$''$ We are interested in solving problems like:
$Ly := (py')'− qy = f$
with boundary conditions
$β_1y(a) + γ_1 y'(a)=0$ , $β_2y(b) + γ_2 y'(b)=0.$
To this end we define the operator $Ly = (py')'− qy$ under the assumption that $λ = 0 $ is not an eigenvalue of $L$ and where $p, p'$ , and $q$ are continuous on $[a, b]$, $p(x) > 0$ on $[a, b]$ and
$|γj| + |βj|\neq0$ for j = 1, 2, in the Hilbert space H = $L^2$(a, b)
One can prove that the solution of this problem is unique and is given by the following type:
$ y(x)=\int_{b}^a g(x,z)f(z) dz $
where $g(x,z)$ is Green's function for operator $L$. $''$
My question is why the assumption that $λ$ IS NOT an eigenvalue of $L$ is necessary for the theory . What could have happen if $ 0 $ was an eigenvalue? Although it seems simple , I've been stuck to this point... I would appreciate any help.
Thanks in advance!!
The operator $L$ defined on a domain $\mathcal{D}(L)$ of twice continuously differentiable functions satisfying the two endpoint conditions is symmetric in the inner product $L^2[a,b]$ where $(h,k)=\int_{a}^{b}h(t)\overline{k(t)}dt$, which is to say that $$ (Lf,g) = (f,Lg),\;\;\; f,g\in\mathcal{D}(L). $$ Because of this, if $Lu = 0$ for $u\in\mathcal{D}(L)$, then $$ (Lf,u) = (f,Lu) = 0,\;\;\; f\in\mathcal{D}(L). $$ That keeps the range from being full, which means you cannot solve $Lf=g$ for all $g$--it is necessary that $(g,u)=0$. Furthermore, even if you can solve $Lf=g$, then $f$ is not unique because $L(f+\alpha u)=0$ for all scalars $\alpha$. On the other hand, if $\lambda=0$ is not an eigenvalue, then $L$ is invertible as stated using a Green function.