Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 105 (Analysis of Partial Differential Equations)

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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For $u,g\in L_{\mathrm{loc}}^1(U)$, the function $g$ is the $i$th weak derivative $D_{i}u$ whenfor every test function $\phi \in C_c^{\infty }(U)$. This is the integration by parts identity with no boundary term and agrees with the ordinary derivative whenever $u$ is classically differentiable.

For $1\le p\le \infty$, the first-order Sobolev space iswith norm, for example,Functions equal almost everywhere represent the same Sobolev element.

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Let $g=Du\in L^p(0,1)$ be the weak derivative and defineThe fundamental theorem of calculus for Lebesgue integration makes $F$ an absolutely continuous function, differentiable almost everywhere, with $F^{\prime }=g$ almost everywhere. The distributional derivative of $u-F$ is zero. A locally integrable function with zero distributional derivative on a connected interval is equal almost everywhere to a constant $C$. Consequentlyis an absolutely continuous representative of $u$, and ${\tilde{u}}^{\prime }=g\in L^p(0,1)$ almost everywhere.

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Write points of $U$ as $(x,y)$ with $x>0$. For $h>0$, translate $u$ into the domain byContinuity of translations in $L^p$ applied to $u,D_{x}u,D_{y}u$ givesas $h\downarrow 0$. Choose a standard mollifier ${\rho }_{\epsilon }$ supported in a ball of radius $\epsilon <h/2$. For $(x,y)\in U$, the convolutiononly samples points with first coordinate greater than $-h$, so it is well-defined and smooth throughout $U$. The approximation-to-the-identity theorem, applied also to each weak derivative, allows $\epsilon (h)<h/2$ to be chosen so thatTaking $h=1/j$ and using the triangle inequality proves the density of smooth functions in a Sobolev space.

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Use the reflection extension from a half-spaceIt is plainly linear and restricts to $u$ on $U$. For a smooth $u$, the chain rule givesA change of variables therefore givesfor $p<\infty$, with the evident equality of essential suprema for $p=\infty$. Approximate a general $u$ by the smooth functions from part c. The estimate makes their reflections Cauchy in $W^{1,p}({\mathbb{R}}^2)$, and their limit defines a bounded Sobolev extension operator with $\Vert E\Vert \le 2^{1/p}$.

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For a smooth $u$ and fixed $y$, the fundamental theorem of calculus gives, for $0<x<1$,Average this inequality over $x\in (0,1)$, use Holder inequality on that unit interval, raise to the power $p$, and integrate in $y$. This proves the estimate behind the W1p trace theorem on a half-space:Use the Sobolev extension operator from part d, approximate $Eu$ in $W^{1,p}({\mathbb{R}}^2)$ by smooth functions, and define $Tu$ as the $L^p(\mathbb{R})$ limit of their restrictions to $x=0$. The trace inequality makes this limit independent of the approximation and proves thatis linear and bounded. For a smooth function that extends continuously to the boundary, $Tu=u{|}_{\partial U}$, so this is the trace operator required.

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The Rellich-Kondrachov compactness theorem says that if $U\subset {\mathbb{R}}^n$ is a bounded Lipschitz domain, thenfor $1\le q<p^{\ast }=np/(n-p)$ when $p<n$. When $p=n$, the embedding is compact into every finite $L^q$, and when $p>n$ it is compact into $C^0(\overline{U})$, hence into every $L^q$.

The boundedness of the domain is essential. Choose a nonzero $\varphi \in C_c^{\infty }(0,1)$ and setTranslation invariance gives $\Vert u_j{\Vert }_{W^{1,1}({\mathbb{R}}_{+})}=\Vert \varphi {\Vert }_{W^{1,1}}$, so after a fixed rescaling these functions lie in the unit ball. Their supports are pairwise disjoint andNo subsequence is Cauchy in $L^1$, so the unit ball is not compact. This is the standard failure of Rellich compactness on an unbounded domain.

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For $s\ge 0$ and $\xi \in \mathbb{R}$, define the characteristic curve $X(t;s,\xi )$ byThe bounded derivative $F_x$ makes $F(t,\cdot )$ globally Lipschitz, uniformly in $t$. On each finite time interval, $|F(t,x)|\le |F(t,0)|+L|x|$, so Gronwall inequality prevents finite-time escape. The Picard-Lindelof theorem therefore gives a unique trajectory for every finite $t\ge 0$. Differentiation in $\xi$ givesso the characteristic flow map is a $C^1$ increasing diffeomorphism.

Along a characteristic, the chain rule changes the equation intoTracing $(t,x)$ backward by the flow therefore givesThe regularity of the flow makes this a $C^1$ classical solution. Conversely, every classical solution obeys the same ordinary differential equation along every characteristic, so the formula also proves uniqueness.

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For $u_0\in L^{\infty }(\mathbb{R})$, call $u\in L^{\infty }({\mathbb{R}}_{+}\times \mathbb{R})$ a weak solution when, for every $\phi \in C_c^1([0,\infty )\times \mathbb{R})$,This follows by multiplying the equation by a test function and applying integration by parts in time and space; $(F\phi {)}_x$ includes the term $F_x\phi$ because the original transport operator is not in divergence form.

If $u$ is $C^1$, test functions supported away from $t=0$ show that $u_t+F{u}_x=g$ as a distributional identity, hence pointwise. Integrating this pointwise equation by parts in the weak identity leavesArbitrary boundary test functions and the fundamental lemma of the calculus of variations give $u(0,x)=u_0(x)$. Thus a $C^1$ weak solution is the unique classical solution from part a.

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With $g=0$, pull the bounded measurable initial value back along the characteristic flow map:The flow is measurable and invertible, so $u$ is measurable and $\Vert u{\Vert }_{\infty }\le \Vert u_0{\Vert }_{\infty }$. Choose smooth $u_0^{(k)}$ converging to $u_0$ in $L_{\mathrm{loc}}^1$ with uniformly bounded essential suprema, and definePart a makes each $u^{(k)}$ a classical, hence weak, solution. On every compact subset of spacetime, the $C^1$ change-of-variables formula for the flow and its locally bounded Jacobian determinant give $u^{(k)}\to u$ in $L^1$. Passing to the limit in the weak identity by dominated convergence proves that $u$ is a bounded weak solution with initial datum $u_0$.

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The characteristic curves are $x=\xi +at$. Along one of them, $U(t)=u(t,\xi +at)$ obeys the separable ordinary differential equationThereforeFor $0\le t<1$, the denominator is positive everywhere. At $t=1$, it first vanishes where $\cos \xi =1$, namely at $\xi =2\pi k$. The solution consequently has finite-time blowup at timealong the points $x=a+2\pi k$. No finite classical solution can continue through that time.

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Write the Inviscid Burgers equation in conservation form asA bounded function $u$ is a weak solution with initial datum $u_0$ whenfor every compactly supported $C^1$ test function $\phi$.

Across a straight discontinuity $x=\sigma t$, integration by parts on its two sides shows that the boundary terms cancel exactly when the Rankine-Hugoniot condition holds:when $u_{-}\ne u_{+}$. For every $c>0$, defineThe three jumps have left and right states $(0,-c)$, $(-c,c)$, and $(c,0)$, so their Rankine-Hugoniot speeds are respectively $-c/2$, $0$, and $c/2$, exactly the speeds of the displayed lines. Hence each $u_c$ satisfies the weak equation away from the origin and across every jump. Moreover, its nonzero support at time $t$ has length $ct$, so $u_c(t,\cdot )\to 0$ in $L_{\mathrm{loc}}^1$ as $t\downarrow 0$; its initial datum is therefore zero in the weak identity.

The zero function and all the distinct functions $u_c$ are bounded weak solutions with the same zero initial datum. Thus weak solutions are not unique. The central jump from $-c$ to $c$ is an expansion shock, which an entropy condition would exclude.

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