what is the ODE that gives rise to the Laplace Transform.

Question

It can be shown that the transform pairs can be obtained from a differential equation with some boundary conditions. See, for example, Keener's book chapter 7.

For example. The Fourier transform can be obtained by starting with the ODE:

\begin{eqnarray} -u''(x) - \lambda u(x) = 0 \quad , \quad \lim_{x \to \pm \infty} u(x)=0. \end{eqnarray} with $u(x) \in L^2(-\infty, \infty)$ .This ODE has a Green function $G(x, \xi, \lambda)$ with the property that

\begin{eqnarray} \delta(x-\xi) = - \frac{1}{2 \pi \mathrm{i} } \int_{C_{\infty}} G(x,\xi, \lambda) d \lambda \end{eqnarray}

where the contour $C_{\infty}$ should enclose all the spectrum of the operator $L=-d^2/dx^2 - \lambda$.

Then the $\delta(x-\xi)$ comes as the function composition of the forward and inverse transforms.

What is the differential equation (ODE) and its boundary conditions such that the integral of its Green function having all its spectrum (a Dirac delta) is a composition of a forward and inverse Laplace transforms?

Thanks.

Answer 1

The Laplace transform comes from considering $Lf=-f'$ and its resolvent $$ -f'(x)-\lambda f(x) = g(x),\;\;\; 0 \le x < \infty. $$ The spectrum of $L$ is in the left half plane.

The Laplace transform is a function $F$ that is holomorphic on a right half plane, and nice enough that you end up with the Cauchy representation. For $\Re s > \gamma$, \begin{align} F(s)& = -\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{F(z)}{z-s}dz \\ & = \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}F(z)\int_{0}^{\infty}e^{t(z-s)}dtdz \\ & = \int_{0}^{\infty}e^{-ts}\left(\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{tz}F(z)dz\right)dt. \end{align} This representation is isometric when you deal with the Hardy class of holomorphic functions $F$ on the right half plane for which $$ \|F\|_{H^2}^2=\sup_{u > 0} \int_{-\infty}^{\infty}|F(u+iv)|^2dv < \infty. $$ The holomorphic function is recovered from its $L^2$ boundary function, and then you can write the Cauchy kernel $1/(z-s)$ as shown above; the above steps become rigorous. So this unusual pairing comes from the integral representation of the Cauchy kernel, which is a $\delta$ function for holomorphic functions: $$ \int_{0}^{\infty}e^{-t(s-z)}dt = \frac{1}{s-z} = \delta_{s}(z). $$ The Laplace transform is inverted by the Bromwich integral.

Answer 2

We choose $x$ to be along all the imaginary axis. That is, or $x \in \mathrm{i} (-\infty, \infty)$. Instead of $x$ we want to think of $\mathrm{i} x$ for $x$ real, and the substitution $s=\mathrm{i} x$.

The operator (ODE) is $Lu=-u''$, with boundary conditions $\lim_{x \to \pm \infty} u(x) = 0$. Then the spectral problem has as a Green function, the solution to the equation

\begin{eqnarray*} -G''(s,\xi,\lambda) - \lambda G(s, \xi, \lambda) = \delta(s - \xi). \end{eqnarray*}

The Green function, in terms of $x$ and $y$, is given by (please see my notes on Green functions)

\begin{equation} G(x,y, \lambda) =- \frac{ \mathrm{e}^{-\mathrm{i} \sqrt{\lambda} |x-y|}} {2 \mathrm{i} \sqrt{\lambda}} . \end{equation} where $x,y$ are in the positive real numbers, and in terms of $s=\mathrm{i}x$, and $t=\mathrm{i} y$,

\begin{eqnarray*} G(x,y, \lambda) = - \frac{ \mathrm{e}^{ \sqrt{\lambda} |s-t|}} {2 \mathrm{i} \sqrt{\lambda}} . \end{eqnarray*}

We want to compute the right hand side of the following equation:

\begin{eqnarray*} \delta(s-t) = -\frac{1}{2 \pi \mathrm{i}} \int_{C_{\infty}} G(s,t,\lambda) d \lambda \end{eqnarray*} where the contour $C_{\infty}$ contains all the spectrum (which is located in the imaginary line). As it is, this function as has a brunch cut at $0$, so we make the change of variables $\lambda=\mu^2$, $d \lambda= 2 \mu d \mu$, and unfold the integral where now $\mu$ runs from $-\infty$ to $\infty$ along the imaginary axis. We have then

\begin{eqnarray*} -\frac{1}{2 \pi \mathrm{i}} \int_{C_{\infty}} G(s,t,\lambda) d \lambda &=& -\frac{1}{2 \pi \mathrm{i}} \int_{-\mathrm{i} \infty}^{\mathrm{i} \infty} 2 \mu \frac{\mathrm{e}^{ \mu |s -t|}}{2 \mathrm{i} \mu} d \mu \\ &=& \frac{1}{2 \pi} \int_{-\mathrm{i} \infty}^{\mathrm{i} \infty} \mathrm{e}^{\mu |s -t|} d \mu. \end{eqnarray*}

We observe that that there is no need for the absolute bars in the exponent (check this by assuming $|s|>|t|$ and then $|t|>|s|$) so we write

\begin{eqnarray*} \delta(s-t) = \frac{1}{2 \pi} \int_{-\mathrm{i} \infty}^{\mathrm{i} \infty} \mathrm{e}^{\mu(s-t)} d \mu. \end{eqnarray*}

At this moment we assume that $u(x)$ is a causal function, or simple a regular (no causal function) multiplied by the Heaviside function $H(x)$. Let us now multiply both sides of this equation by $u(x)$ and integrate over $s$ between $0$ and $\infty$. Then

\begin{eqnarray*} u(t) = \int_0^{\infty} ds \, u(s) \frac{1}{2 \pi} \int_{-\mathrm{i}\infty}^ {\mathrm{i} \infty} \mathrm{e}^{\mu(s - t)} d \mu. = \frac{1}{2 \pi} \int_{-\mathrm{i} \infty}^{\mathrm{i} \infty} d \mu \left ( \int_0^{\infty} ds \, u(s) \mathrm{e}^{\mu s} \right ) \; \mathrm{e}^{-\mu t} \end{eqnarray*}

We define the expression inside parenthesis as $U(\mu)$, so that we have the Laplace transform pair:

\begin{eqnarray*} U(\mu) &=& \int_0^{\infty} ds \, u(s) \mathrm{e}^{\mu s} \\ u(t) &=& \frac{1}{2 \pi} \int_{-\mathrm{i} \infty}^{\mathrm{i} \infty} d \mu U(t) \mathrm{e}^{-t \mu}. \end{eqnarray*}