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

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

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:

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

The Cauchy transform also works:

Therefore the Hilbert transform is given by:

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

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

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:

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:

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):

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

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

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

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 $\I$:

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 $$ h(s) = - \widehat f_{\rm R}(s) = {\I \over s} + {\alpha \over s^2} $$

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 $-1 < \Im s < 0$.

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

We therefore have

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

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:

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 $$ \kappa_+(s) = \kappa_-(s) g(s) $$

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

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

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

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 $$ u(x) = {x \over \gamma^2} - \E^{-x \gamma}(\gamma-1)/\gamma^2 $$

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 $$ g(s) = \kappa_+(s) \kappa_-(s)^{-1} = {(s +3 \I) (s+4 \I) \over s+5 \I} {(s -3 \I) (s-4 \I) \over s-5 \I} $$

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:

Here we check the jump of $Y$:

Here we check the jump of $\Phi$: