Nonlinear Differential Equations in the Teaching Model of Educational Informatisation

Students 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.

We 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.

The 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 ⋯ a_i,c_−i ⋯ c_i 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.

The (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 ⋯ a_i, c_−i ⋯ c_i 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 a_i, c_i, 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 = 0

Now, 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 }

RLW-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}} = 0 The 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‴=0 c{u'} + 2{u'} + Bu{u'} - \mu {u{''}} - c\beta {u{'''}} = 0 Let us suppose that the solution of the RLW-Burgers equation (14) has the following form (16) u(ξ)=∑i=0maiYi+∑i=1ma−iY−i u(\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 ⋯ a_i, c_−i, c_i 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), a_i 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 a_i, α, β, μ can be obtained.

As follows

Y⁵ : 24βa2−2Ba22=0 24\beta {a_2} - 2Ba_2^2 = 0

In 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}}

Now, 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.