M3M6: Methods of Mathematical Physics

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

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

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

Lecture 24: Riemann–Hilbert problems on the real line

Last lecture we saw that integral equation $$ \lambda u(x) + \int_{0}^\infty K(x-t)u(t) \dt = f(x)\qqfor 0 < x < \infty. $$ could be transformmed into a Riemann–Hilbert problem: find $\Phi(z)$ analytic off $\R$ 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.

We now wish to study the solution of this equations, using the Plemelj theorem and Cauchy transform of last lecture.

Outline:

1. Solving variable coefficient Riemann–Hilbert problems
   • Homogeneous problems: $h = 0$
   • Inhomogeneous problems: $h \neq 0$

Homogenous Riemann–Hilbert problems on the real line

Back to the Riemann–Hilbert problem, but let's assume $h$ is zero. Consider $$ \kappa_+(s) = g(s) \kappa_-(s) \qqand \kappa(\infty) = I $$

Formally, taking logs of both sides reduces this to a subtractive RH problem: $$ \log \kappa_+(s) - \log\kappa_-(s) \questionequals \log g(s) $$ Assuming that $g(s) \rightarrow 1$ as $s \rightarrow \pm \infty$ at a sufficient rate, this motivates the guess $$ \kappa(z) = \E^{\CC[\log g](z)} $$

Assuming $\log g(s)$ is "nice", we have guaranteed that

Theorem (Homogeneous solution to RH problem) Suppose $\log g(s)$ satisfies the conditions of Plemelj on the circle. Then $\kappa(z) = \E^{\CC_\R[\log g](z)}$ is the unique solution to the following RH problem:

1. $\kappa(z)$ is analytic off $\R$
2. $\lim_{z\rightarrow \infty}\kappa(z) = 1$
3. $\kappa_+(s) = g(s) \kappa_-(s)$ for $s \in \R$

Proof (1) follows from definition. (2) follows since $\CC[\log g](z) \rightarrow 0$. And (3) follows via: $$ \kappa_+(s) = \E^{\CC_+[\log g](s)} = \E^{\CC_-[\log g](s) + \log g(s)} = \kappa_-(s) g(s) $$ To see uniqueness, observe that $\kappa$ must be invertible, as it is an exponential of something finite. Thus $\kappa(z)^{-1}$ is also analytic off $\R$. Therefore, if we have another solution $\tilde \kappa(z)$ we can consider $r(z) = \tilde\kappa(z) \kappa(z)^{-1}$ which satisfies: $$ r_+(s) = {\tilde\kappa_+(s) \over \kappa_+(s)} = {\tilde\kappa_-(s) g(s) \over \kappa_-(s) g(s)} = r_-(s) $$ Hence $r(z)$ is entire. since both terms tend to $1$, it must be $r(z) = 1$.

⬛️

When is $\log g(s)$ nice? It is sufficient that $g(s) = 1 + O(s^{-1})$ at $\pm \infty$ but also we need to worry about the image: for example, $g(s) \neq 0$ is required to avoid a singularity. We similarly need that the winding number of the image of $g(s)$ to be zero: otherwise, $\log g(s)$ will extend to another sheet. For example, if $g(s) = {z-\I \over z+\I}$ it satisfies the right asymptotics, but surrounds the origin:

Therefore, $\log g(s)$ has a branch cut if we use the standard branch, which breaks the continuity requirement:

We could have analytically continued $\log g(z)$ using $$ \log_1 z = \begin{cases} \log z & \Im z > 0 \\ \log_+ z & z < 0 \\ \log z + 2\pi \I & \Im z < 0 \end{cases} $$ But then $\lim_{s\rightarrow + \infty} \log_1 g(s) = 2 \pi \I$:

Example Consider $$ g(s) = {s^2+3 \over s^2+1} = 1 + O(s^{-1}) $$ Before we do anything Verify that the winding number is zero.

We provide two methods for calculating $\kappa$: one guesses the solution, the other uses the solution formula.

Method 1 (Guess and check / kernel factorization) If we can guess the solution, we can check it satisfies the right criteria. Factoring $g$ we see immediately that $$ g(s) = {s-\I \over s-\sqrt{3} \I} {s+\I \over s+\sqrt{3} \I} $$ Note that the first term is analytic in the upper half plane. The second term is analytic in the lower half plane and invertible. Therefore we can guess the solution is $$ \kappa(z) = \begin{cases} {z+\sqrt 3\I \over z+\I} & \Im z > 0 \\ {z-\I \over z-\sqrt 3\I} & \Im z < 0 \end{cases} $$ This satisfies the three conditions:

1. $\kappa$ is analytic off $\R$
2. $\lim_{z\rightarrow \infty}\kappa(z) = 1$
3. It has the right jump $$ g(s) \kappa_-(s) = {s^2+3 \over s^2+1} {s-\I \over s-\sqrt 3\I} = {s+\I \over s+\sqrt 3 \I} = \kappa_+(s) $$ This function is indeed analytic off the real line.

Let's check that it has the right jump:

Method 2 (evaluate explicit form) This is real valued and positive, hence the winding number of its image is zero. We have $$ \log g(s) = \log({s+\sqrt 3\I \over s+\I}{s-\sqrt 3\I \over s-\I}) $$ Because they are complex conjugates, we know $\log a \bar a = \log a + \log \bar a$ as $[1, \bar a, a]$ lies in the same half plane for $a = {s+\sqrt 3 \I \over s+ \I}$, therefore we can expand: $$ \log g(s) = \log{s+\sqrt 3\I \over s+\I} + \log {s-\sqrt 3\I \over s-\I} $$ Now we note that $\log{s+3\I \over s+\I}$ is analytic in the upper-half plane, therefore it's Cauchy transform, by Plemelj, is $$ \CC\br[\log{s+\sqrt 3\I \over s+\I}](z) = \begin{cases} \log{z+\sqrt 3\I \over z+\I} & \Im z > 0 \\ 0 & \Im z < 0 \end{cases} $$ Similarly, $$ \CC\br[\log{s-\sqrt 3\I \over s-\I}](z) = \begin{cases} -\log{z-\sqrt 3\I \over z-\I} & \Im z > 0 \\ 0 & \Im z < 0 \end{cases} $$ We thus get: $$ \kappa(z) = \E^{\CC\log g(z)} = \E^{\begin{cases} \log{z+\sqrt 3 \I \over z+\I} & \Im z > 0 \\ -\log{z- \sqrt 3\I \over z-\I} & \Im z < 0 \end{cases}} = \begin{cases} {z+ \sqrt 3 \I \over z+\I} & \Im z > 0 \\ {z-\I \over z-\sqrt 3\I} & \Im z < 0 \end{cases} $$

Inhomogenous Riemann–Hilbert problem

Consider now the Riemann–Hilbert problem with zero at infinity: $$ \Phi_+(s) - g(s)\Phi_-(s) = h(s) \qqand \Phi(\infty) =0 $$ Consider writing $\Phi(z) = \kappa(z) \Psi(z)$. Then we can reduce the Riemann–Hilbert problem to a subtractive problem:

$$ \Phi_+(s) - g(s)\Phi_-(s) = \kappa_+(s)(\Psi_+(s) - \Psi_-(s))\qqand \Psi(\infty) = 0 $$

Thus once we have $\kappa$ we can construct $\Phi$.

What if we don't have decay? Just add in a constant times $\kappa$:

Corollary Suppose $\log g$ satisfies the conditions of Plemelj's theorem. Then $$ \Phi(z) = \kappa(z) \CC_\R\br[{h \over \kappa_+}](z) + D \kappa(z) $$ is the unique solution to $$ \Phi_+(s) - g(s)\Phi_-(s) = h(s) \qqand \Phi(\infty) =D $$

Example Suppose $h(s) = {\I \over \I-s}$.

To decompose this as a sum of things analytic in half planes, we just use partial fraction expansion! \begin{align*} {\I \over \I-s} {s + \I \over s+ \I \sqrt{3}} &= {\I \over \I-s}{2\I \over \I(1+ \sqrt{3})} +{\I \over \I(1+\sqrt 3)} {\I(1-\sqrt 3) \over s+ \I \sqrt{3}}\\ &= \underbrace{{-2 \I \over s-\I}{1 \over 1+ \sqrt{3}}}_{-Y_-(s)} + \underbrace{{\I \over 1+\sqrt 3} {1-\sqrt 3 \over s+ \I \sqrt{3}}}_{Y_+(s)} \end{align*} Thus we get $$ Y(z) = \begin{cases} {\I \over 1+\sqrt 3} {1-\sqrt 3 \over z+ \I \sqrt{3}} & \Im z > 0 \\ {2 \I \over z-\I}{1 \over 1+ \sqrt{3}} & \Im z < 0 \end{cases} $$ Let's double check:

We thus have the solution: \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 \sqrt{3}} {z+\sqrt{3}\I \over z+\I} & \Im z > 0 \\ {2 \I \over z-\I}{1 \over 1+ \sqrt{3}} {z-\I \over z-\sqrt{3}\I} & \Im z < 0 \end{cases} \\ &= \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*}

Let's verify it's the right thing:

1. It's analytic off $\R$

2. It goes to zero at infinity

3. It satisfies the right jump: \begin{align*} \Phi_+(s) - g(s) \Phi_-(s) = {\I \over 1+\sqrt 3} {1-\sqrt 3 \over s+\I} - {s^2 + 3 \over s^2 + 1} {2 \I \over 1+ \sqrt{3}} {1 \over s-\sqrt{3}\I} \\ = {\I (s-\I) \over 1+\sqrt 3} {1-\sqrt 3 \over s^2+1} - {s + \sqrt{3} \I \over s^2 + 1} {2 \I \over 1+ \sqrt{3}} \\ = {1\over s^2 + 1} {1 \over 1+ \sqrt{3}} \left(\I(1-\sqrt 3) s+1-\sqrt{3} - 2\I s + 2 \sqrt{3} \right) \\ = {1\over s^2 + 1} \left(-\I s+1 \right) = {\I \over \I - s} \end{align*}