Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 105 / 2

Suppose the claimed Poincare inequality with a partial Dirichlet boundary fails. There are $v_k\in V$ withAfter the normalization $w_k=v_k/\Vert v_k{\Vert }_{L^2(U)}$,Thus $(w_k)$ is bounded in the Sobolev space $H^1(U)$. The Rellich-Kondrashov compactness theorem and the corresponding compact embedding for a bounded $C^1$ domain give a subsequence that converges strongly in $L^2(U)$ and weakly in $H^1(U)$ to some $w$. The Sobolev space with a partial Dirichlet condition $V$ is a closed vector subspace, hence weakly closed, so $w\in V$. Moreover $\nabla w=0$, and connectedness of $U$ makes $w$ a constant function. Its trace vanishes on the positive-measure set ${\Gamma }_1$, so that constant is zero. This contradictsTherefore some $C_P$ satisfiesSince the reverse bound $\Vert \nabla v{\Vert }_2\le \Vert v{\Vert }_{H^1}$ is immediate,The gradient seminorm is a norm on $V$ because equality to zero would make $v$ a constant whose trace on ${\Gamma }_1$ is zero.

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Multiply the Poisson equation by $v\in V$ and apply Green's first identity. The Dirichlet boundary condition makes the trace of $v$ vanish on ${\Gamma }_1$, while the Neumann boundary condition makes the boundary flux vanish on ${\Gamma }_2$. Thus the weak formulation is: find $u\in V$ such thatwhere

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First choose a test function $v\in C_c^{\infty }(U)$. The weak formulation and integration by parts giveThe fundamental lemma of the calculus of variations implies $-\Delta u=f$ pointwise because $u\in C^2(\overline{U})$ and $f$ is continuous. The Dirichlet boundary condition on ${\Gamma }_1$ already follows from $u\in V$ and continuity of $u$.

For arbitrary $v\in V$, Green's first identity and the interior equation now reduce the weak identity toThe traces of smooth members of $V$ can be chosen freely on compact subsets of ${\Gamma }_2$. Another application of the fundamental lemma of the calculus of variations, now on the boundary, gives ${\partial }_{\nu }u=0$ pointwise on ${\Gamma }_2$. Hence $u$ is a classical solution of the complete mixed boundary value problem.

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The restricted trace map is continuous, and $V$ is its kernel, so $V$ is a closed vector subspace of the Hilbert space $H^1(U)$. It is therefore complete in the $H^1$ norm. Part a shows that the gradient normis equivalent to that norm, so it is complete as well. It comes from the inner productConsequently $(V,\Vert \cdot {\Vert }_V)$ is a Hilbert space.

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On the Hilbert space $V$, the form $B$ from part b obeysso it is a bounded bilinear form and a coercive bilinear form. The Cauchy-Schwarz inequality and the Poincare inequality with a partial Dirichlet boundary giveso $\ell$ is a bounded linear functional. The Lax-Milgram theorem now gives a unique weak solution $u\in V$. Taking $v=u$ in the weak identity yieldsand therefore

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