6.1: Poisson's Formula

Last updated
Save as PDF

Page ID: 2156

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$

Assume $u$ is a solution of (6.2), then, since Fourier transform is a linear mapping,

\[\widehat{u_t-\triangle u}=\hat{0}.\]

From properties of the Fourier transform, see Proposition 5.1, we have

\[\widehat{\triangle u}=\sum_{k=1}^n\widehat{\frac{\partial^2 u}{\partial x_k^2}}=\sum_{k=1}^n i^2\xi^2_k\widehat{u}(\xi),\]

provided the transforms exist. Thus we arrive at the ordinary differential equation for the Fourier transform of $u$

\[\frac{d\widehat{u}}{dt}+|\xi|^2\widehat{u}=0,\]

where $\xi$ is considered as a parameter. The solution is

\[\widehat{u}(\xi,t)=\widehat{\phi}(\xi)e^{-|\xi|^2 t}\]

since $\widehat{u}(\xi,0)=\widehat{\phi}(\xi)$. From Theorem 5.1 it follows

\begin{eqnarray*}
u(x,t)&=&(2\pi)^{-n/2}\int_{\mathbb{R}^n}\ \widehat{\phi}(\xi)e^{-|\xi|^2t}e^{i\xi\cdot x}\ d\xi\\
&=&(2\pi)^{-n}\int_{\mathbb{R}^n}\ \phi(y)\left(\int_{\mathbb{R}^n}e^{i\xi\cdot (x-y)-|\xi|^2t}\ d\xi\right)\ dy.
\end{eqnarray*}

Set

$$K(x,y,t)=(2\pi)^{-n}\int_{\mathbb{R}^n}e^{i\xi\cdot (x-y)-|\xi|^2t}\ d\xi.\]

By the same calculations as in the proof of Theorem 5.1, step (vi), we find

\begin{equation}
\label{kernel1}
K(x,y,t)=(4\pi t)^{-n/2}e^{-|x-y|^2/4t}.
\end{equation}

$Kernel $K(x,y,t)$, $\rho=|x-y|$, $t_1<t_2$$

Figure 6.1.1: Kernel $K(x,y,t)$, $\rho=|x-y|$, $t_1<t_2$

Thus we have

\begin{equation}
\label{poisson1}
u(x,t)=\frac{1}{\left(2\sqrt{\pi t}\right)^n}\int_{\mathbb{R}^n}\ \phi (z)e^{-|x-z|^2/4t}\ dz.
\end{equation}

Definition. Formula (\ref{poisson1}) is called Poisson's formula} and the function $K$ defined by (\ref{kernel1}) is called heat kernel or fundamental solution of the heat equation.

Proposition 6.1 The kernel $K$ has following properties:

(i) $K(x,y,t)\in C^\infty(\mathbb{R}^n\times\mathbb{R}^n\times\mathbb{R}^1_+)$,
(ii) $(\partial/\partial t\ -\triangle)K(x,y,t)=0,\ t>0$,
(iii) $K(x,y,t)>0,\ t>0$,
(iv) $\int_{\mathbb{R}^n}\ K(x,y,t)\ dy=1$, $x\in\mathbb{R}^n$, $t>0$
$\delta>0$:(v) For each fixed

$$\lim_{\begin{array}{l}t\to0\\ t>0\end{array}}\int_{\mathbb{R}^n\setminus B_\delta(x)}\ K(x,y,t)\ dy=0$$

uniformly for $x\in\mathbb{R}$.

Proof. (i) and (iii) are obviously, and (ii) follows from the definition of $K$. Equations (iv) and (v) hold since

\begin{eqnarray*}
\int_{\mathbb{R}^n\setminus B_\delta(x)}\ K(x,y,t)\ dy&=& \int_{\mathbb{R}^n\setminus B_\delta(x)}\ (4\pi t)^{-n/2}e^{-|x-y|^2/4t}\ dy\\
&=&\pi^{-n/2}\int_{\mathbb{R}^n\setminus B_{\delta/\sqrt{4t}}(0)}e^{-|\eta|^2}\ d\eta
\end{eqnarray*}
by using the substitution $y=x+(4t)^{1/2}\eta$. For fixed $\delta>0$ it follows (v) and for $\delta:=0$ we obtain (iv).

$\Box$

Theorem 6.1. Assume $\phi\in C(\mathbb{R}^n)$ and $\sup_{\mathbb{R}^n}|\phi(x)|<\infty$. Then $u(x,t)$ given by Poisson's formula (\ref{poisson1}) is in $C^{\infty}(\mathbb{R}^n\times\mathbb{R}^1_+)$, continuous on $\mathbb{R}^n\times[0,\infty)$ and a solution of the initial value problem (6.2), (6.3).

Proof. It remains to show

$$
\lim_{\begin{array}{l}x\to\xi\\
t\to0\end{array}}u(x,t)=\phi(\xi).
\]

$Figure to the proof of Theorem 6.1$

Figure 6.1.2: Figure to the proof of Theorem 6.1

Since $\phi$ is continuous there exists for given $\varepsilon>0$ a $\delta=\delta(\varepsilon)$ such that $|\phi(y)-\phi(\xi)|<\varepsilon$ if $|y-\xi|<2\delta$.
Set $M:=\sup_{\mathbb{R}^n}|\phi(y)|$. Then, see Proposition 6.1,
$$
u(x,t)-\phi(\xi)=\int_{\mathbb{R}^n}\ K(x,y,t)\left(\phi(y)-\phi(\xi)\right)\ dy.
$$
It follows, if $|x-\xi|<\delta$ and $t>0$, that
\begin{eqnarray*}
|u(x,t)-\phi(\xi)|&\le&\int_{B_{\delta}(x)}\ K(x,y,t)\left|\phi(y)-\phi(\xi)\right|\ dy\\
&&+\int_{\mathbb{R}^n\setminus B_{\delta}(x)}\ K(x,y,t)\left|\phi(y)-\phi(\xi)\right|\ dy\\
&\le&\int_{B_{2\delta}(x)}\ K(x,y,t)\left|\phi(y)-\phi(\xi)\right|\ dy\\
&&+2M\int_{\mathbb{R}^n\setminus B_{\delta}(x)}\ K(x,y,t)\ dy\\
&\le&\varepsilon\int_{\mathbb{R}^n}\ K(x,y,t)\ dy+2M\int_{\mathbb{R}^n\setminus B_{\delta}(x)}\ K(x,y,t)\ dy\\
&<&2\varepsilon
\end{eqnarray*}
if $0<t\le t_0$, $t_0$ sufficiently small.

$\Box$

Remarks. 1. Uniqueness follows under the additional growth assumption
$$
|u(x,t)|\le Me^{a|x|^2}\ \ \mbox{in}\ D_T,
$$
where $M$ and $a$ are positive constants,
see Proposition 6.2 below.
In the one-dimensional case, one has uniqueness in the class $u(x,t)\ge 0$ in $D_T$, see [10], pp. 222.

2. $u(x,t)$ defined by Poisson's formula depends on all values $\phi(y)$, $y\in\mathbb{R}^n$. That means, a perturbation of $\phi$, even far from a fixed $x$, has influence to the value $u(x,t)$. This means that heat travels with infinite speed, in contrast to the experience.

Contributors and Attributions

Prof. Dr. Erich Miersemann (Universität Leipzig)
Integrated by Justin Marshall.