M3M6: Methods of Mathematical Physics

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

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

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

Lecture 22: Integral equations on the half-line and Riemann–Hilbert problems

Our goal is to solve the integral equation $$ \lambda u(x) + \int_{0}^\infty K(x-t)u(t) \dt = f(x)\qqfor 0 < x < \infty. $$ We will demonstrate the procedure for the special case $K(x) = \E^{-\gamma |x|}$.

Taking Fourier transforms, we get functions analytic above or below the real axis, giving us a Riemann–Hilbert problem of finding $\Phi(z)$ analytic off $(-\infty,\infty)$ such that $$ \Phi_+(s) - g(s)\Phi_-(s) = h(s) \qqand \Phi(\infty) = C $$ where $\Phi_\pm(s) = \lim_{\epsilon \rightarrow 0} \Phi(s \pm \I \epsilon)$ are the limits from above and below. Here $g$ and $h$ are given, and $g(\pm \infty) = 1$.

Outline:

1. Integral equation to Riemann–Hilbert problem
2. Cauchy transforms on the Real line
   • Application: Calculating error functions

Integral equation to Riemann–Hilbert problem

Recall the notation $$ f_{\rm R}(x) = \begin{cases}f(t) & t \geq 0 \\ 0 & \hbox{otherwise} \end{cases} $$ and $$ f_{\rm L}(x) = \begin{cases}f(t) & t < 0 \\ 0 & \hbox{otherwise} \end{cases} $$

Using this, we can rewrite the integral equation on the half line $$ \lambda u(x) + \int_{0}^\infty K(x-t)u(t) \dt = f(x)\qqfor 0 < x < \infty. $$ as an integral equation on the whole line: $$ \lambda u_{\rm R}(x) + \int_{-\infty}^\infty K(x-t)u_{\rm R}(t) \dt = f_{\rm R}(x) + p_{\rm L}(x)\qqfor -\infty < x < \infty. $$ where $$ p(x) = \int_{-\infty}^\infty K(x-t)u_{\rm R}(t) \dt $$ Taking Fourier transforms, this becomes: $$ (\lambda + \widehat K(s)) \widehat{u_{\rm R}}(s) = \widehat{f_{\rm R}}(s) + \widehat{p_{\rm L}}(s) $$ As discussed last lecture, assuming $u$ is "nice" we are guaranteed that $\widehat{ u_{\rm R}}(s)$ is analytic in the lower half-plane and $\widehat{ p_{\rm L}}(s)$ is analytic in the upper-half plane. Thus introduce the sectionally analytic function: $$ \Phi(z) = \begin{cases} \widehat{p_{\rm L}}(z) & \Im z > 0 \\ \widehat{u_{\rm R}}(z) & \Im z < 0 \end{cases} $$ Then our integral transformed integral equation becomes: $$ \underbrace{\Phi_+(s)}_{\widehat{p_{\rm L}}(s)} - \underbrace{g(s)}_{\lambda + \widehat K(s)} \underbrace{\Phi_-(s)}_{\widehat{u_{\rm R}}(s)} = \underbrace{h(s)}_{-\widehat{f_{\rm R}}(s) } \qqand \Phi(\infty) = 0 $$

Here there is one unknown $\Phi(z)$, and we claim that in certain conditions this has one—and only—one solution. Thus we wish to:

1. Find $\Phi(z)$
2. Recover $u(x)$ via the inverse Fourier transform ${\cal F}^{-1} \Phi_-$