Duhamel principle for the whole-space heat equation

For $(t,x)\in [0,T]\times {\mathbb{R}}^n$ the double integral defining $u$ converges absolutely, because $\iint \mathrm{\Gamma }(x-y,t-s)\mathrm{\mid }f(y,s)\mathrm{\mid }dyds\le M{\int }_0^t\mathrm{\parallel }\mathrm{\Gamma }(\cdot ,t-s){\mathrm{\parallel }}_{1}ds=Mt$ by the unit mass of [F1]; hence $u$ is well defined, $\mathrm{\mid }u(t,x)\mathrm{\mid }\le Mt$, and $u(0,\cdot )=0$ since the $s$-integral is over the empty interval. Substituting $\tau =t-s$, $u(t,x)={\int }_0^t({\mathrm{\Gamma }}_{\tau }\ast f(\cdot ,t-\tau ))(x)d\tau$, the convolution being that of [F1].

In the mild setting, write $Df(t)={\int }_0^T{\mathbf{1}}_{\{s<t\}}{H}_{t-s}f(s)ds$. If $t_j\to t$, the integrands converge in $L^p$ for every $s\mathrm{\ne }t$, by positive-time strong continuity when $s<t$ and eventual vanishing when $s>t$. Their norms are bounded by the integrable function $\mathrm{\parallel }f(s){\mathrm{\parallel }}_p$. Thus Bochner dominated convergence [F4] gives $Df(t_j)\to Df(t)$, including at $t=0$, where $Df(0)=0$.

Fix $t>0$ and $\mathrm{\mid }\alpha \mathrm{\mid }\le 2$, and define the candidate $v_{\alpha }(t,x):={\int }_0^t{\int }_{{\mathbb{R}}^n}{D}^{\alpha }\mathrm{\Gamma }(x-y,\tau )f(y,t-\tau )dyd\tau$. The inner integral converges absolutely with a $\tau$-majorant integrable on $(0,t)$: for $\mathrm{\mid }\alpha \mathrm{\mid }=0$ the unit mass gives the bound $M$, and for $\mathrm{\mid }\alpha \mathrm{\mid }=1$ the Gaussian bound gives $\int \mathrm{\mid }{D}^{\alpha }\mathrm{\Gamma }(z,\tau )\mathrm{\mid }dz\le C_{n,\alpha }(8\pi {)}^{n\mathrm{/}2}{\tau }^{-1\mathrm{/}2}$, so the inner integral is bounded by a constant times $M{\tau }^{-1\mathrm{/}2}$; for $\mathrm{\mid }\alpha \mathrm{\mid }=2$, differentiating ${\int }_{{\mathbb{R}}^n}\mathrm{\Gamma }(\xi +z,\tau )dz=1$ in $\xi$ at $\xi =0$ by [F2] with the Gaussian majorant of [F1] gives ${\int }_{{\mathbb{R}}^n}{D}^{\alpha }\mathrm{\Gamma }(z,\tau )dz=0$, so the inner integral equals $\int D^{\alpha }\mathrm{\Gamma }(z,\tau )[f(x-z,t-\tau )-f(x,t-\tau )]dz$, of absolute value at most $C{C}_{n,\alpha }\int \mathrm{\mid }z{\mathrm{\mid }}^{\gamma }{\tau }^{-(n+2)\mathrm{/}2}{e}^{-\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{/}(8\tau )}dz=C^{\mathrm{\prime }}C{\tau }^{-1+\gamma \mathrm{/}2}$, integrable at $\tau =0$ since $\gamma >0$. Dominated convergence over the parameter $(t,x)$ with these majorants makes each $v_{\alpha }$ continuous on ${\mathbb{R}}^n\times (0,T]$.

In the mild setting $Df$ satisfies the forced semigroup relation: for $0<s<t\le T$, splitting $Df(t)={\int }_0^s{H}_{t-\tau }f(\tau )d\tau +{\int }_s^t{H}_{t-\tau }f(\tau )d\tau$ and using $H_{t-\tau }=H_{t-s}{H}_{s-\tau }$ for $\tau <s$ together with [F4], the first integral equals $H_{t-s}{\int }_0^s{H}_{s-\tau }f(\tau )d\tau =H_{t-s}Df(s)$, so $Df(t)=H_{t-s}Df(s)+{\int }_s^t{H}_{t-\tau }f(\tau )d\tau$.

The candidate formulas of step 1.3 are the spatial derivatives of $u$: for $\delta >0$ put $u_{\delta }(t,x):={\int }_{\delta }^t\int \mathrm{\Gamma }(x-y,\tau )f(y,t-\tau )dyd\tau$. On the strip $\tau \ge \delta$ the kernel derivatives are uniformly dominated by an integrable function, so [F2] gives $D_x^{\alpha }{u}_{\delta }(t,x)={\int }_{\delta }^t\int D^{\alpha }\mathrm{\Gamma }(x-y,\tau )f(y,t-\tau )dyd\tau$ with $u_{\delta }$ spatially $C^{\mathrm{\infty }}$; by the majorants of step 1.3, $u_{\delta }\to u$ and $D_x^{\alpha }{u}_{\delta }\to v_{\alpha }$ uniformly on compact subsets of ${\mathbb{R}}^n\times (0,T]$ as $\delta \downarrow 0$. Applying [F2]'s interval statement along each coordinate direction (the functions $h\mapsto u_{\delta }(t,x+h{e}_i)$ are $C^1$ with derivatives $D_{e_i}{u}_{\delta }(t,x+h{e}_i)\to v_{e_i}(t,x+h{e}_i)$ uniformly on compact $h$-intervals) gives $D_{e_i}u=v_{e_i}$ for every $i$; repeating the argument with the family $h\mapsto v_{e_i}^{\delta }(t,x+h{e}_j)$, whose $h$-derivatives converge uniformly to $v_{e_i+e_j}$, gives ${\mathrm{\partial }}_j{\mathrm{\partial }}_{i}u=v_{e_i+e_j}$. Hence $u$ has continuous spatial derivatives of every order $\mathrm{\mid }\alpha \mathrm{\mid }\le 2$, equal to the corresponding $v_{\alpha }$.

Mild uniqueness: if $w\in C([0,T];L^p({\mathbb{R}}^n))$ with $w(0)=0$ satisfies the forced relation of step 2.1, then $v:=w-Df$ is continuous with $v(0)=0$ and satisfies $v(t)=H_{t-s}v(s)$ for all $0<s<t\le T$; the contraction bound of [F5] gives $\mathrm{\parallel }v(t){\mathrm{\parallel }}_p\le \mathrm{\parallel }v(s){\mathrm{\parallel }}_p$ for every $s\in (0,t)$, and letting $s\downarrow 0$ with continuity at $0$ gives $\mathrm{\parallel }v(t){\mathrm{\parallel }}_p=0$, so $w=Df$ and the mild solution with zero data is unique.

Fix a compact positive-time interval $[a,b]\subset (0,T]$ and $0<\delta <a$. The same truncated potential as in step 2.2 is $u_{\delta }(t,x)={\int }_0^{t-\delta }({\mathrm{\Gamma }}_{t-s}\ast f(\cdot ,s))(x)ds$. Differentiation under the integral and its continuous moving endpoint give ${\mathrm{\partial }}_t{u}_{\delta }=({\mathrm{\Gamma }}_{\delta }\ast f(\cdot ,t-\delta ))(x)+{\int }_0^{t-\delta }(\mathrm{\Delta }{\mathrm{\Gamma }}_{t-s}\ast f(\cdot ,s))(x)ds$. The endpoint derivative follows by splitting the increment into the added interval, whose average tends to the endpoint integrand by continuity, and the old interval, where [F2] applies away from zero depth. The first term tends locally uniformly to $f(x,t)$: the spatial Hölder bound gives $\mathrm{\mid }{\mathrm{\Gamma }}_{\delta }\ast f(\cdot ,t-\delta )-f(\cdot ,t-\delta )\mathrm{\mid }\le C\int {\mathrm{\Gamma }}_{\delta }(z)\mathrm{\mid }z{\mathrm{\mid }}^{\gamma }dz=C_1{\delta }^{\gamma \mathrm{/}2}$, and joint continuity controls $f(x,t-\delta )-f(x,t)$ on compact sets. By the second-derivative cancellation estimate of step 1.3, the second term tends locally uniformly to ${\sum }_i{v}_{2e_i}=\mathrm{\Delta }u$, with omitted tail bounded by $C_2{\delta }^{\gamma \mathrm{/}2}$. Thus $u_{\delta }\to u$ and ${\mathrm{\partial }}_t{u}_{\delta }\to f+\mathrm{\Delta }u$ uniformly on compact space-time sets. The uniform derivative-limit theorem [F2] on $[a,b]$ proves the two-sided time derivative in $(0,T)$ and the left derivative at $T$, with continuous value $u_t=f+\mathrm{\Delta }u$. Together with step 2.2 and $\mathrm{\mid }u\mathrm{\mid }\le Mt$, this proves the classical clause and continuity at zero.

Classical uniqueness: let $u$ be a classical solution with zero initial data and $\mathrm{\mid }u(t,x)\mathrm{\mid }\le C{e}^{a\mathrm{\mid }x{\mathrm{\mid }}^2}$ whose forcing $f$ satisfies the hypotheses of (i), and let $u^{\ast }$ be the potential of (i), which by steps 1.1 and 3.2 is classical with $u_t^{\ast }-\mathrm{\Delta }{u}^{\ast }=f$, $u^{\ast }(0,\cdot )=0$ and $\mathrm{\mid }{u}^{\ast }\mathrm{\mid }\le TM$ on the strip. Then $w:=u-u^{\ast }$ is continuous on the closed strip, $C^{1,2}$ in positive time, solves the homogeneous heat equation with zero initial data, and obeys $\mathrm{\mid }w\mathrm{\mid }\le (C+TM)e^{a\mathrm{\mid }x{\mathrm{\mid }}^2}$ because $e^{a\mathrm{\mid }x{\mathrm{\mid }}^2}\ge 1$ for $a\ge 0$; [F6] gives $w\equiv 0$, so $u=u^{\ast }$ is the potential of (i), the unique classical solution with zero initial data in the Gaussian growth class.