Solution to nonlinear differential equation

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Marcus Rosales
Marcus Rosales le 29 Oct 2019
Commenté : darova le 31 Oct 2019
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:
  1. How do you input initial conditions which are both given in terms of just the function (i.e. not it's derivatives)?
  2. 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 commentaires
Cyrus Tirband
Cyrus Tirband le 29 Oct 2019
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.
Marcus Rosales
Marcus Rosales le 29 Oct 2019
I was uncertain too, but yes I have it correct. I would have expected a negative sign in the expoential, so maybe the actual paper has a typo. I attached a screen shot of the equation.

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darova
darova le 30 Oct 2019
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 commentaires
Marcus Rosales
Marcus Rosales le 31 Oct 2019
Thanks for the reply! This works I beileve (I have plots to compare this to); just had to modify plot(...) to plot([-1,1],sol.y) to get what I wanted.
Could I ask a quick question: why did you choose the intreval [-1,1]? I would have thought I'd need a longer intreval than this to approximate the BCs, so was there a reason or was it more of an arbitrary choice?
darova
darova le 31 Oct 2019
  • 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=(

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