Heat Equation Maximum Principles Duhamel and Smoothing: Examples

The constant curve $s\mapsto f$ belongs to $C([0,T];L^p({\mathbb{R}}^n))$, so [F1] applies to it: $Df\in C([0,T];L^p({\mathbb{R}}^n))$, $Df(0)=0$, and $Df$ satisfies the forced semigroup relation with the constant forcing.

The two representations agree. For each fixed $t$ the curves $s\mapsto H_{t-s}f$ and $\tau \mapsto H_{\tau }f$ are norm continuous by [F2], and they are related by the reflection $s\mapsto t-s$ of $[0,t]$; by [F3] both Bochner integrals are limits of the integrals of simple approximations, and for a finitely-valued approximation the substitution reduces to the equality of the Lebesgue measures of a measurable level set and its reflection, so passing to the limit gives ${\int }_0^t{H}_{t-s}fds={\int }_0^t{H}_{\tau }fd\tau$.

Classical compactly supported case. If $f$ is spatially Hölder continuous with compact support, then as a function of $(x,t)$ it is bounded, jointly continuous and uniformly spatially Hölder on ${\mathbb{R}}^n\times [0,T]$; by [F1] the scalar potential $u(t,x)={\int }_0^t\int \mathrm{\Gamma }(x-y,t-s)f(y)dyds$ is $C^{1,2}$ with $u_t-\mathrm{\Delta }u=f$ and $u(0,\cdot )=0$, and it represents the Bochner potential $Df$ in the sense of [F2]; moreover $\mathrm{\mid }u\mathrm{\mid }\le T\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, so $u$ lies in the Gaussian growth class and is the unique classical solution with zero initial data there. Hence $(Df{)}_t-\mathrm{\Delta }Df=f$ on ${\mathbb{R}}^n\times (0,T]$, the representation of step 2.1 shows that the two displayed formulas for $Df$ coincide, and no commutation of the unbounded Laplacian with the flow on arbitrary $L^p$ data is asserted.

On the bounded cylinder, apply [F1] to the pair $(U,V)=(a,u)$: both lie in $C^{2,1}(\bar{Q})$, $a_t-\mathrm{\Delta }a=0=u_t-\mathrm{\Delta }u$ in $Q$ by [F1], and $a\le u$ on ${\mathrm{\partial }}_{p}Q$ by hypothesis, so $a\le u$ on $\bar{Q}$; applying [F1] to $(U,V)=(u,b)$ in the same way gives $u\le b$ on $\bar{Q}$. Hence $a\le u\le b$ on all of $\bar{Q}$.

The analogous whole-space statement in the bounded-data case $u_0\in L^{\mathrm{\infty }}({\mathbb{R}}^n)$: the constant functions $a$ and $b$ lie in $L^{\mathrm{\infty }}$, and $H_{t}a=a$, $H_{t}b=b$ almost everywhere because the constant $c$ convolves to $c\int \mathrm{\Gamma }(x-y,t)dy=c$ for almost every $x$ by the unit mass of [F2]; since $a\le u_0\le b$ almost everywhere, [F3] applied to the pairs $(a,u_0)$ and $(u_0,b)$ gives $a=H_{t}a\le H_t{u}_0\le H_{t}b=b$ almost everywhere.

For general $u_0\in L^p({\mathbb{R}}^n)$, $1\le p\le \mathrm{\infty }$, the same conclusion follows from the kernel representation: at every $x$ where the defining integral of [F2] converges, $b-H_t{u}_0(x)={\int }_{{\mathbb{R}}^n}\mathrm{\Gamma }(x-y,t)(b-u_0(y))dy$ by linearity of the integral [F4] and $b={\int }_{{\mathbb{R}}^n}\mathrm{\Gamma }(x-y,t)bdy$ by the unit mass of [F2], while the integrand $\mathrm{\Gamma }(x-y,t)(b-u_0(y))$ is nonnegative almost everywhere in $y$; its integral is therefore nonnegative by the monotonicity clause of [F4], so $H_t{u}_0(x)\le b$ at each such $x$, and $H_t{u}_0\le b$ almost everywhere because the defining integral converges almost everywhere [F2]; the inequality $H_t{u}_0\ge a$ follows the same way from $u_0-a\ge 0$.

The function $u_k$ is smooth on the closed rectangle (a product of a smooth exponential and a smooth sine), and [F2] gives ${\mathrm{\partial }}_t{u}_k=-k^2{e}^{-k^{2}t}\sin (kx)=-k^2{u}_k$ together with ${\mathrm{\partial }}_x^2{u}_k=-k^2{e}^{-k^{2}t}\sin (kx)=-k^2{u}_k$; hence $u_{k,t}={\mathrm{\partial }}_x^2{u}_k=\mathrm{\Delta }{u}_k$ on the open rectangle by [F1].

The boundary values are $u_k(0,t)=e^{-k^{2}t}\sin 0=0$ and $u_k(\pi ,t)=e^{-k^{2}t}\sin (k\pi )=0$ for every $t$, while $u_k(x,0)=e^0\sin (kx)=\sin (kx)$; the nodal set at time $t$ is $\{x\in (0,\pi ):\sin (kx)=0\}=\{m\pi \mathrm{/}k:1\le m\le k-1\}$, independent of $t$ because the positive factor $e^{-k^{2}t}$ never vanishes.

By [F3] the squared norm of step 1.1's function is $\mathrm{\parallel }{u}_k(\cdot ,t){\mathrm{\parallel }}_2^2=e^{-2k^{2}t}{\int }_0^{\pi }\sin (kx{)}^{2}dx=e^{-2k^{2}t}\pi \mathrm{/}2$, so $\mathrm{\parallel }{u}_k(\cdot ,t){\mathrm{\parallel }}_2=e^{-k^{2}t}\sqrt{\pi \mathrm{/}2}$ for every $t\ge 0$.

Since $k\mapsto e^{-k^{2}t}$ is strictly decreasing in $k\ge 1$ for every fixed $t>0$, higher modes decay faster at each positive time, with the ratio $e^{-(k^2-l^2)t}$ between the $k$-th and $l$-th modes for $k>l$.

Steps 1.1, 1.2, 2.1 and 3.1 verify that $u_k$ is a classical solution smooth up to $t=0$ with Dirichlet data, decay rate $e^{-k^{2}t}\sqrt{\pi \mathrm{/}2}$, faster decay for higher modes, and a time-independent nodal set.

The function $u$ is smooth on $\bar{Q}$ (a product of the smooth functions $e^t$ and $\sin x$, [F2] and [F3]), with $u_t=e^t\sin x$ and $\mathrm{\Delta }u=u_{xx}=-e^t\sin x$ by [F2] and [F4]; hence $u_t-\mathrm{\Delta }u=2e^t\sin x\ge 0$ on $Q$, because $e^t>0$ by [F3] and $\sin x>0$ for $0<x<\pi$ by [F2].

On the parabolic boundary, $u(0,t)=u(\pi ,t)=e^t\sin \pi =0$ for every $t\in [0,T]$, and $u(x,0)=\sin x\le 1$ with $u(\pi \mathrm{/}2,0)=1$; hence ${\max }_{{\mathrm{\partial }}_{p}Q}u=1$.

For every $(x,t)\in \bar{Q}$ one has $\sin x\le 1$ by [F2] and $e^t\le e^T$: indeed $e^T-e^t=e^c(T-t)>0$ for some $c\in (t,T)$ whenever $t<T$ by [F3] and the mean value theorem, so $u(x,t)\le e^T$; equality holds exactly at $(x,t)=(\pi \mathrm{/}2,T)$, where $u=e^T\sin (\pi \mathrm{/}2)=e^T$. Thus ${\max }_{\bar{Q}}u=e^T$, attained at the point $(\pi \mathrm{/}2,T)$ of the final-time face.

The point $(\pi \mathrm{/}2,T)$ does not belong to ${\mathrm{\partial }}_{p}Q$, because $\pi \mathrm{/}2\in (0,\pi )$ and $T>0$ by [F1]; by step 1.3 it is a point of $\bar{Q}$ at which $u$ attains the value $e^T$, while by step 1.2 the parabolic boundary carries the strictly smaller maximum $1$. Since $e^T>1$ (again by the mean value theorem applied to $\exp$ on $[0,T]$, [F3]), a supersolution has its maximum over $\bar{Q}$ strictly larger than the parabolic-boundary maximum, refuting the proposed supersolution maximum bound. The minimum principle for supersolutions, obtained by applying the weak maximum principle to $-u$, remains valid.

The sign sensitivity is real and the weak maximum principle is not contradicted: $v:=-u$ satisfies $v_t-\mathrm{\Delta }v=-2e^t\sin x\le 0$ on $Q$, and by [F4] its maximum over $\bar{Q}$ equals ${\max }_{{\mathrm{\partial }}_{p}Q}v=0$, attained on the lateral faces, consistently with the theorem being stated for subsolutions.

If $u$ is continuous on $[0,\pi ]\times [0,T]$, then at the corner $(0,0)$ the sequence $(1\mathrm{/}(k+1),0)$ lies in the initial face, converges to $(0,0)$, and $u(1\mathrm{/}(k+1),0)=1$ for every $k\ge 0$; [F1] therefore forces $u(0,0)={\lim }_{k}u(1\mathrm{/}(k+1),0)=1$.

Along the lateral face, the sequence $(0,T\mathrm{/}(k+1))$ converges to $(0,0)$ with $u(0,T\mathrm{/}(k+1))=0$ for every $k\ge 0$, so the same continuity forces $u(0,0)={\lim }_{k}u(0,T\mathrm{/}(k+1))=0$.

Steps 1.1 and 1.2 assign the two different values $1$ and $0$ to $u(0,0)$, a contradiction; hence no function continuous on $[0,\pi ]\times [0,T]$ attains both data sets.

The obstruction is exactly the incompatibility of the limiting initial and lateral values at the parabolic corner, and it is independent of any interior solution theory: even a mild solution of the heat equation with these data cannot be made continuous up to the corner.

For every $k$ the function ${\tilde{u}}_k$ is a classical solution of $u_t=u_{xx}$ on $(0,\pi )\times (0,T]$ with ${\tilde{u}}_k(0,t)={\tilde{u}}_k(\pi ,t)=0$, terminal data $g_k(x):=e^{-k^{2}T}\sin (kx)$ and initial state ${\tilde{u}}_k(x,0)=\sin (kx)$.

By [F2] the terminal data have norm $\mathrm{\parallel }{g}_k{\mathrm{\parallel }}_2=e^{-k^{2}T}\mathrm{\parallel }\sin (k\cdot ){\mathrm{\parallel }}_2=\sqrt{\pi \mathrm{/}2}{e}^{-k^{2}T}$, while the initial states satisfy $\mathrm{\parallel }{\tilde{u}}_k(\cdot ,0){\mathrm{\parallel }}_2=\mathrm{\parallel }\sin (k\cdot ){\mathrm{\parallel }}_2=\sqrt{\pi \mathrm{/}2}$ for every $k$; moreover ${\tilde{u}}_k(\cdot ,0)=e^{k^{2}T}{g}_k$, so the terminal-to-initial amplification factor of the $k$-th mode is exactly $e^{k^{2}T}$.

By [F3] the terminal norms $\mathrm{\parallel }{g}_k{\mathrm{\parallel }}_2=\sqrt{\pi \mathrm{/}2}{e}^{-k^{2}T}$ tend to $0$ while the initial norms remain the fixed constant $\sqrt{\pi \mathrm{/}2}$, so terminal perturbations of arbitrarily small $L^2$ norm come from solutions whose initial states have order-one norm; this refutes continuous dependence on the terminal data, and the amplification is a failure of stability rather than of existence (the solution pair is explicit for every $k$).

For each $k$ the summand $u_k(x,t)=a_k{e}^{k^2(T-t)}\sin (kx)$ satisfies ${\mathrm{\partial }}_t{u}_k=-k^2{u}_k$ by [F2] and ${\mathrm{\partial }}_x^2{u}_k=-k^2{u}_k$ by [F1], so ${\mathrm{\partial }}_t{u}_k-{\mathrm{\partial }}_x^2{u}_k=0$ on $(0,\pi )\times (0,T]$.

Since $u$ is a finite sum of the summands of step 1.1, [F3] gives ${\mathrm{\partial }}_{t}u={\sum }_k{\mathrm{\partial }}_t{u}_k$ and ${\mathrm{\partial }}_x^{2}u={\sum }_k{\mathrm{\partial }}_x^2{u}_k$, so $u_t-u_{xx}={\sum }_k({\mathrm{\partial }}_t{u}_k-{\mathrm{\partial }}_x^2{u}_k)=0$; moreover $u(0,t)={\sum }_k{a}_k\sin 0=0$ and $u(\pi ,t)={\sum }_k{a}_k\sin (k\pi )=0$ for every $t$ by [F4], while $u(x,T)={\sum }_k{a}_k\sin (kx)=g(x)$ because $e^0=1$; the sum is smooth because it has finitely many smooth summands.

Step 2.1 exhibits, for every finite terminal sine sum $g$, the classical backward solution $u$ on the closed rectangle, so existence is unconditional for finite data; the $k$-th summand carries the factor $e^{k^2(T-t)}$, which equals $e^{k^{2}T}$ at $t=0$ and grows without bound as $k$ increases, so no uniform amplification bound over all $k$ is claimed, in agreement with the example's final sentence and the unboundedness of the backward solution map.