Solving shooting method with ode45

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beginner
beginner 2016-3-21
回答： Torsten 2016-3-22
Hi there, I want to solve 2nd order ODE system with the Shooting Method. With my code I can't find any solution...
I have H(inf) = dH/dX(inf) = 0 and dH/dX(0) = 0. What I want to find out is H(0). I used fsolve and ode45.
Is my code correct at all? It doesnt give any error message but it doesn't find any solution...
function dhdt = wihelmy(X,h,Bo)
%h(1) = H
%h(2) = H'
% X is the input.
% tspan 1:3
dhdt_1 = h(2); % dhdt_1 = H'
dhdt_2 = Bo*(h(1)*(1+h(2)))^(3/2); %dhdt_2 = H''
dhdt = [dhdt_1; dhdt_2];
function HdH = test1(h0, theta, Bo)
%Bo = 1; % Bo = rho*g*l^2/gamma
%h0 = ?
% h_inf = 0;
% dh0 = 0;
% dh_inf = 0;
%h0 = 10e-5;
%thetadeg = 90; %theta in degrees
%theta = thetadeg*pi/180;
init = [10e-5 -cot(theta)];
function_handle = @(X,h) wihelmy(X,h,Bo);
[Xout,Hout] = ode45(function_handle, [0:0.1:3], init);
H_x_inf = Hout(end,1);
dH_x_inf = Hout(end,2);
HdH = [H_x_inf; dH_x_inf];
Main function is:
Bo=1;
thetadeg = 85; %theta in degrees
theta = thetadeg*pi/180;
Hopt=fsolve(@(h0) test1(h0, theta, Bo), 10e-5);
Thank you sooo much, if you can help me!

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回答（1 个）

Torsten
Torsten 2016-3-22
There are so many errors in your code that you should start simple.
Your problem is not a standard boundary value problem.
If your equation reads
y''=Bo*(y*(1+y'))^(3/2)
and you know y'(0)=0, I'd first examine some solutions for different values of y(0) using ODE45 alone.
Then, after you've got a feeling on how your function behaves, I'd switch to bvp4c to impose boundary conditions at x=0 and at x=L for a large value of L.
After you've been successful with these two foregoing steps, you can dare to apply your own solution method, namely shooting.
Best wishes
Torsten.

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