Notebook

In [2]:

using Plots, ComplexPhasePortrait, ApproxFun, SingularIntegralEquations, OscillatoryIntegrals, 
        SpecialFunctions
gr();

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:

Analyticity properties of Fourier transforms
- Inverse Fourier transform on shifted contours
Half-Fourier transforms
- Inverting the Half-Fourier transform
- Relationship to Laplace transform
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:

In [5]:

xx = -10:0.01:10
plot(xx,sech.(xx))

Out[5]:

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.

In [6]:

f = Fun(sech, Line())
fourier(f, 2.0)

Out[6]:

0.2710149513994194 + 0.0im

In [7]:

π*sech(π*2.0/2)

Out[7]:

0.2710149513994184

Note that

$\pi \sech {\pi z \over 2} = {2 \pi \over \E^{\pi z \over 2} + \E^{-{\pi z \over 2}}}$ is analytic for

$-1 < \Im z < 1$ .

In [9]:

phaseplot(-3..3, -3..3, z -> π*sech(π*z/2))

Out[9]:

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

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

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

$\phi(z) = -{\E^{-z} {\rm Ei}(z) \over 2 \pi \I }$

This is analytic off $[0,\infty)$
Integrating by parts we have decay at $\infty$ in all directions:

$\E^{-z} {\rm Ei}(z) = {1 \over z} - \int_{-\infty}^z {\E^{\zeta-z} \over \zeta^2} \D \zeta = O(z^{-1})$

$\phi$ has a logarithmic singularity at 0

$\phi_+(x) - \phi_-(x) = \E^{-x}.$

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:

In [20]:

const ei₋₁ = let ζ = Fun(-50 .. -1)
    sum(exp(ζ)/ζ)
end
function ei(z) 
    ζ = Fun(Segment(-1 , z))
    ei₋₁ + sum(exp(ζ)/ζ)
end

Out[20]:

ei (generic function with 1 method)

In [16]:

φ = (z) -> -exp(-z)*ei(z)/(2π*im)

Out[16]:

#9 (generic function with 1 method)

The expression matches the Cauchy transform:

In [18]:

t = Fun(0 .. 50)
s = 2.0
z = 2.0+2.0im
sum(exp(-s*t)/(t-z))/(2π*im)

Out[18]:

0.02534853710699088 + 0.017828329563678025im

In [21]:

φ(s*z)

Out[21]:

0.02534853710699058 + 0.017828329563677824im

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.

In [23]:

t = Fun(0 .. 50)
s = 2.0
z = 2.0+2.0im
f = 1/(t+1)
sum(exp(-s*t)*f/(t-z))/(2π*im)

Out[23]:

0.0174397493357667 + 0.013485355450154557im

In [24]:

(φ(s*z)-φ(-s))/(z+1)

Out[24]:

0.017439749335766603 + 0.01348535545015462im

Therefore,

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

In [25]:

@show sum(exp(-s*t)*f)
@show 2π*im*φ(-s)
@show -exp(s)*ei(-s);

sum(exp(-s * t) * f) = 0.3613286168882254
(2π) * im * φ(-s) = 0.3613286168882283 + 0.0im
-(exp(s)) * ei(-s) = 0.3613286168882283