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Nonlinear Differential Equations in the Teaching Model of Educational Informatisation

Nonlinear Differential Equations in the Teaching Model of Educational Informatisation 1IntroductionThe rapid development of information technology continues to impact the traditional-closed, one-way transmission course teaching mode. This is also rapidly changing the learning lifestyle of students and changing teachers’ traditional teaching concepts and teaching models. How to create a new curriculum teaching model according to the teaching objectives of the curriculum and the characteristics of information technology is an important research topic for teachers in the information age [1]. In the teaching practice in recent years, we have actively explored the research teaching model with the characteristics of the times. The use of information technology to create a research teaching model is an excellent example of the combination of information technology and curriculum teaching reform.Nonlinear phenomena are common in nature and social life. Many significant issues such as engineering technology and natural phenomena can be attributed to nonlinear equations. Therefore, using nonlinear equations to study the objective world is an inevitable way. With the development of social sciences and natural sciences, the solution of nonlinear differential equations, especially the analysis of the exact solutions, is one of the most important ways to solve practical problems [2]. It is widely used in various fields such as physics, engineering technology and applied mathematics. Therefore, it is necessary to strengthen the research on the exact solution of nonlinear differential equations to make it serve the reality. There are many functional methods for solving nonlinear differential equations, including the homogeneous balance method, backscattering method, Darboux transformation method, variable minimum separation method, Painleve expansion method and Tanh-function method. These exact solutions have extensively promoted the development of nonlinear equation theory and practical applications. Tanh-function method is one of the very effective methods to construct exact solutions of nonlinear equations. In this regard, this article takes the nonlinear differential theory as the starting point and focuses on the Tanh-function method and its promotion. At the same time, we use this method to solve the common three kinds of nonlinear differential equations commonly used.2The overall goal of the teaching planAccording to the characteristics of the content structure of the ordinary differential equations course, our overall goal of designing the teaching plan is as follows: first, through the teaching of ordinary differential equations, we must teach students the necessary basic knowledge, and more importantly, cultivate students’ abilities [3]. We want students to learn, learn to think, learn to acquire information, learn to analyse and solve problems. Second, we must select typical materials that are closely related to the textbook's content and use multimedia technology to create a learning situation that can reflect the development and development of mathematical concepts, conclusions and thinking methods. In this way, students can master and use mathematical thinking methods to ‘discover’ new mathematical problems independently to guide students to conduct study and research and cultivate creativity and innovation. Figure 1 shows the basic structure of the information teaching classroom.Fig. 1The basic structure of the informatisation teaching classroom3The specific implementation of the teaching planThe key to research teaching is the design of the research teaching plan. ‘Ordinary differential equations’ is a professional introductory course for mathematics and applied mathematics [4]. To meet the rapid development of higher education, we have to focus on the reforms on the integration of modern information technology and the teaching mode of ordinary differential equations.3.1Analysis of the students’ existing foundationStudents have taken courses such as ‘Mathematical Analysis’, ‘Advanced Algebra’, ‘Analytic Geometry’ and ‘General Physics’ before taking the course of ordinary differential equations [5]. Thus, they have a particular foundation in mathematics and applied mathematics. Students have also studied ‘Computer Application Fundamentals’, ‘Algorithm Language’ and other educational courses and have a thorough understanding of nonlinear differential equation courseware functions.3.2Arrangement of teaching content under the new teaching modelWe use the second edition of ‘Ordinary Differential Equations’ compiled by the Department of Mathematics of Northeast Normal University as the introductory textbook. In the organising of teaching content, the teaching materials are scientifically processed following the actual situation of students majoring in mathematics and applied mathematics [6]. Thus, for example, the methods and theoretical applications of solving differential equations are regarded as the content of the intensive lecture.4Tanh-function method introductionThe Tanh-function expansion method is the abbreviation of the hyperbolic tangent function expansion method. This function method is currently one of the most effective methods for constructing nonlinear differential equations for precise solutions [7]. In solving nonlinear functions, people find that the solitary wave solutions of many nonlinear differential equations, including the Burgers equation, KdV equation, etc., can be represented by polynomials in the form of Tanh functions. This inspired people to use nonlinear differential equations to construct solitary wave solutions with Tanh functions. This is also the origin of the Tanh-function expansion method.This section first briefly introduces the expansion method of the Tanh function. Only after understanding and mastering can the application of accurate solutions be better carried out. Figure 2 shows the expanded form of the Tanh function [8]. The specific description of the Tanh-function expansion method is to give a nonlinear differential equation as shown in equation (1)(1)p(u,ux,uy,ut,uxx,uxy,uxt,uyy,⋯)=0p(u,{u_x},{u_y},{u_t},{u_{xx}},{u_{xy}},{u_{xt}},{u_{yy}}, \cdots ) = 0Among them, P is a polynomial of the derivatives of u.Fig. 2Tanh-function expanded formThe first step assumes that the travelling wave solution of the equation has the following form, and the travelling wave change has (2)u(x,y,t)=u(ξ),ξ=lx+ky+ctu(x,y,t) = u(\xi ),\xi = lx + ky + ctAmong them l,k,c are all undetermined constants. We can use the above formula to reduce the formula (1) to become the ordinary differential equation of u(ξ). As shown in formula (3)(3)p(u,u′,u″,u‴,⋯)=0p(u,{u^{\prime}},{u{''}},{u{'''}}, \cdots ) = 0Among them, u′,u″,u‴, ⋯ represents the derivative of u to ξ, respectively.The second step hypothesis is (4)Y=tanh(ξ), ξ=lx+ky+ctY = \tanh (\xi ),\quad \xi = lx + ky + ctddξ=(1−Y2)ddY{d \over {d\xi }} = (1 - {Y^2}){d \over {dY}}, d2dξ2=(1−Y2)(−2YddY+(1−Y2)d2dY2){{{d^2}} \over {d{\xi ^2}}} = (1 - {Y^2})( - 2Y{d \over {dY}} + (1 - {Y^2}){{{d^2}} \over {d{Y^2}}}), d3dξ3=(1−Y3)(6Y2−2)ddY−6Y(1−Y2)d2dY2+(1−Y2)2d3dY3{{{d^3}} \over {d{\xi ^3}}} = (1 - {Y^3})(6{Y^2} - 2){d \over {dY}} - 6Y(1 - {Y^2}){{{d^2}} \over {d{Y^2}}} + {(1 - {Y^2})^2}{{{d^3}} \over {d{Y^3}}}can be obtained by formula (4). Then suppose that the finite series form of Y (ξ) can be expressed by u(ξ). Then, there is (5)u(ξ)=∑i=0maiYi+∑i=0ma−iY−iu(\xi ) = \sum\limits_{i = 0}^m {a_i}{Y^i} + \sum\limits_{i = 0}^m {a_{ - i}}{Y^{ - i}}Among them m,ai(i = −m ⋯ m) is a constant to be determined.The third step is to determine the value of m. Generally, a positive integer can be obtained by balancing the nonlinear term and higher-order derivative term of the nonlinear differential equation (1).The fourth step is to introduce formula (5) into formula (3) and simplify it to be that all coefficients of Y (ξ) are all zero [9]. In this way, several nonlinear algebraic equations about the coefficients of ai, l, k and c can be obtained.In the fifth step, we use Maple to solve the above algebraic equations. In this way, the precise expression result of each coefficient can be obtained [10]. Finally, these results are re-introduced into the formula (3) and the exact solution of the nonlinear equation (1) can be finally obtained. The exact solution is represented by Y (ξ).5Application examples of Tanh-function method5.1Accurately solve the coupled KdV equationThe expression of the coupled KdV equations is shown in formula (6)(6){ut+6αuux−υυx+βuxxx=0υt+3αuux+αυxxx=0\left\{ {\matrix{ {{u_t} + 6\alpha u{u_x} - \upsilon {\upsilon _x} + \beta {u_{xxx}} = 0} \hfill \cr {{\upsilon _t} + 3\alpha u{u_x} + \alpha {\upsilon _{xxx}} = 0} \hfill \cr } } \right.Assuming that u(x,t) = u(ξ), υ(x,t) = υ(ξ), ξ = x + ct will be substituted into the coupled KdV equation, the u(ξ) ODE can be obtained as follows:(7){cu′+6αuu′−υυ′+βu‴=0cυ′+3αuυ′+αυ‴=0\left\{ {\matrix{ {c{u'} + 6\alpha u{u'} - \upsilon {\upsilon '} + \beta {u{'''}} = 0} \hfill\cr {c{\upsilon '} + 3\alpha u{\upsilon '} + \alpha {\upsilon {'''}} = 0} \hfill\cr } } \right.Among them, u′u″, u″ respectively represents the derivative corresponding to u to ξ. υ′, υ‴ represent the derivative corresponding to υ to ξ. Assume that the solution of equation (1) has the following form (8){u(x,t)=∑i=0maiYi+∑i=1ma−iY−iυ(x,t)=∑i=0nciYi+∑i=1nc−iY−i\left\{ {\matrix{ {u(x,t) = \sum\limits_{i = 0}^m {a_i}{Y^i} + \sum\limits_{i = 1}^m {a_{ - i}}{Y^{ - i}}} \hfill \cr {\upsilon (x,t) = \sum\limits_{i = 0}^n {c_i}{Y^i} + \sum\limits_{i = 1}^n {c_{ - i}}{Y^{ - i}}} \hfill \cr } } \right.Among them a−i ⋯ ai,c−i ⋯ ci are all undetermined coefficients. According to the principle of homogeneous balance, formula (9) can be obtained (9){u(ξ)=a−2Y2+a−1Y+a0+a1Y+a2Y2υ(ξ)=c−2Y2+c−1Y+c0+c1Y+c2Y2\left\{ {\matrix{ {u(\xi ) = {{{a_{ - 2}}} \over {{Y^2}}} + {{{a_{ - 1}}} \over Y} + {a_0} + {a_1}Y + {a_2}{Y^2}} \hfill \cr {\upsilon (\xi ) = {{{c_{ - 2}}} \over {{Y^2}}} + {{{c_{ - 1}}} \over Y} + {c_0} + {c_1}Y + {c_2}{Y^2}} \hfill \cr } } \right.5.2Accurately solve the (2+1)-dimensional Burgers equationThe (2+1)-dimensional Burgers equation is generated in the phenomenon of fluid mechanics mainly to study turbulence in fluids [11]. The equation expression is shown in formula (10)(10){−ut+uuy+αυux+βuyy+αβuxx=0ux−υy=0\left\{ {\matrix{ { - {u_t} + u{u_y} + \alpha \upsilon {u_x} + \beta {u_{yy}} + \alpha \beta {u_{xx}} = 0} \hfill \cr {{u_x} - {\upsilon _y} = 0} \hfill \cr } } \right.Assuming u(x,y,t) = u(ξ)υ(x,y,t) = υ(ξ), ξ = x +ky +ct, and substituting it into the (2+1)-dimensional Burgers equation, the ODE about u(ξ) can be obtained as follows (11){−cu′+kuu′+aυu′+k2βu″+αβu″=0u′−kυ′=0\left\{ {\matrix{ { - c{u'} + ku{u'} + a\upsilon {u'} + {k^2}\beta {u{''}} + \alpha \beta {u{''}} = 0} \hfill \cr {{u'} - k{\upsilon '} = 0} \hfill \cr } } \right.Among them, u′,u″,u‴ represents the derivative of u to ξ, respectively.represents the derivative of υ to ξ. Let us suppose that the (2+1)-dimensional Burgers equation (10) has the following form solution (12){u(x,y,t)=∑i=0maiYi+∑i=1ma−iY−iυ(x,y,t)=∑i=0nciYi+∑i=1nc−iY−i\left\{ {\matrix{ {u(x,y,t) = \sum\limits_{i = 0}^m {a_i}{Y^i} + \sum\limits_{i = 1}^m {a_{ - i}}{Y^{ - i}}} \hfill \cr {\upsilon (x,y,t) = \sum\limits_{i = 0}^n {c_i}{Y^i} + \sum\limits_{i = 1}^n {c_{ - i}}{Y^{ - i}}} \hfill \cr } } \right.Among them a−i ⋯ ai, c−i ⋯ ci are all undetermined coefficients. According to the principle of homogeneous balance, formula (13) can be obtained (13){u(ξ)=a−1Y+a0+a1Yυ(ξ)=c−1Y+c0+c1Y\left\{ {\matrix{ {u(\xi ) = {{{a_{ - 1}}} \over Y} + {a_0} + {a_1}Y} \hfill \cr {\upsilon (\xi ) = {{{c_{ - 1}}} \over Y} + {c_0} + {c_1}Y} \hfill \cr } } \right.Where Y = tanh(ξ), α, β are the coefficient to be determined. According to the third step and the fourth step in the previous steps of the tanh function expansion method, the algebraic equations of each coefficient of ai, ci, α, μ can be obtained [12]. In this regard, the precise expressions of the coefficients of the equations solved with the Maple auxiliary tool are as follows:a0=c−αc0k{a_0} = {{c - \alpha {c_0}} \over k}, c1 = 2β, a−1 = 0, a1 = 2kβ, c−i = 0a0=c−αc0k{a_0} = {{c - \alpha {c_0}} \over k}, a−1 = 2kβ, c−1 = 2β, a1 = 0, c1 = 0a−1 = 2β, a0=c−c0αk{a_0} = {{c - {c_0}\alpha } \over k}, a−1 = 2kβ, c1 = 2β, a1 = 2kβNow, the coefficients solved by each equation system are substituted into equation (13) to get the final accurate solution. The statistics of the specific exact solution are as follows.First set of results u1=−αc0+ck+2kβtanh(x+ky+ct)υ1=c0+2βtanh(x+ky+ct)\matrix{ {{u_1} = {{ - \alpha {c_0} + c} \over k} + 2k\beta \tanh (x + ky + ct)} \cr {{\upsilon _1} = {c_0} + 2\beta \tanh (x + ky + ct)} \cr } The second set of results u2=2kβtanh(x+ct+ky)+c−ac0kυ2=2βtanh(x+ct+ky)+c0\matrix{ {{u_2} = {{2k\beta } \over {\tanh (x + ct + ky)}} + {{c - a{c_0}} \over k}} \cr {{\upsilon _2} = {{2\beta } \over {\tanh (x + ct + ky)}} + {c_0}} \cr } The third set of results u3=2kβtanh(x+ky+ct)+c−ac0k+2kβtanh(x+ct+ky)υ3=2βtanh(x+ct+ky)+c0+2βtanh(x+ct+ky)+c0\matrix{ {{u_3} = {{2k\beta } \over {\tanh (x + ky + ct)}} + {{c - a{c_0}} \over k} + 2k\beta \tanh (x + ct + ky)} \cr {{\upsilon _3} = {{2\beta } \over {\tanh (x + ct + ky)}} + {c_0} + 2\beta \tanh (x + ct + ky) + {c_0}} \cr } 5.3Solve the RLW-Burgers equation accuratelyRLW-Burgers mainly describes the problem of nonlinear dispersion wavelength and the expression of its equations is shown in formula (14)(14)ut+2ux+Buux−μuxx−βuxxt=0{u_t} + 2{u_x} + Bu{u_x} - \mu {u_{xx}} - \beta {u_{xxt}} = 0The solving steps are similar to the previous two types of equations. Also, suppose u(x,t) = u(ξ), ξ = x + ct and substitute it into the RLW-Burgers equation to obtain the ODE of u(ξ):(15)cu′+2u′+Buu′−μu″−cβu‴=0c{u'} + 2{u'} + Bu{u'} - \mu {u{''}} - c\beta {u{'''}} = 0Let us suppose that the solution of the RLW-Burgers equation (14) has the following form (16)u(ξ)=∑i=0maiYi+∑i=1ma−iY−iu(\xi ) = \sum\limits_{i = 0}^m {a_i}{Y^i} + \sum\limits_{i = 1}^m {a_{ - i}}{Y^{ - i}}Among them, u′, u″, u‴ represent the derivative of u to ξ, respectively and note that a−i ⋯ ai, c−i, ci are all undetermined coefficients [13]. According to the principle of homogeneous balance, formula (17) can be obtained (17)u(ξ)=a−2Y2(ξ)+a−1Y(ξ)+a0+a1Y(ξ)+a2Y2(ξ)u(\xi ) = {{{a_{ - 2}}} \over {{Y^2}(\xi )}} + {{{a_{ - 1}}} \over {Y(\xi )}} + {a_0} + {{{a_1}} \over {Y(\xi )}} + {{{a_2}} \over {{Y^2}(\xi )}}where Y = tanh(ξ), ai is the coefficient to be determined. According to the third and fourth steps of the previous Tanh-function expansion method steps, the algebraic equations of each coefficient of ai, α, β, μ can be obtained.As follows Y5 : 24βa2−2Ba22=024\beta {a_2} - 2Ba_2^2 = 0Y4 : − 6μa2 + 6cβ a1 − 3Ba1a2 = 0Y3 : 2Ba22−4a2+2Ba22−2ca2−40cβa2−2Ba0a2−Ba12=02Ba_2^2 - 4{a_2} + 2Ba_2^2 - 2c{a_2} - 40c\beta {a_2} - 2B{a_0}{a_2} - Ba_1^2 = 0Y2 : − Ba0a1 + 2Ba1a2 − 2a1 + 8μa2 − Ba1a2 − 8cβ a1 − ca1 + Ba2(a−1 + a1) = 0Y1 : − Ba−1 + 2Ba0a2 − Ba1(a−1 + a1) + 2μa1 + 4a2 + 16cβ a2 + 2ca2 = 0Y0 : μ(2a−2 + 2a2) + 2a−1 + 2a−1 + c(a−1 + a1) + Ba2a1 + Ba−1a2 − cβ (−2a−1 − 2a1) + Ba−2a1 + Ba0(a−1 + a1) + 2a1 = 0Y−2 : − 8β ca−1 − ca−1 + 2Ba−2a−1 + Ba−2(a−1 + a1) − 2Ba−2a1 − Ba0a−1 − 2a−1 + 8μa−2 = 0Y−3 : − 2Ba0a−2 − 2ca−2 − 2μa−1 − Ba−1 − 40cβ a−2 + 2Ba−2 − 4a−2 = 0Y−4 : − 3Ba−2a−1 + 6β a−1 − 6μa−2 = 0Y−5 : 24βa−2−2Ba−22=024\beta {a_{ - 2}} - 2Ba_{ - 2}^2 = 0In this regard, the precise expressions of the coefficients of the equations solved with the Maple auxiliary tool are as follows a0=−(Ba−1+2c+4)2B{a_0} = {{ - (B{a_{ - 1}} + 2c + 4)} \over {2B}}, a1 = 0, a2 = 0, a−2=12a−1{a_{ - 2}} = {1 \over 2}{a_{ - 1}}, μ=−5Ba−112\mu = - {{5B{a_{ - 1}}} \over {12}}, β=Ba−124c\beta = {{B{a_{ - 1}}} \over {24c}}a0=Ba−1−2c−42B{a_0} = {{B{a_{ - 1}} - 2c - 4} \over {2B}}, a1 = 0, a2 = 0, a−2=12a−1{a_{ - 2}} = {1 \over 2}{a_{ - 1}}, μ=−5Ba−112\mu = - {{5B{a_{ - 1}}} \over {12}}, β=−Ba−124c\beta = {{ - B{a_{ - 1}}} \over {24c}}B=12cβa2B = {{12c\beta } \over {{a_2}}}, a0=−a2(c+2+12βc)12cβ{a_0} = - {{{a_2}(c + 2 + 12\beta c)} \over {12c\beta }}, a1 = 2a2, μ = −10cβ, a−2 = 0, a−1 = 0B=48cβa−1B = {{48c\beta } \over {{a_{ - 1}}}}, a0=−148a−1(24cβ+c+2)cβa2=14a−1{a_0} = - {1 \over {48}}{{{a_{ - 1}}(24c\beta + c + 2)} \over {c\beta }}{a_2} = {1 \over 4}{a_{ - 1}}, μ = −20cβ, a−2=14a−1{a_{ - 2}} = {1 \over 4}{a_{ - 1}}, a1 = a−1Now, the coefficients of each equation system are substituted in equation (17) to obtain the final exact solution. The statistics of the specific 4 exact solutions are as follows u1=a−12tanh2(x+ct)+a−1tanh(x+ct)−12Ba−1+2c+4Bu2=−a−12tanh2(x+ct)+a−1tanh(x+ct)−12Ba−1−2c−4Bu3=−112a2(12cβ+c+2)cβ+2a2tanh(x+ct)+a2tanh2(x+ct)u4=a−14tanh2(x+ct)+a−1tanh(x+ct)−a−1(24cβ+c+2)48cβ+a−1tanh(x+ct)+14a−1tanh2(x+ct)\matrix{ {{u_1} = {{{a_{ - 1}}} \over {2{{\tanh }^2}(x + ct)}} + {{{a_{ - 1}}} \over {\tanh (x + ct)}} - {1 \over 2}{{B{a_{ - 1}} + 2c + 4} \over B}} \hfill \cr {{u_2} = - {{{a_{ - 1}}} \over {2{{\tanh }^2}(x + ct)}} + {{{a_{ - 1}}} \over {\tanh (x + ct)}} - {1 \over 2}{{B{a_{ - 1}} - 2c - 4} \over B}} \hfill \cr {{u_3} = - {1 \over {12}}{{{a_2}(12c\beta + c + 2)} \over {c\beta }} + 2{a_2}\tanh (x + ct) + {a_2}\mathop {\tanh }\nolimits^2 (x + ct)} \hfill \cr {{u_4} = {{{a_{ - 1}}} \over {4{{\tanh }^2}(x + ct)}} + {{{a_{ - 1}}} \over {\tanh (x + ct)}} - {{{a_{ - 1}}(24c\beta + c + 2)} \over {48c\beta }} + {a_{ - 1}}\tanh (x + ct) + {1 \over 4}{a_{ - 1}}\mathop {\tanh }\nolimits^2 (x + ct)} \hfill \cr } Two issues need to be considered at this stage: one is to provide students with the conditions for practice; the other is to create an appropriate learning and inquiry context – the problem scenario we gave. First, for the initial value problem of a first-order differential equation system, find the third approximate solution, and design courseware to solve this problem (completed after class). Then, students can complete the assigned homework and actively explore the existence and uniqueness of the solutions of some differential equations.6ConclusionThe exact solution of nonlinear differential functions has always been a severe and essential physics and mathematics content. We use the Tanh-function method to solve the exact solution concisely and quickly. We apply the Tanh-function method to the coupled KdV process and (2+1)-dimensional Burgers equation and other types of nonlinear differential equations to solve specific examples to confirm affirmatively the Tanh-function method's value in the application of nonlinear differential functions. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Applied Mathematics and Nonlinear Sciences de Gruyter

Nonlinear Differential Equations in the Teaching Model of Educational Informatisation

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References (11)

Publisher
de Gruyter
Copyright
© 2021 L.V. Shengnan et al., published by Sciendo
eISSN
2444-8656
DOI
10.2478/amns.2021.2.00100
Publisher site
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Abstract

1IntroductionThe rapid development of information technology continues to impact the traditional-closed, one-way transmission course teaching mode. This is also rapidly changing the learning lifestyle of students and changing teachers’ traditional teaching concepts and teaching models. How to create a new curriculum teaching model according to the teaching objectives of the curriculum and the characteristics of information technology is an important research topic for teachers in the information age [1]. In the teaching practice in recent years, we have actively explored the research teaching model with the characteristics of the times. The use of information technology to create a research teaching model is an excellent example of the combination of information technology and curriculum teaching reform.Nonlinear phenomena are common in nature and social life. Many significant issues such as engineering technology and natural phenomena can be attributed to nonlinear equations. Therefore, using nonlinear equations to study the objective world is an inevitable way. With the development of social sciences and natural sciences, the solution of nonlinear differential equations, especially the analysis of the exact solutions, is one of the most important ways to solve practical problems [2]. It is widely used in various fields such as physics, engineering technology and applied mathematics. Therefore, it is necessary to strengthen the research on the exact solution of nonlinear differential equations to make it serve the reality. There are many functional methods for solving nonlinear differential equations, including the homogeneous balance method, backscattering method, Darboux transformation method, variable minimum separation method, Painleve expansion method and Tanh-function method. These exact solutions have extensively promoted the development of nonlinear equation theory and practical applications. Tanh-function method is one of the very effective methods to construct exact solutions of nonlinear equations. In this regard, this article takes the nonlinear differential theory as the starting point and focuses on the Tanh-function method and its promotion. At the same time, we use this method to solve the common three kinds of nonlinear differential equations commonly used.2The overall goal of the teaching planAccording to the characteristics of the content structure of the ordinary differential equations course, our overall goal of designing the teaching plan is as follows: first, through the teaching of ordinary differential equations, we must teach students the necessary basic knowledge, and more importantly, cultivate students’ abilities [3]. We want students to learn, learn to think, learn to acquire information, learn to analyse and solve problems. Second, we must select typical materials that are closely related to the textbook's content and use multimedia technology to create a learning situation that can reflect the development and development of mathematical concepts, conclusions and thinking methods. In this way, students can master and use mathematical thinking methods to ‘discover’ new mathematical problems independently to guide students to conduct study and research and cultivate creativity and innovation. Figure 1 shows the basic structure of the information teaching classroom.Fig. 1The basic structure of the informatisation teaching classroom3The specific implementation of the teaching planThe key to research teaching is the design of the research teaching plan. ‘Ordinary differential equations’ is a professional introductory course for mathematics and applied mathematics [4]. To meet the rapid development of higher education, we have to focus on the reforms on the integration of modern information technology and the teaching mode of ordinary differential equations.3.1Analysis of the students’ existing foundationStudents have taken courses such as ‘Mathematical Analysis’, ‘Advanced Algebra’, ‘Analytic Geometry’ and ‘General Physics’ before taking the course of ordinary differential equations [5]. Thus, they have a particular foundation in mathematics and applied mathematics. Students have also studied ‘Computer Application Fundamentals’, ‘Algorithm Language’ and other educational courses and have a thorough understanding of nonlinear differential equation courseware functions.3.2Arrangement of teaching content under the new teaching modelWe use the second edition of ‘Ordinary Differential Equations’ compiled by the Department of Mathematics of Northeast Normal University as the introductory textbook. In the organising of teaching content, the teaching materials are scientifically processed following the actual situation of students majoring in mathematics and applied mathematics [6]. Thus, for example, the methods and theoretical applications of solving differential equations are regarded as the content of the intensive lecture.4Tanh-function method introductionThe Tanh-function expansion method is the abbreviation of the hyperbolic tangent function expansion method. This function method is currently one of the most effective methods for constructing nonlinear differential equations for precise solutions [7]. In solving nonlinear functions, people find that the solitary wave solutions of many nonlinear differential equations, including the Burgers equation, KdV equation, etc., can be represented by polynomials in the form of Tanh functions. This inspired people to use nonlinear differential equations to construct solitary wave solutions with Tanh functions. This is also the origin of the Tanh-function expansion method.This section first briefly introduces the expansion method of the Tanh function. Only after understanding and mastering can the application of accurate solutions be better carried out. Figure 2 shows the expanded form of the Tanh function [8]. The specific description of the Tanh-function expansion method is to give a nonlinear differential equation as shown in equation (1)(1)p(u,ux,uy,ut,uxx,uxy,uxt,uyy,⋯)=0p(u,{u_x},{u_y},{u_t},{u_{xx}},{u_{xy}},{u_{xt}},{u_{yy}}, \cdots ) = 0Among them, P is a polynomial of the derivatives of u.Fig. 2Tanh-function expanded formThe first step assumes that the travelling wave solution of the equation has the following form, and the travelling wave change has (2)u(x,y,t)=u(ξ),ξ=lx+ky+ctu(x,y,t) = u(\xi ),\xi = lx + ky + ctAmong them l,k,c are all undetermined constants. We can use the above formula to reduce the formula (1) to become the ordinary differential equation of u(ξ). As shown in formula (3)(3)p(u,u′,u″,u‴,⋯)=0p(u,{u^{\prime}},{u{''}},{u{'''}}, \cdots ) = 0Among them, u′,u″,u‴, ⋯ represents the derivative of u to ξ, respectively.The second step hypothesis is (4)Y=tanh(ξ), ξ=lx+ky+ctY = \tanh (\xi ),\quad \xi = lx + ky + ctddξ=(1−Y2)ddY{d \over {d\xi }} = (1 - {Y^2}){d \over {dY}}, d2dξ2=(1−Y2)(−2YddY+(1−Y2)d2dY2){{{d^2}} \over {d{\xi ^2}}} = (1 - {Y^2})( - 2Y{d \over {dY}} + (1 - {Y^2}){{{d^2}} \over {d{Y^2}}}), d3dξ3=(1−Y3)(6Y2−2)ddY−6Y(1−Y2)d2dY2+(1−Y2)2d3dY3{{{d^3}} \over {d{\xi ^3}}} = (1 - {Y^3})(6{Y^2} - 2){d \over {dY}} - 6Y(1 - {Y^2}){{{d^2}} \over {d{Y^2}}} + {(1 - {Y^2})^2}{{{d^3}} \over {d{Y^3}}}can be obtained by formula (4). Then suppose that the finite series form of Y (ξ) can be expressed by u(ξ). Then, there is (5)u(ξ)=∑i=0maiYi+∑i=0ma−iY−iu(\xi ) = \sum\limits_{i = 0}^m {a_i}{Y^i} + \sum\limits_{i = 0}^m {a_{ - i}}{Y^{ - i}}Among them m,ai(i = −m ⋯ m) is a constant to be determined.The third step is to determine the value of m. Generally, a positive integer can be obtained by balancing the nonlinear term and higher-order derivative term of the nonlinear differential equation (1).The fourth step is to introduce formula (5) into formula (3) and simplify it to be that all coefficients of Y (ξ) are all zero [9]. In this way, several nonlinear algebraic equations about the coefficients of ai, l, k and c can be obtained.In the fifth step, we use Maple to solve the above algebraic equations. In this way, the precise expression result of each coefficient can be obtained [10]. Finally, these results are re-introduced into the formula (3) and the exact solution of the nonlinear equation (1) can be finally obtained. The exact solution is represented by Y (ξ).5Application examples of Tanh-function method5.1Accurately solve the coupled KdV equationThe expression of the coupled KdV equations is shown in formula (6)(6){ut+6αuux−υυx+βuxxx=0υt+3αuux+αυxxx=0\left\{ {\matrix{ {{u_t} + 6\alpha u{u_x} - \upsilon {\upsilon _x} + \beta {u_{xxx}} = 0} \hfill \cr {{\upsilon _t} + 3\alpha u{u_x} + \alpha {\upsilon _{xxx}} = 0} \hfill \cr } } \right.Assuming that u(x,t) = u(ξ), υ(x,t) = υ(ξ), ξ = x + ct will be substituted into the coupled KdV equation, the u(ξ) ODE can be obtained as follows:(7){cu′+6αuu′−υυ′+βu‴=0cυ′+3αuυ′+αυ‴=0\left\{ {\matrix{ {c{u'} + 6\alpha u{u'} - \upsilon {\upsilon '} + \beta {u{'''}} = 0} \hfill\cr {c{\upsilon '} + 3\alpha u{\upsilon '} + \alpha {\upsilon {'''}} = 0} \hfill\cr } } \right.Among them, u′u″, u″ respectively represents the derivative corresponding to u to ξ. υ′, υ‴ represent the derivative corresponding to υ to ξ. Assume that the solution of equation (1) has the following form (8){u(x,t)=∑i=0maiYi+∑i=1ma−iY−iυ(x,t)=∑i=0nciYi+∑i=1nc−iY−i\left\{ {\matrix{ {u(x,t) = \sum\limits_{i = 0}^m {a_i}{Y^i} + \sum\limits_{i = 1}^m {a_{ - i}}{Y^{ - i}}} \hfill \cr {\upsilon (x,t) = \sum\limits_{i = 0}^n {c_i}{Y^i} + \sum\limits_{i = 1}^n {c_{ - i}}{Y^{ - i}}} \hfill \cr } } \right.Among them a−i ⋯ ai,c−i ⋯ ci are all undetermined coefficients. According to the principle of homogeneous balance, formula (9) can be obtained (9){u(ξ)=a−2Y2+a−1Y+a0+a1Y+a2Y2υ(ξ)=c−2Y2+c−1Y+c0+c1Y+c2Y2\left\{ {\matrix{ {u(\xi ) = {{{a_{ - 2}}} \over {{Y^2}}} + {{{a_{ - 1}}} \over Y} + {a_0} + {a_1}Y + {a_2}{Y^2}} \hfill \cr {\upsilon (\xi ) = {{{c_{ - 2}}} \over {{Y^2}}} + {{{c_{ - 1}}} \over Y} + {c_0} + {c_1}Y + {c_2}{Y^2}} \hfill \cr } } \right.5.2Accurately solve the (2+1)-dimensional Burgers equationThe (2+1)-dimensional Burgers equation is generated in the phenomenon of fluid mechanics mainly to study turbulence in fluids [11]. The equation expression is shown in formula (10)(10){−ut+uuy+αυux+βuyy+αβuxx=0ux−υy=0\left\{ {\matrix{ { - {u_t} + u{u_y} + \alpha \upsilon {u_x} + \beta {u_{yy}} + \alpha \beta {u_{xx}} = 0} \hfill \cr {{u_x} - {\upsilon _y} = 0} \hfill \cr } } \right.Assuming u(x,y,t) = u(ξ)υ(x,y,t) = υ(ξ), ξ = x +ky +ct, and substituting it into the (2+1)-dimensional Burgers equation, the ODE about u(ξ) can be obtained as follows (11){−cu′+kuu′+aυu′+k2βu″+αβu″=0u′−kυ′=0\left\{ {\matrix{ { - c{u'} + ku{u'} + a\upsilon {u'} + {k^2}\beta {u{''}} + \alpha \beta {u{''}} = 0} \hfill \cr {{u'} - k{\upsilon '} = 0} \hfill \cr } } \right.Among them, u′,u″,u‴ represents the derivative of u to ξ, respectively.represents the derivative of υ to ξ. Let us suppose that the (2+1)-dimensional Burgers equation (10) has the following form solution (12){u(x,y,t)=∑i=0maiYi+∑i=1ma−iY−iυ(x,y,t)=∑i=0nciYi+∑i=1nc−iY−i\left\{ {\matrix{ {u(x,y,t) = \sum\limits_{i = 0}^m {a_i}{Y^i} + \sum\limits_{i = 1}^m {a_{ - i}}{Y^{ - i}}} \hfill \cr {\upsilon (x,y,t) = \sum\limits_{i = 0}^n {c_i}{Y^i} + \sum\limits_{i = 1}^n {c_{ - i}}{Y^{ - i}}} \hfill \cr } } \right.Among them a−i ⋯ ai, c−i ⋯ ci are all undetermined coefficients. According to the principle of homogeneous balance, formula (13) can be obtained (13){u(ξ)=a−1Y+a0+a1Yυ(ξ)=c−1Y+c0+c1Y\left\{ {\matrix{ {u(\xi ) = {{{a_{ - 1}}} \over Y} + {a_0} + {a_1}Y} \hfill \cr {\upsilon (\xi ) = {{{c_{ - 1}}} \over Y} + {c_0} + {c_1}Y} \hfill \cr } } \right.Where Y = tanh(ξ), α, β are the coefficient to be determined. According to the third step and the fourth step in the previous steps of the tanh function expansion method, the algebraic equations of each coefficient of ai, ci, α, μ can be obtained [12]. In this regard, the precise expressions of the coefficients of the equations solved with the Maple auxiliary tool are as follows:a0=c−αc0k{a_0} = {{c - \alpha {c_0}} \over k}, c1 = 2β, a−1 = 0, a1 = 2kβ, c−i = 0a0=c−αc0k{a_0} = {{c - \alpha {c_0}} \over k}, a−1 = 2kβ, c−1 = 2β, a1 = 0, c1 = 0a−1 = 2β, a0=c−c0αk{a_0} = {{c - {c_0}\alpha } \over k}, a−1 = 2kβ, c1 = 2β, a1 = 2kβNow, the coefficients solved by each equation system are substituted into equation (13) to get the final accurate solution. The statistics of the specific exact solution are as follows.First set of results u1=−αc0+ck+2kβtanh(x+ky+ct)υ1=c0+2βtanh(x+ky+ct)\matrix{ {{u_1} = {{ - \alpha {c_0} + c} \over k} + 2k\beta \tanh (x + ky + ct)} \cr {{\upsilon _1} = {c_0} + 2\beta \tanh (x + ky + ct)} \cr } The second set of results u2=2kβtanh(x+ct+ky)+c−ac0kυ2=2βtanh(x+ct+ky)+c0\matrix{ {{u_2} = {{2k\beta } \over {\tanh (x + ct + ky)}} + {{c - a{c_0}} \over k}} \cr {{\upsilon _2} = {{2\beta } \over {\tanh (x + ct + ky)}} + {c_0}} \cr } The third set of results u3=2kβtanh(x+ky+ct)+c−ac0k+2kβtanh(x+ct+ky)υ3=2βtanh(x+ct+ky)+c0+2βtanh(x+ct+ky)+c0\matrix{ {{u_3} = {{2k\beta } \over {\tanh (x + ky + ct)}} + {{c - a{c_0}} \over k} + 2k\beta \tanh (x + ct + ky)} \cr {{\upsilon _3} = {{2\beta } \over {\tanh (x + ct + ky)}} + {c_0} + 2\beta \tanh (x + ct + ky) + {c_0}} \cr } 5.3Solve the RLW-Burgers equation accuratelyRLW-Burgers mainly describes the problem of nonlinear dispersion wavelength and the expression of its equations is shown in formula (14)(14)ut+2ux+Buux−μuxx−βuxxt=0{u_t} + 2{u_x} + Bu{u_x} - \mu {u_{xx}} - \beta {u_{xxt}} = 0The solving steps are similar to the previous two types of equations. Also, suppose u(x,t) = u(ξ), ξ = x + ct and substitute it into the RLW-Burgers equation to obtain the ODE of u(ξ):(15)cu′+2u′+Buu′−μu″−cβu‴=0c{u'} + 2{u'} + Bu{u'} - \mu {u{''}} - c\beta {u{'''}} = 0Let us suppose that the solution of the RLW-Burgers equation (14) has the following form (16)u(ξ)=∑i=0maiYi+∑i=1ma−iY−iu(\xi ) = \sum\limits_{i = 0}^m {a_i}{Y^i} + \sum\limits_{i = 1}^m {a_{ - i}}{Y^{ - i}}Among them, u′, u″, u‴ represent the derivative of u to ξ, respectively and note that a−i ⋯ ai, c−i, ci are all undetermined coefficients [13]. According to the principle of homogeneous balance, formula (17) can be obtained (17)u(ξ)=a−2Y2(ξ)+a−1Y(ξ)+a0+a1Y(ξ)+a2Y2(ξ)u(\xi ) = {{{a_{ - 2}}} \over {{Y^2}(\xi )}} + {{{a_{ - 1}}} \over {Y(\xi )}} + {a_0} + {{{a_1}} \over {Y(\xi )}} + {{{a_2}} \over {{Y^2}(\xi )}}where Y = tanh(ξ), ai is the coefficient to be determined. According to the third and fourth steps of the previous Tanh-function expansion method steps, the algebraic equations of each coefficient of ai, α, β, μ can be obtained.As follows Y5 : 24βa2−2Ba22=024\beta {a_2} - 2Ba_2^2 = 0Y4 : − 6μa2 + 6cβ a1 − 3Ba1a2 = 0Y3 : 2Ba22−4a2+2Ba22−2ca2−40cβa2−2Ba0a2−Ba12=02Ba_2^2 - 4{a_2} + 2Ba_2^2 - 2c{a_2} - 40c\beta {a_2} - 2B{a_0}{a_2} - Ba_1^2 = 0Y2 : − Ba0a1 + 2Ba1a2 − 2a1 + 8μa2 − Ba1a2 − 8cβ a1 − ca1 + Ba2(a−1 + a1) = 0Y1 : − Ba−1 + 2Ba0a2 − Ba1(a−1 + a1) + 2μa1 + 4a2 + 16cβ a2 + 2ca2 = 0Y0 : μ(2a−2 + 2a2) + 2a−1 + 2a−1 + c(a−1 + a1) + Ba2a1 + Ba−1a2 − cβ (−2a−1 − 2a1) + Ba−2a1 + Ba0(a−1 + a1) + 2a1 = 0Y−2 : − 8β ca−1 − ca−1 + 2Ba−2a−1 + Ba−2(a−1 + a1) − 2Ba−2a1 − Ba0a−1 − 2a−1 + 8μa−2 = 0Y−3 : − 2Ba0a−2 − 2ca−2 − 2μa−1 − Ba−1 − 40cβ a−2 + 2Ba−2 − 4a−2 = 0Y−4 : − 3Ba−2a−1 + 6β a−1 − 6μa−2 = 0Y−5 : 24βa−2−2Ba−22=024\beta {a_{ - 2}} - 2Ba_{ - 2}^2 = 0In this regard, the precise expressions of the coefficients of the equations solved with the Maple auxiliary tool are as follows a0=−(Ba−1+2c+4)2B{a_0} = {{ - (B{a_{ - 1}} + 2c + 4)} \over {2B}}, a1 = 0, a2 = 0, a−2=12a−1{a_{ - 2}} = {1 \over 2}{a_{ - 1}}, μ=−5Ba−112\mu = - {{5B{a_{ - 1}}} \over {12}}, β=Ba−124c\beta = {{B{a_{ - 1}}} \over {24c}}a0=Ba−1−2c−42B{a_0} = {{B{a_{ - 1}} - 2c - 4} \over {2B}}, a1 = 0, a2 = 0, a−2=12a−1{a_{ - 2}} = {1 \over 2}{a_{ - 1}}, μ=−5Ba−112\mu = - {{5B{a_{ - 1}}} \over {12}}, β=−Ba−124c\beta = {{ - B{a_{ - 1}}} \over {24c}}B=12cβa2B = {{12c\beta } \over {{a_2}}}, a0=−a2(c+2+12βc)12cβ{a_0} = - {{{a_2}(c + 2 + 12\beta c)} \over {12c\beta }}, a1 = 2a2, μ = −10cβ, a−2 = 0, a−1 = 0B=48cβa−1B = {{48c\beta } \over {{a_{ - 1}}}}, a0=−148a−1(24cβ+c+2)cβa2=14a−1{a_0} = - {1 \over {48}}{{{a_{ - 1}}(24c\beta + c + 2)} \over {c\beta }}{a_2} = {1 \over 4}{a_{ - 1}}, μ = −20cβ, a−2=14a−1{a_{ - 2}} = {1 \over 4}{a_{ - 1}}, a1 = a−1Now, the coefficients of each equation system are substituted in equation (17) to obtain the final exact solution. The statistics of the specific 4 exact solutions are as follows u1=a−12tanh2(x+ct)+a−1tanh(x+ct)−12Ba−1+2c+4Bu2=−a−12tanh2(x+ct)+a−1tanh(x+ct)−12Ba−1−2c−4Bu3=−112a2(12cβ+c+2)cβ+2a2tanh(x+ct)+a2tanh2(x+ct)u4=a−14tanh2(x+ct)+a−1tanh(x+ct)−a−1(24cβ+c+2)48cβ+a−1tanh(x+ct)+14a−1tanh2(x+ct)\matrix{ {{u_1} = {{{a_{ - 1}}} \over {2{{\tanh }^2}(x + ct)}} + {{{a_{ - 1}}} \over {\tanh (x + ct)}} - {1 \over 2}{{B{a_{ - 1}} + 2c + 4} \over B}} \hfill \cr {{u_2} = - {{{a_{ - 1}}} \over {2{{\tanh }^2}(x + ct)}} + {{{a_{ - 1}}} \over {\tanh (x + ct)}} - {1 \over 2}{{B{a_{ - 1}} - 2c - 4} \over B}} \hfill \cr {{u_3} = - {1 \over {12}}{{{a_2}(12c\beta + c + 2)} \over {c\beta }} + 2{a_2}\tanh (x + ct) + {a_2}\mathop {\tanh }\nolimits^2 (x + ct)} \hfill \cr {{u_4} = {{{a_{ - 1}}} \over {4{{\tanh }^2}(x + ct)}} + {{{a_{ - 1}}} \over {\tanh (x + ct)}} - {{{a_{ - 1}}(24c\beta + c + 2)} \over {48c\beta }} + {a_{ - 1}}\tanh (x + ct) + {1 \over 4}{a_{ - 1}}\mathop {\tanh }\nolimits^2 (x + ct)} \hfill \cr } Two issues need to be considered at this stage: one is to provide students with the conditions for practice; the other is to create an appropriate learning and inquiry context – the problem scenario we gave. First, for the initial value problem of a first-order differential equation system, find the third approximate solution, and design courseware to solve this problem (completed after class). Then, students can complete the assigned homework and actively explore the existence and uniqueness of the solutions of some differential equations.6ConclusionThe exact solution of nonlinear differential functions has always been a severe and essential physics and mathematics content. We use the Tanh-function method to solve the exact solution concisely and quickly. We apply the Tanh-function method to the coupled KdV process and (2+1)-dimensional Burgers equation and other types of nonlinear differential equations to solve specific examples to confirm affirmatively the Tanh-function method's value in the application of nonlinear differential functions.

Journal

Applied Mathematics and Nonlinear Sciencesde Gruyter

Published: Jan 1, 2022

Keywords: Information technology; nonlinear differential equations; research teaching mode; teaching plan; computing software Maple; 34A34

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