Suppose $u_1$ and $u_2$ are <weak solutions> with the same <Sobolev trace theorem>[trace], and put $w=u_1-u_2\in H_0^1(\Omega)$. The <weak formulation> permits $w$ itself as a <test function>, giving
$$
0=\int_\Omega Dw\mathbin\cdot Dw=\int_\Omega|Dw|^2.
$$
Thus $w$ is <almost everywhere> constant, and its zero trace makes that constant zero. This proves uniqueness.

The weak identity also says that $\Delta u=0$ in the sense of <distribution theory>[distributions]. The <Weyl lemma> therefore gives $u\in C^\infty(\Omega)$ and $\Delta u=0$ pointwise. The assumed <continuous function>[continuity] on $\overline\Omega$ retains the prescribed boundary values, so the weak solution is the unique classical solution in $C^\infty(\Omega)\cap C^0(\overline\Omega)$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 107 / 1 / b / Solution

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The Weyl lemma says that every locally integrable distributional solution of $\Delta u=0$ agrees almost everywhere with a smooth harmonic function. Convolution with a <mollifier> and the harmonic mean value property provide a standard proof.

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 Incoming links: Weyl lemma