Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 105 (Analysis of Partial Differential Equations)

Write $S=x_1+\cdots +x_n$. The Taylor series of the cosine isAfter expanding each power by the multinomial theorem, the coefficient of $x^{\alpha }$ has absolute value $1/\alpha !$ when $|\alpha |$ is even and is zero when $|\alpha |$ is odd. The hyperbolic cosinetherefore has exactly the absolute values of the coefficients of $f$. Thus $g$ is an entire majorant series for $f$.

The principal symbol ofis $p(x,t;\xi ,\tau )=(1-x^2){\xi }^2$. For a regular curve $s\mapsto (x(s),t(s))$, the characteristic curve equation isAway from $x=\pm 1$ this gives $t=\text{constant}$. The two degenerate lines $x=1$ and $x=-1$ are also characteristic. These are the characteristic curves, apart from reparametrization and pieces joined at the degenerate lines.

Along $x=\sin s$ and $t=s^2$, the characteristic expression isIt vanishes exactly when

The initial line $x=0$ has conormal $dx$, and the principal symbol on this conormal is $1$. It is therefore a non-characteristic hypersurface at every $(0,t_0)$. The equation's coefficients and the prescribed Cauchy data $u(0,t)=\cos t$ and $u_x(0,t)=0$ are real analytic functions. The Cauchy-Kovalevskaya theorem consequently gives a unique analytic solution in a neighbourhood of $(0,t_0)$ for every

In polar coordinates, setaway from the origin, assigning any value at the origin. This is unbounded as $r=\sqrt{x^2+y^2}\downarrow 0$. It belongs to $L^2(U)$ becauseMoreoverand henceThus $u\in H^1(U)$ but $u\notin L^{\infty }(U)$, exhibiting the failure of first-order Sobolev embedding into Linfinity in two dimensions.

The boundedness in the Sobolev space $H^1(U)$ and the weak sequential compactness of bounded sequences in a reflexive Banach space give a subsequence converging weakly to some $u\in H^1(U)$. For each integer $m\ge 2$, the Rellich-Kondrachov compactness theorem makes $H^1(U)\hookrightarrow L^m(U)$ compact because the dimension is two. Repeated extraction followed by the diagonal argument gives one subsequence $u_{n_k}$ converging strongly to $u$ in every $L^m(U)$ with integral $m\ge 2$.

For any finite real $p\ge 1$, choose an integer $m\ge \max \{2,p\}$. Since $U$ has finite measure, the Lp inclusion on a finite measure space givesThe same subsequence therefore works for every finite $p$.

The Holder inequality interpolates between $L^2(U)$ and $L^6(U)$:Taking cube roots and applying the three-dimensional Sobolev inequality $H^1(U)\hookrightarrow L^6(U)$ givesThis is the H1 L3 interpolation inequality in three dimensions.

Linearity in $u$ follows from linearity of the weak derivative and Lebesgue integration. By the Holder inequality, the three-dimensional Sobolev inequality, and the preceding $L^3$ interpolation estimate,Thus $\Phi$ is a linear functional and a continuous linear map on $H^1(U)$.

After passing to a subsequence, weak compactness and the Rellich-Kondrachov compactness theorem giveFor fixed $w\in H^1(U)$, the Sobolev inequality gives $w\in L^6(U)$, soMeanwhile ${\partial }_j{u}_{n_k}\rightharpoonup {\partial }_{j}u$ in $L^2(U)$. Pairing this weak convergence with the strong convergence of the products, or equivalently using the weak-strong product convergence lemma, yields

In three dimensions the Sobolev embedding theorem givesConsequently, for $w\in H^2(U)\cap H_0^1(U)$,Thus $f+|Dw{|}^{2}w\in L^2(U)$. The Dirichlet Poisson regularity theorem on a bounded $C^2$ domain says thathas a unique $v\in H^2(U)\cap H_0^1(U)$ andHence $\Phi (w)=v$ is well defined.

The preceding estimate gives constants $A,B>0$ such that, whenever $\Vert w{\Vert }_{H^2}\le R$,Choose $R>0$ so that $B{R}^2\le 1/2$, and then choose $r>0$ so that $Ar\le R/2$. If $\Vert f{\Vert }_2\le r$, the closed ball $B_R(0)$ is mapped into itself.

For $w,z\in B_R(0)$, factor the cubic gradient nonlinearity asThe same Sobolev embedding theorem and the Holder inequality implyThe elliptic estimate therefore yieldsShrinking $R$ further makes $C^{\prime }{R}^2<1$, so $\Phi$ is a contraction mapping of the closed ball. This ball is complete because $H^2(U)\cap H_0^1(U)$ is a Banach space. The contraction mapping theorem gives a fixed point $u=\Phi (u)$, and its defining equation isThus the nonlinear elliptic boundary value problem has a solution for sufficiently small $\Vert f{\Vert }_2$.