Solve ODE for the eigenvalue

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Alexander Kimbley
Alexander Kimbley 2020 年 2 月 15 日
コメント済み: darova 2020 年 2 月 15 日
Hi,
I'm trying to solve for the eigenvalue X; for specific values M and Mf, with boundary conditions at infinity, that is, y(-inf)=y(inf)=0, which is why ive used ya(1)-10^10 etc as the conditions, im not sure if this is approprate or not though.
Any help would be greatly appreciated. Ive attached the file.
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darova
darova 2020 年 2 月 15 日
Please attach the original equation
Alexander Kimbley
Alexander Kimbley 2020 年 2 月 15 日
The original equation is given by
y'' + [M * sqrt(1-(2*Mf^2)/X) *(X+1-(2*Mf^2)/X -2*Mf^2) -Y^2]y=0, where y=y(Y).

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回答 (1 件)

darova
darova 2020 年 2 月 15 日
I changed limits and the call of bvp4c
xlow = -2;
xhigh = 2;
X0 =8 ;
y10 = 1;
y20 = 1;
sol = bvp4c(@bvp4ode, @bvp4bc, solinit, options); %
Boundary function
function res = bvp4bc(ya,yb,p)
res = [ya(1) % y(a) = 0
yb(1) % y(b) = 0
ya(2)-1]; % additional (not sure if it's correct)
end
The result
Look here and here
  2 件のコメント
Alexander Kimbley
Alexander Kimbley 2020 年 2 月 15 日
Hi, thanks for your help, ive had a look and it seems very close to the correct answer when Mf=0, but not quite, ive attached the new file, im not sure if ive entered everthing right; to check its right when Mf=0, the eigenvalue c should equal -1+ 1/M.
darova
darova 2020 年 2 月 15 日
For interval [-2 2]
>> mainbetarot
c =
-0.2322
>> -1+1/1.4
ans =
-0.2857
Try to change interval

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