Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 3 / 30E / a / Solution

Use a Fourier transform convention for which $-\Delta$ has symbol $|\xi {|}^2$. The distributional equation gives $\hat{u}=\hat{f}/(|\xi {|}^2+m^2)$ for real nonzero $m$, since this smooth positive multiplier has no zeros. Consequentlywhere $\boxed{C_m=\max (1,m^{-2})}$. This bound is independent of $s$. For $f\in H^s$, the same multiplier defines $u\in H^{s+2}$; alternatively approximate $f$ by Schwartz functions and pass to the limit using the estimate. This is the massive Laplacian isomorphism on Sobolev spaces.

For $a=1+|\xi {|}^2$, the inequality $a\le ϵa^2+1/(4ϵ)$, integrated against $|\hat{v}{|}^2$, gives the interpolation estimateFor a finite global bound in the last request, assume $\Vert u{\Vert }_{L^2}<\infty$ as well as boundedness. This assumption is necessary: the smooth bounded solution $u\equiv m$ has infinite $H^2$ norm on ${\mathbb{R}}^n$. Put $M=\Vert u{\Vert }_{\infty }$, $U=\Vert u{\Vert }_2$. Then $u^3\in L^2$ with norm at most $M^{2}U$. Incorporate the constant drift on the left:The constant-drift massive elliptic estimate follows because the modulus of this symbol is at least $|\xi {|}^2+m^2$, so the same multiplier argument first establishes $u\in H^2$ and then gives the stronger a priori boundIt is allowed to depend on $\alpha$ but in fact does not need to. Once regularity has been established, the earlier estimate and interpolation also give the conventional absorption proof: $\Vert u{\Vert }_{H^2}\le C_m(M^{2}U+|\alpha |\Vert u{\Vert }_{H^1})$, and choosing $ϵ$ small absorbs the $H^2$ term.