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Mathematics LibreTexts

6.1: Poisson's Formula

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:

  1. (i) \(K(x,y,t)\in C^\infty(\mathbb{R}^n\times\mathbb{R}^n\times\mathbb{R}^1_+)\),
  2. (ii) \((\partial/\partial t\ -\triangle)K(x,y,t)=0,\ t>0\),
  3. (iii) \(K(x,y,t)>0,\ t>0\),
  4. (iv) \(\int_{\mathbb{R}^n}\ K(x,y,t)\ dy=1\), \(x\in\mathbb{R}^n\), \(t>0\)
  5. \(\delta>0\):(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.

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