Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 105 / 1

The scalar Cauchy-Kovalevskaya theorem says that a partial differential equation solved for its highest derivative normal to a real-analytic non-characteristic hypersurface, with real-analytic coefficients and Cauchy data, has a unique local real-analytic solution. In coordinates, an equationhas such a solution near the origin when $F$ and the prescribed values of ${\partial }_t^j\varphi (0,x)$ for $0\le j<m$ are real analytic.

Choose a real-analytic primitive $F_0$ of $f$ near zero and apply the theorem to the scalar Laplace equationThe line $t=0$ is non-characteristic because the coefficient of ${\varphi }_{tt}$ is one. DefineThen $u(0,x)=f(x)$ and $v(0,x)=g(x)$, while equality of mixed derivatives and the Laplace equation giveThus $(u,v)$ is the required local real-analytic solution.

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Suppose a $C^1$ classical solution existed and set $w=u+iv$. The system is preciselywhich are the Cauchy-Riemann equations for $w$ as a function of the complex number $z=x+it$. Hence $w$ is holomorphic near $z=0$. Every holomorphic function is real analytic, so its restrictionto the real axis is real analytic near zero. Its real part and imaginary part show that both $f$ and $g$ must be real analytic, contradicting the hypothesis. Therefore no such $C^1$ solution exists.

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Write $q_i=a^{ij}{D}_{j}u$. If $u$ is a classical solution, multiply $-{\partial }_i{q}_i=f$ by a smooth function $v$ and apply the divergence theorem. The Neumann boundary condition $q_i{N}_i=0$ removes the boundary term and givesThe density of smooth functions in a Sobolev space and boundedness of the coefficients extend this identity to every $v\in H^1(U)$, so $u$ is a weak solution.

Conversely, take $v$ to be a test function compactly supported in $U$. The weak formulation saysThe fundamental lemma of the calculus of variations gives the equation in $U$. Under the regularity implicit in the stated notion of a classical solution, it holds pointwise. Applying integration by parts again with arbitrary $v\in H^1(U)$ leaveswhere $T$ is the trace operator. Traces of smooth functions can be chosen arbitrarily on the boundary, so the boundary fundamental lemma of the calculus of variations gives $(a^{ij}{\partial }_{j}u)N_i=0$. Thus $u$ is a classical solution.

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Testing the weak formulation with the constant function $1$ proves the necessary compatibility conditionAssume first that $U$ is connected and this condition holds. On the mean-zero Sobolev spaceuse the norm $\Vert v{\Vert }_{†}=\Vert Dv{\Vert }_{L^2(U)}$. To prove the needed Poincare-Wirtinger inequality, suppose it failed. There would be $v_k\in H_{†}^1(U)$ with $\Vert v_k{\Vert }_2=1$ and $\Vert D{v}_k{\Vert }_2\to 0$. The Rellich-Kondrachov compactness theorem gives a subsequence converging strongly in $L^2$ and weakly in $H^1$ to a function $v$. Its weak derivative vanishes, so connectedness makes $v$ constant; its zero mean makes it zero. This contradicts $\Vert v{\Vert }_2=1$. Hence $\Vert Dv{\Vert }_2$ is an equivalent Hilbert space norm on $H_{†}^1(U)$.

DefineBoundedness of $a^{ij}$ makes $B$ a bounded bilinear form, while uniform ellipticity givesso it is a coercive bilinear form. The Cauchy-Schwarz inequality and the Poincare-Wirtinger inequality make $\ell$ a bounded linear functional. The Lax-Milgram theorem supplies a unique $u\in H_{†}^1(U)$ satisfying $B(u,v)=\ell (v)$ for every mean-zero $v$. For arbitrary $v\in H^1(U)$, subtract its mean; the omitted constant contributes zero on both sides because $D1=0$ and ${\int }_{U}f=0$. Thus $u$ solves the original problem.

If two solutions exist, their difference $w$ satisfies $B(w,w)=0$, so uniform ellipticity gives $Dw=0$. It is therefore constant on $U$. The solution is unique up to an additive constant, and its mean-zero representative is unique. If $U$ is disconnected, the precise condition is ${\int }_{U_j}f=0$ on every connected component $U_j$, and one independent additive constant remains on each component.

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