The Heat Kernel and the Cauchy Problem — Examples

Comparing the two formulas, $f(x)=(2\pi {\sigma }^2{)}^{-n\mathrm{/}2}{e}^{-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}(2{\sigma }^2)}=\mathrm{\Gamma }(x,{\sigma }^2\mathrm{/}2)$ for every $x$, because $(4\pi \cdot {\sigma }^2\mathrm{/}2{)}^{-n\mathrm{/}2}=(2\pi {\sigma }^2{)}^{-n\mathrm{/}2}$ and $4\cdot ({\sigma }^2\mathrm{/}2)=2{\sigma }^2$.

Hence $f\in L^1({\mathbb{R}}^n)$ with $\mathrm{\parallel }f{\mathrm{\parallel }}_1=1$ by unit mass in [F1], and $f\in L^{\mathrm{\infty }}({\mathbb{R}}^n)$ because $f$ is continuous with finite supremum $(2\pi {\sigma }^2{)}^{-n\mathrm{/}2}$ attained at $0$; so $f$ belongs to $L^1\cap L^{\mathrm{\infty }}$.

Since $f$ is bounded, [F2] gives $H_{t}f(x)=\int \mathrm{\Gamma }(x-y,t)f(y)dy$ for every $x$, and step 1.1 turns this into the convolution $({\mathrm{\Gamma }}_t\ast {\mathrm{\Gamma }}_{{\sigma }^2\mathrm{/}2})(x)={\mathrm{\Gamma }}_{t+{\sigma }^2\mathrm{/}2}(x)$ by the semigroup identity of [F1].

Substituting $s=t+{\sigma }^2\mathrm{/}2$ in the explicit formula of [F1] gives ${\mathrm{\Gamma }}_{t+{\sigma }^2\mathrm{/}2}(x)=(4\pi (t+{\sigma }^2\mathrm{/}2){)}^{-n\mathrm{/}2}{e}^{-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}(4(t+{\sigma }^2\mathrm{/}2))}=(2\pi ({\sigma }^2+2t){)}^{-n\mathrm{/}2}{e}^{-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}(2({\sigma }^2+2t))}$, which is the density of the centred Gaussian law with covariance $({\sigma }^2+2t)I_n$ by the first and second moments in [F3] with $s=t+{\sigma }^2\mathrm{/}2$.

Steps 1.1, 2.1, 2.2 and 3.1 show $f\in L^1\cap L^{\mathrm{\infty }}$ and identify the heat evolution pointwise with the centred Gaussian density of covariance $({\sigma }^2+2t)I_n$, which is the example.

Membership and setup: by [F2] the class $f={\mathbf{1}}_{(a,b)}$ is bounded and measurable with ${\int }_{\mathbb{R}}\mathrm{\mid }f{\mathrm{\mid }}^p=b-a$ for every $1\le p<\mathrm{\infty }$ and $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}=1$, so $f$ lies in every $L^p(\mathbb{R})$, $1\le p\le \mathrm{\infty }$; $H_{t}f$ is the everywhere-defined representative $H_{t}f(x)=(4\pi t{)}^{-1\mathrm{/}2}{\int }_a^b{e}^{-(x-y{)}^2\mathrm{/}(4t)}dy$ of [F2].

Smoothness and strict positivity of $\mathrm{\Phi }$: for every real $u$ the identity $\mathrm{\Phi }(u)=\frac{1}{2}+{\int }_0^u\phi (s)ds$ holds with $\phi$ as in [F5], because $\phi$ is even and has total mass $\sqrt{2\pi }(2\pi {)}^{-1\mathrm{/}2}=1$ by [F4]; the fundamental theorem [F6] gives ${\mathrm{\Phi }}^{\mathrm{\prime }}=\phi >0$ for every real $u$, while [F5] and induction give ${\mathrm{\Phi }}^{(m)}={\phi }^{(m-1)}$ for every $m\ge 1$, so $\mathrm{\Phi }$ is $C^{\mathrm{\infty }}$ and strictly increasing on $\mathbb{R}$.

Substitution: the map $s\mapsto y=x+\sqrt{2t}s$ is a $C^1$ diffeomorphism of $\mathbb{R}$ onto itself with $dy=\sqrt{2t}ds$ and $(x-y{)}^2\mathrm{/}(4t)=s^2\mathrm{/}2$, so the nonnegative-function substitution [F3] turns the interval $a<y<b$ into $(a-x)\mathrm{/}\sqrt{2t}<s<(b-x)\mathrm{/}\sqrt{2t}$ and gives $H_{t}f(x)=(4\pi t{)}^{-1\mathrm{/}2}\sqrt{2t}{\int }_{(a-x)\mathrm{/}\sqrt{2t}}^{(b-x)\mathrm{/}\sqrt{2t}}{e}^{-s^2\mathrm{/}2}ds=\mathrm{\Phi }((b-x)\mathrm{/}\sqrt{2t})-\mathrm{\Phi }((a-x)\mathrm{/}\sqrt{2t})$, since $(4\pi t{)}^{-1\mathrm{/}2}\sqrt{2t}=(2\pi {)}^{-1\mathrm{/}2}$.

Consequences: since $a<b$ implies $(a-x)\mathrm{/}\sqrt{2t}<(b-x)\mathrm{/}\sqrt{2t}$, strict monotonicity of $\mathrm{\Phi }$ in step 1.2 gives $H_{t}f(x)=\mathrm{\Phi }(\cdot )-\mathrm{\Phi }(\cdot )>0$ at every $x$, and the affine maps $x\mapsto (a-x)\mathrm{/}\sqrt{2t}$ and $x\mapsto (b-x)\mathrm{/}\sqrt{2t}$ are $C^{\mathrm{\infty }}$, so the composite $H_{t}f$ is $C^{\mathrm{\infty }}$ on $\mathbb{R}$ by the closure of smooth maps under composition in [F5]; the indicator $f$ is discontinuous at $a$ and $b$.

Steps 1.1, 2.1 and 2.2 give the membership $f\in L^p$ for all $1\le p\le \mathrm{\infty }$, the displayed difference-of-Gaussian-tails formula, strict positivity of $H_{t}f$ at every point of every positive time, and smoothness of $H_{t}f$ despite the discontinuity of $f$.

Solving the heat equation: by the smoothness and heat-equation clauses of [F1], the function $u(x,t)=\mathrm{\Gamma }(x,t)$ is $C^{\mathrm{\infty }}$ on ${\mathbb{R}}^n\times (0,\mathrm{\infty })$ and satisfies ${\mathrm{\partial }}_{t}u(x,t)={\mathrm{\partial }}_t\mathrm{\Gamma }(x,t)={\mathrm{\Delta }}_x\mathrm{\Gamma }(x,t)={\mathrm{\Delta }}_{x}u(x,t)$ at every point.

Parabolic self-similarity: the scaling clause of [F1] reads $\mathrm{\Gamma }(\lambda x,{\lambda }^{2}t)={\lambda }^{-n}\mathrm{\Gamma }(x,t)$ for every $\lambda >0$; multiplying both sides by ${\lambda }^n$ gives the equivalent form ${\lambda }^{n}u(\lambda x,{\lambda }^{2}t)=u(x,t)$, equivalently $u(x,{\lambda }^{2}t)={\lambda }^{-n}u(x\mathrm{/}\lambda ,t)$, so the profile at time ${\lambda }^{2}t$ has spatial scale multiplied by $\lambda$ and amplitude multiplied by ${\lambda }^{-n}$.

Conserved mass: the unit-mass clause of [F1] gives ${\int }_{{\mathbb{R}}^n}u(x,t)dx={\int }_{{\mathbb{R}}^n}\mathrm{\Gamma }(x,t)dx=1$ for every $t>0$, independently of $t$.

Steps 1.1, 2.1 and 2.2 show that $u=\mathrm{\Gamma }$ solves the heat equation, satisfies the stated parabolic dilation law with amplitude ${\lambda }^n$, and has unit total mass at every positive time; the semigroup identity [F2] records the equivalent convolution form of the same one-parameter family.

The half-line datum $f$ is bounded and measurable, with $f(0)=1$, but ${\int }_{\mathbb{R}}\mathrm{\mid }f{\mathrm{\mid }}^p=\mathrm{\infty }$ for every finite $p$. Evenness and unit mass give $H_{t}f(0)={\int }_0^{\mathrm{\infty }}\mathrm{\Gamma }(y,t)dy=1\mathrm{/}2$, so its pointwise supremum distance is at least $1\mathrm{/}2$. This also gives an essential supremum bound: for any $0<\eta <1\mathrm{/}2$, continuity of $H_{t}f$ at $0$ supplies $\delta >0$ such that $\mathrm{\mid }{H}_{t}f(x)-1\mathrm{/}2\mathrm{\mid }<1\mathrm{/}2-\eta$ for $0<x<\delta$. On that interval $f(x)=1$, so $\mathrm{\mid }{H}_{t}f(x)-f(x)\mathrm{\mid }>\eta$. As the interval has positive measure and $\eta$ is arbitrary, $\mathrm{\parallel }{H}_{t}f-f{\mathrm{\parallel }}_{\mathrm{\infty }}\ge 1\mathrm{/}2$ for every $t>0$.

Uniform continuity is necessary for an actual supremum-norm convergence claim on bounded real data: for each fixed $t>0$, [F5] and the mean value theorem make $H_{t}g$ globally Lipschitz, hence uniformly continuous. If ${\sup }_x\mathrm{\mid }{H}_{t}g(x)-g(x)\mathrm{\mid }\to 0$, the sequence $H_{1\mathrm{/}(k+1)}g$ converges uniformly to $g$, which is uniformly continuous by [F5]. For convergence in the essential supremum norm the corresponding necessity concerns the class: the continuous differences $H_{1\mathrm{/}(k+1)}g-H_{1\mathrm{/}(\mathrm{\ell }+1)}g$ have equal supremum and essential supremum, so the sequence is uniformly Cauchy; real completeness gives a pointwise limit $v$. For any $\epsilon >0$, a uniform Cauchy bound $\mathrm{\mid }{H}_{1\mathrm{/}(k+1)}g(x)-H_{1\mathrm{/}(\mathrm{\ell }+1)}g(x)\mathrm{\mid }<\epsilon$ for all $x$ and sufficiently large $k,\mathrm{\ell }$, followed by $\mathrm{\ell }\to \mathrm{\infty }$, gives ${\sup }_x\mathrm{\mid }{H}_{1\mathrm{/}(k+1)}g(x)-v(x)\mathrm{\mid }\le \epsilon$. Thus the limit is uniform and $v$ is uniformly continuous by [F5]. Finally $\mathrm{\parallel }v-g{\mathrm{\parallel }}_{\mathrm{\infty }}\le {\sup }_x\mathrm{\mid }v-H_{1\mathrm{/}(k+1)}g\mathrm{\mid }+\mathrm{\parallel }{H}_{1\mathrm{/}(k+1)}g-g{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. Thus that class has a uniformly continuous representative.

The interval datum $f_0$ has $\int \mathrm{\mid }{f}_0{\mathrm{\mid }}^p=1$ for every finite $p$ and $\mathrm{\parallel }{f}_0{\mathrm{\parallel }}_{\mathrm{\infty }}=1$. By positivity and step 1.1, $H_t{f}_0(0)={\int }_0^1\mathrm{\Gamma }(y,t)dy<1\mathrm{/}2$, since the omitted integral on $(1,\mathrm{\infty })$ is positive. Put $d=1-H_t{f}_0(0)>1\mathrm{/}2$ and $c=(d+1\mathrm{/}2)\mathrm{/}2$, so $1\mathrm{/}2<c<d$. Continuity at $0$ supplies $0<\delta <1$ with $\mathrm{\mid }{H}_t{f}_0(x)-H_t{f}_0(0)\mathrm{\mid }<d-c$ for $0<x<\delta$. There $f_0(x)=1$ and $\mathrm{\mid }{H}_t{f}_0(x)-1\mathrm{\mid }\ge d-\mathrm{\mid }{H}_t{f}_0(x)-H_t{f}_0(0)\mathrm{\mid }>c$. Hence $\mathrm{\parallel }{H}_t{f}_0-f_0{\mathrm{\parallel }}_{\mathrm{\infty }}\ge c>1\mathrm{/}2$ on a set of positive measure, and the pointwise supremum is also greater than $1\mathrm{/}2$.

For this same $f_0$, [F4] gives $\mathrm{\parallel }{H}_t{f}_0-f_0{\mathrm{\parallel }}_p\to 0$ for every $1\le p<\mathrm{\infty }$. Nevertheless each compact set containing $0$ has pointwise supremum error at least $\mathrm{\mid }{H}_t{f}_0(0)-f_0(0)\mathrm{\mid }>1\mathrm{/}2$, so locally uniform convergence fails there. The continuity hypothesis in the bounded-data theorem cannot be discarded.

Steps 1.1–3.1 preserve the exact half-line value $1\mathrm{/}2$ and supply compactly supported data in $L^{\mathrm{\infty }}\cap {\bigcap }_{1\le p<\mathrm{\infty }}{L}^p$ with essential and pointwise supremum errors bounded away from zero, while every finite-$p$ norm converges. Step 1.2 justifies the uniform-continuity qualification with the distinction between representatives and classes explicit. These witnesses refute the claimed endpoint extension and locally uniform convergence at the jump.

The datum $f={\mathbf{1}}_{[-1,1]}$ is measurable by [F1], is nonnegative everywhere with $f=1$ on a set of measure $2$, is not the zero class, has ${\int }_{\mathbb{R}}\mathrm{\mid }f\mathrm{\mid }=2<\mathrm{\infty }$, hence lies in $L^1(\mathbb{R})$, and has compact support $[-1,1]$.

Infinite propagation: since $f\in L^1(\mathbb{R})$, $f\ge 0$ and $f\mathrm{\ne }0$, the infinite-propagation corollary [F2] gives $H_{t}f(x)>0$ at every $x\in \mathbb{R}$ and every $t>0$; consequently the set on which the solution is nonzero is all of $\mathbb{R}$, so $\mathrm{supp}(H_{t}f)=\mathbb{R}$ for every $t>0$.

Fix any finite speed $c\ge 0$ and any $t>0$. At $x=2+ct$ one has $\mathrm{dist}(x,[-1,1])=1+ct>ct$, but $H_{t}f(x)>0$ by step 2.1. Thus the required support inclusion fails for every proposed finite speed.

Steps 1.1, 2.1 and 3.1 exhibit a nonzero nonnegative compactly supported $L^1$ datum whose heat flow is strictly positive at every point at every positive time, so the support of the solution is the whole line although the support of the datum is $[-1,1]$; the finite-propagation claim is refuted.

By [F1] and [F2] the kernel satisfies ${\mathrm{\Gamma }}_t(x)=(4\pi t{)}^{-n\mathrm{/}2}{e}^{-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}(4t)}$ with ${\mathrm{\Gamma }}_t\in L^1({\mathbb{R}}^n)$, so its transform of [F3] is defined at every $\xi$ by the absolutely convergent integral $\mathcal{F}{\mathrm{\Gamma }}_t(\xi )=(4\pi t{)}^{-n\mathrm{/}2}{\int }_{{\mathbb{R}}^n}{e}^{-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}(4t)}{e}^{-2\pi ix\cdot \xi }dx$.

Writing $s:=1\mathrm{/}(4\pi t)>0$, the identity $-\frac{\mathrm{\mid }x{\mathrm{\mid }}^2}{4t}=-\pi s\mathrm{\mid }x{\mathrm{\mid }}^2$ holds, so ${\mathrm{\Gamma }}_t(x)=(4\pi t{)}^{-n\mathrm{/}2}{e}^{-\pi s\mathrm{\mid }x{\mathrm{\mid }}^2}$ and $(4\pi t{)}^{-n\mathrm{/}2}=s^{n\mathrm{/}2}$.

Applying the Gaussian transform [F4] with this $s$ and factoring the constant out of the integral gives $\mathcal{F}{\mathrm{\Gamma }}_t(\xi )=(4\pi t{)}^{-n\mathrm{/}2}{s}^{-n\mathrm{/}2}{e}^{-\pi \mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{/}s}=e^{-\pi \mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{/}s}$, and substituting $s=1\mathrm{/}(4\pi t)$ yields $-\pi \mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{/}s=-4{\pi }^{2}t\mathrm{\mid }\xi {\mathrm{\mid }}^2$, so $\mathcal{F}{\mathrm{\Gamma }}_t(\xi )=e^{-4{\pi }^{2}t\mathrm{\mid }\xi {\mathrm{\mid }}^2}$ for every $\xi$.

For the unnormalised convention, the definition gives $\mathcal{G}{\mathrm{\Gamma }}_t(\xi )=\int {\mathrm{\Gamma }}_t(x)e^{-ix\cdot \xi }dx=\mathcal{F}{\mathrm{\Gamma }}_t(\xi \mathrm{/}(2\pi ))$ because $e^{-ix\cdot \xi }=e^{-2\pi ix\cdot (\xi \mathrm{/}(2\pi ))}$; substituting $\xi \mathrm{/}(2\pi )$ in step 3.1 gives $\mathcal{G}{\mathrm{\Gamma }}_t(\xi )=e^{-4{\pi }^{2}t\mathrm{\mid }\xi \mathrm{/}(2\pi ){\mathrm{\mid }}^2}=e^{-t\mathrm{\mid }\xi {\mathrm{\mid }}^2}$.

Steps 1.1, 2.1, 3.1 and 4.1 establish $\mathcal{F}{\mathrm{\Gamma }}_t(\xi )=e^{-4{\pi }^{2}t\mathrm{\mid }\xi {\mathrm{\mid }}^2}$ in the normalisation of [F3] and $\mathcal{G}{\mathrm{\Gamma }}_t(\xi )=e^{-t\mathrm{\mid }\xi {\mathrm{\mid }}^2}$ in the unnormalised convention, which is the whole example.

Absolute convergence: put $q(z)=\mathrm{\Gamma }(z,t)P(x-z)$. For $P\equiv 1$ the integrand is $\mathrm{\Gamma }(z,t)$, integrable with integral $1$ by [F1]; for $P(y)=y_i$ the substituted integrand is $x_i\mathrm{\Gamma }(z,t)-z_i\mathrm{\Gamma }(z,t)$, a sum of integrable terms by [F1] and the first-moment clause of [F2]; for $P(y)=y_i{y}_j$ it is the finite expansion $x_i{x}_j\mathrm{\Gamma }-x_i{z}_j\mathrm{\Gamma }-x_j{z}_i\mathrm{\Gamma }+z_i{z}_j\mathrm{\Gamma }$, integrable by [F1] and the second-moment clause of [F2]; and for $P(y)=\mathrm{\mid }y{\mathrm{\mid }}^2$ it is $\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{\Gamma }-2{\sum }_i{x}_i{z}_i\mathrm{\Gamma }+\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{\Gamma }$, integrable by the same clauses. Thus $q\in L^1$ in all four cases; applying [F3] to $\mathrm{\mid }q\mathrm{\mid }$ and $q$ shows that $q(x-y)=\mathrm{\Gamma }(x-y,t)P(y)$ is absolutely integrable and $H_{t}P(x)=\int q(z)dz$.

Constant and affine data: by [F1], $H_{t}1(x)=\int \mathrm{\Gamma }(z,t)dz=1$; by [F1] and the vanishing first moments of [F2], $H_t{y}_i(x)=\int \mathrm{\Gamma }(z,t)(x_i-z_i)dz=x_i\int \mathrm{\Gamma }(z,t)dz-\int z_i\mathrm{\Gamma }(z,t)dz=x_i$.

Quadratic data: expanding as in step 1.1 and using [F1] and the covariance clause of [F2], $H_t(y_i{y}_j)(x)=x_i{x}_j\int \mathrm{\Gamma }-x_i\int z_j\mathrm{\Gamma }-x_j\int z_i\mathrm{\Gamma }+\int z_i{z}_j\mathrm{\Gamma }=x_i{x}_j+2t{\delta }_{ij}$, and, summing the diagonal identities, $H_t\mathrm{\mid }y{\mathrm{\mid }}^2(x)=\mathrm{\mid }x{\mathrm{\mid }}^2\int \mathrm{\Gamma }-2{\sum }_i{x}_i\int z_i\mathrm{\Gamma }+\int \mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{\Gamma }=\mathrm{\mid }x{\mathrm{\mid }}^2+2nt$.

Steps 1.1, 2.1 and 3.1 show that all four moment integrals converge absolutely for every $x$ and have the stated values, which is the example.