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 个评论
Torsten
2018-4-6
The problem is not clear from your description.
Dereje Beyene
2018-4-6
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
2018-4-6
编辑:Walter Roberson
2018-4-7
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
2018-4-7
Dereje
2018-4-7
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
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