ode45 for the shooting method.

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Dereje 2018-4-6

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https://ww2.mathworks.cn/matlabcentral/answers/392984-ode45-for-the-shooting-method

评论： Dereje 2018-4-9

I want to predict a constant for the target height for the given ode problem. The target height is highly dependent on the constant alpha. Some one told me to use shooting /iterative methods but I am new for such a method. I need your help.

  zspan=[0,400];
v0mat = [1 0.01 1];
zsol = {};
v1sol = {};
v2sol = {};
v3sol = {};
for k=1:size(v0mat,1)
v0=v0mat(k,:);
[z,v]=ode45(@rhs,zspan,v0);
zsol{k}=z;
v1sol{k}=v(:,1);
v2sol{k}=v(:,2);
v3sol{k}=v(:,3);
end
for r=1:length(v2sol)
q(r)=r;
end
for k1 = 1:length(v2sol)
  zsol04(k1) = interp1(v2sol{k1}, zsol{k1}, 0.4);
end
figure()
scatter(q,zsol04,'p')
xlabel('q')
ylabel('Height')
function parameters=rhs(z,v)
          alpha=0.08116; 
          db= 2*alpha-(v(1).*v(3))./(2*v(2).^2);
          dw= (v(3)./v(2))-(2*alpha*v(2)./v(1));
          dgmark= -(2*alpha*v(3)./v(1)); 
 parameters=[db;dw;dgmark];
end

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Torsten 2018-4-9

Please read my answer again:

Use "bvp4c" with three boundary conditions at h=0, one boundary condition as v2(height)=0.4 and a free parameter alpha.

Best wishes

Torsten.

Dereje 2018-4-9

Hi Torsten, Yes I get it now. I was mixing the two boundary conditions. Thanks a lot!! It was better if u put your comments in the 'Answer this question' section so that I could be able to accept the answer.

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