M3M6: Methods of Mathematical Physics

$$ \def\dashint{{\int\!\!\!\!\!\!-\,}} \def\infdashint{\dashint_{\!\!\!-\infty}^{\,\infty}} \def\D{\,{\rm d}} \def\E{{\rm e}} \def\dx{\D x} \def\dt{\D t} \def\dz{\D z} \def\C{{\mathbb C}} \def\R{{\mathbb R}} \def\H{{\mathbb H}} \def\CC{{\cal C}} \def\HH{{\cal H}} \def\FF{{\cal F}} \def\I{{\rm i}} \def\qqqquad{\qquad\qquad} \def\qqand{\qquad\hbox{for}\qquad} \def\qqfor{\qquad\hbox{for}\qquad} \def\qqwhere{\qquad\hbox{where}\qquad} \def\Res_#1{\underset{#1}{\rm Res}}\, \def\sech{{\rm sech}\,} \def\acos{\,{\rm acos}\,} \def\vc#1{{\mathbf #1}} \def\ip<#1,#2>{\left\langle#1,#2\right\rangle} \def\norm#1{\left\|#1\right\|} \def\half{{1 \over 2}} \def\fL{f_{\rm L}} \def\fR{f_{\rm R}} $$

Dr Sheehan Olver
s.olver@imperial.ac.uk

Website: https://github.com/dlfivefifty/M3M6LectureNotes

Lecture 21: Laplace transforms

Our goal is to solve the integral equation $$ \lambda u(x) + \int_{0}^\infty K(x-t)u(t) \dt = f(x)\qqfor 0 < x < \infty. $$ A key tool will be the half-Fourier transforms $$ \int_{-\infty}^0 u(t) \E^{-\I s t} \dt \qqand \int_0^\infty u(t) \E^{-\I s t} \dt $$ and the Laplace transform $$ \int_0^\infty u(t) \E^{-z t} \dt $$ in particular we are interested in the analyticity properties with respect to $s$/$z$.

Outline:

1. Analyticity properties of Fourier transforms
   • Inverse Fourier transform on shifted contours
2. Half-Fourier transforms
   • Inverting the Half-Fourier transform
   • Relationship to Laplace transform
3. Application: solving differential equations on the half-line

Analyticity properties of Fourier transforms

Consider the Fourier transform of $$ \sech x = {2 \over \E^x + \E^{-x}} $$ This function has exponential decay in both directions:

Now the Fourier transform of $\sech x$ is $$ \FF\sech(s) = \int_{-\infty}^\infty \sech t \, \E^{-\I s t} \dt = \pi \sech{\pi s \over 2} $$ This is calculated via Residue theorem with a bit of work.

Note that $$ \pi \sech {\pi z \over 2} = {2 \pi \over \E^{\pi z \over 2} + \E^{-{\pi z \over 2}}} $$ is analytic for $-1 < \Im z < 1$.

This is because of the expontial decay.

Theorem (Analyticity of Fourier transforms) Suppose $|f(x) \E^{\gamma x}| < |M(x)|$ where $M$ is absolutely integrable for all $a < \gamma < b$. Then $$ \widehat f(z) = \int_{-\infty}^\infty f(t) \E^{-\I z t} \dt $$ is analytic for $a < \Im z < b$.

Proof Let $z = s + \I \gamma$ and note that $|f(t) \E^{\I z t}| = |f(t) \E^{\gamma t}|$. Thus for $a < \gamma < b$, we can exchange differentiation and integration to get $$ {\D \widehat f \over \dz} = -\I z \hat f(z) $$ ⬛️

Remark We don't need $f$ to be analytic at all! Decay in $f$ gives analyticity.

In the case of $\sech x$, we get exponential decay in both directions: that is $\sech x \E^{|\gamma| x}$ is absolutely integrable for $\gamma < 1$.

Another example is $\E^{-x^2/2}$, which is absolutely integrable for any $\gamma$. Therefore, it's Fourier transform is in fact entire: $$ \FF[\E^{-\diamond^2/2}](z) = \sqrt{2\pi} \E^{-{z^2 \over 2}} $$

Inverse Fourier transform on shifted contours

A neglected fact of the Fourier transform is that we can think of $\hat f(z)$ living on any line $(-\infty + \I \gamma, \infty+\I \gamma)$, and in fact we can recover $f$ from the Fourier transform only on this line. This works even if $\hat f(s)$ is not defined on the real-axis, the real-axis is NOT special!

Theorem Suppose $f(x) \E^{\gamma x}$ is square integrable. Then $$ f(x) = {1 \over 2 \pi} \int_{-\infty+\I \gamma}^{\infty + \I \gamma} \widehat f(\zeta) \E^{\I x \zeta} \D \zeta $$

Proof Note for $g(x) = f(x) \E^{\gamma x}$ $$ \widehat g(s) = \int_{-\infty}^\infty f(t) \E^{\gamma t - \I s t} \dt = \widehat f(s + \I \gamma). $$ Therefore we have \begin{align*} \E^{\gamma x} f(x) &= g(x) = \FF^{-1} \widehat g(x) = {1 \over 2 \pi} \int_{-\infty}^{\infty} \widehat g(s) \E^{\I x s} \D s ={1 \over 2 \pi} \int_{-\infty}^{\infty} \widehat f(s + \I \gamma) \E^{\I x s} \D s \\ & = {1 \over 2 \pi} \int_{-\infty+ \I \gamma}^{\infty+ \I \gamma} \widehat f(\zeta) \E^{\I x (\zeta - \I \gamma)} \D \zeta \\ &= {1 \over 2 \pi} \E^{\gamma x} \int_{-\infty+ \I \gamma}^{\infty+ \I \gamma} \widehat f(\zeta) \E^{\I x \zeta} \D \zeta \end{align*} Which shows the result by cancelling out $\E^{\gamma x}$.

Half-Fourier transforms

Consider now $$ \int_0^\infty f(t) \E^{-\I s t} \dt $$ This is in fact the Fourier transform of $f$ extended to the negative real axis by zero: $$ \int_{-\infty}^\infty \begin{cases}f(t) & t \geq 0 \\ 0 & \hbox{otherwise} \end{cases} \E^{-\I s t} \dt $$ To make sure we remember the domain of definition, we introduce the notation: $$ f_{\rm R}(x) = \begin{cases}f(t) & t \geq 0 \\ 0 & \hbox{otherwise} \end{cases} $$ and $$ f_{\rm L}(x) = \begin{cases}f(t) & t < 0 \\ 0 & \hbox{otherwise} \end{cases} $$ Therefore $$ \widehat\fR(s) = \int_0^\infty f(t) \E^{-\I s t} \dt \qqand \widehat\fL(s) = \int_{-\infty}^0 f(t) \E^{-\I s t} \dt $$

Because it is identically zero on the negative real axis, we immediately get the following:

Corollary (analyticity of Half-Fourier transform) Suppose $f(x)$ is bounded for $x \geq 0$. Then $\widehat{\fR}(z)$ is analytic in the lower half plane $$\H_+ = \{ z : \Im z < 0 \}.$$

More generally, $f$ can even have exponential decay: if $f(x) \E^{\gamma x}$ is bounded then $\widehat\fR(z)$ is analytic in $\{z : \Im z < \gamma \}$. As before, the same inversion formula follows:

Corollary (inverting Half-Fourier transform) Suppose $f(x) \E^{\gamma x}$ is square integrable for $x \geq 0$. Then $$ f(x) = {1 \over 2 \pi} \int_{-\infty + \I M}^{\infty + \I M} \widehat{\fR}(\zeta) \E^{\I x \zeta} \D \zeta $$ for any choice of $-\infty < M \leq \gamma$.

Example Consider $f(x) = x \E^{-x}$ for $0 \leq x < \infty$. Note that $f(x) \E^{\gamma x}$ is square integrable for any $\gamma < 1$, and we have $$ \widehat\fR(z) = \int_0^\infty t \E^{-t -\I s t} \dt = {1 \over (1+\I s)^2} $$ is analytic for $\Im z < 1$. Thus for any $M < 1$ we have \begin{align*} f(x) &= {1 \over 2 \pi} \int_{-\infty+\I M}^{\infty +\I M}\widehat\fR(\zeta) \E^{\I x \zeta}\D \zeta \\ &={1 \over 2 \pi} \int_{-\infty+\I M}^{\infty +\I M}{1 \over (1+\I \zeta)^2} \E^{\I x \zeta}\D \zeta \end{align*}

Since $x > 0$, we can use Residue calculus in the upper-half plane, which confirms the result: $$ \Res_{z = -\I} {\E^{\I x z} \over (1+\I z)^2} = \Res_{z = -\I} {\E^{- x} + \I x \E^{-x} (z +\I) + O(z+\I)^2 \over -(z-\I)^2} = -\I x \E^{-x}. $$ Example Consider $f(x) = x$. This function is not square-integrable, but we have $f(x) \E^{\gamma x}$ is square integrable for any $\gamma < 0$, and we find for $\Im z < 0$ $$ \widehat{\fR}(z) = -{1 \over z^2} $$ Thus we can still use the result to say, for any $M < 0$, $$ f(x) =-{1 \over 2 \pi} \int_{-\infty+\I M}^{\infty +\I M} {1 \over \zeta^2} \E^{\I x \zeta}\D \zeta $$

Note that the results have corresponding analogues for $\fL$:

Corollary (analyticity of left Half-Fourier transform) Suppose $f(x) \E^{\gamma x}$ is bounded for $x \leq 0$. Then $\widehat{\fL}(z)$ is analytic for $\{z : \Im z > \gamma \}$.

Corollary (inverting left Half-Fourier transform) Suppose $f(x) \E^{\gamma x}$ is square integrable for $x < 0$. Then $$ f(x) = {1 \over 2 \pi} \int_{-\infty + \I M}^{\infty + \I M} \widehat{\fL}(\zeta) \E^{\I x \zeta} \D \zeta $$ for any choice of $\gamma \leq M < \infty $.

Laplace transforms

Now consider the Laplace transform $$ \check f(z) = \int_0^\infty f(t) \E^{-z t} \dt $$ but this is just the half Fourier transform evaluated on the negative imaginary axis! $$ \check f(z) = \widehat\fR (-\I z) $$ Thus if $f(x) \E^{\gamma x}$ is square integrable, then $\check f(z)$ is well-defined for $\Re z \geq \gamma$.

NEVER think of the Laplace transform as a real-valued object: it only makes sense as a complex object. This is seen from the inverse Laplace transform $$ f(x) = {1 \over 2 \pi \I} \int_{-\I \infty - M}^{\I \infty - M} \check f(\zeta) \E^{\zeta x} \D \zeta $$ which is of course just the inverse Fourier transform in disguise.

Application: solving differential equations on the half-line

Consider the following ODE for $x \geq 0$: \begin{align*} u''(x) + 2u'(x) + u(x) = f(x) \end{align*} with initial conditions $u(0) = u'(0) = 0$. Note that we have by integration-by-parts \begin{align*} \check{u'}(z) = \int_0^\infty u'(t) \E^{-z t} \dt = u(0) + z \int_0^\infty u(t) \E^{-z t} \dt = u(0) + z \check u(z) \\ \check{u''}(z) = u'(0) + z \check{u'}(z) = u'(0) + z u(0) + z^2 \check u(0) \end{align*} Thus taking into account the initial conditions, are equation in Laplace space becomes $$ (z^2 + 2z + 1) \check u(z) = \check f(z) $$ Hence we have $$ \check u(z) = {1 \over 2 \pi \I}\int_{-\I\infty-M}^{\I \infty-M} {\check f(\zeta) \over \zeta^2 + 2\zeta+1} \E^{x \zeta} \D \zeta $$ Consider the case $f(x) = x$, so that $$ \check f(z) = {1 \over z^2} $$ Here we need $M < 0$ hence we are integrating on a contour in the right-half plane. Using Residue calculus, we have $$ u(z) = \left(\Res_{z = -1} + \Res_{z=0}\right) {\E^{z x} \over z^2 (z +1)^2} = \Res_{z = -1} {\E^{-x} +\E^{-x}(x+2) (z+1) +O(z+1)^2 \over (z +1)^2} + \Res_{z=0} {1 + (x-2) z +O(z)^2 \over z^2} \\ = (x+2)\E^{-x} + x-2 $$

Laplace transform of rational functions

We now consider the question of calculating Laplace transforms (or equivalently, half-Fourier transforms) $$ \check f(s) = \int_0^\infty f(t) \E^{-s t} \D t $$ where $f$ is rational. We're going to do something seemingly crazy: we'll first calculate the Cauchy transform $$ \CC[f \E^{-s \diamond}](z) = {1 \over 2 \pi \I} \int_0^\infty {f(t) \E^{-s t} \over t- z} \D t $$ so that $$ \check f(s) = -2 \pi \I\lim_{z \rightarrow \infty} z \CC[f \E^{-s \diamond}](z) $$ Note that the exponential decay in the integrand allows us to use Plemelj's lemma: if we find a function $\phi(z)$ such that

1. $\phi(z)$ is analytic off $[0,\infty)$
2. $\lim_{z \rightarrow \infty} \phi(z) = 0$ for any angle of approach
3. $\phi$ has weaker than pole singularities at 0
4. $\phi_+(x) - \phi_-(x) = f(x) \E^{-s x}$

Then we have calculated the Cauchy transform: $$\phi(z) = \CC[f \E^{-s \diamond}](z).$$

Let's start with $f(x) = 1$ and $s = 1$, that is, what is the Cauchy transform of $\E^{-x}$? Consider the exponential integral: $$ {\rm Ei}(z) = \int_{-\infty}^z {\E^\zeta \over \zeta} \D \zeta $$ Without loss of generality, the contour of integration is $$ (\infty,-1) \cup [-1, z) $$ that is, a straight line to $-1$ and a straightline from $-1$ to $z$. Thus we have a branch cut on $[0,\infty)$ which has the jump $$ {\rm Ei}^+(x) - {\rm Ei}^-(x) = -\oint {\E^\zeta \over \zeta} \D \zeta = -2 \pi \I $$ where $$ {\rm Ei}^\pm(x) = \lim_{\epsilon \rightarrow 0} {\rm Ei}(x\pm\I \epsilon) $$ Consider $$ \phi(z) = -{\E^{-z} {\rm Ei}(z) \over 2 \pi \I } $$

1. This is analytic off $[0,\infty)$
2. Integrating by parts we have decay at $\infty$ in all directions: $$ \E^{-z} {\rm Ei}(z) = {1 \over z} - \int_{-\infty}^z {\E^{\zeta-z} \over \zeta^2} \D \zeta = O(z^{-1}) $$
3. $\phi$ has a logarithmic singularity at 0
4. $$\phi_+(x) - \phi_-(x) = \E^{-x}.$$

Thus $$ \CC[ \E^{-s \diamond}](z) = \phi(s z) = - {\E^{-s z} {\rm Ei}(s z) \over 2 \pi \I} $$

Let's make sure I didn't make a mistake. Here we first define Ei:

The expression matches the Cauchy transform:

We then recover the Laplace transform by taking the limit: $$ \check 1(s) = -2 \pi \I \lim_{z\rightarrow \infty} z \phi(s z) = \lim_{z\rightarrow \infty} {z \over s z} = {1 \over s} $$

What about rational $f$? Do the same trick of subtracting off the singularities. For example, consider $f(z) = 1/(z+1)$. Then $$ {\phi(s z) - \phi(-s) \over z+1} $$ satisfies all the necessary properties.

Therefore, $$ \check f(s) = -2 \pi \I \lim_{z\rightarrow \infty} z {\phi(s z) - \phi(-s) \over z+1} = 2 \pi \I \phi(-s) = -\E^{s} {\rm Ei}(-s) $$