So I found the general solution for this DE with these BCs: $$\frac{d^2u}{dx^2}+u=1$$ $$u(0)=u(L)=0$$ by doing: $$u(x)=Acos(x)+Bsin(x)+1$$ $$u(0)=0=A+1 \therefore A=-1$$ $$u(L)=0=-cos(L)+Bsin(L)+1 \therefore B=\frac{cos(L)-1}{sin(L)}$$ $$\implies u(x)=-cos(x)+1+sin(x)\frac{cos(L)-1}{sin(L)}=1-cos(x)-sin(x)tan(\frac{L}{2})$$
So then I wanted to check the Fredholm Alternative. So the homogeneous solution would be: $$u_h(x)=Acos(x)+Bsin(x)$$ $$u_h(0)=0=A \therefore A=0$$ $$u_h(L)=0=Bsin(L) \implies L=n\pi \implies B=1$$ $$\implies u_h(x)=sin(x)$$ so checking Fredholm... $$\int_{0}^{L} (1)sin(x) dx=-cos(x)\Big|_0^L=1-cos(L)\ne0$$
Could anyone give me insight what's going on here? Doesn't this mean that the Fredholm Alternative is telling me there's no solution? What's going on here? Any help in understanding this would be greatly appreciated.
edit:
The Fredholm alternative summarizes the results for nonhomogeneous problems: $$L(u)=f(x)$$ subject to homogeneous BCs of the self-adjoint type. Either:
- $u=0$ is the only homogeneous solution (i.e. $\lambda \ne 0$) in which case the nonhomogeneous problem has a unqiue solutions, or
- there are nontrivial homogeneous solutions $\phi_h(x)$ (i.e. $\lambda = 0$ is an eigenvalue) in which case the nonhomogeneous problem has no solutions or an infinite number of solutions
For the nonhomogeneous problem with homogeneous BCs, then, solutions exist only if the forcing function is orthogonal to all homogeneous solutions.
The forcing function being orthogonal to the homogeneous solution (with weight 1) is represented by: $$\int_{a}^{b}f(x) \phi_h(x) dx=0$$
The number of solutions of $L(u)=f(x)$ subject to homogeneous BCs is given by:
- if $\phi_h=0$ ($\lambda\ne0$) and $\int_{a}^{b}f(x) \phi_h(x) dx=0$ then the number of solutions is $=1$
- if $\phi_h\ne0$ ($\lambda=0$) and $\int_{a}^{b}f(x) \phi_h(x) dx=0$ then the number of solutions is $=\infty$
- if $\phi_h\ne0$ ($\lambda=0$) and $\int_{a}^{b}f(x) \phi_h(x) dx\ne0$ then the number of solutions is $=0$
When you do your first solution, you have to be a bit careful. If $L=n\pi$ with $n\in\mathbb Z$ you cannot divide by $\sin(L)$ since it is zero.
If $L\neq n\pi$ with $n\in\mathbb Z$ you have exactly one solution.
If $L=2n\pi$ you have $\cos(L)=1$ so the second boundary condition is satisfied, and there will be infinitely many solutions, $$ u(x)=1-\cos x+B\sin x. $$
If $L=(2n-1)\pi$ you have $\cos(L)=-1$, so the second boundary condition is not satisfied. Thus, no solutions.
I suggest that you try to fit this into the theory of Fredholm alternatives.