Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 30B (Partial Differential Equations)

There is a missing hypothesis in the printed uniqueness claim: on an unbounded domain it is false without a condition at infinity. For example, on $\Omega =\{x_d>0\}$ take $f=0$ and $u_D=0$. Both $u=0$ and $u=x_d$ are classical solutions, although $f_u=0$.

For the intended bounded-domain statement, let $u,v\in C^2(\Omega )\cap C(\overline{\Omega })$ be solutions and put $w=u-v$. The derivative assumption makes $f$ nondecreasing in its first argument. On the open set $D=\{x\in \Omega :w(x)>0\}$,The difference $w$ vanishes on $\partial \Omega$, and continuity makes it zero on every interior boundary point of $D$ as well. Since $D$ is bounded, the weak maximum principle for subharmonic functions gives ${\sup }_{\overline{D}}w\le {\sup }_{\partial D}w=0$, contradicting $w>0$ there unless $D$ is empty. Thus $u\le v$. Interchanging $u,v$ gives the reverse inequality, proving $\boxed{u=v}$. An appropriate bound or decay at infinity can replace boundedness, but the printed hypotheses alone do not include it.

Use the Fourier transform convention $\hat{\varphi }(\xi )={\int }_{{\mathbb{R}}^n}{e}^{-ix\cdot \xi }\varphi (x)dx$. For every orthogonal matrix $Q$, changing variables $x=Qy$ givessince the Jacobian has absolute determinant one and a radial function satisfies $\varphi (Qy)=\varphi (y)$. Any two vectors of the same length are related by an orthogonal transformation, so $\boxed{\hat{\varphi }\text{ is radial}}$.

Taking the Fourier transform gives $(1+|\xi {|}^2)\hat{u}(\xi )=\hat{\varphi }(\xi )$. Thus the canonical solution isA smooth compactly supported function is a Schwartz function, as is its Fourier transform. The multiplier $(1+|\xi {|}^2{)}^{-1}$ and all its derivatives have polynomial bounds, so their product with $\hat{\varphi }$ is again a Schwartz function. The inverse integral can therefore be differentiated to verify $-\Delta u+u=\varphi$ classically. By part (b), its Fourier-space integrand factor $\hat{\varphi }/(1+|\xi {|}^2)$ is radial. Changing variables by any orthogonal transformation in the inverse integral gives $u(Qx)=u(x)$, proving radiality.

This is also the unique tempered-distribution solution: the nonvanishing multiplier $1+|\xi {|}^2$ can be inverted on that space. Unrestricted classical solutions need not be unique, since one may add, for example, $e^{x_1}$; the Fourier construction selects the rapidly decaying solution.

The real Lax-Milgram theorem states that if $V$ is a Hilbert space, $B:V\times V\to \mathbb{R}$ is a continuous bilinear form with $B(v,v)\ge \alpha \Vert v{\Vert }_V^2$ for some $\alpha >0$, and $\ell \in V^{\ast }$ is continuous, then there is a unique $u\in V$ such that $B(u,v)=\ell (v)$ for all $v\in V$. Moreover $\Vert u{\Vert }_V\le \Vert \ell \Vert /\alpha$.

Take the Sobolev space $V=H_0^1(\Omega )$ with norm $\Vert v{\Vert }_V=\Vert \nabla v{\Vert }_2$. Since $\Omega$ is bounded, the Poincaré inequality gives $\Vert v{\Vert }_2\le C_P\Vert \nabla v{\Vert }_2$, making this norm equivalent to the full Sobolev norm. For the given bounded measurable coefficient defineThen $|B(u,v)|\le (1+\bar{a}{C}_P^2)\Vert u{\Vert }_V\Vert v{\Vert }_V$, $B(v,v)\ge \Vert v{\Vert }_V^2$ since $a\ge 0$, and $|\ell (v)|\le C_P\Vert f{\Vert }_2\Vert v{\Vert }_V$. The Lax-Milgram theorem therefore gives a unique $u\in H_0^1(\Omega )$ satisfying this identity, which is exactly the required weak solution. Strict positivity of the lower coefficient bound is unnecessary: the zero boundary condition and boundedness supply coercivity through the gradient term.