M3M6: Methods of Mathematical Physics¶

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

Dr Sheehan Olver

s.olver@imperial.ac.uk
Website: https://github.com/dlfivefifty/M3M6LectureNotes

Lecture 25: The Wiener–Hopf method¶

We can now employ the Wiener–Hopf method to solve

$$\lambda u(x) + \int_{0}^\infty K(x-t)u(t) \dt = f(x)\qqfor 0 < x < \infty.$$ For simplicity, take $\lambda = 1$ (note we can divide through).

The Wiener–Hopf method consists of the following steps:

1. Calculate Fourier transforms $\widehat K(s)$ and $\widehat{f_{\rm R}}(s)$ to reduce to RH problem $\Phi_+(s) -g(s) \Phi_-(s) = h(s)$ with $\lim \Phi(z) = 0$
2. Find homogenous solution $\kappa_+(s) = g(s) \kappa_-(s)$ with $\lim \kappa(z) = 1$
3. Solve Cauchy transform problem $Y_+(s) - Y_-(s) = {h(s) \over \kappa_+(s)}$
4. Find $u(x)$ by inverting Fourier transform $\widehat{u_{\rm R}}(s) = \Phi_-(s) = \kappa_-(s) Y_-(s)$

We will demonstrate these four steps using the example $K(x) = \E^{-|x|}$ and $f(x) = \E^{-x}$.

Step 1: Calculate Fourier transforms¶

We have

$$\widehat{K}(s) = \int_{-\infty}^0 \E^{t}\E^{-\I s t} \dt + \int_0^\infty \E^{-t} \E^{-\I s t} \dt = {1 \over 1-\I s} - {1 \over -1-\I s} = {2 \over 1+s^2}$$ Because we have exponential decay at $\pm \infty$, $\widehat{K}$ is analytic in a strip. Furthermore, $$\widehat{f_{\rm R}}(s) = \int_0^\infty \E^{-t} \E^{-\I t s} \dt = -{1 \over -1-\I s} = {\I \over \I-s}$$ As predicted, $\widehat{\fR}(s)$ is analytic in the lower half-plane.

We thus have the RH problem:

$$\underbrace{\Phi_+(s)}_{\widehat{p_{\rm L}}(s)} - \underbrace{g(s)}_{1 + \widehat K(s)} \underbrace{\Phi_-(s)}_{\widehat{u_{\rm R}}(s)} = \underbrace{h(s)}_{-\widehat{\fR}(s)} \qqand \lim\Phi(z) = 0$$ where $$p(x) = \int_{-\infty}^\infty K(x-t)u_{\rm R}(t) \dt$$ is unknown, or in our case $$\Phi_+(s) - {3 + s^2 \over 1 + s^2} \Phi_-(s) = {\I \over s - \I} \qqand \lim\Phi(z) = 0$$

Here we confirm our calculations are correct for the Fourier transforms:

Step 2: Find homogeneous solution¶

Last lecture we employed the "guess and check" kernel factorization method to find

$$g(s) = {3 + s^2 \over 1 + s^2} = \underbrace{s + \I \sqrt{3} \over s + \I}_{\kappa_+(s)} \underbrace{s - \I \sqrt{3} \over s-\I }_{\kappa_-(s)^{-1}}$$ That is, we have the solution $$\kappa(z) = \begin{cases} {z + \I \sqrt{3} \over z + \I} & \Im z > 0 \\ {z - \I \over z - \I \sqrt{3}} & \Im z < 0 \end{cases}$$ Always verify this solution satisfies the right conditions:

1. $\kappa(z)$ is analytic off $\R$, since the pole of ${z + \I \sqrt{3} \over z + \I}$ is in the lower-half plane and the pole of ${z - \I \over z - \I \sqrt{3}}$ is in the upper-half plane
2. $\lim \kappa(z) = 1$ by L'Hopital's rule or similar.
3. It has the right jump:

$$\kappa_+(s) = {s + \I \sqrt{3} \over s + \I} = {3 + s^2 \over 1 + s^2}{s - \I \over s - \I \sqrt{3}} = g(s)\kappa_-(s)$$

Here we confirm that we have the right jump:

Step 3: Solve Cauchy transform problem¶

Writing $\Phi(z) = \kappa(z) Y(z)$ we have

$$\Phi_+(s) - g(s) \Phi_-(s) = \kappa_+(s) (Y_+(s) - Y_-(s))$$ hence we need to solve $$Y_+(s) - Y_-(s) = {h(s) \over \kappa_+(s)} = {\I \over s- \I} {s + \I \over s+ \I \sqrt{3}}$$ As in last lecture, we found: $$Y(z) = \begin{cases} {-\I \over 1+\sqrt 3} {1-\sqrt 3 \over z+ \I \sqrt{3}} & \Im z > 0 \\ {-2 \I \over z-\I}{1 \over 1+ \sqrt{3}} & \Im z < 0 \end{cases}$$ Let's double check that $Y$ has the right jump:

Thus we get

$$\begin{align*} \Phi(z) &= \kappa(z) \CC_\R\br[{h \over \kappa_+}](z) = \begin{cases} {-\I \over 1+\sqrt 3} {1-\sqrt 3 \over z+\I} & \Im z > 0 \\ {-2 \I \over 1+ \sqrt{3}} {1 \over z-\sqrt{3}\I} & \Im z < 0 \end{cases} \end{align*}$$

We can confirm it satisfies the right RHP:

Step 4: Invert the Fourier transform¶

In particular, we have

$$\widehat{u_{\rm R}}(s) = \Phi_-(s) = \kappa_-(s) Y_-(s) = {-2 \I \over (1+\sqrt{3})(s-\I \sqrt 3)}$$ This is analytic in the lower half plane, and for $x > 0$ we can use Jordan's lemma hence use Residue calculus in the upper-half plane: $$u(x) = {1 \over 2 \pi} \int_{-\infty}^\infty \widehat{u_{\rm R}}(s) \E^{\I s x} \ds = {-\I \over \pi(1+\sqrt{3})} \int_{-\infty}^\infty {1 \over s-\I \sqrt 3} \E^{\I s x} \ds = {2 \over (1+\sqrt{3})} \Res_{z=\I \sqrt 3} {1 \over z-\I \sqrt 3} \E^{\I z x} = {2 \E^{-\sqrt{3}x} \over 1+\sqrt{3}}$$

Step 5: Check I didn't make a mistake¶

Did this work? Amazingly, yes (and only took me four tries to get the sign right! 🕺):

$$\begin{align*} u(x) + \int_0^\infty K(t -x) u(t) \dt &= {2 \over 1+\sqrt{3}} \left( \E^{-\sqrt{3}x} + \int_0^x \E^{t-x} \E^{-\sqrt{3}t} \dt + \int_x^\infty \E^{x-t} \E^{-\sqrt{3}t}\dt \right) \\ &={2 \over 1+\sqrt{3}}\left( \E^{-\sqrt{3}x} + \E^{-x} \int_0^x \E^{(1-\sqrt{3})t} \dt + \E^x \int_x^\infty \E^{-(1+\sqrt{3})t}\dt \right) \\ &={2 \over 1+\sqrt{3}}\left( \E^{-\sqrt{3}x} + {1 \over 1-\sqrt 3} (\E^{-\sqrt{3}x} - \E^{-x}) + { \E^{-\sqrt 3 x } \over 1+\sqrt{3}} \right) \\ &=\E^{-x} + {2 \E^{-\sqrt{3}x} \over 1+\sqrt{3}}\left({-2 \over -2} + {1+ \sqrt 3 \over -2} + {1 - \sqrt 3 \over -2} \right) \\ &= \E^{-x} \end{align*}$$

We can verify it numerically: