M3M6: Methods of Mathematical Physics

$$ \def\dashint{{\int\!\!\!\!\!\!-\,}} \def\infdashint{\dashint_{\!\!\!-\infty}^{\,\infty}} \def\D{\,{\rm d}} \def\dx{\D x} \def\dt{\D t} \def\C{{\mathbb C}} \def\CC{{\cal C}} \def\HH{{\cal H}} \def\I{{\rm i}} \def\qqfor{\qquad\hbox{for}\qquad} $$

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

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

Lecture 11: Riemann–Hilbert problems

1. Constant coefficient Riemann–Hilbert problems
   • Subtractive Riemann–Hilbert problem $\phi_+(x) - \phi_-(x) = f(x)$
   • Additive Riemann–Hilbert problem $\phi_+(x) + \phi_-(x) = f(x)$

Let $\Gamma$ be a set of contours, and for now assume $\Gamma \subset {\mathbb R}$. Given functions $f$ and $g$ defined on $\Gamma$, a (scalar) Riemann–Hilbert problem consists of finding a function $\phi(z)$ with left/right limits $$\phi_\pm(x) = \lim_{\epsilon \rightarrow 0} \phi(x \pm \I \epsilon),$$ satisfying the following conditions:

1. $\phi(z)$ analytic in $\bar\C \backslash \Gamma$
2. $\phi(\infty) = C$
3. $\phi(z)$ has weaker than pole singularities everywhere
4. $\phi_+(x) - g(x) \phi_-(x) = f(x)$ for $x \in \Gamma$

Numerous applications! See [Trogdon & Olver 2015]. Here are some classical applications:

1. Ideal fluid flow
2. Solving integral equations via Weiner–Hopf factorization
3. Spectral analysis of Schrödinger operators

More recently, non-classical applications have arisen from integrable systems:

1. Solutions to Painlevé equations
2. Random matrix eigenvalue statistics
3. Solving partial differential equations like the Korteweg–de Vries (KdV) equation describing shallow water waves $$ u_t + 6u u_x + u_{xxx} = 0 $$

Constant coefficient Riemann–Hilbert problems

Consider the Riemann–Hilbert problem $$ \phi+(x) - c \phi_-(x) = f(x) $$ on the interval. General $c$ will be looked at in the problem sheets, so we only consider $c \pm 1$.

Subtractive Riemann–Hilbert problems

The simplest example is when $c = 1$, in which case, assuming $f(x)$ satisfies the conditions of Plemelj theorem, the solution to

1. $\phi_+(x) - \phi_-(x) = f(x)$ for $x \in \Gamma$
2. $\phi(\infty) = C$
3. $\phi(z)$ has weaker than pole singularities

is simply

$$ \phi(z) = \CC_\Gamma f(z) + C $$

Analytic functions are continuous, hence if $\Gamma$ is in the domain of analyticity, we have $\phi_+(x) - \phi_-(x) = 0$. This is why the constant $C$ does not impact the jump. This makes subtractive Riemann–Hilbert problems particularly nice as we can solve each component separately: that is $$ \CC_{\Gamma_1 \cup \Gamma_2} f(z) = \CC_{\Gamma_1} f(z) + \CC_{\Gamma_2} f(z) $$ We can also fix up analyticity:

Example Solve the Riemann–Hilbert problem

1. $\phi(z)$ analytic in $\bar\C \backslash [-1,1]$
2. $\phi(\infty) = -{\I \over 2}$
3. $\phi(z)$ has weaker than pole singularities everywhere
4. $\phi_+(x) - \phi_-(x) = {\sqrt{1-x^2} \over x +2}$ for $-1 < x < 1$

Consider the first guess $$ \psi(z) = - \I {\sqrt{z-1} \sqrt{z+1} \over 2(z + 2)} $$ This satisfies 2, 3, and 4: it has the right jump on $-1 < x < 1$: $$ \psi_+(x) - \psi_-(x) = {\sqrt{1-x^2} \over x+2} $$

The catch: it has a pole at $z = - 2$ 😩:

In particular, $$ \psi(z) = {\I \sqrt 3 \over 2} {1 \over z + 2} + O(1) $$ near $z = -2$. Note that this takes a bit of care to work out since we are on the cancelling out branch cuts of $\psi(z)$. But we do know away from $z = -2$ it is analytic, hence continuous, that is, $$ \psi(x) = \lim_{\epsilon \rightarrow 0} \psi(x \pm \I \epsilon) = \psi_\pm(x) $$ for real valued $x < -1$. In particular, we have $$ \psi(x) = \psi_+(x) = - \I { \I \sqrt{1-x} \I \sqrt{-x-1} \over 2( x+2)} = \I {\sqrt{x^2 - 1} \over 2( x+2)} $$

Good thing it's a subtractive Riemann–Hilbert problem: we can subtract out the pole without impacting the jump. Therefore, we have $$ \phi(z) = \psi(z) - {\I \sqrt 3 \over 2} {1 \over z + 2}=- \I {\sqrt{z-1} \sqrt{z+1} \over 2(z + 2)} - {\I \sqrt 3 \over 2} {1 \over z + 2} $$ This is analytic off $[-2,2]$ (including at $z = -2$) and has the right jump:

Additive Riemann–Hilbert problem

We now introduce the case $c = -1$, in other words, we want to solve $$ \phi_+(x) + \phi_-(x) = f(x) \qquad \phi(\infty) = 0 $$ and take $\Gamma = [-1,1]$. This is equivalent to finding an unknown function $u(x)$ such that $$ -\I \HH u(x) = f(x) $$

Reduction to Cauchy transform

Consider $\phi(z) = {\psi(z) \over \sqrt{z-1}\sqrt{z+1} }$, so that $$ f(x) = \phi_+(x) + \phi_-(x) = { \psi_+(x) \over \I \sqrt{1-x^2}} - {\psi_-(x) \over \I \sqrt{1-x^2}} = {\psi_+(x) - \psi_-(x) \over \I \sqrt{1-x^2}} $$ Thus we have reduced an additive Riemann–Hilbert problem to a subtractive one: if $$\psi_+(x) - \psi_-(x) = {f(x) \I \sqrt{1-x^2}},$$ we satisfy the correct jump. Therefore, we have the solution $$ \phi(z) = {\I \over \sqrt{z-1}\sqrt{z+1}} \CC_{[-1,1]} \left[{ f \sqrt{1-\diamond^2}}\right](z) $$

Is this the only solution? No! consider $$ \kappa(z) = {1 \over \sqrt{z-1}\sqrt{z+1}} $$ which satisfies $\kappa_+(x) + \kappa_-(x) = 0$.

General solution on $[-1,1]$ built from a Cauchy transform

Here we propose the general solution to an additive Riemann–Hilbert problem:

Theorem (Cauchy transform solution to Additive Riemann–Hilbert problem) Suppose $\phi(z)$ analytic in $\C \backslash [-1,1]$ satisfies

1. $\phi_+(x) + \phi_-(x) = f(x) $ for $-1 < x < 1$, where $f$ is smooth on $[-1,1]$.
2. $\phi(z)$ has weaker than pole singularities,
3. $\phi(\infty) = 0$.

Then, for some constant $C$, $$ \phi(z) = {\I \over \sqrt{z-1}\sqrt{z+1}} \CC_{[-1,1]} \left[{ f(\diamond) \sqrt{1-\diamond^2} }\right](z) + {C \over \sqrt{z-1} \sqrt{z+1}} $$

On other intervals $(a,b)$, we get the same forms of solution:

$$ \phi(z) = {\I \over \sqrt{z-b}\sqrt{z-a}} \CC_{[a,b]} \left[{ f(\diamond) \sqrt{b-\diamond}\sqrt{\diamond-a} }\right](z) + {C \over \sqrt{z-b} \sqrt{z-a}} $$

General solution on $[-1,1]$ expressed as a Cauchy transform

Note the expression above is not a Cauchy transform of a function. We can determine a Cauchy transform expression by taking the difference: on $[-1,1]$ we have

\begin{align*} u(x) &= \phi_+(x) - \phi_-(x) = {1 \over \sqrt{1 - x^2}}\left( \CC_{[-1,1]}^+ + \CC_{[-1,1]}^- \right) \left[{ f(\diamond) \sqrt{1-\diamond^2} }\right](x) - {2C \I \over \sqrt{1-x^2}} \\ &= {-\I \over \sqrt{1 - x^2}}\HH_{[-1,1]} \left[{ f(\diamond) \sqrt{1-\diamond^2} }\right](x) - {2C \I \over \sqrt{1-x^2}} \end{align*}

BY (Plemelj II), we guarantee that $$ \phi(z) = \CC_{[-1,1]} u(z) $$

On other intervals we have \begin{align*} u(x) &= {-\I \over \sqrt{b - x}\sqrt{x-a}}\HH_{[a,b]} \left[{ f(\diamond) \sqrt{b-\diamond}\sqrt{\diamond-a} }\right](x) - {2C \I \over \sqrt{b-x} \sqrt{x-a}} \end{align*}

In the special case where we can calculate $\CC \left[ f(\diamond) \sqrt{b-\diamond}\sqrt{\diamond-a}\right](z)$ exactly, we can work out the precise formula for $u(x)$ by taking the difference.

Example Consider solving

1. $\phi_+(x) + \phi_-(x) = x $ for $-b < x < b$
2. $\phi(z)$ has weaker than pole singularities,
3. $\phi(\infty) = 0$.

Note that $$ \kappa(z) = {z \sqrt{z-b} \sqrt{z+b} - z^2 + b^2/2 \over 2 \I} $$ satisfies $$ \kappa^+(x) - \kappa^-(x) = x \sqrt{b^2 - x^2} $$ and $\kappa(\infty) = 0$. The second property follows since $\kappa$ is analytic off $[-1,1]$ and has at most polynomial growth at $\infty$. To understand the order of the polynomial growth, it's sufficient to study the limit on the real axis as $x \rightarrow \infty$, where we use the Taylor series of $\sqrt x$ near $1$ to determine: \begin{align*} x \sqrt{x-b}\sqrt{x + b} &= x^2 \sqrt{1-b/x} \sqrt{1 + b/x} = x^2 \left(1 - {b\over 2x} - {b^2 \over 8x^2}\right) \left(1 + {b\over 2x} - {b^2 \over 8x^2}\right)\\ = x^2 - {b^2\over 2} + O(x^{-2}) \end{align*}

Hence every solution has the form \begin{align*} \phi(z) = {1 \over \sqrt{z-b} \sqrt{z+b}} \left({z \sqrt{z-b} \sqrt{z+b} - z^2 + b^2/2 \over 2 } \right) + {C \over \sqrt{z-b} \sqrt{z+b}} \\ = {z \over 2 } - {z^2 \over 2 \sqrt{z-b} \sqrt{z+b}} + {D \over \sqrt{z-b} \sqrt{z+b}} \end{align*} where $D$ is again a arbitrary constant.

Inspection reveals that $$ u(x) = \phi_+(x) - \phi_-(x) = \I {x^2 - 2D \over \sqrt{b^2-x^2}} $$

Therefore, $\CC u(z)$ satsifies $\CC^+u(x) + \CC^- u(x) = x$:

$D$ is a free paremeter, hence, similar to ODEs, we can add a boundary condition. In other words we can ask for the solution satisfying, for example $u(0) = 0$. This is precisely when $D = 0$ and $u(x) = \I x^2 / \sqrt{b^2-x^2}$.

Often we ask for the solution that is bounded at the left/right endpoint. Because of the symmetry in the problem (since $x$ is an odd function), we can actually get a solution that is bounded at both $\pm b$ by choosing $D = b^2/2$: $$ \phi_+(x) - \phi_-(x) = -\I \sqrt{b^2 - x^2} $$

In other words, $$ \phi(z) = - \I \CC_{[-b,b]} \sqrt{b^2 - \diamond^2}(z) $$ is a bounded solution to the additive Riemann–Hilbert problem.

Inverting the Hilbert transform

We can use this along with Plemelj's lemma to invert the Hilbert transform. That is, we want to find $v(x)$ such that $$ \HH v(x) = f(x) $$ Based on the above construction of $u(x)$, consider $$ v(x) =-\I u(x) = {-1 \over \sqrt{1 - x^2}}\HH_{[-1,1]} \left[{ f(\diamond) \sqrt{1-\diamond^2} }\right](x) - {2C \over \sqrt{1-x^2}} $$

This construction works since $\phi(z) = \CC u(z) = \I \CC v(z)$, and $$ \HH v(x) = \I (\CC^+ + \CC^-) v(x) = \phi^+(x) + \phi^-(x) = f(x) $$

Example

Consider $\HH_{[-b,b]} v(x) = x$. We have $$ -\I \HH_{[-b,b]} \sqrt{b^2 - \diamond^2}(x) = \CC_{[-b,b]}^+ \sqrt{b^2 - \diamond^2}(z) + \CC_{[-b,b]}^- \sqrt{b^2 - \diamond^2}(z) = \I (\phi^+(x) + \phi^-(x) ) = \I x $$ which shows that the solution is $$ v(x) = - \sqrt{b^2 - \diamond^2} $$