# Weak solution for equation $-\Delta u = f$.

Suppose $$f \in L^{2}$$. I know that

$$\left\{\begin{array}{c} −\Delta u = f(x) & \text{on }\Omega \\ u(x)=0 & \text{on } \partial\Omega \end{array}\right.,$$

where $$\Omega \subset \mathbb{R}^{N}$$ is open and bounded, has a weak solution $$u \in H^{1}_{0}(\Omega)$$. I can obtain this solution using the Riesz Representation Theorem.

But, how obtain a weak soution for this equation with Neumann boundary conditions? And in the case where $$\Omega = \mathbb{R}^{N}$$?

In both cases I have been tried use the Lax Milgran Theorem, but I can't show the coercitive condition.

Another question: What the regularity of these weak solutions? I know that if $$f$$ is $$C^{0,\alpha}$$ and the problem has Dirichlet conditions, then the solution is $$C^{2}$$, am I right?

Let us start by discussing the following Neumann problem: \begin{align} - \Delta u &= f \quad \textrm{ in } \Omega, \\ \frac{\partial u}{\partial \nu} &= 0 \quad \textrm{ on } \partial \Omega. \end{align} Here, $\Omega \subset \mathbb{R}^n$ is a bounded Lipschitz domain. As Michal has already mentioned, a solution $u$ of this boundary value problem is not unique; you can always add a constant. Note that \begin{equation} \int_{\Omega} f = 0 \end{equation} is a necessary condition on $f$ to ensure existence of a solution to the boundary value problem. (Why?)
The tool to prove existence of a weak solution is the Lax-Milgram theorem. We choose the following subspace \begin{equation} H = \left\{ u \in H^1(\Omega) \; \colon \, \int_{\Omega} u = 0 \right\} \end{equation} of $H^1(\Omega)$. Note that the only constant function $c \in H$ is $c = 0$. Moreover, for functions $u \in H$ the following Poincare inequality holds: \begin{equation} \int_{\Omega} \lvert u \rvert^2 \leq C_P \int_{\Omega} \lvert \nabla u \rvert^2. \end{equation} Now, define $b \: \colon H \times H \to \mathbb{R}$ by \begin{equation} b(u, v) = \int_{\Omega} \langle \nabla u , \nabla v \rangle. \end{equation} Similar to the Dirichlet problem, you can show that $b$ is bounded; that is, there exists a constant $C> 0$ such that for all $u, v \in H$ we find that \begin{equation} \lvert b(u, v)\rvert \leq C \|u \| \cdot \|v\|. \end{equation} Coercivity of $b$ follows from the Poincare inequality. Thus, we can apply the Lax-Milgram theorem. For every $g \in H'$ we obtain a unique $u \in H$ such that \begin{equation} b(u, v) = \langle g, v \rangle_{H', H} \end{equation} for all $v \in H$. Here, $\langle \cdot, \cdot \rangle_{H', H}$ denotes the dual pairing of $H'$ and $H$.
To find a solution $u \in H$ to the boundary value problem above, we have to choose the functional $g$ appropriately. Set $g_0 \: \colon H \to \mathbb{R}$, \begin{equation} \langle g_0, v \rangle_{H', H} = \int_{\Omega} f v. \end{equation} Note that the boundary value $\partial_{\nu} u = 0$ is encoded in $g_0$, since there is no integral over the boundary $\partial \Omega$. We claim that the function $u \in H$ that satisfies \begin{equation} b(u, v) = \langle g_0, v \rangle \end{equation} for all $v \in H$ is a solution of the boundary value problem. To see this, we choose $\varphi \in C_c^{\infty}(\Omega)$. Then \begin{equation} \tilde{\varphi} = \varphi - \int_{\Omega} \varphi \in H. \end{equation} Moreover, we find that \begin{equation} \int_{\Omega} \langle \nabla u, \nabla \varphi \rangle = b(u, \tilde{\varphi}) = \langle g_0, \tilde{\varphi} \rangle = \int_{\Omega} f \varphi - \bigg( \int_{\Omega} f \bigg) \bigg( \int_{\Omega} \varphi\bigg) = \int_{\Omega} f \varphi. \end{equation} As $\varphi \in C_c^{\infty}(\Omega)$ was chosen arbitrarily, we have proved that $- \Delta u = f$ in $\Omega$.