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Mathematics LibreTexts

6.1: Poisson's Formula

  • Page ID
    2156
  • [ "article:topic", "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}

    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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