Solution to nonlinear differential equation
6 次查看(过去 30 天)
显示 更早的评论
Hello, I am trying to solve the following differential equation, but I am running into some issues when using ODE45:

subject to the boundary conditions:


I broke this up into a system of two first order differential equations, but I am having trouble with the following aspects of the problem:
- How do you input initial conditions which are both given in terms of just the function (i.e. not it's derivatives)?
- ODE45 will not be able to go too far past |t|=10, so should I just set my bc's at this point?
Any help would be greatly appreciated or a indication of the correct direction for solution. Thanks in advance!
2 个评论
Cyrus Tirband
2019-10-29
If you're dealing with boundary conditions rather than initial conditions, it's better to use a boundary value problem solver like bvp4c. It allows you to set the values of your function at the boundaries, like you want.
However, I'm not sure your particular equations will work. Can you double check your equation? I tried putting it in the solver but the solution is a mess and very unstable. If a is not perfectly pi/2 for t>0, it will never have a limit value.
采纳的回答
darova
2019-10-30
Here is an attempt
function bvp4c_mathworks
solinit = bvpinit([-1 1],[0 0]); %[-inf, inf] replaces with [-1, 1]
sol = bvp4c(@myode,@mybc,solinit);
plot(sol.x, sol.y), grid on
% legend('y(t)','x(t)')
% xlabel('t')
end
function dy = myode(t,y)
dy(1,1) = y(2);
dy(2,1) = 1/2*(1-exp(2*t))*sin(y(1));
end
function res = mybc(ya,yb)
res = [ya(1)
yb(1)-pi/2];
end
2 个评论
darova
2019-10-31
- why did you choose the intreval [-1,1]?
It was just a shot in the dark. I like how it looks so why not. [-10 10] doesn't work for me=(
更多回答(0 个)
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Ordinary Differential Equations 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!