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Let $w$ be a bounded harmonic function in a bounded connected smooth planar region, with zero continuous boundary limits except at finitely many points. Then $w=0$. This extends zero-boundary harmonic uniqueness to bounded solutions with isolated boundary exceptions, without assuming their initial Dirichlet energy is finite.

Choose ${\eta }_{ϵ}=0$ within distance ${ϵ}^2$ of each exceptional point, ${\eta }_{ϵ}=1$ beyond distance $ϵ$, and interpolate linearly in the logarithm of distance. Each annular transition has $\int |\nabla {\eta }_{ϵ}{|}^2=O(1/|\log ϵ|)$. Apply the divergence theorem to the equation multiplied by ${\eta }_{ϵ}^{2}w$. The ordinary boundary terms vanish, and the Cauchy-Schwarz inequality givesEvery fixed compact subset eventually has ${\eta }_{ϵ}=1$, so the gradient vanishes everywhere. The nonexceptional boundary data then fix the constant to zero. Differences of bounded harmonic extensions with identical piecewise continuous data satisfy this hypothesis. The individual extensions can have infinite Dirichlet energy at a jump; the cutoff argument avoids applying an unjustified finite-energy identity to them.

For a real-valued function on a bounded regular region, apply the divergence theorem to $\varphi \nabla \varphi$. The product rule yields $\nabla \cdot (\varphi \nabla \varphi )=|\nabla \varphi {|}^2+\varphi {\nabla }^2\varphi$. The prescribed zero boundary values and Laplace equation therefore implyThe continuous nonnegative integrand must vanish everywhere. Hence $\varphi$ is constant on each connected component, and its boundary value makes each constant zero. This proves zero-boundary harmonic uniqueness. If complex-valued functions are allowed, apply the same argument separately to their real and imaginary parts. If ${\psi }_1,{\psi }_2$ solve the same Poisson equation and Dirichlet boundary data, their difference is harmonic with zero boundary value, so is zero. Thus there is at most one solution. An exterior unbounded region would require an additional condition at infinity; the preceding bounded-domain proof does not assert uniqueness without one.

For the radial calculation use Cartesian coordinates with $r=(x_i{x}_i{)}^{1/2}>0$. The chain rule gives ${\partial }_{i}r=x_i/r$ andDifferentiate again and sum the three coordinates:This derives the radial Laplacian directly in Cartesian coordinates. For ${\nabla }^2\psi =c$, set $u=r\psi$, so $u^{\prime \prime }=cr$. Integrating twice gives $u=c{r}^3/6+ar+b$, and consequentlyIf the solution must extend boundedly through the origin, $b=0$. The remaining polynomial is smooth there.