$$\def\E{{\rm e}} \def\I{{\rm i}} \def\D{{\rm d}}$$

Master Solutions (2018)¶

Problem 1.2¶

Following the mastery material, we know that we can re-cast the Cauchy transform in terms of a two-sheeted version of the Cauchy transform. In particular, define

$$\begin{align*} C^1 u(z) &= {C u(z) + i \sqrt{-z} C[u/\sqrt{\diamond}](z) \over 2} \\ C^2 u(z) &= {C u(z) - i \sqrt{-z} C[u/\sqrt{\diamond}](z) \over 2} \end{align*}$$ This satisfies a 2-sheeted version of Plemelj, where the sheets are cut along $[0,\infty)$. From sheet 1 to sheet 2 we have the right jump: $$C_+^1 u(x) - C_-^2 u(x) = u(x)$$ From sheet 2 back to sheet 1 we have continuity/analyticity: $$C_+^2 u(x) - C_-^1 u(x) = 0$$

By summing over all inverses of the map $z \mapsto z^{3/2}$ we can recover the Cauchy transform, with some care on which sheet we map to. In the upper-half plane, $z^{2/3}$ and $\E^{4 \pi \I \over 3} z^{2/3}$ are still on the first sheet as the phase has not crossed $2\pi$, while the third inverse $\E^{8 \pi \I \over 3} z^{2/3}$ lives on the second sheet. Therefore, for $g(x) = f(x^3/2)$ we have

$$C f(z) = C_1 g (z^{2/3}) + C_1 g (\E^{4 \pi \I \over 3} z^{2/3}) + C_2 g(\E^{8 \pi \I \over 3} z^{2/3})$$

As we pass to the lower half plane over the negative real axis, the second root passes to the second sheet. However, simealtaneously the roots swap plasses: for $x < 0$ we have for $\lambda_1(z) := z^{2/3}$, $\lambda_2(z) := \E^{4 \I \pi/3} z^{2/3}$, $\lambda_3(z) = \E^{8 \I \pi /3} z^{2/3}$:

$$\begin{align*} \lambda_1^+(x) = x_+^{2/3} &= \E^{4 \I \pi/3} x_-^{2/3} = \lambda_2^-(x) \\ \lambda_2^+(x) =\E^{4 \I \pi/3} x_+^{2/3} &= \E^{8 \I \pi/3} x_-^{2/3} = \lambda_3^-(x) \\ \lambda_3^+(x) = \E^{8 \I \pi/3} x_+^{2/3} &= \ x_-^{2/3} = \lambda_1^-(x) \end{align*}$$

and we therefore get:

$$C f(z) = C_1 g (\E^{4 \I \pi/3} z^{2/3}) + C_2 g (\E^{8 \pi \I \over 3} z^{2/3}) + C_2 g( z^{2/3})$$

In other words, in terms of $g$ and $h(x) = g(x)/\sqrt(x)$ we have:

$$C f(z) = {1 \over 2} \begin{cases} C g (\lambda_1(z)) + \I \sqrt{-\lambda_1(z)} C h(\lambda_1(z)) + C g (\lambda_2(z)) + \I \sqrt{-\lambda_2(z)} C h(\lambda_2(z)) + C g (\lambda_3(z)) - \I \sqrt{-\lambda_3(z))} C h(\lambda_3(z))) & \Im z > 0 \\ C g (\lambda_1(z)) - \I \sqrt{-\lambda_1(z)} C h(\lambda_1(z)) + C g (\lambda_2(z)) + \I \sqrt{-\lambda_2(z)} C h(\lambda_2(z)) + C g (\lambda_3(z)) - \I \sqrt{-\lambda_3(z))} C h(\lambda_3(z))) & \Im z < 0 \\ \end{cases}$$ We can verify this satisfies Plemel.

1. It decays at infinity because each of the Cauchy transforms decay
2. It has weaker than pole singularities Because $C h(z)$ has at worse a $|z|^{-1/2}$ singularity, hence $\sqrt{-z} C h(z^{2/3})$ has at worse a $|z|^{1/6}$ singularity
3. It is analytic off $[0,\infty)$. For $x < 0$ we have, noting that $\lambda_2(x)^{2/3} = \E^{4 \I \pi /3} * \E^{2 \I \pi /3} |x|^{2/3} = |x|^{2/3}$ is right on the branch cut of $\sqrt{-z}$:

$$\begin{align*} 2C_+ f(x) &= C g (\lambda_1^+(x)) + \I \sqrt{-\lambda_1^+(x)} C h(\lambda_1^+(x)) + C g (\lambda_2^+(x)) + \I \sqrt{-\lambda_2^+(x)} C h(\lambda_2^+(x)) + C g (\lambda_3^+(x)) - \I \sqrt{-\lambda_3^+(x))} C h(\lambda_3^+(x))) \\ & = C g (\lambda_2^-(x)) + \I \sqrt{-\lambda_2^-(x)} C h(\lambda_2^-(x)) + C g (\lambda_3^-(x)) - \I \sqrt{-\lambda_3^-(x)} C h(\lambda_3^-(x)) + C g (\lambda_1^-(x)) - \I \sqrt{-\lambda_1^-(x))} C h(\lambda_1^-(x))) \\ &= 2 C_- f(x) \end{align*}$$

4. It has the right jump on $[0,\infty)$: for $x>0$ we have, noting only $\lambda_1(x)$ is in $[0,\infty)$ and other terms cancel:

$$2C_+ f(x) - 2C_-f(x) = (C^+ - C^-) g(\lambda_1(x)) + \sqrt{x} C h(\lambda_1(x)) = 2g(x^{2/3}) = 2f(x)$$

Problem 1.2¶

We specialize the above for $f(x) = \E^{-x^{2/3}}$ so that $g(x) = \E^{-x}$ and $h(x) = \E^{-x} / \sqrt{x}$. From Problem Sheet 3, Problem 4.2 we have for $0 < \alpha < 1$

$$C[\diamond^\alpha \E^{-\diamond}](z) = {1 \over \Gamma(-\alpha)} {(-z)^\alpha \E^{-z} \Gamma(-\alpha, - z) \over   \E^{-\I\pi\alpha} - \E^{\I\pi\alpha}}$$ Integrating by parts we can relate $\Gamma(a,z)$ to $\Gamma(a+1,z)$, that is: $$\Gamma(a,z) = \int_z^\infty \zeta^{a-1} \E^{-\zeta} \D \zeta = {1 \over a} \int_z^\infty {\D \over \D \zeta}[\zeta^{a}] \E^{-\zeta} \D \zeta = -{z^a \E^{-z} \over a} + {1 \over a} \int_z^ \infty \zeta^a \E^{-\zeta} \D \zeta = {\Gamma(a+1,z) - z^a \E^{-z} \over a}$$ This allows us to deduce the above formula also works for $\alpha = -1/2$, in other words (using $\Gamma(1/2) = \sqrt{\pi})$, $$C h(z) = C[\diamond^{-1/2} \E^{-\diamond}](z) = {1 \over \sqrt{\pi}} {(-z)^(-1/2) \E^{-z} \Gamma(1/2, - z) \over 2 \I}$$

For $\alpha = 0$, we use the alternative for $x < 0$

$$\Gamma(0,x_+) - \Gamma(0,x_-) = -\oint {\E^{-\zeta} \over \zeta} \D \zeta = -2 \pi \I$$ implying that $$C g(z) = C[\E^{-\diamond}](z) = {\E^{-z} \Gamma(0,-z) \over - 2 \pi \I}$$