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J. Evol. Equ. 21 (2021), 2047–2080 © 2021 The Author(s) Journal of Evolution 1424-3199/21/022047-34, published online January 18, 2021 Equations https://doi.org/10.1007/s00028-020-00666-y Equivalence of viscosity and weak solutions for a p-parabolic equation Jarkko Siltakoski Abstract. We study the relationship of viscosity and weak solutions to the equation ∂ u − Δ u = f (Du), t p where p > 1and f ∈ C (R ) satisfies suitable assumptions. Our main result is that bounded viscosity supersolutions coincide with bounded lower semicontinuous weak supersolutions. Moreover, we prove the lower semicontinuity of weak supersolutions when p ≥ 2. 1. Introduction A classical solution to a partial differential equation is a smooth function that satis- fies the equation pointwise. Since many equations that appear in applications admit no such solutions, a more general class of solutions is needed. One such class is the exten- sively studied distributional weak solutions defined by integration by parts. Another is the celebrated viscosity solutions based on generalized pointwise derivatives. When both classes of solutions can be meaningfully defined, it is naturally crucial that they coincide. This has been profusely studied starting from [10]. In [12], the equivalence of solutions was proved for the parabolic p-Laplacian. The objective of the present work is to prove this equivalence in a different way while also allowing the equation to depend on a first-order term. To the best of our knowledge, the proof is new even in the homogeneous case, at least when 1 < p < 2. More precisely, we study the parabolic equation ∂ u − Δ u = f (Du), (1.1) t p where 1 < p < ∞ and f ∈ C (R ) satisfies a certain growth condition, for details see Sect. 2. We show that bounded viscosity supersolutions to (1.1) coincide with bounded lower semicontinuous weak supersolutions. Moreover, we prove the lower Mathematics Subject Classification: 35K92, 35J60, 35D40, 35D30, 35B51 Keywords: Comparison principle, Gradient term, Parabolic p-Laplacian, Viscosity solution, Weak solution. 2048 J. Siltakoski J. Evol. Equ. semicontinuity of weak supersolutions in the range p ≥ 2 under slightly stronger assumptions on f . To show that viscosity supersolutions are weak supersolutions, we apply the tech- nique introduced by Julin and Juutinen [11]. In contrast to [12], we do not employ the uniqueness machinery of viscosity solutions. Instead, our strategy is to approximate a viscosity supersolution u by its inf-convolution u . It is straightforward to show that u is still a viscosity supersolution in a smaller set. This and the pointwise proper- ties of the inf-convolution imply that u is also a weak supersolution in the smaller set. Furthermore, it follows from Caccioppoli’s estimates that u converges to u in a suitable Sobolev space. It then remains to pass to the limit to see that u is a weak supersolution. To show that weak supersolutions are viscosity supersolutions, we apply the argu- ment from [12] that is based on the comparison principle of weak solutions. However, we could not find a reference for comparison principle for Eq. (1.1). Therefore, we give a detailed proof of such a result. To prove the lower semicontinuity of weak supersolutions, we adapt the strategy of [17]. First, we prove estimates for the essential supremum of a subsolution using Moser’s iteration technique. Then, we use those estimates to deduce that a supersolu- tion is lower semicontinuous at its Lebesgue points. The equivalence of viscosity and weak solutions for the p-Laplace equation and its parabolic version was first proven in [12]. A different proof in the elliptic case was found in [11]. Recently the equivalence of solutions has been studied for various equations. These include the normalized p-Poisson equation [1], a non-homogeneous p-Laplace equation [22] and the normalized p(x )-Laplace equation [25]. Moreover, in [24] the equivalence is shown for the radial solutions of a parabolic equation. We also mention that an unpublished version of [18] applies [11] to sketch the equivalence of solutions to (1.1) in the homogeneous case when p ≥ 2. Comparison principles for quasilinear parabolic equations have been studied by several authors. In [13], comparison is proven for ∂ u − Δ u + f (u, x , t ) = 0 when t p p > 2 and f is a continuous function such that | f (u, x , t )| ≤ g(u) for some g ∈ C . The homogeneous case for the p-parabolic equation is considered also in [16] and the general equation ∂ u − div A(x , t, Du) = 0in [15]. Equations with gradient terms are studied for example in [2], where comparison principle is shown for the equation ∂ u − Δ u − |Du| = 0 when p > 2 and β> p − 1. In the recent papers [4,5], both t p positive results and counter examples are provided for the comparison, strong compari- p−2 son, and maximum principles for the equation ∂ u − Δ u − λ |u| u − f (x , t ) = 0. t p | | Furthermore, according to [3], the equation ∂ u − Δ u = q(x ) u can admit multi- t p ple solutions with zero boundary values when 0 <α < 1. The paper is organized as follows. Section 2 contains the precise definitions of weak and viscosity solutions. In Sect. 3, we show that weak supersolutions are viscosity supersolutions, and the converse is shown in Sect. 4. Finally, the lower semicontinuity of weak supersolutions is considered in Sect. 5. Vol. 21 (2021) Equivalence of viscosity and weak solutions 2049 2. Preliminaries N N The symbols Ξ and Ω are reserved for bounded domains in R × R and R , respectively. For t < t , we define the cylinder Ω := Ω × (t , t ) and its para- 1 2 t ,t 1 2 1 2 bolic boundary ∂ Ω := (Ω × {t }) ∪ (∂Ω × (t , t ]). Moreover, for T > 0we p t ,t 1 1 2 1 2 set Ω := Ω . T 0,T 1,p p The Sobolev space W (Ω) contains the functions u ∈ L (Ω) for which the dis- tributional gradient Du exists and belongs in L (Ω). It is equipped with the norm u 1,p := u p + Du p . W (Ω) L (Ω) L (Ω) A Lebesgue measurable function u : Ω → R belongs to the parabolic Sobolev t ,t 1 2 p 1,p 1,p space L (t , t ; W (Ω)) if u(·, t ) ∈ W (Ω) for almost every t ∈ (t , t ) and the 1 2 1 2 norm p p |u| + |Du| dz t ,t 1 2 is finite. By dz, we mean integration with respect to space and time variables, i.e., dz = dx dt. Integral average is denoted by − u dz := u dz. |Ω | Ω T Ω T T Growth condition Unless otherwise stated, the function f ∈ C (R ) is assumed to satisfy the growth condition β N | f (ξ )| ≤ C (1 + |ξ | ) for all ξ ∈ R , (G1) where C > 0 and 1 ≤ β< p. Definition 2.1. (Weak solution) A function u : Ξ → R is a weak supersolution to p 1,p (1.1)in Ξ if u ∈ L (t , t ; W (Ω)) whenever Ω Ξ, and 1 2 t ,t 1 2 p−2 −u∂ ϕ + |Du| Du · Dϕ − ϕ f (Du) dz ≥ 0 for all non-negative test functions ϕ ∈ C (Ω ).For weak subsolutions, the inequal- t ,t 1 2 ity is reversed and a function is a weak solution if it is both a super- and subsolution. To define viscosity solutions to (1.1), we set for all ϕ ∈ C with Dϕ = 0 p − 2 p−2 2 Δ ϕ := |Dϕ| Δϕ + D ϕ Dϕ, Dϕ . | | Dϕ 2050 J. Siltakoski J. Evol. Equ. Definition 2.2. (Viscosity solution) A lower semicontinuous and bounded function u : Ξ → R is a viscosity supersolution to (1.1)in Ξ if whenever ϕ ∈ C (Ξ ) and (x , t ) ∈ Ξ are such that 0 0 ϕ(x , t ) = u(x , t ), 0 0 0 0 ϕ(x , t)< u(x , t ) when (x , t ) = (x , t ), 0 0 Dϕ(x , t ) = 0 when x = x , then lim sup ∂ ϕ(x , t ) − Δ ϕ(x , t ) − f (Dϕ(x , t )) ≥ 0. t p (x ,t )→(x ,t ) 0 0 x =x An upper semicontinuous and bounded function u : Ξ → R is a viscosity subsolution to (1.1)in Ξ if whenever ϕ ∈ C (Ξ ) and (x , t ) ∈ Ξ are such that 0 0 ⎪ ϕ(x , t ) = u(x , t ), 0 0 0 0 ϕ(x , t)> u(x , t ) when (x , t ) = (x , t ), 0 0 Dϕ(x , t ) = 0 when x = x 0, then lim inf ∂ ϕ(x , t ) − Δ ϕ(x , t ) − f (Dϕ(x , t )) ≤ 0. t p (x ,t )→(x ,t ) 0 0 x =x A function that is both a viscosity sub- and supersolution is a viscosity solution. If a function ϕ is like in the definition of viscosity supersolution, we say that ϕ touches u from below at (x , t ). The limit superior in the definition is needed because 0 0 the operator Δ is singular when 1 < p < 2. When p ≥ 2, the operator is degenerate and the limit superior disappears. 3. Weak solutions are viscosity solutions We show that bounded, lower semicontinuous weak supersolutions to (1.1) are vis- cosity supersolutions when 1 < p < ∞ and f ∈ C (R ) satisfies the growth condition (G1). One way to prove this kind of results is by applying the comparison principle [12]. However, we could not find the comparison principle for Eq. (1.1) in the literature and therefore we prove it first. To this end, we first prove comparison Lemmas 3.2 and 3.3 for locally Lipschitz continuous f . The local Lipschitz continuity allows us to absorb the first-order terms into the terms that appear due to the p-Laplacian, see Step 2 in proof of Lemma 3.2. To deal with general f , we take a locally Lipschitz continuous approximant f such that f − f ∞ N <δ/4T . Then, for sub- and δ δ L (R ) supersolutions u and v, we consider the functions δ δ u := u − and v := v + . δ δ T − t /2 T − t /2 Vol. 21 (2021) Equivalence of viscosity and weak solutions 2051 These functions will be sub- and supersolutions to (1.1) where f is replaced by f . Since f is locally Lipschitz continuous, it follows from Lemmas 3.2 and 3.3 that u ≤ v . Letting δ → 0 then yields that u ≤ v. δ δ For similar comparison results, see [2, Proposition 2.1] and [13]. See also Chapters 3.5 and 3.6 in [23] for the elliptic case. A minor difference in our results is that instead of requiring that both the subsolution and the supersolution have uniformly bounded gradients, we only require this for the subsolution. To prove the comparison principle, we need to use a test function that depends on the supersolution itself. However, supersolutions do not necessarily have a time derivative. One way to deal with this is to use mollifications in the time direction. For a compactly supported ϕ ∈ L (Ω ), we define its time-mollification by ϕ (x , t ) = ϕ(x , t − s)ρ (s) ds, where ρ is a standard mollifier whose support is contained in (− , ). Then, ϕ has time derivative and ϕ → ϕ in L (Ω ). Furthermore, the time-mollification of a supersolution satisfies a regularized equation in the sense of the following lemma. Lemma 3.1. Let v ∈ L (Ω ) be a weak supersolution (subsolution) to (1.1) in Ω . T T Then, we have p−2 −v ∂ ϕ + |Dv| Dv · Dϕ − ϕ ( f (Dv)) dz ≥ (≤) 0 (3.1) 1,p ∞ for all ϕ ∈ W (Ω ) ∩ L (Ω ) with compact support in Ω . Moreover, if the T T T stronger growth condition p−1 | f (ξ )| ≤ C 1 + |ξ | (G2) holds, then the assumption ϕ ∈ L (Ω ) is not needed. 1,p Observe that in the above lemma ϕ is in the usual Sobolev space W (Ω ) so it has a weak time derivative ∂ ϕ ∈ L (Ω ). To prove the lemma, one first assumes that t T ϕ is smooth. Then, testing the weak formulation of (1.1) with ϕ , changing variables and using Fubini’s theorem yields (3.1). The general case follows by approximating 1,p ϕ in W (Ω ) with the standard mollification. We omit the details. Lemma 3.2. Let 1 < p < 2 and let f be locally Lipschitz. Let u, v ∈ L (Ω ), respectively, be weak sub- and supersolutions to (1.1) in Ω . Assume that for all (x , t ) ∈ ∂ Ω 0 0 p T ess lim sup u(x , t ) ≤ ess lim inf v(x , t ). (x ,t )→(x ,t ) 0 0 (x ,t )→(x ,t ) 0 0 Suppose also that Du ∈ L (Ω ). Then, u ≤ v a.e. in Ω . T T 2052 J. Siltakoski J. Evol. Equ. Proof. (Step 1) Let l > 0 and set w := (u − v − l) . Let also s ∈ (0, T ). We want to use w · χ as a test function, but since it is not smooth, we must perform molli- [0,s] fications. Let h > 0 and define ϕ := η (u − v − l) , where ⎪ 1, t ∈ (0, s − h], η(t ) = (−t + s + h)/2h, t ∈ (s − h, s + h), 0, t ∈[s + h, T ). 1,p The function ϕ is compactly supported and belongs in W (Ω ). Therefore, by Lemma 3.1 we have −(u − v) ∂ ϕ dz p−2 p−2 ≤ |Dv| Dv − |Du| Du · Dϕ + ϕ f (Du) − f (Dv) dz. (3.2) We use the linearity of convolution and integration by parts to eliminate the time derivative. We obtain −(u − v) ∂ ϕ dz =− (u − v) (u − v − l) ∂ η + η(u − v) ∂ (u − v − l) dz t t + + =− (u − v − l) ((u − v − l) ) ∂ η + l (u − v − l) ∂ η + t t + η(u − v − l) ∂ (u − v − l) + lη∂ (u − v − l) dz t t + + 2 2 =− ((u − v − l) ) ∂ η + η∂ ((u − v − l) ) dz t t + + =− ((u − v − l) ) ∂ η dz → − (u − v − l) ∂ η dz. →0 2 Moreover, by the Lebesgue differentiation theorem for a.e. s ∈ (0, T ) it holds s+h 1 1 1 2 2 2 − (u − v − l) ∂ η dz = w (x , t ) dx dt → w (x , s) dx . 2 4h h→0 2 Ω s−h Ω Ω The terms at the right-hand side of (3.2) converge similarly. Hence, for a.e. s ∈ (0, T ) we have Vol. 21 (2021) Equivalence of viscosity and weak solutions 2053 w (x , s) dx p−2 p−2 ≤ | f (Du) − f (Dv)| w dz − |Du| Du − |Dv| Dv · Dw dz Ω Ω s s =: I − I . (3.3) 1 2 (Step 2) We seek to absorb some of I into I so that we can conclude from Grönwall’s 1 2 inequality that w ≡ 0 almost everywhere. Since f is locally Lipschitz continuous, there are constants M ≥ max(2 Du ∞ , 1) and L = L(M ) such that L (Ω ) | f (ξ ) − f (η)| ≤ L |ξ − η| when |ξ | , |η| < M. (3.4) We denote Ω := {x ∈ Ω : w ≥ 0}, + + A := Ω ∩ {|Dv| < M } and B := Ω ∩ {|Dv| ≥ M } . s s Observe that in B we have by the growth condition (G1), choice of M and the assump- tion that β ≥ 1 β β β β | f (Du)| ≤ C (1 + |Du| ) ≤ C (M + M ) ≤ 2C M ≤ 2C |Dv| (3.5) f f f f and β β | f (Dv)| ≤ C (1 + |Dv| ) ≤ 2C |Dv| . (3.6) f f It follows from (3.4), (3.5), (3.6) and Young’s inequality that I ≤ L |Du − Dv| w dz + (| f (Du)| + | f (Dv)|) w dz A B ≤ L |Du − Dv| w dz + 4C |Dv| w dz A B βp p 2 2 β p−β ≤ |Du − Dv| + C ( , L)w dz + |Dv| + C ( , p,β, L , C )w dz A B 2 p 2 ≤ |Du − Dv| dz + |Dv| dz + C ( , p,β, L , C , w ∞ ) w dz, A B Ω (3.7) where in the last step we used that > 2 to estimate p−β p/( p−β)−2 p/( p−β) p/( p−β)−2 2 2 w dz = w w dz ≤ w w dz. L (Ω ) Ω Ω Ω s s s Using the vector inequality p−2 p−2 p−2 2 2 2 |a| a −|b| b · (a − b) ≥ (p − 1) |a − b| 1 + |a| +|b| , (3.8) 2054 J. Siltakoski J. Evol. Equ. which holds when 1 < p < 2[19, p98], we get p−2 p−2 I = |Du| Du − |Dv| Dv · Dw dz |Du − Dv| ≥ (p − 1) dz 2−p 2 2 Ω 2 1 + |Du| + |Dv| 2 2 |Du − Dv| (|Dv| − |Du|) ≥ (p − 1) dz + (p − 1) dz 2−p 2−p A 2 2 2 B 2 1 + M + M 3 |Dv| |Dv| − M 2 2 ≥ C (p, M ) |Du − Dv| dz + p − 1 dz ( ) 2−p A B 2 3 |Dv| |Dv| 2 2 ≥ C (p, M ) |Du − Dv| dz + (p − 1) dz 2−p A B 2 3 |Dv| 2 p = C (p, M ) |Du − Dv| dz + C (p) |Dv| dz, (3.9) A B where C (p, M ), C (p)> 0. Combining the estimates (3.7) and (3.9), we arrive at 2 p I − I ≤ ( − C (p, M )) |Du − Dv| dz + ( − C (p)) |Dv| dz 1 2 A B + C w dz, where C = C ( , p,β, L , C , w ∞ ). Recalling (3.3) and taking small enough 0 f yields 2 2 w (x , s) dx ≤ 2C w dz. Ω Ω Since this holds for a.e. s ∈ (0, T ), Grönwall’s inequality implies that w ≡ 0a.e.in Ω . Finally, letting l → 0 yields that u − v ≤ 0a.e. in Ω . T T Lemma 3.3. Let p ≥ 2 and let f be locally Lipschitz. Let v ∈ L (Ω ) be a weak supersolution to (1.1) and let u ∈ L (Ω ) be a weak subsolution to ∂ u − Δ u − f (Du) ≤−δ in Ω t p T for some δ> 0. Assume that for all (x , t ) ∈ ∂ Ω 0 0 p T ess lim sup u(x , t ) ≤ ess lim inf v(x , t ). (x ,t )→(x ,t ) 0 0 (x ,t )→(x ,t ) 0 0 Suppose also that Du ∈ L (Ω ). Then, u ≤ v a.e. in Ω . T T Vol. 21 (2021) Equivalence of viscosity and weak solutions 2055 Proof. Let l > 0 and set w := (u − v − l) . Let also s ∈ (0, T ). Repeating the first step of the proof of Lemma 3.2, we arrive at the inequality w (x , s) dx p−2 p−2 ≤ | f (Du) − f (Dv)| w dz − |Du| Du − |Dv| Dv · Dw dz Ω Ω s s − δw dz =: I − I − δw dz. (3.10) 1 2 Moreover, we define the constants M and L, and the sets A and B, exactly in the same way as in the proof of Lemma 3.2. Then, by (3.4), (3.5), (3.6) and Young’s inequality I ≤ L |Du − Dv| w dz + (| f (Du)| + | f (Dv)|) w dz A B p β p−1 ≤ |Du − Dv| + C ( , L)w dz + 4C |Dv| w dz A B p p p−1 ≤ |Du − Dv| dz + |Dv| dz + C ( , p,β, L , C ) w A B Ω p−β + w dz. (3.11) Using the vector inequality p−2 p−2 2−p p |a| a − |b| b · a − b ≥ 2 |a − b| , (3.12) ( ) which holds when p ≥ 2[19, p. 95], we get p p I ≥ C (p) |Du − Dv| dz + C (p) |Du − Dv| dz. A B Furthermore, since in B it holds p p p |Du − Dv| ≥ (|Dv| − |Du|) ≥ |Dv| − M ≥ C (p) |Dv| , we arrive at p p I ≥ C (p) |Du − Dv| dz + C (p) |Dv| dz. (3.13) A B Combining (3.11) and (3.13) with (3.10), we get 2 p p w dx ≤ ( − C (p)) |Du − Dv| dz + |Dv| dz Ω A B p p p−1 p−β + C ( , p,β, L , C ) w + w − δw dz. s 2056 J. Siltakoski J. Evol. Equ. By taking small enough = (p), the above becomes p p p−1 p−β w (x , s) dx ≤ C (p,β, L , C ) w + w − δw dz. (3.14) Ω Ω p p Observe that since w is bounded and , > 1, the integrand at the right-hand p−1 p−β side is bounded by some constant times w . To argue this rigorously, we write down the following algebraic fact. If a ,δ,γ > 0 and α> 1, then there exists C(α,γ,δ, a )> 0 such that 0 0 α 2 γ a ≤ δa + C(α,γ,δ, a )a for all a ∈[0, a ). 0 0 To see this, let first α< 2. Then, by Young’s inequality δ 2 2 α α−1 (α−1)· 3−α α−1 γ a = γ a · a ≤ a + C(α,γ,δ, a )a 3−α 1 + a ≤ δa + C(α,γ,δ, a )a . If α ≥ 2, then α−2 α α−2 2 2 γ a = γ a · a ≤ γ a a . Applying the algebraic fact on (3.14), we get 2 2 w (x , s) dx ≤ C (p,β, L , C ,δ, w ∞ ) w dz. Ω Ω The conclusion now follows from Grönwall’s inequality and letting l → 0. Next, we use the previous comparison results to prove the comparison principle for general continuous f . Theorem 3.4. Let 1 < p < ∞. Let u,v ∈ L (Ω ), respectively, be weak sub- and supersolutions to (1.1) in Ω . Assume that for all (x , t ) ∈ ∂ Ω T 0 0 p T ess lim sup u(x , t ) ≤ ess lim inf v(x , t ). (x ,t )→(x ,t ) (x ,t )→(x ,t ) 0 0 0 0 Assume also that Du ∈ L (Ω ). Then, u ≤ v a.e. in Ω . T T Proof. For δ> 0, define u := u − . T − t /2 Then, for any non-negative test function ϕ ∈ C (Ω ) we have by integration by parts −u ∂ ϕ dz = −u∂ ϕ + ∂ ϕ dz δ t t t T − t /2 Ω Ω T T Vol. 21 (2021) Equivalence of viscosity and weak solutions 2057 = −u∂ ϕ − ϕ dz 2 (T − t /2) ≤ −u∂ ϕ − ϕ dz. 2T Since f is continuous, there is a locally Lipschitz continuous function f such that f − f ∞ N ≤ (see, e.g., [21]). Then, since u is a weak subsolution, we have L (R ) 4T p−2 −u ∂ ϕ + |Du | Du · Dϕ − ϕ f (Du ) dz δ t δ δ δ δ p−2 ≤ −u∂ ϕ + |Du| Du · Dϕ − ϕ f (Du) + ϕ f − f − ϕ dz ∞ N t δ L (R ) 2T ≤ − ϕ dz. 4T Hence, u is a weak subsolution to ∂ u − Δ u − f (Du ) ≤− in Ω . t δ p δ δ δ T 4T Similarly, since v is a weak supersolution, we define v := v + T − t /2 and deduce that v is a weak supersolution to ∂ v − Δ v − f (Dv ) ≥0in Ω . t δ p δ δ δ T Now it follows from the comparison Lemmas 3.2 and 3.3 that u ≤ v a.e. in Ω . δ δ T Thus, 2δ u ≤ v + a.e. in Ω . T − t /2 Letting δ → 0 finishes the proof. Now that the comparison principle is proven, we are ready to show that weak solutions are viscosity solutions. To state this part of the equivalence, we define the lower semicontinuous regularization of a function u : Ξ → R by u (x , t ) := ess lim inf u(y, s) := lim ess inf u. p p R→0 (y,s)→(x ,t ) B (x )×(t −R ,t +R ) The time scaling R is technically convenient in Sect. 5. We have included it here for notational consistency. Theorem 3.5. Let 1 < p < ∞. Let u ∈ L (Ξ ) be a weak supersolution to (1.1) in loc Ξ for which u = u almost everywhere in Ξ. Then, u is a viscosity supersolution to ∗ ∗ (1.1) in Ξ. 2058 J. Siltakoski J. Evol. Equ. Proof. Assume on the contrary that there is φ ∈ C (Ξ ) touching u from below at (x , t ) ∈ Ξ, Dφ(x , t ) = 0for x = x and 0 0 0 lim sup ∂ φ(x , t ) − Δ φ(x , t ) − f (Dφ(x , t )) < 0. (3.15) t p (x ,t )→(x ,t ) 0 0 x =x Denote Q := B (x ) × (t − r, t + r ). It follows from above that there are r > 0 r r 0 0 0 and δ> 0 such that ∂ φ − Δ φ − f (Dφ) < −δ in Q \ {x = x } . (3.16) t p r 0 Indeed, otherwise there would be a sequence (x , t ) → (x , t ) such that x = x n n 0 0 n 0 and ∂ φ(x , t ) − Δ φ(x , t ) − f (Dφ(x , t )) > − , t n n p n n n n but this contradicts (3.15). Using Gauss’s theorem and (3.16), we obtain for any non- negative test function ϕ ∈ C (Q ) that p−2 −φ∂ ϕ + |Dφ| Dφ · Dϕ − ϕ f (Dφ) dz p−2 = lim −φ∂ ϕ + |Dφ| Dφ · Dϕ − ϕ f (Dφ) dz ρ→0 Q \{|x −x |≤ρ} r 0 p−2 = lim ϕ∂ φ − ϕ div(|Dφ| Dφ) − ϕ f (Dφ) dz ρ→0 Q \{|x −x |≤ρ} r 0 t +r (x − x ) p−2 + ϕ |Dφ| Dφ · dS dt t −r {|x −x |=ρ} 0 0 = lim ϕ ∂ φ − Δ φ − f (Dφ) dz t p ρ→0 Q \{|x −x |≤ρ} r 0 ≤ −δϕ dz. Let l := min (u − φ) > 0 and set φ := φ + l. Then, by the above inequality, φ ∂ Q ∗ p r is a weak subsolution to ∂ φ − Δ φ − f (Dφ) ≤−δ in Q t p r and on ∂ Q it holds φ = φ + l ≤ φ + u − φ = u . Hence, Theorem 3.4 implies p r ∗ ∗ that φ ≤ u almost everywhere in Q . By the definition of u ,itfollows that r ∗ φ ≤ u everywhere in Q , (3.17) ∗ r which is a contradiction since in particular φ(x , t ) = φ(x , t ) + l > u (x , t ). 0 0 0 0 ∗ 0 0 To see (3.17), fix (y , s ) ∈ Q and let ε> 0. By continuity of φ and the definition 0 0 r of u , there is R > 0 such that φ(y, s) − φ(y , s ) ≤ ε for all (y, s) ∈ Q 0 0 R Vol. 21 (2021) Equivalence of viscosity and weak solutions 2059 and ess inf u − u (y , s ) <ε, ∗ 0 0 p p where we denoted Q := B (y ) × (s − R , s + R ). In particular, R 0 0 0 u (y , s ) ≥ ess inf u − ε. ∗ 0 0 By the definition of ess inf u, there is A ⊂ Q with |A| > 0 such that ess inf u + ε> u(y, s) for all (y, s) ∈ A. Moreover, since φ ≤ u almost everywhere in Q , we can take (y, s) ∈ A such that φ(y, s) ≤ u(y, s). Now we have by the last three displays u (y s ) ≥ ess inf u − ε> u(y, s) − 2ε ≥ φ(y, s) − 2ε ≥ φ(y , s ) − 3ε. ∗ 0, 0 0 0 Since ε> 0 was arbitrary, this implies that u (y , s ) ≥ φ(y , s ). ∗ 0 0 0 0 4. Viscosity solutions are weak solutions We show that bounded viscosity supersolutions to (1.1) are weak supersolutions when 1 < p < ∞ and f ∈ C (R ) satisfies the growth condition (G1). We use the method developed in [11]. The method of [11] was previously applied to parabolic equations in [24], but for radially symmetric solutions. The idea is to approximate a viscosity supersolution u to (1.1)bythe inf-convolution q 2 |x − y| |t − s| u (x , t ) := inf u(y, s) + + , q−1 (y,s)∈Ξ qε 2ε q−2 where ε> 0 and q ≥ 2 is a fixed constant so large that p − 2 + > 0. If u q−1 is bounded, it is straightforward to show that the inf-convolution u is a viscosity supersolution in the smaller set Ξ = (x , t ) ∈ Ξ : B (x ) × (t − t (ε), t + t (ε)) Ξ , ε r (ε) where r (ε), t (ε) → 0as ε → 0. Moreover, u is semi-concave by definition and therefore it has a second derivative almost everywhere. It follows from these pointwise properties that u is a weak supersolution to (1.1)in Ξ . Caccioppoli-type estimates ε ε then imply that u converges to u in a parabolic Sobolev space and consequently u is a weak supersolution. 2060 J. Siltakoski J. Evol. Equ. The standard properties of the inf-convolution are postponed to the end of this section. Instead, we begin by proving the key observation: that the inf-convolution of a viscosity supersolution is a weak supersolution in the smaller set Ξ . When p ≥ 2, the idea is the following. Since u is a viscosity supersolution to (1.1) that is twice differentiable almost everywhere, it satisfies the equation pointwise almost everywhere. Hence, we may multiply the equation by a non-negative test function ϕ and integrate over Ξ so that the integral will be non-negative. Then, we approximate this expression through smooth functions u defined via the standard mollification. ε, j Since u is smooth, we may integrate by parts to reach the weak formulation of ε, j the equation, see (4.1). It then remains to let j →∞ to conclude that u is a weak supersolution. The range 1 < p < 2 is more delicate because of the singularity of the p-Laplace operator (p − 2) p−2 2 Δ u := |Du| Δu + D uDu, Du , |Du| and therefore we consider the case p ≥ 2first. Lemma 4.1. Let p ≥ 2. Let u be a bounded viscosity supersolution to (1.1) in Ξ. Then, u is a weak supersolution to (1.1) in Ξ . ε ε Proof. Fix a non-negative test function ϕ ∈ C (Ξ ). By Remark 4.8, the function 2 2 φ(x , t ) := u (x , t ) − C (q,ε, u) |x | + t is concave in Ξ and we can approximate it by smooth concave functions φ so that ε j 2 2 φ ,∂ φ , Dφ , D φ → φ, ∂ φ, Dφ, D φ a.e. in Ξ . We define j t j j j t ε 2 2 u (x , t ) := φ (x , t ) + C (q,ε, u) |x | + t . ε, j j Since u is smooth and ϕ is compactly supported in Ξ , we integrate by parts to get ε, j ε (p − 2) p−2 ϕ ∂ u − Du Δu + D u Du , Du t ε, j ε, j ε, j ε, j ε, j ε, j ε Du ε, j − f (Du ) dz ε, j p−2 = ϕ∂ u − ϕ div Du Du − ϕ f (Du ) dz t ε, j ε, j ε, j ε, j p−2 = −u ∂ ϕ + Du Du · Dϕ − ϕ f (Du ) dz. (4.1) ε, j t ε, j ε, j ε, j This implies that (p − 2) p−2 lim inf ϕ ∂ u − Du Δu + D u Du , Du t ε, j ε, j ε, j ε, j ε, j ε, j j →∞ ε Du ε, j Vol. 21 (2021) Equivalence of viscosity and weak solutions 2061 − f (Du ) dz ε, j p−2 ≤ lim −u ∂ ϕ + Du Du · Dϕ − ϕ f (Du ) dz. ε, j t ε, j ε, j ε, j j →∞ We intend to use Fatou’s lemma at the left-hand side and dominated convergence at the right-hand side. Once we verify their assumptions, we arrive at the inequality ϕ ∂ u − Δ u − f (Du ) dz ≤ −u ∂ ϕ t ε p ε ε ε t Ξ Ξ ε ε p−2 + |Du | Du · Dϕ − ϕ f (Du ) dz. ε ε ε The left-hand side is non-negative since by Lemma 4.7 the inf-convolution u is still a viscosity supersolution in Ξ . Consequently u is a weak supersolution in Ξ as ε ε ε desired. It remains to justify our use of Fatou’s lemma and the dominated convergence theorem. It follows from Remark 4.8 that u , ∂ u and Du are uniformly ε, j t ε, j ε, j bounded by some constant M > 0 in the support of ϕ with respect to j. This justifies our use of the dominated convergence theorem. Observe then that since φ is concave, we have D u ≤ C (q,ε, u)I . Hence, ε, j (p − 2) p−2 ∂ u − Du Δu + D u Du , Du − f (Du ) t ε, j ε, j ε, j ε, j ε, j ε, j ε, j Du ε, j p−2 ≥−M − C (q,ε, u)M (N + p − 2) − sup | f (ξ )| . |ξ |≤M The integrand at the left-hand side of (4.1) is therefore bounded from below with respect to j, justifying our use of Fatou’s lemma. Next, we consider the singular case 1 < p < 2. We cannot directly repeat the previ- ous proof because Δ u no longer has a clear meaning at the points where Du = 0. p ε ε To deal with this, we consider the regularized terms p−2 p − 2 Δ u := δ + |Du| Δu + Δ u , (4.2) p,δ ∞ δ + |Du| where Δ u = D uDu, Du . Lemma 4.2. Let 1 < p < 2 . Let u be a bounded viscosity supersolution to (1.1) in Ξ. Then, u is a weak supersolution to (1.1) in Ξ . ε ε Proof. (Step 1) Let ϕ ∈ C (Ξ ) be a non-negative test function. We set 2 2 φ(x , t ) := u (x , t ) − C (q,ε, u) |x | + t , where C (q,ε, u) is the semi-concavity constant of u in Ξ . Then, by Remark 4.8 we ε ε can approximate φ by smooth concave functions φ so that φ ,∂ φ , Dφ , D φ → j j t j j j φ, ∂ φ, Dφ, D φ a.e. in Ξ . We define t ε 2 2 u (x , t ) := φ (x , t ) + C (q,ε, u) |x | + t . ε, j j 2062 J. Siltakoski J. Evol. Equ. Let δ ∈ (0, 1). Since u is smooth and ϕ is compactly supported in Ξ , we calculate ε, j ε via integration by parts p−2 p − 2 2 2 ϕ ∂ u − δ + Du Δu + Δ u − f (Du ) dz t ε, j ε, j ε, j ∞ ε, j ε, j δ + Du ε, j p−2 2 2 = ϕ∂ u − ϕ div δ + Du Du − ϕ f (Du ) dz t ε, j ε, j ε, j ε, j p−2 2 2 = −u ∂ ϕ + δ + Du Du · Dϕ − ϕ f (Du ) dz. ε, j t ε, j ε, j ε, j Recalling the shorthand Δ defined in (4.2), we deduce from the above that p,δ lim inf ϕ ∂ u − Δ u − f (Du ) dz t ε, j p,δ ε, j ε, j j →∞ p−2 2 2 ≤ lim −u ∂ ϕ + δ + Du Du · Dϕ − ϕ f (Du ) dz. (4.3) ε, j t ε, j ε, j ε, j j →∞ We use Fatou’s lemma at the left-hand side and the dominated convergence at the right- hand side. Once we verify their assumptions, we arrive at the auxiliary inequality ϕ ∂ u − Δ u − f (Du ) dz t ε p,δ ε ε p−2 | | ≤ −u ∂ ϕ + δ + Du Du · Dϕ − ϕ f (Du ) dz. (4.4) ε t ε ε ε Next, we verify the assumptions of Fatou’s lemma and the dominated convergence theorem. By Remark 4.8, the functions u , ∂ u and Du are uniformly ε, j t ε, j ε, j bounded by some constant M > 1 in the support of ϕ with respect to j. Hence, the assumptions of the dominated convergence theorem are satisfied. Observe then that the concavity of φ implies that D u ≤ C (q,ε, u)I . Thus, the integrand at the j ε, j left-hand side of (4.3) has a lower bound independent of j when Du = 0. When ε, j Du = 0, we have ε, j p−2 p − 2 2 2 ∂ u − δ + Du Δu + Δ u − f (Du ) t ε, j ε, j ε, j ∞ ε, j ε, j δ + Du ε, j p−2 2 2 δ + Du ε, j p − 2 = ∂ u − Du Δu + Δ u + δΔu t ε, j ε, j ε, j ∞ ε, j ε, j 2 2 δ + Du Du ε, j ε, j − f (Du ) ε, j p−2 2 2 δ + Du ε, j ≥−∂ u − C (q,ε, u) Du (N + p − 2) + δN − f (Du ) t ε, j ε, j ε, j δ + Du ε, j p−2 2 2 ≥−∂ u − C (q,ε, u) δ + Du (2N + p − 2) − f (Du ) t ε, j ε, j ε, j p−2 ≥−M − C (q,ε, u)δ 2N + p − 2 − sup | f (ξ )| , ( ) |ξ |≤M Vol. 21 (2021) Equivalence of viscosity and weak solutions 2063 so that our use of Fatou’s lemma is justified. (Step 2) We let δ → 0 in the auxiliary inequality (4.4). Since u is Lipschitz contin- uous, the dominated convergence theorem implies lim inf ϕ ∂ u − Δ u − f (Du ) dz t ε p,δ ε ε δ→0 p−2 ≤ −u ∂ ϕ + |Du | Du · Dϕ − ϕ f (Du ) dz. (4.5) ε t ε ε ε Applying Fatou’s lemma (we verify assumptions at the end), we get lim inf ϕ ∂ u − Δ u − f (Du ) dz t ε p,δ ε ε δ→0 ≥ lim inf ϕ ∂ u − Δ u − f (Du ) dz t ε p,δ ε ε δ→0 = lim inf ϕ ∂ u − Δ u − f (Du ) dz t ε p,δ ε ε δ→0 Ξ ∩{Du =0} ε ε p−2 + lim inf ϕ(∂ u − δ Δu − f (0)) dz t ε ε δ→0 Ξ ∩{Du =0} ε ε = ϕ ∂ u − Δ u − f (Du ) dz t ε p ε ε Ξ ∩{Du =0} ε ε + ϕ (∂ u − f (0)) dz ≥ 0, (4.6) t ε Ξ ∩{Du =0} ε ε where the last inequality follows from Lemma 4.7 since u is twice differentiable almost everywhere. Combining (4.5) and (4.6), we find that u is a weak supersolution in Ξ . It remains to verify the assumptions of Fatou’s lemma, i.e., that the integrand at the left-hand side of (4.5) has a lower bound independent of δ. When Du = 0, this follows directly from the inequality q−2 q − 1 q−1 D u ≤ |Du | I, ε ε which holds by Lemma 4.6. When Du = 0, we recall that by Lipschitz continuity ∂ u and Du are uniformly bounded in Ξ , and estimate t ε ε ε p−2 p − 2 − δ + |Du | Δu + Δ u ε ε ∞ ε | | δ + Du p−2 δ + |Du | p − 2 =− |Du | Δu + Δ u + δΔu ε ε ∞ ε ε 2 2 δ + |Du | |Du | ε ε p−2 2 2 q−2 q−2 δ + |Du | (q − 1) q−1 q−1 ≥− |Du | (N + p − 2) + |Du | δN ε ε δ + |Du | p−2 q−2 (q − 1) q−1 ≥− δ + |Du | |Du | (2N + p − 2) ε ε ε 2064 J. Siltakoski J. Evol. Equ. q−2 (q − 1) p−2+ q−1 ≥− |Du | (2N + p − 2) q−2 p−2+ (q − 1) q−1 ≥− Du (2N + p − 2) , ε ∞ L (Ξ ) q−2 where we used that p − 2 + > 0. Hence, the assumptions of Fatou’s lemma hold. q−1 If u is the sequence of inf-convolutions of a viscosity supersolution to (1.1), then by next Caccioppoli’s inequality the sequence Du converges weakly in L (Ξ ) up loc to a subsequence. However, we need stronger convergence to pass to the limit under the integral sign of p−2 −u ∂ ϕ + |Du | Du · Dϕ − ϕ f (Du ) dz ≥ 0. ε t ε ε ε For this end, we show in Lemma 4.4 that Du converges in L (Ξ ) for all 1 < r < p. loc Lemma 4.3. (Caccioppoli’s inequality) Let 1 < p < ∞. Assume that u is a locally Lipschitz continuous weak supersolution to (1.1) in Ξ. Then, there is a constant C = C (p,β, C ) such that for any test function ξ ∈ C (Ξ ) we have p p 2 p p p p p−β ξ |Du| dz ≤ C M ∂ ξ + M |Dξ | + (M + M )ξ dz, Ξ Ξ where M = u ∞ . L (spt ξ) Proof. Since u is locally Lipschitz continuous, the function ϕ := (M − u) ξ is an admissible test function. Testing the weak formulation of (1.1) with ϕ yields p p p−1 p−1 ξ |Du| dz ≤ u∂ ϕ + pξ (M − u) |Du| |Dξ | + ϕ f (Du) dz. (4.7) Ξ Ξ We have by integration by parts p p u∂ ϕ dz = −ξ u∂ u + u(M − u)∂ ξ dz t t t Ξ Ξ p 2 p = − ξ ∂ u + u(M − u)∂ ξ dz t t 2 p p 2 p = u ∂ ξ + u(M − u)∂ ξ dz ≤ CM ∂ ξ dz. t t t Ξ Ξ By Young’s inequality, p−1 p−1 p p p p pξ (M − u) |Du| |Dξ | dz ≤ ξ |Du| dz + C (p) M |Dξ | dz. Ξ Ξ Ξ Vol. 21 (2021) Equivalence of viscosity and weak solutions 2065 Using the growth condition (G1) and Young’s inequality, we get ϕ f (Du) dz p β ≤ (M − u) ξ C 1 + |Du| dz p−β β β p = C (M − u) ξ ξ |Du| + C (M − u)ξ dz f f p p p p p−β ≤ ξ |Du| + C (p,β, C ) (M − u) ξ + C (M − u) ξ dz f f p p p p−β ≤ ξ |Du| + C (p,β, C ) M + M ξ dz. Combining these estimates with (4.7) and absorbing the terms with Du to the left-hand side yields the desired inequality. The proof of Lemma 4.4 is based on that of Lemma 5 in [20], see also Theorem 5.3 in [15]. For the convenience of the reader, we give the full details. Lemma 4.4. Let 1 < p < ∞. Suppose that u is a sequence of locally Lipschitz continuous weak supersolutions to (1.1) such that u → uinL (Ξ ). Then, Du j j loc is a Cauchy sequence in L (Ξ ) for any 1 < r < p. loc Proof. Let U Ξ and take a cutoff function θ ∈ C (Ξ ) such that 0 ≤ θ ≤ 1 and θ ≡ 1in U.For δ> 0, we set δ, u − u >δ, j k w = u − u , u − u ≤ δ, jk j k j k −δ, u − u < −δ. j k Then, the function (δ − w )θ is an admissible test function with a time derivative jk since it is Lipschitz continuous. Since u is a weak supersolution, testing the weak formulation of (1.1) with (δ − w )θ yields jk p−2 0 ≤ −u ∂ ((δ − w )θ ) + Du Du · D((δ − w )θ ) − (δ − w )θ f (Du ) dz j t jk j j jk jk j p−2 p−2 = −θ Du Du · Dw + (δ − w ) Du Du · Dθ − (δ − w )θ f (Du ) j j jk jk j j jk j + u ∂ (w )θ − (δ − w )u ∂ θ dz. j t jk jk j t Since w ≤ δ and Dw = χ Du − Du , the above becomes jk jk j k u −u <δ {| | } j k p−2 θ Du Du · Du − Du dz j j j k u −u <δ {| | } j k p−1 ≤ 2δ Du |Dθ | + 2δθ f (Du ) + u ∂ (w )θ + 2δ u |∂ θ | dz. j j j t jk j t Ξ 2066 J. Siltakoski J. Evol. Equ. Since u is a weak supersolution, the same arguments as above but testing this time with (δ + w )θ yield the analogous estimate jk p−2 −θ |Du | Du · Du − Du dz k k j k {|u −u |<δ} j k p−1 ≤ 2δ |Du | |Dθ | + 2δθ | f (Du )| − u ∂ w θ + 2δ |u ||∂ θ | dz. k k k t jk k t Summing up these two inequalities, we arrive at p−2 p−2 θ Du Du − |Du | Du · Du − Du dz j j k k j k {|u −u |<δ} j k p−1 p−1 ≤ 2δ |Dθ | Du + |Du | dz + 2δ θ f (Du ) + | f (Du )| dz j k j k Ξ Ξ + (u − u )∂ w θ dz + 2δ u + |u | |∂ θ | dz j k t jk j k t Ξ Ξ =: I + I + I + I . (4.8) 1 2 3 4 We proceed to estimate these integrals. Denoting M := sup u < ∞,we j ∞ L (spt θ) have by the Caccioppoli’s inequality Lemma 4.3 sup Du dz ≤ C (p,β, C ,θ, M ). (4.9) j f j spt θ The estimate (4.9) and Hölder’s inequality imply that I ≤ δC (p,β, C ,θ, M ). 1 f To estimate I , we also use the growth condition (G1) and the assumption β< p.We get I ≤ 2δ θC (2 + Du + |Du | ) dz ≤ δC (p,β, C ,θ, M ). 2 f j k f The integral I is estimated using integration by parts and that w ≤ δ 3 jk 1 1 2 2 I = θ(u − u )∂ w dz = θ∂ w dz = − w ∂ θ dz ≤ δC (θ , M ). 3 j k t jk t t jk jk 2 2 Ξ Ξ Ξ For the last integral, we have directly I ≤ δC (θ , M ). Combining these estimates with (4.8), we arrive at p−2 p−2 θ Du Du − |Du | Du · Du − Du dz ≤ δC , j j k k j k 0 {|u −u |<δ} j k (4.10) where C = C (p,β, C ,θ, M ).If1 < p < 2, Hölder’s inequality and the algebraic 0 f inequality (3.8) give the estimate (recall that 1 < r < p and θ ≡ 1in U) Du − Du dz j k U ∩{|u −u |<δ} j k Vol. 21 (2021) Equivalence of viscosity and weak solutions 2067 2−r r (2−p) 2(2−r ) ≤ 1 + Du + |Du | dz j k U ∩ u −u <δ {| j k | } Du − Du j k · dz 2−p U ∩{|u −u |<δ} 2 2 j k 1 + Du + |Du | j k ≤ C (p,β, r, C ,θ, M ) p−2 p−2 | | · θ Du Du − Du Du · Du − Du dz , j j k k j k {|u −u |<δ} j k r (2−p) p(2−p) where in the last inequality we also used (4.9) with the knowledge ≤ = (2−r ) 2−p p. If p ≥ 2, Hölder’s inequality and the algebraic inequality (3.12)imply Du − Du dz j k U ∩{|u −u |<δ} j k p−r r p p ≤ 1dz Du − Du dz j k Ξ U ∩{|u −u |<δ} j k p−2 p−2 ≤ C ( p, r ) θ Du Du − |Du | Du · Du − Du dz . j j k k j k u −u <δ {| j k | } Hence, (4.10) leads to max(2,p) Du − Du dz ≤ δ C (p,β, r, C ,θ, M ). j k f U ∩{|u −u |<δ} j k On the other hand, Hölder’s and Tchebysheff’s inequalities with (4.9)imply Du − Du dz j k U ∩{|u −u |≥δ} j k p−r p ≤ U ∩ u − u ≥ δ Du − Du dz j k j k U ∩ u −u ≥δ {| j k | } p−r r − p ≤ δ u − u C (p,β, r, C ,θ, M ). j k p f L (U ) So we arrive at r p−r r −p max(2,p) Du − Du dz ≤ (δ + δ u − u )C (p,β, r, C ,θ, M ). j k j k f L (U ) Taking first small δ> 0 and then large j, k, we can make the right-hand side arbitrarily small. Now we are ready to prove the main result of this section which states that bounded viscosity supersolutions are weak supersolutions. Theorem 4.5. Let 1 < p < ∞. Let u be a bounded viscosity supersolution to (1.1) in Ξ. Then, u is a weak supersolution to (1.1) in Ξ. 2068 J. Siltakoski J. Evol. Equ. Proof. Fix a non-negative test function ϕ ∈ C (Ξ ) and take an open cylinder Ω Ξ such that spt ϕ Ω .Let ε> 0 be so small that Ω Ξ . Then, t ,t t ,t t ,t ε 1 2 1 2 1 2 Lemma 4.2 implies that u is a weak supersolution to (1.1)in Ξ . Therefore, by the ε ε Caccioppoli’s inequality Lemma 4.3, Du is bounded in L (Ω ). Hence, Du con- ε t ,t ε 1 2 p p verges weakly in L (Ω ) up to a subsequence. Since also u → u in L (Ω ) by t ,t ε t ,t 1 2 1 2 dominated convergence and the fact that u → u pointwise in Ω , it follows that ε t ,t 1 2 p 1,p u ∈ L (t , t ; W (Ω)). 1 2 Since u is a weak supersolution, it remains to show that up to a subsequence p−2 p−2 lim u ∂ ϕ + |Du | Du · Dϕ dz = u∂ ϕ + |Du| Du · Dϕ dz ε t ε ε t ε→0 Ω Ω t ,t t ,t 1 2 1 2 (4.11) and lim ϕ f (Du ) dz = ϕ f (Du) dz. (4.12) ε→0 Ω Ω t ,t t ,t 1 2 1 2 p r Since u → u in L (Ω ) and Du → Du in L (Ω ) for any 1 < r < p by ε t ,t ε t ,t 1 2 1 2 Lemma 4.4, the claim (4.11) follows by applying the vector inequality (see [19, pp. 95–96]) 2−p p−1 | | 2 a − b when p < 2, p−2 p−2 |a| a − |b| b ≤ −1 p−2 p−2 2 |a| + |b| |a − b| when p ≥ 2. To show (4.12), let M ≥ 1 and write using the growth condition (G1) | f (Du ) − f (Du)| dz t ,t 1 2 β β ≤ | f (Du ) − f (Du)| dz + C (2 + |Du | + |Du| ) dz ε f ε {|Du |<M } {|Du |≥M } ε ε =: I + I . 1 2 Then, by Hölder’s inequality p p β p−β 2 |Du | |Du | |Du| |Du | ε ε ε I = C + + dz 2 f p−β p−β | | Du |Du | |Du | {|Du |≥M } ε ε ε ε 2 1 1 p β p−β ≤ C + Du + C Du Du p p p f ε f ε L (Ω ) L (Ω ) L (Ω ) p p−β t ,t p−β t ,t t ,t M M 1 2 M 1 2 1 2 ≤ C ( p,β, C , Du p , sup Du p ). f ε L (Ω ) L (Ω ) p−β t ,t t ,t 1 2 1 2 On the other hand, we have | f (Du ) − f (Du)| → 0a.e.in Ω up to a subse- ε t ,t 1 2 quence and the integrand in I is dominated by an integrable function since the growth condition (G1) implies β β | f (Du ) − f (Du)| ≤ C (2 + |M | + |Du| ) when |Du | < M. ε f ε Vol. 21 (2021) Equivalence of viscosity and weak solutions 2069 Hence, for any M ≥ 1, we have I → 0as ε → 0 by the dominated convergence theorem. By taking first large M ≥ 1 and then small ε> 0, we can make I + I 1 2 arbitrarily small. The rest of this section is devoted to the properties of the inf-convolution. The facts in the following lemma are well known, see, e.g., [6,11,14]or[24]. Lemma 4.6. Assume that u : Ξ → R is lower semicontinuous and bounded. Then, u has the following properties. (i) We have u ≤ uin Ξ and u → u pointwise as ε → 0. ε ε q−1 N +1 (ii) Denote r (ε) := qε osc u ,t (ε) := (2ε osc u) .For (x , t ) ∈ R , Ξ Ξ set Ξ := (x , t ) ∈ Ξ : B (x ) × (t − t (ε), t + t (ε)) Ξ . ε r (ε) Then, for any (x , t ) ∈ Ξ there exists (x , t ) ∈ B (x ) ×[t − t (ε), t + t (ε)] ε ε ε r (ε) such that q 2 |x − x | |t − t | ε ε u (x , t ) = u(x , t ) + + . ε ε ε q−1 qε 2ε (iii) The function u is semi-concave in Ξ with a semi-concavity constant depending ε ε only on u, q and ε. (iv) Assume that u is differentiable in time and twice differentiable in space at (x , t ) ∈ Ξ . Then, t − t ∂ u (x , t ) = , t ε q−2 |x − x | Du (x , t ) = (x − x ) , ε ε q−1 q−2 q − 1 q−1 D u (x , t ) ≤ |Du | I. ε ε Next, we show that the inf-convolution of a viscosity supersolution to (1.1) is still a supersolution in the smaller set Ξ . Since the inf-convolution is “flat enough,” that is, since q > p/(p − 1), the inf-convolution essentially cancels the singularity of the p-Laplace operator. This allows us to extract information on the time derivative at those points of differentiability where Du vanishes. Lemma 4.7. Let 1 < p < ∞. Let u be a bounded viscosity supersolution to (1.1) in Ξ. Then, the inf-convolution u is also a viscosity supersolution to (1.1) in Ξ . ε ε Moreover, if u is differentiable in time and twice differentiable in space at (x , t ) ∈ Ξ and Du (x , t ) = 0, then ∂ u (x , t ) − f (0) ≥ 0. ε ε t ε Proof. Assume that ϕ touches u from below at (x , t ) ∈ Ξ .Let (x , t ) be likeinthe ε ε ε ε property (ii) of Lemma 4.6. Then, q 2 |x − x | |t − t | ε ε ϕ(x , t ) = u (x , t ) = u(x , t ) + + , (4.13) ε ε ε q−1 qε 2ε 2070 J. Siltakoski J. Evol. Equ. q 2 |y − z| |τ − s| ϕ(y,τ) ≤ u (y,τ) ≤ u(z, s) + + for all (y,τ), (z, s) ∈ Ξ. q−1 qε 2ε (4.14) Set q 2 |x − x | |t − t | ε ε ψ(z, s) := ϕ(z + x − x , s + t − t ) − − . ε ε q−1 qε 2ε Then, ψ touches u from below at (x , t ) since by (4.13) ε ε q 2 |x − x | |t − t | ε ε ψ(x t ) =ϕ(x , t ) − − = u(x , t ) ε, ε ε ε q−1 qε 2ε and selecting (y,τ) = (z + x − x , s + t − t ) in (4.14)gives ε ε q 2 |x − x | |t − t | ε ε ψ(z, s) = ϕ(z + x − x , s + t − t ) − − ≤ u(z, s). ε ε q−1 qε 2ε Since u is a viscosity supersolution, it follows that 0 ≤ lim sup ∂ ψ(z, s) − Δ ψ(z, s) − f (Dψ(z, s)) s p (z,s)→(x ,t ) ε ε z =x = lim sup ∂ ϕ(z, s) − Δ ϕ(z, s) − f (Dϕ(z, s)) , s p (z,s)→(x ,t ) z =x and the first claim is proven. To prove the second claim, assume that u is differentiable in time and twice differentiable in space at (x , t ) ∈ Ξ and Du (x , t ) = 0. By the ε ε property (iv) in Lemma 4.6,wehave x = x , so that |t − t | u (x , t ) = u(x , t ) + . ε ε 2ε Hence, by the definition of inf-convolution q 2 2 |x − y| |t − s| |t − t | u(y, s) + + ≥ u (x , t ) = u(x , t ) + for all (y, s) ∈ Ξ. ε ε q−1 qε 2ε 2ε Arranging the terms as q 2 2 |x − y| |t − s| |t − t | u(y, s) ≥ u(x , t ) − − + =: φ(y, s), q−1 qε 2ε 2ε we see that the function φ touches u from below at (x , t ). Since u is a viscosity supersolution and Dφ(y, s) = 0 when y = x,wehave lim sup ∂ φ(y, s) − Δ φ(y, s) − f (Dφ(y, s)) ≥ 0. s p (y,s)→(x ,t ) y =x Vol. 21 (2021) Equivalence of viscosity and weak solutions 2071 On the other hand, since q > p/(p − 1),wehave Δ φ(y, s) → 0as y → x. Hence, we get t − t 0 ≤ ∂ φ(x , t ) − f (0) = − f (0) = ∂ u (x , t ) − f (0), s ε t ε where the last equality follows from the property (iv) in Lemma 4.6. Remark 4.8. Semi-concavity implies that the inf-convolution u is locally Lipschitz in Ξ (see [8, p. 267]). Therefore, u is differentiable almost everywhere in Ξ , ε ε ε ∞ ∞ 1,∞ ∂ u ∈ L (Ξ ) and u ∈ L (t , t ; W (Ω)) for any Ω Ξ (see [8, p. 266]). t ε ε ε 1 2 t ,t ε 1 2 loc 2 2 Moreover, since the function φ(x , t ) := u (x , t )−C (q,ε, u)(|x | +|t | ) is concave, Alexandrov’s theorem implies that u is twice differentiable almost everywhere in Ξ . ε ε Furthermore, the proof of Alexandrov’s theorem in [8, p. 273] establishes that if φ is 2 2 the standard mollification of φ, then D φ → D φ almost everywhere in Ξ . j ε 5. Lower semicontinuity of supersolutions We show the lower semicontinuity of weak supersolutions when p ≥ 2 and the function f ∈ C (R ) satisfies that f (0) = 0 as well as the stronger growth condition p−1 | f (ξ )| ≤ C 1 + |ξ | . (G2) Our proof follows the method of Kuusi [17], but the first-order term causes some modifications. In particular, our essential supremum estimate is slightly different, see Theorem 5.3 and the brief discussion before it. The assumption f (0) = 0is used to ensure that the positive part u of a subsolution is still a subsolution. We begin by proving estimates for the essential supremum of a subsolution using Moser’s iteration technique. We first need the following Caccioppoli’s inequalities. Lemma 5.1. (Caccioppoli’s inequalities) Assume that p ≥ 2 and that (G2) holds. Sup- p−1+λ pose that u is a non-negative weak subsolution to (1.1) in Ω and u ∈ L (Ω ) t ,t t ,t 1 2 1 2 for some λ ≥ 1. Then, there exists a constant C = C (p, C ) that satisfies the estimates 1+λ p ess sup u (x,τ)ζ (x,τ) dx t <τ <t 1 2 Ω p−1+λ p 1+λ p−1 λ p−1+λ p ≤ C λu |Dζ | + u |∂ ζ | ζ + λ u + u ζ dz t ,t 1 2 and p−1+λ D(u ζ) dz t ,t 1 2 p p−1+λ p p−1 1+λ p−1 p λ p−1+λ p ≤ C λ u |Dζ | + λ u |∂ ζ | ζ + λ u + u ζ dz t ,t 1 2 for all non-negative ζ ∈ C (Ω ×[t , t ]) such that spt ζ(·, t ) Ω and ζ(x , t ) = 0. 1 2 1 λ−1 Proof. We test the regularized equation in Lemma (3.1) with ϕ := min(u , k) u ζ η, where η is the following cutoff function 2072 J. Siltakoski J. Evol. Equ. 0, t ∈ (t , s − h), ⎪ 1 ⎪ (t − s + h)/2h, t ∈[s − h, s + h], η(t ) = 1, t ∈ (s + h,τ − h), (−t + τ + h)/2h, t ∈[τ − h,τ + h], 0, t ∈ (τ + h, t ), λ−1 and t < s <τ < t , h > 0. We denote g(l) := min(r, k) r dr. Then, integration 1 2 by parts and Lebesgue’s differentiation theorem yield for a.e. s,τ ∈ (t , t ) 1 2 λ−1 p ∂ (u ) min(u , k) u ζ η dz t ,t 1 2 = ∂ g(u )ζ η dz t ,t 1 2 p p = −ηg(u )∂ (ζ ) − ζ g(u )∂ η dz t t t ,t 1 2 p p → −g(u)∂ (ζ ) dz − ζ (x , s)g(u(x , s)) dx →0,h→0 Ω Ω s,τ + ζ (x,τ)g(u(x,τ)) dx . Letting s → t and observing that the other terms of (3.1) converge as well, we obtain for a.e. τ ∈ (t , t ) that 1 2 g(u(x , τ ))ζ (x,τ) dx λ−1 λ−1 p p−2 p p ≤ g(u)∂ (ζ ) − |Du| Du · D(u uζ ) + u uζ f (Du) dz, k k t ,τ where we have denoted u := min(u, k). Since λ−1 λ−2 Du = χ (λ − 1)u Du, {u<k} we have by Young’s inequality λ−1 λ−1 p−2 p p λ−1 p − |Du| Du · D(u uζ ) ≤−ζ (λ − 1)χ u + u |Du| {u<k} k k p−1 λ−1 p−1 + pζ u u |Du| |Dζ | p λ−1 p p−1+λ p ≤− ζ u |Du| + C (p)u |Dζ | . Moreover, by the growth condition (G2) and Young’s inequality λ−1 p p λ−1 p λ−1 p−1 u uζ f (Du) ≤ C ζ u u + C ζ u u |Du| f f k k k p λ−1 p p−1+λ p λ−1 p ≤ C ζ u + C (p, C )ζ u + ζ u |Du| . f f 4 Vol. 21 (2021) Equivalence of viscosity and weak solutions 2073 Collecting the estimates, moving the terms with Du to the left-hand side and letting k →∞, we arrive at −1 λ+1 p p λ−1 p λ u ζ (x,τ) dx + ζ u |Du| dz Ω Ω t ,τ −1 λ+1 p p−1+λ p p λ−1 p−1+λ ≤ C (p, C ) λ u ∂ ζ + u |Dζ | + ζ (u + u ) dz. f t t ,τ (5.1) Since the integrals are positive, this yields the first inequality of the lemma by taking essential supremum over τ . The second inequality follows from (5.1) by using that p−1+λ p−1+λ p p p λ−1 p D(u ζ) dz ≤ C (p) u |Dζ | + λ ζ u |Du| dz. Ω Ω t ,t t ,t 1 2 1 2 We first prove the following essential supremum estimate where we assume that the subsolution is bounded away from zero. Lemma 5.2. Assume that p ≥ 2 and that (G2) holds. Suppose that u is a weak subsolution to (1.1) in Ξ and B (x ) × (t − T , t ) Ξ where R, T < 1 are such R 0 0 0 that p p p−1 R R ≤ 1 and u ≥ . (5.2) T T Then, there exists a constant C (N , p, C ) such that 1/δ −N −p p−2+δ ess sup u ≤ C (1 − σ) − u dz B p B (x )×(t −T ,t ) R 0 0 0 σ R(x )×(t −σ T ,t ) 0 0 0 for every 1/2 ≤ σ< 1 and 1 <δ < 2. Proof. Let σ R ≤ s < S < R.For j ∈ 0, 1, 2,...,weset − j R := S − (S − s) (1 − 2 ) and U := B × Γ := B (x ) × (t − (R /S) T , t ). j j j R 0 0 j 0 We choose test functions ϕ ∈ C (U ) such that spt ϕ (·, t ) B (x ), j j j R 0 0 ≤ ϕ ≤ 1,ϕ ≡ 0on ∂ U ,ϕ ≡ 1in U j j p j j j +1 and C R C j jp Dϕ ≤ 2 , ∂ ϕ ≤ 2 . j t j S − s T (S − s) 2074 J. Siltakoski J. Evol. Equ. We set γ := 1 + p/N and λ := 2γ − 1, j = 0, 1, 2,... . p−1+λ Assuming that we already know that u ∈ L (U ), then we have by a parabolic Sobolev’s inequality (see [7, p7]) κp β/p κα α/p u dz ≤ u ϕ dz U U j +1 j p/N p (κ−1)N β/p β/p α/p α/p ≤ C (N , p) D(u ϕ ) dz ess sup u ϕ dx , j j U Γ B j j j (5.3) where p(1 + λ ) p(p − 1 + λ ) j j α = p − 1 + λ ,κ = 1 + ,β = . N (p − 1 + λ ) 1 + λ j j ThefirstestimateinLemma 5.1 gives (κ−1)N β/p p α/p 1+λ ess sup u ϕ dx = ess sup u ϕ dx j j Γ B Γ B j j j j p−1 p p−1+λ 1+λ λ p−1+λ j j j j ≤ C λ u Dϕ + u ∂ ϕ ϕ + u + u ϕ dz. j j t j j j (5.4) β/p Using the second estimate with ζ = ϕ , we obtain β/p α/p D(u ϕ ) dz p p p−1 p p−1+λ 1+λ λ p−1+λ j j j j ≤ C λ u Dϕ + u ∂ ϕ ϕ + u + u ϕ dz. j t j j j j (5.5) Combining (5.3) with (5.4) and (5.5), we arrive at jp p jp 2 R 2 κα p−1+λ 1+λ λ j j j u dz ≤ C λ u + u + u dz, j p (S − s) T (S − s) U U j +1 j (5.6) where γ = 1 + p/N . We wish to iterate this inequality, but having multiple terms at the right-hand side is a problem. This is where the assumption (5.2) comes into play. p 1/( p−1) Since u ≥ (R /T ) ,wehave p−1 p−1 p−1 1 T 1 λ p−1+λ p−1+λ p−1+λ j j j j u = u ≤ u ≤ u p p u R (S − s) Vol. 21 (2021) Equivalence of viscosity and weak solutions 2075 and since T /R ≥ 1, we have also p−2 p−2 p−1 1 T T 1+λ p−1+λ p−1+λ p−1+λ j j j j u = u ≤ u ≤ u . p p u R R Using these estimates it follows from (5.6) that jp C λ 2 κα p−1+λ u dz ≤ u dz. (5.7) (S − s) U U j +1 j Observe that κα = p − 1 + λ (1 + p/N ) + p/N = p − 1 + λ . j j +1 −p Hence, by denoting Y := C (S − s) , the inequality (5.7) becomes p−1+λ jp p−1+λ j +1 j u dz ≤ Y (2γ) u dz. U U j +1 j We iterate this inequality. When j = 0, it reads as p−1+λ p u dz ≤ Y u dz. U U 1 0 Then, when j = 1, we have 1 1 2 γ 1 1 1 1 p 1+ p p−1+λ p−1+λ p 2 1 γ γ γ γ u dz ≤ Y (2γ) u dz ≤ Y (2γ) u dz. U U U 2 1 0 Continuing this way, we arrive at j +1 γ 1 1 1 2 1+ +...+ p( + +...+ ) γ j γ 2 j p−1+λ p j +1 γ γ γ u dz ≤ Y (2γ) u dz U U j +1 0 ≤ CY u dz, so that j +1 p−1+λ j +1 p−1+λ j +1 p−1+λ p j +1 u dz ≤ CY u dz . U U j +1 0 j +1 Since γ /(p − 1 + λ ) → 1/2 and p − 1 + λ →∞ as j →∞, we obtain j +1 j +1 that 1/2 −N −p p ess sup u ≤ C (S − s) u dz , Q(s) Q(S) 2076 J. Siltakoski J. Evol. Equ. where Q(s) = B(x , s) × (t − (s/S) T , t ). By Young’s inequality, we have for 0 0 0 every 1 <δ < 2 that 1/2 2−δ −N −p p−2+δ ess sup u ≤ ess sup u (S − s) u dz Q(s) Q(S) Q(S) 1/δ −N −p p−2+δ ≤ ess sup u + (S − s) u dz . (5.8) Q(S) B (x )×(t −T ,t ) R 0 0 0 A standard iteration argument such as [9, Lemma 1.1] now finishes the proof. Indeed, if f :[T , T ]→ R is a non-negative bounded function such that all T ≤ t ≤ τ ≤ T 0 1 0 1 satisfy −η f (t ) ≤ θ f (τ ) + (τ − t ) A, (5.9) where A,θ,η ≥ 0 with θ< 1, then −η f (T ) ≤ C (η, θ )(T − T ) A. 0 1 0 Selecting T := σ R, T := (σ R + R) /2 and the other variables so that (5.8) implies 0 1 (5.9), we get the desired estimate. Next, we consider the case where the non-negative subsolution is not necessarily bounded away from zero. Observe that the estimate differs from the usual estimate for the p-Laplacian because of the power 1/(p − 1) in the first term (cf. [7, Theorem 4.1] or [17, Theorem 3.4]). However, we have the additional assumption (5.10). Theorem 5.3. Assume that p ≥ 2 and that (G2) holds. Suppose that u is a non- negative weak subsolution to (1.1) in Ξ and B (x ) × (t − T , t ) Ξ with R, T < 1 R 0 0 0 such that ≤ 1. (5.10) Then, there exists a constant C = C (N , p, C ,δ) such that we have the estimate ess sup u B(x ,R/2)×(t −T /2 ,t ) 0 0 0 1 δ−1 1 p t 0 δ p−1 δ R T p−2+δ ≤ C + C − − u dx dt T R t −T B (x ) 0 R 0 for all 1 <δ < 2. Proof. We denote p−1 −N −p Λ := (1 − σ) ,θ := . T Vol. 21 (2021) Equivalence of viscosity and weak solutions 2077 Using Lemma 5.2 on the subsolution v := θ + u, we get the estimate p−2+δ ess sup u ≤ C Λ − (θ + u) dz B p B (x )×(t −T ,t ) R 0 0 0 σ R(x )×(t −σ T ,t ) 0 0 0 1 T p−2+δ ≤ C Λ θ 1 T p−2+δ + C Λ − u dz , B (x )×(t −T ,t ) 0 0 0 where δ−1 p−2+δ p(p−2+δ) p−1 T R p−1 p−2+δ 1− −p+ 1−δ p δ−1 ( ) p−1 p−1 θ = T R = T R = . R T Taking σ = 1/2 now yields the desired inequality. Lemma 5.4. Assume that p ≥ 2 and that f (0) = 0. Let u be a weak subsolution to (1.1) in Ω . Then, u = max(u, 0) is also a weak subsolution. t ,t + 1 2 Proof. Fix a non-negative test function ζ ∈ C (Ω ). We test the regularized t ,t 0 1 2 equation in Lemma 3.1 with min {k(u ) , 1} ζ . Then, by similar arguments as in the proof of Lemma 5.1 we get the estimate p−2 min {ku , 1} (−u∂ ζ + |Du| Du · Dζ − ζ f (Du)) dz + t t ,t 1 2 2 p ≤− (min {ku , 1}) ∂ ζ dz − k ζ |Du| dz. + t 2k Ω {0<ku<1} t ,t 1 2 Letting k →∞ this implies p−2 −u∂ ζ + |Du| Du · Dζ − ζ f (Du) dz ≤ 0. {u>0} Since f (0) = 0 and u ∂ ζ = 0 = Du a.e. in {u ≤ 0}, we get that + t + p−2 | | −u ∂ ζ + Du Du · Dζ − ζ f (Du ) dz ≤ 0. + t + + + t ,t 1 2 Theorem 5.5. Assume that p ≥ 2, (G2) holds and that f (0) = 0. Suppose that u is a weak supersolution to (1.1) in Ξ. Let u denote the lower semicontinuous regularization of u, that is, u (x , t ) := ess lim inf u(y, s) := lim ess inf u. p p (y,s)→(x ,t ) R→0 B (x )×(t −R ,t +R ) Then, u = u almost everywhere. ∗ 2078 J. Siltakoski J. Evol. Equ. Proof. For all M ∈ N, we define the cylinders M p p Q (x , t ) := B (x ) × (t − MR , t + MR ). We denote by E the set of Lebesgue points with respect to the basis {Q }, that is, p− E := (x , t ) ∈ Ξ : lim − |u(x , t ) − u(y, s)| dy ds = 0 . R→0 Q (x ,t ) Then, E ⊂ E so that M M +1 E := E = E . M 1 M ∈N Moreover, we have |E | = |Ξ |, which follows from [26, p. 13] by a simple argument, see for example [8,p.54]. We now claim that if (x , t ) ∈ E, then 0 0 u(x , t ) ≤ ess lim inf u(x , t ). (5.11) 0 0 (x ,t )→(x ,t ) 0 0 We make the counter assumption u(x , t ) − ess lim inf u(x , t ) = ε> 0. 0 0 (x ,t )→(x ,t ) 0 0 Let R be a radius such that ess lim inf u(x , t ) − ess inf u ≤ ε/2 (x ,t )→(x ,t ) 0 0 Q (x ,t ) 0 0 for all 0 < R ≤ R . For such R,wehave u(x , t ) − ess inf u ≥ ε/2. (5.12) 0 0 Q (x ,t ) 0 0 We set v := (u(x , t ) − u) . Since (x , t ) ∈ E, we find for any M ∈ N a radius 0 0 + 0 0 R = R (M ) such that 1 1 1 1 1 p− p− 2 2 − v dx dt ≤ − |u(x , t ) − u| dx dt ≤ . (5.13) 0 0 M M Q (x ,t ) Q (x ,t ) 0 0 0 0 R R 1 1 On the other hand, by Lemma 5.4 the function v is a weak subsolution to ∂ v + Δ v − g(Dv) ≤ 0, t p where g(ξ ) =− f (−ξ). Observe also that the cylinder Q (x , t ) satisfies the con- 0 0 p p dition (5.10) since R /(MR ) ≤ 1. Hence, we may apply Theorem 5.3 with δ = 3/2 1 1 and then use (5.13) to get 1 2 p 3(p−1) p 3 R R M 1 1 1 p− ess sup v ≤ C + C − v dx dt p p M R M R Q (x ,t ) 0 0 Q (x ,t ) 1 1 0 0 R (R )/2 1 1 Vol. 21 (2021) Equivalence of viscosity and weak solutions 2079 C 1 ≤ + C M · 3( p−1) 2 M M ≤ C . Now we first fix M so large that C/M ≤ ε/4 and this will also fix R . Then, we 1 M take R ∈ (0, R ] so small that Q (x , t ) ⊂ Q (x , t ). Then, (5.12) leads to a 0 0 0 0 0 R (R )/2 contradiction since ε/4 ≥ ess sup v ≥ ess sup v ≥ u(x , t ) − ess inf u ≥ ε/2. 0 0 M 1 Q (x ,t ) 0 0 Q (x ,t ) Q (x ,t ) 0 0 0 0 (R )/2 R Hence, (5.11) holds and we have u(x , t ) ≤ ess lim inf u(x , t ) ≤ lim − u(x , t ) dx dt = u(x , t ). 0 0 0 0 R→0 1 (x ,t )→(x ,t ) 0 0 Thus, u = u almost everywhere and it is easy to show that u is lower semicontinuous. ∗ ∗ Funding Open Access funding provided by University of Jyväskylä (JYU). Open Access. This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/ by/4.0/. Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. REFERENCES 1,α [1] A. Attouchi, M. Parviainen, and E. Ruosteenoja. C regularity for the normalized p-Poisson problem. J. Math. Pures Appl., 108(4):553–591, 2017. [2] A. Attouchi. Well-posedness and gradient blow-up estimate near the boundary for a Hamilton-Jacobi equation with degenerate diffusion. J. Diff. Eq., 253(8):2474–2492, 2012. [3] J. Benedikt, P. Girg, L. 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[23] P. Pucci and J. Serrin. The maximum principle, volume 73 of Progress in non-linear differential equations and their applications. Birkhäuser, Boston, 2007. [24] M. Parviainen and J. L. Vázquez. Equivalence between radial solutions of different parabolic gradient-diffusion equations and applications. To appear in Ann. Scuola Norm. Sup. Pisa Cl. Sci. [25] J. Siltakoski. Equivalence of viscosity and weak solutions for the normalized p(x )-Laplacian. Calc. Var. Partial Differential Equations, 57(95), 2018. [26] E. M. Stein. Harmonic analysis: Real-variable methods, orthogonality and oscillatory integrals. Princeton University Press, 1993. Jarkko Siltakoski Department of Mathematics and Statistics University of Jyväskylä P.O. Box 35 40014 Jyvaskyla Finland E-mail: jarkko.j.m.siltakoski@jyu.fi Accepted: 19 December 2020
Journal of Evolution Equations – Springer Journals
Published: Jan 18, 2021
Keywords: Comparison principle; Gradient term; Parabolic p-Laplacian; Viscosity solution; Weak solution; 35K92; 35J60; 35D40; 35D30; 35B51
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