ode45 for the shooting method.
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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
7 Comments
Torsten
on 6 Apr 2018
The problem is not clear from your description.
Dereje Beyene
on 6 Apr 2018
The above code gives height at a point where v2sol is 0.4. Now this result is highly dependent on alpha which I used constant value. Now I want to go the other way. Let’s I don’t have alpha, but I have height. My question is for what value of alpha will I get the given height. I am not sure if I explained it well.
Torsten
on 6 Apr 2018
Edited: Walter Roberson
on 7 Apr 2018
Use "bvp4c" with three boundary conditions at h=0, one boundary condition as v2(height)=0.4 and a free parameter alpha.
The example
"Compute Fourth Eigenvalue of Mathieu’s Equation"
under
will show you how to proceed.
Here, lambda plays the role of your alpha.
Best wishes
Torsten.
Dereje
on 7 Apr 2018
Dereje
on 7 Apr 2018
Torsten
on 9 Apr 2018
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
on 9 Apr 2018
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