beta }{4}\right)| \tau \left(r){| }^{2}+\frac{1}{3}{r}^{\beta }| \tau \left(r){| }^{3}.By (3.12), we have τ′(r)≤0\tau ^{\prime} \left(r)\le 0, then by multiplying (3.2) by τ′(r)\tau ^{\prime} \left(r), we obtain (3.14)ddrI[τ](r)=τ′τ″+1+β2ττ′+rβτ2τ′+β3rβ−1∣τ∣3=−N−1r+r2∣τ′∣2+(τ′)3τ+β3rβ−1∣τ∣3≤β3rβ−1∣τ∣3≤βrI[τ](r).\begin{array}{rcl}\frac{{\rm{d}}}{{\rm{d}}r}I\left[\tau ]\left(r)& =& \tau ^{\prime} {\tau }^{^{\prime\prime} }+\left(1+\frac{\beta }{2}\right)\tau \tau ^{\prime} +{r}^{\beta }{\tau }^{2}\tau ^{\prime} +\frac{\beta }{3}{r}^{\beta -1}| \tau {| }^{3}\\ & =& -\left(\frac{N-1}{r}+\frac{r}{2}\right)| \tau ^{\prime} {| }^{2}+\frac{{\left(\tau ^{\prime} )}^{3}}{\tau }+\frac{\beta }{3}{r}^{\beta -1}| \tau {| }^{3}\\ & \le & \frac{\beta }{3}{r}^{\beta -1}| \tau {| }^{3}\\ & \le & \frac{\beta }{r}I\left[\tau ]\left(r).\end{array}Then we have (3.15)I[τ](r)≤Crβ.I\left[\tau ]\left(r)\le C{r}^{\beta }.It is apparent that ∣τ(r)∣≤C| \tau \left(r)| \le Cfrom the definition of I[τ](r)I\left[\tau ]\left(r). Therefore, we obtain that the solution of (3.2) exists globally in time. Hence, the proof of Proposition 3.3 is completed.□3.2Asymptotic behavior of forward self-similar solutionsIn this subsection, we study the asymptotic behavior of the solution for (3.2) and obtain the decay rate of the solution for (3.2) as r→∞r\to \infty . The proofs of Lemmas 3.4 and 3.5, which follows the ideas of Haraux and Weissler [16, Proposition 3.1], are also enclosed in this section.Lemma 3.4Let N≥1N\ge 1, 0<β<20\lt \beta \lt 2, α>0\alpha \gt 0, and τ\tau be the solution of (3.2) satisfying τ(0)=α\tau \left(0)=\alpha . Assume that there exist constants C>0C\gt 0and σ≥0\sigma \ge 0such that 0<τ(r)≤C(1+r)−σ0\lt \tau \left(r)\le C{\left(1+r)}^{-\sigma }for r>0r\gt 0. Then there exists a constant C˜>0\tilde{C}\gt 0such that ∣τ′(r)∣≤C˜(1+r)β−2σ−1| \tau ^{\prime} \left(r)| \le \tilde{C}{\left(1+r)}^{\beta -2\sigma -1}if β≥σ\beta \ge \sigma , and ∣τ′(r)∣≤C˜(1+r)−σ−1| \tau ^{\prime} \left(r)| \le \tilde{C}{\left(1+r)}^{-\sigma -1}if β<σ\beta \lt \sigma for all r>0r\gt 0.ProofWe introduce g(r)≔1+β2τ(r)+rβτ(r)2−τ′(r)2τ(r).g\left(r):= \left(1+\frac{\beta }{2}\right)\tau \left(r)+{r}^{\beta }\tau {\left(r)}^{2}-\frac{\tau ^{\prime} {\left(r)}^{2}}{\tau \left(r)}.Then (3.2) is equivalent to (3.16)rN−1er24τ′(r)′=−g(r)rN−1er24,\left({r}^{N-1}{e}^{\tfrac{{r}^{2}}{4}}\tau ^{\prime} \left(r)\right)^{\prime} =-g\left(r){r}^{N-1}{e}^{\tfrac{{r}^{2}}{4}},and since τ′(0)=0\tau ^{\prime} \left(0)=0, integration of (3.16) gives (3.17)τ′(r)=−r1−Ne−r24∫0rg(t)tN−1et24dt.\tau ^{\prime} \left(r)=-{r}^{1-N}{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{r}{\int }}g\left(t){t}^{N-1}{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t.Note that ∣g(r)∣≤C1(1+r)−σ+C2(1+r)β−2σ.| g\left(r)| \le {C}_{1}{\left(1+r)}^{-\sigma }+{C}_{2}{\left(1+r)}^{\beta -2\sigma }.Next we split the proof into three cases.Case A: If β−2σ≤−σ<0\beta -2\sigma \le -\sigma \lt 0, then ∣g(r)∣≤C(1+r)−σ| g\left(r)| \le C{\left(1+r)}^{-\sigma }. Thus, ∣τ′(r)∣≤r1−Ne−r24∫0r∣g(t)∣tN−1et24dt≤Ce−r24∫0r(1+t)−σet24dt≤Ce−r24∫0r2et24dt+Ce−r24∫r2r(1+t)−σet24dt≤Cr2e−3r216+1+r2−σ−1e−r24∫r2r(1+t)et24dt.\begin{array}{rcl}| \tau ^{\prime} \left(r)| & \le & {r}^{1-N}{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{r}{\displaystyle \int }}| g\left(t)| {t}^{N-1}{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\\ & \le & C{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{r}{\displaystyle \int }}{\left(1+t)}^{-\sigma }{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\\ & \le & C{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{\frac{r}{2}}{\displaystyle \int }}{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t+C{e}^{-\tfrac{{r}^{2}}{4}}\underset{\frac{r}{2}}{\overset{r}{\displaystyle \int }}{\left(1+t)}^{-\sigma }{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\\ & \le & C\left[\frac{r}{2}{e}^{-\tfrac{3{r}^{2}}{16}}+{\left(1+\frac{r}{2}\right)}^{-\sigma -1}{e}^{-\tfrac{{r}^{2}}{4}}\underset{\frac{r}{2}}{\overset{r}{\displaystyle \int }}\left(1+t){e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\right].\end{array}If r≤2r\le 2, τ′(r)\tau ^{\prime} \left(r)is obviously bounded. If r≥2r\ge 2, the integral in the aforementioned formula is controlled by ∫r2r2tet24dt=4er24−4er216≤4er24.\underset{\frac{r}{2}}{\overset{r}{\int }}2t{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t=4{e}^{\tfrac{{r}^{2}}{4}}-4{e}^{\tfrac{{r}^{2}}{16}}\le 4{e}^{\tfrac{{r}^{2}}{4}}.Then, ∣τ′(r)∣≤C˜(1+r)−σ−1| \tau ^{\prime} \left(r)| \le \tilde{C}{\left(1+r)}^{-\sigma -1}.Case B: If −σ≤β−2σ<0-\sigma \le \beta -2\sigma \lt 0, then ∣g(r)∣≤C(1+r)β−2σ| g\left(r)| \le C{\left(1+r)}^{\beta -2\sigma }. Thus, by the same argument in Case A, we obtain ∣τ′(r)∣≤C˜(1+r)β−2σ−1| \tau ^{\prime} \left(r)| \le \tilde{C}{\left(1+r)}^{\beta -2\sigma -1}.Case C: If −σ<0≤β−2σ-\sigma \lt 0\le \beta -2\sigma , then ∣g(r)∣≤C(1+r)β−2σ| g\left(r)| \le C{\left(1+r)}^{\beta -2\sigma }. Thus, ∣τ′(r)∣≤r1−Ne−r24∫0r∣g(t)∣tN−1et24dt≤Ce−r24∫0r(1+t)β−2σet24dt=Ce−r24∫0r2(1+t)β−2σet24dt+Ce−r24∫r2r(1+t)β−2σet24dt≤Ce−r241+r2β−2σer216∫0r2dt+Ce−r24∫r2r(1+t)β−2σ−1(1+t)et24dt≤Ce−3r2161+r2β−2σ+1+C1+r2β−2σ−1e−r24∫r2r(1+t)et24dt\begin{array}{rcl}| \tau ^{\prime} \left(r)| & \le & {r}^{1-N}{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{r}{\displaystyle \int }}| g\left(t)| {t}^{N-1}{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\\ & \le & C{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{r}{\displaystyle \int }}{\left(1+t)}^{\beta -2\sigma }{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\\ & =& C{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{\frac{r}{2}}{\displaystyle \int }}{\left(1+t)}^{\beta -2\sigma }{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t+C{e}^{-\tfrac{{r}^{2}}{4}}\underset{\frac{r}{2}}{\overset{r}{\displaystyle \int }}{\left(1+t)}^{\beta -2\sigma }{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\\ & \le & C{e}^{-\tfrac{{r}^{2}}{4}}{\left(1+\frac{r}{2}\right)}^{\beta -2\sigma }{e}^{\tfrac{{r}^{2}}{16}}\underset{0}{\overset{\frac{r}{2}}{\displaystyle \int }}{\rm{d}}t+C{e}^{-\tfrac{{r}^{2}}{4}}\underset{\frac{r}{2}}{\overset{r}{\displaystyle \int }}{\left(1+t)}^{\beta -2\sigma -1}\left(1+t){e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\\ & \le & C{e}^{-\tfrac{3{r}^{2}}{16}}{\left(1+\frac{r}{2}\right)}^{\beta -2\sigma +1}+C{\left(1+\frac{r}{2}\right)}^{\beta -2\sigma -1}{e}^{-\tfrac{{r}^{2}}{4}}\underset{\frac{r}{2}}{\overset{r}{\displaystyle \int }}\left(1+t){e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\end{array}for rrlarge. And CCdenotes a positive generic constant. Next we can use the similar method in Case A to prove Case C. Hence, the proof of Lemma 3.4 is completed.□Lemma 3.5Let N≥1N\ge 1, 0<β<20\lt \beta \lt 2, α>0\alpha \gt 0, and τ\tau be the solution of (3.2) satisfying τ(0)=α\tau \left(0)=\alpha . Then there exists a constant C>0C\gt 0such that0<τ(r)≤C(1+r)−2−β2,∣τ′(r)∣≤C(1+r)−3−β20\lt \tau \left(r)\le C{\left(1+r)}^{-2-\tfrac{\beta }{2}},\hspace{1.0em}| \tau ^{\prime} \left(r)| \le C{\left(1+r)}^{-3-\tfrac{\beta }{2}}for all r>0r\gt 0.ProofMultiplying (3.2) by r−β−1τ(r){r}^{-\beta -1}\tau \left(r), we obtain r−β−11+β2τ2+r−1τ3=−r−βddrτ24+ττ′r+2r−β−1(τ′)2−r−β−2Nττ′.{r}^{-\beta -1}\left(1+\frac{\beta }{2}\right){\tau }^{2}+{r}^{-1}{\tau }^{3}=-{r}^{-\beta }\frac{{\rm{d}}}{{\rm{d}}r}\left(\frac{{\tau }^{2}}{4}+\frac{\tau \tau ^{\prime} }{r}\right)+2{r}^{-\beta -1}{\left(\tau ^{\prime} )}^{2}-{r}^{-\beta -2}N\tau \tau ^{\prime} .This together with (3.13) implies r−β−1I[τ](r)=12r−β−1(τ′)2+12+β4r−β−1τ2+13r−1τ3≤12r−β−1(τ′)2+1+β2r−β−1τ2+r−1τ3=−12r−βddrτ24+ττ′r−12r−β−2Nττ′+32r−β−1(τ′)2.\begin{array}{rcl}{r}^{-\beta -1}I\left[\tau ]\left(r)& =& \frac{1}{2}{r}^{-\beta -1}{\left(\tau ^{\prime} )}^{2}+\left(\frac{1}{2}+\frac{\beta }{4}\right){r}^{-\beta -1}{\tau }^{2}+\frac{1}{3}\hspace{0.33em}{r}^{-1}{\tau }^{3}\\ & \le & \frac{1}{2}\left({r}^{-\beta -1}{\left(\tau ^{\prime} )}^{2}+\left(1+\frac{\beta }{2}\right){r}^{-\beta -1}{\tau }^{2}+{r}^{-1}{\tau }^{3}\right)\\ & =& -\frac{1}{2}{r}^{-\beta }\frac{{\rm{d}}}{{\rm{d}}r}\left(\frac{{\tau }^{2}}{4}+\frac{\tau \tau ^{\prime} }{r}\right)-\frac{1}{2}{r}^{-\beta -2}N\tau \tau ^{\prime} +\frac{3}{2}{r}^{-\beta -1}{\left(\tau ^{\prime} )}^{2}.\end{array}Since ∣τ′(r)∣≤C(1+r)β−1| \tau ^{\prime} \left(r)| \le C{\left(1+r)}^{\beta -1}by Lemma 3.4 (with σ=0\sigma =0), we have ττ′rβ+1→0\frac{\tau \tau ^{\prime} }{{r}^{\beta +1}}\to 0as r→∞r\to \infty . And τ28≤I[τ](r)4+2β\frac{{\tau }^{2}}{8}\le \frac{I\left[\tau ]\left(r)}{4+2\beta }by (3.13), τ′(r)<0\tau ^{\prime} \left(r)\lt 0by (3.12). Thus, we obtain ∫r∞t−β−1I[τ](t)dt=−12∫r∞t−βdτ24+ττ′t−N2∫r∞t−β−2ττ′dt+32∫r∞t−β−1(τ′)2dt=r−βτ28+ττ′2r−limr→∞r−βτ28+ττ′2r−β2∫r∞t−β−1τ24+ττ′tdt−N2∫r∞t−β−2ττ′dt+32∫r∞t−β−1(τ′)2dt≤r−βI[τ](r)4+2β+β+N2∫r∞t−β−2∣ττ′∣dt+32∫r∞t−β−1(τ′)2dt.\begin{array}{rcl}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -1}I\left[\tau ]\left(t){\rm{d}}t& =& -\frac{1}{2}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta }d\left(\frac{{\tau }^{2}}{4}+\frac{\tau \tau ^{\prime} }{t}\right)-\frac{N}{2}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -2}\tau \tau ^{\prime} {\rm{d}}t+\frac{3}{2}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -1}{\left(\tau ^{\prime} )}^{2}{\rm{d}}t\\ & =& {r}^{-\beta }\left(\frac{{\tau }^{2}}{8}+\frac{\tau \tau ^{\prime} }{2r}\right)-\mathop{\mathrm{lim}}\limits_{r\to \infty }\left[{r}^{-\beta }\left(\frac{{\tau }^{2}}{8}+\frac{\tau \tau ^{\prime} }{2r}\right)\right]-\frac{\beta }{2}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -1}\left(\frac{{\tau }^{2}}{4}+\frac{\tau \tau ^{\prime} }{t}\right){\rm{d}}t-\frac{N}{2}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -2}\tau \tau ^{\prime} {\rm{d}}t+\frac{3}{2}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -1}{\left(\tau ^{\prime} )}^{2}{\rm{d}}t\\ & \le & {r}^{-\beta }\frac{I\left[\tau ]\left(r)}{4+2\beta }+\frac{\beta +N}{2}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -2}| \tau \tau ^{\prime} | {\rm{d}}t+\frac{3}{2}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -1}{\left(\tau ^{\prime} )}^{2}{\rm{d}}t.\end{array}Assume now (3.18)I[τ](r)≤Crβ−σ,r≥1I\left[\tau ]\left(r)\le C{r}^{\beta -\sigma },\hspace{1.0em}r\ge 1for some fixed σ≥0\sigma \ge 0. By (3.13), we have ∣τ(r)∣≤Cr−σ3,r≥1.| \tau \left(r)| \le C{r}^{-\tfrac{\sigma }{3}},\hspace{1.0em}r\ge 1.First, we consider the case where β≥σ3\beta \ge \frac{\sigma }{3}, then ∣τ′(r)∣≤Crβ−2σ3−1| \tau ^{\prime} \left(r)| \le C{r}^{\beta -\tfrac{2\sigma }{3}-1}by Lemma 3.4. Therefore, (3.19)∫r∞t−β−1I[τ](t)dt≤r−βI[τ](r)4+2β+C∫r∞t−σ−3dt+C∫r∞tβ−4σ3−3dt≤r−βI[τ](r)4+2β+Cr−σ−2+Crβ−4σ3−2≤r−βI[τ](r)4+2β+Crβ−4σ3−2,\begin{array}{rcl}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -1}I\left[\tau ]\left(t){\rm{d}}t& \le & {r}^{-\beta }\frac{I\left[\tau ]\left(r)}{4+2\beta }+C\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\sigma -3}{\rm{d}}t+C\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{\beta -\tfrac{4\sigma }{3}-3}{\rm{d}}t\\ & \le & {r}^{-\beta }\frac{I\left[\tau ]\left(r)}{4+2\beta }+C{r}^{-\sigma -2}+C{r}^{\beta -\tfrac{4\sigma }{3}-2}\\ & \le & {r}^{-\beta }\frac{I\left[\tau ]\left(r)}{4+2\beta }+C{r}^{\beta -\tfrac{4\sigma }{3}-2},\end{array}which is equivalent to (3.20)(4+2β)∫r∞t−β−1I[τ](t)dt≤r−βI[τ](r)+Crβ−4σ3−2,r≥1.\left(4+2\beta )\underset{r}{\overset{\infty }{\int }}{t}^{-\beta -1}I\left[\tau ]\left(t){\rm{d}}t\le {r}^{-\beta }I\left[\tau ]\left(r)+C{r}^{\beta -\tfrac{4\sigma }{3}-2},\hspace{1.0em}r\ge 1.Define F(r)≔r−βI[τ](r),G(r)≔∫r∞t−1F(t)dt.F\left(r):= {r}^{-\beta }I\left[\tau ]\left(r),\hspace{1.0em}G\left(r):= \underset{r}{\overset{\infty }{\int }}{t}^{-1}F\left(t){\rm{d}}t.It is easy to see that (3.21)(4+2β)∫r∞t−1F(t)dt≤F(r)+Crβ−4σ3−2,\left(4+2\beta )\underset{r}{\overset{\infty }{\int }}{t}^{-1}F\left(t){\rm{d}}t\le F\left(r)+C{r}^{\beta -\tfrac{4\sigma }{3}-2},and (3.22)dG(r)dr=−r−1F(r).\frac{{\rm{d}}G\left(r)}{{\rm{d}}r}=-{r}^{-1}F\left(r).Combining (3.21) with (3.22), we obtain (3.23)(4+2β)G(r)+rdG(r)dr≤Crβ−4σ3−2,\left(4+2\beta )G\left(r)+r\frac{{\rm{d}}G\left(r)}{{\rm{d}}r}\le C{r}^{\beta -\tfrac{4\sigma }{3}-2},that is, (3.24)d[r4+2βG(r)]dr≤Cr3β−4σ3+1,r≥1.\frac{{\rm{d}}{[}{r}^{4+2\beta }G\left(r)]}{{\rm{d}}r}\le C{r}^{3\beta -\tfrac{4\sigma }{3}+1},\hspace{1.0em}r\ge 1.We divided the remainder of proof into two cases.Case A: If 3β−4σ3+1≠−13\beta -\frac{4\sigma }{3}+1\ne -1, by integrating (3.24) from 1 to rr, we yield (3.25)G(r)≤Cr−4−2β+Crβ−4σ3−2,r≥1.G\left(r)\le C{r}^{-4-2\beta }+C{r}^{\beta -\tfrac{4\sigma }{3}-2},\hspace{1.0em}r\ge 1.By (3.14), dF(r)dr=rβdI[τ](r)dr−βr−1I[τ](r)≤0,\frac{{\rm{d}}F\left(r)}{{\rm{d}}r}={r}^{\beta }\left[\frac{{\rm{d}}I\left[\tau ]\left(r)}{{\rm{d}}r}-\beta {r}^{-1}I\left[\tau ]\left(r)\right]\le 0,then G(r)=∫r∞t−1F(t)dt≥∫r2rt−1F(t)dt≥F(2r)2,G\left(r)=\underset{r}{\overset{\infty }{\int }}{t}^{-1}F\left(t){\rm{d}}t\ge \underset{r}{\overset{2r}{\int }}{t}^{-1}F\left(t){\rm{d}}t\ge \frac{F\left(2r)}{2},or (3.26)F(r)≤2Gr2.F\left(r)\le 2G\left(\frac{r}{2}\right).Thus, (3.27)F(r)≤Cr−4−2β+Crβ−4σ3−2.F\left(r)\le C{r}^{-4-2\beta }+C{r}^{\beta -\tfrac{4\sigma }{3}-2}.That is, (3.28)I[τ](r)≤Cr−4−β+Cr2β−4σ3−2,r≥2.I\left[\tau ]\left(r)\le C{r}^{-4-\beta }+C{r}^{2\beta -\tfrac{4\sigma }{3}-2},\hspace{1.0em}r\ge 2.From (3.15), we know that I[τ](r)≤CrβI\left[\tau ]\left(r)\le C{r}^{\beta }for all r≥0r\ge 0, then (3.28) holds for r≥1r\ge 1.Clearly (3.28) holds with σ0=0{\sigma }_{0}=0by (3.15), then (3.18) holds with σ1=min{4+2β,2−β}=2−β{\sigma }_{1}=\min \{4+2\beta ,2-\beta \}=2-\beta . We can continue recursively letting σi+1=min4+2β,4σi3+2−β{\sigma }_{i+1}=\min \left\{4+2\beta ,\frac{4{\sigma }_{i}}{3}+2-\beta \right\}until (3.29)I[τ](r)≤Crβ−σj,β<σj3.I\left[\tau ]\left(r)\le C{r}^{\beta -{\sigma }_{j}},\hspace{1.0em}\beta \lt \frac{{\sigma }_{j}}{3}.Then ∣τ(r)∣≤Cr−σj3| \tau \left(r)| \le C{r}^{-\tfrac{{\sigma }_{j}}{3}}, and ∣τ′(r)∣≤Cr−σj3−1| \tau ^{\prime} \left(r)| \le C{r}^{-\tfrac{{\sigma }_{j}}{3}-1}by Lemma 3.4. Thus, (3.30)∫r∞t−β−1I[τ](t)dt≤r−βI[τ](r)4+2β+Cr−β−2σj3−2.\underset{r}{\overset{\infty }{\int }}{t}^{-\beta -1}I\left[\tau ]\left(t){\rm{d}}t\le {r}^{-\beta }\frac{I\left[\tau ]\left(r)}{4+2\beta }+C{r}^{-\beta -\tfrac{2{\sigma }_{j}}{3}-2}.That is, (3.31)(4+2β)∫r∞t−β−1I[τ](t)dt≤r−βI[τ](r)+Cr−β−2σj3−2.\left(4+2\beta )\underset{r}{\overset{\infty }{\int }}{t}^{-\beta -1}I\left[\tau ]\left(t){\rm{d}}t\le {r}^{-\beta }I\left[\tau ]\left(r)+C{r}^{-\beta -\tfrac{2{\sigma }_{j}}{3}-2}.We can obtain (3.32)d[r4+2βG(r)]dr≤Crβ−2σj3+1.\frac{{\rm{d}}{[}{r}^{4+2\beta }G\left(r)]}{{\rm{d}}r}\le C{r}^{\beta -\tfrac{2{\sigma }_{j}}{3}+1}.If β−2σj3+1≠−1\beta -\frac{2{\sigma }_{j}}{3}+1\ne -1, then I[τ](r)≤Cr−4−β+Cr−2σj3−2I\left[\tau ]\left(r)\le C{r}^{-4-\beta }+C{r}^{-\tfrac{2{\sigma }_{j}}{3}-2}. Letting σj+1=min4+2β,β+2σj3+2{\sigma }_{j+1}=\min \left\{4+2\beta ,\beta +\frac{2{\sigma }_{j}}{3}+2\right\}, we continue recursively until (3.29) holds with σj+1=4+2β{\sigma }_{j+1}=4+2\beta . If β−2σj3+1=−1\beta -\frac{2{\sigma }_{j}}{3}+1=-1, we can infer a similar formula as (3.25), then a same recursive argument with σ0=0{\sigma }_{0}=0and σj+1=min4+2β,β+2σj3+1{\sigma }_{j+1}=\min \left\{4+2\beta ,\beta +\frac{2{\sigma }_{j}}{3}+1\right\}shows the expected result. Thus, I[τ](r)≤Cr−4−βI\left[\tau ]\left(r)\le C{r}^{-4-\beta }.Case B: If 3β−4σ3+1=−13\beta -\frac{4\sigma }{3}+1=-1, we can derive a formula similar to (3.25). A same recursive argument with σ0=0{\sigma }_{0}=0and σi+1=min4+2β,4σi3+1−β{\sigma }_{i+1}=\min \left\{4+2\beta ,\frac{4{\sigma }_{i}}{3}+1-\beta \right\}can be used to obtain I[τ](r)≤Cr−4−βI\left[\tau ]\left(r)\le C{r}^{-4-\beta }for r≥1r\ge 1.In both cases, we have I[τ](r)≤Cr−4−β,r≥1.I\left[\tau ]\left(r)\le C{r}^{-4-\beta },\hspace{1.0em}r\ge 1.Together with I[τ](r)≤CrβI\left[\tau ]\left(r)\le C{r}^{\beta }for all r≥0r\ge 0, it turns out I[τ](r)≤C(1+r)−4−β,r≥0.I\left[\tau ]\left(r)\le C{\left(1+r)}^{-4-\beta },\hspace{1.0em}r\ge 0.Thus, by (3.13), ∣τ(r)∣≤C(1+r)−2−β2,r≥0.| \tau \left(r)| \le C{\left(1+r)}^{-2-\tfrac{\beta }{2}},\hspace{1.0em}r\ge 0.From Lemma 3.4, we obtain the estimate for ∣τ′(r)∣| \tau ^{\prime} \left(r)| , i.e., ∣τ′(r)∣≤C(1+r)−3−β2,r≥0.| \tau ^{\prime} \left(r)| \le C{\left(1+r)}^{-3-\tfrac{\beta }{2}},\hspace{1.0em}r\ge 0.The letter CCdenotes a positive generic constant. Thus, Lemma 3.5 holds.□On account of aforementioned lemmas, we have the following corollary.Corollary 3.6Let N≥1N\ge 1, α>0\alpha \gt 0, 0<β<20\lt \beta \lt 2, and τ\tau be the solution of (3.2) with initial data τ(0)=α\tau \left(0)=\alpha . Then the limitd(α)≔limr→∞r2+βτ(r)d\left(\alpha ):= \mathop{\mathrm{lim}}\limits_{r\to \infty }{r}^{2+\beta }\tau \left(r)exists in R{\mathbb{R}}and d(α)>0d\left(\alpha )\gt 0for any α>0\alpha \gt 0.ProofBy (3.3), we have r2+βτ(r)=r2+βαexp−∫0rσN(s)−1∫0stβτ(t)+1+β2σN(t)dtds=αexp(2+β)logr−1+β2γN(r)−∫0rσN(s)−1∫0stβτ(t)σN(t)dtds.\begin{array}{rcl}{r}^{2+\beta }\tau \left(r)& =& {r}^{2+\beta }\alpha \exp \left\{-\underset{0}{\overset{r}{\displaystyle \int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\displaystyle \int }}\left[{t}^{\beta }\tau \left(t)+1+\frac{\beta }{2}\right]{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\}\\ & =& \alpha \exp \left\{\left(2+\beta )\log r-\left(1+\frac{\beta }{2}\right){\gamma }_{N}\left(r)-\underset{0}{\overset{r}{\displaystyle \int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\displaystyle \int }}{t}^{\beta }\tau \left(t){\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\}.\end{array}First, for s>0s\gt 0is small, we have σN(s)−1∫0stβτ(t)σN(t)dt≤s1−Ne−s24⋅sβ⋅α⋅sN−1es24⋅s=αsβ+1.{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}{t}^{\beta }\tau \left(t){\sigma }_{N}\left(t){\rm{d}}t\right)\le {s}^{1-N}{e}^{-\tfrac{{s}^{2}}{4}}\cdot {s}^{\beta }\cdot \alpha \cdot {s}^{N-1}{e}^{\tfrac{{s}^{2}}{4}}\cdot s=\alpha {s}^{\beta +1}.Next, for s>0s\gt 0large enough, by Lemmas 3.2 and 3.5, we see that there exist constants C1>0{C}_{1}\gt 0and C2>0{C}_{2}\gt 0such that σN(s)−1∫0stβτ(t)σN(t)dt≤C1σN(s)−1∫0stβ(1+t)−2−β2σN(t)dt≤C1σN(s)−1(1+s)β2∫0s(1+t)−2σN(t)dt≤C2σN(s)−1(1+s)β2⋅sN−4es24≤C2(1+s)β2−3.\begin{array}{rcl}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\displaystyle \int }}{t}^{\beta }\tau \left(t){\sigma }_{N}\left(t){\rm{d}}t\right)& \le & {C}_{1}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\displaystyle \int }}{t}^{\beta }{\left(1+t)}^{-2-\tfrac{\beta }{2}}{\sigma }_{N}\left(t){\rm{d}}t\right)\\ & \le & {C}_{1}{\sigma }_{N}{\left(s)}^{-1}{\left(1+s)}^{\tfrac{\beta }{2}}\left(\underset{0}{\overset{s}{\displaystyle \int }}{\left(1+t)}^{-2}{\sigma }_{N}\left(t){\rm{d}}t\right)\\ & \le & {C}_{2}\hspace{0.33em}{\sigma }_{N}{\left(s)}^{-1}{\left(1+s)}^{\tfrac{\beta }{2}}\cdot {s}^{N-4}{e}^{\tfrac{{s}^{2}}{4}}\\ & \le & {C}_{2}{\left(1+s)}^{\tfrac{\beta }{2}-3}.\end{array}Thus, we obtain ∫0∞σN(s)−1∫0stβτ(t)σN(t)dtds<∞.\underset{0}{\overset{\infty }{\int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}{t}^{\beta }\tau \left(t){\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\lt \infty .Therefore, by Lemma 3.1, we derive that the limit d(α)d\left(\alpha )exists and (3.33)d(α)=αexplimr→∞(2+β)logr−1+β2γN(r)−∫0∞σN(s)−1∫0stβτ(t)σN(t)dtds.d\left(\alpha )=\alpha \exp \left\{\mathop{\mathrm{lim}}\limits_{r\to \infty }\left[\left(2+\beta )\log r-\left(1+\frac{\beta }{2}\right){\gamma }_{N}\left(r)\right]-\underset{0}{\overset{\infty }{\int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}{t}^{\beta }\tau \left(t){\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\}.Then we complete the proof of Corollary 3.6.□The next Lemma is about the continuity of d(α)d\left(\alpha )with respect to α\alpha .Lemma 3.7Let N≥1N\ge 1, 0<β<20\lt \beta \lt 2, a>0a\gt 0, and b>0b\gt 0. Let τa{\tau }_{a}and τb{\tau }_{b}be the solutions of (3.2) and satisfy the condition τa(0)=a{\tau }_{a}\left(0)=aand τb(0)=b{\tau }_{b}\left(0)=bseparately. Then there exists a constant C>0C\gt 0, which depends only on NN, such thatsupr≥0(1+r)2+β∣τa(r)−τb(r)∣≤C∣a−b∣\mathop{\sup }\limits_{r\ge 0}{\left(1+r)}^{2+\beta }| {\tau }_{a}\left(r)-{\tau }_{b}\left(r)| \le C| a-b| for all a,b>0a,b\gt 0. In particular, d(α)d\left(\alpha )is continuous with respect to α\alpha .ProofBy (3.3), we have τa(r)−τb(r)=aexp−∫0rσN(s)−1∫0stβτa(t)+1+β2σN(t)dtds−exp−∫0rσN(s)−1∫0stβτb(t)+1+β2σN(t)dtds+(a−b)exp−∫0rσN(s)−1∫0stβτb(t)+1+β2σN(t)dtds.\begin{array}{rcl}{\tau }_{a}\left(r)-{\tau }_{b}\left(r)& =& a\left[\exp \left\{-\underset{0}{\overset{r}{\displaystyle \int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\displaystyle \int }}\left[{t}^{\beta }{\tau }_{a}\left(t)+1+\frac{\beta }{2}\right]{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\}-\exp \left\{-\underset{0}{\overset{r}{\displaystyle \int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\displaystyle \int }}\left[{t}^{\beta }{\tau }_{b}\left(t)+1+\frac{\beta }{2}\right]{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\}\right]\\ & & +\left(a-b)\exp \left\{-\underset{0}{\overset{r}{\displaystyle \int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\displaystyle \int }}\left[{t}^{\beta }{\tau }_{b}\left(t)+1+\frac{\beta }{2}\right]{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\}.\end{array}Since τa,τb{\tau }_{a},{\tau }_{b}and σN{\sigma }_{N}are positive functions, by using (3.11), we have ∣τa(r)−τb(r)∣≤e−1+β2γN(r)a∫0rσN(s)−1∫0stβ∣τa(t)−τb(t)∣σN(t)dtds+∣a−b∣,| {\tau }_{a}\left(r)-{\tau }_{b}\left(r)| \le {e}^{-\left(1+\tfrac{\beta }{2}\right){\gamma }_{N}\left(r)}\left[a\underset{0}{\overset{r}{\int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}{t}^{\beta }| {\tau }_{a}\left(t)-{\tau }_{b}\left(t)| {\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s+| a-b| \right],where γN{\gamma }_{N}is the function defined by (3.4). Hence, we have (1+r)2+β∣τa(r)−τb(r)∣≤(1+r)2+βe−1+β2γN(r)a∫0rsup0≤t≤s(1+t)2+β∣τa(t)−τb(t)∣×σN(s)−1∫0s(1+t)−2σN(t)dtds+∣a−b∣.{\left(1+r)}^{2+\beta }| {\tau }_{a}\left(r)-{\tau }_{b}\left(r)| \le {\left(1+r)}^{2+\beta }{e}^{-\left(1+\tfrac{\beta }{2}\right){\gamma }_{N}\left(r)}\left[a\underset{0}{\overset{r}{\int }}\mathop{\sup }\limits_{0\le t\le s}{\left(1+t)}^{2+\beta }| {\tau }_{a}\left(t)-{\tau }_{b}\left(t)| \times {\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}{\left(1+t)}^{-2}{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s+| a-b| \right].By Lemma 3.1, we have supr≥0(1+r)2+βe−1+β2γN(r)=supr≥0exp(2+β)log(1+r)−1+β2γN(r)=supr≥0exp1+β2(2log(1+r)−γN(r))<∞.\begin{array}{rcl}\mathop{\sup }\limits_{r\ge 0}{\left(1+r)}^{2+\beta }{e}^{-\left(1+\tfrac{\beta }{2}\right)}{\gamma }_{N}\left(r)& =& \mathop{\sup }\limits_{r\ge 0}\exp \left[\left(2+\beta )\log \left(1+r)-\left(1+\frac{\beta }{2}\right){\gamma }_{N}\left(r)\right]\\ & =& \mathop{\sup }\limits_{r\ge 0}\exp \left[\left(1+\frac{\beta }{2}\right)(2\log \left(1+r)-{\gamma }_{N}\left(r))\right]\lt \infty .\end{array}We thus deduce that there exists a constant C>0C\gt 0, which depends only on NN, such that h1(r)≤Ca∫0rh1(s)h2(s)ds+C∣a−b∣,{h}_{1}\left(r)\le Ca\underset{0}{\overset{r}{\int }}{h}_{1}\left(s){h}_{2}\left(s){\rm{d}}s+C| a-b| ,where h1(s)≔sup0≤t≤s(1+t)2+β∣τa(t)−τb(t)∣,h2(s)≔σN(s)−1∫0s(1+t)−2σN(t)dt.{h}_{1}\left(s):= \mathop{\sup }\limits_{0\le t\le s}{\left(1+t)}^{2+\beta }| {\tau }_{a}\left(t)-{\tau }_{b}\left(t)| ,\hspace{1.0em}{h}_{2}\left(s):= {\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}{\left(1+t)}^{-2}{\sigma }_{N}\left(t){\rm{d}}t\right).Then, by using the Gronwall inequality, we obtain h1(r)≤C∣a−b∣CaexpCa∫0∞h2(r)dr∫0∞h2(s)ds+1.{h}_{1}\left(r)\le C| a-b| \left[Ca\exp \left(Ca\underset{0}{\overset{\infty }{\int }}{h}_{2}\left(r)dr\right)\underset{0}{\overset{\infty }{\int }}{h}_{2}\left(s){\rm{d}}s+1\right].From Lemma 3.2, we know that h2∈L1(0,∞){h}_{2}\in {L}^{1}\left(0,\infty ). Therefore, h1(r)≤C∣a−b∣.{h}_{1}\left(r)\le C| a-b| .The proof of Lemma 3.7 is completed.□4Multiplicity of forward self-similar solutionsIn this section, we study the multiplicity of the solutions for the problem (1.8) by an ODE approach, which follows from [22]. For α∈R\alpha \in {\mathbb{R}}, let ωα{\omega }_{\alpha }be the solution of (1.8) with ω(0)=α\omega \left(0)=\alpha . Set (4.1)zα(s)≔ωα(r)−α,s≔eα2+βr.{z}_{\alpha }\left(s):= {\omega }_{\alpha }\left(r)-\alpha ,\hspace{1.0em}s:= {e}^{\tfrac{\alpha }{2+\beta }}r.Then a direct computation shows that (4.2)(2+β)logs+zα(s)=(2+β)logr+ωα(r),\left(2+\beta )\log s+{z}_{\alpha }\left(s)=\left(2+\beta )\log r+{\omega }_{\alpha }\left(r),and zα{z}_{\alpha }satisfies (4.3)zα″+N−1s+e−2α2+βs2z′+sβezα+1+β2e−2α2+β=0,s>0,zα(0)=0,zα′(0)=0.\left\{\begin{array}{l}{z}_{\alpha }^{^{\prime\prime} }+\left(\frac{N-1}{s}+{e}^{\tfrac{-2\alpha }{2+\beta }}\frac{s}{2}\right)z^{\prime} +{s}^{\beta }{e}^{{z}_{\alpha }}+\left(1+\frac{\beta }{2}\right){e}^{\tfrac{-2\alpha }{2+\beta }}=0,\hspace{1.0em}s\gt 0,\hspace{1.0em}\\ {z}_{\alpha }\left(0)=0,\hspace{1.0em}{z}_{\alpha }^{\prime} \left(0)=0.\hspace{1.0em}\end{array}\right.Let ZZbe the solution of (4.4)Z″+N−1sZ′+sβeZ=0in(0,∞){Z}^{^{\prime\prime} }+\frac{N-1}{s}Z^{\prime} +{s}^{\beta }{e}^{Z}=0\hspace{1.0em}{\rm{in}}\hspace{0.33em}\left(0,\infty )with Z(0)=0Z\left(0)=0and Z′(0)=0Z^{\prime} \left(0)=0. One can easily compute that Z∗(s)≔−(2+β)logs+log[(2+β)(N−2)]{Z}_{\ast }\left(s):= -\left(2+\beta )\log s+\log {[}\left(2+\beta )\left(N-2)]is a singular solution of equation (4.4). We can use the similar method in [31, Theorem 1.1, Lemma 2.1] to study intersection properties of solutions for (4.4) and obtain the following lemma.Lemma 4.1Let N≥3N\ge 3, 0<β<20\lt \beta \lt 2, and Z be the solution of (4.4). Thenlims→∞[(2+β)logs+Z(s)]=log[(2+β)(N−2)].\mathop{\mathrm{lim}}\limits_{s\to \infty }{[}\left(2+\beta )\log s+Z\left(s)]=\log {[}\left(2+\beta )\left(N-2)].Furthermore, Z intersects the singular solution Z∗{Z}_{\ast }infinitely many times in (0,∞)\left(0,\infty )when 3≤N<10+4β3\le N\lt 10+4\beta .The solution of (4.4) would converge to the one of (4.3) when α\alpha is large. Indeed, we have the following lemma by [16, Proposition 3.1] and [22, Lemma 4.3].Lemma 4.2Let N≥1N\ge 1, 0<β<20\lt \beta \lt 2, α>0\alpha \gt 0, and ωα{\omega }_{\alpha }be the solution of (1.8) with ω(0)=α\omega \left(0)=\alpha . Let the function zα=zα(s){z}_{\alpha }={z}_{\alpha }\left(s)defined as (4.1). Then, for any S>0S\gt 0, there holdslimα→∞sup0≤s≤S∣zα(s)−Z(s)∣=0,\mathop{\mathrm{lim}}\limits_{\alpha \to \infty }\mathop{\sup }\limits_{0\le s\le S}| {z}_{\alpha }\left(s)-Z\left(s)| =0,where Z is the solution of (4.4).With the aforementioned lemmas, we could analyze the behavior of ωα{\omega }_{\alpha }as α→∞\alpha \to \infty near the origin. The following Propositions 4.3 and 4.4 can be proved by using similar method as in the proof of [22, Proposition 4.1] with the help of Lemmas 4.1 and 4.2.Proposition 4.3Let N≥1N\ge 1, 0<β<20\lt \beta \lt 2, α>0\alpha \gt 0, and ωα{\omega }_{\alpha }be the solution of (1.8) satisfying ω(0)=α\omega \left(0)=\alpha . For any ε>0\varepsilon \gt 0, there exists a constant α0∈R{\alpha }_{0}\in {\mathbb{R}}, and for any α≥α0\alpha \ge {\alpha }_{0}, there exists a constant rα>0{r}_{\alpha }\gt 0such that(4.5)∣(2+β)logrα+ωα(rα)−log[(2+β)(N−2)]∣<ε,| \left(2+\beta )\log {r}_{\alpha }+{\omega }_{\alpha }\left({r}_{\alpha })-\log {[}\left(2+\beta )\left(N-2)]| \lt \varepsilon ,where rα=e−α2+βs0{r}_{\alpha }={e}^{-\tfrac{\alpha }{2+\beta }}{s}_{0}and s0>0{s}_{0}\gt 0is some constant, which depends on ε\varepsilon . Furthermore, if 3≤N<10+4β3\le N\lt 10+4\beta , then there exists a positive constant rα{r}_{\alpha }satisfying (4.5) and(4.6)ddr[(2+β)logr+ωα(r)]r=rα=0{\left.\frac{{\rm{d}}}{{\rm{d}}r}{[}\left(2+\beta )\log r+{\omega }_{\alpha }\left(r)]\right|}_{r={r}_{\alpha }}=0for all α≥α0\alpha \ge {\alpha }_{0}.In the case of 3≤N<10+4β3\le N\lt 10+4\beta , we present the following intersection results.Proposition 4.4Let 3≤N<10+4β3\le N\lt 10+4\beta , 0<β<20\lt \beta \lt 2, α>0\alpha \gt 0, and ωα{\omega }_{\alpha }be the solution of (1.8) satisfying ω(0)=α\omega \left(0)=\alpha . Then for any r0>0{r}_{0}\gt 0and n0∈N{n}_{0}\in N, there exists a constant α0∈R{\alpha }_{0}\in {\mathbb{R}}such that (2+β)logr+ωα(r)\left(2+\beta )\log r+{\omega }_{\alpha }\left(r)intersects log[(2+β)(N−2)]\log {[}\left(2+\beta )\left(N-2)]at least n0{n}_{0}times on [0,r0]\left[0,{r}_{0}]for any α≥α0\alpha \ge {\alpha }_{0}.Next we study the structure of the set of solution SD{S}_{D}for equation (1.8). Recall that (4.7)limr→∞[(2+β)logr+ω(r)]=D,\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\omega \left(r)]=D,where ω(r)\omega \left(r)is a solution of (1.8), and SD≔{ω∈C2([0,∞)):ωis a solution of (1.8) satisfying (4.7)}.{S}_{D}:= \left\{\omega \in {C}^{2}\left(\left[0,\infty )):\omega \hspace{0.33em}\hspace{0.1em}\text{is a solution of (1.8) satisfying (4.7)}\hspace{0.1em}\right\}.By Proposition 3.3, we know that there exists a solution τ\tau of (3.2) with τ(0)=eα\tau \left(0)={e}^{\alpha }for any α∈R\alpha \in {\mathbb{R}}. Hence, ω(r)=logτ(r)\omega \left(r)=\log \tau \left(r)is a solution of (1.8) satisfying ω(0)=α\omega \left(0)=\alpha . Moreover, since (2+β)logr+ω(r)=logr2+βτ(r)\left(2+\beta )\log r+\omega \left(r)=\log {r}^{2+\beta }\tau \left(r), we can deduce that limr→∞[(2+β)logr+ω(r)]=logd(eα),\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\omega \left(r)]=\log d\left({e}^{\alpha }),where d(α)d\left(\alpha )was shown in Corollary 3.6. So we have that SD≠∅{S}_{D}\ne \varnothing for some D∈RD\in {\mathbb{R}}.Now set (4.8)D(α)≔limr→∞[(2+β)logr+ωα(r)],D\left(\alpha ):= \mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+{\omega }_{\alpha }\left(r)],where ωα{\omega }_{\alpha }is the solution of (1.8) with ω(0)=α\omega \left(0)=\alpha . By (3.33), we know limα→−∞d(eα)≤limα→−∞expα+limr→∞(2+β)logr−1+β2γN(r)=0,\mathop{\mathrm{lim}}\limits_{\alpha \to -\infty }d\left({e}^{\alpha })\le \mathop{\mathrm{lim}}\limits_{\alpha \to -\infty }\exp \left\{\phantom{\rule[-1.25em]{}{0ex}},\alpha +\mathop{\mathrm{lim}}\limits_{r\to \infty }\left[\left(2+\beta )\log r-\left(1+\frac{\beta }{2}\right){\gamma }_{N}\left(r)\right]\right\}=0,and therefore, limα→−∞D(α)=limα→−∞logd(eα)=−∞.\mathop{\mathrm{lim}}\limits_{\alpha \to -\infty }D\left(\alpha )=\mathop{\mathrm{lim}}\limits_{\alpha \to -\infty }\log d\left({e}^{\alpha })=-\infty .Then we begin by researching on the specific behavior of D(α)D\left(\alpha )as α→∞\alpha \to \infty and divide the argument into two cases: 3≤N<10+4β3\le N\lt 10+4\beta and N≥10+4βN\ge 10+4\beta .We first concentrate on the case where 3≤N<10+4β3\le N\lt 10+4\beta , and we obtain the following proposition.Proposition 4.5Let 3≤N<10+4β3\le N\lt 10+4\beta , 0<β<20\lt \beta \lt 2, and D(α)D\left(\alpha )be the constant defined by (4.8). Then the following results hold. (i)There exists a sequence {αk}k∈N{\left\{{\alpha }_{k}\right\}}_{k\in {\mathbb{N}}}such that α1<α2<⋯{\alpha }_{1}\lt {\alpha }_{2}\lt \cdots \hspace{0.33em}andD(α2k)<log[(2+β)(N−2)]<D(α2k−1)D\left({\alpha }_{2k})\lt \log {[}\left(2+\beta )\left(N-2)]\lt D\left({\alpha }_{2k-1})for k∈Nk\in {\mathbb{N}}.(ii)limα→∞D(α)=log[(2+β)(N−2)]{\mathrm{lim}}_{\alpha \to \infty }D\left(\alpha )=\log {[}\left(2+\beta )\left(N-2)].In particular, log[(2+β)(N−2)]<supα∈RD(α)<∞\log {[}\left(2+\beta )\left(N-2)]\lt {\sup }_{\alpha \in {\mathbb{R}}}D\left(\alpha )\lt \infty and SD=∅{S}_{D}=\varnothing for D>supα∈RD(α)D\gt {\sup }_{\alpha \in {\mathbb{R}}}D\left(\alpha ).For the purpose of proving Proposition 4.5, we first give some lemmas that follow the ideas of [22, Lemmas 4.7 and 4.9].Lemma 4.6Let 3≤N<10+4β3\le N\lt 10+4\beta , 0<β<20\lt \beta \lt 2, then for any τ∈R\tau \in {\mathbb{R}}, there exists a sequence {αk}k∈N{\left\{{\alpha }_{k}\right\}}_{k\in {\mathbb{N}}}such that α1<α2<⋯{\alpha }_{1}\lt {\alpha }_{2}\lt \cdots \hspace{0.33em}andgαk(τ)=0,gαk′(τ)gαk+1′(τ)<0{g}_{{\alpha }_{k}}\left(\tau )=0,\hspace{1.0em}{g}_{{\alpha }_{k}}^{\prime} \left(\tau ){g}_{{\alpha }_{k+1}}^{\prime} \left(\tau )\lt 0for any k∈Nk\in {\mathbb{N}}, where(4.9)gα(t)=(2+β)logr+ωα(r)−log[(2+β)(N−2)],t=logr.{g}_{\alpha }\left(t)=\left(2+\beta )\log r+{\omega }_{\alpha }\left(r)-\log {[}\left(2+\beta )\left(N-2)],\hspace{1.0em}t=\log r.ProofStep 1: By direct computation, we obtain (4.10)gα″+N−2+12e2tgα′+(2+β)(N−2)ϑα(t)gα=0,t∈R,{{g}_{\alpha }}^{^{\prime\prime} }+\left(N-2+\frac{1}{2}{e}^{2t}\right){g}_{\alpha }^{^{\prime} }+\left(2+\beta )\left(N-2){{\vartheta }}_{\alpha }\left(t){g}_{\alpha }=0,\hspace{1.0em}t\in {\mathbb{R}},where (4.11)ϑα(t)≔egα(t)−1gα(t)ifgα(t)≠0,1ifgα(t)=0.{{\vartheta }}_{\alpha }\left(t):= \left\{\begin{array}{ll}\frac{{e}^{{g}_{\alpha }\left(t)}-1}{{g}_{\alpha }\left(t)}\hspace{2.0em}\hspace{1.0em}& {\rm{if}}\hspace{1.0em}{g}_{\alpha }\left(t)\ne 0,\\ 1\hspace{1.0em}& {\rm{if}}\hspace{1.0em}{g}_{\alpha }\left(t)=0.\end{array}\right.It is clear that ϑα∈C(R){{\vartheta }}_{\alpha }\in C\left(R)and ϑα(t)>0{{\vartheta }}_{\alpha }\left(t)\gt 0.Moreover, we rewrite (4.10) as the following form: (4.12)(p0(t)gα′)′+(2+β)(N−2)p0(t)ϑα(t)gα=0,t∈R,({p}_{0}\left(t){g}_{\alpha }^{^{\prime} })^{\prime} +\left(2+\beta )\left(N-2)\hspace{0.33em}{p}_{0}\left(t){{\vartheta }}_{\alpha }\left(t){g}_{\alpha }=0,\hspace{1.0em}t\in {\mathbb{R}},where p0(t)≔exp(N−2)t+14e2t.{p}_{0}\left(t):= \exp \left(\left(N-2)t+\frac{1}{4}{e}^{2t}\right).Next we adapt the Prüfer transformation and set ρα(t){\rho }_{\alpha }\left(t)and θα(t){\theta }_{\alpha }\left(t)by p0(t)gα′(t)=ρα(t)cosθα(t),gα(t)=ρα(t)sinθα(t).{p}_{0}\left(t){g}_{\alpha }^{^{\prime} }\left(t)={\rho }_{\alpha }\left(t)\cos {\theta }_{\alpha }\left(t),\hspace{1.0em}{g}_{\alpha }\left(t)={\rho }_{\alpha }\left(t)\sin {\theta }_{\alpha }\left(t).Namely, ρα(t)=(gα(t)2+p0(t)2gα′(t)2)1/2,θα(t)=tan−1gα(t)p0(t)gα′(t),{\rho }_{\alpha }\left(t)={({{g}_{\alpha }\left(t)}^{2}+{{p}_{0}\left(t)}^{2}{{g}_{\alpha }^{^{\prime} }\left(t)}^{2})}^{1\text{/}2},\hspace{1.0em}{\theta }_{\alpha }\left(t)={\tan }^{-1}\frac{{g}_{\alpha }\left(t)}{{p}_{0}\left(t)\hspace{0.33em}{g}_{\alpha }^{^{\prime} }\left(t)},which combined with (4.12) gives θα′(t)=1p0(t)cos2θα(t)+(2+β)(N−2)p0(t)ϑα(t)sin2θα(t)>0,t∈R.{\theta }_{\alpha }^{\prime} \left(t)=\frac{1}{{p}_{0}\left(t)}{\cos }^{2}{\theta }_{\alpha }\left(t)+\left(2+\beta )\left(N-2){p}_{0}\left(t){{\vartheta }}_{\alpha }\left(t){\sin }^{2}{\theta }_{\alpha }\left(t)\gt 0,\hspace{1.0em}t\in {\mathbb{R}}.It is easy to see p0(t)→0,gα′(t)=2+β+rωα′(r)→2+β,gα(t)→∞ast→−∞,{p}_{0}\left(t)\to 0,{g}_{\alpha }^{^{\prime} }\left(t)=2+\beta +r{\omega }_{\alpha }^{^{\prime} }\left(r)\to 2+\beta ,\hspace{0.33em}{g}_{\alpha }\left(t)\to \infty \hspace{0.33em}{\rm{as}}\hspace{0.33em}t\to -\infty ,and we can assume limt→−∞θα(t)=−π2.\mathop{\mathrm{lim}}\limits_{t\to -\infty }{\theta }_{\alpha }\left(t)=-\frac{\pi }{2}.Then gα(t){g}_{\alpha }\left(t)has kkzeros in (−∞,τ]\left(-\infty ,\tau ]which is equivalent to (k−1)π≤θα(τ)≤kπ\left(k-1)\pi \le {\theta }_{\alpha }\left(\tau )\le k\pi .Step 2: We claim that (4.13)limα→∞θα(τ)=∞.\mathop{\mathrm{lim}}\limits_{\alpha \to \infty }{\theta }_{\alpha }\left(\tau )=\infty .Let n0{n}_{0}be any positive integer and set r0=eτ{r}_{0}={e}^{\tau }. By Proposition 4.4, there exists some α0>0{\alpha }_{0}\gt 0that satisfies α≥α0\alpha \ge {\alpha }_{0}, (2+β)logr+ωα(r)\left(2+\beta )\log r+{\omega }_{\alpha }\left(r)intersects log[(2+β)(N−2)]\log {[}\left(2+\beta )\left(N-2)]at least n0{n}_{0}times on [0,r0]\left[0,{r}_{0}]. Hence, gα(t){g}_{\alpha }\left(t)has at least n0{n}_{0}zeros on (−∞,τ)\left(-\infty ,\tau )for α≥α0\alpha \ge {\alpha }_{0}, and θα(τ)≥(n0−1)π{\theta }_{\alpha }\left(\tau )\ge \left({n}_{0}-1)\pi for α≥α0\alpha \ge {\alpha }_{0}. Therefore, (4.13) holds.We can find some k0{k}_{0}such that k0π≤θα0(τ)<(k0+1)π{k}_{0}\pi \le {\theta }_{{\alpha }_{0}}\left(\tau )\lt \left({k}_{0}+1)\pi . By (4.13) and the continuity of θα(t){\theta }_{\alpha }\left(t)with respect to α>0\alpha \gt 0, there exists a sequence {αk}k∈N{\left\{{\alpha }_{k}\right\}}_{k\in {\mathbb{N}}}with α0≤α1<α2<⋯{\alpha }_{0}\le {\alpha }_{1}\lt {\alpha }_{2}\lt \cdots \hspace{0.33em}such that θαk(τ)=(k0+k)π,k=1,2,…{\theta }_{{\alpha }_{k}}\left(\tau )=\left({k}_{0}+k)\pi ,\hspace{0.33em}k=1,2,\ldots . Then gαk(τ)=ραk(τ)sinθαk(τ)=0,{g}_{{\alpha }_{k}}\left(\tau )={\rho }_{{\alpha }_{k}}\left(\tau )\sin {\theta }_{{\alpha }_{k}}\left(\tau )=0,and gαk′(τ)=ραk′(τ)sinθαk(τ)+ραk(τ)cosθαk(τ)⋅θαk′(τ)=C0cosθαk(τ)=C0cos(k0+k)π,{g}_{{\alpha }_{k}}^{\prime} \left(\tau )={\rho }_{{\alpha }_{k}}^{\prime} \left(\tau )\sin {\theta }_{{\alpha }_{k}}\left(\tau )+{\rho }_{{\alpha }_{k}}\left(\tau )\cos {\theta }_{{\alpha }_{k}}\left(\tau )\cdot {\theta }_{{\alpha }_{k}}^{\prime} \left(\tau )={C}_{0}\cos {\theta }_{{\alpha }_{k}}\left(\tau )={C}_{0}\cos \left({k}_{0}+k)\pi ,where C0=ραk(τ)θαk′(τ)>0{C}_{0}={\rho }_{{\alpha }_{k}}\left(\tau ){\theta }_{{\alpha }_{k}}^{\prime} \left(\tau )\gt 0. Thus, Lemma 4.6 holds.□Lemma 4.7There exists T0∈R{T}_{0}\in {\mathbb{R}}, which does not depend on α>0\alpha \gt 0such that, if gα∈C2(R){g}_{\alpha }\in {C}^{2}\left({\mathbb{R}})is a solution of (4.10) satisfying gα(t˜)=0{g}_{\alpha }\left(\tilde{t})=0and gα′(t˜)>0{g}_{\alpha }^{^{\prime} }\left(\tilde{t})\gt 0with some t˜≥T0\tilde{t}\ge {T}_{0}, then gα(t)>0{g}_{\alpha }\left(t)\gt 0for t>T0t\gt {T}_{0}and limt→∞gα(t)>0{\mathrm{lim}}_{t\to \infty }{g}_{\alpha }\left(t)\gt 0.ProofSet I[gα(t)]≔12(gα′(t))2+(2+β)(N−2)(egα(t)−gα(t)).I\left[{g}_{\alpha }\left(t)]:= \frac{1}{2}{\left({g}_{\alpha }^{^{\prime} }\left(t))}^{2}+\left(2+\beta )\left(N-2)({e}^{{g}_{\alpha }\left(t)}-{g}_{\alpha }\left(t)).A direct computation yields (4.14)ddtI[gα(t)]=gα′[gα″+(2+β)(N−2)(egα(t)−1)]=−N−2+12e2tgα′(t)2≤0\frac{{\rm{d}}}{{\rm{d}}t}I\left[{g}_{\alpha }\left(t)]={g}_{\alpha }^{^{\prime} }{[}{g}_{\alpha }^{^{\prime\prime} }+\left(2+\beta )\left(N-2)({e}^{{g}_{\alpha }\left(t)}-1)]=-\left(N-2+\frac{1}{2}{e}^{2t}\right){{g}_{\alpha }^{^{\prime} }\left(t)}^{2}\le 0for any t∈Rt\in {\mathbb{R}}. Since gα(t)→∞{g}_{\alpha }\left(t)\to \infty as t→−∞t\to -\infty , there exists some constants t0∈R{t}_{0}\in {\mathbb{R}}satisfy gα(t)≤0{g}_{\alpha }\left(t)\le 0for all t≤t0t\le {t}_{0}.In the case of t≥t0t\ge {t}_{0}, from (4.14), we know that I[gα(t)]≤I[gα(t0)].I\left[{g}_{\alpha }\left(t)]\le I\left[{g}_{\alpha }\left({t}_{0})].And this implies that we can find a constant C>1C\gt 1such that egα(t)−gα(t)≤C{e}^{{g}_{\alpha }\left(t)}-{g}_{\alpha }\left(t)\le Cfor t≥t0t\ge {t}_{0}. Due to limt→±∞(et−t)=∞{\mathrm{lim}}_{t\to \pm \infty }\left({e}^{t}-t)=\infty and inft∈R(et−t)=1{\inf }_{t\in {\mathbb{R}}}\left({e}^{t}-t)=1, we have just proved that gα{g}_{\alpha }is bounded in [t0,∞)\left[{t}_{0},\infty ). Hence, we obtain that gα{g}_{\alpha }is bounded from above in R{\mathbb{R}}, and from (4.11) 0<ϑα(t)≤M,t∈R0\lt {{\vartheta }}_{\alpha }\left(t)\le M,\hspace{1.0em}t\in {\mathbb{R}}for some constant M>0M\gt 0. Therefore, we can complete the proof of Lemma 4.7 by [22, Proposition B.1].□Now we can begin to prove Proposition 4.5.Proof of Proposition 4.5Put T0∈R{T}_{0}\in {\mathbb{R}}be the constant in Lemma 4.7, and let t≥T0t\ge {T}_{0}. By Lemma 4.6, there exists {αk}k∈N{\left\{{\alpha }_{k}\right\}}_{k\in {\mathbb{N}}}with α1<α2<⋯{\alpha }_{1}\lt {\alpha }_{2}\lt \cdots \hspace{0.33em}such that gαk(t)=0,gαk′(t)gαk+1′(t)<0{g}_{{\alpha }_{k}}\left(t)=0,\hspace{1.0em}{g}_{{\alpha }_{k}}^{\prime} \left(t)\hspace{0.33em}{g}_{{\alpha }_{k+1}}^{\prime} \left(t)\lt 0for any k∈Nk\in {\mathbb{N}}. We could assume that (−1)k−1gαk′(t)>0{\left(-1)}^{k-1}{g}_{{\alpha }_{k}}^{\prime} \left(t)\gt 0for k=1,2,⋯k=1,2,\cdots \hspace{0.33em}.Note that (−1)k−1gαk(t){\left(-1)}^{k-1}{g}_{{\alpha }_{k}}\left(t)can be regarded as a solution of (4.10). Then, by Lemma 4.7, we know that limt→∞(−1)k−1gαk(t){\mathrm{lim}}_{t\to \infty }{\left(-1)}^{k-1}{g}_{{\alpha }_{k}}\left(t)is positive for k=1,2,…k=1,2,\ldots . It is obvious that limt→∞gαk(t)=limr→∞[(2+β)logr+ωαk(r)−log[(2+β)(N−2)]]=D(αk)−log[(2+β)(N−2)].\mathop{\mathrm{lim}}\limits_{t\to \infty }{g}_{{\alpha }_{k}}\left(t)=\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+{\omega }_{{\alpha }_{k}}\left(r)-\log {[}\left(2+\beta )\left(N-2)]]=D\left({\alpha }_{k})-\log {[}\left(2+\beta )\left(N-2)].Thus, we obtain D(α2k)−log[(2+β)(N−2)]=limt→∞gα2k(t)<0,i.e.D(α2k)<log[(2+β)(N−2)],D(α2k−1)−log[(2+β)(N−2)]=limt→∞gα2k−1(t)>0,i.e.D(α2k−1)>log[(2+β)(N−2)].\begin{array}{rcl}D\left({\alpha }_{2k})-\log {[}\left(2+\beta )\left(N-2)]& =& \mathop{\mathrm{lim}}\limits_{t\to \infty }{g}_{{\alpha }_{2k}}\left(t)\lt 0,\hspace{1.0em}{\rm{i.e.}}\hspace{0.33em}D\left({\alpha }_{2k})\lt \log {[}\left(2+\beta )\left(N-2)],\\ D\left({\alpha }_{2k-1})-\log {[}\left(2+\beta )\left(N-2)]& =& \mathop{\mathrm{lim}}\limits_{t\to \infty }{g}_{{\alpha }_{2k-1}}\left(t)\gt 0,\hspace{1.0em}{\rm{i.e.}}\hspace{0.33em}D\left({\alpha }_{2k-1})\gt \log {[}\left(2+\beta )\left(N-2)].\end{array}Then Proposition 4.5(i) holds.Let ε>0\varepsilon \gt 0. By Proposition 4.3, there exists a constant α0>0{\alpha }_{0}\gt 0such that for any α≥α0\alpha \ge {\alpha }_{0}, there holds ∣gα(tα)∣<ε,gα′(tα)=2+β+rαωα′(rα)=rα⋅ddr[(2+β)logr+ωα(r)]r=rα=0,| {g}_{\alpha }\left({t}_{\alpha })| \lt \varepsilon ,\hspace{1.0em}{g}_{\alpha }^{^{\prime} }\left({t}_{\alpha })=2+\beta +{r}_{\alpha }{\omega }_{\alpha }^{^{\prime} }\left({r}_{\alpha })={\left.{r}_{\alpha }\cdot \frac{{\rm{d}}}{{\rm{d}}r}{[}\left(2+\beta )\log r+{\omega }_{\alpha }\left(r)]\right|}_{r={r}_{\alpha }}=0,where rα{r}_{\alpha }is the constant given in Proposition 4.3 and tα=logrα{t}_{\alpha }=\log {r}_{\alpha }. Then, by (4.14), we obtain (4.15)(2+β)(N−2)(egα(t)−gα(t))≤I[gα(t)]≤I[gα(tα)]=(2+β)(N−2)(egα(tα)−gα(tα))\left(2+\beta )\left(N-2)\left({e}^{{g}_{\alpha }\left(t)}-{g}_{\alpha }\left(t))\le I\left[{g}_{\alpha }\left(t)]\le I\left[{g}_{\alpha }\left({t}_{\alpha })]=\left(2+\beta )\left(N-2)({e}^{{g}_{\alpha }\left({t}_{\alpha })}-{g}_{\alpha }\left({t}_{\alpha }))for any t≥tαt\ge {t}_{\alpha }. The definition of gα(t){g}_{\alpha }\left(t)leads us to the equation that limt→∞gα(t)=limr→∞[(2+β)logr+ωα(r)−log[(2+β)(N−2)]]=D(α)−log[(2+β)(N−2)].\mathop{\mathrm{lim}}\limits_{t\to \infty }{g}_{\alpha }\left(t)=\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+{\omega }_{\alpha }\left(r)-\log {[}\left(2+\beta )\left(N-2)]]=D\left(\alpha )-\log {[}\left(2+\beta )\left(N-2)].This together with (4.15) implies that 1≤eD(α)−log[(2+β)(N−2)]−(D(α)−log[(2+β)(N−2)])=elimt→∞gα(t)−limt→∞gα(t)≤egα(tα)−gα(tα)≤eε+ε.\begin{array}{rcl}1& \le & {e}^{D\left(\alpha )-\log {[}\left(2+\beta )\left(N-2)]}-(D\left(\alpha )-\log {[}\left(2+\beta )\left(N-2)])\\ & =& {e}^{\mathop{\mathrm{lim}}\limits_{t\to \infty }{g}_{\alpha }\left(t)}-\mathop{\mathrm{lim}}\limits_{t\to \infty }{g}_{\alpha }\left(t)\\ & \le & {e}^{{g}_{\alpha }\left({t}_{\alpha })}-{g}_{\alpha }\left({t}_{\alpha })\\ & \le & {e}^{\varepsilon }+\varepsilon .\end{array}Since et−t>1{e}^{t}-t\gt 1if t≠0t\ne 0and et−t=1{e}^{t}-t=1if t=0t=0, ε>0\varepsilon \gt 0is arbitrary, we obtain limα→∞[eD(α)−log[(2+β)(N−2)]−(D(α)−log[(2+β)(N−2)])]=1,\mathop{\mathrm{lim}}\limits_{\alpha \to \infty }{[}{e}^{D\left(\alpha )-\log {[}\left(2+\beta )\left(N-2)]}-(D\left(\alpha )-\log {[}\left(2+\beta )\left(N-2)])]=1,and then limα→∞[D(α)−log[(2+β)(N−2)]]=0{\mathrm{lim}}_{\alpha \to \infty }{[}D\left(\alpha )-\log {[}\left(2+\beta )\left(N-2)]]=0. Note that D(α)D\left(\alpha )is continuous with regard to α\alpha according to Lemma 3.7. Thus, Proposition 4.5(ii) holds.□Next we can obtain the following proposition when we consider the case where N≥10+4βN\ge 10+4\beta .Proposition 4.8Let N≥10+4βN\ge 10+4\beta , 0<β<20\lt \beta \lt 2. Denote the solution of (1.8) satisfying ω(0)=α\omega \left(0)=\alpha by ωα{\omega }_{\alpha }and define D(α)D\left(\alpha )by (4.8) with α∈R\alpha \in {\mathbb{R}}. Then for any r>0r\gt 0, the following results hold. (i)Let α>0\alpha \gt 0, then ωα(r)<−(2+β)logr+log[(2+β)(N−2)]{\omega }_{\alpha }\left(r)\lt -\left(2+\beta )\log r+\log {[}\left(2+\beta )\left(N-2)].(ii)Let 0<a<b0\lt a\lt b, then ωa(r)<ωb(r){\omega }_{a}\left(r)\lt {\omega }_{b}\left(r).(iii)limα→∞D(α)=log[(2+β)(N−2)]{\mathrm{lim}}_{\alpha \to \infty }D\left(\alpha )=\log {[}\left(2+\beta )\left(N-2)].(iv)For any r0>0{r}_{0}\gt 0, it holdslimα→∞supr≥r0∣ωα(r)+(2+β)logr−log[(2+β)(N−2)]∣=0.\mathop{\mathrm{lim}}\limits_{\alpha \to \infty }\mathop{\sup }\limits_{r\ge {r}_{0}}| {\omega }_{\alpha }\left(r)+\left(2+\beta )\log r-\log {[}\left(2+\beta )\left(N-2)]| =0.In particular, SD=∅{S}_{D}=\varnothing for any D>log[(2+β)(N−2)]D\gt \log {[}\left(2+\beta )\left(N-2)].Following the idea of [22, Lemma 4.10], we present an auxiliary lemma.Lemma 4.9Let N≥10+4βN\ge 10+4\beta , 0<β<20\lt \beta \lt 2, and T∈RT\in {\mathbb{R}}. Then there is no function x=x(t)x=x\left(t)satisfying x∈C2((−∞,T))∩C1((−∞,T])x\in {C}^{2}\left(\left(-\infty ,T))\cap {C}^{1}\left(\left(-\infty ,T])and(4.16)x″+(N−2)+12e2tx′+(2+β)(N−2)x<0,t<T,{x}^{^{\prime\prime} }+\left(\left(N-2)+\frac{1}{2}{e}^{2t}\right)x^{\prime} +\left(2+\beta )\left(N-2)x\lt 0,\hspace{1.0em}t\lt T,(4.17)x(t)<0,t<Tandx(T)=0,x\left(t)\lt 0,\hspace{0.33em}t\lt T\hspace{1.0em}and\hspace{1.0em}x\left(T)=0,(4.18)limsupt→−∞x(t)t<∞andlimsupt→−∞∣x′(t)∣<∞.\mathop{\mathrm{limsup}}\limits_{t\to -\infty }\left|\frac{x\left(t)}{t}\right|\lt \infty \hspace{1.0em}and\hspace{1.0em}\mathop{\mathrm{limsup}}\limits_{t\to -\infty }| x^{\prime} \left(t)| \lt \infty .Proof of Proposition 4.8Set gα(t)≔(2+β)logr+ωα(r)−log[(2+β)(N−2)],t=logr.{g}_{\alpha }\left(t):= \left(2+\beta )\log r+{\omega }_{\alpha }\left(r)-\log {[}\left(2+\beta )\left(N-2)],\hspace{1.0em}t=\log r.Then gα(t){g}_{\alpha }\left(t)satisfies (4.10) with (4.11). Moreover, (4.19)limsupt→−∞gα(t)t<∞,limsupt→−∞∣gα′(t)∣<∞,\mathop{\mathrm{limsup}}\limits_{t\to -\infty }\left|\frac{{g}_{\alpha }\left(t)}{t}\right|\lt \infty ,\hspace{1.0em}\mathop{\mathrm{limsup}}\limits_{t\to -\infty }| {g}_{\alpha }^{^{\prime} }\left(t)| \lt \infty ,and gα(t)→−∞{g}_{\alpha }\left(t)\to -\infty as t→−∞t\to -\infty . Suppose that there exists a constant T∈RT\in {\mathbb{R}}such that gα(t)<0{g}_{\alpha }\left(t)\lt 0for all t∈(−∞,T)t\in \left(-\infty ,T)and gα(T)=0{g}_{\alpha }\left(T)=0. Then ϑα(t)<1{{\vartheta }}_{\alpha }\left(t)\lt 1for all t∈(−∞,T)t\in \left(-\infty ,T)according to (4.11). This leads to gα″+(N−2)+12e2tgα′+(2+β)(N−2)gα<0,t<T,{g}_{\alpha }^{^{\prime\prime} }+\left(\left(N-2)+\frac{1}{2}{e}^{2t}\right){g}_{\alpha }^{^{\prime} }+\left(2+\beta )\left(N-2){g}_{\alpha }\lt 0,\hspace{1.0em}t\lt T,which contradicts to Lemma 4.9 and (4.19). Hence, we have gα(t)<0{g}_{\alpha }\left(t)\lt 0for all t∈Rt\in {\mathbb{R}}. Thus, assertion (i) holds.Then we present the proof of assertion (ii). Let 0<a<b0\lt a\lt band set h(t)=ga(t)−gb(t)h\left(t)={g}_{a}\left(t)-{g}_{b}\left(t). Then we obtain h″+(N−2)+12e2th′+(2+β)(N−2)(ega(t)−egb(t))=0,t∈R.{h}^{^{\prime\prime} }+\left(\left(N-2)+\frac{1}{2}{e}^{2t}\right)h^{\prime} +\left(2+\beta )\left(N-2)({e}^{{g}_{a}\left(t)}-{e}^{{g}_{b}\left(t)})=0,\hspace{1.0em}t\in {\mathbb{R}}.Here, note that h(t)→a−b<0h\left(t)\to a-b\lt 0and h′(t)→0h^{\prime} \left(t)\to 0as t→−∞t\to -\infty . Suppose that there exists a constant T∈RT\in {\mathbb{R}}such that h(t)<0h\left(t)\lt 0for all t∈(−∞,T)t\in \left(-\infty ,T)and h(T)=0h\left(T)=0. Due to ega(t)−egb(t)>h(t){e}^{{g}_{a}\left(t)}-{e}^{{g}_{b}\left(t)}\gt h\left(t)for all t∈(−∞,T)t\in \left(-\infty ,T), we obtain h″+(N−2)+12e2th′+(2+β)(N−2)h<0,t<T.{h}^{^{\prime\prime} }+\left(\left(N-2)+\frac{1}{2}{e}^{2t}\right)h^{\prime} +\left(2+\beta )\left(N-2)h\lt 0,\hspace{1.0em}t\lt T.This contradicts Lemma 4.9. Thus, assertion (ii)\left(ii)holds.Next is the proof of assertion (iii). We can deduce that (p0(t)gα′)′>0\left({p}_{0}\left(t){g}_{\alpha }^{^{\prime} })^{\prime} \gt 0in R{\mathbb{R}}from (4.12) and assertion (i)\left(i). By (4.19), we have p0(t)gα′(t)→0{p}_{0}\left(t){g}_{\alpha }^{^{\prime} }\left(t)\to 0as t→−∞t\to -\infty . Thus, p0(t)gα′(t)>0{p}_{0}\left(t){g}_{\alpha }^{^{\prime} }\left(t)\gt 0for any t∈Rt\in {\mathbb{R}}and (4.20)gα′(t)>0,t∈R.{g}_{\alpha }^{^{\prime} }\left(t)\gt 0,\hspace{1.0em}t\in {\mathbb{R}}.Recall assertion (i), we see (2+β)logr+ωα(r)<log[(2+β)(N−2)]\left(2+\beta )\log r+{\omega }_{\alpha }\left(r)\lt \log {[}\left(2+\beta )\left(N-2)]for all r∈(0,∞)r\in \left(0,\infty ), and hence, (4.21)D(α)≤log[(2+β)(N−2)].D\left(\alpha )\le \log {[}\left(2+\beta )\left(N-2)].On the other hand, for any ε>0\varepsilon \gt 0, we can find constants α0∈R{\alpha }_{0}\in {\mathbb{R}}and t0∈R{t}_{0}\in {\mathbb{R}}satisfying gα0(t0)>−ε{g}_{{\alpha }_{0}}\left({t}_{0})\gt -\varepsilon by Proposition 4.3. This together with (4.20) gives (4.22)−ε<D(α0)−log[(2+β)(N−2)].-\varepsilon \lt D\left({\alpha }_{0})-\log {[}\left(2+\beta )\left(N-2)].By assertion (ii), we obtain that D(α)D\left(\alpha )is nondecreasing in α\alpha . By combining (4.21) and (4.22), we have limα→∞D(α)=log[(2+β)(N−2)]{\mathrm{lim}}_{\alpha \to \infty }D\left(\alpha )=\log {[}\left(2+\beta )\left(N-2)]. Thus, assertion (iii) holds.Finally, we prove assertion (iv). It suffices to prove that, for any t0∈R{t}_{0}\in {\mathbb{R}}, there holds (4.23)limα→∞supt≥t0∣gα(t)∣=0.\mathop{\mathrm{lim}}\limits_{\alpha \to \infty }\mathop{\sup }\limits_{t\ge {t}_{0}}| {g}_{\alpha }\left(t)| =0.Let t0∈R{t}_{0}\in {\mathbb{R}}and ε>0\varepsilon \gt 0. By Proposition 4.3 and (4.9), we can choose constants tε{t}_{\varepsilon }and αε{\alpha }_{\varepsilon }such that −ε<gαεtε−α2≤0-\varepsilon \lt {g}_{{\alpha }_{\varepsilon }}\left({t}_{\varepsilon }-\frac{\alpha }{2}\right)\le 0for all α≥αε\alpha \ge {\alpha }_{\varepsilon }. Keeping in mind assertions (i) and (ii) and (4.20), for any α≥αε\alpha \ge {\alpha }_{\varepsilon }with tε−α2≤t0{t}_{\varepsilon }-\frac{\alpha }{2}\le {t}_{0}and t≥t0t\ge {t}_{0}, there holds −ε<gαεtε−α2≤gαε(t)<gα(t)<0,t≥t0.-\varepsilon \lt {g}_{{\alpha }_{\varepsilon }}\left({t}_{\varepsilon }-\frac{\alpha }{2}\right)\le {g}_{{\alpha }_{\varepsilon }}\left(t)\lt {g}_{\alpha }\left(t)\lt 0,\hspace{1.0em}t\ge {t}_{0}.Then we have (4.23) as ε\varepsilon is arbitrary. Thus, assertion (iv) holds. The proof of Proposition 4.8 is completed.□Proof of Theorem 1.1By Propositions 4.5 and 4.8, we can directly obtain the conclusion of Theorem 1.1, which demonstrates the structure of the solution set SD{S}_{D}.□5Proof of Theorem 1.2In this section, our purpose is to prove Theorem 1.2. Assume by contradiction that the function vvis a global in time solution for (1.1) satisfying the assumptions of Theorem 1.2. Then by using Lemma 5.1, we can construct a radially symmetric continuous weak sub-solution ω1̲=ω1̲(∣y∣)\underline{{\omega }_{1}}=\underline{{\omega }_{1}}\left(| y| )for (1.7) such that ω1̲(∣y∣)<v0(y)inRN,liminf∣y∣→∞(ω1̲(∣y∣)+(2+β)log∣y∣)>D∗.\underline{{\omega }_{1}}\left(| y| )\lt {v}_{0}(y)\hspace{1.0em}{\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N},\hspace{1.0em}\mathop{\mathrm{liminf}}\limits_{| y| \to \infty }(\underline{{\omega }_{1}}\left(| y| )+\left(2+\beta )\log | y| )\gt {D}^{\ast }.Thus, the solution ψ∗{\psi }^{\ast }for (1.12) with ψ0∗(y)=ω1̲(∣y∣){\psi }_{0}^{\ast }(y)=\underline{{\omega }_{1}}\left(| y| )is nondecreasing in ss. Moreover, by Lemma 2.6, ψ∗{\psi }^{\ast }would converge to the solution ω˜\tilde{\omega }for (1.8) as s→∞s\to \infty , which maybe a singular solution. Next we assume the existence of ω˜\tilde{\omega }and construct a continuous weak super-solution for (1.7) using ω˜\tilde{\omega }, which yields a contradiction. Finally we can obtain a solution ω=ω(∣y∣)\omega =\omega \left(| y| )of (1.8) satisfying ω∈SD0\omega \in {S}_{{D}_{0}}for some D0>D∗{D}_{0}\gt {D}^{\ast }, contradicting the definition of D∗{D}^{\ast }.Before starting the proof, we present the following lemma, which can be find in [8, Lemma 5.1].Lemma 5.1Let N≥3N\ge 3, 0<β<20\lt \beta \lt 2. Assume that f∈C(RN)f\in C\left({{\mathbb{R}}}^{N})satisfiesliminf∣y∣→∞[(2+β)log∣y∣+f(y)]=D\mathop{\mathrm{liminf}}\limits_{| y| \to \infty }{[}\left(2+\beta )\log | y| +f(y)]=Dfor some D∈RD\in {\mathbb{R}}. Then for any δ>0\delta \gt 0, there exists a continuous weak sub-solution ω̲=ω̲(∣y∣)\underline{\omega }=\underline{\omega }\left(| y| )of (1.7) such that(5.1)limr→∞[(2+β)logr+ω̲(r)]=D−δ,\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\underline{\omega }\left(r)]=D-\delta ,(5.2)ω̲(∣y∣)<f(y)inRN,\underline{\omega }\left(| y| )\lt f(y)\hspace{1.0em}in\hspace{0.33em}{{\mathbb{R}}}^{N},where ω̲\underline{\omega }is radially symmetric and nonincreasing in ∣y∣| y| .Now we can begin to prove Theorem 1.2.ProofRecall that D∗{D}^{\ast }is the constant defined in Section 1 and D∗>log[(2+β)(N−2)]{D}^{\ast }\gt \log {[}\left(2+\beta )\left(N-2)]if 3≤N<10+4β3\le N\lt 10+4\beta , D∗=log[(2+β)(N−2)]{D}^{\ast }=\log {[}\left(2+\beta )\left(N-2)]if N≥10+4βN\ge 10+4\beta by Theorem 1.1. Then take a constant D>0D\gt 0such that D∗<D≤liminf∣x∣→∞[(2+β)log∣x∣+v0(x)].{D}^{\ast }\lt D\le \mathop{\mathrm{liminf}}\limits_{| x| \to \infty }{[}\left(2+\beta )\log | x| +{v}_{0}\left(x)].Assume by contradiction that vvis a global in time solution for (1.1). Then we have that ψ\psi is global in time, that is, the solution of (1.12), which is defined by (1.11), and holds D∗<D≤liminf∣y∣→∞[(2+β)log∣y∣+ψ(y,0)],{D}^{\ast }\lt D\le \mathop{\mathrm{liminf}}\limits_{| y| \to \infty }{[}\left(2+\beta )\log | y| +\psi (y,0)],here ψ0(y)=ψ(y,0)=v0(y){\psi }_{0}(y)=\psi (y,0)={v}_{0}(y). Take a constant δ>0\delta \gt 0such that D−δ>D∗D-\delta \gt {D}^{\ast }, then we can construct a continuous weak sub-solution ω1̲\underline{{\omega }_{1}}for (1.7) according to Lemma 5.1. The function ω1̲\underline{{\omega }_{1}}satisfies ω1̲(∣y∣)≤ψ0(y)=v0(y)\underline{{\omega }_{1}}\left(| y| )\le {\psi }_{0}(y)={v}_{0}(y)in RN{{\mathbb{R}}}^{N}and (5.3)lim∣y∣→∞(ω1̲(∣y∣)+(2+β)log∣y∣)=D−δ.\mathop{\mathrm{lim}}\limits_{| y| \to \infty }(\underline{{\omega }_{1}}\left(| y| )+\left(2+\beta )\log | y| )=D-\delta .Let ψ∗{\psi }^{\ast }be a solution of (1.12) with ψ∗(⋅,0)=ω1̲{\psi }^{\ast }\left(\cdot ,0)=\underline{{\omega }_{1}}. Then we note that ψ∗{\psi }^{\ast }is nondecreasing in ssby Lemma 2.3. Thus, the function (5.4)ω˜(r)≔lims→∞ψ∗(r,s)\tilde{\omega }\left(r):= \mathop{\mathrm{lim}}\limits_{s\to \infty }{\psi }^{\ast }\left(r,s)exists.We claim that ω˜\tilde{\omega }is not bounded from above, otherwise, by Lemma 2.6, we obtain that ω˜∈C2([0,∞))\tilde{\omega }\in {C}^{2}\left(\left[0,\infty ))and ω˜\tilde{\omega }satisfies (1.8) with ω˜′(0)=0\tilde{\omega }^{\prime} \left(0)=0. We also have D−δ=lim∣y∣→∞[(2+β)log∣y∣+ψ∗(y,0)]≤limr→∞[(2+β)logr+ω˜(r)]≕D˜,D-\delta =\mathop{\mathrm{lim}}\limits_{| y| \to \infty }{[}\left(2+\beta )\log | y| +{\psi }^{\ast }(y,0)]\le \mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\tilde{\omega }\left(r)]\hspace{0.33em}=: \hspace{0.33em}\tilde{D},that is, ω˜∈SD˜\tilde{\omega }\in {S}_{\tilde{D}}. Thus, we obtain D˜≥D−δ>D∗\tilde{D}\ge D-\delta \gt {D}^{\ast }, which contradicts the definition of D∗{D}^{\ast }. Hence ω˜\tilde{\omega }is not bounded from above. Then by using Lemma 2.6 again, we have that ω˜∈C2((0,∞))\tilde{\omega }\in {C}^{2}\left(\left(0,\infty ))and ω˜\tilde{\omega }satisfies (1.8) with ω˜(r)→∞\tilde{\omega }\left(r)\to \infty as r→0r\to 0. Again, we define D˜≔limr→∞[(2+β)logr+ω˜(r)].\tilde{D}:= \mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\tilde{\omega }\left(r)].First, we consider the case where 3≤N<10+4β3\le N\lt 10+4\beta . From [8, Lemma 5.2], one can find a constant r2>0{r}_{2}\gt 0such that (5.5)ω˜(r2)=−(2+β)logr2+log[(2+β)(N−2)],ω˜′(r2)<−2+βr2.\tilde{\omega }\left({r}_{2})=-\left(2+\beta )\log {r}_{2}+\log {[}\left(2+\beta )\left(N-2)],\hspace{1.0em}{\tilde{\omega }}^{^{\prime} }\left({r}_{2})\lt -\frac{2+\beta }{{r}_{2}}.Moreover, by Proposition 4.4, we can find constants r0∈(0,r2){r}_{0}\in \left(0,{r}_{2})and α0∈R{\alpha }_{0}\in {\mathbb{R}}such that (5.6)ω˜α0(r0)=−(2+β)logr0+log[(2+β)(N−2)],ω˜α0′(r0)>−2+βr0.{\tilde{\omega }}_{{\alpha }_{0}}\left({r}_{0})=-\left(2+\beta )\log {r}_{0}+\log {[}\left(2+\beta )\left(N-2)],\hspace{1.0em}{\tilde{\omega }}_{{\alpha }_{0}}^{\prime} \left({r}_{0})\gt -\frac{2+\beta }{{r}_{0}}.Thus, combining the aforementioned equation and Lemma 2.4, we know that ω¯(r)\overline{\omega }\left(r)defined by ω¯(r)≔ω˜α0(r)forr∈[0,r0],−(2+β)logr+log[(2+β)(N−2)]forr∈[r0,r2],ω˜(r)forr∈[r2,∞]\overline{\omega }\left(r):= \left\{\begin{array}{ll}{\tilde{\omega }}_{{\alpha }_{0}}\left(r)\hspace{1.0em}& {\rm{for}}\hspace{1.0em}r\in \left[0,{r}_{0}],\\ -\left(2+\beta )\log r+\log {[}\left(2+\beta )\left(N-2)]\hspace{1.0em}& {\rm{for}}\hspace{1.0em}r\in \left[{r}_{0},{r}_{2}],\\ \tilde{\omega }\left(r)\hspace{1.0em}& {\rm{for}}\hspace{1.0em}r\in \left[{r}_{2},\infty ]\end{array}\right.is a continuous weak super-solution of (1.7) and (5.7)limr→∞[(2+β)logr+ω¯(r)]=limr→∞[(2+β)logr+ω˜(r)]=D˜≥D−δ>D∗.\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\overline{\omega }\left(r)]=\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\tilde{\omega }\left(r)]=\tilde{D}\ge D-\delta \gt {D}^{\ast }.Then, by using Lemma 5.1, we could construct a continuous weak sub-solution ω2̲\underline{{\omega }_{2}}of (1.7) such that ω2̲<ω¯\underline{{\omega }_{2}}\lt \overline{\omega }and limr→∞[(2+β)logr+ω2̲(r)]=D−δ.\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\underline{{\omega }_{2}}\left(r)]=D-\delta .Let ψ˜\tilde{\psi }be a solution of (1.12) with ψ˜(⋅,0)=ω2̲(∣⋅∣)\tilde{\psi }\left(\cdot ,0)=\underline{{\omega }_{2}}\left(| \cdot | ). Then we can deduce from Lemma 2.3 that ψ˜\tilde{\psi }is nondecreasing in ss. Hence, the limit ω(r)=lims→∞ψ˜(r,s)\omega \left(r)=\mathop{\mathrm{lim}}\limits_{s\to \infty }\tilde{\psi }\left(r,s)exists and ω2̲(r)≤ψ˜(r,s)≤ω¯(r)\underline{{\omega }_{2}}\left(r)\le \tilde{\psi }\left(r,s)\le \overline{\omega }\left(r)by Lemma 2.1(ii), which shows that (5.8)ω2̲(r)≤ω(r)≤ω¯(r).\underline{{\omega }_{2}}\left(r)\le \omega \left(r)\le \overline{\omega }\left(r).Due to ω¯(0)<∞\overline{\omega }\left(0)\lt \infty and Lemma 2.6, we know that ω\omega is a regular solution of (1.8) with ω′(0)=0\omega ^{\prime} \left(0)=0. Moreover, we obtain limr→∞[(2+β)logr+ω2̲(r)]=D−δ≤limr→∞[(2+β)logr+ω(r)]≕D0,\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\underline{{\omega }_{2}}\left(r)]=D-\delta \le \mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\omega \left(r)]\hspace{0.33em}=: \hspace{0.33em}{D}_{0},by (5.8) and ω∈SD0\omega \in {S}_{{D}_{0}}for some D0≥D−δ>D∗{D}_{0}\ge D-\delta \gt {D}^{\ast }. Hence, there is a contradiction to the definition of D∗{D}^{\ast }.Thus, we obtain that vvblows up locally in time. Therefore, we proved Theorem 1.2 when 3≤N<10+4β3\le N\lt 10+4\beta .The proof for the case N≥10+4βN\ge 10+4\beta can be completed by the method analogous to that used earlier, the key is to create a continuous weak super-solution ω¯\overline{\omega }of (1.7), which satisfies (5.9)limr→∞[(2+β)logr+ω¯(r)]>D∗.\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\overline{\omega }\left(r)]\gt {D}^{\ast }.And the aforementioned inequality could be achieved with the aid of Proposition 4.8. We omit the details; thus, the proof of Theorem 1.2 is completed.□ http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Advances in Nonlinear Analysis de Gruyter

Existence and blow-up of solutions in Hénon-type heat equation with exponential nonlinearity

Gao, Dongmei; Wang, Jun; Wang, Xuan
Advances in Nonlinear Analysis , Volume 12 (1): 1 – Jan 1, 2023
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de Gruyter
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© 2023 the author(s), published by De Gruyter
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2191-950X
DOI
10.1515/anona-2022-0290
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Abstract

1IntroductionIn this article, we study the Hénon-type heat equation with exponential growth: (1.1)vt=Δv+∣x∣βev,x∈RN,t>0,v(x,0)=v0(x),x∈RN,\left\{\phantom{\rule[-1.25em]{}{0ex}}\begin{array}{ll}{v}_{t}=\Delta v+| x{| }^{\beta }{e}^{v},\hspace{1.0em}& x\in {{\mathbb{R}}}^{N},\hspace{1em}t\gt 0,\\ v\left(x,0)={v}_{0}\left(x),\hspace{1.0em}& x\in {{\mathbb{R}}}^{N},\end{array}\right.where N≥3N\ge 3, 0<β<20\lt \beta \lt 2, and v0∈C(RN){v}_{0}\in C\left({{\mathbb{R}}}^{N}). In this whole article, we make the assumption that there exist constants σ∈(0,2)\sigma \in \left(0,2)and C>0C\gt 0, such that (1.2)−Ce∣x∣2−σ≤v0(x)≤CinRN.-C{e}^{| x{| }^{2-\sigma }}\le {v}_{0}\left(x)\le C\hspace{1.0em}{\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N}.We denote T=T(v0)>0T=T\left({v}_{0})\gt 0by the maximal existence time of the unique classical solution of problem (1.1). It is obvious that under the constraint (1.2), the solution vvof (1.1) on RN{{\mathbb{R}}}^{N}is always bounded below as long as vvexists. If T(v0)<∞T\left({v}_{0})\lt \infty , then limt→T(v0)supx∈RNv(t,x)=+∞{\mathrm{lim}}_{t\to T\left({v}_{0})}{\sup }_{x\in {{\mathbb{R}}}^{N}}v\left(t,x)=+\infty . We say that such solution vvblows up in finite time and its blow-up time is T(v0)T\left({v}_{0}). Otherwise, we call that vvis the global solution if T=∞T=\infty .To make a clear statement of results for (1.1) with exponential nonlinearity, we begin by reviewing some established findings with regard to the global existence and blow-up of the following heat equation: (1.3)ut−Δu=∣x∣β∣u∣p−1u,x∈RN,t>0,u(x,0)=u0(x),x∈RN,\left\{\phantom{\rule[-1.25em]{}{0ex}}\begin{array}{ll}{u}_{t}-\Delta u=| x{| }^{\beta }| u{| }^{p-1}u,\hspace{1.0em}& x\in {{\mathbb{R}}}^{N},\hspace{1em}t\gt 0,\\ u\left(x,0)={u}_{0}\left(x),\hspace{1.0em}& x\in {{\mathbb{R}}}^{N},\end{array}\right.where p>1p\gt 1. It can be proved that for β≠0\beta \ne 0, N≥3N\ge 3, p>N+βN−2p\gt \frac{N+\beta }{N-2}, the function ϕ∗(x)=Bp,N,β1p−1∣x∣−2+βp−1,Bp,N,β=2+βp−1N−2−2+βp−1.{\phi }_{\ast }\left(x)={B}_{p,N,\beta }^{\frac{1}{p-1}}| x{| }^{-\tfrac{2+\beta }{p-1}},\hspace{1em}{B}_{p,N,\beta }=\frac{2+\beta }{p-1}\left(N-2-\frac{2+\beta }{p-1}\right).is a positive singular stationary solution of (1.3). We will find that whether the solutions of the problem (1.3) exist globally in time or blow up in finite time strongly depended on the decay rate of u0{u}_{0}. Precisely, the critical decay rate of the u0{u}_{0}is given by ∣x∣−2+βα−1| x{| }^{-\tfrac{2+\beta }{\alpha -1}}.We start by taking the case β=0\beta =0in (1.3) into account. Let pF=1+2/N{p}_{F}=1+2\hspace{0.1em}\text{/}\hspace{0.1em}Nbe the Fujita exponent (see [9,10]). It has been proved that all solutions of (1.3) with u0≥0{u}_{0}\ge 0, u0≢0{u}_{0}\not\equiv 0, blow up in finite time if and only if p≤pFp\le {p}_{F}(see [10,11]). On the other hand, for sufficient small initial data, there exist global in time solutions of (1.3) when p>pFp\gt {p}_{F}. Lee and Ni [20] proved that if p>pFp\gt {p}_{F}and u0(x)=λψ(x)(λ>0){u}_{0}\left(x)=\lambda \psi \left(x)\left(\lambda \gt 0), then the solution uuof (1.3) exists globally in time (T(u0)=∞)\left(T\left({u}_{0})=\infty )for λ>0\lambda \gt 0small enough, where ψ≢0\psi \not\equiv 0is a nonnegative function satisfying limsup∣x∣→∞∣x∣2p−1ψ(x)<∞{\mathrm{limsup}}_{| x| \to \infty }| x{| }^{\tfrac{2}{p-1}}\psi \left(x)\lt \infty . On the contrary, if liminf∣x∣→∞∣x∣2p−1ψ(x){\mathrm{liminf}}_{| x| \to \infty }| x{| }^{\tfrac{2}{p-1}}\psi \left(x)large enough, then the solution uuof (1.3) blows up in finite time (T(u0)<∞)\left(T\left({u}_{0})\lt \infty ). Equation (1.3) has one parameter family {Vτ}τ>0{\left\{{V}_{\tau }\right\}}_{\tau \gt 0}of radially symmetric regular stationary solutions for the supercritical case p≥N+2N−2p\ge \frac{N+2}{N-2}with N≥3N\ge 3, where each Vτ{V}_{\tau }satisfies lim∣x∣→∞∣x∣2p−1Vτ(∣x∣)=Bp,N,0{\mathrm{lim}}_{| x| \to \infty }| x{| }^{\tfrac{2}{p-1}}{V}_{\tau }\left(| x| )={B}_{p,N,0}. The articles [14,15] showed the stability results of Vτ{V}_{\tau }.Recently, Quittner and Souplet [28] considered the model (1.4)ut−Δu=λu+∣u∣p−1u,x∈RN,t>0,u(x,0)=u0(x),x∈RN,\left\{\phantom{\rule[-1.25em]{}{0ex}}\begin{array}{ll}{u}_{t}-\Delta u=\lambda u+| u{| }^{p-1}u,\hspace{1.0em}& x\in {{\mathbb{R}}}^{N},\hspace{1em}t\gt 0,\\ u\left(x,0)={u}_{0}\left(x),\hspace{1.0em}& x\in {{\mathbb{R}}}^{N},\end{array}\right.with p>1p\gt 1and λ∈R\lambda \in {\mathbb{R}}. They derived that T(u0)<∞T\left({u}_{0})\lt \infty under the assumptions 0≤u0∈L∞(RN),∫RNu0(x)π−n2e−∣x∣2dx>c≔(max(0,2n−λ))1/(p−1),0\le {u}_{0}\in {L}^{\infty }\left({{\mathbb{R}}}^{N}),\hspace{1.0em}\mathop{\int }\limits_{{{\mathbb{R}}}^{N}}{u}_{0}\left(x){\pi }^{-\tfrac{n}{2}}{e}^{-| x{| }^{2}}{\rm{d}}x\gt c:= {({\rm{\max }}\left(0,2n-\lambda ))}^{1\text{/}\left(p-1)},or 0≤u0∈L∞(RN),liminf∣x∣→∞∣x∣2/(p−1)u0(x)>μ1/(p−1),0\le {u}_{0}\in {L}^{\infty }\left({{\mathbb{R}}}^{N}),\hspace{1.0em}\mathop{\mathrm{liminf}}\limits_{| x| \to \infty }| x{| }^{2\text{/}\left(p-1)}{u}_{0}\left(x)\gt {\mu }^{1\text{/}\left(p-1)},where μ>0\mu \gt 0be the first eigenvalue of the Dirichlet Laplacian in the unit ball of RN{{\mathbb{R}}}^{N}. Wang [33] obtained the optimal bound of u0{u}_{0}, which classifies the global existence and blow-up of solutions for large pp. The Joseph-Lundgren exponent pJL≔1+4N−4−2N−1ifN≥11,∞ifN≤10,{p}_{{\rm{JL}}}:= \left\{\begin{array}{ll}1+\frac{4}{N-4-2\sqrt{N-1}}\hspace{1.0em}& {\rm{if}}\hspace{1.0em}N\ge 11,\\ \infty \hspace{1.0em}& {\rm{if}}\hspace{1.0em}N\le 10,\end{array}\right.plays a key role in the study of the stability of stationary solutions. The previous article [33] showed that if p≥pJLp\ge {p}_{{\rm{JL}}}, and u0{u}_{0}satisfies liminf∣x∣→∞∣x∣2p−1u0(x)>Bp,N,0{\mathrm{liminf}}_{| x| \to \infty }| x{| }^{\tfrac{2}{p-1}}{u}_{0}\left(x)\gt {B}_{p,N,0}, then solutions of (1.3) blow up in finite time. Otherwise, if p≥pJLp\ge {p}_{{\rm{JL}}}and liminf∣x∣→∞∣x∣2p−1u0(x)=Bp,N,0{\mathrm{liminf}}_{| x| \to \infty }| x{| }^{\tfrac{2}{p-1}}{u}_{0}\left(x)={B}_{p,N,0}, then there exists a global in time solution of (1.3). Later, the previous article [23] generalized the results of [33] to the case p≥pFp\ge {p}_{F}. We refer to previous articles [1–3,17,19,34] and the references therein for more results.For the case β≠0\beta \ne 0, the Joseph-Lundgren exponent is defined by p˜JL≔(N−2)2−2(β+2)(β+N)+2(β+2)(N+β)2−(N−2)2(N−2)(N−10−4β),ifN≥11+4β,+∞,ifN≤10+4β.{\tilde{p}}_{{\rm{JL}}}:= \left\{\begin{array}{ll}\frac{{\left(N-2)}^{2}-2\left(\beta +2)\left(\beta +N)+2\left(\beta +2)\sqrt{{\left(N+\beta )}^{2}-{\left(N-2)}^{2}}}{\left(N-2)\left(N-10-4\beta )},\hspace{1.0em}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{1em}N\ge 11+4\beta ,\\ +\infty ,\hspace{1.0em}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{1em}N\le 10+4\beta .\end{array}\right.The previous article [33] proved that when N≥3N\ge 3, N+βN−2<p<p˜JL\frac{N+\beta }{N-2}\lt p\lt {\tilde{p}}_{{\rm{JL}}}and β>−2\beta \gt -2, if the initial value u0(x){u}_{0}\left(x)satisfies, 0≤u0(x)≤ϕ∗(x),0\le {u}_{0}\left(x)\le {\phi }_{\ast }\left(x),then (1.3) with the nonnegative initial value u0(x)∈C(RN){u}_{0}\left(x)\in C\left({{\mathbb{R}}}^{N})has a global solution u(x)u\left(x)such that 0≤u(x)≤ϕ∗(x)and∣∣u(⋅,t)∣∣L∞(RN)→0ast→∞.0\le u\left(x)\le {\phi }_{\ast }\left(x)\hspace{1.0em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1.0em}| | u\left(\cdot ,\hspace{0.33em}t)| {| }_{{L}^{\infty }\left({{\mathbb{R}}}^{N})}\to 0\hspace{1em}\hspace{0.1em}\text{as}\hspace{0.1em}\hspace{1em}t\to \infty .Moreover, the aforementioned result holds with 0≤u0(x)≤η˜ϕ∗(x)0\le {u}_{0}\left(x)\le \tilde{\eta }{\phi }_{\ast }\left(x)if p>N+βN−2p\gt \frac{N+\beta }{N-2}for some η˜∈(0,1)\tilde{\eta }\in \left(0,1). Later, the previous article [27] proved that the Cauchy problem for (1.3) possesses the Fujita exponent N+2+βN\frac{N+2+\beta }{N}with β>−1\beta \gt -1. Furthermore, the solutions of (1.3) blow up in finite time for all nontrivial nonnegative initial value u0{u}_{0}if 1<p≤N+2+βN1\lt p\le \frac{N+2+\beta }{N}, and there are both global and blowing-up solutions if p>N+2+βNp\gt \frac{N+2+\beta }{N}. Recently, the previous articles [12,13] proved the existence of sign-changing solutions of (1.3) with singular initial value u0{u}_{0}. For more general results on this direction, one can see the previous articles [4,5,7,18,21,25,26,29,30,35,36] and the references therein.The previous article [32] investigated the existence, uniqueness, and blow-up solutions of the reaction-diffusion equation with the exponential growth: (1.5)ut=Δu+λeu,x∈RN,t>0,λ>0.{u}_{t}=\Delta u+\lambda {e}^{u},\hspace{1.0em}x\in {{\mathbb{R}}}^{N},\hspace{1em}t\gt 0,\hspace{0.33em}\lambda \gt 0.The author proved that if u0(x)=U3(x)+c{u}_{0}\left(x)={U}_{3}\left(x)+cis a continuous function away from 0 and assume that lim∣x∣→0(u0(x)−U3(x))<c∗{\mathrm{lim}}_{| x| \to 0}\left({u}_{0}\left(x)-{U}_{3}\left(x))\lt {c}^{\ast }, then the Cauchy problem (1.5) admits a weak solution, at least locally in time, where c∈Rc\in {\mathbb{R}}, U3(x){U}_{3}\left(x)is the singular solution of (1.5): U3(x)≔−2log∣x∣+log(2N−4)λ.{U}_{3}\left(x):= -2\log | x| +\log \frac{\left(2N-4)}{\lambda }.On the contrary, if lim∣x∣→0(u0(x)−U3(x))>c∗{\mathrm{lim}}_{| x| \to 0}\left({u}_{0}\left(x)-{U}_{3}\left(x))\gt {c}^{\ast }, then there is no solution even locally in time, and instantaneous blow-up occurs. Recently, Fujishima [8] considered the problem (1.1) with β=0\beta =0. Precisely, the author proved that if N≥3,u0∈C(RN),liminf∣x∣→∞(2log∣x∣+u0(x))>L∗N\ge 3,\hspace{1.0em}{u}_{0}\in C\left({{\mathbb{R}}}^{N}),\hspace{1.0em}\mathop{\mathrm{liminf}}\limits_{| x| \to \infty }\left(2\log | x| +{u}_{0}\left(x))\gt {L}^{\ast }and −Ce∣x∣2−ε≤u0(x)≤CinRN-C{e}^{| x{| }^{2-\varepsilon }}\le {u}_{0}\left(x)\le C\hspace{1.0em}{\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N}for some constants ε\varepsilon in (0,2)\left(0,2)and C>0C\gt 0, then T(u0)<∞T\left({u}_{0})\lt \infty , where L∗{L}^{\ast }can be described in two cases: L∗>log(2N−4){L}^{\ast }\gt \log \left(2N-4)with 3≤N≤93\le N\le 9and L∗=log(2N−4){L}^{\ast }=\log \left(2N-4)with N≥10N\ge 10. The previous article [8] obtained the optimal bound of u0{u}_{0}, which classifies the global existence and blow-up of solutions of (1.1). For more results on this direction, one can also refer to [6,24] and the references therein.Inspired by previous works [8,33], the purpose of this article is to consider the existence and blow-up of solutions of the Hénon-type heat equation with exponential nonlinearity. As in [31], it is obvious that for any initial value v0{v}_{0}satisfying (1.2), there exists T=T(v0)>0T=T\left({v}_{0})\gt 0such that (1.1) has a unique classical solution in C2,1(RN×(0,T))∩C(RN×[0,T)){C}^{2,1}\left({{\mathbb{R}}}^{N}\times \left(0,T))\cap C\left({{\mathbb{R}}}^{N}\times \left[0,T)). Moreover, it is clear that the function V∗(x)=−(2+β)log∣x∣+log[(2+β)(N−2)]{V}_{\ast }\left(x)=-\left(2+\beta )\log | x| +\log {[}\left(2+\beta )\left(N-2)]is a singular stationary solution of (1.1). Hence, we guess that the critical decay rate at the space infinity for the existence of global in time solutions of (1.1) is given by −(2+β)log∣x∣-\left(2+\beta )\log | x| . More specifically, we investigate the case where v0{v}_{0}decays to −∞-\infty at space infinity with the rate −(2+β)log∣x∣-\left(2+\beta )\log | x| and obtain the optimal bound of lim∣x∣→∞[(2+β)log∣x∣+v0(x)]{\mathrm{lim}}_{| x| \to \infty }{[}\left(2+\beta )\log | x| +{v}_{0}\left(x)], which is the crucial indicator to classify the global in time solutions and blow-up solutions of the problem (1.1). To accomplish this, we study the properties of forward self-similar solutions of (1.1), which is defined by (1.6)v(x,t)=−1+β2logt+ωxt.v\left(x,t)=-\left(1+\frac{\beta }{2}\right)\log t+\omega \left(\frac{x}{\sqrt{t}}\right).A direct computation shows that if vvis a solution of (1.1), then ω\omega satisfies the elliptic equation: (1.7)Δω+12y⋅∇ω+∣y∣βeω+1+β2=0inRN.\Delta \omega +\frac{1}{2}y\cdot \nabla \omega +| y{| }^{\beta }{e}^{\omega }+1+\frac{\beta }{2}=0\hspace{1.0em}{\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N}.In particular, if ω=ω(r)\omega =\omega \left(r)(r=∣y∣r=| y| ) is radially symmetric about the origin, then ω\omega satisfies ω′(0)=0\omega ^{\prime} \left(0)=0and (1.8)ω″+N−1r+r2ω′+rβeω+1+β2=0,r>0.{\omega }^{^{\prime\prime} }+\left(\frac{N-1}{r}+\frac{r}{2}\right)\omega ^{\prime} +{r}^{\beta }{e}^{\omega }+1+\frac{\beta }{2}=0,\hspace{1.0em}r\gt 0.We will look into the specific behavior of the function ω\omega . Comparing to the previous work [8], we encounter a new difficulty here to obtain the precise behavior of the function ω\omega . That is, since the energy functional is not monotone decreasing, we cannot obtain the global and decay estimates of the function ω\omega directly. Motivated by the previous works [12,13,16], we can overcome the difficulty by deducing the new inequality of the energy functional.One refers to ω\omega as a solution of (1.7) if ω∈C2([0,∞))\omega \in {C}^{2}\left(\left[0,\infty ))is a classical solution of (1.7). We are interested in solutions ω\omega of (1.7) satisfying (1.9)limr→∞[(2+β)logr+ω(r)]=D,D∈R.\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\omega \left(r)]=D,\hspace{1em}D\in {\mathbb{R}}.Set SD≔{ω∈C2([0,∞)):ωis a solution of (1.8) satisfying (1.9)},{S}_{D}:= \left\{\omega \in {C}^{2}\left(\left[0,\infty )):\omega \hspace{0.33em}\hspace{0.1em}\text{is a solution of (1.8) satisfying (1.9)}\hspace{0.1em}\right\},and D∗≔sup{D∈R:SD≠∅}.{D}^{\ast }:= \sup \left\{D\in R:{S}_{D}\ne \varnothing \right\}.Then we have the following main results.Theorem 1.1Let N≥3N\ge 3, 0<β<20\lt \beta \lt 2. Then the constant D∗{D}^{\ast }defined byD∗≔sup{D∈R:SD≠∅}{D}^{\ast }:= \sup \left\{D\in R:{S}_{D}\ne \varnothing \right\}is finite. Furthermore, D∗>log[(2+β)(N−2)]andSD∗≠∅,if3≤N<10+4β,D∗=log[(2+β)(N−2)]andSD∗=∅,ifN≥10+4β.\left\{\phantom{\rule[-1.25em]{}{0ex}}\begin{array}{ll}{D}^{\ast }\gt \log {[}\left(2+\beta )\left(N-2)]\hspace{0.33em}and\hspace{0.33em}{S}_{{D}^{\ast }}\ne \varnothing ,\hspace{1.0em}& if\hspace{0.33em}3\le N\lt 10+4\beta ,\\ {D}^{\ast }=\log {[}\left(2+\beta )\left(N-2)]\hspace{0.33em}and\hspace{0.33em}{S}_{{D}^{\ast }}=\varnothing ,\hspace{1.0em}& if\hspace{0.33em}N\ge 10+4\beta .\end{array}\right.Next we state the sufficient condition for the solutions of (1.1) to blow up in finite time.Theorem 1.2Let N≥3N\ge 3, 0<β<20\lt \beta \lt 2, and v0∈C(RN){v}_{0}\in C\left({{\mathbb{R}}}^{N})be an initial function satisfying (1.2). If v0{v}_{0}satisfiesliminf∣x∣→∞[(2+β)log∣x∣+v0(x)]>D∗,\mathop{\mathrm{liminf}}\limits_{| x| \to \infty }{[}\left(2+\beta )\log | x| +{v}_{0}\left(x)]\gt {D}^{\ast },then T(v0)<∞T\left({v}_{0})\lt \infty .Remark 1.3Consider solutions of the following equation: (1.10)v″+N−1rv′+rβev=0,{v}^{^{\prime\prime} }+\frac{N-1}{r}v^{\prime} +{r}^{\beta }{e}^{v}=0,with v(0)=αv\left(0)=\alpha , v′(0)=0v^{\prime} \left(0)=0, α∈R\alpha \in {\mathbb{R}}, that is, the radially symmetric stationary solution vvof (1.1). According to Lemma 4.1, we show that limr→∞[(2+β)logr+v(r)]=log[(2+β)(N−2)].\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+v\left(r)]=\log {[}\left(2+\beta )\left(N-2)].Thus, every solution of (1.10) has the same decay bound with V∗(x)=−(2+β)log∣x∣+log[(2+β)(N−2)]{V}_{\ast }\left(x)=-\left(2+\beta )\log | x| +\log {[}\left(2+\beta )\left(N-2)]. Then by Theorem 1.1, we know that for any D≤D∗D\le {D}^{\ast }, there exists a global in time solution of (1.1) with lim∣x∣→∞[(2+β)log∣x∣+v0(x)]=D.\mathop{\mathrm{lim}}\limits_{| x| \to \infty }{[}\left(2+\beta )\log | x| +{v}_{0}\left(x)]=D.Therefore, one can observe that the optimal decay bound is −(2+β)log∣x∣+D∗-\left(2+\beta )\log | x| +{D}^{\ast }by Theorem 1.2.Remark 1.4To prove Theorem 1.2, we define the function ψ\psi as follows: (1.11)ψ(y,s)≔1+β2log(1+t)+v(x,t),y=x1+t,s=log(1+t).\psi (y,s):= \left(1+\frac{\beta }{2}\right)\log \left(1+t)+v\left(x,t),\hspace{1.0em}y=\frac{x}{\sqrt{1+t}},\hspace{1.0em}s=\log \left(1+t).Then ψ\psi satisfies (1.12)ψs=Δψ+12y⋅∇ψ+∣y∣βeψ+1+β2inRN×(0,∞),ψ(y,0)=ψ0(y)inRN,\left\{\begin{array}{ll}{\psi }_{s}=\Delta \psi +\frac{1}{2}y\cdot \nabla \psi +| y{| }^{\beta }{e}^{\psi }+1+\frac{\beta }{2}\hspace{1.0em}& {\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N}\times \left(0,\infty ),\\ \psi (y,0)={\psi }_{0}(y)\hspace{1.0em}& {\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N},\end{array}\right.where ψ0=v0{\psi }_{0}={v}_{0}.The structure of this article is as follows: In Section 2, we present some preliminary results. In Section 3, we obtain the existence of radially symmetric forward self-similar solutions for (1.1) as well as its asymptotic behavior at space infinity. In Section 4, we investigate the multiplicity of the forward self-similar solutions. In Section 5, we prove Theorem 1.2 by using the previous results.2Preliminaries2.1Super- and sub-solution methodsIn this section, we review some definitions of continuous weak super-solution and sub-solution. We introduce the comparison principle for problems (1.1), (1.7), and (1.12).Definition 1(i)Function vvis a continuous weak super-solution of (1.1) in RN×[0,T]{{\mathbb{R}}}^{N}\times \left[0,T]if v∈C(RN×[0,T])v\in C\left({{\mathbb{R}}}^{N}\times \left[0,T])satisfies v(x,0)≥v0(x)v\left(x,0)\ge {v}_{0}\left(x)in RN{{\mathbb{R}}}^{N}and ∫RNv(x,t)ξ(x,t)t=0t=T′≥∬RN×[0,T′][v(x,t)(∂tξ(x,t)+Δξ(x,t))+∣x∣βev(x,t)ξ(x,t)]dxdt{\left.\underset{{\mathbb{R}}}{\overset{N}{\int }}v\left(x,t)\xi \left(x,t)\right|}_{t=0}^{t=T^{\prime} }\ge \mathop{\iint }\limits_{{{\mathbb{R}}}^{N}\times \left[0,T^{\prime} ]}{[}v\left(x,t)\left({\partial }_{t}\xi \left(x,t)+\Delta \xi \left(x,t))+| x{| }^{\beta }{e}^{v\left(x,t)}\xi \left(x,t)]{\rm{d}}x{\rm{d}}tfor all T′∈[0,T]T^{\prime} \in \left[0,T]and all nonnegative test functions ξ∈C2,1(RN×[0,T])\xi \in {C}^{2,1}\left({{\mathbb{R}}}^{N}\times \left[0,T])such that supp ξ(⋅,t)\xi \left(\cdot ,t)is compact in RN{{\mathbb{R}}}^{N}for all t∈[0,T]t\in \left[0,T]. If the reverse inequalities of the aforementioned inequalities hold, then function vvis referred to as a continuous weak sub-solution of (1.1) with the initial value v0{v}_{0}.(ii)Function ω\omega is a continuous weak super-solution of (1.7) if ω∈C(RN)\omega \in C\left({{\mathbb{R}}}^{N})satisfies ∫RNωΔη−12y⋅∇η−N2η+∣y∣βeω+1+β2ηdy≤0\mathop{\int }\limits_{{{\mathbb{R}}}^{N}}\left[\omega \left(\Delta \eta -\frac{1}{2}y\cdot \nabla \eta -\frac{N}{2}\eta \right)+\left(| y{| }^{\beta }{e}^{\omega }+1+\frac{\beta }{2}\right)\eta \right]{\rm{d}}y\le 0for all nonnegative test functions η∈C2(RN)\eta \in {C}^{2}\left({{\mathbb{R}}}^{N})with compact support in RN{{\mathbb{R}}}^{N}. If the reverse inequalities of the aforementioned inequalities hold, then function ω\omega is referred to as a continuous weak sub-solution of (1.7).(iii)Function ψ\psi is a continuous weak super-solution of (1.12) in RN×[0,S]{{\mathbb{R}}}^{N}\times \left[0,S]if ψ∈C(RN×[0,S])\psi \in C\left({{\mathbb{R}}}^{N}\times \left[0,S])satisfies ψ(y,0)≥ψ0(y)\psi (y,0)\ge {\psi }_{0}(y)in RN{{\mathbb{R}}}^{N}and ∫RNψ(y,s)η(y,s)s=0s=S′≥∬RN×[0,S′]ψ∂sη+Δη−12y⋅∇η−N2η+∣y∣βeψ+1+β2ηdyds{\left.\underset{{\mathbb{R}}}{\overset{N}{\int }}\psi (y,s)\eta (y,s)\right|}_{s=0}^{s=S^{\prime} }\ge \mathop{\iint }\limits_{{{\mathbb{R}}}^{N}\times \left[0,S^{\prime} ]}\left[\psi \left({\partial }_{s}\eta +\Delta \eta -\frac{1}{2}y\cdot \nabla \eta -\frac{N}{2}\eta \right)+\left(| y{| }^{\beta }{e}^{\psi }+1+\frac{\beta }{2}\right)\eta \right]{\rm{d}}y{\rm{d}}sfor all S′∈[0,S]S^{\prime} \in \left[0,S]and all nonnegative test functions η∈C2,1(RN×[0,S])\eta \in {C}^{2,1}\left({{\mathbb{R}}}^{N}\times \left[0,S]), which satisfies supp η(⋅,s)\eta \left(\cdot ,s)is compact in RN{{\mathbb{R}}}^{N}for all s∈[0,S]s\in \left[0,S]. If the reverse inequalities of the aforementioned inequalities hold, then function ψ\psi is referred to as a continuous weak sub-solution of (1.12).On account of [23, Lemmas 2.1, 2.3] and [33, Lemma 1.3], we can obtain the comparison principle as follows.Lemma 2.1(i)Suppose that v¯\overline{v}and v̲\underline{v}are bounded above and satisfy v¯−v̲≥−AeB∣x∣2\overline{v}-\underline{v}\ge -A{e}^{B| x{| }^{2}}in RN×[0,T]{{\mathbb{R}}}^{N}\times \left[0,T]for some constants A,B>0A,B\gt 0. Then one has that v̲(x,t)≤v¯(x,t)\underline{v}\left(x,t)\le \overline{v}\left(x,t)in RN×[0,T]{{\mathbb{R}}}^{N}\times \left[0,T], and there exists a classical solution vvof (1.1) satisfying v̲≤v≤v¯\underline{v}\le v\le \overline{v}in RN×[0,T]{{\mathbb{R}}}^{N}\times \left[0,T], where v¯\overline{v}and v̲\underline{v}are continuous weak super-solution and sub-solution of (1.1) in RN×[0,T]{{\mathbb{R}}}^{N}\times \left[0,T], respectively.(ii)If ψ¯\overline{\psi }and ψ̲\underline{\psi }are bounded above and satisfy ψ¯(y,s)−ψ̲(y,s)≥−AeB∣y∣2\overline{\psi }(y,s)-\underline{\psi }(y,s)\ge -A{e}^{B| y{| }^{2}}in RN×[0,S]{{\mathbb{R}}}^{N}\times \left[0,S]for some constants A,B>0A,B\gt 0, then one gets that ψ̲≤ψ¯\underline{\psi }\le \overline{\psi }in RN×[0,S]{{\mathbb{R}}}^{N}\times \left[0,S]and there exists a classical solution ψ\psi of (1.12) satisfying ψ̲≤ψ≤ψ¯\underline{\psi }\le \psi \le \overline{\psi }in RN×[0,S]{{\mathbb{R}}}^{N}\times \left[0,S], where ψ¯\overline{\psi }and ψ̲\underline{\psi }be continuous weak super-solution and sub-solution of (1.12) in RN×[0,S]{{\mathbb{R}}}^{N}\times \left[0,S], respectively.By using the method of [23, Lemmas 2.4, 2.5], with the help of Lemma 2.1(ii), we obtain some results on the monotonicity.Lemma 2.2If ψ̲0(ψ¯0){\underline{\psi }}_{0}\left({\overline{\psi }}_{0})is a continuous weak sub-solution (super-solution) of (1.7) and ψ\psi is a global in time solution of (1.12) with ψ0(y)≥ψ̲0(y)(≤ψ¯0(y)){\psi }_{0}(y)\ge {\underline{\psi }}_{0}(y)\left(\le {\overline{\psi }}_{0}(y))for y∈RNy\in {{\mathbb{R}}}^{N}, then one obtains ψ(y,s)≥ψ̲0(y)(≤ψ¯0(y))\psi (y,s)\ge {\underline{\psi }}_{0}(y)\left(\le {\overline{\psi }}_{0}(y))for y∈RN,s≥0y\in {{\mathbb{R}}}^{N},s\ge 0.Lemma 2.3If ψ0{\psi }_{0}is a continuous weak sub-solution (super-solution) of (1.7) and ψ\psi is a global in time solution of (1.12), then one deduces that ψ\psi is nondecreasing (nonincreasing) in ss.Next we prepare a lemma about the construction of comparison functions. For a rigorous proof of this lemma, the readers can refer to [23, Lemma 2.6].Lemma 2.4(i)Assume that n∈Nn\in {\mathbb{N}}(N={1,2,3,…}{\mathbb{N}}=\left\{1,2,3,\ldots \right\}) with n≥3n\ge 3and ωi=ωi(∣y∣)(i=1,2,…,n){\omega }_{i}={\omega }_{i}\left(| y| )\left(i=1,2,\ldots ,n)is regular radially symmetric sub-solutions of (1.7). If there exist constants R1<R2<⋯<Rn−1{R}_{1}\lt {R}_{2}\hspace{0.33em}\lt \cdots \lt {R}_{n-1}such that ωi(Ri)=ωi+1(Ri){\omega }_{i}\left({R}_{i})={\omega }_{i+1}\left({R}_{i})and ωi′(Ri)≤ωi+1′(Ri){\omega }_{i}^{^{\prime} }\left({R}_{i})\le {\omega }_{i+1}^{^{\prime} }\left({R}_{i})for i=1,2,…,n−1i=1,2,\ldots ,n-1, then the function ω̲\underline{\omega }defined byω̲(r)≔ω1(r)forr∈[0,R1],ωi+1(r)forr∈[Ri,Ri+1],(i=1,2,…,n−2)ωn(r)forr∈[Rn−1,∞),\underline{\omega }\left(r):= \left\{\begin{array}{ll}{\omega }_{1}\left(r)\hspace{1.0em}& for\hspace{1.0em}r\in \left[0,{R}_{1}],\\ {\omega }_{i+1}\left(r)\hspace{1.0em}& for\hspace{1.0em}r\in \left[{R}_{i},{R}_{i+1}],\left(i=1,2,\ldots ,n-2)\\ {\omega }_{n}\left(r)\hspace{1.0em}& for\hspace{1.0em}r\in \left[{R}_{n-1},\infty ),\end{array}\right.is a continuous weak sub-solution of (1.7).(ii)Suppose that n∈Nn\in {\mathbb{N}}with n≥3n\ge 3and ωi=ωi(∣y∣)(i=1,2,…,n){\omega }_{i}={\omega }_{i}\left(| y| )\left(i=1,2,\ldots ,n)is regular radially symmetric super-solutions of (1.7). If there exist constants R1<R2<⋯<Rn−1{R}_{1}\lt {R}_{2}\hspace{0.33em}\lt \cdots \lt {R}_{n-1}such that ωi(Ri)=ωi+1(Ri){\omega }_{i}\left({R}_{i})={\omega }_{i+1}\left({R}_{i})and ωi′(Ri)≥ωi+1′(Ri){\omega }_{i}^{^{\prime} }\left({R}_{i})\ge {\omega }_{i+1}^{^{\prime} }\left({R}_{i})for i=1,2,…,n−1i=1,2,\ldots ,n-1, then one obtains that the function ω¯\overline{\omega }defined byω¯(r)≔ω1(r)forr∈[0,R1],ωi+1(r)forr∈[Ri,Ri+1],(i=1,2,…,n−2)ωn(r)forr∈[Rn−1,∞),\overline{\omega }\left(r):= \left\{\begin{array}{ll}{\omega }_{1}\left(r)\hspace{1.0em}& for\hspace{1.0em}r\in \left[0,{R}_{1}],\\ {\omega }_{i+1}\left(r)\hspace{1.0em}& for\hspace{1.0em}r\in \left[{R}_{i},{R}_{i+1}],\left(i=1,2,\ldots ,n-2)\\ {\omega }_{n}\left(r)\hspace{1.0em}& for\hspace{1.0em}r\in \left[{R}_{n-1},\infty ),\end{array}\right.is a continuous weak super-solution of (1.7).2.2Convergence properties of solutionsIn this subsection, we study the asymptotic behavior of solution for (1.12) and obtain that the solution of (1.12) converges to a solution of (1.8). We first prove the following results for the convergence.Proposition 2.5Let ψ\psi be a global solution of (1.12) such that ψ=ψ(r,s)\psi =\psi \left(r,s)is spatially radially symmetric about the origin where r=∣y∣r=| y| . Assume that ψ(r,s)\psi \left(r,s)is either nonincreasing or nondecreasing in s≥0s\ge 0for each fixed r≥0r\ge 0, and put ω(r)≔lims→∞ψ(r,s)\omega \left(r):= {\mathrm{lim}}_{s\to \infty }\psi \left(r,s)for r≥0r\ge 0. Then we have the following results: (i)If ω∈L∞([0,∞))\omega \in {L}^{\infty }\left(\left[0,\infty )), then ω∈C2([0,∞))\omega \in {C}^{2}\left(\left[0,\infty ))and satisfies (1.8) with ω′(0)=0\omega ^{\prime} \left(0)=0.(ii)If ψ(r,s)\psi \left(r,s)is nondecreasing in s≥0s\ge 0and nonincreasing in r≥0r\ge 0, ω∉L∞([0,∞))\omega \notin {L}^{\infty }\left(\left[0,\infty )), then ω∈C2((0,∞))\omega \in {C}^{2}\left(\left(0,\infty ))and satisfies (1.8) with limr→0ω(r)=∞{\mathrm{lim}}_{r\to 0}\omega \left(r)=\infty .We first show the following lemma.Lemma 2.6Suppose that ψ\psi is a global solution of (1.12), and ψ=ψ(r,s)\psi =\psi \left(r,s), where r=∣y∣r=| y| is radially symmetric and ψ(r,s)\psi \left(r,s)is either nonincreasing or nondecreasing in s≥0s\ge 0for each fixed r≥0r\ge 0, and put ω(r)≔lims→∞ψ(r,s)\omega \left(r):= {\mathrm{lim}}_{s\to \infty }\psi \left(r,s)for r≥0r\ge 0. Then ω∈Lloc1(RN)\omega \in {L}_{{\rm{loc}}}^{1}\left({{\mathbb{R}}}^{N})is a solution of (1.8) under the mean of the distribution. Furthermore, (i)If ω∈L∞(RN)\omega \in {L}^{\infty }\left({{\mathbb{R}}}^{N}), then ω∈C2(RN)\omega \in {C}^{2}\left({{\mathbb{R}}}^{N})and is a classical solution of (1.8).(ii)If ω∉L∞(RN)\omega \notin {L}^{\infty }\left({{\mathbb{R}}}^{N})and nonincreasing in r>0r\gt 0, then ω∈C2(0,∞)\omega \in {C}^{2}\left(0,\infty )and is a solution of (1.8) satisfying limr→0ω(r)=∞{\mathrm{lim}}_{r\to 0}\omega \left(r)=\infty .ProofOur first goal is to show that ω∈Lloc1(RN)\omega \in {L}_{{\rm{loc}}}^{1}\left({{\mathbb{R}}}^{N}). It is easy to verify the case where ψ(y,s)\psi (y,s)is nonincreasing in s≥0s\ge 0, then we need to only consider that ψ(y,s)\psi (y,s)is nondecreasing in s≥0s\ge 0.Let R>0R\gt 0, and set BR≔{x∈RN:∣x∣<R}{B}_{R}:= \left\{x\in {{\mathbb{R}}}^{N}:| x| \lt R\right\}. Consider the eigenvalue problem (2.1)∇⋅(ρ∇ϕ)+λϕρ=0inBR,ϕ=0on∂BR,\left\{\phantom{\rule[-1.25em]{}{0ex}}\begin{array}{ll}\nabla \cdot \left(\rho \nabla \phi )+\lambda \phi \rho =0\hspace{1.0em}& {\rm{in}}\hspace{1.0em}{B}_{R},\\ \phi =0\hspace{1.0em}& {\rm{on}}\hspace{1.0em}\partial {B}_{R},\end{array}\right.where ρ(x)≔exp∣x∣24\rho \left(x):= \exp \left(\frac{| x{| }^{2}}{4}\right). Let λ0∈R{\lambda }_{0}\in {\mathbb{R}}denote the first eigenvalue of (2.1) and ϕ0{\phi }_{0}be the corresponding eigenfunction with ∫BRϕ0ρdx=1{\int }_{{B}_{R}}{\phi }_{0}\rho {\rm{d}}x=1. By [23, Lemma A.1], we know λ0>0{\lambda }_{0}\gt 0. Let us first define Ψ(s)≔∫BRψ(∣y∣,s)ϕ0(y)ρ(y)∣y∣βdy.\Psi \left(s):= \mathop{\int }\limits_{{B}_{R}}\psi \left(| y| ,s){\phi }_{0}(y)\rho (y)| y{| }^{\beta }{\rm{d}}y.We rewrite (1.12) in the form (2.2)ψsρ=∇⋅(ρ∇ψ)+eψρ∣y∣β+1+β2ρinRN×(0,∞).{\psi }_{s}\rho =\nabla \cdot \left(\rho \nabla \psi )+{e}^{\psi }\rho | y{| }^{\beta }+\left(1+\frac{\beta }{2}\right)\rho \hspace{1.0em}{\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N}\times \left(0,\infty ).By multiplying (2.2) with ϕ0∣y∣β{\phi }_{0}| y{| }^{\beta }and integrating on BR{B}_{R}, we obtain Ψ′(s)=∫BRψs(∣y∣,s)ϕ0(y)ρ∣y∣βdy=−∫BR∇ψ⋅∇(ϕ0∣y∣β)ρdy+∫BReψϕ0ρ∣y∣2βdy+∫BR1+β2ϕ0ρ∣y∣βdy.\begin{array}{rcl}\Psi ^{\prime} \left(s)& =& \mathop{\displaystyle \int }\limits_{{B}_{R}}{\psi }_{s}\left(| y| ,s){\phi }_{0}(y)\rho | y{| }^{\beta }{\rm{d}}y\\ & =& -\mathop{\displaystyle \int }\limits_{{B}_{R}}\nabla \psi \cdot \nabla ({\phi }_{0}| y{| }^{\beta })\rho {\rm{d}}y+\mathop{\displaystyle \int }\limits_{{B}_{R}}{e}^{\psi }{\phi }_{0}\rho | y{| }^{2\beta }{\rm{d}}y+\mathop{\displaystyle \int }\limits_{{B}_{R}}\left(1+\frac{\beta }{2}\right){\phi }_{0}\rho | y{| }^{\beta }{\rm{d}}y.\end{array}By multiplying (2.1) with ψ∣y∣β\psi | y{| }^{\beta }and integrating by parts, we obtain ∫∂BRψ∂ϕ0∂νρ∣y∣βdS−∫BR∇(ψ∣y∣β)⋅∇ϕ0ρdy+λ0∫BRψϕ0ρ∣y∣βdy=0.\mathop{\int }\limits_{\partial {B}_{R}}\psi \frac{\partial {\phi }_{0}}{\partial \nu }\rho | y{| }^{\beta }{\rm{d}}S-\mathop{\int }\limits_{{B}_{R}}\nabla (\psi | y{| }^{\beta })\cdot \nabla {\phi }_{0}\rho {\rm{d}}y+{\lambda }_{0}\mathop{\int }\limits_{{B}_{R}}\psi {\phi }_{0}\rho | y{| }^{\beta }{\rm{d}}y=0.It follows from maximum principle that ∂ϕ0∂ν<0\frac{\partial {\phi }_{0}}{\partial \nu }\lt 0on ∂BR\partial {B}_{R}.From the aforementioned analysis, we obtain Ψ′(s)=−∫∂BRψ∂ϕ0∂νρ∣y∣βdS−λ0∫BRψϕ0ρ∣y∣βdy+∫BReψϕ0ρ∣y∣2βdy+∫BR1+β2ϕ0ρ∣y∣βdy.\Psi ^{\prime} \left(s)=-\mathop{\int }\limits_{\partial {B}_{R}}\psi \frac{\partial {\phi }_{0}}{\partial \nu }\rho | y{| }^{\beta }{\rm{d}}S-{\lambda }_{0}\mathop{\int }\limits_{{B}_{R}}\psi {\phi }_{0}\rho | y{| }^{\beta }{\rm{d}}y+\mathop{\int }\limits_{{B}_{R}}{e}^{\psi }{\phi }_{0}\rho | y{| }^{2\beta }{\rm{d}}y+\mathop{\int }\limits_{{B}_{R}}\left(1+\frac{\beta }{2}\right){\phi }_{0}\rho | y{| }^{\beta }{\rm{d}}y.Using Jensen’s inequality yields (2.3)Ψ′(s)≥−∫∂BR∣R∣βψ∂ϕ0∂νρdS−λ0Ψ+∫BReψϕ0ρ∣y∣βdy≥−∫∂BR∣R∣βψ∂ϕ0∂νρdS−λ0Ψ+eΨ\Psi ^{\prime} \left(s)\ge -\mathop{\int }\limits_{\partial {B}_{R}}| R{| }^{\beta }\psi \frac{\partial {\phi }_{0}}{\partial \nu }\rho {\rm{d}}S-{\lambda }_{0}\Psi +\mathop{\int }\limits_{{B}_{R}}{e}^{\psi }{\phi }_{0}\rho | y{| }^{\beta }{\rm{d}}y\ge -\mathop{\int }\limits_{\partial {B}_{R}}| R{| }^{\beta }\psi \frac{\partial {\phi }_{0}}{\partial \nu }\rho {\rm{d}}S-{\lambda }_{0}\Psi +{e}^{\Psi }for all s≥0s\ge 0. Next we consider two cases: ω(r0)=∞\omega \left({r}_{0})=\infty for some r0>0{r}_{0}\gt 0or ω(r)<∞\omega \left(r)\lt \infty for any r>0r\gt 0.Let us assume initially that there exists some r0>0{r}_{0}\gt 0satisfying ω(r0)=∞\omega \left({r}_{0})=\infty . Then we could choose some s1>0{s}_{1}\gt 0such that ψ(r0,s)>0\psi \left({r}_{0},s)\gt 0for all s≥s1s\ge {s}_{1}. Thus, by (2.3), we have (2.4)Ψ′(s)≥eΨ−λ0Ψ,s≥s1.\Psi ^{\prime} \left(s)\ge {e}^{\Psi }-{\lambda }_{0}\Psi ,\hspace{1.0em}s\ge {s}_{1}.If lims→∞Ψ(s)=∞{\mathrm{lim}}_{s\to \infty }\Psi \left(s)=\infty , we may choose some s2≥s1{s}_{2}\ge {s}_{1}such that eΨ−λ0Ψ≥eΨ2for anys≥s2.{e}^{\Psi }-{\lambda }_{0}\Psi \ge {e}^{\tfrac{\Psi }{2}}\hspace{1.0em}\hspace{0.1em}\text{for any}\hspace{0.1em}\hspace{0.33em}s\ge {s}_{2}.In view of (2.4), we see Ψ′(s)≥eΨ2for alls≥s2,\Psi ^{\prime} \left(s)\ge {e}^{\tfrac{\Psi }{2}}\hspace{1.0em}\hspace{0.1em}\text{for all}\hspace{0.1em}\hspace{0.33em}s\ge {s}_{2},and we have Ψ(s)≥−2loge−Ψ(s2)2−12(s−s2),s≥s2,\Psi \left(s)\ge -2\log \left({e}^{-\tfrac{\Psi \left({s}_{2})}{2}}-\frac{1}{2}\left(s-{s}_{2})\right),\hspace{1.0em}s\ge {s}_{2},which leads to a contradiction; thus, Ψ\Psi exists for all s≥0s\ge 0.Next we assume that ω(r)<∞\omega \left(r)\lt \infty for any r>0r\gt 0. From (2.3), there exist some C≥0C\ge 0such that Ψ′(s)≥eΨ−λ0Ψ−C\Psi ^{\prime} \left(s)\ge {e}^{\Psi }-{\lambda }_{0}\Psi -Cfor all s≥0s\ge 0. If lims→∞Ψ(s)=∞Ψ(s)=∞{\mathrm{lim}}_{s\to \infty }\Psi \left(s)=\infty \Psi \left(s)=\infty , we may choose some s3≥0{s}_{3}\ge 0such that eΨ−λ0Ψ−C≥eΨ2for anys≥s3.{e}^{\Psi }-{\lambda }_{0}\Psi -C\ge {e}^{\tfrac{\Psi }{2}}\hspace{1.0em}\hspace{0.1em}\text{for any}\hspace{0.1em}\hspace{0.33em}s\ge {s}_{3}.Then we yield a contradiction in the same way as mentioned earlier.Thus, we infer the boundedness of the function Ψ\Psi in [0,∞)\left[0,\infty ). Since ψ\psi is nondecreasing in ss, so is Ψ\Psi . According to the monotone convergence theorem, we have ∫BRω(∣y∣,s)ϕ0(y)ρ(y)∣y∣βdy<∞.\mathop{\int }\limits_{{B}_{R}}\omega \left(| y| ,s){\phi }_{0}(y)\rho (y)| y{| }^{\beta }{\rm{d}}y\lt \infty .Since R>0R\gt 0is arbitrarily, we obtain ω∈Lloc1(RN)\omega \in {L}_{{\rm{loc}}}^{1}\left({{\mathbb{R}}}^{N}).It remains to show that ω\omega is a solution of (1.7) in the sense of distribution. For ς>0\varsigma \gt 0, we have ∫RNψ(∣y∣,s+ς)η(y)dy∣s=0s=1=∫01∫RNψ(∣y∣,s+ς)Δη−12y⋅∇η−N2η+∣y∣βeψ+1+β2ηdyds,\underset{{\mathbb{R}}}{\overset{N}{\int }}\psi \left(| y| ,s+\varsigma )\eta (y){\rm{d}}y{| }_{s=0}^{s=1}=\underset{0}{\overset{1}{\int }}\mathop{\int }\limits_{{{\mathbb{R}}}^{N}}\psi \left(| y| ,s+\varsigma )\left(\Delta \eta -\frac{1}{2}y\cdot \nabla \eta -\frac{N}{2}\eta \right)+\left(| y{| }^{\beta }{e}^{\psi }+1+\frac{\beta }{2}\right)\eta {\rm{d}}y{\rm{d}}s,where η∈C2(RN)\eta \in {C}^{2}\left({{\mathbb{R}}}^{N})with compact support. Letting ς→∞\varsigma \to \infty in the aforementioned equation, we obtain 0=∫RNω(y)Δη−12y⋅∇η−N2η+∣y∣βeψ+1+β2ηdy.0=\mathop{\int }\limits_{{{\mathbb{R}}}^{N}}\omega (y)\left(\Delta \eta -\frac{1}{2}y\cdot \nabla \eta -\frac{N}{2}\eta \right)+\left(| y{| }^{\beta }{e}^{\psi }+1+\frac{\beta }{2}\right)\eta {\rm{d}}y.It turns out that ω\omega is a solution of (1.7) in the sense of distribution.If ω∈L∞(RN)\omega \in {L}^{\infty }\left({{\mathbb{R}}}^{N}), we have ω∈C2(RN)\omega \in {C}^{2}\left({{\mathbb{R}}}^{N})by applying the regularity theory of elliptic equations; therefore, ω\omega is a classical solution of (1.8). Thus, (i)\left(i)follows.If ω∉L∞(RN)\omega \notin {L}^{\infty }\left({{\mathbb{R}}}^{N})and nonincreasing in r>0r\gt 0for any s≥0s\ge 0and by ω∈Lloc1(RN)\omega \in {L}_{{\rm{loc}}}^{1}\left({{\mathbb{R}}}^{N}), we have limr→0ω(r)=∞{\mathrm{lim}}_{r\to 0}\omega \left(r)=\infty and ω∈L∞(RN\Br)\omega \in {L}^{\infty }\left({{\mathbb{R}}}^{N}\backslash {B}_{r})for any r>0r\gt 0. Therefore, ω∈C2(RN\0)\omega \in {C}^{2}\left({{\mathbb{R}}}^{N}\backslash 0)by the regularity theory and ω\omega satisfies (1.8) in (RN\0)\left({{\mathbb{R}}}^{N}\backslash 0). Thus, (ii)\left(ii)follows.□Proof of Proposition 2.5(i)Owing to Lemma 2.6(i), we have that ω=ω(r)\omega =\omega \left(r)is a classical radial solution of (1.8). Thus, ω∈C2([0,∞))\omega \in {C}^{2}\left(\left[0,\infty ))and satisfies (1.8) with ω′(0)=0\omega ^{\prime} \left(0)=0.(ii)Since ψ(r,s)\psi \left(r,s)is nonincreasing in r≥0r\ge 0, ω(r)\omega \left(r)is nonincreasing in r≥0r\ge 0. Then Lemma 2.6(ii) implies that ω∈C2((0,∞))\omega \in {C}^{2}\left(\left(0,\infty ))and satisfies (1.8) with limr→0ω(r)=∞{\mathrm{lim}}_{r\to 0}\omega \left(r)=\infty .□3Existence and asymptotic of forward self-similar solutionsIn this section, we prove the existence of forward self-similar solutions for (1.1), that is, the solutions for (1.7). We also obtain the asymptotic behavior of solutions at space infinity.3.1Existence of forward self-similar solutionsIn this subsection, we prove the existence of the solution for (1.7) by using the contraction mapping argument. Let ω\omega be a solution of (1.7) and define function τ(y)≔eω(y)\tau (y):= {e}^{\omega (y)}, then τ\tau satisfies (3.1)Δτ+12y⋅∇τ+1+β2τ+∣y∣βτ2−∣∇τ2∣τ=0inRN,τ>0inRN.\Delta \tau +\frac{1}{2}y\cdot \nabla \tau +\left(1+\frac{\beta }{2}\right)\tau +| y{| }^{\beta }{\tau }^{2}-\frac{| {\nabla \tau }^{2}| }{\tau }=0\hspace{1.0em}{\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N},\hspace{1.0em}\tau \gt 0\hspace{0.33em}{\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N}.Conversely, if τ\tau is a positive solution of (3.1), then the function defined by ω(y)≔logτ(y)\omega (y):= \log \tau (y)is a solution of (1.7). Studying solutions of (1.7) is therefore equivalent to analyzing the solutions of (3.1). In particular, if τ=τ(r)\tau =\tau \left(r)is radially symmetric about the origin, where r=∣y∣r=| y| , then τ\tau satisfies (3.2)τ″+N−1r+r2τ′+1+β2τ+rβτ2−(τ′)2τ=0in(0,∞),τ′(0)=0.{\tau }^{^{\prime\prime} }+\left(\frac{N-1}{r}+\frac{r}{2}\right)\tau ^{\prime} +\left(1+\frac{\beta }{2}\right)\tau +{r}^{\beta }{\tau }^{2}-\frac{{\left(\tau ^{\prime} )}^{2}}{\tau }=0\hspace{1.0em}{\rm{in}}\hspace{0.33em}\left(0,\infty ),\hspace{1.0em}\tau ^{\prime} \left(0)=0.Now consider the following integral equation: (3.3)τ(r)=αexp−∫0rσN(s)−1∫0stβτ(t)+1+β2σN(t)dtds,\tau \left(r)=\alpha \exp \left\{-\underset{0}{\overset{r}{\int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}\left[{t}^{\beta }\tau \left(t)+1+\frac{\beta }{2}\right]{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\},where α>0\alpha \gt 0. One can easily to check that (3.3) is equivalent to problem (3.2) with τ(0)=α\tau \left(0)=\alpha . We first introduce two functions σN=σN(r){\sigma }_{N}={\sigma }_{N}\left(r)and γN=γN(r){\gamma }_{N}={\gamma }_{N}\left(r)defined by (3.4)σN(r)≔rN−1er24,γN(r)≔∫0rσN(s)−1∫0sσN(t)dtds,r>0.{\sigma }_{N}\left(r):= {r}^{N-1}{e}^{\tfrac{{r}^{2}}{4}},\hspace{1.0em}{\gamma }_{N}\left(r):= \underset{0}{\overset{r}{\int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s,\hspace{1.0em}r\gt 0.Functions σN=σN(r){\sigma }_{N}={\sigma }_{N}\left(r)and γN=γN(r){\gamma }_{N}={\gamma }_{N}\left(r)are useful to study the radially symmetric solutions ω=ω(∣y∣)\omega =\omega \left(| y| )of (1.7) as ∣y∣→∞| y| \to \infty . We state following lemmas, which describe the asymptotic behavior of functions σN=σN(r){\sigma }_{N}={\sigma }_{N}\left(r)and γN=γN(r){\gamma }_{N}={\gamma }_{N}\left(r). The proof of the following lemmas can be found in [8, Lemmas 2.1 and 2.2].Lemma 3.1For N≥1N\ge 1, the limit(3.5)limr→∞(γN(r)−2logr)\mathop{\mathrm{lim}}\limits_{r\to \infty }\left({\gamma }_{N}\left(r)-2\log r)exists in R{\mathbb{R}}. In particular, it holds(3.6)supr≥0∣γN(r)−2log(1+r)∣<∞.\mathop{\sup }\limits_{r\ge 0}| {\gamma }_{N}\left(r)-2\log \left(1+r)| \lt \infty .Lemma 3.2For N≥1N\ge 1, inequality∫0s(1+t)−2σN(t)dt≤CsN−4es24\left|\underset{0}{\overset{s}{\int }}{\left(1+t)}^{-2}{\sigma }_{N}\left(t){\rm{d}}t\right|\le C{s}^{N-4}{e}^{\tfrac{{s}^{2}}{4}}holds for all sufficiently large s>0s\gt 0and C>0C\gt 0is a constant.We are now turning to the existence of solutions for (3.2).Proposition 3.3Let N≥1N\ge 1, 0<β<20\lt \beta \lt 2. For any α>0\alpha \gt 0, there exists a unique positive solution τ∈C2([0,∞))\tau \in {C}^{2}\left(\left[0,\infty ))of (3.2) satisfying τ(0)=α\tau \left(0)=\alpha .ProofFirst, we will prove the local existence and uniqueness. Let g(r)≔rβeω+1+β2,g\left(r):= {r}^{\beta }{e}^{\omega }+1+\frac{\beta }{2},then (1.8) is equivalent to (3.7)rN−1er24ω′(r)′=−g(r)rN−1er24.\left({r}^{N-1}{e}^{\tfrac{{r}^{2}}{4}}\omega ^{\prime} \left(r)\right)^{\prime} =-g\left(r){r}^{N-1}{e}^{\tfrac{{r}^{2}}{4}}.Since ω′(0)=0\omega ^{\prime} \left(0)=0, integrating of (3.7) with respect of rrgives (3.8)ω′(r)=−r1−Ne−r24∫0rg(t)tN−1et24dt=−σN(r)−1∫0rg(t)σN(t)dt.\omega ^{\prime} \left(r)=-{r}^{1-N}{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{r}{\int }}g\left(t){t}^{N-1}{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t=-{\sigma }_{N}{\left(r)}^{-1}\underset{0}{\overset{r}{\int }}g\left(t){\sigma }_{N}\left(t){\rm{d}}t.Integrating once more, we obtain (3.9)ω(r)=logα−∫0rσN(s)−1∫0sg(t)σN(t)dtds.\omega \left(r)=\log \alpha -\underset{0}{\overset{r}{\int }}{\sigma }_{N}{\left(s)}^{-1}\underset{0}{\overset{s}{\int }}g\left(t){\sigma }_{N}\left(t){\rm{d}}t{\rm{d}}s.Therefore, τ\tau satisfies the following integral equation: (3.10)τ(r)=αexp−∫0rσN(s)−1∫0stβτ(t)+1+β2σN(t)dtds.\tau \left(r)=\alpha \exp \left\{-\underset{0}{\overset{r}{\int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}\left[{t}^{\beta }\tau \left(t)+1+\frac{\beta }{2}\right]{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\}.Next we use a contraction mapping argument to prove existence and uniqueness of solution τ∈C([0,δ])\tau \in C\left(\left[0,\delta ])of (3.10) for δ>0\delta \gt 0small enough. To accomplish that, we choose δ>0\delta \gt 0and define Jα,δ≔{τ∈C([0,δ]):τ(0)=α}.{J}_{\alpha ,\delta }:= \left\{\tau \in C\left(\left[0,\delta ]):\tau \left(0)=\alpha \right\}.For τ∈Jα,δ\tau \in {J}_{\alpha ,\delta }, H[τ](r)≔αexp−∫0rσN(s)−1∫0stβτ(t)+1+β2σN(t)dtds,H\left[\tau ]\left(r):= \alpha \exp \left\{-\underset{0}{\overset{r}{\int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}\left[{t}^{\beta }\tau \left(t)+1+\frac{\beta }{2}\right]{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\},Then we obtain 0<H[τ](r)≤α0\lt H\left[\tau ]\left(r)\le \alpha , H[τ](0)=αH\left[\tau ]\left(0)=\alpha , and obvious H[τ](r)∈Jα,δH\left[\tau ]\left(r)\in {J}_{\alpha ,\delta }. Furthermore, by the elementary inequality (3.11)∣e−A−e−B∣≤∣A−B∣forA,B>0,| {e}^{-A}-{e}^{-B}| \le | A-B| \hspace{1.0em}{\rm{for}}\hspace{0.33em}A,B\gt 0,we have sup0<r<δ∣H[τ1](r)−H[τ2](r)∣≤αsup0<r<δ∫0rσN(s)−1∫0stβ∣τ1(t)−τ2(t)∣σN(t)dtds≤α∫0δσN(s)−1sβ⋅σN(s)⋅sds⋅sup0<r<δ∣τ1(r)−τ2(r)∣=α∫0δs1+βds⋅sup0<r<δ∣τ1(r)−τ2(r)∣=α2+βδ2+βsup0<r<δ∣τ1(r)−τ2(r)∣\begin{array}{rcl}\mathop{\sup }\limits_{0\lt r\lt \delta }| H\left[{\tau }_{1}]\left(r)-H\left[{\tau }_{2}]\left(r)| & \le & \alpha \mathop{\sup }\limits_{0\lt r\lt \delta }\underset{0}{\overset{r}{\displaystyle \int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\displaystyle \int }}{t}^{\beta }| {\tau }_{1}\left(t)-{\tau }_{2}\left(t)| {\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\\ & \le & \alpha \underset{0}{\overset{\delta }{\displaystyle \int }}{\sigma }_{N}{\left(s)}^{-1}{s}^{\beta }\cdot {\sigma }_{N}\left(s)\cdot s{\rm{d}}s\cdot \mathop{\sup }\limits_{0\lt r\lt \delta }| {\tau }_{1}\left(r)-{\tau }_{2}\left(r)| \\ & =& \alpha \underset{0}{\overset{\delta }{\displaystyle \int }}{s}^{1+\beta }{\rm{d}}s\cdot \mathop{\sup }\limits_{0\lt r\lt \delta }| {\tau }_{1}\left(r)-{\tau }_{2}\left(r)| \\ & =& \frac{\alpha }{2+\beta }{\delta }^{2+\beta }\mathop{\sup }\limits_{0\lt r\lt \delta }| {\tau }_{1}\left(r)-{\tau }_{2}\left(r)| \end{array}for all τ1,τ2∈Jα,δ{\tau }_{1},{\tau }_{2}\in {J}_{\alpha ,\delta }. Thus, by selecting δ=α−12+β\delta ={\alpha }^{-\tfrac{1}{2+\beta }}, we obtain the unique positive solution τ∈C([0,δ])\tau \in C\left(\left[0,\delta ])of (3.3). And τ\tau satisfies 0<τ(r)≤α0\lt \tau \left(r)\le \alpha in [0,δ]\left[0,\delta ]. It is easy to see that τ∈C1([0,δ])\tau \in {C}^{1}\left(\left[0,\delta ]), τ′(0)=0\tau ^{\prime} \left(0)=0, and (3.12)τ′(r)=−τ(r)σN(r)−1∫0rtβτ(t)+1+β2σN(t)dt,\tau ^{\prime} \left(r)=-\tau \left(r){\sigma }_{N}{\left(r)}^{-1}\underset{0}{\overset{r}{\int }}\left[{t}^{\beta }\tau \left(t)+1+\frac{\beta }{2}\right]{\sigma }_{N}\left(t){\rm{d}}t,by (3.10). Therefore, τ∈C2((0,δ])\tau \in {C}^{2}\left(\left(0,\delta ])and satisfies (3.2).Finally, we infer from (3.12) that τ′(r)−τ′(0)r=τ′(r)r=−τ(r)r−Ne−r24∫0rtβτ(t)+1+β2tN−1et24dt=−τ(r)r−Ne−r24tNNtβτ(t)+1+β2et24t=0t=r−∫0rtNN⋅tβτ′(t)+βtβ−1τ(t)+t2tβτ(t)+1+β2et24dt=−τ(r)r−Ne−r24rNNrβτ(r)+1+β2er24+O(rN+1)\begin{array}{rcl}\frac{\tau ^{\prime} \left(r)-\tau ^{\prime} \left(0)}{r}& =& \frac{\tau ^{\prime} \left(r)}{r}=-\tau \left(r){r}^{-N}{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{r}{\displaystyle \int }}\left[{t}^{\beta }\tau \left(t)+1+\frac{\beta }{2}\right]{t}^{N-1}{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\\ & =& -\tau \left(r){r}^{-N}{e}^{-\tfrac{{r}^{2}}{4}}\left\{{\left[\frac{{t}^{N}}{N}\left({t}^{\beta }\tau \left(t)+1+\frac{\beta }{2}\right){e}^{\tfrac{{t}^{2}}{4}}\right]}_{t=0}^{t=r}-\underset{0}{\overset{r}{\displaystyle \int }}\frac{{t}^{N}}{N}\cdot \left[{t}^{\beta }\tau ^{\prime} \left(t)+\beta {t}^{\beta -1}\tau \left(t)+\frac{t}{2}\left({t}^{\beta }\tau \left(t)+1+\frac{\beta }{2}\right)\right]{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\right\}\\ & =& -\tau \left(r){r}^{-N}{e}^{-\tfrac{{r}^{2}}{4}}\left\{\phantom{\rule[-1.25em]{}{0ex}},\frac{{r}^{N}}{N}\left[{r}^{\beta }\tau \left(r)+1+\frac{\beta }{2}\right]{e}^{\tfrac{{r}^{2}}{4}}+O\left({r}^{N+1})\right\}\end{array}for r>0r\gt 0small enough, which indicates that τ″(0){\tau }^{^{\prime\prime} }\left(0)exists and equals to −αN1+β2-\frac{\alpha }{N}\left(1+\frac{\beta }{2}\right).On the other hand, since τ\tau satisfies (3.3) in [0,δ]\left[0,\delta ], we observe that τ\tau also satisfies (3.2) for r>0r\gt 0small enough. Moreover, we infer from (3.2) that limr→0τ″(r)=−(N−1)limr→0τ′(r)r−α1+β2=−(N−1)τ″(0)+Nτ″(0)=τ″(0).\mathop{\mathrm{lim}}\limits_{r\to 0}{\tau }^{^{\prime\prime} }\left(r)=-\left(N-1)\mathop{\mathrm{lim}}\limits_{r\to 0}\frac{\tau ^{\prime} \left(r)}{r}-\alpha \left(1+\frac{\beta }{2}\right)=-\left(N-1){\tau }^{^{\prime\prime} }\left(0)+N{\tau }^{^{\prime\prime} }\left(0)={\tau }^{^{\prime\prime} }\left(0).Thus, we obtain τ∈C2([0,δ])\tau \in {C}^{2}\left(\left[0,\delta ]), which shows the local existence and uniqueness.For the purpose of proving the global existence, we introduce the energy of the solution τ\tau (3.13)I[τ](r)≔12∣τ′(r)∣2+12+β4∣τ(r)∣2+13rβ∣τ(r)∣3.I\left[\tau ]\left(r):= \frac{1}{2}| \tau ^{\prime} \left(r){| }^{2}+\left(\frac{1}{2}+\frac{\beta }{4}\right)| \tau \left(r){| }^{2}+\frac{1}{3}{r}^{\beta }| \tau \left(r){| }^{3}.By (3.12), we have τ′(r)≤0\tau ^{\prime} \left(r)\le 0, then by multiplying (3.2) by τ′(r)\tau ^{\prime} \left(r), we obtain (3.14)ddrI[τ](r)=τ′τ″+1+β2ττ′+rβτ2τ′+β3rβ−1∣τ∣3=−N−1r+r2∣τ′∣2+(τ′)3τ+β3rβ−1∣τ∣3≤β3rβ−1∣τ∣3≤βrI[τ](r).\begin{array}{rcl}\frac{{\rm{d}}}{{\rm{d}}r}I\left[\tau ]\left(r)& =& \tau ^{\prime} {\tau }^{^{\prime\prime} }+\left(1+\frac{\beta }{2}\right)\tau \tau ^{\prime} +{r}^{\beta }{\tau }^{2}\tau ^{\prime} +\frac{\beta }{3}{r}^{\beta -1}| \tau {| }^{3}\\ & =& -\left(\frac{N-1}{r}+\frac{r}{2}\right)| \tau ^{\prime} {| }^{2}+\frac{{\left(\tau ^{\prime} )}^{3}}{\tau }+\frac{\beta }{3}{r}^{\beta -1}| \tau {| }^{3}\\ & \le & \frac{\beta }{3}{r}^{\beta -1}| \tau {| }^{3}\\ & \le & \frac{\beta }{r}I\left[\tau ]\left(r).\end{array}Then we have (3.15)I[τ](r)≤Crβ.I\left[\tau ]\left(r)\le C{r}^{\beta }.It is apparent that ∣τ(r)∣≤C| \tau \left(r)| \le Cfrom the definition of I[τ](r)I\left[\tau ]\left(r). Therefore, we obtain that the solution of (3.2) exists globally in time. Hence, the proof of Proposition 3.3 is completed.□3.2Asymptotic behavior of forward self-similar solutionsIn this subsection, we study the asymptotic behavior of the solution for (3.2) and obtain the decay rate of the solution for (3.2) as r→∞r\to \infty . The proofs of Lemmas 3.4 and 3.5, which follows the ideas of Haraux and Weissler [16, Proposition 3.1], are also enclosed in this section.Lemma 3.4Let N≥1N\ge 1, 0<β<20\lt \beta \lt 2, α>0\alpha \gt 0, and τ\tau be the solution of (3.2) satisfying τ(0)=α\tau \left(0)=\alpha . Assume that there exist constants C>0C\gt 0and σ≥0\sigma \ge 0such that 0<τ(r)≤C(1+r)−σ0\lt \tau \left(r)\le C{\left(1+r)}^{-\sigma }for r>0r\gt 0. Then there exists a constant C˜>0\tilde{C}\gt 0such that ∣τ′(r)∣≤C˜(1+r)β−2σ−1| \tau ^{\prime} \left(r)| \le \tilde{C}{\left(1+r)}^{\beta -2\sigma -1}if β≥σ\beta \ge \sigma , and ∣τ′(r)∣≤C˜(1+r)−σ−1| \tau ^{\prime} \left(r)| \le \tilde{C}{\left(1+r)}^{-\sigma -1}if β<σ\beta \lt \sigma for all r>0r\gt 0.ProofWe introduce g(r)≔1+β2τ(r)+rβτ(r)2−τ′(r)2τ(r).g\left(r):= \left(1+\frac{\beta }{2}\right)\tau \left(r)+{r}^{\beta }\tau {\left(r)}^{2}-\frac{\tau ^{\prime} {\left(r)}^{2}}{\tau \left(r)}.Then (3.2) is equivalent to (3.16)rN−1er24τ′(r)′=−g(r)rN−1er24,\left({r}^{N-1}{e}^{\tfrac{{r}^{2}}{4}}\tau ^{\prime} \left(r)\right)^{\prime} =-g\left(r){r}^{N-1}{e}^{\tfrac{{r}^{2}}{4}},and since τ′(0)=0\tau ^{\prime} \left(0)=0, integration of (3.16) gives (3.17)τ′(r)=−r1−Ne−r24∫0rg(t)tN−1et24dt.\tau ^{\prime} \left(r)=-{r}^{1-N}{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{r}{\int }}g\left(t){t}^{N-1}{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t.Note that ∣g(r)∣≤C1(1+r)−σ+C2(1+r)β−2σ.| g\left(r)| \le {C}_{1}{\left(1+r)}^{-\sigma }+{C}_{2}{\left(1+r)}^{\beta -2\sigma }.Next we split the proof into three cases.Case A: If β−2σ≤−σ<0\beta -2\sigma \le -\sigma \lt 0, then ∣g(r)∣≤C(1+r)−σ| g\left(r)| \le C{\left(1+r)}^{-\sigma }. Thus, ∣τ′(r)∣≤r1−Ne−r24∫0r∣g(t)∣tN−1et24dt≤Ce−r24∫0r(1+t)−σet24dt≤Ce−r24∫0r2et24dt+Ce−r24∫r2r(1+t)−σet24dt≤Cr2e−3r216+1+r2−σ−1e−r24∫r2r(1+t)et24dt.\begin{array}{rcl}| \tau ^{\prime} \left(r)| & \le & {r}^{1-N}{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{r}{\displaystyle \int }}| g\left(t)| {t}^{N-1}{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\\ & \le & C{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{r}{\displaystyle \int }}{\left(1+t)}^{-\sigma }{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\\ & \le & C{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{\frac{r}{2}}{\displaystyle \int }}{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t+C{e}^{-\tfrac{{r}^{2}}{4}}\underset{\frac{r}{2}}{\overset{r}{\displaystyle \int }}{\left(1+t)}^{-\sigma }{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\\ & \le & C\left[\frac{r}{2}{e}^{-\tfrac{3{r}^{2}}{16}}+{\left(1+\frac{r}{2}\right)}^{-\sigma -1}{e}^{-\tfrac{{r}^{2}}{4}}\underset{\frac{r}{2}}{\overset{r}{\displaystyle \int }}\left(1+t){e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\right].\end{array}If r≤2r\le 2, τ′(r)\tau ^{\prime} \left(r)is obviously bounded. If r≥2r\ge 2, the integral in the aforementioned formula is controlled by ∫r2r2tet24dt=4er24−4er216≤4er24.\underset{\frac{r}{2}}{\overset{r}{\int }}2t{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t=4{e}^{\tfrac{{r}^{2}}{4}}-4{e}^{\tfrac{{r}^{2}}{16}}\le 4{e}^{\tfrac{{r}^{2}}{4}}.Then, ∣τ′(r)∣≤C˜(1+r)−σ−1| \tau ^{\prime} \left(r)| \le \tilde{C}{\left(1+r)}^{-\sigma -1}.Case B: If −σ≤β−2σ<0-\sigma \le \beta -2\sigma \lt 0, then ∣g(r)∣≤C(1+r)β−2σ| g\left(r)| \le C{\left(1+r)}^{\beta -2\sigma }. Thus, by the same argument in Case A, we obtain ∣τ′(r)∣≤C˜(1+r)β−2σ−1| \tau ^{\prime} \left(r)| \le \tilde{C}{\left(1+r)}^{\beta -2\sigma -1}.Case C: If −σ<0≤β−2σ-\sigma \lt 0\le \beta -2\sigma , then ∣g(r)∣≤C(1+r)β−2σ| g\left(r)| \le C{\left(1+r)}^{\beta -2\sigma }. Thus, ∣τ′(r)∣≤r1−Ne−r24∫0r∣g(t)∣tN−1et24dt≤Ce−r24∫0r(1+t)β−2σet24dt=Ce−r24∫0r2(1+t)β−2σet24dt+Ce−r24∫r2r(1+t)β−2σet24dt≤Ce−r241+r2β−2σer216∫0r2dt+Ce−r24∫r2r(1+t)β−2σ−1(1+t)et24dt≤Ce−3r2161+r2β−2σ+1+C1+r2β−2σ−1e−r24∫r2r(1+t)et24dt\begin{array}{rcl}| \tau ^{\prime} \left(r)| & \le & {r}^{1-N}{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{r}{\displaystyle \int }}| g\left(t)| {t}^{N-1}{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\\ & \le & C{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{r}{\displaystyle \int }}{\left(1+t)}^{\beta -2\sigma }{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\\ & =& C{e}^{-\tfrac{{r}^{2}}{4}}\underset{0}{\overset{\frac{r}{2}}{\displaystyle \int }}{\left(1+t)}^{\beta -2\sigma }{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t+C{e}^{-\tfrac{{r}^{2}}{4}}\underset{\frac{r}{2}}{\overset{r}{\displaystyle \int }}{\left(1+t)}^{\beta -2\sigma }{e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\\ & \le & C{e}^{-\tfrac{{r}^{2}}{4}}{\left(1+\frac{r}{2}\right)}^{\beta -2\sigma }{e}^{\tfrac{{r}^{2}}{16}}\underset{0}{\overset{\frac{r}{2}}{\displaystyle \int }}{\rm{d}}t+C{e}^{-\tfrac{{r}^{2}}{4}}\underset{\frac{r}{2}}{\overset{r}{\displaystyle \int }}{\left(1+t)}^{\beta -2\sigma -1}\left(1+t){e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\\ & \le & C{e}^{-\tfrac{3{r}^{2}}{16}}{\left(1+\frac{r}{2}\right)}^{\beta -2\sigma +1}+C{\left(1+\frac{r}{2}\right)}^{\beta -2\sigma -1}{e}^{-\tfrac{{r}^{2}}{4}}\underset{\frac{r}{2}}{\overset{r}{\displaystyle \int }}\left(1+t){e}^{\tfrac{{t}^{2}}{4}}{\rm{d}}t\end{array}for rrlarge. And CCdenotes a positive generic constant. Next we can use the similar method in Case A to prove Case C. Hence, the proof of Lemma 3.4 is completed.□Lemma 3.5Let N≥1N\ge 1, 0<β<20\lt \beta \lt 2, α>0\alpha \gt 0, and τ\tau be the solution of (3.2) satisfying τ(0)=α\tau \left(0)=\alpha . Then there exists a constant C>0C\gt 0such that0<τ(r)≤C(1+r)−2−β2,∣τ′(r)∣≤C(1+r)−3−β20\lt \tau \left(r)\le C{\left(1+r)}^{-2-\tfrac{\beta }{2}},\hspace{1.0em}| \tau ^{\prime} \left(r)| \le C{\left(1+r)}^{-3-\tfrac{\beta }{2}}for all r>0r\gt 0.ProofMultiplying (3.2) by r−β−1τ(r){r}^{-\beta -1}\tau \left(r), we obtain r−β−11+β2τ2+r−1τ3=−r−βddrτ24+ττ′r+2r−β−1(τ′)2−r−β−2Nττ′.{r}^{-\beta -1}\left(1+\frac{\beta }{2}\right){\tau }^{2}+{r}^{-1}{\tau }^{3}=-{r}^{-\beta }\frac{{\rm{d}}}{{\rm{d}}r}\left(\frac{{\tau }^{2}}{4}+\frac{\tau \tau ^{\prime} }{r}\right)+2{r}^{-\beta -1}{\left(\tau ^{\prime} )}^{2}-{r}^{-\beta -2}N\tau \tau ^{\prime} .This together with (3.13) implies r−β−1I[τ](r)=12r−β−1(τ′)2+12+β4r−β−1τ2+13r−1τ3≤12r−β−1(τ′)2+1+β2r−β−1τ2+r−1τ3=−12r−βddrτ24+ττ′r−12r−β−2Nττ′+32r−β−1(τ′)2.\begin{array}{rcl}{r}^{-\beta -1}I\left[\tau ]\left(r)& =& \frac{1}{2}{r}^{-\beta -1}{\left(\tau ^{\prime} )}^{2}+\left(\frac{1}{2}+\frac{\beta }{4}\right){r}^{-\beta -1}{\tau }^{2}+\frac{1}{3}\hspace{0.33em}{r}^{-1}{\tau }^{3}\\ & \le & \frac{1}{2}\left({r}^{-\beta -1}{\left(\tau ^{\prime} )}^{2}+\left(1+\frac{\beta }{2}\right){r}^{-\beta -1}{\tau }^{2}+{r}^{-1}{\tau }^{3}\right)\\ & =& -\frac{1}{2}{r}^{-\beta }\frac{{\rm{d}}}{{\rm{d}}r}\left(\frac{{\tau }^{2}}{4}+\frac{\tau \tau ^{\prime} }{r}\right)-\frac{1}{2}{r}^{-\beta -2}N\tau \tau ^{\prime} +\frac{3}{2}{r}^{-\beta -1}{\left(\tau ^{\prime} )}^{2}.\end{array}Since ∣τ′(r)∣≤C(1+r)β−1| \tau ^{\prime} \left(r)| \le C{\left(1+r)}^{\beta -1}by Lemma 3.4 (with σ=0\sigma =0), we have ττ′rβ+1→0\frac{\tau \tau ^{\prime} }{{r}^{\beta +1}}\to 0as r→∞r\to \infty . And τ28≤I[τ](r)4+2β\frac{{\tau }^{2}}{8}\le \frac{I\left[\tau ]\left(r)}{4+2\beta }by (3.13), τ′(r)<0\tau ^{\prime} \left(r)\lt 0by (3.12). Thus, we obtain ∫r∞t−β−1I[τ](t)dt=−12∫r∞t−βdτ24+ττ′t−N2∫r∞t−β−2ττ′dt+32∫r∞t−β−1(τ′)2dt=r−βτ28+ττ′2r−limr→∞r−βτ28+ττ′2r−β2∫r∞t−β−1τ24+ττ′tdt−N2∫r∞t−β−2ττ′dt+32∫r∞t−β−1(τ′)2dt≤r−βI[τ](r)4+2β+β+N2∫r∞t−β−2∣ττ′∣dt+32∫r∞t−β−1(τ′)2dt.\begin{array}{rcl}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -1}I\left[\tau ]\left(t){\rm{d}}t& =& -\frac{1}{2}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta }d\left(\frac{{\tau }^{2}}{4}+\frac{\tau \tau ^{\prime} }{t}\right)-\frac{N}{2}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -2}\tau \tau ^{\prime} {\rm{d}}t+\frac{3}{2}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -1}{\left(\tau ^{\prime} )}^{2}{\rm{d}}t\\ & =& {r}^{-\beta }\left(\frac{{\tau }^{2}}{8}+\frac{\tau \tau ^{\prime} }{2r}\right)-\mathop{\mathrm{lim}}\limits_{r\to \infty }\left[{r}^{-\beta }\left(\frac{{\tau }^{2}}{8}+\frac{\tau \tau ^{\prime} }{2r}\right)\right]-\frac{\beta }{2}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -1}\left(\frac{{\tau }^{2}}{4}+\frac{\tau \tau ^{\prime} }{t}\right){\rm{d}}t-\frac{N}{2}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -2}\tau \tau ^{\prime} {\rm{d}}t+\frac{3}{2}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -1}{\left(\tau ^{\prime} )}^{2}{\rm{d}}t\\ & \le & {r}^{-\beta }\frac{I\left[\tau ]\left(r)}{4+2\beta }+\frac{\beta +N}{2}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -2}| \tau \tau ^{\prime} | {\rm{d}}t+\frac{3}{2}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -1}{\left(\tau ^{\prime} )}^{2}{\rm{d}}t.\end{array}Assume now (3.18)I[τ](r)≤Crβ−σ,r≥1I\left[\tau ]\left(r)\le C{r}^{\beta -\sigma },\hspace{1.0em}r\ge 1for some fixed σ≥0\sigma \ge 0. By (3.13), we have ∣τ(r)∣≤Cr−σ3,r≥1.| \tau \left(r)| \le C{r}^{-\tfrac{\sigma }{3}},\hspace{1.0em}r\ge 1.First, we consider the case where β≥σ3\beta \ge \frac{\sigma }{3}, then ∣τ′(r)∣≤Crβ−2σ3−1| \tau ^{\prime} \left(r)| \le C{r}^{\beta -\tfrac{2\sigma }{3}-1}by Lemma 3.4. Therefore, (3.19)∫r∞t−β−1I[τ](t)dt≤r−βI[τ](r)4+2β+C∫r∞t−σ−3dt+C∫r∞tβ−4σ3−3dt≤r−βI[τ](r)4+2β+Cr−σ−2+Crβ−4σ3−2≤r−βI[τ](r)4+2β+Crβ−4σ3−2,\begin{array}{rcl}\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\beta -1}I\left[\tau ]\left(t){\rm{d}}t& \le & {r}^{-\beta }\frac{I\left[\tau ]\left(r)}{4+2\beta }+C\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{-\sigma -3}{\rm{d}}t+C\underset{r}{\overset{\infty }{\displaystyle \int }}{t}^{\beta -\tfrac{4\sigma }{3}-3}{\rm{d}}t\\ & \le & {r}^{-\beta }\frac{I\left[\tau ]\left(r)}{4+2\beta }+C{r}^{-\sigma -2}+C{r}^{\beta -\tfrac{4\sigma }{3}-2}\\ & \le & {r}^{-\beta }\frac{I\left[\tau ]\left(r)}{4+2\beta }+C{r}^{\beta -\tfrac{4\sigma }{3}-2},\end{array}which is equivalent to (3.20)(4+2β)∫r∞t−β−1I[τ](t)dt≤r−βI[τ](r)+Crβ−4σ3−2,r≥1.\left(4+2\beta )\underset{r}{\overset{\infty }{\int }}{t}^{-\beta -1}I\left[\tau ]\left(t){\rm{d}}t\le {r}^{-\beta }I\left[\tau ]\left(r)+C{r}^{\beta -\tfrac{4\sigma }{3}-2},\hspace{1.0em}r\ge 1.Define F(r)≔r−βI[τ](r),G(r)≔∫r∞t−1F(t)dt.F\left(r):= {r}^{-\beta }I\left[\tau ]\left(r),\hspace{1.0em}G\left(r):= \underset{r}{\overset{\infty }{\int }}{t}^{-1}F\left(t){\rm{d}}t.It is easy to see that (3.21)(4+2β)∫r∞t−1F(t)dt≤F(r)+Crβ−4σ3−2,\left(4+2\beta )\underset{r}{\overset{\infty }{\int }}{t}^{-1}F\left(t){\rm{d}}t\le F\left(r)+C{r}^{\beta -\tfrac{4\sigma }{3}-2},and (3.22)dG(r)dr=−r−1F(r).\frac{{\rm{d}}G\left(r)}{{\rm{d}}r}=-{r}^{-1}F\left(r).Combining (3.21) with (3.22), we obtain (3.23)(4+2β)G(r)+rdG(r)dr≤Crβ−4σ3−2,\left(4+2\beta )G\left(r)+r\frac{{\rm{d}}G\left(r)}{{\rm{d}}r}\le C{r}^{\beta -\tfrac{4\sigma }{3}-2},that is, (3.24)d[r4+2βG(r)]dr≤Cr3β−4σ3+1,r≥1.\frac{{\rm{d}}{[}{r}^{4+2\beta }G\left(r)]}{{\rm{d}}r}\le C{r}^{3\beta -\tfrac{4\sigma }{3}+1},\hspace{1.0em}r\ge 1.We divided the remainder of proof into two cases.Case A: If 3β−4σ3+1≠−13\beta -\frac{4\sigma }{3}+1\ne -1, by integrating (3.24) from 1 to rr, we yield (3.25)G(r)≤Cr−4−2β+Crβ−4σ3−2,r≥1.G\left(r)\le C{r}^{-4-2\beta }+C{r}^{\beta -\tfrac{4\sigma }{3}-2},\hspace{1.0em}r\ge 1.By (3.14), dF(r)dr=rβdI[τ](r)dr−βr−1I[τ](r)≤0,\frac{{\rm{d}}F\left(r)}{{\rm{d}}r}={r}^{\beta }\left[\frac{{\rm{d}}I\left[\tau ]\left(r)}{{\rm{d}}r}-\beta {r}^{-1}I\left[\tau ]\left(r)\right]\le 0,then G(r)=∫r∞t−1F(t)dt≥∫r2rt−1F(t)dt≥F(2r)2,G\left(r)=\underset{r}{\overset{\infty }{\int }}{t}^{-1}F\left(t){\rm{d}}t\ge \underset{r}{\overset{2r}{\int }}{t}^{-1}F\left(t){\rm{d}}t\ge \frac{F\left(2r)}{2},or (3.26)F(r)≤2Gr2.F\left(r)\le 2G\left(\frac{r}{2}\right).Thus, (3.27)F(r)≤Cr−4−2β+Crβ−4σ3−2.F\left(r)\le C{r}^{-4-2\beta }+C{r}^{\beta -\tfrac{4\sigma }{3}-2}.That is, (3.28)I[τ](r)≤Cr−4−β+Cr2β−4σ3−2,r≥2.I\left[\tau ]\left(r)\le C{r}^{-4-\beta }+C{r}^{2\beta -\tfrac{4\sigma }{3}-2},\hspace{1.0em}r\ge 2.From (3.15), we know that I[τ](r)≤CrβI\left[\tau ]\left(r)\le C{r}^{\beta }for all r≥0r\ge 0, then (3.28) holds for r≥1r\ge 1.Clearly (3.28) holds with σ0=0{\sigma }_{0}=0by (3.15), then (3.18) holds with σ1=min{4+2β,2−β}=2−β{\sigma }_{1}=\min \{4+2\beta ,2-\beta \}=2-\beta . We can continue recursively letting σi+1=min4+2β,4σi3+2−β{\sigma }_{i+1}=\min \left\{4+2\beta ,\frac{4{\sigma }_{i}}{3}+2-\beta \right\}until (3.29)I[τ](r)≤Crβ−σj,β<σj3.I\left[\tau ]\left(r)\le C{r}^{\beta -{\sigma }_{j}},\hspace{1.0em}\beta \lt \frac{{\sigma }_{j}}{3}.Then ∣τ(r)∣≤Cr−σj3| \tau \left(r)| \le C{r}^{-\tfrac{{\sigma }_{j}}{3}}, and ∣τ′(r)∣≤Cr−σj3−1| \tau ^{\prime} \left(r)| \le C{r}^{-\tfrac{{\sigma }_{j}}{3}-1}by Lemma 3.4. Thus, (3.30)∫r∞t−β−1I[τ](t)dt≤r−βI[τ](r)4+2β+Cr−β−2σj3−2.\underset{r}{\overset{\infty }{\int }}{t}^{-\beta -1}I\left[\tau ]\left(t){\rm{d}}t\le {r}^{-\beta }\frac{I\left[\tau ]\left(r)}{4+2\beta }+C{r}^{-\beta -\tfrac{2{\sigma }_{j}}{3}-2}.That is, (3.31)(4+2β)∫r∞t−β−1I[τ](t)dt≤r−βI[τ](r)+Cr−β−2σj3−2.\left(4+2\beta )\underset{r}{\overset{\infty }{\int }}{t}^{-\beta -1}I\left[\tau ]\left(t){\rm{d}}t\le {r}^{-\beta }I\left[\tau ]\left(r)+C{r}^{-\beta -\tfrac{2{\sigma }_{j}}{3}-2}.We can obtain (3.32)d[r4+2βG(r)]dr≤Crβ−2σj3+1.\frac{{\rm{d}}{[}{r}^{4+2\beta }G\left(r)]}{{\rm{d}}r}\le C{r}^{\beta -\tfrac{2{\sigma }_{j}}{3}+1}.If β−2σj3+1≠−1\beta -\frac{2{\sigma }_{j}}{3}+1\ne -1, then I[τ](r)≤Cr−4−β+Cr−2σj3−2I\left[\tau ]\left(r)\le C{r}^{-4-\beta }+C{r}^{-\tfrac{2{\sigma }_{j}}{3}-2}. Letting σj+1=min4+2β,β+2σj3+2{\sigma }_{j+1}=\min \left\{4+2\beta ,\beta +\frac{2{\sigma }_{j}}{3}+2\right\}, we continue recursively until (3.29) holds with σj+1=4+2β{\sigma }_{j+1}=4+2\beta . If β−2σj3+1=−1\beta -\frac{2{\sigma }_{j}}{3}+1=-1, we can infer a similar formula as (3.25), then a same recursive argument with σ0=0{\sigma }_{0}=0and σj+1=min4+2β,β+2σj3+1{\sigma }_{j+1}=\min \left\{4+2\beta ,\beta +\frac{2{\sigma }_{j}}{3}+1\right\}shows the expected result. Thus, I[τ](r)≤Cr−4−βI\left[\tau ]\left(r)\le C{r}^{-4-\beta }.Case B: If 3β−4σ3+1=−13\beta -\frac{4\sigma }{3}+1=-1, we can derive a formula similar to (3.25). A same recursive argument with σ0=0{\sigma }_{0}=0and σi+1=min4+2β,4σi3+1−β{\sigma }_{i+1}=\min \left\{4+2\beta ,\frac{4{\sigma }_{i}}{3}+1-\beta \right\}can be used to obtain I[τ](r)≤Cr−4−βI\left[\tau ]\left(r)\le C{r}^{-4-\beta }for r≥1r\ge 1.In both cases, we have I[τ](r)≤Cr−4−β,r≥1.I\left[\tau ]\left(r)\le C{r}^{-4-\beta },\hspace{1.0em}r\ge 1.Together with I[τ](r)≤CrβI\left[\tau ]\left(r)\le C{r}^{\beta }for all r≥0r\ge 0, it turns out I[τ](r)≤C(1+r)−4−β,r≥0.I\left[\tau ]\left(r)\le C{\left(1+r)}^{-4-\beta },\hspace{1.0em}r\ge 0.Thus, by (3.13), ∣τ(r)∣≤C(1+r)−2−β2,r≥0.| \tau \left(r)| \le C{\left(1+r)}^{-2-\tfrac{\beta }{2}},\hspace{1.0em}r\ge 0.From Lemma 3.4, we obtain the estimate for ∣τ′(r)∣| \tau ^{\prime} \left(r)| , i.e., ∣τ′(r)∣≤C(1+r)−3−β2,r≥0.| \tau ^{\prime} \left(r)| \le C{\left(1+r)}^{-3-\tfrac{\beta }{2}},\hspace{1.0em}r\ge 0.The letter CCdenotes a positive generic constant. Thus, Lemma 3.5 holds.□On account of aforementioned lemmas, we have the following corollary.Corollary 3.6Let N≥1N\ge 1, α>0\alpha \gt 0, 0<β<20\lt \beta \lt 2, and τ\tau be the solution of (3.2) with initial data τ(0)=α\tau \left(0)=\alpha . Then the limitd(α)≔limr→∞r2+βτ(r)d\left(\alpha ):= \mathop{\mathrm{lim}}\limits_{r\to \infty }{r}^{2+\beta }\tau \left(r)exists in R{\mathbb{R}}and d(α)>0d\left(\alpha )\gt 0for any α>0\alpha \gt 0.ProofBy (3.3), we have r2+βτ(r)=r2+βαexp−∫0rσN(s)−1∫0stβτ(t)+1+β2σN(t)dtds=αexp(2+β)logr−1+β2γN(r)−∫0rσN(s)−1∫0stβτ(t)σN(t)dtds.\begin{array}{rcl}{r}^{2+\beta }\tau \left(r)& =& {r}^{2+\beta }\alpha \exp \left\{-\underset{0}{\overset{r}{\displaystyle \int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\displaystyle \int }}\left[{t}^{\beta }\tau \left(t)+1+\frac{\beta }{2}\right]{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\}\\ & =& \alpha \exp \left\{\left(2+\beta )\log r-\left(1+\frac{\beta }{2}\right){\gamma }_{N}\left(r)-\underset{0}{\overset{r}{\displaystyle \int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\displaystyle \int }}{t}^{\beta }\tau \left(t){\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\}.\end{array}First, for s>0s\gt 0is small, we have σN(s)−1∫0stβτ(t)σN(t)dt≤s1−Ne−s24⋅sβ⋅α⋅sN−1es24⋅s=αsβ+1.{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}{t}^{\beta }\tau \left(t){\sigma }_{N}\left(t){\rm{d}}t\right)\le {s}^{1-N}{e}^{-\tfrac{{s}^{2}}{4}}\cdot {s}^{\beta }\cdot \alpha \cdot {s}^{N-1}{e}^{\tfrac{{s}^{2}}{4}}\cdot s=\alpha {s}^{\beta +1}.Next, for s>0s\gt 0large enough, by Lemmas 3.2 and 3.5, we see that there exist constants C1>0{C}_{1}\gt 0and C2>0{C}_{2}\gt 0such that σN(s)−1∫0stβτ(t)σN(t)dt≤C1σN(s)−1∫0stβ(1+t)−2−β2σN(t)dt≤C1σN(s)−1(1+s)β2∫0s(1+t)−2σN(t)dt≤C2σN(s)−1(1+s)β2⋅sN−4es24≤C2(1+s)β2−3.\begin{array}{rcl}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\displaystyle \int }}{t}^{\beta }\tau \left(t){\sigma }_{N}\left(t){\rm{d}}t\right)& \le & {C}_{1}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\displaystyle \int }}{t}^{\beta }{\left(1+t)}^{-2-\tfrac{\beta }{2}}{\sigma }_{N}\left(t){\rm{d}}t\right)\\ & \le & {C}_{1}{\sigma }_{N}{\left(s)}^{-1}{\left(1+s)}^{\tfrac{\beta }{2}}\left(\underset{0}{\overset{s}{\displaystyle \int }}{\left(1+t)}^{-2}{\sigma }_{N}\left(t){\rm{d}}t\right)\\ & \le & {C}_{2}\hspace{0.33em}{\sigma }_{N}{\left(s)}^{-1}{\left(1+s)}^{\tfrac{\beta }{2}}\cdot {s}^{N-4}{e}^{\tfrac{{s}^{2}}{4}}\\ & \le & {C}_{2}{\left(1+s)}^{\tfrac{\beta }{2}-3}.\end{array}Thus, we obtain ∫0∞σN(s)−1∫0stβτ(t)σN(t)dtds<∞.\underset{0}{\overset{\infty }{\int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}{t}^{\beta }\tau \left(t){\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\lt \infty .Therefore, by Lemma 3.1, we derive that the limit d(α)d\left(\alpha )exists and (3.33)d(α)=αexplimr→∞(2+β)logr−1+β2γN(r)−∫0∞σN(s)−1∫0stβτ(t)σN(t)dtds.d\left(\alpha )=\alpha \exp \left\{\mathop{\mathrm{lim}}\limits_{r\to \infty }\left[\left(2+\beta )\log r-\left(1+\frac{\beta }{2}\right){\gamma }_{N}\left(r)\right]-\underset{0}{\overset{\infty }{\int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}{t}^{\beta }\tau \left(t){\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\}.Then we complete the proof of Corollary 3.6.□The next Lemma is about the continuity of d(α)d\left(\alpha )with respect to α\alpha .Lemma 3.7Let N≥1N\ge 1, 0<β<20\lt \beta \lt 2, a>0a\gt 0, and b>0b\gt 0. Let τa{\tau }_{a}and τb{\tau }_{b}be the solutions of (3.2) and satisfy the condition τa(0)=a{\tau }_{a}\left(0)=aand τb(0)=b{\tau }_{b}\left(0)=bseparately. Then there exists a constant C>0C\gt 0, which depends only on NN, such thatsupr≥0(1+r)2+β∣τa(r)−τb(r)∣≤C∣a−b∣\mathop{\sup }\limits_{r\ge 0}{\left(1+r)}^{2+\beta }| {\tau }_{a}\left(r)-{\tau }_{b}\left(r)| \le C| a-b| for all a,b>0a,b\gt 0. In particular, d(α)d\left(\alpha )is continuous with respect to α\alpha .ProofBy (3.3), we have τa(r)−τb(r)=aexp−∫0rσN(s)−1∫0stβτa(t)+1+β2σN(t)dtds−exp−∫0rσN(s)−1∫0stβτb(t)+1+β2σN(t)dtds+(a−b)exp−∫0rσN(s)−1∫0stβτb(t)+1+β2σN(t)dtds.\begin{array}{rcl}{\tau }_{a}\left(r)-{\tau }_{b}\left(r)& =& a\left[\exp \left\{-\underset{0}{\overset{r}{\displaystyle \int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\displaystyle \int }}\left[{t}^{\beta }{\tau }_{a}\left(t)+1+\frac{\beta }{2}\right]{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\}-\exp \left\{-\underset{0}{\overset{r}{\displaystyle \int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\displaystyle \int }}\left[{t}^{\beta }{\tau }_{b}\left(t)+1+\frac{\beta }{2}\right]{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\}\right]\\ & & +\left(a-b)\exp \left\{-\underset{0}{\overset{r}{\displaystyle \int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\displaystyle \int }}\left[{t}^{\beta }{\tau }_{b}\left(t)+1+\frac{\beta }{2}\right]{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s\right\}.\end{array}Since τa,τb{\tau }_{a},{\tau }_{b}and σN{\sigma }_{N}are positive functions, by using (3.11), we have ∣τa(r)−τb(r)∣≤e−1+β2γN(r)a∫0rσN(s)−1∫0stβ∣τa(t)−τb(t)∣σN(t)dtds+∣a−b∣,| {\tau }_{a}\left(r)-{\tau }_{b}\left(r)| \le {e}^{-\left(1+\tfrac{\beta }{2}\right){\gamma }_{N}\left(r)}\left[a\underset{0}{\overset{r}{\int }}{\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}{t}^{\beta }| {\tau }_{a}\left(t)-{\tau }_{b}\left(t)| {\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s+| a-b| \right],where γN{\gamma }_{N}is the function defined by (3.4). Hence, we have (1+r)2+β∣τa(r)−τb(r)∣≤(1+r)2+βe−1+β2γN(r)a∫0rsup0≤t≤s(1+t)2+β∣τa(t)−τb(t)∣×σN(s)−1∫0s(1+t)−2σN(t)dtds+∣a−b∣.{\left(1+r)}^{2+\beta }| {\tau }_{a}\left(r)-{\tau }_{b}\left(r)| \le {\left(1+r)}^{2+\beta }{e}^{-\left(1+\tfrac{\beta }{2}\right){\gamma }_{N}\left(r)}\left[a\underset{0}{\overset{r}{\int }}\mathop{\sup }\limits_{0\le t\le s}{\left(1+t)}^{2+\beta }| {\tau }_{a}\left(t)-{\tau }_{b}\left(t)| \times {\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}{\left(1+t)}^{-2}{\sigma }_{N}\left(t){\rm{d}}t\right){\rm{d}}s+| a-b| \right].By Lemma 3.1, we have supr≥0(1+r)2+βe−1+β2γN(r)=supr≥0exp(2+β)log(1+r)−1+β2γN(r)=supr≥0exp1+β2(2log(1+r)−γN(r))<∞.\begin{array}{rcl}\mathop{\sup }\limits_{r\ge 0}{\left(1+r)}^{2+\beta }{e}^{-\left(1+\tfrac{\beta }{2}\right)}{\gamma }_{N}\left(r)& =& \mathop{\sup }\limits_{r\ge 0}\exp \left[\left(2+\beta )\log \left(1+r)-\left(1+\frac{\beta }{2}\right){\gamma }_{N}\left(r)\right]\\ & =& \mathop{\sup }\limits_{r\ge 0}\exp \left[\left(1+\frac{\beta }{2}\right)(2\log \left(1+r)-{\gamma }_{N}\left(r))\right]\lt \infty .\end{array}We thus deduce that there exists a constant C>0C\gt 0, which depends only on NN, such that h1(r)≤Ca∫0rh1(s)h2(s)ds+C∣a−b∣,{h}_{1}\left(r)\le Ca\underset{0}{\overset{r}{\int }}{h}_{1}\left(s){h}_{2}\left(s){\rm{d}}s+C| a-b| ,where h1(s)≔sup0≤t≤s(1+t)2+β∣τa(t)−τb(t)∣,h2(s)≔σN(s)−1∫0s(1+t)−2σN(t)dt.{h}_{1}\left(s):= \mathop{\sup }\limits_{0\le t\le s}{\left(1+t)}^{2+\beta }| {\tau }_{a}\left(t)-{\tau }_{b}\left(t)| ,\hspace{1.0em}{h}_{2}\left(s):= {\sigma }_{N}{\left(s)}^{-1}\left(\underset{0}{\overset{s}{\int }}{\left(1+t)}^{-2}{\sigma }_{N}\left(t){\rm{d}}t\right).Then, by using the Gronwall inequality, we obtain h1(r)≤C∣a−b∣CaexpCa∫0∞h2(r)dr∫0∞h2(s)ds+1.{h}_{1}\left(r)\le C| a-b| \left[Ca\exp \left(Ca\underset{0}{\overset{\infty }{\int }}{h}_{2}\left(r)dr\right)\underset{0}{\overset{\infty }{\int }}{h}_{2}\left(s){\rm{d}}s+1\right].From Lemma 3.2, we know that h2∈L1(0,∞){h}_{2}\in {L}^{1}\left(0,\infty ). Therefore, h1(r)≤C∣a−b∣.{h}_{1}\left(r)\le C| a-b| .The proof of Lemma 3.7 is completed.□4Multiplicity of forward self-similar solutionsIn this section, we study the multiplicity of the solutions for the problem (1.8) by an ODE approach, which follows from [22]. For α∈R\alpha \in {\mathbb{R}}, let ωα{\omega }_{\alpha }be the solution of (1.8) with ω(0)=α\omega \left(0)=\alpha . Set (4.1)zα(s)≔ωα(r)−α,s≔eα2+βr.{z}_{\alpha }\left(s):= {\omega }_{\alpha }\left(r)-\alpha ,\hspace{1.0em}s:= {e}^{\tfrac{\alpha }{2+\beta }}r.Then a direct computation shows that (4.2)(2+β)logs+zα(s)=(2+β)logr+ωα(r),\left(2+\beta )\log s+{z}_{\alpha }\left(s)=\left(2+\beta )\log r+{\omega }_{\alpha }\left(r),and zα{z}_{\alpha }satisfies (4.3)zα″+N−1s+e−2α2+βs2z′+sβezα+1+β2e−2α2+β=0,s>0,zα(0)=0,zα′(0)=0.\left\{\begin{array}{l}{z}_{\alpha }^{^{\prime\prime} }+\left(\frac{N-1}{s}+{e}^{\tfrac{-2\alpha }{2+\beta }}\frac{s}{2}\right)z^{\prime} +{s}^{\beta }{e}^{{z}_{\alpha }}+\left(1+\frac{\beta }{2}\right){e}^{\tfrac{-2\alpha }{2+\beta }}=0,\hspace{1.0em}s\gt 0,\hspace{1.0em}\\ {z}_{\alpha }\left(0)=0,\hspace{1.0em}{z}_{\alpha }^{\prime} \left(0)=0.\hspace{1.0em}\end{array}\right.Let ZZbe the solution of (4.4)Z″+N−1sZ′+sβeZ=0in(0,∞){Z}^{^{\prime\prime} }+\frac{N-1}{s}Z^{\prime} +{s}^{\beta }{e}^{Z}=0\hspace{1.0em}{\rm{in}}\hspace{0.33em}\left(0,\infty )with Z(0)=0Z\left(0)=0and Z′(0)=0Z^{\prime} \left(0)=0. One can easily compute that Z∗(s)≔−(2+β)logs+log[(2+β)(N−2)]{Z}_{\ast }\left(s):= -\left(2+\beta )\log s+\log {[}\left(2+\beta )\left(N-2)]is a singular solution of equation (4.4). We can use the similar method in [31, Theorem 1.1, Lemma 2.1] to study intersection properties of solutions for (4.4) and obtain the following lemma.Lemma 4.1Let N≥3N\ge 3, 0<β<20\lt \beta \lt 2, and Z be the solution of (4.4). Thenlims→∞[(2+β)logs+Z(s)]=log[(2+β)(N−2)].\mathop{\mathrm{lim}}\limits_{s\to \infty }{[}\left(2+\beta )\log s+Z\left(s)]=\log {[}\left(2+\beta )\left(N-2)].Furthermore, Z intersects the singular solution Z∗{Z}_{\ast }infinitely many times in (0,∞)\left(0,\infty )when 3≤N<10+4β3\le N\lt 10+4\beta .The solution of (4.4) would converge to the one of (4.3) when α\alpha is large. Indeed, we have the following lemma by [16, Proposition 3.1] and [22, Lemma 4.3].Lemma 4.2Let N≥1N\ge 1, 0<β<20\lt \beta \lt 2, α>0\alpha \gt 0, and ωα{\omega }_{\alpha }be the solution of (1.8) with ω(0)=α\omega \left(0)=\alpha . Let the function zα=zα(s){z}_{\alpha }={z}_{\alpha }\left(s)defined as (4.1). Then, for any S>0S\gt 0, there holdslimα→∞sup0≤s≤S∣zα(s)−Z(s)∣=0,\mathop{\mathrm{lim}}\limits_{\alpha \to \infty }\mathop{\sup }\limits_{0\le s\le S}| {z}_{\alpha }\left(s)-Z\left(s)| =0,where Z is the solution of (4.4).With the aforementioned lemmas, we could analyze the behavior of ωα{\omega }_{\alpha }as α→∞\alpha \to \infty near the origin. The following Propositions 4.3 and 4.4 can be proved by using similar method as in the proof of [22, Proposition 4.1] with the help of Lemmas 4.1 and 4.2.Proposition 4.3Let N≥1N\ge 1, 0<β<20\lt \beta \lt 2, α>0\alpha \gt 0, and ωα{\omega }_{\alpha }be the solution of (1.8) satisfying ω(0)=α\omega \left(0)=\alpha . For any ε>0\varepsilon \gt 0, there exists a constant α0∈R{\alpha }_{0}\in {\mathbb{R}}, and for any α≥α0\alpha \ge {\alpha }_{0}, there exists a constant rα>0{r}_{\alpha }\gt 0such that(4.5)∣(2+β)logrα+ωα(rα)−log[(2+β)(N−2)]∣<ε,| \left(2+\beta )\log {r}_{\alpha }+{\omega }_{\alpha }\left({r}_{\alpha })-\log {[}\left(2+\beta )\left(N-2)]| \lt \varepsilon ,where rα=e−α2+βs0{r}_{\alpha }={e}^{-\tfrac{\alpha }{2+\beta }}{s}_{0}and s0>0{s}_{0}\gt 0is some constant, which depends on ε\varepsilon . Furthermore, if 3≤N<10+4β3\le N\lt 10+4\beta , then there exists a positive constant rα{r}_{\alpha }satisfying (4.5) and(4.6)ddr[(2+β)logr+ωα(r)]r=rα=0{\left.\frac{{\rm{d}}}{{\rm{d}}r}{[}\left(2+\beta )\log r+{\omega }_{\alpha }\left(r)]\right|}_{r={r}_{\alpha }}=0for all α≥α0\alpha \ge {\alpha }_{0}.In the case of 3≤N<10+4β3\le N\lt 10+4\beta , we present the following intersection results.Proposition 4.4Let 3≤N<10+4β3\le N\lt 10+4\beta , 0<β<20\lt \beta \lt 2, α>0\alpha \gt 0, and ωα{\omega }_{\alpha }be the solution of (1.8) satisfying ω(0)=α\omega \left(0)=\alpha . Then for any r0>0{r}_{0}\gt 0and n0∈N{n}_{0}\in N, there exists a constant α0∈R{\alpha }_{0}\in {\mathbb{R}}such that (2+β)logr+ωα(r)\left(2+\beta )\log r+{\omega }_{\alpha }\left(r)intersects log[(2+β)(N−2)]\log {[}\left(2+\beta )\left(N-2)]at least n0{n}_{0}times on [0,r0]\left[0,{r}_{0}]for any α≥α0\alpha \ge {\alpha }_{0}.Next we study the structure of the set of solution SD{S}_{D}for equation (1.8). Recall that (4.7)limr→∞[(2+β)logr+ω(r)]=D,\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\omega \left(r)]=D,where ω(r)\omega \left(r)is a solution of (1.8), and SD≔{ω∈C2([0,∞)):ωis a solution of (1.8) satisfying (4.7)}.{S}_{D}:= \left\{\omega \in {C}^{2}\left(\left[0,\infty )):\omega \hspace{0.33em}\hspace{0.1em}\text{is a solution of (1.8) satisfying (4.7)}\hspace{0.1em}\right\}.By Proposition 3.3, we know that there exists a solution τ\tau of (3.2) with τ(0)=eα\tau \left(0)={e}^{\alpha }for any α∈R\alpha \in {\mathbb{R}}. Hence, ω(r)=logτ(r)\omega \left(r)=\log \tau \left(r)is a solution of (1.8) satisfying ω(0)=α\omega \left(0)=\alpha . Moreover, since (2+β)logr+ω(r)=logr2+βτ(r)\left(2+\beta )\log r+\omega \left(r)=\log {r}^{2+\beta }\tau \left(r), we can deduce that limr→∞[(2+β)logr+ω(r)]=logd(eα),\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\omega \left(r)]=\log d\left({e}^{\alpha }),where d(α)d\left(\alpha )was shown in Corollary 3.6. So we have that SD≠∅{S}_{D}\ne \varnothing for some D∈RD\in {\mathbb{R}}.Now set (4.8)D(α)≔limr→∞[(2+β)logr+ωα(r)],D\left(\alpha ):= \mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+{\omega }_{\alpha }\left(r)],where ωα{\omega }_{\alpha }is the solution of (1.8) with ω(0)=α\omega \left(0)=\alpha . By (3.33), we know limα→−∞d(eα)≤limα→−∞expα+limr→∞(2+β)logr−1+β2γN(r)=0,\mathop{\mathrm{lim}}\limits_{\alpha \to -\infty }d\left({e}^{\alpha })\le \mathop{\mathrm{lim}}\limits_{\alpha \to -\infty }\exp \left\{\phantom{\rule[-1.25em]{}{0ex}},\alpha +\mathop{\mathrm{lim}}\limits_{r\to \infty }\left[\left(2+\beta )\log r-\left(1+\frac{\beta }{2}\right){\gamma }_{N}\left(r)\right]\right\}=0,and therefore, limα→−∞D(α)=limα→−∞logd(eα)=−∞.\mathop{\mathrm{lim}}\limits_{\alpha \to -\infty }D\left(\alpha )=\mathop{\mathrm{lim}}\limits_{\alpha \to -\infty }\log d\left({e}^{\alpha })=-\infty .Then we begin by researching on the specific behavior of D(α)D\left(\alpha )as α→∞\alpha \to \infty and divide the argument into two cases: 3≤N<10+4β3\le N\lt 10+4\beta and N≥10+4βN\ge 10+4\beta .We first concentrate on the case where 3≤N<10+4β3\le N\lt 10+4\beta , and we obtain the following proposition.Proposition 4.5Let 3≤N<10+4β3\le N\lt 10+4\beta , 0<β<20\lt \beta \lt 2, and D(α)D\left(\alpha )be the constant defined by (4.8). Then the following results hold. (i)There exists a sequence {αk}k∈N{\left\{{\alpha }_{k}\right\}}_{k\in {\mathbb{N}}}such that α1<α2<⋯{\alpha }_{1}\lt {\alpha }_{2}\lt \cdots \hspace{0.33em}andD(α2k)<log[(2+β)(N−2)]<D(α2k−1)D\left({\alpha }_{2k})\lt \log {[}\left(2+\beta )\left(N-2)]\lt D\left({\alpha }_{2k-1})for k∈Nk\in {\mathbb{N}}.(ii)limα→∞D(α)=log[(2+β)(N−2)]{\mathrm{lim}}_{\alpha \to \infty }D\left(\alpha )=\log {[}\left(2+\beta )\left(N-2)].In particular, log[(2+β)(N−2)]<supα∈RD(α)<∞\log {[}\left(2+\beta )\left(N-2)]\lt {\sup }_{\alpha \in {\mathbb{R}}}D\left(\alpha )\lt \infty and SD=∅{S}_{D}=\varnothing for D>supα∈RD(α)D\gt {\sup }_{\alpha \in {\mathbb{R}}}D\left(\alpha ).For the purpose of proving Proposition 4.5, we first give some lemmas that follow the ideas of [22, Lemmas 4.7 and 4.9].Lemma 4.6Let 3≤N<10+4β3\le N\lt 10+4\beta , 0<β<20\lt \beta \lt 2, then for any τ∈R\tau \in {\mathbb{R}}, there exists a sequence {αk}k∈N{\left\{{\alpha }_{k}\right\}}_{k\in {\mathbb{N}}}such that α1<α2<⋯{\alpha }_{1}\lt {\alpha }_{2}\lt \cdots \hspace{0.33em}andgαk(τ)=0,gαk′(τ)gαk+1′(τ)<0{g}_{{\alpha }_{k}}\left(\tau )=0,\hspace{1.0em}{g}_{{\alpha }_{k}}^{\prime} \left(\tau ){g}_{{\alpha }_{k+1}}^{\prime} \left(\tau )\lt 0for any k∈Nk\in {\mathbb{N}}, where(4.9)gα(t)=(2+β)logr+ωα(r)−log[(2+β)(N−2)],t=logr.{g}_{\alpha }\left(t)=\left(2+\beta )\log r+{\omega }_{\alpha }\left(r)-\log {[}\left(2+\beta )\left(N-2)],\hspace{1.0em}t=\log r.ProofStep 1: By direct computation, we obtain (4.10)gα″+N−2+12e2tgα′+(2+β)(N−2)ϑα(t)gα=0,t∈R,{{g}_{\alpha }}^{^{\prime\prime} }+\left(N-2+\frac{1}{2}{e}^{2t}\right){g}_{\alpha }^{^{\prime} }+\left(2+\beta )\left(N-2){{\vartheta }}_{\alpha }\left(t){g}_{\alpha }=0,\hspace{1.0em}t\in {\mathbb{R}},where (4.11)ϑα(t)≔egα(t)−1gα(t)ifgα(t)≠0,1ifgα(t)=0.{{\vartheta }}_{\alpha }\left(t):= \left\{\begin{array}{ll}\frac{{e}^{{g}_{\alpha }\left(t)}-1}{{g}_{\alpha }\left(t)}\hspace{2.0em}\hspace{1.0em}& {\rm{if}}\hspace{1.0em}{g}_{\alpha }\left(t)\ne 0,\\ 1\hspace{1.0em}& {\rm{if}}\hspace{1.0em}{g}_{\alpha }\left(t)=0.\end{array}\right.It is clear that ϑα∈C(R){{\vartheta }}_{\alpha }\in C\left(R)and ϑα(t)>0{{\vartheta }}_{\alpha }\left(t)\gt 0.Moreover, we rewrite (4.10) as the following form: (4.12)(p0(t)gα′)′+(2+β)(N−2)p0(t)ϑα(t)gα=0,t∈R,({p}_{0}\left(t){g}_{\alpha }^{^{\prime} })^{\prime} +\left(2+\beta )\left(N-2)\hspace{0.33em}{p}_{0}\left(t){{\vartheta }}_{\alpha }\left(t){g}_{\alpha }=0,\hspace{1.0em}t\in {\mathbb{R}},where p0(t)≔exp(N−2)t+14e2t.{p}_{0}\left(t):= \exp \left(\left(N-2)t+\frac{1}{4}{e}^{2t}\right).Next we adapt the Prüfer transformation and set ρα(t){\rho }_{\alpha }\left(t)and θα(t){\theta }_{\alpha }\left(t)by p0(t)gα′(t)=ρα(t)cosθα(t),gα(t)=ρα(t)sinθα(t).{p}_{0}\left(t){g}_{\alpha }^{^{\prime} }\left(t)={\rho }_{\alpha }\left(t)\cos {\theta }_{\alpha }\left(t),\hspace{1.0em}{g}_{\alpha }\left(t)={\rho }_{\alpha }\left(t)\sin {\theta }_{\alpha }\left(t).Namely, ρα(t)=(gα(t)2+p0(t)2gα′(t)2)1/2,θα(t)=tan−1gα(t)p0(t)gα′(t),{\rho }_{\alpha }\left(t)={({{g}_{\alpha }\left(t)}^{2}+{{p}_{0}\left(t)}^{2}{{g}_{\alpha }^{^{\prime} }\left(t)}^{2})}^{1\text{/}2},\hspace{1.0em}{\theta }_{\alpha }\left(t)={\tan }^{-1}\frac{{g}_{\alpha }\left(t)}{{p}_{0}\left(t)\hspace{0.33em}{g}_{\alpha }^{^{\prime} }\left(t)},which combined with (4.12) gives θα′(t)=1p0(t)cos2θα(t)+(2+β)(N−2)p0(t)ϑα(t)sin2θα(t)>0,t∈R.{\theta }_{\alpha }^{\prime} \left(t)=\frac{1}{{p}_{0}\left(t)}{\cos }^{2}{\theta }_{\alpha }\left(t)+\left(2+\beta )\left(N-2){p}_{0}\left(t){{\vartheta }}_{\alpha }\left(t){\sin }^{2}{\theta }_{\alpha }\left(t)\gt 0,\hspace{1.0em}t\in {\mathbb{R}}.It is easy to see p0(t)→0,gα′(t)=2+β+rωα′(r)→2+β,gα(t)→∞ast→−∞,{p}_{0}\left(t)\to 0,{g}_{\alpha }^{^{\prime} }\left(t)=2+\beta +r{\omega }_{\alpha }^{^{\prime} }\left(r)\to 2+\beta ,\hspace{0.33em}{g}_{\alpha }\left(t)\to \infty \hspace{0.33em}{\rm{as}}\hspace{0.33em}t\to -\infty ,and we can assume limt→−∞θα(t)=−π2.\mathop{\mathrm{lim}}\limits_{t\to -\infty }{\theta }_{\alpha }\left(t)=-\frac{\pi }{2}.Then gα(t){g}_{\alpha }\left(t)has kkzeros in (−∞,τ]\left(-\infty ,\tau ]which is equivalent to (k−1)π≤θα(τ)≤kπ\left(k-1)\pi \le {\theta }_{\alpha }\left(\tau )\le k\pi .Step 2: We claim that (4.13)limα→∞θα(τ)=∞.\mathop{\mathrm{lim}}\limits_{\alpha \to \infty }{\theta }_{\alpha }\left(\tau )=\infty .Let n0{n}_{0}be any positive integer and set r0=eτ{r}_{0}={e}^{\tau }. By Proposition 4.4, there exists some α0>0{\alpha }_{0}\gt 0that satisfies α≥α0\alpha \ge {\alpha }_{0}, (2+β)logr+ωα(r)\left(2+\beta )\log r+{\omega }_{\alpha }\left(r)intersects log[(2+β)(N−2)]\log {[}\left(2+\beta )\left(N-2)]at least n0{n}_{0}times on [0,r0]\left[0,{r}_{0}]. Hence, gα(t){g}_{\alpha }\left(t)has at least n0{n}_{0}zeros on (−∞,τ)\left(-\infty ,\tau )for α≥α0\alpha \ge {\alpha }_{0}, and θα(τ)≥(n0−1)π{\theta }_{\alpha }\left(\tau )\ge \left({n}_{0}-1)\pi for α≥α0\alpha \ge {\alpha }_{0}. Therefore, (4.13) holds.We can find some k0{k}_{0}such that k0π≤θα0(τ)<(k0+1)π{k}_{0}\pi \le {\theta }_{{\alpha }_{0}}\left(\tau )\lt \left({k}_{0}+1)\pi . By (4.13) and the continuity of θα(t){\theta }_{\alpha }\left(t)with respect to α>0\alpha \gt 0, there exists a sequence {αk}k∈N{\left\{{\alpha }_{k}\right\}}_{k\in {\mathbb{N}}}with α0≤α1<α2<⋯{\alpha }_{0}\le {\alpha }_{1}\lt {\alpha }_{2}\lt \cdots \hspace{0.33em}such that θαk(τ)=(k0+k)π,k=1,2,…{\theta }_{{\alpha }_{k}}\left(\tau )=\left({k}_{0}+k)\pi ,\hspace{0.33em}k=1,2,\ldots . Then gαk(τ)=ραk(τ)sinθαk(τ)=0,{g}_{{\alpha }_{k}}\left(\tau )={\rho }_{{\alpha }_{k}}\left(\tau )\sin {\theta }_{{\alpha }_{k}}\left(\tau )=0,and gαk′(τ)=ραk′(τ)sinθαk(τ)+ραk(τ)cosθαk(τ)⋅θαk′(τ)=C0cosθαk(τ)=C0cos(k0+k)π,{g}_{{\alpha }_{k}}^{\prime} \left(\tau )={\rho }_{{\alpha }_{k}}^{\prime} \left(\tau )\sin {\theta }_{{\alpha }_{k}}\left(\tau )+{\rho }_{{\alpha }_{k}}\left(\tau )\cos {\theta }_{{\alpha }_{k}}\left(\tau )\cdot {\theta }_{{\alpha }_{k}}^{\prime} \left(\tau )={C}_{0}\cos {\theta }_{{\alpha }_{k}}\left(\tau )={C}_{0}\cos \left({k}_{0}+k)\pi ,where C0=ραk(τ)θαk′(τ)>0{C}_{0}={\rho }_{{\alpha }_{k}}\left(\tau ){\theta }_{{\alpha }_{k}}^{\prime} \left(\tau )\gt 0. Thus, Lemma 4.6 holds.□Lemma 4.7There exists T0∈R{T}_{0}\in {\mathbb{R}}, which does not depend on α>0\alpha \gt 0such that, if gα∈C2(R){g}_{\alpha }\in {C}^{2}\left({\mathbb{R}})is a solution of (4.10) satisfying gα(t˜)=0{g}_{\alpha }\left(\tilde{t})=0and gα′(t˜)>0{g}_{\alpha }^{^{\prime} }\left(\tilde{t})\gt 0with some t˜≥T0\tilde{t}\ge {T}_{0}, then gα(t)>0{g}_{\alpha }\left(t)\gt 0for t>T0t\gt {T}_{0}and limt→∞gα(t)>0{\mathrm{lim}}_{t\to \infty }{g}_{\alpha }\left(t)\gt 0.ProofSet I[gα(t)]≔12(gα′(t))2+(2+β)(N−2)(egα(t)−gα(t)).I\left[{g}_{\alpha }\left(t)]:= \frac{1}{2}{\left({g}_{\alpha }^{^{\prime} }\left(t))}^{2}+\left(2+\beta )\left(N-2)({e}^{{g}_{\alpha }\left(t)}-{g}_{\alpha }\left(t)).A direct computation yields (4.14)ddtI[gα(t)]=gα′[gα″+(2+β)(N−2)(egα(t)−1)]=−N−2+12e2tgα′(t)2≤0\frac{{\rm{d}}}{{\rm{d}}t}I\left[{g}_{\alpha }\left(t)]={g}_{\alpha }^{^{\prime} }{[}{g}_{\alpha }^{^{\prime\prime} }+\left(2+\beta )\left(N-2)({e}^{{g}_{\alpha }\left(t)}-1)]=-\left(N-2+\frac{1}{2}{e}^{2t}\right){{g}_{\alpha }^{^{\prime} }\left(t)}^{2}\le 0for any t∈Rt\in {\mathbb{R}}. Since gα(t)→∞{g}_{\alpha }\left(t)\to \infty as t→−∞t\to -\infty , there exists some constants t0∈R{t}_{0}\in {\mathbb{R}}satisfy gα(t)≤0{g}_{\alpha }\left(t)\le 0for all t≤t0t\le {t}_{0}.In the case of t≥t0t\ge {t}_{0}, from (4.14), we know that I[gα(t)]≤I[gα(t0)].I\left[{g}_{\alpha }\left(t)]\le I\left[{g}_{\alpha }\left({t}_{0})].And this implies that we can find a constant C>1C\gt 1such that egα(t)−gα(t)≤C{e}^{{g}_{\alpha }\left(t)}-{g}_{\alpha }\left(t)\le Cfor t≥t0t\ge {t}_{0}. Due to limt→±∞(et−t)=∞{\mathrm{lim}}_{t\to \pm \infty }\left({e}^{t}-t)=\infty and inft∈R(et−t)=1{\inf }_{t\in {\mathbb{R}}}\left({e}^{t}-t)=1, we have just proved that gα{g}_{\alpha }is bounded in [t0,∞)\left[{t}_{0},\infty ). Hence, we obtain that gα{g}_{\alpha }is bounded from above in R{\mathbb{R}}, and from (4.11) 0<ϑα(t)≤M,t∈R0\lt {{\vartheta }}_{\alpha }\left(t)\le M,\hspace{1.0em}t\in {\mathbb{R}}for some constant M>0M\gt 0. Therefore, we can complete the proof of Lemma 4.7 by [22, Proposition B.1].□Now we can begin to prove Proposition 4.5.Proof of Proposition 4.5Put T0∈R{T}_{0}\in {\mathbb{R}}be the constant in Lemma 4.7, and let t≥T0t\ge {T}_{0}. By Lemma 4.6, there exists {αk}k∈N{\left\{{\alpha }_{k}\right\}}_{k\in {\mathbb{N}}}with α1<α2<⋯{\alpha }_{1}\lt {\alpha }_{2}\lt \cdots \hspace{0.33em}such that gαk(t)=0,gαk′(t)gαk+1′(t)<0{g}_{{\alpha }_{k}}\left(t)=0,\hspace{1.0em}{g}_{{\alpha }_{k}}^{\prime} \left(t)\hspace{0.33em}{g}_{{\alpha }_{k+1}}^{\prime} \left(t)\lt 0for any k∈Nk\in {\mathbb{N}}. We could assume that (−1)k−1gαk′(t)>0{\left(-1)}^{k-1}{g}_{{\alpha }_{k}}^{\prime} \left(t)\gt 0for k=1,2,⋯k=1,2,\cdots \hspace{0.33em}.Note that (−1)k−1gαk(t){\left(-1)}^{k-1}{g}_{{\alpha }_{k}}\left(t)can be regarded as a solution of (4.10). Then, by Lemma 4.7, we know that limt→∞(−1)k−1gαk(t){\mathrm{lim}}_{t\to \infty }{\left(-1)}^{k-1}{g}_{{\alpha }_{k}}\left(t)is positive for k=1,2,…k=1,2,\ldots . It is obvious that limt→∞gαk(t)=limr→∞[(2+β)logr+ωαk(r)−log[(2+β)(N−2)]]=D(αk)−log[(2+β)(N−2)].\mathop{\mathrm{lim}}\limits_{t\to \infty }{g}_{{\alpha }_{k}}\left(t)=\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+{\omega }_{{\alpha }_{k}}\left(r)-\log {[}\left(2+\beta )\left(N-2)]]=D\left({\alpha }_{k})-\log {[}\left(2+\beta )\left(N-2)].Thus, we obtain D(α2k)−log[(2+β)(N−2)]=limt→∞gα2k(t)<0,i.e.D(α2k)<log[(2+β)(N−2)],D(α2k−1)−log[(2+β)(N−2)]=limt→∞gα2k−1(t)>0,i.e.D(α2k−1)>log[(2+β)(N−2)].\begin{array}{rcl}D\left({\alpha }_{2k})-\log {[}\left(2+\beta )\left(N-2)]& =& \mathop{\mathrm{lim}}\limits_{t\to \infty }{g}_{{\alpha }_{2k}}\left(t)\lt 0,\hspace{1.0em}{\rm{i.e.}}\hspace{0.33em}D\left({\alpha }_{2k})\lt \log {[}\left(2+\beta )\left(N-2)],\\ D\left({\alpha }_{2k-1})-\log {[}\left(2+\beta )\left(N-2)]& =& \mathop{\mathrm{lim}}\limits_{t\to \infty }{g}_{{\alpha }_{2k-1}}\left(t)\gt 0,\hspace{1.0em}{\rm{i.e.}}\hspace{0.33em}D\left({\alpha }_{2k-1})\gt \log {[}\left(2+\beta )\left(N-2)].\end{array}Then Proposition 4.5(i) holds.Let ε>0\varepsilon \gt 0. By Proposition 4.3, there exists a constant α0>0{\alpha }_{0}\gt 0such that for any α≥α0\alpha \ge {\alpha }_{0}, there holds ∣gα(tα)∣<ε,gα′(tα)=2+β+rαωα′(rα)=rα⋅ddr[(2+β)logr+ωα(r)]r=rα=0,| {g}_{\alpha }\left({t}_{\alpha })| \lt \varepsilon ,\hspace{1.0em}{g}_{\alpha }^{^{\prime} }\left({t}_{\alpha })=2+\beta +{r}_{\alpha }{\omega }_{\alpha }^{^{\prime} }\left({r}_{\alpha })={\left.{r}_{\alpha }\cdot \frac{{\rm{d}}}{{\rm{d}}r}{[}\left(2+\beta )\log r+{\omega }_{\alpha }\left(r)]\right|}_{r={r}_{\alpha }}=0,where rα{r}_{\alpha }is the constant given in Proposition 4.3 and tα=logrα{t}_{\alpha }=\log {r}_{\alpha }. Then, by (4.14), we obtain (4.15)(2+β)(N−2)(egα(t)−gα(t))≤I[gα(t)]≤I[gα(tα)]=(2+β)(N−2)(egα(tα)−gα(tα))\left(2+\beta )\left(N-2)\left({e}^{{g}_{\alpha }\left(t)}-{g}_{\alpha }\left(t))\le I\left[{g}_{\alpha }\left(t)]\le I\left[{g}_{\alpha }\left({t}_{\alpha })]=\left(2+\beta )\left(N-2)({e}^{{g}_{\alpha }\left({t}_{\alpha })}-{g}_{\alpha }\left({t}_{\alpha }))for any t≥tαt\ge {t}_{\alpha }. The definition of gα(t){g}_{\alpha }\left(t)leads us to the equation that limt→∞gα(t)=limr→∞[(2+β)logr+ωα(r)−log[(2+β)(N−2)]]=D(α)−log[(2+β)(N−2)].\mathop{\mathrm{lim}}\limits_{t\to \infty }{g}_{\alpha }\left(t)=\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+{\omega }_{\alpha }\left(r)-\log {[}\left(2+\beta )\left(N-2)]]=D\left(\alpha )-\log {[}\left(2+\beta )\left(N-2)].This together with (4.15) implies that 1≤eD(α)−log[(2+β)(N−2)]−(D(α)−log[(2+β)(N−2)])=elimt→∞gα(t)−limt→∞gα(t)≤egα(tα)−gα(tα)≤eε+ε.\begin{array}{rcl}1& \le & {e}^{D\left(\alpha )-\log {[}\left(2+\beta )\left(N-2)]}-(D\left(\alpha )-\log {[}\left(2+\beta )\left(N-2)])\\ & =& {e}^{\mathop{\mathrm{lim}}\limits_{t\to \infty }{g}_{\alpha }\left(t)}-\mathop{\mathrm{lim}}\limits_{t\to \infty }{g}_{\alpha }\left(t)\\ & \le & {e}^{{g}_{\alpha }\left({t}_{\alpha })}-{g}_{\alpha }\left({t}_{\alpha })\\ & \le & {e}^{\varepsilon }+\varepsilon .\end{array}Since et−t>1{e}^{t}-t\gt 1if t≠0t\ne 0and et−t=1{e}^{t}-t=1if t=0t=0, ε>0\varepsilon \gt 0is arbitrary, we obtain limα→∞[eD(α)−log[(2+β)(N−2)]−(D(α)−log[(2+β)(N−2)])]=1,\mathop{\mathrm{lim}}\limits_{\alpha \to \infty }{[}{e}^{D\left(\alpha )-\log {[}\left(2+\beta )\left(N-2)]}-(D\left(\alpha )-\log {[}\left(2+\beta )\left(N-2)])]=1,and then limα→∞[D(α)−log[(2+β)(N−2)]]=0{\mathrm{lim}}_{\alpha \to \infty }{[}D\left(\alpha )-\log {[}\left(2+\beta )\left(N-2)]]=0. Note that D(α)D\left(\alpha )is continuous with regard to α\alpha according to Lemma 3.7. Thus, Proposition 4.5(ii) holds.□Next we can obtain the following proposition when we consider the case where N≥10+4βN\ge 10+4\beta .Proposition 4.8Let N≥10+4βN\ge 10+4\beta , 0<β<20\lt \beta \lt 2. Denote the solution of (1.8) satisfying ω(0)=α\omega \left(0)=\alpha by ωα{\omega }_{\alpha }and define D(α)D\left(\alpha )by (4.8) with α∈R\alpha \in {\mathbb{R}}. Then for any r>0r\gt 0, the following results hold. (i)Let α>0\alpha \gt 0, then ωα(r)<−(2+β)logr+log[(2+β)(N−2)]{\omega }_{\alpha }\left(r)\lt -\left(2+\beta )\log r+\log {[}\left(2+\beta )\left(N-2)].(ii)Let 0<a<b0\lt a\lt b, then ωa(r)<ωb(r){\omega }_{a}\left(r)\lt {\omega }_{b}\left(r).(iii)limα→∞D(α)=log[(2+β)(N−2)]{\mathrm{lim}}_{\alpha \to \infty }D\left(\alpha )=\log {[}\left(2+\beta )\left(N-2)].(iv)For any r0>0{r}_{0}\gt 0, it holdslimα→∞supr≥r0∣ωα(r)+(2+β)logr−log[(2+β)(N−2)]∣=0.\mathop{\mathrm{lim}}\limits_{\alpha \to \infty }\mathop{\sup }\limits_{r\ge {r}_{0}}| {\omega }_{\alpha }\left(r)+\left(2+\beta )\log r-\log {[}\left(2+\beta )\left(N-2)]| =0.In particular, SD=∅{S}_{D}=\varnothing for any D>log[(2+β)(N−2)]D\gt \log {[}\left(2+\beta )\left(N-2)].Following the idea of [22, Lemma 4.10], we present an auxiliary lemma.Lemma 4.9Let N≥10+4βN\ge 10+4\beta , 0<β<20\lt \beta \lt 2, and T∈RT\in {\mathbb{R}}. Then there is no function x=x(t)x=x\left(t)satisfying x∈C2((−∞,T))∩C1((−∞,T])x\in {C}^{2}\left(\left(-\infty ,T))\cap {C}^{1}\left(\left(-\infty ,T])and(4.16)x″+(N−2)+12e2tx′+(2+β)(N−2)x<0,t<T,{x}^{^{\prime\prime} }+\left(\left(N-2)+\frac{1}{2}{e}^{2t}\right)x^{\prime} +\left(2+\beta )\left(N-2)x\lt 0,\hspace{1.0em}t\lt T,(4.17)x(t)<0,t<Tandx(T)=0,x\left(t)\lt 0,\hspace{0.33em}t\lt T\hspace{1.0em}and\hspace{1.0em}x\left(T)=0,(4.18)limsupt→−∞x(t)t<∞andlimsupt→−∞∣x′(t)∣<∞.\mathop{\mathrm{limsup}}\limits_{t\to -\infty }\left|\frac{x\left(t)}{t}\right|\lt \infty \hspace{1.0em}and\hspace{1.0em}\mathop{\mathrm{limsup}}\limits_{t\to -\infty }| x^{\prime} \left(t)| \lt \infty .Proof of Proposition 4.8Set gα(t)≔(2+β)logr+ωα(r)−log[(2+β)(N−2)],t=logr.{g}_{\alpha }\left(t):= \left(2+\beta )\log r+{\omega }_{\alpha }\left(r)-\log {[}\left(2+\beta )\left(N-2)],\hspace{1.0em}t=\log r.Then gα(t){g}_{\alpha }\left(t)satisfies (4.10) with (4.11). Moreover, (4.19)limsupt→−∞gα(t)t<∞,limsupt→−∞∣gα′(t)∣<∞,\mathop{\mathrm{limsup}}\limits_{t\to -\infty }\left|\frac{{g}_{\alpha }\left(t)}{t}\right|\lt \infty ,\hspace{1.0em}\mathop{\mathrm{limsup}}\limits_{t\to -\infty }| {g}_{\alpha }^{^{\prime} }\left(t)| \lt \infty ,and gα(t)→−∞{g}_{\alpha }\left(t)\to -\infty as t→−∞t\to -\infty . Suppose that there exists a constant T∈RT\in {\mathbb{R}}such that gα(t)<0{g}_{\alpha }\left(t)\lt 0for all t∈(−∞,T)t\in \left(-\infty ,T)and gα(T)=0{g}_{\alpha }\left(T)=0. Then ϑα(t)<1{{\vartheta }}_{\alpha }\left(t)\lt 1for all t∈(−∞,T)t\in \left(-\infty ,T)according to (4.11). This leads to gα″+(N−2)+12e2tgα′+(2+β)(N−2)gα<0,t<T,{g}_{\alpha }^{^{\prime\prime} }+\left(\left(N-2)+\frac{1}{2}{e}^{2t}\right){g}_{\alpha }^{^{\prime} }+\left(2+\beta )\left(N-2){g}_{\alpha }\lt 0,\hspace{1.0em}t\lt T,which contradicts to Lemma 4.9 and (4.19). Hence, we have gα(t)<0{g}_{\alpha }\left(t)\lt 0for all t∈Rt\in {\mathbb{R}}. Thus, assertion (i) holds.Then we present the proof of assertion (ii). Let 0<a<b0\lt a\lt band set h(t)=ga(t)−gb(t)h\left(t)={g}_{a}\left(t)-{g}_{b}\left(t). Then we obtain h″+(N−2)+12e2th′+(2+β)(N−2)(ega(t)−egb(t))=0,t∈R.{h}^{^{\prime\prime} }+\left(\left(N-2)+\frac{1}{2}{e}^{2t}\right)h^{\prime} +\left(2+\beta )\left(N-2)({e}^{{g}_{a}\left(t)}-{e}^{{g}_{b}\left(t)})=0,\hspace{1.0em}t\in {\mathbb{R}}.Here, note that h(t)→a−b<0h\left(t)\to a-b\lt 0and h′(t)→0h^{\prime} \left(t)\to 0as t→−∞t\to -\infty . Suppose that there exists a constant T∈RT\in {\mathbb{R}}such that h(t)<0h\left(t)\lt 0for all t∈(−∞,T)t\in \left(-\infty ,T)and h(T)=0h\left(T)=0. Due to ega(t)−egb(t)>h(t){e}^{{g}_{a}\left(t)}-{e}^{{g}_{b}\left(t)}\gt h\left(t)for all t∈(−∞,T)t\in \left(-\infty ,T), we obtain h″+(N−2)+12e2th′+(2+β)(N−2)h<0,t<T.{h}^{^{\prime\prime} }+\left(\left(N-2)+\frac{1}{2}{e}^{2t}\right)h^{\prime} +\left(2+\beta )\left(N-2)h\lt 0,\hspace{1.0em}t\lt T.This contradicts Lemma 4.9. Thus, assertion (ii)\left(ii)holds.Next is the proof of assertion (iii). We can deduce that (p0(t)gα′)′>0\left({p}_{0}\left(t){g}_{\alpha }^{^{\prime} })^{\prime} \gt 0in R{\mathbb{R}}from (4.12) and assertion (i)\left(i). By (4.19), we have p0(t)gα′(t)→0{p}_{0}\left(t){g}_{\alpha }^{^{\prime} }\left(t)\to 0as t→−∞t\to -\infty . Thus, p0(t)gα′(t)>0{p}_{0}\left(t){g}_{\alpha }^{^{\prime} }\left(t)\gt 0for any t∈Rt\in {\mathbb{R}}and (4.20)gα′(t)>0,t∈R.{g}_{\alpha }^{^{\prime} }\left(t)\gt 0,\hspace{1.0em}t\in {\mathbb{R}}.Recall assertion (i), we see (2+β)logr+ωα(r)<log[(2+β)(N−2)]\left(2+\beta )\log r+{\omega }_{\alpha }\left(r)\lt \log {[}\left(2+\beta )\left(N-2)]for all r∈(0,∞)r\in \left(0,\infty ), and hence, (4.21)D(α)≤log[(2+β)(N−2)].D\left(\alpha )\le \log {[}\left(2+\beta )\left(N-2)].On the other hand, for any ε>0\varepsilon \gt 0, we can find constants α0∈R{\alpha }_{0}\in {\mathbb{R}}and t0∈R{t}_{0}\in {\mathbb{R}}satisfying gα0(t0)>−ε{g}_{{\alpha }_{0}}\left({t}_{0})\gt -\varepsilon by Proposition 4.3. This together with (4.20) gives (4.22)−ε<D(α0)−log[(2+β)(N−2)].-\varepsilon \lt D\left({\alpha }_{0})-\log {[}\left(2+\beta )\left(N-2)].By assertion (ii), we obtain that D(α)D\left(\alpha )is nondecreasing in α\alpha . By combining (4.21) and (4.22), we have limα→∞D(α)=log[(2+β)(N−2)]{\mathrm{lim}}_{\alpha \to \infty }D\left(\alpha )=\log {[}\left(2+\beta )\left(N-2)]. Thus, assertion (iii) holds.Finally, we prove assertion (iv). It suffices to prove that, for any t0∈R{t}_{0}\in {\mathbb{R}}, there holds (4.23)limα→∞supt≥t0∣gα(t)∣=0.\mathop{\mathrm{lim}}\limits_{\alpha \to \infty }\mathop{\sup }\limits_{t\ge {t}_{0}}| {g}_{\alpha }\left(t)| =0.Let t0∈R{t}_{0}\in {\mathbb{R}}and ε>0\varepsilon \gt 0. By Proposition 4.3 and (4.9), we can choose constants tε{t}_{\varepsilon }and αε{\alpha }_{\varepsilon }such that −ε<gαεtε−α2≤0-\varepsilon \lt {g}_{{\alpha }_{\varepsilon }}\left({t}_{\varepsilon }-\frac{\alpha }{2}\right)\le 0for all α≥αε\alpha \ge {\alpha }_{\varepsilon }. Keeping in mind assertions (i) and (ii) and (4.20), for any α≥αε\alpha \ge {\alpha }_{\varepsilon }with tε−α2≤t0{t}_{\varepsilon }-\frac{\alpha }{2}\le {t}_{0}and t≥t0t\ge {t}_{0}, there holds −ε<gαεtε−α2≤gαε(t)<gα(t)<0,t≥t0.-\varepsilon \lt {g}_{{\alpha }_{\varepsilon }}\left({t}_{\varepsilon }-\frac{\alpha }{2}\right)\le {g}_{{\alpha }_{\varepsilon }}\left(t)\lt {g}_{\alpha }\left(t)\lt 0,\hspace{1.0em}t\ge {t}_{0}.Then we have (4.23) as ε\varepsilon is arbitrary. Thus, assertion (iv) holds. The proof of Proposition 4.8 is completed.□Proof of Theorem 1.1By Propositions 4.5 and 4.8, we can directly obtain the conclusion of Theorem 1.1, which demonstrates the structure of the solution set SD{S}_{D}.□5Proof of Theorem 1.2In this section, our purpose is to prove Theorem 1.2. Assume by contradiction that the function vvis a global in time solution for (1.1) satisfying the assumptions of Theorem 1.2. Then by using Lemma 5.1, we can construct a radially symmetric continuous weak sub-solution ω1̲=ω1̲(∣y∣)\underline{{\omega }_{1}}=\underline{{\omega }_{1}}\left(| y| )for (1.7) such that ω1̲(∣y∣)<v0(y)inRN,liminf∣y∣→∞(ω1̲(∣y∣)+(2+β)log∣y∣)>D∗.\underline{{\omega }_{1}}\left(| y| )\lt {v}_{0}(y)\hspace{1.0em}{\rm{in}}\hspace{0.33em}{{\mathbb{R}}}^{N},\hspace{1.0em}\mathop{\mathrm{liminf}}\limits_{| y| \to \infty }(\underline{{\omega }_{1}}\left(| y| )+\left(2+\beta )\log | y| )\gt {D}^{\ast }.Thus, the solution ψ∗{\psi }^{\ast }for (1.12) with ψ0∗(y)=ω1̲(∣y∣){\psi }_{0}^{\ast }(y)=\underline{{\omega }_{1}}\left(| y| )is nondecreasing in ss. Moreover, by Lemma 2.6, ψ∗{\psi }^{\ast }would converge to the solution ω˜\tilde{\omega }for (1.8) as s→∞s\to \infty , which maybe a singular solution. Next we assume the existence of ω˜\tilde{\omega }and construct a continuous weak super-solution for (1.7) using ω˜\tilde{\omega }, which yields a contradiction. Finally we can obtain a solution ω=ω(∣y∣)\omega =\omega \left(| y| )of (1.8) satisfying ω∈SD0\omega \in {S}_{{D}_{0}}for some D0>D∗{D}_{0}\gt {D}^{\ast }, contradicting the definition of D∗{D}^{\ast }.Before starting the proof, we present the following lemma, which can be find in [8, Lemma 5.1].Lemma 5.1Let N≥3N\ge 3, 0<β<20\lt \beta \lt 2. Assume that f∈C(RN)f\in C\left({{\mathbb{R}}}^{N})satisfiesliminf∣y∣→∞[(2+β)log∣y∣+f(y)]=D\mathop{\mathrm{liminf}}\limits_{| y| \to \infty }{[}\left(2+\beta )\log | y| +f(y)]=Dfor some D∈RD\in {\mathbb{R}}. Then for any δ>0\delta \gt 0, there exists a continuous weak sub-solution ω̲=ω̲(∣y∣)\underline{\omega }=\underline{\omega }\left(| y| )of (1.7) such that(5.1)limr→∞[(2+β)logr+ω̲(r)]=D−δ,\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\underline{\omega }\left(r)]=D-\delta ,(5.2)ω̲(∣y∣)<f(y)inRN,\underline{\omega }\left(| y| )\lt f(y)\hspace{1.0em}in\hspace{0.33em}{{\mathbb{R}}}^{N},where ω̲\underline{\omega }is radially symmetric and nonincreasing in ∣y∣| y| .Now we can begin to prove Theorem 1.2.ProofRecall that D∗{D}^{\ast }is the constant defined in Section 1 and D∗>log[(2+β)(N−2)]{D}^{\ast }\gt \log {[}\left(2+\beta )\left(N-2)]if 3≤N<10+4β3\le N\lt 10+4\beta , D∗=log[(2+β)(N−2)]{D}^{\ast }=\log {[}\left(2+\beta )\left(N-2)]if N≥10+4βN\ge 10+4\beta by Theorem 1.1. Then take a constant D>0D\gt 0such that D∗<D≤liminf∣x∣→∞[(2+β)log∣x∣+v0(x)].{D}^{\ast }\lt D\le \mathop{\mathrm{liminf}}\limits_{| x| \to \infty }{[}\left(2+\beta )\log | x| +{v}_{0}\left(x)].Assume by contradiction that vvis a global in time solution for (1.1). Then we have that ψ\psi is global in time, that is, the solution of (1.12), which is defined by (1.11), and holds D∗<D≤liminf∣y∣→∞[(2+β)log∣y∣+ψ(y,0)],{D}^{\ast }\lt D\le \mathop{\mathrm{liminf}}\limits_{| y| \to \infty }{[}\left(2+\beta )\log | y| +\psi (y,0)],here ψ0(y)=ψ(y,0)=v0(y){\psi }_{0}(y)=\psi (y,0)={v}_{0}(y). Take a constant δ>0\delta \gt 0such that D−δ>D∗D-\delta \gt {D}^{\ast }, then we can construct a continuous weak sub-solution ω1̲\underline{{\omega }_{1}}for (1.7) according to Lemma 5.1. The function ω1̲\underline{{\omega }_{1}}satisfies ω1̲(∣y∣)≤ψ0(y)=v0(y)\underline{{\omega }_{1}}\left(| y| )\le {\psi }_{0}(y)={v}_{0}(y)in RN{{\mathbb{R}}}^{N}and (5.3)lim∣y∣→∞(ω1̲(∣y∣)+(2+β)log∣y∣)=D−δ.\mathop{\mathrm{lim}}\limits_{| y| \to \infty }(\underline{{\omega }_{1}}\left(| y| )+\left(2+\beta )\log | y| )=D-\delta .Let ψ∗{\psi }^{\ast }be a solution of (1.12) with ψ∗(⋅,0)=ω1̲{\psi }^{\ast }\left(\cdot ,0)=\underline{{\omega }_{1}}. Then we note that ψ∗{\psi }^{\ast }is nondecreasing in ssby Lemma 2.3. Thus, the function (5.4)ω˜(r)≔lims→∞ψ∗(r,s)\tilde{\omega }\left(r):= \mathop{\mathrm{lim}}\limits_{s\to \infty }{\psi }^{\ast }\left(r,s)exists.We claim that ω˜\tilde{\omega }is not bounded from above, otherwise, by Lemma 2.6, we obtain that ω˜∈C2([0,∞))\tilde{\omega }\in {C}^{2}\left(\left[0,\infty ))and ω˜\tilde{\omega }satisfies (1.8) with ω˜′(0)=0\tilde{\omega }^{\prime} \left(0)=0. We also have D−δ=lim∣y∣→∞[(2+β)log∣y∣+ψ∗(y,0)]≤limr→∞[(2+β)logr+ω˜(r)]≕D˜,D-\delta =\mathop{\mathrm{lim}}\limits_{| y| \to \infty }{[}\left(2+\beta )\log | y| +{\psi }^{\ast }(y,0)]\le \mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\tilde{\omega }\left(r)]\hspace{0.33em}=: \hspace{0.33em}\tilde{D},that is, ω˜∈SD˜\tilde{\omega }\in {S}_{\tilde{D}}. Thus, we obtain D˜≥D−δ>D∗\tilde{D}\ge D-\delta \gt {D}^{\ast }, which contradicts the definition of D∗{D}^{\ast }. Hence ω˜\tilde{\omega }is not bounded from above. Then by using Lemma 2.6 again, we have that ω˜∈C2((0,∞))\tilde{\omega }\in {C}^{2}\left(\left(0,\infty ))and ω˜\tilde{\omega }satisfies (1.8) with ω˜(r)→∞\tilde{\omega }\left(r)\to \infty as r→0r\to 0. Again, we define D˜≔limr→∞[(2+β)logr+ω˜(r)].\tilde{D}:= \mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\tilde{\omega }\left(r)].First, we consider the case where 3≤N<10+4β3\le N\lt 10+4\beta . From [8, Lemma 5.2], one can find a constant r2>0{r}_{2}\gt 0such that (5.5)ω˜(r2)=−(2+β)logr2+log[(2+β)(N−2)],ω˜′(r2)<−2+βr2.\tilde{\omega }\left({r}_{2})=-\left(2+\beta )\log {r}_{2}+\log {[}\left(2+\beta )\left(N-2)],\hspace{1.0em}{\tilde{\omega }}^{^{\prime} }\left({r}_{2})\lt -\frac{2+\beta }{{r}_{2}}.Moreover, by Proposition 4.4, we can find constants r0∈(0,r2){r}_{0}\in \left(0,{r}_{2})and α0∈R{\alpha }_{0}\in {\mathbb{R}}such that (5.6)ω˜α0(r0)=−(2+β)logr0+log[(2+β)(N−2)],ω˜α0′(r0)>−2+βr0.{\tilde{\omega }}_{{\alpha }_{0}}\left({r}_{0})=-\left(2+\beta )\log {r}_{0}+\log {[}\left(2+\beta )\left(N-2)],\hspace{1.0em}{\tilde{\omega }}_{{\alpha }_{0}}^{\prime} \left({r}_{0})\gt -\frac{2+\beta }{{r}_{0}}.Thus, combining the aforementioned equation and Lemma 2.4, we know that ω¯(r)\overline{\omega }\left(r)defined by ω¯(r)≔ω˜α0(r)forr∈[0,r0],−(2+β)logr+log[(2+β)(N−2)]forr∈[r0,r2],ω˜(r)forr∈[r2,∞]\overline{\omega }\left(r):= \left\{\begin{array}{ll}{\tilde{\omega }}_{{\alpha }_{0}}\left(r)\hspace{1.0em}& {\rm{for}}\hspace{1.0em}r\in \left[0,{r}_{0}],\\ -\left(2+\beta )\log r+\log {[}\left(2+\beta )\left(N-2)]\hspace{1.0em}& {\rm{for}}\hspace{1.0em}r\in \left[{r}_{0},{r}_{2}],\\ \tilde{\omega }\left(r)\hspace{1.0em}& {\rm{for}}\hspace{1.0em}r\in \left[{r}_{2},\infty ]\end{array}\right.is a continuous weak super-solution of (1.7) and (5.7)limr→∞[(2+β)logr+ω¯(r)]=limr→∞[(2+β)logr+ω˜(r)]=D˜≥D−δ>D∗.\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\overline{\omega }\left(r)]=\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\tilde{\omega }\left(r)]=\tilde{D}\ge D-\delta \gt {D}^{\ast }.Then, by using Lemma 5.1, we could construct a continuous weak sub-solution ω2̲\underline{{\omega }_{2}}of (1.7) such that ω2̲<ω¯\underline{{\omega }_{2}}\lt \overline{\omega }and limr→∞[(2+β)logr+ω2̲(r)]=D−δ.\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\underline{{\omega }_{2}}\left(r)]=D-\delta .Let ψ˜\tilde{\psi }be a solution of (1.12) with ψ˜(⋅,0)=ω2̲(∣⋅∣)\tilde{\psi }\left(\cdot ,0)=\underline{{\omega }_{2}}\left(| \cdot | ). Then we can deduce from Lemma 2.3 that ψ˜\tilde{\psi }is nondecreasing in ss. Hence, the limit ω(r)=lims→∞ψ˜(r,s)\omega \left(r)=\mathop{\mathrm{lim}}\limits_{s\to \infty }\tilde{\psi }\left(r,s)exists and ω2̲(r)≤ψ˜(r,s)≤ω¯(r)\underline{{\omega }_{2}}\left(r)\le \tilde{\psi }\left(r,s)\le \overline{\omega }\left(r)by Lemma 2.1(ii), which shows that (5.8)ω2̲(r)≤ω(r)≤ω¯(r).\underline{{\omega }_{2}}\left(r)\le \omega \left(r)\le \overline{\omega }\left(r).Due to ω¯(0)<∞\overline{\omega }\left(0)\lt \infty and Lemma 2.6, we know that ω\omega is a regular solution of (1.8) with ω′(0)=0\omega ^{\prime} \left(0)=0. Moreover, we obtain limr→∞[(2+β)logr+ω2̲(r)]=D−δ≤limr→∞[(2+β)logr+ω(r)]≕D0,\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\underline{{\omega }_{2}}\left(r)]=D-\delta \le \mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\omega \left(r)]\hspace{0.33em}=: \hspace{0.33em}{D}_{0},by (5.8) and ω∈SD0\omega \in {S}_{{D}_{0}}for some D0≥D−δ>D∗{D}_{0}\ge D-\delta \gt {D}^{\ast }. Hence, there is a contradiction to the definition of D∗{D}^{\ast }.Thus, we obtain that vvblows up locally in time. Therefore, we proved Theorem 1.2 when 3≤N<10+4β3\le N\lt 10+4\beta .The proof for the case N≥10+4βN\ge 10+4\beta can be completed by the method analogous to that used earlier, the key is to create a continuous weak super-solution ω¯\overline{\omega }of (1.7), which satisfies (5.9)limr→∞[(2+β)logr+ω¯(r)]>D∗.\mathop{\mathrm{lim}}\limits_{r\to \infty }{[}\left(2+\beta )\log r+\overline{\omega }\left(r)]\gt {D}^{\ast }.And the aforementioned inequality could be achieved with the aid of Proposition 4.8. We omit the details; thus, the proof of Theorem 1.2 is completed.□

Journal

Advances in Nonlinear Analysis – de Gruyter

Published: Jan 1, 2023

Keywords: Hénon type equation; heat equation; self-similar solutions; blow-up solutions; 35K58; 35J15; 35J67; 35J61

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