Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 105 / 1

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 $C$.