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

Office Hours: 3-4pm Mondays, Huxley 6M40
Website: https://github.com/dlfivefifty/M3M6LectureNotes

Lecture 19: Logarithmic singular integrals

1. Evaluating logarithmic singular integrals
2. Logarithmic singular integrals and orthogonal polynomials
3. Solving a logarithmic singular integral equation
4. Application: electrostatic potentials in 2D
   • Potential arising from a point charge and a single plate

The motivation behind this lecture is to calculate electrostatic potentials. An example is the Faraday cage: imagine a series of metal plates connected together so that they have the same charge. If configured to surround a region, this configuration will shield the interior from an external charge:

Here, the coloured lines are equipotential lines, and there is a point source at $x = 2$, which corresponds to a forcing of $$\log\| (x,y) - (2,0) \| = \log|z - 2|$$ where $z = x + \I y$.

Logarithmic singular integrals

From the Green's function of the Laplacian, it is natural to consider logarithmic singular integrals, and we focus yet again on $[-1,1]$:

$$v(z) = {1 \over \pi }\int_{-1}^1 f(t) \log | z - t| \dt$$

Note that off $[-1,1]$, $v(z)$ solves Laplace's equation, and is continuous on $[-1,1]$:

Continuity follows since $\log|x-t|$ is integrable for $-1 \leq x \leq 1$ provided $f(t)$ has weaker than pole singularities.

For $z \notin (-\infty,1]$ this can also be seen since $v$ is the real part of an analytic function: $$ v(z) = \Re {1 \over \pi} \int_{-1}^1 f(t) \log ( z - t) \dt $$ Note that the integrand avoids the branch cut of $\log z$. To extend this to $z\in (-\infty,-1]$ (or more generally, $z \notin [-1,\infty)$), we can use the alternative expression $$ v(z) = \Re {1 \over \pi} \int_{-1}^1 f(t) \log (t-z) \dt $$ which follows from $\log|z-t| = \log|t-z|$.

Evaluating logarithmic singular integrals

How can we actually evaluate $$ Lu(z) = {1 \over \pi} \int_{-1}^1 u(t) \log (z-t) \dt $$ for $z$ in the complex plane? We relate this to the indefinite integral of the Cauchy transform, and then use that to reduce it to a Cauchy transform itself.

Integrated Cauchy transform

For $z$ away from the branch cut our integrand is "nice" (assuming $u$ itself is "nice") and we can exchange integration and differentiation to determine $$ {\D \over \D z} Lu(z) = {1 \over \pi} \int_{-1}^1 u(t) {1\over z -t} \dt = -2 \I \CC u(z) $$ thus if we can calculate $\int^z Cu(\zeta) \D \zeta$, we are in good shape.

Example If $u(x) = {1 \over \sqrt{1-x^2}}$, then recall $$ Cu(z) = {-1 \over 2 \I \sqrt{z-1} \sqrt{z+1}} $$ We therefore have $$ {\D \over \D z} Lu(z) = {1 \over \sqrt{z-1} \sqrt{z+1}} $$ and therefore for some constant of integration $D$ we have $$ Lu(z) + D = \pi \int^z {\D z \over \sqrt{z-1} \sqrt{z+1}} = 2 \pi \log\left(\sqrt{z-1} + \sqrt{z+1}\right) $$ We need the right constant of integration. we can find that via the behaviour as $x \rightarrow \infty$: $$ Lu(x) ={1 \over \pi} \int_{-1}^1 u(t) \log(x-t) \dt = {\log x \over \pi} \int_{-1}^1 u(t) \dt + {1 \over \pi} \int_{-1}^1 u(t) \log(1-t/x) \dt = {\log x \over \pi}\int_{-1}^1 u(t) \dt + O(x^{-1}) $$ Since we have $$ \int_{-1}^1 {1 \over \sqrt{1-x^2}} \dx = \pi $$ and $$ 2 \log(\sqrt{x-1} + \sqrt{x+1}) = 2 \log \sqrt x + 2 \log(\sqrt{1-1/x} + \sqrt{1 + 1/x}) = \pi \log x + 2\pi \log(2) + O(1/x) $$ therefore, $$ Lu(z) = 2 \log\left(\sqrt{z-1} + \sqrt{z+1}\right) - 2 \log 2 $$

Demonstration We want to calculate the log kernel of $1/\sqrt{1-x^2}$:

We saw that its derivative is given in terms of the Cauchy transform, here we test it using finite difference approximation:

But we already know the Cauchy transform in closed form:

This we can integrate in closed form to determine:

Reduction to a Cauchy transform

The representation is not ideal as it involves indefinite integration in the complex plane. However, we will see that we can actually express $L u (z)$ as a Cauchy transform of $\int_x^1 u(t) \dt$.

Lemma (Log kernel as Cauchy transform) Suppose $u$ is "nice" (as in Plemelj). Then $$ Lu(z) = {\log(z-a) \over \pi} \int_a^b u(x) \dx +2 \I C U(z) $$ where $$ U(x) = \int_x^b u(t) \dt $$

Proof

We use Plemelj to prove this result (of course!) and assume $[a,b] = [-1,1]$. Note that $Lu$ satisfies the following Riemann–Hilbert problem:

1. $Lu(z) \sim {1 \over \pi} \int_{-1}^1 u(t) \dt \log z + o(1)$
2. $L^+ u(x) - L^- u(x) = 2 \I \int_x^1 u(t) \dt $ for $-1 < x < 1$
3. $L^+ u(x) - L^- u(x) = 2\I \int_{-1}^1 u(t) \dt $ for $x < -1$

The first part follows immediately by noting for $z \rightarrow + \infty$ we have $$\log(z - x) = \log(z (1 - x/z)) = \log z + \log(1-x/z) = \log z + o(1)$$

We now see where the jumps come from. Since $Lu'(z) = -2 \I Cu(z)$ and This let's us think of $Lu (z)$ as (2 is arbitrary here) $$ L u(z) = Lu(2) + \int_2^z C u (\zeta) \D\zeta $$ where the contour is a straight line (just like in the problem sheet for $\log z$ that I'm sure everyone has done by now), but we can think of it also as an arc that avoids $[-1,1]$.

Consider $x < -1$. Given an ellipse $\gamma$ surrounding $[-1,1]$ we can write the integral representations and exchange orders to get: $$ L u^+(x) - L u^-(x) = -2 \I \oint_\gamma C u(\zeta) \D\zeta = -2 \I \int_{-1}^1 u(t) {1\over 2 \pi \I} \oint_\gamma {1 \over t-z} \dt \D \zeta = 2 \I \int_{-1}^1 u(t) \dt $$

For $-1 < x < 1$, note that $ L u(1) $ is well-defined as there is only a logarithmic singularity times an algebraic one. Thus deforming the contour onto the interval we have $$ L^\pm u(x) = -2 \I \int_1^x C^\pm u(t) \dt + L u(1), $$ hence $$ L^+ u(x) -L^- u(x) = 2 \I \int_x^1 (C^+ u(t) - C^- u(t)) \dt =2 \I \int_x^1 u(t) \dt $$ ⬛️

Example For the problem above, we have that $\int_x^1 {1 \over \sqrt{1-t^2}}\dt = {\rm arccos}\, x$, which gives us:

Log transforms of weighted orthogonal polynomials

Now consider ${1 \over \pi} \int_a^b p_k(x) w(x) \log |z-x| \dx$, which we write in terms of the real part of $$ L_k(z) = L[p_k w](z) = {1 \over \pi } \int_a^b p_k(x) w(x) \log (z-x) \dx $$

For $k > 0$ we have $\int_a^b p_k(x) w(x) dx = 0$ due to orthogonality, and hence we actually have no branch cut:

Example: Weighted Chebyshev log transform

For classical orthogonal polynomials we can go a step further and relate the indefinite integrals to other orthogonal polynomials.

For example, recall that $$ {\D\over \dx}[\sqrt{1-x^2} U_n(x)] = -{n+1 \over \sqrt{1-x^2}} T_{n+1}(x) $$ in other words, $$ \int_x^1 {T_k(t) \over \sqrt{1-t^2}} \dt = -{\sqrt{1-x^2} U_{k-1}(x) \over k} $$ Thus for $k=1,2,\ldots$, $$ L_k(z) = -{1 \over n+1} \CC[\sqrt{1-\diamond^2} U_{k-1}](z) $$ and $$ {1 \over \pi} \int_{-1}^1 {T_k(x) \over \sqrt{1-x^2}}\log(z-x) \dx = {2 \I \over k} \CC[\sqrt{1-\diamond^2} U_{k-1}](z) $$

As we saw last lecture, Cauchy transforms of OPs satisfy simple recurrences, and this relationship renders log transforms equally calculable.

Solving logarithmic singular integral equations

Oddly enough, it is easier to solve a singular integral equation involving the logarithmic kernel than to actually calculate the logarithmic singular integral, so we begin here. Consider the problem of calculating $u(x)$ such that $$ {1 \over \pi} \int_{-1}^1 u(t) \log | x-t| \dt = f(x) \qqfor -1 < x < 1 $$

To tackle this problem we first need to understand the additive jump $Re L^\pm(x) = {L^+(x) + L^-(x) \over 2}$. Thinking in terms of integrals of Cauchy transforms we see that $$ L^+u(x) + L^-u(x) = -2\I (2\int_2^1 \CC u(x) \dx + \int_1^x [C^+ u(t) + C^-u(t)] \dt) = D -2 \I \int_1^x (-\I H u(t)) \dt = D + 2\int_x^1 H u(t) dt $$ That is, it is expressed as an indefinite integral of the Hilbert transform. We therefore have: $$ {\D \over \dx} {1 \over \pi} \int_{-1}^1 u(t) \log | x-t| \dt = {\D \over \dx} \int_x^1 H u(t) dt = - Hu(x) $$ where the minus sign comes from the fact that $x$ is the lower limit. In other words, our original SIE becomes equivalent to inverting the Hilbert transform: $$ \HH u(x) = -f'(x) $$ recall, we can express the solution as $$ u = {1 \over \sqrt{1-x^2}} \HH[\sqrt{1-\diamond^2} f'] + {C\over \sqrt{1-x^2}} $$

Let's do a numerical example:

Remember: for inverting the Hilbert transform we had a degree of freedom: every solution plus $C/\sqrt{1-x^2}$ was also a solution. But here, since we differentiated, we use that degree of freedom to ensure that we have arrived at the right solution.

To choose $C$, we use the fact that $$ f(0) = {1 \over \pi} \int_{-1}^1 u(t) \log |t| \dt $$ If we can calculate the right integral, we can use this to choose $C$:

Example We now do an example which can be solved by hand. Find $u(x)$ so that: $$ \int_{-1}^1 u(t) \log | x-t| \dt = 1. $$ Differentiating, we know that $$ \int_{-1}^1 {u(t) \over x-t} \dt = 0 $$ hence $u(x)$ must be of the form ${C \over \sqrt{1-x^2}}$. Which $C$? Make it work for $x = 0$: $$ 1 = \int_{-1}^1 u(t) \log | 0 - t| \dt = C \int_{-1}^1 {\log | t| \over \sqrt{1-t^2}} \dt $$ To evaluate the integral, we can use trigonmetric variables: $$ \int_0^1 {\log t \over \sqrt{1-t^2}} \dt = \int_0^{\pi \over 2} {\cos \theta \log \sin \theta \over \sqrt{1-\sin^2 \theta}} \D\theta = \int_0^{\pi \over 2} \log \sin \theta \D\theta = - {\pi \log 2\over 2} $$ (The last identity takes some work: I'll leave it as an excercise.)

Thus we have $ C = -{1 \over \pi \log 2}$

Physically, this solution gives us the potential field $$ \int_{-1}^1 u(t) \log|z-t| \dt $$ corresponding to holding a metal plate at constant potential:

Application: Potential arising from a point charge and a single plate

Now imagine we put a point source at $x = 2$, and a metal plate on $[-1,1]$. We know the potential on the plate must be constant, but we don't know what constant. This is equivalent to the following problem (see e.g. [Chapman, Hewett & Trefethen 2015]): \begin{align*} v_{xx} + v_{yy} = 0 &\qqfor \hbox{off $[-1,1]$ and $2$} \\ v(z) \sim \log |z - 2| + O(1) &\qqfor z \rightarrow 2 \\ v(z) \sim \log|z| + o(1) &\qqfor z \rightarrow \infty \\ v(x) = \kappa &\qqfor -1 < x < 1 \end{align*} where $\kappa$ is an unknown constant. We write the solution as $$ v(z) = {1 \over \pi} \int_{-1}^1 u(t) \log|t-z| \dt + \log|z-2| $$ for a to-be-determined $u$. On $-1 < x < 1$ this satisfies $$ {1 \over \pi} \int_{-1}^1 u(t) \log|t-x| \dt = \kappa - \log(2-x) $$

We can calculate the solution numerically as follows:

We can solve this equation explicitly. By differentiating we see that $u$ satisfies the following: $$ Hu(x) = -f'(x) = {1 \over x-2} $$ We now use the inverse Hilbert formula to determine: $$ u(x) = -{1\over\sqrt{1-x^2}} H[{1 \over \diamond-2} \sqrt{1-\diamond^2}](x) + {D \over \sqrt{1-x^2}} $$ Using the usual procedure of taking an obvious ansatz and subtracting off the singularities at poles and $\infty$ we find that: $$ C[{\sqrt{1-\diamond^2} \over x -2}](z) = \underbrace{\sqrt{z-1} \sqrt{z+1} \over 2 \I (z-2)}_{\hbox{ansatz}} - \underbrace{\sqrt 3\over 2 \I (z-2)}_{\hbox{remove pole near $z=2$}} - \underbrace{{1 \over 2 \I}}_{\hbox{remove constant at $\infty$}} $$ This implies that $$ H[{1 \over\diamond-2} \sqrt{1-\diamond^2}](x) = \I(C^+ + C^-)[{1 \over\diamond-2} \sqrt{1-\diamond^2}](x) = -{\sqrt 3 \over x -2} - 1 $$ in other words, $$ u = {\sqrt{3} \over (x - 2) \sqrt{1-x^2}} + {D \over \sqrt{1-x^2}} $$ We still need to determine $D$. This is chosen so that we tend to $\log |z|$ near $\infty$. In particular, we know that $$ {1 \over \pi} \int_{-1}^1 \log|z-x| u(x) \dx \sim {\int_{-1}^1 u(x) \dx \over \pi} \log |z| $$ so we need to choose $D$ so that $\int_{-1}^1 u(x) \dx = 0$ and only the $\log|z-2|$ singularity appears near $\infty$. We first note that \begin{align*} C[{1 \over (\diamond - 2) \sqrt{1-\diamond^2}}](z) &= \underbrace{{\I \over 2 \sqrt{z-1} \sqrt{z+1} (z-2)}}_{\hbox{ansatz}} - \underbrace{{\I \over 2\sqrt{3} (z-2)}}_{\hbox{remove pole}} \\ & = -{\I \over 2 \sqrt{3} z} + O(z^{-2}) \end{align*} near $\infty$. Since $$ Cw(z) ={ \int_{-1}^1 w(x) \dx \over -2\pi \I z} + O(z^{-2}) $$ we can infer the integral from the Cauchy transform and find that $$ \int_{-1}^1 {1 \over (x - 2) \sqrt{1-x^2}} \dx = -{ \pi \over \sqrt{3}} $$ On the other hand, $$ \int_{-1}^1 {\dx \over \sqrt{1-x^2}} = \pi $$ Thus we require $D = 1$ and have the solution: $$ u(x) = {\sqrt{3} \over (x - 2) \sqrt{1-x^2}} - {1 \over \sqrt{1-x^2}} $$

Demonstration We first check that differentiation reduces to the Hilbert transform:

In the inverse Hilbert transform we calculated the following Cauchy transform:

Which means that:

Thus we have the inverse Hilbert transform:

We still need to determine the constant $D$. What goes wrong if we have the wrong choice is that we don't tend to precisely $\log z$ near $\infty$:

We determined the integral

Therefore:

We thus found $u$: