# Soling a boundary value problem

4 vues (au cours des 30 derniers jours)
Sharon le 2 Fév 2024
Commenté : Sharon le 3 Fév 2024
I am trying to solve the following elliptic equation
on the interval [0,1] with Neumann Boundary condition, I want to get a positive solution φ with the maximum of φ is 1.
The following is the code I am using
h=0.001;
L=1;
M=L/h;
x=linspace(0,L,M+1);
gamma = @(x)(1+0.1*sin(2*pi*x)).*(1+0.1*sin(2*pi*x));
beta=@(x)(1+0.1*sin(2*pi*x));
%v0 = @(x)(1+0.5*cos(pi*x));
%v0x=@(x)-0.5*pi*sin(pi*x);
bvpfcn = @(x,v)[v(2);(gamma(x)-(beta(x)).*v(1)];
bcfcn = @(va,vb)[va(2);vb(2)];
guess = @(x)[1;0];
%guess = @(x)[v0(x);v0x(x)];
solinit = bvpinit(x,guess);
sol = bvp4c(bvpfcn, bcfcn, solinit);
phi1=sol.y(1,:);
M=max(phi1);
phi=phi1/M;
plot(sol.x,phi)
I have difficulity in selecting the initial function. I tried [1;0] and [v0(x);v0x(x)] which is defined above, I got different solutions. I cannot figure out the problem. Any help would be appreciated!
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### Réponses (1)

Torsten le 3 Fév 2024
Modifié(e) : Torsten le 3 Fév 2024
Prove the following:
If u is a solution of your differential equation, so is c*u for every c in IR.
So it does not surprise that bvp4c gives different solutions dependent on your initial guess: the solution is not unique.
If you normalize the solution as you did after solving, it seems it becomes unique. You can see this if you solve your equation for your two initial guess functions and plot the solutions in one graph.
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Torsten le 3 Fév 2024
Modifié(e) : Torsten le 3 Fév 2024
It seems that bvp4c as a numerical solver always converges to the trivial solution phi = 0.
I tried "dsolve", but it is unable to find an analytical solution (s.b.).
I'm still interested in your source that the equation has a solution different from 0.
syms x y(x)
eqn = diff(y,x,2)-((1+0.1*sin(2*pi*x))^2-(1+0.1*sin(2*pi*x)))*y==0
eqn(x) =
dsolve(eqn)
Warning: Unable to find symbolic solution.
ans = [ empty sym ]
Sharon le 3 Fév 2024
Thank you very much for your time and efforts!

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