Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 3 / 30E (Partial Differential Equations)

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.

Let $m_0={\min }_{x}u(x,0)>0$, which exists by periodicity and continuity. Fix any finite interval $[0,T]$ on which the classical solution is smooth. Choose $ϵ>0$ with $ϵ(1+T)<m_0/2$ and consider the strictly positive decreasing barrier $b(t)=m_0-ϵ(1+t)$.

Initially $u>b$. If this strict inequality first failed, compactness of the spatial circle would give a first contact $(x_{\ast },t_{\ast })$ with $u=b>0$. At that contact, the spatial minimum of $u-b$ has $u_x=0$, $u_{xx}\ge 0$, and the first-contact condition gives $(u-b{)}_t\le 0$. But the equation yieldsa contradiction. Therefore $u>b$ throughout the interval. Letting $ϵ\downarrow 0$ gives $\boxed{u(x,t)\ge m_0>0}$ on $[0,T]$. Since $T$ was arbitrary, positivity persists for every time for which the assumed smooth solution exists. This argument is a parabolic comparison principle proved directly and does not assume uniform parabolicity before positivity has been established.