Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 105 (Analysis of Partial Differential Equations)

The principal symbol of the equation isIn the standard notation $A{u}_{tt}+2B{u}_{tx}+C{u}_{xx}$, we have $A=tx$, $B=0$, and $C=-1$, soA second-order equation is a hyperbolic partial differential equation exactly where this discriminant is positive. Hence the hyperbolic set isthe union of the first and third open quadrants.

A characteristic hypersurface of the form $\varphi (t,x)=p(t)\pm q(x)$ must satisfyOn either connected component of $tx>0$, separation givesOne choice valid on both components isIndeed, $p^{\prime }(t)=|t{|}^{-1/2}$ and $q^{\prime }(x)=|x{|}^{1/2}$ away from the axes, so $tx{p}^{\prime 2}=q^{\prime 2}$. Thusare the two families of characteristics, and $p\pm q$ are characteristic coordinates.

The initial line is $t=t_0$, whose conormal is $dt$. Evaluating the principal symbol on it givesIt is therefore a non-characteristic hypersurface at $(t_0,x_0)$ exactly when $t_0{x}_0\ne 0$. At such a point the equation can be writtenwhose right-hand side is real analytic locally, and the prescribed Cauchy data are also real analytic. The Cauchy-Kovalevskaya theorem consequently gives a unique local analytic solution exactly at the points

For $u\in L_{\mathrm{loc}}^1(\mathbb{R})$, a function $g\in L_{\mathrm{loc}}^1(\mathbb{R})$ is its weak derivative whenfor every test function $\phi \in C_c^{\infty }(\mathbb{R})$. Equivalently, the distributional derivative of $u$ is represented by the locally integrable function $g$.

For $u(x)=\sin |x|$, the ordinary derivative away from zero isThis bounded function is locally integrable. For a test function $\phi$, integration by parts on $(-\pi ,0)$ and $(0,\pi )$ produces boundary terms at zero which cancel because $u$ is continuous there and $u(0)=0$. HenceTherefore $u$ has the weak derivativeThe jump in the ordinary derivative creates no Dirac delta function; such a term would arise from a jump in the function itself.

For $1\le p<n$, define the Sobolev conjugate exponentThe Sobolev inequality, also called the Gagliardo--Nirenberg--Sobolev inequality, states that there is a constant $C=C(n,p)$ such thatfor every $u\in C_c^{\infty }({\mathbb{R}}^n)$, and hence by completion for every $u\in W^{1,p}({\mathbb{R}}^n)$ for which the right formulation applies.

Take $u\in W_0^{1,p}(U)$ and extend it by zero outside $U$. The zero extension of W01 belongs to $W^{1,p}({\mathbb{R}}^n)$ and the Sobolev inequality givesBecause $U$ has finite measure, the Holder inequality givesThusThe reverse estimate follows directly from $\Vert Du{\Vert }_{L^p}\le \Vert u{\Vert }_{W^{1,p}}$. After adjusting constants,This is the Poincare inequality on $W_0^{1,p}(U)$.

The Fredholm alternative for an elliptic Dirichlet problem says that either the homogeneous adjoint problem has only the zero solution, in which case $Lu=f$ has a unique solution for every admissible $f$, or the homogeneous kernels are nontrivial and finite-dimensional. In the latter case,is solvable exactly whenand any two solutions differ by an element of $\ker L$. Moreover $\dim \ker L=\dim \ker L^{\ast }$.

For $L=d^2/d{x}^2+1$ with homogeneous Dirichlet data at $0$ and $2\pi$, integration by parts shows that $L^{\ast }=L$. Its homogeneous kernel isbecause $A\cos x+B\sin x$ vanishes at both endpoints exactly when $A=0$.

The forcing obeys the orthogonality conditionThe Fredholm alternative for an elliptic Dirichlet problem therefore says that solutions exist, though they are not unique. Indeed,satisfies $u^{\prime \prime }+u=\cos x$ and both boundary conditions for every constant $C$.

SetThe Holder inequality givesUsing $x^{-p}dx=-(p-1{)}^{-1}d(x^{1-p})$ and integration by parts, while discarding the nonpositive boundary term at $x=1$, yieldsAnother application of Hölder's inequality givesAfter cancellation, with the zero case immediate,This is the Hardy averaging inequality.

The one-dimensional Sobolev representative of $u$ is absolutely continuous, and its zero trace givesConsequently $u(x)/x=A(u^{\prime })(x)$ for the Hardy operator. Applying the Hardy averaging inequality to $u^{\prime }$ gives the Hardy inequality on an interval:

The Sobolev trace theorem makes evaluation at zero a continuous linear map $T_0:H^1(0,1)\to \mathbb{R}$. Henceis a closed vector subspace of the Hilbert space $H^1(0,1)$. Every closed vector subspace of a Hilbert space is complete with the restricted inner product, so $H$ is a Hilbert space with the standard $H^1$ inner product.

For $u\in H$, the trace $u(0)$ vanishes. Apply the Hardy inequality on an interval with $p=2$ to obtainThus $u/x\in L^2(0,1)$.

Multiply the differential equation by $v\in H$ and integrate. The Hardy inequality on an interval makes the terms containing $u/x$ and $v/x$ integrable. For smooth $v$, integration by parts givesbecause $v(0)=0$ and the Neumann boundary condition is $u^{\prime }(1)=0$. ThereforeThe density of smooth functions in a Sobolev space and continuity of all four terms extend this weak formulation to every $v\in H$.

Define on $H$ the bilinear formand the linear functionalThe Hardy inequality on an interval, Cauchy-Schwarz inequality, and the one-sided Poincare inequality show that $a$ is a bounded bilinear form and that

For $v\in H$, $v(x)={\int }_0^x{v}^{\prime }(t)dt$, so Cauchy--Schwarz and Fubini's theorem giveHenceThus $a$ is a coercive bilinear form. The Lax-Milgram theorem supplies a unique $u\in H$ satisfying $a(u,v)=\ell (v)$ for every $v\in H$. This is precisely the unique weak solution described by the weak boundary value problem with an inverse-square potential.

The condition ${\sum }_{i=1}^{n}1/p_i=1$ and positivity imply $p_i\ge 1$. We prove the Generalized Holder inequality by induction on $n$. The case $n=2$ is the usual Holder inequality. For the induction step, setThe exponents $q$ and $p_n$ are conjugate, so Hölder followed by the induction hypothesis givesThis proves the claim.

Since $U\subset {\mathbb{R}}^3$ is bounded and smooth, the Sobolev embedding theorem gives continuous embeddings $H^1(U)\hookrightarrow L^4(U)$ and $H^1(U)\hookrightarrow L^6(U)$. Because $|\sin s|\le 1$,Taking the $L^2$ norm in time gives

For $F(s)=s^2\sin s$, the mean value theorem and $|F^{\prime }(s)|\le 2|s|+s^2$ implyApply the Generalized Holder inequality in space, using $L^4$ for the quadratic products and $L^6\cdot L^3$ for the cubic products, and then use the two Sobolev embeddings. Pointwise in time this yieldsTaking the $L^2$ norm in time proves the required estimate with the factor $\gamma T^{1/2}$.

PutThe assumed energy estimate for the linear wave equation and the first nonlinear estimate give, for $w\in X_{b,\tau }$,Choose $b\ge 2C_{0}D$ and then choose $\tau >0$ so small that $C_0\beta {\tau }^{1/2}{b}^2\le b/2$. Then $A$ maps the closed ball $X_{b,\tau }$ into itself.

For $w,\tilde{w}\in X_{b,\tau }$, the difference $Aw-A\tilde{w}$ solves the linear equation with zero Cauchy data and forcing $F(w)-F(\tilde{w})$. The second nonlinear estimate therefore givesShrinking $\tau$ once more makes the coefficient strictly smaller than one. Since $X_{\tau }$ is a Banach space and its closed ball is complete, $A$ is a contraction on $X_{b,\tau }$.

By the contraction mapping theorem, the contraction $A:X_{b,\tau }\to X_{b,\tau }$ has a unique fixed point $u$. The identity $Au=u$ says exactly that $u$ is the weak solution of the linear initial boundary value problem whose forcing is $u^2\sin u$. Consequently $u\in X_{\tau }$ satisfieswith the prescribed initial and homogeneous Dirichlet data. Thus a sufficiently small $\tau >0$ gives a local weak solution of the semilinear wave equation, as summarized by the local weak solution by contraction for a semilinear wave equation.