Fourier heat flow on the continuum torus

Since $0\le {\rho }_0\le 1$ and $dx({\mathbb{T}}^d)=1$, ${\rho }_0\in L^1$ and $\mathrm{\mid }{\widehat{\rho }}_0(k)\mathrm{\mid }\le 1$ for every $k$. For any smooth $G$ and integer $m\ge 1$, periodic integration by parts gives $(1+4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^2{)}^m\widehat{G}(k)=\widehat{(1-\mathrm{\Delta }{)}^{m}G}(k)$, hence $\mathrm{\mid }\widehat{G}(k)\mathrm{\mid }\le \mathrm{\parallel }(1-\mathrm{\Delta }{)}^{m}G{\mathrm{\parallel }}_{L^1}(1+4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^2{)}^{-m}$. If $q\ge 0$ is fixed, choose $m>(d+q)\mathrm{/}2$; grouping lattice points by $r=\mathrm{\parallel }k{\mathrm{\parallel }}_{\mathrm{\infty }}$ gives at most $(2r+1{)}^d-(2r-1{)}^d\le C_d{r}^{d-1}$ points in shell $r$. The shell contribution to ${\sum }_k(1+\mathrm{\mid }k\mathrm{\mid }{)}^q\mathrm{\mid }\widehat{G}(k)\mathrm{\mid }$ is at most $C{r}^{d-1+q-2m}$. Grouping shells into dyadic blocks $2^j\le r<2^{j+1}$ bounds the $j$th block by $C^{\mathrm{\prime }}{2}^{j(d+q-2m)}$, a convergent geometric series since $2m>d+q$. Thus ${\sum }_k(1+\mathrm{\mid }k\mathrm{\mid }{)}^q\mathrm{\mid }\widehat{G}(k)\mathrm{\mid }<\mathrm{\infty }$.

For each $t>0$ and each fixed derivative order, multiplication of the coefficients by any corresponding polynomial in $k$ still leaves an absolutely summable sequence, because $\mathrm{\mid }{\widehat{\rho }}_0(k)\mathrm{\mid }\le 1$ and the Gaussian factor $e^{-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^{2}t}$ dominates every inverse power on lattice shells. Therefore the heat series and all fixed-order differentiated series converge uniformly on ${\mathbb{T}}^d$ for $t\ge \epsilon >0$. Pairing the $k$ and $-k$ terms shows ${\rho }_t$ is real. Termwise differentiation is valid on every such compact time interval; both ${\mathrm{\partial }}_t$ and $\mathrm{\Delta }$ multiply the $k$th term by $-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^2$, so ${\mathrm{\partial }}_t{\rho }_t=\mathrm{\Delta }{\rho }_t$.

The decay estimate in step 1.1 makes the Fourier series for $G$ and for each fixed space or time derivative of $P_{t}G$ uniformly absolutely convergent even at $t=0$. By [F2], the series at $t=0$ equals $G$ in $L^2$; both are continuous, so equality almost everywhere implies equality everywhere. Uniform convergence of the differentiated series proves smoothness and uniform derivative bounds on $[0,T]\times {\mathbb{T}}^d$. Also $\mathrm{\parallel }{P}_{t}G-G{\mathrm{\parallel }}_{\mathrm{\infty }}\le {\sum }_k\mathrm{\mid }\widehat{G}(k)\mathrm{\mid }\mathrm{\mid }{e}^{-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^{2}t}-1\mathrm{\mid }\to 0$ by splitting into a finite set of modes and an absolutely summable tail. Termwise differentiation gives ${\mathrm{\partial }}_t{P}_{t}G=\mathrm{\Delta }{P}_{t}G$.

Absolute convergence permits termwise integration against ${\rho }_0\in L^1$. Reindexing $k\mapsto -k$ and using the even multiplier $e^{-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^{2}t}$ gives $\int {\rho }_{t}G={\sum }_k{e}^{-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^{2}t}{\widehat{\rho }}_0(k)\widehat{G}(-k)=\int {\rho }_0{P}_{t}G$. By step 2.2 this pairing tends to $\int {\rho }_{0}G$ as $t\downarrow 0$. Differentiating the pairing for $t>0$ and using $\mathrm{\Delta }{P}_{t}G=P_t\mathrm{\Delta }G$ yields $\frac{d}{dt}\int {\rho }_{t}G=\int {\rho }_t\mathrm{\Delta }G$; integration from $0$ proves the weak identity in [F3]. Thus the constructed family is a weak heat solution.

Parseval and the heat coefficients give $\mathrm{\parallel }{\rho }_t-{\rho }_0{\mathrm{\parallel }}_2^2={\sum }_k\mathrm{\mid }{\widehat{\rho }}_0(k){\mathrm{\mid }}^2\mathrm{\mid }{e}^{-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^{2}t}-1{\mathrm{\mid }}^2\to 0$ as $t\downarrow 0$: choose a finite set whose tail in the convergent series ${\sum }_k\mathrm{\mid }{\widehat{\rho }}_0(k){\mathrm{\mid }}^2$ is small, then take the limit on that finite set. For $ϵ>0$ put $a_{ϵ}(z)=\frac{1}{2}(1-z\mathrm{/}\sqrt{z^2+{ϵ}^2})$ and ${\mathrm{\Phi }}_{ϵ}(z)={\int }_0^z{\int }_0^r{a}_{ϵ}(q)dqdr$. Then ${\mathrm{\Phi }}_{ϵ}$ is smooth and convex, ${\mathrm{\Phi }}_{ϵ}(0)={\mathrm{\Phi }}_{ϵ}^{\mathrm{\prime }}(0)=0$, and $0\le {\mathrm{\Phi }}_{ϵ}(z)\le z^2\mathrm{/}2$. On any interval $[a,b]\subset (0,T]$, the heat equation, chain rule, and periodic integration by parts give $\int {\mathrm{\Phi }}_{ϵ}({\rho }_b)-\int {\mathrm{\Phi }}_{ϵ}({\rho }_a)=-{\int }_a^b\int {\mathrm{\Phi }}_{ϵ}^{\mathrm{\prime }\mathrm{\prime }}({\rho }_s){\sum }_{i=1}^d\mathrm{\mid }{\mathrm{\partial }}_{x_i}{\rho }_s{\mathrm{\mid }}^{2}dxds\le 0$. For each real $z$, the defining double integral tends to $\frac{1}{2}(z^{-}{)}^2$, where $z^{-}:=\max \{-z,0\}$; dominated convergence on the bounded range of $\rho$ over $[a,b]\times {\mathbb{T}}^d$ gives the same monotonicity for $\mathrm{\parallel }{\rho }_t^{-}{\mathrm{\parallel }}_2^2$. Letting $a\downarrow 0$, the $L^2$ continuity just proved and ${\rho }_0\ge 0$ imply ${\rho }_t^{-}=0$. The assumed datum already gives the required bound when $T=0$.

For the upper bound put $b_{ϵ}(z)=\frac{1}{2}(1+(z-1)\mathrm{/}\sqrt{(z-1{)}^2+{ϵ}^2})$ and ${\mathrm{\Psi }}_{ϵ}(z)={\int }_1^z{\int }_1^r{b}_{ϵ}(q)dqdr$. This is smooth and convex with ${\mathrm{\Psi }}_{ϵ}(1)={\mathrm{\Psi }}_{ϵ}^{\mathrm{\prime }}(1)=0$ and $0\le {\mathrm{\Psi }}_{ϵ}(z)\le (z-1{)}^2\mathrm{/}2$. The same integration by parts gives $\int {\mathrm{\Psi }}_{ϵ}({\rho }_b)\le \int {\mathrm{\Psi }}_{ϵ}({\rho }_a)$ for $0<a<b$, and ${\mathrm{\Psi }}_{ϵ}(z)\to \frac{1}{2}((z-1{)}^{+}{)}^2$, where $r^{+}:=\max \{r,0\}$. Dominated convergence and $L^2$ continuity at zero, together with ${\rho }_0\le 1$, give $({\rho }_t-1{)}^{+}=0$. The zero Fourier coefficient is unchanged by the heat multiplier, so $\int {\rho }_t={\widehat{\rho }}_0(0)=\int {\rho }_0$. The assumed datum already gives the bound when $T=0$.

Let $(u_t)$ be any weak heat solution in the sense of [F3]. Testing its weak identity with $e_{-k}$ shows $g_k(t):={\widehat{u}}_t(k)$ is continuous and satisfies $g_k(t)={\widehat{\rho }}_0(k)-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^2{\int }_0^t{g}_k(s)ds$. Hence $g_k$ is differentiable and solves $g_k^{\mathrm{\prime }}=-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^2{g}_k$, so $g_k(t)=e^{-4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^{2}t}{\widehat{\rho }}_0(k)$, the coefficient of the constructed ${\rho }_t$. For each $t$, both functions are in $L^2$; equality of all Fourier coefficients and completeness in [F2] imply $u_t={\rho }_t$ almost everywhere. Thus the weak heat solution is unique.