Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 105 / 2 / b

The principal coefficients satisfy the positive uniformly elliptic operator condition if there is $\theta >0$, independent of $x$, such thatThe smooth coefficients on the compact closure also give a uniform upper bound $\sum a^{ij}{\xi }_i{\xi }_j\le \Theta |\xi {|}^2$. The first-order and zeroth-order coefficients do not enter uniform ellipticity.

Let $T:H^1(U)\to L^2(\partial U)$ be the trace map from the Sobolev trace theorem. The weak Robin problem for a uniformly elliptic operator uses all of $H^1(U)$ as the test space, since a Robin boundary condition does not require a zero trace. DefineThe weak formulation isFor a classical solution, integration by parts produces the boundary term $-{\int }_{\partial U}(\sum a^{ij}{u}_{x_i}{\nu }_j)v$. Substituting the Robin boundary condition replaces it by ${\int }_{\partial U}\beta uv$, giving the displayed identity.

Conversely, if $u\in C^2(\overline{U})$ satisfies the identity, tests $v\in C_c^{\infty }(U)$ first show $(L+\lambda )u=f$ in the sense of distributions. Subtracting this interior equality after integration by parts givesEvery smooth boundary function is the trace of a smooth function on $\overline{U}$. The boundary residual is continuous, so this proves the boundary equation pointwise. Thus the formulations agree for $C^2$ solutions. If $f$ is continuous the interior equation holds pointwise; for merely $f\in L^2$, equality with that given representative of $f$ is understood almost everywhere.

The positive uniform ellipticity condition from part (i) is needed here. It is not explicitly imposed in the printed preamble. Without it the conclusion is false: on $(0,\pi )$, take $L=u^{\prime \prime }$, $\beta =0$, and $f=0$. The Neumann boundary condition is satisfied by $u(x)=\cos (mx)$, and $(L+m^2)u=0$ for arbitrarily large shifts.

There is also an actual error in the PDF's supplied trace inequality: a nonzero constant has nonzero trace and zero gradient. Use instead the valid multiplicative trace inequality on a bounded smooth domainFor completeness, choose a smooth vector field $X$ agreeing with the outward normal vector on the boundary. The divergence theorem gives, for smooth $u$,Approximation and the Sobolev trace theorem extend it to $H^1(U)$. Thus the intended existence result remains true after correcting the hint.

Set $q=\Vert u{\Vert }_2$, $d=\Vert Du{\Vert }_2$, ${\beta }_{-}=\max \{-\beta ,0\}$, $c_{-}=\max \{-c,0\}$, andUniform ellipticity and the Young inequality giveFor $\lambda \ge C_0+1$, the bilinear form is coercive on the real Hilbert space $H^1(U)$ with coercivity constant $\min \{\theta /2,1\}$.

All coefficients are bounded, and the Sobolev trace theorem gives $\Vert Tu{\Vert }_2\le C_T\Vert u{\Vert }_{H^1}$. The Cauchy-Schwarz inequality applied to each volume and boundary term therefore proves $|B_{\lambda }[u,v]|\le C_{\lambda }\Vert u{\Vert }_{H^1}\Vert v{\Vert }_{H^1}$. Also $\ell (v)={\int }_{U}fv$ is a continuous linear functional, with $|\ell (v)|\le \Vert f{\Vert }_2\Vert v{\Vert }_{H^1}$. Every hypothesis of the Lax-Milgram theorem is now verified, and it gives