invertibiliy of $I + {{{\partial ^2}} \over {\partial {x^2}}}$

Question

We consider the operator $$T=I + {{{\partial ^2}} \over {\partial {x^2}}}:{H^2}(0,L) \cap H_0^1(0,L) \to {L^2}(0,L)$$. We shall prove that $T$ is invertible if and only if $L = n\pi $. and for this purpose we must show that $T$ is injective and surjective.. $T$ is Injective: $${\rm{Ker\{ T\} = \{ u}} \in {H^2}(0,L) \cap H_0^1(0,L),sinL = 0\} = \{ 0\} $$ implies $L = n\pi $. $T$ is surjectivr: the solution can be calculated explicitly for all $f$ in ${L^2}(0,L)$. My question is: How can we prove the Only if statement. Thank you.

Answer 1

Note that $\{\sin(\frac{n\pi x}{L})\}_{n=1}^\infty$ is dense in ${H^2}(0,L) \cap H_0^1(0,L)$. Let $$u=\sum_{n=1}^\infty u_n\sin(\frac{n\pi x}{L})$$ be such that $Tu=0$. Then $$ Tu=\sum_{n=1}^\infty(1-\frac{n^2\pi^2}{L^2})u_n\sin(\frac{n\pi x}{L}). $$ Thus $Tu=0$ if and only if $L=n\pi$ for some $n$ and then $\ker(T)=\text{span}\{\sin(x)\}$. Thus if $L\neq n\pi$ for any $n$, $\ker(T)=\{0\}$.

Answer 2

If $L=\pi,2\pi,3\pi,\cdots$, then $f(x)=\sin(x)$ is in $\mathcal{D}(T)$ because $f(0)=f(L)=0$, and $Tf=0$. So $T$ invertible only if $L \ne \pi,2\pi,3\pi$.

Conversely, if $L\ne \pi,2\pi,3\pi$, and if $g\in L^2$, then you can write down a solution $f\in\mathcal{D}(T)$ of $Tf=g$, and the solution is unique because $\mathcal{N}(T)=\{0\}$.