M3M6: Methods of Mathematical Physics 2018

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

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

Solution Sheet 2

Problem 1.1

1.

Take as an initial guess $$ \phi_1(z) = {\sqrt{z-1} \sqrt{z+1} \over 2\I(1 + z^2)} $$ This satisfies for $-1 < x < 1$ $$ \phi_1^+(x) -\phi_1^-(x) = { \sqrt{1-x^2} \over 2(1 + x^2)} - {- \sqrt{1-x^2} \over 2(1 + x^2)} = {\sqrt{1 - x^2} \over 1+x^2} $$

Further, as $z \rightarrow \infty$, $$ \phi_1(z) \sim {z \over \I(1+ z^2)} \rightarrow 0 $$

The catch is that it has poles at $\pm \I$: \begin{align*} \phi_1(z) = -{\sqrt{\I -1} \sqrt{\I+1} \over 4} { 1 \over z - \I} + O(1) \\ \phi_1(z) = {\sqrt{-\I -1} \sqrt{-\I+1} \over 4} { 1 \over z + \I} + O(1) \end{align*} Thus it follows that $$ \phi(z) = \phi_1(z) + {\sqrt{\I -1} \sqrt{\I+1} \over 4} { 1 \over z - \I} - {\sqrt{-\I -1} \sqrt{-\I+1} \over 4} { 1 \over z + \I} $$ is

1. Analytic at $\pm \I$ and off $[-1,1]$ (Analyticity)
2. $\phi(\infty) = 0$ (Decay)
3. Has weaker than pole singularities (Regularity)
4. Satisfies $$ \phi_+(x) - \phi_-(x) = {\sqrt{1-x^2} \over 1+x^2} $$

By Plemelj II, this must be the Cauchy transform.

Demonstration We will see experimentally that it correct. First we do a phase plot to make sure we satisfy (Analyticity):

We can also see from the phase plot (Regularity): we have weaker than pole singularities, otherwise we would have at least a full,counter clockwise colour wheel. We can check decay as well:

Finally, we compare it numerically it to cauchy(f, z) which is implemented in SingularIntegralEquations.jl:

2.

Recall that $$ \psi(z) = {\log(z-1) - \log(z+1) \over 2 \pi \I} $$ satisfies $$ \psi_+(x) - \psi_-(x) = 1 $$ Therefore, consider $$ \phi_1(z) = {\psi(z) \over 2 + z} $$ This has the right jump, but has an extra pole at $z = -2$: for $x < -1$ we have $$ \phi_1(x) = {\log_+(x-1) - \log_+(x+1) \over 2 \pi \I} {1 \over 2 + x} = {\log(1-x) - \log(-1-x) \over 2 \pi \I} {1 \over 2 + x} $$ hence we arrive at the solution $$ \phi_1(z) - {\log 3 \over 2 \pi \I (2+z)} $$ We can verify that $\phi_1(\infty) = 0$.

4.

We first calculate the Cauchy transform of $f(x) = x/\sqrt{1-x^2}$: $$ \phi(z) = {\I z \over 2 \sqrt{z -1} \sqrt{z+1}} - {\I \over 2} $$ This vanishes at $\infty$ and has the correct jump. We then have $$ -\I{\cal H}f(x) = \phi^+(x) + \phi^-(x) = -\I $$ This implies that $$ \dashint_{-1}^1 {t \over (t-x) \sqrt{1-t^2}} \dt = \pi \HH f(x) = \pi $$

Problem 1.2

1.2.1

From Problem 1.1 part 4, we have a solution: $$ \phi(z) = -{ z \over 2 \sqrt{z -1} \sqrt{z+1}} + {1 \over 2} $$ All other solutions are then of the form: $$ \phi(z) + {C \over \sqrt{z-1} \sqrt{z+1}} $$

1.2.2

$\psi(z) = \phi(z) - 1$ satifies $$ \psi_+(x) + \psi_-(x) = -2, \psi(\infty) = 0 $$ hence we know that $\psi(z) = { z \over \sqrt{z -1} \sqrt{z+1}} - 1 + {C \over \sqrt{z-1} \sqrt{z+1}}$, giving $$ \psi(z) = { z \over \sqrt{z -1} \sqrt{z+1}} + {C \over \sqrt{z-1} \sqrt{z+1}} + 1 $$

1.2.3

For $f(x) = \sqrt{1-\diamond^2}$, we use the formula $$ \phi(z) = {\I \over \sqrt{z - 1} \sqrt{z+1}} \CC[\sqrt{1-\diamond^2} f](z) + {C \over \sqrt{z-1}\sqrt{z+1}} = {\I \over \sqrt{z - 1} \sqrt{z+1}} \CC[1-\diamond^2](z) + {C \over \sqrt{z-1}\sqrt{z+1}} $$ We already know $\CC1(z)$, and we can deduce $\CC[\diamond^2]$ as follows: try $$ \phi_1(z) = z^2 \CC1(z) = z^2 {\log(z-1) - \log(z+1) \over 2 \pi \I} $$ this has the right jump, but blows up at $\infty$ like: $$ x^2 (\log(x-1) - \log(x+1)) = x^2 (\log(1-1/x) - \log(1+1/x) ) = -2 x + O(x^{-1}) $$ using $$ \log z = (z-1) - {1\over 2}(z-1)^2 + O(z-1)^3 $$ Thus we have $$ \CC[\diamond^2](z) = {z^2 (\log(z-1) - \log(z+1)) + 2 z \over 2 \pi \I} $$ and $$ \phi(z) = {\I \over \sqrt{z - 1} \sqrt{z+1}} {(1-z^2)(\log(z-1) - \log(z+1)) - 2 z \over 2 \pi \I} + {C \over \sqrt{z-1}\sqrt{z+1}} $$

Demonstration Here we see that the Cauchy transform of $x^2$ has the correct formula:

We now see that $\phi$ has the right jumps:

Finally, it vanishes at infinity:

1.2.4

Let $f(x) = {1 \over 1+x^2}$. From Problem 1.1 part 1 we know $$ \CC[\sqrt{1-\diamond^2}] f(z) = {\sqrt{z-1} \sqrt{z+1} \over 2\I(1 + z^2)} + {\sqrt{\I -1} \sqrt{\I+1} \over 4} { 1 \over z - \I} - {\sqrt{-\I -1} \sqrt{-\I+1} \over 4} { 1 \over z + \I} $$ hence from the solution formula we have $$ \phi(z) = {1 \over 2(1 + z^2)} + {\sqrt{\I -1} \sqrt{\I+1} \I \over 4\sqrt{z-1} \sqrt{z+1}} { 1 \over z - \I} - {\sqrt{-\I -1} \sqrt{-\I+1} \I \over 4\sqrt{z-1} \sqrt{z+1}} { 1 \over z + \I} + {C \over \sqrt{z-1} \sqrt{z+1}} $$

But we want something stronger: that $\phi(z) = O(z^{-2})$. To accomplish this, we need to choose $C$. Fortunately, I made the problem easy as every term apart from the last one is already $O(z^{-2})$, so choose $C = 0$:

We see also that it has the right jump:

Problem 1.3

1. From the Hilbert formula, we know that the general solution of $\HH u = f$ is $$ u(x) = {-1 \over \sqrt{1 - x^2}}\HH \left[{ f(\diamond) \sqrt{1-\diamond^2} }\right](x) - {C \over \sqrt{1-x^2}} $$
   Plugging in $f(x) = x/\sqrt{1-x^2}$ means we need to calculate $$ \HH \left[{\diamond}\right](x) $$ We do so by first finding the Cauchy transform. Consider $$ \phi_1(z) = z \CC 1(z) = z {\log(z-1) - \log(z+1) \over 2 \pi \I} $$ This has the right jump: $$ \phi_1^+(x) - \phi_1^-(x) = x $$ but doesn't decay at $\infty$: \begin{align*} x {\log(x-1) - log(x+1) \over 2 \pi \I} = x {\log x + \log(1-1/x) - \log x -\log(1+1/x) \over 2 \pi \I} \\ = -x { 2 \over x 2 \pi \I} = -{1 \over \I \pi} \end{align*} But this means that $$ \phi(z) = \phi_1(z) + {1 \over \I \pi} = z {\log(z-1) - \log(z+1) \over 2 \pi \I} + {1 \over \I \pi} $$ Decays and has the right jump, hence is $\CC[\diamond](z)$.

Therefore, we have $$ \HH[\diamond](x) = \I (\CC^+ + \CC^-) \diamond(x) = x {\log(1-x) - \log(1+x) \over \pi} +{2 \over \pi} $$

Therefore, we get $$ u(x) = - {x(\log(1-x) - \log(1+x))+2 \over \pi\sqrt{1-x^2}} - {C \over \sqrt{1-x^2}} $$ This can be verified in Mathematica via

NIntegrate[-((
   x (Log[1 - x] - Log[1 + x]) + 
    2)/(π Sqrt[1 - x^2] (x - 0.1))), {x, -1, 0.1, 1}, 
  PrincipalValue -> True]/π

1. Following the procedure of multiplying $\CC[\sqrt{1-\diamond^2}](z)$ by $1/(2+z)$ and subtracting off the pole at $z=-2$, we first find: $$ \CC\left[{\sqrt{1-\diamond^2} \over 2+\diamond}\right](z) = {\sqrt{z-1}\sqrt{z+1} - z \over 2\I(2+z)} -{\sqrt{-2-1}_+ \sqrt{-2+1}_+ +2 \over 2\I(2+z)} = {\sqrt{z-1}\sqrt{z+1} - z \over 2\I(2+z)} +{ \sqrt{3} -2 \over 2\I(2+z)} $$

Therefore, calculating $\I(\CC^+ + \CC^-)$ we find that $$ \HH\left[{\sqrt{1-\diamond^2} \over 2+\diamond}\right](z) = -{x\over 2+x} +{ \sqrt{3} -2 \over 2+x} $$

Thus the general solution is $$ u(x) = {1 \over \sqrt{1-x^2}} \left( {x\over 2+x} -{ \sqrt{3} -2 \over 2+x} + C \right) $$ We need to choose $C$ so this is bounded at the right-endpoint: In other words, $$ u(x) = {1 \over \sqrt{1-x^2}} \left( {x\over 2+x} -{ \sqrt{3} -2 \over 2+x} +{ \sqrt{3} -3 \over 3} \right) $$

Problem 2.1

Doing the change of variables $\zeta = b s$ we have $$ \log(ab) = \int_1^{ab} {\D \zeta \over \zeta} = \int_{1/b}^a {\D s \over s} $$ if $\gamma$ does not surround the origin, we have $$ 0 = \oint_\gamma {\D s \over s} = \left[\int_1^{1/b} + \int_{1/b}^a + \int_a^1\right] {\D s \over s} $$ which implies $$ \log(ab) = \left[-\int_a^1 -\int_1^{1/b} \right] {\D s \over s} = \log a - \log {1 \over b} = \log a + \log b $$ Here's a picture:

If it surrounds the origin counbter-clockwise, that is, it has positive orientation, we have $2\pi \I = \oint_\gamma {\D s \over s}$, which shoes that $$ \log(ab) = 2 \pi \I - \left[\int_a^1 +\int_1^{1/b} \right] {\D s \over s} = \log a + \log b + 2\pi \I $$ and a similar result when counter clockwise.

If the contour passes through the origin, there are three possibility:

1. $[a,1]$ contains zero, hence $a < 0$
2. $[1,1/b]$ contains zero, hence $b < 0$
3. $[1/b, a]$ contains zero, which can only be true if $a b < 0$ by considering the equation of the line segment.

1. In the case where $a < 0$ and $b < 0$ (and hence $a b > 0$), perturbing $a$ above and $b$ below or vice versa avoids $\gamma$ winding around zero, so we have $$ \log(a b) = \log_+ a + \log_- b = \log_- a + \log_+ b = \log_+ a + \log_+ b - 2 \pi \I = \log_- a + \log_- b + 2 \pi \I $$

In the case where $a < 0$ and $b > 0$, then $ a b < 0$, but we can perturb $a$ above/below to get $$ \log_\pm(a b) = \log_\pm a + \log b $$ (and by symmetry, the equivalent holds for $b < 0$ and $a > 0$.)

In the case where $a < 0$, if $\Im b > 0$ we can perturb $a$ below so that $\gamma$ does not contain zero, giving us $$ \log(ab) = \log_- a + \log b $$ similarly, if $\Im b < 0$ we can perturb $a$ above.

2. In this case, swap the role of $a$ and $b$ and use the answers for $a < 0$.

3. Finally, we have the case $a b < 0$ and neither $a$ nor $b$ is real. Note that $$ ab = (a_x + \I a_y) (b_x + \I b_y) = a_x b_x - a_y b_y + \I(a_x b_y + a_yb_x) $$ It follows if $b_x > 0$ we have $$ (ab)_+ = a_+ b $$ and if $b_x < 0$ we have $$ (ab)_+ = a_- b $$

We can use this perturbation to reduce to the previous cases. For example, if $a = 1 + \I$ and $b = -1 + \I$, pertubing $ab$ above causes $a$ to be perturbed above, which causes the contour to surround the origin clockwise, hence we have $$ \log_+(ab) = \log(a)_+b = \log a b - 2 \pi \I $$

Problem 2.2

Use the contour $\gamma(t) = 1 + t(1-z)$ to reduce it to a normal integral: $$\overline{\log z } = \overline{\int_1^z {1 \over \zeta} \D \zeta} = \overline{\int_0^1 {(z-1) \over 1+(z-1) t} \dt} = \int_0^1 {(\bar z-1) \over 1+(\bar z-1) t} \dt = \int_1^{\bar z} {\D \zeta \over \zeta} = \log \bar z.$$ We then have, since the contour from $1$ to $1/(\bar z)$ to $z$ never surrounds the origin since both $\Im z$ and $\Im 1/(\bar z)$ have the same sign, we have $$ 2 \Re \log z = \log z + \overline{\log z} = \log z + \log \bar z = \log z \bar z = \log |z|^2 = 2 \log |z| $$ On the other hand, we have, where the contour of integration is chosen to be to the right of zero and then we do the change of variables $\zeta = |z| \E^{\I \theta}$ $$ 2 \Im \log z = \log z - \log \bar z = \int_{\bar z}^z {\D \zeta \over \zeta} = \I \int_{-\arg z}^{\arg z} \D \theta = 2 \I \arg z $$

Problem 2.3

We first show that it is analytic on $(-\infty,0)$. To do this, we need to show that the limit from above equals the limit from below: for $x < 0$ we have $$\log_1^+ x -\log_1^- x = \log_+x - \log_- x +2 \pi \I = 0$$ Then for $x > 0$ and using $\log_1^\pm(x) = \lim_{\epsilon\rightarrow 0} \log(x\pm \I \epsilon)$ we find $$ \log_1^+(x) - \log_1^-x = \log x- \log x- 2\pi \I = -2 \pi \I $$

Demonstration Here we see that the following is the analytic continuation:

Problem 3.1

1. Because it's absolutely integrable, we can exchange derivatives and integrals to determine $$ {\D \CC f \over \D z} = {1 \over 2 \pi \I} \oint {f(\zeta)\over (\zeta-z)^2} \D \zeta $$
2. There are two different possible approaches:
   • the subtract and add back in technique: since $f$ is analytic for $z$ near $\zeta$, we can write $$ \CC f(z) = {1 \over 2 \pi \I} \oint {f(\zeta)-f(z) \over (\zeta-z)^2} \D \zeta + f(z) \CC 1(z) $$ Therefore $$ \CC^+ f(\zeta) - \CC^- f(\zeta) = f(\zeta)( \CC^+ 1(\zeta) - \CC^- 1(\zeta)) $$ But we know (using Cauchy's integral formula / Residue calculus) $$ \CC 1(z) = \begin{cases} 1 & |z| < 1 \

                        0 & |z| > 0
                        \end{cases}
     

     $$ hence $(\CC^+-\CC^-)1(\zeta) = 1$
   • Since $f$ is analytic, we have for any radius $R > 1$ but inside the annulus $$ \CC^+ f(\zeta) = {1 \over 2 \pi \I} \oint_{|z| = R} {f(\zeta) \over \zeta -z} \D\zeta $$ Similarly, for $\CC^-f(\zeta)$ with any radius $r < 1$ but inside the annulus. Therfore, $$ \CC^+ f(\zeta) - \CC^- f(\zeta) = {1 \over 2 \pi \I} \left[\oint_{|z| = R} - \oint_{|z| = r} \right] {f(\zeta) \over \zeta -z} \D\zeta $$ Deforming the contour and using Cauchy-integral formula gives the result.
3. This follow since ${1 \over \zeta - z} \rightarrow 0$ uniformly.

Problem 3.2

Suppose we have another solution $\phi$ and consider $\psi(z) = \phi(z) - \CC f(z)$. Then on the circle we have $$ \psi_+(\zeta) - \psi_-(\zeta) = \phi_+(\zeta) - \CC_+f(\zeta) - \phi_-(\zeta) + \CC_+f(\zeta) = f(\zeta)-f(\zeta) = 0 $$ Thus $\psi$ is entire, and since it decays at infinity, it must be zero by Liouville's theorem.

Problem 3.3

When $k \geq 0$, we have from 3.1 and 3.2 $$ \CC[\diamond^k](z) =\begin{cases} z^k & |z| < 0 \\ 0 & |z| > 0 \end{cases} $$ when $k < 0$ since $$\CC[\diamond^k]^+(\zeta) - \CC[\diamond^k]^-(\zeta) = \zeta^k - 0 = \zeta^k$$. we similarly have $$ \CC[\diamond^k](z) =\begin{cases} 0 & |z| < 0 \\ -z^k & |z| > 0 \end{cases} $$

Therefore, $$ \Im \CC^-[\diamond^k](\zeta) =\begin{cases} 0 & k \geq 0 \\ -{\zeta^k - \zeta^{-k} \over 2 \I} & k < 0 \end{cases} $$ and $$ \Re \CC^-[\diamond^k](\zeta) =\begin{cases} 0 & k \geq 0 \\ -{\zeta^k + \zeta^{-k} \over 2} & k < 0 \end{cases} $$

Problem 3.4

Express the solution outside the circle as $$ v(x,y) = \Im ( \E^{-\I \theta} z + \CC f(z)) $$ for a to-be-determined $f$. On the circle, this reduces to $$ \Im \CC^- f(\zeta) = -\cos \theta {\zeta - \zeta^{-1} \over 2 \I} + \sin \theta {\zeta + \zeta^{-1} \over 2} $$ Unlike the real case, we can include imaginary coefficients, thus the solution is $$ f(\zeta) = (\cos \theta + \I \sin \theta) \zeta^{-1} $$ and thus the full solution is $$ v(x,y) = \Im ( \E^{-\I \theta} z - \E^{\I \theta} z^{-1})) $$

 Problem 4.1

$z^\alpha$ has the limits $z_\pm^\alpha = \E^{\pm \I \pi \alpha} |z|^\alpha$, thus choose $\alpha = -{\theta \over 2\pi}$ where if we take $0 < \theta < 2\pi$ we have $0 < \alpha < 1$ (the case $\theta = 0$ and $\theta = \pi$ are covered by the Cauchy transform, that is ). Then consider $$ \kappa(z) = (z-1)^{-\alpha} (z+1)^{\alpha-1} $$ which has weaker than pole singularities and satisfies $\kappa(z) \sim z^{-1}$. For $-1 < x < 1$ it has the right jump $$ \kappa_+(x) = (x-1)_+^\alpha (x+1)^{1 - \alpha} = \E^{\I \pi \alpha} (1-x)^\alpha (x+1)^{1 - \alpha} = \E^{2\I \pi \alpha} (x-1)_-^\alpha (x+1)^{1 - \alpha} = \E^{2 \I \pi \alpha} \kappa_-(x)= \E^{\I \theta} \kappa_-(x) $$ and for $x < -1$ it has the jump $$ \kappa_+(x) = (x-1)_+^\alpha (x+1)_+^{1 - \alpha} = \E^{\I \pi \alpha}\E^{\I \pi (1-\alpha)} (1-x)^\alpha (-1-x)^{1 - \alpha} = \kappa_-(x) $$ hence $\kappa$ is analytic.

We need to show this times a constant spans the entire space. Suppose we have another solution $\tilde \kappa$ and consider $r(z) = {\tilde \kappa(z) \over \kappa(z)}$. Note by construction that $\kappa$ has no zeros. Then $$ r_+(x) = {\tilde\kappa_+(x) \over \kappa_+(x)} = {\tilde\kappa_-(x) \over \kappa_-(x)} = r_-(x) $$ hence $r$ is analytic on $(-1,1)$. It has weaker than pole singularities because $\kappa(z)^{-1}$ is actually bounded at $\pm 1$. Therefore $r$ is bounded and entire, and thus must be a constant $r(z) \equiv r$, and thence $\tilde \kappa(z) = r \kappa(z)$.

Problem 4.2

We want to mimic the solution of $\phi_+(x) + \phi_-(x)$. So take $$ \phi(z) = \kappa(z) \CC\br[{f \over \kappa_+}](z) =\E^{-\I \theta/2} (z-1)^{-\alpha}(z+1)^{\alpha-1} \CC[f (1-x)^{\alpha}(1+x)^{1-\alpha}](z) $$ This has the jump \begin{align*} \phi_+(x) - \E^{\I \theta}\phi_-(x) = \kappa_+(z) \CC_+\br[{f \over \kappa_+}](x) - \E^{\I \theta}\kappa_-(z)\CC_-\br[{f \over \kappa_+}](x) = \kappa_+(x) \pr({\CC_+\br[{f \over \kappa_+}](x) - \CC_-\br[{f \over \kappa_+}](x) }) = f(x) \end{align*}

Thus the general solution is $\phi(z) + C \kappa(z)$.

Problem 4.3

Note for $x < 0$ $$ x_+^{\I \beta} = \E^{\I \beta \log_+ x} = \E^{\I \beta \log_- x - 2\pi y} = \E^{-2 \pi \beta} x_-^{\I \beta} $$

We actually have bounded (oscillatory) growth near zero since $$ |\E^{\I \beta \log z}| = |\E^{\I \beta \log |z|} \E^{-\beta \arg z}| = \E^{-\beta \arg z} $$

Thus if we write $c = r \E^{\I \theta}$ for $0 < \theta < 2 \pi$ and define $\alpha = -{\theta \over 2 \pi} + \I{\log r \over 2 \pi}$ we can write the solution to 4.1 as $$ \kappa(z) = (z-1)^{-\alpha} (z+1)^{\alpha-1} $$

The same arguments as before then proceed. and the solution to 4.3 is $$ \phi(z) = \kappa(z) \CC\br[{f \over \kappa_+}](z)+ C \kappa(z) $$

Problem 5.1

1. It is a product of functions analytic off $(-\infty,1]$ hence is analytic off $(-\infty,1]$, and we just have to check that it has no jump on $(-\infty,-1)$ and $(-a,a)$. This follows via, for $x < -1$: $$ \kappa_+(x) = {1 \over \I^4 \sqrt{1-x}\sqrt{-1-x}\sqrt{a-x}\sqrt{-a-x}} = {1 \over (-\I)^4 \sqrt{1-x}\sqrt{-1-x}\sqrt{a-x}\sqrt{-a-x}} = \kappa_-(x) $$ and for $-a < x < a$ we have $$ \kappa_+(x) = {1 \over \I^2 \sqrt{1-x}\sqrt{x+1}\sqrt{a-x}\sqrt{x+a}} = {1 \over (-\I)^2 \sqrt{1-x}\sqrt{1+x}\sqrt{a-x}\sqrt{-a-x}} = \kappa_-(x) $$
2. This follows via the usual arguments: for $a < x < 1$ we have: $$ \kappa_+(x) = {1 \over \I \sqrt{1-x}\sqrt{x+1}\sqrt{x-a}\sqrt{x+a}} = -\kappa_-(x) $$ and for $-1 < x < -a$ we have $$ \kappa_+(x) = {1 \over \I^3 \sqrt{1-x}\sqrt{x+1}\sqrt{x-a}\sqrt{x+a}} = -\kappa_-(x) $$
3. This has at most square singularitie4s which are weaker than poles
4. $\kappa(z) = {1 \over z^2 \sqrt{1-1/z}\sqrt{1-a/z} \sqrt{1+a/z} \sqrt{1+1/z}} \sim {1 \over z^2} \rightarrow 0$.

Problem 5.2

Ah, this is a trick question! Note that $z \kappa(z) \sim z^{-1} = O(z)$ and satisfies all the other properties. Thus consider any other solution $\tilde \kappa(z)$ and write $$ r(z) = {\tilde \kappa(z) \over \kappa(z) } $$ This has trivial jumps and hence is entire: for example, on $(a,1)$ we have $$ r_+(x) = {\tilde \kappa_+(x) \over \kappa_+(x) } = {-\tilde \kappa_-(x) \over -\kappa_-(x) } = r_-(x) $$ But since $\kappa \sim O(z^{-2})$ we only know that $\kappa$ has at most $O(z)$ growth, hence it can be any first degree polynomial. Therefore, the space of all solutions is in fact two-dimensional: $\psi(z) = (A + Bz) \kappa(z)$.

Problem 5.3

Here we mimick the usual solution techniques and propose: $$ \phi(z) = \kappa(z) \CC\br[{f \over \kappa_+}](z) + (A+Bz) \kappa(z) $$ A quick check confirms it has the right jumps: $$ \phi_+(x) = \kappa_+(x) \CC_+\br[{f \over \kappa_+}](x) + (A+Bx) \kappa_+(x) =\kappa_+(x) \left({f(x) \over \kappa_+(x)} + \CC_-\br[{f \over \kappa_+}](x)\right) - (A+Bx) \kappa_-(x) = -\phi_-(x) + f(x) $$

Problem 6.1

Let's first do a plot and histogram. Here we see the right scaling is $N^{1/4}$:

The limiting distribution has the following form:

We want to solve $$ {\D \lambda_k \over \D t} = \sum_{j=1 \atop j \neq k}^N {1 \over \lambda_k -\lambda_j} - 4\lambda_k^3 $$ Rescale via $\mu_k = {\lambda_k \over N^{1/4}}$ gives $$ 0 = N^{1/4} {\D \mu_k \over \D t} = N^{-1/4} \sum_{j=1 \atop j \neq k}^N {1 \over \mu_k -\mu_j} - 4 N^{3/4} \mu_k^3 $$ or in other words $$ 0 = N^{-1/2} {\D \mu_k \over \D t} = {1 \over N} \sum_{j=1 \atop j \neq k}^N {1 \over \mu_k -\mu_j} - 4 \mu_k^3 $$ We can now formally let $N\rightarrow \infty$ to get our equation $$ \dashint_{-b}^b {w(t) \over x-t} \dt = 4 x^3 $$ where I've used symmetry to assume that the interval is symmetric. We want to find $w$ and $b$ so that this equations holds true and $w$ is a bounded probability density:

1. $w(x) >0$ for $-b < x < b$
2. $\int w(x) \dx = 1$
3. $w$ is bounded

Our equation is equivalent to $$ \HH_{[-b,b]} w(x) = -{4 x^3 \over \pi} $$ recall the inversion formula $$ u(x) = {-1 \over \sqrt{b^2 - x^2}}\HH \left[{ f(\diamond) \sqrt{b^2-\diamond^2} }\right](x) - {C \over \sqrt{b^2-x^2}} $$
In our case $f(x) = -{4 x^3 \over \pi}$ and we use $$ \sqrt{z-b}\sqrt{z+b} = z\sqrt{1-b/z}\sqrt{1+b/z} = z - {b^2 \over 2 z} -{b^4 \over 8z^3} + O(z^{-4}) $$ to determine $$ 2 \I \CC \left[{ \diamond^3 \sqrt{b^2-\diamond^2} }\right](x) = z^3(\sqrt{z-b}\sqrt{z+b} -z + {b^2 \over 2 z} + {b^4 \over 8z^3}) = z^3\sqrt{z-b}\sqrt{z+b} -z^4 + {b^2z^2 \over 2} + {b^4 \over 8} $$

Therefore, \begin{align*} u(x) = {4\I \over \pi \sqrt{b^2 - x^2}}(\CC^+ + \CC^-) \left[{ \diamond^3 \sqrt{b^2-\diamond^2} }\right](x) - {C \over \sqrt{b^2-x^2}} \\ {4 \over \pi \sqrt{b^2 - x^2}}(-x^4+{b^2 x^2 \over 2} + {b^4 \over 8})- {C \over \sqrt{b^2-x^2}} \end{align*} We choose $C$ so this is bounded, in particular, we get get the solution $$ u(x) = {4 \over \pi \sqrt{b^2 - x^2}}(-x^4+{b^2 x^2 \over 2} + {b^4 \over 2}) $$

At least it looks right, we just need to get the right b:

We want to choose $b$ now so that this integrates to $1$. For example, this choice of $b$ is horrible:

There's a nice trick: If $-2\pi \I \CC u(z) \sim {1 \over z}$ then $\int_{-b}^b u(x) \dx = 1$. We know since $$ {1 \over \sqrt{1+z}} = 1 - {z \over 2} + {3z^2 \over 8} - {5z^3 \over 16} + O(z^4) $$ \begin{align*} {-z^4 + {b^2z^2 \over 2} + {b^4 \over 2} \over \sqrt{z-b}\sqrt{z+b}} &= {-z^3 + {b^2z \over 2} + {b^4 \over 2 z} \over \sqrt{1-b/z}\sqrt{1+b/z}} \\ &= (-z^3 + {b^2z \over 2} )(1 + {b \over 2 z} + {3 b^2 \over 8 z^2} + {5 b^3 \over 16z^3} )(1 - {b \over 2 z} + {3 b^2 \over 8 z^2} - {5 b^3 \over 16z^3}) + O(z^{-1}) \\ &=-z^3 + \pr({b^2 \over 2} +{b^2 \over 4} + {b^2 \over 2} - {3 b^2 \over 8} -{3 b^2 \over 8} ) z +O(z^{-1})\\ &= -z^3 + O(z^{-1}) \end{align*} Thus we know $$ \CC u(z) = {2\I \over \pi} z^3 +{2 \I \over \pi} {-z^4 + {b^2z^2 \over 2} + {b^4 \over 2} \over \sqrt{z-b}\sqrt{z+b}} $$

taking this one term further we find
$$ {-z^4 + {b^2z^2 \over 2} + {b^4 \over 2} \over \sqrt{z-b}\sqrt{z+b}} = -z^3 +{3 b^4 \over 8 z} + O(z^{-2}) $$ Hence we want to choose $b$ so that $$ -2\I\pi \CC u(z) = {3 b^4 \over 2 z} \sim {1 \over z} $$ in other words, $b = ({2 \over 3})^{1/4}$

And it worked!