Finite difference method solving boundary condition

Hi try to solve the nonlinear ordinary difference equation by finite difference method with boundary conditions: U(-1) = 0, U(1)=0. The code below.

Function [r, sol] =res(U)

n = 10;

r = ((1:n)-1)/(n-1);

r0=zeros(n+1,1);

color={'r','m','c','b');

For delta = 2:5

Sol = solve(@(r)fun(r, n), r0);

norm(fun(sol, n));

Plot(r, sol(1:n), color{delta-1});hold on;

end

legend('delta=2', 'delta=3','delta=4','delta=5');

Function res=fun(z, n)

lambda =0.1;

beta = 0.1;

eps = 0.1;

m=0;

r = ((1:n)-1)/(n-1);

h=1/(n-1);

U = z(1:n);

res_U = zeros(n, 1);

res_U(1)=0;

For i=2:n-1

res_U(i)=((1+beta*U(i))/h^2)*(U(i+1)-2*U(i)+U(i-1))+((1+beta*U(i))/(r*h))*(U(i+1)-U(i-1))+beta*((U(i+1)-U(i-1))/h)^2+lambda*((1+eps*U(i))^m)*exp(U(i)/(1+eps*U(i)))-delta*U(i);

end

res_U(n)=0;

res = res_U;

end

Outcome

>> res

Warning: Trust-region-dogleg algorithm of Fsolve can not handle non-square systems; using levenberg-marquardt algorithm instead.

Equation solved. Fsolve completed because the vector of function values is near zero as measured by the default value of the function tolerance, and the problem appears regular as measured by the gradient.

'Please help me correct this errors'

[Torsten]

The message you got is not an error message ; Fsolve finished successfully.

Do the values returned in the Sol vector make sense for you ?

[Oluwaseun Adisa]

But how can I convert boundary condition: U(-1) = 0; U(1) = 0; to initial value r0 =?

[Oluwaseun Adisa]

About the codes above not satisfy my result very well and what I discovered is that the boundary conditions not function in equations because if I delete the boundary conditions and run the program the program will still plot the same graph out. Anybody help me out

[Torsten]

Is it necessary to write your own code ? If not, you should use sophisticated matlab code especially designed for this problem class, i.e. bvp4c in your case.

[Oluwaseun Adisa]

How can I get that one?