Changing boundary conditions for ODE

My code below plots the solution of the equation of Mathieu with the initial condition: y(0) = 1, y'(0) = 1

Now I still want the same solution to this problem, but with new boundary conditions: y(1) = -1, y(10) = 1

I tried solving it with dsolve(eq, y(1) == -1, y(10) == 1) but then I couldn't implement it into my function.

syms t y(t)
syms a q
tm = [0 75]; %time intervall
figure(1)
clf
hold on
y0=[-1;1]; %initial conditions
[t,y1] = ode23(@Mathieu, tm, y0);
plot(t,y1(:,2))%y(t)
xlim([0 75])

function dydt = Mathieu(t,y)
a = 2;
q = 0.5;
dydt = [y(2); -(a-2*q*cos(2*t))*y(1)];
end

[Ameer Hamza]

Your limits of integrations are [0 75]. However, you specified the conditions at t=1 and t=10. Are you trying to solve a multiple boundary problem?

[Borjan Trajanoski]

Thanks for the hint, I did not think of that, but the limits are not necessary, I just used them for the initial condition. If I delete them, should I then be able to apply the boundary conditions?

[Ameer Hamza]

Yes, you can apply boundary conditions, but you will need to use bvp4c or bvp5c (if multiple boundaries), instead of ode45. If you just care about boundary conditions at [1 10], then see the code in my answer.

[Borjan Trajanoski]

I just did it with bvp4c, really cool function! And thanks for the additional code :)

[Ameer Hamza]

I am glad to be of help.

Answer 1

[Ameer Hamza]

Try this code with bvp4c in the limits [1 10], with the boundary condition given in the question. For more details, see the documentation of bvp4c

tm = [1 10]; %time intervall
t = linspace(tm(1), tm(2), 50);
guess = bvpinit(t, [1; 1]);

sol = bvp4c(@Mathieu, @bcFun, guess);
plot(sol.x, sol.y(1,:)) % y(t) is the row of solution
xlim(tm)

function dydt = Mathieu(t,y) %ODE
a = 2;
q = 0.5;
dydt = [y(2); -(a-2*q*cos(2*t))*y(1)];
end

function res = bcFun(ya, yb) % boundary function
    res = [ya(1)+1;  % y(1)=-1
           yb(1)-1]; % y(10) = 1
end

Answer 2

[Nagaraja Shamsundar]

Your goal is to solve a boundary value problem (BVP). Some BVPs can be converted into equivalent initial value problems (IVP), but in general it is more appropriate to use a BVP solver such as Matlab's BVP4C instead of an IVP solver such as ODE23.

In Matlab, type help bvp4c or doc bvp4c.

[Borjan Trajanoski]

Thank you! I didn't know there is an extra function for this kind of problem. I am looking it up right now.