solution of coupled differential equation
3 次查看(过去 30 天)
显示 更早的评论
i am trying to solve some couples differential equation , the equations are given in the attachment .
can these equations be solved using [ode45 command ]
if i get some reference for similar type of equation that will be quiet helpful
0 个评论
采纳的回答
Torsten
2022-8-28
编辑:Torsten
2022-8-28
syms z11(x) z12(x) z21(x) z22(x) z31(x) z32(x)
eqn1 = diff(z11) == z12;
eqn2 = diff(z12) + z12 + z22 + z31 - 5 == 0;
eqn3 = diff(z21) == z22;
eqn4 = diff(z22) + z12 + z22 + z32 - 4 == 0;
eqn5 = diff(z31) == z32;
eqn6 = diff(z32) - z22 + z11 + z31 - 3 == 0;
eqns = [eqn1,eqn2,eqn3,eqn4,eqn5,eqn6];
conds = [z11(0)==0,z21(0)==0,z31(0)==0,z11(1)==0,z21(1)==0,z31(1)==0];
sol = dsolve(eqns,conds)
bvpfcn = @(x,y)[y(2);-y(2)-y(4)-y(5)+5;y(4);-y(2)-y(4)-y(6)+4;y(6);y(4)-y(1)-y(5)+3];
bcfcn = @(ya,yb) [ya(1);ya(3);ya(5);yb(1);yb(3);yb(5)];
xmesh = linspace(0,1,100);
solinit = bvpinit(xmesh, [0 0 0 0 0 0]);
sol = bvp4c(bvpfcn, bcfcn, solinit);
plot(sol.x, [sol.y(1,:);sol.y(3,:);sol.y(5,:)])
2 个评论
更多回答(1 个)
Sam Chak
2022-8-27
Hi @asim asrar
The second-order coupled system looks like a boundary value problem, where you need to find the initial values of
,
,
to satisfy the boundary values requirements.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1109665/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1109670/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1109675/image.png)
2 个评论
Torsten
2022-8-28
but i think bvp4c is meant for single differential equation, but mine is coupled differential equation
If you had looked at the examples provided, you would have seen that this is not true.
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Function Creation 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!