Changing boundary conditions for ODE
4 次查看(过去 30 天)
显示 更早的评论
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
5 个评论
采纳的回答
Ameer Hamza
2020-5-9
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
0 个评论
更多回答(1 个)
Nagaraja Shamsundar
2020-5-9
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.
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!