solution of coupled differential equation
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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
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Torsten
2022 年 8 月 28 日
編集済み: Torsten
2022 年 8 月 28 日
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,:)])
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その他の回答 (1 件)
Sam Chak
2022 年 8 月 27 日
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 件のコメント
Torsten
2022 年 8 月 28 日
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.
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