Exploring the Structure of the Poisson Equation

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Homework Statement

please see below

Relevant Equations

please see below

Theorem 1: There exists at most one solution $u\in C^2(\bar{\Omega})$ of the Dirichlet boundary value problem.

Proof:
(1) We assume there is a second solution $\tilde{u}$ of the Dirichlet boundary value problem. We compute $$\Delta v=\Delta (u-\tilde{u})\Rightarrow -\Delta u + \Delta \tilde{u}\Rightarrow f-f=0$$
and $$v=u-\tilde{u}\Rightarrow g-g=0$$
(2) So the Dirichlet problem satisfied by $v=u-\tilde{u}$ is
\begin{equation}
\begin{cases}
-\Delta v=0 \quad \text{in}\quad \Omega\\
\quad v=0 \quad\text{on}\quad\partial \Omega\\
\end{cases}
\end{equation}(3) To deduce $\int_\Omega |Dv(x)|^2dx=0$ we integrate $v\Delta v$ over $\Omega$
$$\int_\Omega v\Delta v dx= \int_\Omega (u-\tilde{u})\Delta (u - \tilde{u})dx$$
using integration by parts
$$=(u-\tilde{u})\int_\Omega \Delta (u-\tilde{u})dx-\int_\Omega\Big[\int_\Omega D(u-\tilde{u})dx\times D(u-\tilde{u})\Big]dx$$
$$\Rightarrow vDv-\int_\Omega v\Delta v dx= \int_\Omega |Dv(x)|^2dx$$ Then setting $v= 0$ and $\Delta v = 0$
$$\int_\Omega |Dv(x)|^2dx=0$$ (4) $v=0$ implies uniqueness for the Dirichlet problem because $u=\tilde{u}$

(5) we need the assumption that $\partial \Omega\in C^1$ to define $\nu$ which be evaluated at a cusp or a discontinuity.

(6) We know that $$v\in C^{\infty}_c(\Omega)$$ and $v$ has compact support in $\Omega$. $$u+sv\in C^2(\bar{\Omega})$$ and in $\partial \Omega$ $$u+sv=g$$ It follows $$u_s=u+sv\in A:=\{w\in C^2(\bar{\Omega})|w=g\quad \text{on} \quad \partial \Omega\}$$
(7) To expand the energy functional $I[u_s]$ in terms of $s$ we make exchanges using $u_s=u+sv$ $$I[u+sv]=\frac{1}{2}\int_\Omega |D(u+sv)(x)|^2dx-\int_\Omega f(u+sv)dx$$
(8) We first differentiate with respect to $s$ and then evaluate at $s=0$ $$\frac{d}{ds}\Big|_{s=0}I[u_s]$$ $$=\frac{d}{ds}\Big[\frac{1}{2}\int_\Omega |D(u+sv)(x)|^2dx-\int_\Omega f(u+sv)dx\Big]_{s=0}$$
$$=\Big[\frac{1}{2}\int_\Omega \frac{d}{ds}|D(u+sv)(x)|^2dx-\int_\Omega \frac{d}{ds}f(u+sv)dx\Big]_{s=0}$$
$$=\Big[\int_\Omega D(u+sv)(x)\Delta (u+sv)v dx-\int_\Omega fvdx\Big]_{s=0}$$ $$\Rightarrow \int_\Omega fv dx-\int_\Omega fvdx=0$$(9) $u$ being a minimizer implies that $$\frac{d}{ds}\Big|_{s=0}I[u_s]=0$$ because $s=0$ is a point of extremum of $I[u_s]$ so the derivative w.r.t $s$ is zero at $s=0$ which is also where $u_s=u$.

(10) We know from part (4) $$v=0$$ and from the Dirichlet problem $$-\Delta u-f=0$$ in $\Omega$. It follows that
$$0=\int_\Omega (-\Delta u - f)vdx$$
\textbf{Lemma 3:} suppose $w\in C^0(\Omega)$. If we have for all $\varphi \in C^\infty_c(\Omega)$ the relation
$$\int_\Omega w\varphi dx=0$$then we have $w=0$ in $\Omega$.

(11) from part (10) we have
$$0=\int_\Omega (-\Delta u - f)vdx$$
which indicates, by Lemma 3
$$-\Delta u-f=0 \quad \text{in} \quad \Omega$$
it follows that
$$-\Delta u = f\quad \text{in} \quad \Omega$$