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Suppose ${\varphi }_1,{\varphi }_2$ are two sufficiently regular solutions with the same Dirichlet boundary condition, and put $w={\varphi }_1-{\varphi }_2$. Then $\Delta w=0$ in $V$ and $w=0$ on its boundary. Apply the divergence theorem to $w\nabla w$:Thus $\nabla w=0$ throughout $V$, so $w$ is constant on each connected component. Its zero boundary values force every such constant to be zero. This proves uniqueness of the Dirichlet solution, without assuming its existence.

For a prescribed Neumann boundary condition, the difference instead has ${\partial }_{n}w=0$. The same energy argument still gives $\nabla w=0$, but does not fix the constant. Consequently the solution is unique up to an additive constant on each connected component. A value at one point, or the integral mean, fixes the constant on a connected region. There is also a necessary compatibility condition: if $g={\partial }_n\varphi$ on $S$, the divergence theorem givesIt is required on each component if the region is disconnected. Data failing this condition admit no harmonic solution. This is the homogeneous-source instance of the Neumann Poisson problem.

Testing a weak Neumann Poisson problem with the constant function one gives ${\int }_{\Omega }f=0$, proving necessity. For sufficiency, use the closed Hilbert space of mean-zero $H^1$ functions. The given Poincaré inequality makes $\Vert \nabla v{\Vert }_{L^2}$ an equivalent norm there. SetThe negative sign corresponds to $\Delta u=f$. Poincare's inequality gives boundedness of $\ell$ and coercivity of $a$. The Lax-Milgram theorem supplies a unique mean-zero solution. Every $v\in H^1$ is the sum of its mean-zero part and a constant; since $\int f=0$, the weak identity extends to every $v\in H^1$. This is precisely the weak equation with zero normal derivative as its natural boundary condition. Existence and uniqueness in the mean-zero space hold exactly when $\bar{f}=0$.

For this Neumann Poisson problem, interpret the forcing in $L^2(U)$, as is automatic if it is smooth up to the boundary. Literal interior smoothness alone does not ensure the integrals or bounded functionals required in this question: for example, $f(x)=x^{-2}$ on $(0,1)$ is interior smooth but even $\int f\cdot 1$ diverges. Classical regularity in the converse is likewise understood up to the boundary.

Suppose the weak solution is smooth on $\overline{U}$. Testing against compactly supported test functions gives $-\Delta u=f$ in distributions and hence pointwise. Now the weak identity and Green's first identity implyfor every smooth $v$ on $\overline{U}$. Every smooth boundary function has such an extension, so ${\partial }_{n}u=0$ on $\partial U$. This proves both the interior equation and the boundary condition.

Conversely, for $u\in C^2(\overline{U})$ satisfying the equation and zero normal derivative, Green's first identity gives the weak identity for all smooth $v$ on $\overline{U}$. The density of smooth functions in a Sobolev space and the Cauchy-Schwarz inequality extend it continuously to every $v\in H^1(U)$. Thus the classical solution is a weak solution, and the smooth weak solution is classical.

The constant function $v=1$ is an admissible $H^1(U)$ test for the Neumann Poisson problem. Its weak gradient is zero, so the weak identity immediately givesThis is necessary, and the preceding Hilbert-space construction proves sufficiency for $f\in L^2(U)$. It is the balance condition corresponding to zero total boundary flux.

The difference $w$ of two weak solutions lies in the clamped second-order Sobolev space. Testing with $w$ yields ${\int }_U(\Delta w{)}^2=0$, so $\Delta w=0$. Since $w\in H_0^1(U)$, integration by parts givesThe Poincaré inequality for zero boundary values now implies $\Vert w{\Vert }_2=0$. The clamped biharmonic problem has at most one weak solution. Unlike the Neumann Poisson problem, no additive constant is allowed by these boundary traces.

Expand the divergence of $\Phi \nabla \Psi -\Psi \nabla \Phi$. The cross terms cancel, leaving $\Phi \Delta \Psi -\Psi \Delta \Phi$. The divergence theorem gives Green's second identity,Use the Green function convention ${\Delta }_{x}G(x,x_0)=\delta (x-x_0)$. Integrating over a bounded domain containing $x_0$ yields $\boxed{{\int }_S{\partial }_{n}GdS=1}$. Thus zero Neumann boundary data would be incompatible with this unit Dirac delta source. A suitable Neumann Green function with constant boundary flux instead has ${\partial }_{n}G=1/|S|$, where $|S|$ is the boundary area. Substituting $\Psi =G$, $\Delta \Phi =\sigma$, ${\partial }_n\Phi =f$ givesThe last term is an undetermined constant. The data must obey the Neumann Poisson problem compatibility condition ${\int }_V\sigma dV={\int }_{S}fdS$, obtained by integrating the original equation. Under this condition an additive constant in $G$ cancels from the first two terms.

For the upper half-space, reflect the source to $x_0^{\ast }=(x_{0,1},x_{0,2},-x_{0,3})$. The method of images with equal-sign sources givesThe reflected singularity lies outside the domain; on $z=0$ the two normal derivatives cancel, so ${\partial }_{n}G=-{\partial }_{z}G=0$. Its total unit flux passes through a large hemisphere at infinity, where $G\sim -1/(2\pi R)$. This is why the bounded-domain condition on the boundary flux cannot be applied to the infinite plane alone.

With zero interior source and the given point boundary source, the boundary representation is $\Phi (x_0)=-G((0,0,0),x_0)$. The terms at infinity vanish for this resulting $1/R$ field. ThereforeTo check the singular boundary condition, $-{\partial }_z\Phi =z/[2\pi (x^2+y^2+z^2{)}^{3/2}]$ is the Poisson kernel for the upper half-space. Its integral over a horizontal plane is one and its mass concentrates at the origin as $z\downarrow 0$, giving ${\partial }_n\Phi =\delta (x)\delta (y)$ in the distributional sense. This is the Neumann point source on a half-space boundary.

On a connected open set, a Sobolev space function with zero weak gradient is constant almost everywhere. Local mollification gives constant smooth representatives on interior balls; overlaps and connectedness identify their constants. Without connectedness, the constant may differ between components. This explains the constant ambiguity in a Neumann Poisson problem.