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Under the usual bounded-domain hypotheses, with piecewise smooth boundary and a sufficiently smooth vector field, the divergence theorem stateswhere $\mathbf{n}$ is outward. Apply it to $g\mathbf{F}$ and use the product rule $\nabla \cdot (g\mathbf{F})=\mathbf{F}\cdot \nabla g+g\nabla \cdot \mathbf{F}$ to obtainFor uniqueness, let $w=u_1-u_2$ for two classical solutions with the same Dirichlet boundary data. Then $\Delta w=0$ and $w=0$ on the boundary. Taking $\mathbf{F}=\nabla w$, $g=w$ gives Green's first identity in the formThe continuous nonnegative integrand must vanish, so $w$ is constant on each connected component. The zero boundary data make every such constant zero. Thus the solution of the Dirichlet problem for the Laplace equation is unique, if it exists. This establishes Uniqueness of the Dirichlet problem; existence is assumed. Boundedness, or suitable conditions at infinity, is needed: on an unbounded half-space the harmonic function $w=z$ has zero boundary values but is not identically zero.

Use the Fourier transform convention $\hat{F}(k)={\int }_{\mathbb{R}}F(x)e^{-ikx}dx$, with inverse factor $1/(2\pi )$. Splitting the inverse integral at zero givesThis is the Poisson kernel for the upper half-plane at height $a$.

Take the Fourier transform in $x$ of the Laplace equation. The transformed boundary-value problem isFor $k\ne 0$ the general solution is a linear combination of $e^{|k|y}$ and $e^{-|k|y}$; the two conditions set their coefficients to zero and one respectively. At $k=0$, the solution is the constant one. Inverting therefore givesThe upper boundary holds pointwise. At the lower boundary the Dirac delta function must be interpreted distributionally: the kernel has integral one and concentrates at zero, so for every continuous compactly supported test function $h$, $\int \psi (x,y)h(x)dx\to h(0)$. This also specifies the solution in the assumed class admitting the stated Fourier transforms.