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Let $\Omega$ be an open set in $\mathbb{R}^n$ and now consider the weak formulation of Poisson's equation $$\int_{\Omega} \langle Du,Dv \rangle = \int_{\Omega}{fv}$$ for $v \in H_0^1$ and $u \in H_0^1$, $f \in L^2(\Omega)$

Does this problem have a unique weak solution $u$ such that the preceding weak formulation holds for all $v$?

It is quite some time ago that I covered this, but I think I can remember that one can actually show this by considering the convex energy functional. Unfortuantely, I think one needs that $\Omega$ is bounded or something like that. So could anybody help me turning my question in a correct statement(i.e. regarding the definition of $\Omega$, existence and uniqueness?)?

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    $\begingroup$ For $\Omega$ bounded the LHS defines an inner product in $H_0^1$ (Poincare's inequality) while the RHS defines a bounded linear functional in the same space. Riesz representation gives existence and uniqueness. $\endgroup$
    – Jose27
    Jul 10, 2015 at 17:41

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Short answer. If $\Omega$ is unbounded, you still have uniqueness but you might lose existence in $H^1_0(\Omega)$.

Long answer. You need boundedness of $\Omega$ to use the typical argument based on Riesz's representation of linear functionals (that's the argument that Jose mentions in his comment). This is because Poincaré's inequality $c\int_{\Omega}u^2\, dx \le \int_\Omega \lvert \nabla u\rvert^2\, dx$ is not available on unbounded domains.

However, you still have uniqueness in $H^1_0(\Omega)$, regardless of whether $\Omega$ is bounded or not. If $u_1$ and $u_2$ are two $H^1_0(\Omega)$ weak solutions to $$\tag{$\star$} \begin{cases}-\Delta u = f, & \Omega \\ u=0, & \partial \Omega\end{cases} $$ then the difference $v=u_1-u_2$ satisfies $$ \int_\Omega \lvert \nabla v\rvert^2\, dx =0$$ and so it is a constant. (I am here assuming that domain means connected open set). The only constant function in $H^1_0(\Omega)$ is the null function. Therefore $v=0$.

Remark 1. If you do not impose the condition $u\in H^1_0(\Omega)$, but rather look for classical solutions to $(\star)$, then you do lose uniqueness in unbounded domains. For example, in $\Omega=\mathbb{R}^n$ you can add any entire harmonic function to a solution and obtain another solution. The point is that $(\star)$ is underdetermined in unbounded domains, because no boundary condition is imposed at infinity. By insisting that the solution be in $H^1_0(\Omega)$ one forces some decay at infinity and thus recovers uniqueness.

Remark 2. The existence of solutions to $(\star)$ in unbounded domains is another story. Under the sole assumption that $f\in L^2(\Omega)$ it might happen that no $H^1_0(\Omega)$ weak solution exist. For a concrete example, fix $n= 3$ and $\Omega=\mathbb{R}^3$. Let $$ u(x)=\left(\frac{1}{1+\lvert x \rvert^2}\right)^\frac{1}{2}$$ and $$ f(x)=3\left(\frac{1}{1+\lvert x \rvert^2}\right)^\frac{5}{2}.$$ (The function $u$ appears in the elliptic theory of PDEs. It is called standard bubble). You can check that $$-\Delta u = f$$ but $u\notin L^2(\mathbb{R}^3)$ (hence $u\notin H^1(\mathbb{R}^3)$) even if $f\in L^2(\mathbb{R}^3)$. And $u$ is the unique solution to $(\star)$ with the given source term $f$, at least in the class of tempered distributions. In particular, no $H^1(\mathbb{R}^3)$ solutions to $(\star)$ with the given source term $f$ exist.

To see things clearly in $\Omega=\mathbb{R}^n$ you can solve $(\star)$ by means of the Fourier transform (or by convolution against the fundamental solution of the Laplace operator). Then the solution is formally written at Fourier side as $$\hat{u}=\frac{\hat{f}}{\lvert\xi\rvert^2}, $$ and this is the unique solution to $(\star)$ in the class of tempered distributions, as announced previously. Now it happens that $u\in H^1_0(\mathbb{R}^n)=H^1(\mathbb{R}^n)$ if and only if $$ \frac{(1+\lvert\xi\rvert^2)^\frac{1}{2}}{\lvert\xi\rvert^2} \hat{f}\in L^2(\mathbb{R}^n).$$ This is not guaranteed by the condition $f\in L^2(\mathbb{R}^n)$ alone: to have this, $\hat{f}$ must vanish at $\xi=0$ to second order at least.

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