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# 6.1: Poisson's Formula

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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} 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 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 , 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

• Integrated by Justin Marshall.