Let
- $D:=(0,1)$
- $U:=L^2(D)$ and $$\phi_n(x):=\sqrt 2\sin(n\pi x)\;\;\;\text{for }n\in\mathbb N\text{ and }x\in D$$
- $H:=H^2(D)\cap H_0^1(D)$ and $$A:=-\frac{\partial^2}{\partial x^2}u\;\;\;\text{for }u\in H$$
Since $(\phi_n)_{n\in\mathbb N}\subseteq C^\infty(D)$ is an orthonormal basis of $U$ with $$\left.\phi_n\right|_{\partial D}=0\;\;\;\text{for all }n\in\mathbb N\;,$$ $(\phi_n)_{n\in\mathbb N}\subseteq H$ with $$A\phi_n=\underbrace{\pi^2n^2}_{=:\lambda_n}\phi_n\;\;\;\text{for all }n\in\mathbb N\tag 1$$ and $$\left\|A\phi_n\right\|_U\stackrel{(1)}=\lambda_n\stackrel{n\to\infty}\infty\;.\tag 2$$ Thus, $A$ is an unbounded operator.
Now, let $y\in D$ and $G:D\times D\to\mathbb R$ be the solution of $$\left\{\begin{matrix}AG(\;\cdot\;,y)&=&\delta(\;\cdot\;-y)&&\text{in}&D\\ G(\;\cdot\;,y)&=&0&&\text{on}&\partial D\end{matrix}\right.\;.\tag 3$$ How can we compute $G$?
I know that the solution will be $$G(x,y)=\begin{cases}x(1-y)&\text{, if }x<y\\y(1-x)&\text{, if }x\ge y\end{cases}\tag 4$$ and I've read that we can obtain it by using $(1)$. However, how can we do that?