In [12]:
using Plots, ComplexPhasePortrait, ApproxFun, SingularIntegralEquations

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

Office Hours: 3-4pm Mondays, Huxley 6M40

Lecture 9: Cauchy transforms

  1. Cauchy transforms on the interval
    • Plemelj's theorem
  2. Hilbert transform

Consider a function $\phi(z)$ analytic in $\bar\C \backslash [-1,1]$ such that $\phi(\infty) = 0$. Surrounding it by a positively oriented simple closed contour $\Gamma$ (e.g., an ellipse) we have from Cauchy's exterior integral formula $$ \phi(z) = -{1 \over 2 \pi \I} \oint_\Gamma {\phi(\zeta) \over \zeta - z} \D\zeta $$ Assuming the limits $$ \phi^\pm(x) = \lim_{\epsilon \rightarrow 0} \phi(x \pm \I \epsilon) $$ are "nice", we can deform $\Gamma$ to be on the contour itself, giving $$ \phi(z) = {1 \over 2 \pi \I} \int_{-1}^1 {\phi^+(x) - \phi^-(x) \over x - z} \dx $$ This lecture studies the relationship between the subtractive jump $\phi^+(x) - \phi^-(x)$ and $\phi(z)$.

Cauchy transform on the interval

We can consistently construct a function with a prescribed subtractive jump using the Cauchy transform, which really is just Cauchy's integral formula but where we know nothing about $f(\zeta)$: we don't assume it is analytic.

Definition (Cauchy transform) $$ \CC_\gamma f(z) := {1 \over 2 \pi \I} \int_\gamma {f(\zeta) \over \zeta - z} \D\zeta $$

We focus on the case of an interval $[a,b]$: $$ \CC_{[a,b]} f(z) := {1 \over 2 \pi \I} \int_a^b {f(x) \over x - z} \dx $$ Here is a phase portrait of the Cauchy transform of a simple function:

In [13]:
x = Fun(-1 .. 1)
f = exp(x)*sqrt(1-x^2)
portrait(-3..3, -3..3, z -> cauchy(f,z))


Remark As an aside, these integrals are computationally difficult because of the singularity in the integrand, hence standard integration methods become slow as $z$ approaches the interval. There are other specialised routines (as implemented in cauchy(f,z)) that are much more efficient:

In [14]:
z = 0.1 +0.001im
@time cauchy(f, z )
@time sum(f/(x-z))/(2π*im)

  0.000018 seconds (13 allocations: 640 bytes)
  0.013608 seconds (287 allocations: 12.823 MiB)
0.5493442175548764 - 0.2181261671282511im

It turns out that the Cauchy transform has a very simple subtractive jump. Here we denote $$ \CC_{[a,b]}^+ f(x) = \lim_{\epsilon \rightarrow 0} \CC_{[a,b]} f( x+ \I \epsilon)\\ \CC_{[a,b]}^- f(x) = \lim_{\epsilon \rightarrow 0} \CC_{[a,b]} f( x- \I \epsilon) $$

Theorem (Plemelj on the interval I) Suppose $(b-x)^\alpha (x-a)^\beta f(x)$ is differentiable on $[a,b]$, for $\alpha, \beta < 1$. Then the Cauchy transform has the following properties:

  1. $\CC_{[a,b]} f(z)$ is analytic in $\bar \C \backslash [a,b]$
  2. $\CC_{[a,b]} f(\infty) = 0$
  3. It has the subtractive jump: $$ \CC_{[a,b]}^+ f(x) - \CC_{[a,b]}^- f(x) = f(x) \qqfor a < x < b $$
  4. $\CC_{[a,b]} f(z)$ has weaker than pole singularities at $a$ and $b$

Demonstration We can evaluate the Cauchy transform using cauchy, including the limit from above and below. Here we see numerically that we recover $f$ from taking the difference:

In [ ]:
cauchy(f, 0.1+0.0im)-cauchy(f, 0.1-0.0im) , f(0.1)

Sketch of Proof We show the proof for $[-1,1]$.

  1. From the dominated convergence theorem, we know that $\CC f(z)$ is complex-differentiable off $[-1,1]$: $$ {\D \over \D z} {\cal C} f(z) = {1 \over 2 \pi \I}\int_{-1}^1 {\D \over \D z} {f(x) \over x - z} \dx = {1 \over 2 \pi \I}\int_{-1}^1 {f(x) \over (x - z)^2} \dx $$ We know it is analytic at $\infty$ because $$ {\cal C} f(z^{-1}) = z {1 \over 2 \pi \I}\int_{-1}^1 {f(x) \over z x - 1} \dx $$ is differentiable at zero.
  2. ${\cal C} f(\infty) = 0$ follows from uniform convergence of $1 \over z - x$ to zero as $z \rightarrow \infty$.
  3. For the constant function, which is analytic, this follows by considering a contour $\gamma_x^+$ perturbed above $x$ and $\gamma_x^-$ perturbed below $x$, see plots below. Therefore, by Cauchy integral formula we have $$ \CC^+ 1 (x) - \CC^- 1 (x) = {1 \over 2 \pi \I}\int_{\gamma_x^+} {1 \over x - z} \dx - {1 \over 2 \pi \I}\int_{\gamma_x^-} {1 \over x - z} \dx = {1 \over 2 \pi \I} \oint {1 \over x -z } \dx = 1. $$ For other functions, we consider, for $z = x + \I \epsilon$, $$ {\cal C} f(z) = {1 \over 2 \pi \I} \int_{-1}^1 {f(t) - f(x) \over t - z} \dt + f(x) \CC 1(z) $$ For $\epsilon = 0$, the first integral exists because the singularity at $t = x$ is removable: $$ \lim_{t \rightarrow x} {f(t) - f(x) \over t - x} = f'(x) $$ We leave it as an excercise (or see [Trogdon & Olver 2015, Lemma 2.7]) to show that $\int_{-1}^1 {f(t) - f(x) \over t - z} \dt$ converges to $\int_{-1}^1 {f(t) - f(x) \over t - x} \dt$ as $z \rightarrow x$. It follows that $$ {\cal C}^\pm f(x) = {1 \over 2 \pi \I} \int_{-1}^1 {f(t) - f(x) \over t - x} \dt + f(x) \CC^\pm 1(x) $$ and in particular $$ {\cal C}^+ f(x) - {\cal C}^- f(x) = f(x) (\CC^+ 1(x) - \CC^- 1(x)) = f(x) $$
  4. We show that it has a weaker than pole singularity at $+1$, with $-1$ following by the same argument. First note that $f$ is absolutely integrable. If we assume we approach $1$ at an angle of $ -\pi + \delta \leq \theta \leq \pi - \delta$, the uniform convergence of $(z - 1) \CC f(z)$ to zero follows from observing that ${z-1 \over z - t}$ can be made arbitrarily small in a larger and larger interval. This is easiest to see for real $x > 1$, where for $1 \leq x \leq 1 + \epsilon^2$ we have $$ \left|{x -1 \over x-t }\right| \leq \epsilon $$ for all $t \leq 1 + \epsilon^2 - \epsilon$, or more generously, $t \leq 1 - \epsilon$. Therefore, $$ | (x-1) {\cal C} f(x) | \leq {1 \over 2 \pi}\int_{-1}^{1-\epsilon} |f(t) | \left|{x -1 \over x-t }\right| \dt + \int_{1-\epsilon}^1 |f(t) | \dt \leq \epsilon \int_{-1}^1 |f(t)| \dt + \int_{1-\epsilon}^1 |f(t) | \dt $$ Both terms tends to zero as $\epsilon \rightarrow 0$, hence so does $| (x-1) {\cal C} f(x) |$. To extend this to the interval itself (that is, $\delta = 0$), we use the stronger requirement that $(1-x)^\alpha(1+x)^\beta f(x)$ is differentiable. For $\alpha = \beta = 0$, this follows from the expression in condition (3) and the fact that (found via direct integration) $$ \CC 1(z) = {\log(z-1) - \log(z+1) \over 2 \pi \I} $$
    has only logarithmic singularities, and $f(x)$ is bounded.


Here is a plot of ${x - 1 \over x- t}$ showing that it is small on an increasing portion of the interval as $x \rightarrow 1$ from the right:

In [ ]:
x = 1 + 0.01
tt = linspace(-1.,1.,1000)
plot(tt, abs.((x - 1) ./ (x .- tt)); legend=false)

Here is a plot of $\gamma_x^\pm$:

In [ ]:
x = 0.4
r = 0.1
tt = linspace(π,0.,100)
plot([-1.; x + r*cos.(tt);1.0], [0.; r*sin.(tt); 0.0];ylims=(-0.5,0.5),label="g_x^+")
plot!([-1.; x + r*cos.(tt);1.0], [0.; -r*sin.(tt); 0.0];ylims=(-0.5,0.5),label="g_x^-")

We can use the results of the previous results to show that it is in fact unique.

Theorem (Plemelj on the interval II) Suppose $\phi(z)$ satsfies the following properties:

  1. $\phi(z)$ is analytic in $\bar \C \backslash [a,b]$
  2. $\phi(z)$ has weaker than pole singularities at $a$ and $b$
  3. $\phi(\infty) = 0$
  4. It has the subtractive jump: $$ \phi^+(x) - \phi^-(x) = f(x) \qqfor a < x < b $$ where $(b-x)^\alpha (x-a)^\beta f(x)$ is differentiable in $[a,b]$ for $\alpha,\beta < 1$.

Then $\phi(z) = {\cal C}_{[a,b]} f(z)$.

Sketch of Proof Consider $$A(z) = \phi(z) - {\cal C}_{[a,b]} f(z)$$ This is continuous (hence analytic) on $(a,b)$ as $$A^+(x) - A^-(x) = \phi^+(x) -\phi^-(x) - {\cal C}_{[a,b]}^+ f(x) + {\cal C}_{[a,b]}^- f(x) =f(x) - f(x) = 0$$ Also, $A$ has weaker than pole singularities at $a$ and $b$, hence is analytic there as well: it's entire. Only entire functions that are bounded are constant, since it vanishes at $\infty$ the constant must be zero.


Example 1 We can use this theorem to prove the following relationships (using $\diamond$ for the dummy variable):

$$ {1 \over \sqrt{z-1} \sqrt{z+1}} = -2 \I {\cal C}\left[{1 \over \sqrt{1-\diamond^2}}\right](z) = -{1 \over \pi}\int_{-1}^1 {\dx \over \sqrt{1-x^2} (x-z)} $$

(1) follows because the jumps cancel. (2 and 3) are immediate. (4) follows from a simple calculation. Here we show that it has the correct jump:

In [ ]:
x = Fun()
z = 2 +2im

Example 2 Now consider a problem of reducing $$ \phi(z) = \sqrt{z-1} \sqrt{z+1} $$ to its behaviour near its singularities. It has two singularities: it blows up at $\infty$ and has a branch cut on $[-1,1]$

We can subtract out the singularity at infinity first to determine

$$\phi(z) = z + 2 \I {\cal C}[\sqrt{1-\diamond^2}](z)$$

Note this works because, as $z \rightarrow \infty$, we have $$ \phi(z) = z (\sqrt{1-{1/z}}\sqrt{1 + {1/z}}) = z (1 + O({1/z}))(1+O(1/z)) = z + O(1/z) $$ hence $\phi(z) - z$ vanishes at the origin. This is an example of summing over the behaviour at each singularity to recover the function (in this case, $\phi$ has a singularity along the cut $[-1,1]$ and polynomial growth at $\infty$).

Because $\phi(z)-z$ decays, we can now deploy Plemelj II to determine: $$ \phi(z) -z = \CC[\phi_+-\phi_-](z) $$ where $$ \phi_+(x) - \phi_-(x) = 2\I \sqrt{1-x^2} $$ Here we see that our derived expression matches $phi(z)$:

In [ ]:
sqrt(z-1)sqrt(z+1), z +2im*cauchy(sqrt(1-x^2),z)

Example 3 Finally, we have the following (also verifiable using indefinite integration):

$${\log(z-1) - \log(z+1) \over 2 \pi \I} = {\cal C}[1](z) = {1 \over 2 \pi \I} \int_{-1}^1 {\dx \over x -z} $$

In [ ]: