In [43]:

using Plots, ComplexPhasePortrait, ApproxFun, SingularIntegralEquations, OscillatoryIntegrals, QuadGK
gr();



# M3M6: Methods of Mathematical Physics 2018

$$\def\dashint{{\int\!\!\!\!\!\!-\,}} \def\infdashint{\dashint_{\!\!\!-\infty}^{\,\infty}} \def\D{\,{\rm d}} \def\E{{\rm e}} \def\dx{\D x} \def\dt{\D t} \def\dz{\D z} \def\C{{\mathbb C}} \def\R{{\mathbb R}} \def\rR{{\rm R}} \def\rL{{\rm L}} \def\CC{{\cal C}} \def\HH{{\cal H}} \def\I{{\rm i}} \def\qqqquad{\qquad\qquad} \def\qqfor{\qquad\hbox{for}\qquad} \def\qqwhere{\qquad\hbox{where}\qquad} \def\Res_#1{\underset{#1}{\rm Res}}\, \def\sech{{\rm sech}\,} \def\vc#1{{\mathbf #1}} \def\Ei{{\rm Ei}\,} \def\pr(#1){\left({#1}\right)} \def\br[#1]{\left[{#1}\right]} \def\set#1{\left\{{#1}\right\}} \def\ip<#1>{\left\langle{#1}\right\rangle} \def\iip<#1>{\left\langle\!\langle{#1}\right\rangle\!\rangle}$$

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

# Solution Sheet 4

## Problem 1.1

We want to solve $$\int_{-1}^1 \log |x-t| u(t) \dt = {1 \over x^2 + 1}$$ differentiating and multiplying though by $1/\pi$ we have $${1 \over \pi} \dashint_{-1}^1 {u(t) \over t-x} u(t) = {2 x \over \pi (1+x^2)^2}$$

This is an inverse Hilbert problem so we know $$u(x) = -{1 \over \sqrt{1-x^2} } H[\sqrt{1-t^2} f(t)] + {C \over \sqrt{1-x^2}}$$ for $$f(x) = {2 x \over \pi (1+x^2)^2}.$$

We determine the Cauchy transform of $\sqrt{1-x^2} f(x)$ using the usual methods: start with the ansatz $$\phi(z) = {\sqrt{z-1} \sqrt{z+1} \over 2 \I} {2 z \over \pi (1+z^2)^2}$$ This decays at infinity so we just need to remove the poleas at $\pm \I$. Here we determine: $$\phi(z) = {\sqrt{\I -1 } \sqrt{\I + 1} \over \pi} \left(-{1 \over 4 (z-\I)^2} + {\I \over 8 (z-\I)} + O(1) \right)$$ and $$\phi(z) = {\sqrt{-\I -1 } \sqrt{1-\I } \over \pi} \left({1 \over 4 (z+\I)^2} + {\I \over 8 (z+\I)} + O(1) \right)$$ Telling us that $$C[\sqrt{1-t^2} f(t)](z) = {\sqrt{z-1} \sqrt{z+1} \over 2 \I} {2 z \over \pi (1+z^2)^2} - {\sqrt{\I -1 } \sqrt{\I + 1} \over \pi} \left(-{1 \over 4 (z-\I)^2} + {\I \over 8 (z-\I)} \right) - {\sqrt{-\I -1 } \sqrt{1-\I } \over \pi} \left({1 \over 4 (z+\I)^2} + {\I \over 8 (z+\I)} \right)$$ Thus we have $$H[\sqrt{1-t^2} f](x) = \I (C^+ + C^-)[\sqrt{1-t^2} f](x) = - {2\I \sqrt{\I -1 } \sqrt{\I + 1} \over \pi} \left(-{1 \over 4 (x-\I)^2} + {2\I \over 8 (x-\I)} \right) - {2\I \sqrt{-\I -1 } \sqrt{1-\I } \over \pi} \left({1 \over 4 (x+\I)^2} + {\I \over 8 (x+\I)} \right)$$

In other words, for $$\tilde u(x) = {2\I \sqrt{\I -1 } \sqrt{\I + 1} \over \pi \sqrt{1-x^2}} \left(-{1 \over 4 (x-\I)^2} + {2\I \over 8 (x-\I)} \right) - {2\I \sqrt{-\I -1 } \sqrt{1-\I } \over \pi \sqrt{1-x^2}} \left({1 \over 4 (x+\I)^2} + {\I \over 8 (x+\I)} \right)$$ we have $$u(x) = \tilde u(x) + {C \over \sqrt{1-x^2}}.$$

To find $C$ we impose the condition that $$\int_{-1}^1 \log|t| u(t) \dt = 1$$ We thus need to determine $C$. Recall from Lecture 19 that $${1 \over \pi} \int_{-1}^1 {\log(z-x) \over \sqrt{1-x^2}} \dx = 2 \log(\sqrt{z-1}+\sqrt{z+1}) - 2 \log 2$$ Thus for $x \in [-1,1]$ we have $${1 \over \pi} \int_{-1}^1 {\log|x-t| \over \sqrt{1-t^2}} \dt = 2 \Re(\log(\I \sqrt{1-x}+\sqrt{x+1}) - 2 \log 2)$$ or in particular $${1 \over \pi} \int_{-1}^1 {\log|x| \over \sqrt{1-x^2}} \dx = \log(1/2)$$ Thus we want to solve $$\int_{-1}^1 \log|t| \tilde u(t) \dt + C {\log(1/2) \over \pi} = 1$$ i.e. $$C = (1 - \int_{-1}^1 \log|t| \tilde u(t) \dt) {1 \over \pi \log(1/2)}$$

Check derivation

Let's check the derivation. First we can calculate $u$ numerically:



In [5]:

L = SingularIntegral(0) : JacobiWeight(-0.5,-0.5,Chebyshev())

x = Fun()
g = (1/(x^2+1))
u = (π*L) \ g

π*logkernel(u, 0.1) - g(0.1)




Out[5]:

-2.220446049250313e-16



Differentiating it satisfies $H u = g'/(-\pi)$:



In [6]:

hilbert(u,0.1) ≈ g'(0.1)/(-π) ≈ (2*0.1)/(π*(1+0.1^2)^2)




Out[6]:

true



The Cauchy transform also works:



In [8]:

f = g'/(-π)

z = 1+im
ψ = z -> sqrt(z-1)sqrt(z+1)/(2im) * 2z/(π*(1+z^2)^2) -
sqrt(im-1)sqrt(im+1)/π * (-1/(4*(z-im)^2) + im/(8(z-im))) -
sqrt(-im-1)sqrt(1-im)/π * (1/(4*(z+im)^2) + im/(8(z+im)))

cauchy(sqrt(1-x^2)*f, z) ≈ ψ(z)




Out[8]:

true



Therefore the Hilbert transform is given by:



In [12]:

H = x-> - 2im*sqrt(im-1)sqrt(im+1)/π * (-1/(4*(x-im)^2) + im/(8(x-im))) -
2im*sqrt(-im-1)sqrt(1-im)/π * (1/(4*(x+im)^2) + im/(8(x+im)))

hilbert(sqrt(1-x^2)*f)(0.1) ≈ H(0.1)




Out[12]:

true



We thus have an expression for $u$, we are just missing the constant:



In [81]:

ũ = x -> -H(x)/sqrt(1-x^2)

C = u(0.0) + H(0.0)
u(0.1) ≈ ũ(0.1) + C/sqrt(1-0.1^2)




Out[81]:

true



We can find $C$ interms of the relevant integral, which we call $D$:



In [92]:

D = quadgk(x -> x == 0 ? 0.0 : ũ(x)*log(abs(x)),-1,1)[1]
C, (1-D)/(π*log(1/2))




Out[92]:

(-0.3247204711377926 + 0.0im, -0.32472047136213555 + 0.0im)



### Problem 1.2

Since $\int_{-1}^1 x \dx = 0$, we have no logarithmic growth at infinity. Thus by the formula in Lecture 19 we find for $$U(x) = \int_x^1 t \dt = {1 - x^2 \over 2}$$ that $$\int_{-1}^1 \log|z-x| x \dx = \Re{-2 \I \pi C U(z)}$$

So we just have to work out the Cauchy transform. Try as an ansatz $$(1 - z^2 \over 2) {\log(z-1) - \log(z+1) \over 2 \pi \I}$$ We only need to remove the growth at infinity. We do so via: $$\log(z-1) - \log(z+1) = -{2 \over z} + O(z^{-3})$$ telling us $$C U(z) = {1 - z^2 \over 2} {\log(z-1) - \log(z+1) \over 2 \pi \I} + {z \over 2 \pi \I}$$ and $$\int_{-1}^1 \log|z-x| x \dx = \Re \left({1 - z^2 \over 2} (\log(z-1) - \log(z+1)) - z \right)$$

We can confirm the formula:



In [93]:

z = 2+im; π*logkernel(x,z) ≈ real((1-z^2)/2 * (log(z-1)-log(z+1)) - z)




Out[93]:

true



## Problem 2

### Problem 2.1

From Lecture 19 we know that we want to solve

1. $v_{xx} + v_{yy} =0$ for $z \notin [-1,1] \cup \I$
2. $v(z) \sim \log|z|$ as $z \rightarrow \infty$
3. $v(z) \sim \log|z-\I|$ as $z \rightarrow \I$
4. $v(x,0) = D$ for some unknown constant $D$ on $[-1,1]$.

### Problem 2.2

We write $$v(x,y) = \int_{-1}^1 u(t) \log|z-t| \dt + \log|z-\I|$$ The behaviour at infinity requires that $\int_{-1}^1 u(t) \dt = 0$. We further have the singular integral equation $$\int_{-1}^1 u(t) \log|x-t| \dt = C - \log|x-\I|$$ which follows from $D = v(x,0)$

We can actually solve this numerically:



In [150]:

ũ = SingularIntegral(0) \ (-log(abs(x-im))/π)
u = ũ - sum(ũ)/(π*sqrt(1-x^2)) # ensure integrates to zero
v = z -> π*logkernel(u, z) + log(abs(z-im))
D = v(0)




Out[150]:

0.18822640645958963




In [151]:

xx = yy = -4:0.011:4
V = v.(xx' .+ im*yy)

contour(xx, yy, V; nlevels=50)
plot!(domain(x); color=:black, legend=false)




Out[151]:

-4

-2

0

2

4

-4

-2

0

2

4

-

5

-

4

-

3

-

2

-

1

0

1



#### Problem 2.3

As usual for logarithmic singular integral equations we want to solve $$\dashint_{-1}^1 {u(t) \over t-x} \dt = f'(x)$$ we need to be a bit careful differentiating $f$: $${\D \over \dx} \log|x-\I| = {\D \over \dx} \log\sqrt{x^2 + 1} = { x \over x^2 + 1}$$ Now we do our usual game and solve for $u$ using the inverse Hilbert transform formula. That is, first calculate $$C[\sqrt{1-x^2} {x \over x^2+1}](z) = {z \sqrt{z-1} \sqrt{z+1} \over 2\I (z^2+1)} - {1 \over 2 \I} + {\I \sqrt{\I - 1} \sqrt{\I+1} \over 4 (z-\I)} + {\I \sqrt{-\I - 1} \sqrt{-\I+1} \over 4 (z+\I)}$$ Therefore, we know that $$u(x) = {\sqrt{\I - 1} \sqrt{\I+1} \over 2 \pi (x-\I)\sqrt{1-x^2}} + {\sqrt{-\I - 1} \sqrt{-\I+1} \over 2 \pi (x+\I) \sqrt{1-x^2}} + {C \over \sqrt{1-x^2}}$$ We can show (e.g. by finding its Cauchy transform and lookin at the asymptotic behaviour) that $$\int_{-1}^1\left[ {\sqrt{\I - 1} \sqrt{\I+1} \over 2 \pi (x-\I)\sqrt{1-x^2}} + {\sqrt{-\I - 1} \sqrt{-\I+1} \over 2 \pi (x+\I) \sqrt{1-x^2}} \right] = -1$$ Combined with the fact that $$\int_{-1}^1 {1 \over \sqrt{1-x^2}} = \pi$$ we find that $C = 1/\pi$.

Verification Let's check our work. First we see that the Hilbert transform of $u$ does indeed satisfy the specified equation (using the numerical u as calculated above):



In [154]:

fp = x -> x/(x^2+1)

π*hilbert(u, 0.1) ≈ fp(0.1)




Out[154]:

true



And the Cauchy transform of $\sqrt{1-x^2} f'(x)$ satisfies the derived formula:



In [156]:

C = z -> z * sqrt(z-1)sqrt(z+1)/(2im*(z^2+1)) - 1/(2im) + im*sqrt(im-1)sqrt(im+1)/(4(z-im)) +
im*sqrt(-im-1)sqrt(-im+1)/(4(z+im))
cauchy(sqrt(1-x^2)fp(x), z) ≈ C(z)




Out[156]:

true



Therefore we can invert to the Hilbert transform for $f'$:



In [166]:

w = -hilbert(sqrt(1-x^2)fp(x))/sqrt(1-x^2)
w̃ = sqrt(im-1)sqrt(im+1)/(2(x-im) * sqrt(1-x^2)) +
sqrt(-im-1)sqrt(-im+1)/(2(x+im) * sqrt(1-x^2))
hilbert(w,0.1) ≈ hilbert(w̃,0.1) ≈ fp(0.1)




Out[166]:

true



And we have recovered $u$ up to $C/\sqrt{1-x^2}$:



In [180]:

C = 1/π

u(0.1) ≈ w̃(0.1)/π + C/sqrt(1-0.1^2)




Out[180]:

true



## Problem 3

### Problem 3.1

To be analytic at all we need decay at either $\pm \infty$, this has neither so is not defined.

### Problem 3.2

It has exponential decay in the right-half plane, therefore $$\E^{\gamma x} f(x) = {\E^{\gamma x } \over 1 + \E^x}$$ has exponential decay at both $\pm \infty$, provided $0 < \gamma < 1$. Therefore, we can take the strip $0 < \Im s < 1$. (Note in each case the contour for the inverse Fourier transform can be any contour in the domain of analyticity.)

We can verify this by exact computation using Residue calculus: for $0 < \Im s < 1$, we can integrate over a rectangle to get: $$\left(\int_{-R}^R + \intR^{2\I \pi + R} + \int{2 \I \pi + R}^{2\I \pi - R} + \int{2 \I \pi - R}^{-R} \right) {\E^{-\I s x} \over 1 + \E^x} \dx = 2 \pi \I \Res{z = \I \pi } {\E^{-\I s z} \over 1 + \E^z} = • 2 \pi \I \E^{\pi s}$$ Note that $${\E^{-\I s (R + \I t)} \over 1 + \E^{R + \I t}} = {\E^{-\I R \Re s + R \Im s + t} \over 1 + \E^{R + \I t}} \rightarrow 0$$ and $${\E^{-\I s (-R + \I t)} \over 1 + \E^{R + \I t}} = {\E^{\I R \Re s - R \Im s + t} \over 1 + \E^{R + \I t}} \rightarrow 0$$ uniformly in $t$ as $R \rightarrow \infty$, hence we deduce that $$\left(\int{-\infty}^\infty + \int{2 \I \pi + \infty}^{2\I \pi - \infty}\right) {\E^{-\I s x} \over 1 + \E^x} \dx = • 2 \pi \I \E^{\pi s}$$ Now note that $$\int{2 \I \pi + \infty}^{2\I \pi - \infty} {\E^{-\I s t} \over 1 + \E^t} \dt = \int{\infty}^{-\infty} {\E^{-\I s (x+2 \I \pi)} \over 1 + \E^x} \dx = -\E^{2 \pi s} \int{-\infty}^\infty {\E^{-\I s x} \over 1 + \E^x} \dx$$ Therefore, we have $$\int{-\infty}^\infty {\E^{-\I s x} \over 1 + \E^x} \dx = - 2 \I \pi {\E^{\pi s} \over 1 -\E^{2 \pi s}} = \I \pi {\rm csch}\, \pi x$$ which has poles at $0$ and $\I$:


In [9]:

phaseplot(-3..3, -10..10, z -> 1/(1+exp(z))) #integrand




Out[9]:

-3

-2

-1

0

1

2

3

-10

-5

0

5

10




In [12]:

phaseplot(-3..3, -3..3, z -> im*π*csch(π*z)) # transform




Out[12]:

-3

-2

-1

0

1

2

3

-3

-2

-1

0

1

2

3



### Problem 3.3

Here $\E^{\gamma x } f(x) = \E^{(\gamma+2) x}$ has decay at $+\infty$ proved $\gamma < -2$, hence we have the strip $\Im s < -2$.

Indeed, its Fourier transform is $$-{\I \over 2 \I +s}$$ by integration by parts.

### Problem 3.4

Here it's $\Im s > 0$: unlike 1.1, we now have decay at $x \rightarrow \infty$ since $f_{\rm L}(x)$ is identically zero.

It's Fourier transform is determinable by integration-by-parts: $$\hat f(s) = \int_{-\infty}^0 x \E^{-\I s x} \dx = {1 \over \I s} \int_{-\infty}^0\E^{-\I s x} \dx = {1 \over s^2}$$

### Problem 3.5

The Fourier transforms are given above.

### Problem 3.6

$$\int_{-\infty}^\infty \delta(x) \E^{\I s x} \dx = 1$$

It's actually an entire function, but non-decaying. This is hinting at the relationship between smoothness of a function and decay of its Fourier transform, and vice-versa: since $\delta(x)$ "decays" to all orders, we expect its Fourier transform to be entire, but since its n ot smooth at all, we expect no decay, so on a formal level we can predict the analyticity properties.

## Problem 4

### Problem 4.1

Note that $$K(z) = {3\over 2} \E^{-|x|} \Rightarrow \hat K(s) = {3 \over 1+s^2}$$ Provided $-1 < \Im s < 1$, and $$\widehat f_{\rm R}(s) = -{\I \over s} - {\alpha \over s^2}$$ for $\Im s < 0$. Define $$h(s) = - \widehat f_{\rm R}(s) = {\I \over s} + {\alpha \over s^2}$$



In [11]:

α = 0.3
x = Fun(0..100)
f = 1  + α*x
h = s -> (im/s + α/s^2)

γ = -0.5  # we take the Fourier transform on R + im*γ
s = -0.5 + im*γ

-sum(f*exp(-im*s*x)) - h(s)




Out[11]:

5.773159728050814e-15 + 7.549516567451064e-15im



Transforming the equation, we have $$\Phi_+(s) - (1 + \hat K(s)) \Phi_-(s) = {\I \over s} + {\alpha \over s^2}$$ where $$1 + \hat K(s) = {4 + s^2 \over 1 + s^2} = {(s-2 \I) (s+2 \I) \over (s+\I)(s-\I)}$$ This is very close to the the example we did in lectures, so we already know the homogenous solution: $$\kappa(z) = \begin{cases} {z + 2\I \over z + \I} & \Im z > \gamma \\ {z - \I \over z - 2\I} & \Im z < \gamma \end{cases}$$ which is valid for $-1 < \Im s < 0$.



In [7]:

g = s -> (4+s^2)/(1+s^2)

κ = z -> imag(z) > γ ? (z+im*2)/(z+im) :
(z-im)/(z-im*2)

phaseplot(-3..3, -3..3, κ)




Out[7]:

-3

-2

-1

0

1

2

3

-3

-2

-1

0

1

2

3




In [8]:

s = 0.1 + γ*im
κ₊ = κ(s + eps()*im)
κ₋ = κ(s - eps()*im)

κ₊ - κ₋*g(s)




Out[8]:

-1.3322676295501878e-15 - 2.7755575615628914e-16im



We thus get the RH problem $$Y_+(s) - Y_-(s) = h(s)/\kappa_+(s) = ({\I \over s} + {\alpha \over s^2}) {s + \I \over s + 2 \I}$$ We see this has poles at $0$ and $-2 \I$, so using partial fraction expansion we get $$({\I \over s} + {\alpha \over s^2}) {s + \I \over s + \sqrt{3} \I} = {\alpha \over 2 s^2}- {\I (\alpha-2) \over 4 s} +{\I (2+\alpha) \over 4 (s+2 \I)}$$ Therefore, splitting the poles between those above and below $\gamma$, we have $$Y(z) = \begin{cases} {\I (2+\alpha) \over 4 (z+2 \I)} & \Im z > \gamma \\ -{\alpha \over 2 z^2}+{\I (\alpha-2) \over 4 z} & \Im z < \gamma \end{cases}$$



In [91]:

s = 0.1 + γ*im
Y = z -> imag(z) > γ ? im*(2+α)/(4*(z+2im)) :
- α/(2z^2) + im*(α-2)/(4z)

Y₊ = Y(s + eps()*im)
Y₋ = Y(s - eps()*im)

Y₊ - Y₋  - h(s)/κ₊




Out[91]:

5.551115123125783e-16 - 4.440892098500626e-16im



We therefore have

$$\Phi(z) = \kappa(z) Y(z) = \begin{cases} {\I (2+\alpha) \over 4 (z+ \I)} & \Im z > \gamma \\ (-{\alpha \over 2 z^2}+{\I (\alpha-2) \over 4 z} ) {z - \I \over z - 2\I} & \Im z < \gamma \end{cases}$$


In [92]:

Φ = z -> imag(z) > γ ? im*(2+α)/(4*(z+im)) :
(-α/(2z^2) + im*(α-2)/(4z))*(z-im)/(z-2im)
Φ₊ = Φ(s+eps()im)
Φ₋ = Φ(s-eps()im)

Φ₊ - Φ₋*g(s) - h(s)




Out[92]:

8.881784197001252e-16 - 1.1102230246251565e-15im



Finally, we recover the solution by inverting $\Phi_-$, using Residue calculus in the upper half plane: for $x > 0$ we have \begin{align*} u(x) &= {1 \over 2 \pi} \int_{-\infty+ \I \gamma}^{\infty + \I \gamma} (-{\alpha \over 2 z^2}+{\I (\alpha-2) \over 4 z} ) {z - \I \over z - 2\I} \E^{\I z x} \dz \\ &= \I (\Res_{z = 0} + \Res_{z = 2\I}) (-{\alpha \over 2 z^2}+{\I (\alpha-2) \over 4 z} ) {z - \I \over z - 2\I} \E^{\I z x} = {1+x \alpha \over 4} - {\alpha+1 \over 4} \E^{-2 x} \end{align*}

Did it work? yes:



In [93]:

t = Fun(0 .. 50)

u = (1+t*α)/4 - (α-1)/4*exp(-2t)

x = 0.1

u(x) + 3/2*sum(exp(-abs(t-x))*u) , f(x)




Out[93]:

(1.0300000000000011, 1.03)



### Problem 4.2

Setting up the problem as above, we arrive at a degenerate RH problem: $$\Phi_+(s) - g(s) \Phi_-(s) = h(s)$$ where $$g(s) = \widehat K(s) = {2\alpha \over \alpha^2 +s^2}= {2 \alpha \over (s-\I \alpha)(s+\I \alpha)}$$ and $$h(s) = {\I \over s} + {\alpha \over s^2} = \I {s -\I \alpha \over s^2}$$

Suppose we allow $\kappa_-(s) \sim s$ to have growth, then we can write $$\kappa(z) = \begin{cases} {1 \over z+\I \alpha }& \Im z > \gamma \\ {z-\I \alpha \over 2 \alpha} & \Im z < \gamma \end{cases}$$ so that $$\kappa_+(s) = \kappa_-(s) g(s)$$



In [9]:

α = 0.3

g = s -> (2α)/(α^2+s^2)
h = s -> (im/s + α/s^2)

κ = z -> imag(z) > γ ? 1/(z + im*α) :
(z-im*α)/(2α)

phaseplot(-3..3, -3..3, κ)




Out[9]:

-3

-2

-1

0

1

2

3

-3

-2

-1

0

1

2

3




In [18]:

s = 0.1 + γ*im
κ₊ = κ(s + eps()*im)
κ₋ = κ(s - eps()*im)

κ₊ - κ₋*g(s)




Out[18]:

2.6645352591003757e-15 + 1.7763568394002505e-15im



Then we have $$h(s)/\kappa_+(s) = \I {s^2 + \alpha^2 \over s^2} = \I + \I {\alpha^2 \over s^2}$$ and then we can write $$Y(z) = \begin{cases} \I & \Im z > \gamma \\ -{\I \alpha^2 \over z^2} & \Im z < \gamma \end{cases}$$



In [25]:

s = 0.1 + γ*im
Y = z -> imag(z) > γ ? im :
-im*α^2/s^2

Y₊ = Y(s + eps()*im)
Y₋ = Y(s - eps()*im)

Y₊ - Y₋  - h(s)/κ₊




Out[25]:

1.3877787807814457e-16 + 7.771561172376096e-16im



Putting things together, we get $$\Phi(z) = \kappa(z) Y(z) = \begin{cases} {\I \over z+\I \alpha }& \Im z > \gamma \\ -\I {\alpha^2 \over z^2} {z-\I \alpha \over 2 \alpha} & \Im z < \gamma \end{cases}$$



In [26]:

Φ = z -> imag(z) > γ ? im/(z + im*α)  :
-im*α^2/z^2* (z-im*α)/(2α)
Φ₊ = Φ(s+eps()im)
Φ₋ = Φ(s-eps()im)

Φ₊ - Φ₋*g(s) - h(s)




Out[26]:

-3.9968028886505635e-15 + 4.218847493575595e-15im



We now invert the Fourier transform of $\Phi_-(s)$ using Jordan's lemma: $$u(x) = {1 \over 2 \pi} \int_{-\infty + \I \gamma}^{\infty + \I \gamma} \Phi_-(s) \E^{\I s x} \D s = {\alpha \over 2}\Res_{z = 0} {z- \I \alpha \over z^2} \E^{\I z x} = {\alpha \over 2} (1+x \alpha)$$



In [122]:

t = Fun(0 .. 200)

u = α*(1+t*α)/2

x = 0.1

sum(exp(-α*abs(t-x))*u)  - (1 + α*x)




Out[122]:

1.199040866595169e-14



## 4.3

1. From the same logic as 2.2, we know we need to solve $$\Phi_+(s) - g(s) \Phi_-(s) = h(s)$$ where $$g(s) = 1 - {2 \lambda \over s^2 + 1} = {s^2 +1-2 \lambda \over s^2 + 1} = {(s- \I\gamma)(s+\I \gamma) \over (s+\I)(s-\I)}$$ and $$h(s) = {1 \over s^2}$$ where $-1 < \Im s < 0$, let's say $\Im s = \delta$ because I annoyingly used $\gamma$ in the statement of the problem. Writing $s = t + \I \delta$, we see that $$g(s) = {t^2 +2 \I \delta t -\delta^2 +\gamma^2 \over s^2 + 1}$$ By ensuring its real part is positive, this has trivial winding number provided $\gamma^2 = 1 - 2\lambda > 0$, which is true for $0 < \lambda < {1 \over 2}$, and restricting the contour $s$ lives on to be $- {\gamma} < \delta < 0$. Factorizing the kernel we get $$\kappa(z) = \begin{cases} {z+\I \gamma \over z + \I} & \Im z > \delta\\ {z-\I \over z-\I \gamma} & \Im z < \delta \end{cases}$$ Thus we want to solve $$Y_+(s) - Y_-(s) = h(s) \kappa_+(s)^{-1} = {s + \I \over s + \I \gamma} { 1 \over s^2} = {1 \over \gamma s^2} - {\I ( \gamma - 1) \over \gamma^2 s} + {\I \over \gamma^2 }{ \gamma - 1 \over s+ \I \gamma}$$ Which has solution, (since $\delta > - \gamma$), $$Y(z) = \begin{cases} {\I \over \gamma^2 }{ \gamma - 1 \over s+ \I \gamma} & \Im z > \delta\\ {\I ( \gamma - 1) \over \gamma^2 z} - {1 \over \gamma z^2} & \Im z < \delta \end{cases}$$ We thus get $$\Phi_-(z) = ({\I ( \gamma - 1) \over \gamma^2 z} - {1 \over \gamma z^2}) {z-\I \over z-\I \gamma}$$ and Jordan's lemma gives us $$u(x) = {x \over \gamma^2} - \E^{-x \gamma}(\gamma-1)/\gamma^2$$


In [55]:

t = Fun(0 .. 200)
λ = 0.1
γ = sqrt(1-2λ)

u = t/γ^2 - exp(-t*γ)*(γ-1)/γ^2

x = 0.1

u(x) - λ*sum(exp(-abs(t-x))*u) - x




Out[55]:

-3.239075674343894e-14



Oddly, this is definitely a solution, but not in the form the question asked for. To get the other solution, consider now the bad winding number case of $-1 < \delta < - \gamma$. Motivated by 2.2, what if we allow $\kappa$ to have different behaviour? Consider $$\kappa(z) = \begin{cases} {1 \over z + \I} & \Im z > \delta\\ {(z-\I) \over (z-\I \gamma) (z+\I \gamma)} & \Im z < \delta \end{cases}$$ Chosen so that both $\kappa_+$ and $\kappa_+^{-1}$ are analytic.

Thus we want to solve $$Y_+(s) - Y_-(s) = h(s) \kappa_+(s)^{-1} = { s + \I \over s^2} = {1 \over s} + {\I \over s^2}$$ but now we only need $Y_+(s) = O(1)$ and $Y_-(s) = O(1)$. Here is where the non-uniqueness comes in, as we can add an arbitrary constant: $$Y(z) = \begin{cases} A & \Im z > 0 \\ A - {1 \over z} - {\I \over z^2} & \Im z < 0 \end{cases}$$ Thus we have $$\Phi_-(z) = Y_-(z) \kappa_-(z) = -( A + {1 \over z} + {\I \over z^2} ){(z-\I) \over (z-\I \gamma) (z+\I \gamma)}$$ Using Jordan's lemma, and now since $\delta < - \gamma$, we get \begin{align*} u(x) &= \I (\Res_{z = 0} + \Res_{z = \I \gamma} + \Res_{z = - \I \gamma}) \Phi_-(z) \E^{\I x z} \\ &= {x \over \gamma^2} - \E^{-x \gamma}( {\gamma^2-1 \over 2 \gamma^3} + {\gamma-1 \over 2\gamma^3} A) - \E^{x \gamma} ({1-\gamma^2 \over 2 \gamma^3} + {\gamma+1 \over 2\gamma^3} A)\\ &= {x \over \gamma^2} + {\E^{x \gamma} - \E^{-x \gamma} \over 2} {\gamma-\gamma^{-1} \over 2 \gamma^2} - {A \over \gamma^3} ( {\E^{x \gamma} - \E^{-x \gamma} \over 2} + \gamma {\E^{x \gamma} + \E^{-x \gamma} \over 2} ) \end{align*} Redefining $A$ and using the definition of $\sinh$ and $\cosh$ gives the form in the assignment.

What's the moral of the story?

1. Different choices of contours can give different solutions
2. When the winding number is non-trivial, the solution may not be unique

### 4.4

1. Integrating by parts we have \begin{align*} \widehat{u_\rR'}(s) &= \I s \widehat{u_\rR}(s) - u(0) = \I s \widehat{u_\rR}(s) \\ \widehat{u_\rR''}(s) &= \I s \widehat{u_\rR'}(s) - u'(0) = -s^2 \widehat{u_\rR}(s) - u'(0) \end{align*}
2. Our integral equation when cast on the whole real line is: $$u_\rR''(x) - {72 \over 5} \int_{-\infty}^\infty \E^{-5 |x-t|} u_\rR(t)\dt = 1_\rR(x) + p_\rL(x)$$ where $$p(x) = {72 \over 5} \int_{-\infty}^\infty \E^{-5 |x-t|} u_\rR(t)\dt = {72 \over 5} \int_0^\infty \E^{-5 |x-t|} u_\rR(t)\dt.$$ Note that, for $-5 <\Im s < 5$, $$\hat K(s) = {10 \over s^2 + 25}$$ provided $s$ is in the lower half plane, $$\widehat{1_\rR}(s) = \int_0^\infty \E^{-\I s x} \dx = {1 \over \I s}$$ Thus our integral equation in frequency space is \begin{align*} -\alpha - s^2 \widehat{u_\rR}(s) - {72 \over 5} \widehat K(s) \widehat{u_\rR}(s) &=\widehat{p_\rL}(x) + \widehat{1_\rR}(s)\\ \Phi_+(s) - (s^2 + {144 \over s^2+25}) \Phi_-(s) &= \alpha + {1 \over \I s}\\ \Phi_+(s) - {(s^2 + 9) (s^2+16) \over s^2+25} \Phi_-(s) &= \alpha + {1 \over \I s} \end{align*} where $s \in \R +\I \gamma$ for any $-5 < \gamma < 0$.
3. We can factorize this to construct $g(s)$ as $$g(s) = \kappa_+(s) \kappa_-(s)^{-1} = {(s +3 \I) (s+4 \I) \over s+5 \I} {(s -3 \I) (s-4 \I) \over s-5 \I}$$


In [33]:

κ = z -> imag(z) > γ ?
(z+3im)*(z+4im)/(z+5im) :
(z-5im)/((z-3im)*(z-4im))

γ  = -1.0
s = 0.1+γ*im
g = s -> (s^2+9)*(s^2+16)/(s^2+25)

κ(s+eps()im) - g(s)κ(s-eps()im)




Out[33]:

-5.551115123125783e-17 - 2.220446049250313e-16im



Writing $\Phi(z) = \kappa(z) Y(z)$ we get the subtractive RH problem $$Y_+(s) - Y_-(s) = {h(s) \over \kappa_+(s)} = (\alpha + {1 \over \I s}) {s + 5 \I \over (s + 3 \I)(s + 4 \I)}$$ We use partial fraction expansion to write $${h(s) \over \kappa_+(s)} = -{\alpha + 1/4 \over s+4 \I} + {2/3 + 2 \alpha \over s+3 \I} - {5 \over 12 s}$$ Therefore we have $$Y(z) = \begin{cases} -{\alpha + 1/4 \over s+4 \I} + {2/3 + 2 \alpha \over s+3 \I} & 2 \\ {5 \over 12 s} &1 \end{cases}$$ and hence $$\Phi(z) = \begin{cases} {(z+3\I)(z+4\I) \over z+5\I} (-{α+1/4 \over z+4\I} + (2/3 + 2α)/(z+3\I)) & \Im z > \gamma \\ {z-5\I \over (z-3\I)(z-4\I)} {5 \over 12z}& \Im z < \gamma \end{cases}$$

We can now invert the Fourier transform of $$\Phi_-(s) = {s-5\I \over (s-3\I)(s-4\I)} {5 \over 12s}$$ This actually decays so fast that we don't need Jordan's lemma to justify here. This has three poles above our contour, so we sum over each residue to get $$u(x) = \I (\Res_{z = 0} +\Res_{z = 3 \I } +\Res_{z = 4\I} ) \E^{\I z x } {z-5\I \over (z-3\I)(z-4\I)} {5 \over 12z} = -{25 \over 144} - {5 \E^{-4 x} \over 48} + {5 \E^{-3 x} \over 18}$$

Here's we check the solution:



In [42]:

t = Fun(0 .. 200)

u = -25/144 - 5exp(-4t)/48 + 5exp(-3t)/18

x = 1.1
u''(x) - 72/5*sum(exp(-5abs(x-t))*u)




Out[42]:

1.0000000000000018



Here we check the jump of $Y$:



In [45]:

α = u'(0)

h = s -> α + 1/(im*s)

Y = z -> imag(z) > γ ?
(-(α+1/4)/(z+4im) + (2/3 + 2α)/(z+3im)) :
5/(12z)

Y(s+eps()im) - Y(s-eps()im) - h(s)/κ(s + eps()im)




Out[45]:

2.7755575615628914e-17 + 5.551115123125783e-17im



Here we check the jump of $\Phi$:



In [46]:

γ  = -1.0

Φ = z -> imag(z) > γ ?
(z+3im)*(z+4im)/(z+5im) * (-(α+1/4)/(z+4im) + (2/3 + 2α)/(z+3im)) :
(z-5im)/((z-3im)*(z-4im)) * 5/(12z)

Φ(s + eps()*im) - g(s)*Φ(s - eps()*im) - h(s)




Out[46]:

-1.1102230246251565e-16 + 1.3877787807814457e-17im