I have found (in lecture notes) a method to calculate the spectrum of the operator $$ A:D(A)\subset L^2([0,\pi])\longrightarrow L^2([0,\pi])\text{, such that} $$ $$ Au=\dfrac{\partial^2u}{\partial^2x^2}\, \text{, with the domain } D(A)=\{u\in H^{2}([0,\pi]):u(0)=u(\pi)=0\}. $$ The method is: take $v\in L^2([0,\pi])$. we will show that for $\lambda\neq n^2$($n\in\mathbb{N}^{*})$, the system $$(1)\begin{cases} \lambda u(x)-u''(x)=v(x), \text{ for } 0<x<\pi\\ u(0)=u(\pi)=0 \end{cases} $$ admits a unique solution $u\in D(A)$. Indeed, denoting by $$ v(x)=\sum_{n\geq 1}v_{n}\sin(nx),\, \text{ for }x\in[0,\pi], $$ the Fourier series of $v$. We seek a solution $u$ such that $$ u(x)=\sum_{n\geq 1}u_{n}\sin(nx),\, \text{ for }x\in[0,\pi]. $$ To satisfy the system $(1)$, we must have $$ (\lambda+n^2)u_{n}=v_{n},\, \text{ for }n\geq 1. $$ So, for $\lambda\neq -n^2$ the system $(1)$ admits a unique solution given by $$ u(x)=\sum_{n\geq 1}\dfrac{v_{n}}{\lambda+n^2}\sin(nx),\, \text{ for }x\in[0,\pi].$$ Thus $\sigma(A)=\{-n^2:\,n\in\mathbb{N}\}$.
My question is: how can I calculate the spectrum of the following operator $$ A:D(A)\subset L^1([0,\pi])\longrightarrow L^1([0,\pi])\text{, such that} $$ $$ Au=\dfrac{\partial^2u}{\partial^2x^2}\, \text{, with the domain } D(A)=\{u\in W^{2,1}([0,\pi]):u(0)=u(\pi)=0\}? $$ Thank you in advance