Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 105 (Analysis of Partial Differential Equations)

WriteAs a differential operator of order two, its principal symbol isIf $\nu =({\nu }_t,{\nu }_x,{\nu }_y)$ is a unit normal to the real-analytic hypersurface $\Sigma$, the non-characteristic hypersurface condition is thereforeUnder this condition the equation can be solved for the second derivative normal to $\Sigma$. The Cauchy-Kovalevskaya theorem then gives a unique real-analytic solution in a neighbourhood of $\Sigma$ for the prescribed analytic Cauchy data $u{|}_{\Sigma }$ and ${\partial }_{\nu }u{|}_{\Sigma }$.

The Lax-Milgram theorem says that if $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is a bounded bilinear form satisfyingfor some $\alpha >0$, and $F\in H^{\ast }$, then there is a unique $u\in H$ such that $B(u,v)=F(v)$ for every $v\in H$.

For this problem take $H=H_0^1(U)$ andThe Cauchy-Schwarz inequality and the continuous embedding $H_0^1(U)\hookrightarrow L^2(U)$ make both maps bounded. For smooth zero-boundary functions, integration by parts givesdensity extends this identity to $H_0^1(U)$. Consequentlyso $B$ is coercive. Lax–Milgram supplies the unique weak solution.

The Sobolev fundamental theorem of calculus on lines gives, for almost every $x$,Apply the Minkowski integral inequality and translation invariance of Lebesgue measure:The restriction on $h$ ensures that every translated copy used above lies in $U$.

Choose $h_j\to 0$. The assumed bound and the weak subsequence of a bounded Hilbert-space sequence result give a subsequence for whichFor every test function $\phi \in C_c^{\infty }(V)$, a change of variables yields the difference-quotient integration-by-parts identityThe right side converges to $-{\int }_{V}u{D}_i\phi$, while the left side converges to ${\int }_V{g}_i\phi$. Thus $g_i$ is the $i$th weak derivative of $u$. This holds for every $i$, so

The Rellich-Kondrachov compactness theorem states in particular that for a bounded open set $U\subset {\mathbb{R}}^n$ with smooth boundary, the embeddingis compact. More generally it is compact into $L^q(U)$ for $1\le q<2n/(n-2)$ when $n\ge 3$, for every finite $q$ when $n=2$, and into $C^0(\overline{U})$ in dimension one.

Boundedness is essential. Choose a nonzero $\phi \in C_c^{\infty }({\mathbb{R}}^n)$ and set $u_k(x)=\phi (x-k{e}_1)$ with pairwise disjoint supports. Their $H^1({\mathbb{R}}^n)$ norms are equal, whilefor $j\ne k$. Thus no subsequence converges in $L^2$.

The functionalis bounded by the Cauchy-Schwarz inequality. Hence $H=\ker I$ is a closed vector subspace of the Hilbert space $H^1(U)$. Every closed subspace of a Hilbert space is complete in the inherited norm, so $H$ is a Hilbert space with the standard $H^1$ inner product.

If the asserted Poincare inequality failed, after rescaling there would be $u_j\in H$ withThe sequence is bounded in $H^1(U)$. By the Rellich-Kondrachov compactness theorem, a subsequence converges strongly in $L^2(U)$ to some $u$ and weakly in $H^1(U)$. Its weak gradient is zero, so connectedness of $U$ makes $u$ almost everywhere constant. Continuity of the integral under $L^2$ convergence gives ${\int }_{U}u=0$, hence $u=0$. Strong convergence would then imply $\Vert u_j{\Vert }_2\to 0$, contradicting the normalization. Therefore the required constant exists.

The assumed inequality is equivalent toand equality holds at $w$. For $v\in H$ and $t\in \mathbb{R}$, expand $J(w+tv)\ge 0$ and use $J(w)=0$:Since this holds for both signs of arbitrarily small $t$, the linear coefficient vanishes:Every $v\in H^1(U)$ is a mean-zero function plus a constant. The same identity holds for constants because ${\int }_{U}w=0$, so it holds for all $v\in H^1(U)$. This is precisely the weak formulation ofwhere the Neumann boundary condition is the natural boundary condition encoded by the weak formulation.

Let $B$ and $F$ satisfy the hypotheses of the Lax-Milgram theorem. By the Riesz representation theorem, there are a bounded linear operator $T:H\to H$ and $z\in H$ such thatCoercivity and the Cauchy-Schwarz inequality implyso $\Vert Tu{\Vert }_H\ge \alpha \Vert u{\Vert }_H$. Thus $T$ is injective and its range is closed. If $y$ is orthogonal to its range, then $B(x,y)=0$ for every $x$; taking $x=y$ and using coercivity gives $y=0$. The range is therefore dense as well as closed, hence all of $H$. There is a unique $u=T^{-1}z$, and it satisfies $B(u,v)=F(v)$ for all $v$. The lower bound also gives $\Vert u{\Vert }_H\le \Vert F\Vert /\alpha$.

Set $\mathcal{H}=H_0^1(U{)}^2$. A weak solution is a vector $\varphi \in \mathcal{H}$ such thatwhere $D\varphi :Dv={\sum }_{j=1}^{2}D{\varphi }_j\cdot D{v}_j$ and the natural data space is $F\in H^{-1}(U{)}^2=(H_0^1(U{)}^2{)}^{\ast }$.

If $\varphi \in C^2(\overline{U}{)}^2$, take $v\in C_c^{\infty }(U{)}^2$ and integrate the gradient term by parts. The weak identity becomesThe fundamental lemma of the calculus of variations gives $-\Delta \varphi +A\varphi =F$ pointwise. Membership in $H_0^1(U{)}^2$, together with continuity up to the boundary, says that the boundary trace is zero, so the Dirichlet condition also holds classically.

For compactly supported smooth vector fields, two integrations by parts show that $-\Delta$ is formally self-adjoint. Pointwise,ThereforeIt follows that $L=L^{†}$ exactly when $A=A^T$, that is, when $A$ is a symmetric matrix at every point.

Use the bilinear formon $H_0^1(U{)}^2$. Smoothness on the compact set $\overline{U}$ makes $A$ bounded, so $B$ is bounded. Positive semidefiniteness givesThe Poincare inequality makes the right side coercive for the $H_0^1(U{)}^2$ norm. Hence the Lax-Milgram theorem gives a unique weak solution for everyIn particular, every $F\in L^2(U{)}^2$ is admissible.

In three dimensions, the Sobolev inequality gives $\Vert v{\Vert }_{L^6(U)}\le C_U\Vert v{\Vert }_{H_0^1(U)}$. Thereforeand hence

Using $w^3-{\tilde{w}}^3=(w-\tilde{w})(w^2+w\tilde{w}+{\tilde{w}}^2)$, the Holder inequality and the same Sobolev embedding give at each timeTaking the $L^2$ norm in time supplies the required estimate with the factor $T^{1/2}$.

The assumed linear estimate and the first bound giveChoose $b>2\alpha \Vert \psi {\Vert }_{H^1}$ and then choose $\tau >0$ so small that $\alpha \beta {\tau }^{1/2}{b}^3\le b/2$. This proves that $A$ maps $X_{b,\tau }$ into itself.

For $w,\tilde{w}\in X_{b,\tau }$, linearity of the heat equation and the second cubic estimate giveShrinking $\tau$ again makes $2\alpha \gamma {\tau }^{1/2}{b}^2<1$. Thus $A$ is a contraction of the closed ball $X_{b,\tau }$ in the Banach space $L^{\infty }((0,\tau );H_0^1(U))$.

The Banach fixed-point theorem gives a unique $u\in X_{b,\tau }$ with $A(u)=u$. By the definition of $A$, this fixed point is a weak solution of the linear heat problem with forcing $f=-u^3$. It therefore satisfieswith the prescribed initial and homogeneous Dirichlet data. Hence the nonlinear heat equation has a local weak solution in $L^{\infty }((0,\tau );H_0^1(U))$.