M3M6: Methods of Mathematical Physics 2018

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

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

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

Solution Sheet 3

Problem 1.1

We actually start by showing the second properties of Problem 1.2, for all $\alpha$: $$ {\D \over \dx} \br[x^\alpha \E^{-x} L_n^{(\alpha)}(x)] = {1 \over n!} {\D^{n+1} \over \dx^{n+1}}\br[x^{\alpha+n}\E^{-x}] = (n+1)x^{\alpha-1}\E^{-x} {x^{1-\alpha} \E^{x} \over (n+1)!}{ \D^{n+1} \over \dx^{n+1}}\br[x^{\alpha+n}\E^{-x}] = (n+1) x^{\alpha-1}\E^{-x} L_{n+1}^{(\alpha-1)}(x). $$ Expanding out the derivative we see $$ x^{\alpha-1} \E^{-x}\pr({(\alpha -x)L_n^{(\alpha)}(x) + (L_n^{(\alpha)})'(x)}) = (n+1) x^{\alpha-1}\E^{-x} L_{n+1}^{(\alpha-1)}(x) $$ or in other words $$ (\alpha -x)L_n^{(\alpha)}(x) + (L_n^{(\alpha)})'(x) = (n+1) L_{n+1}^{(\alpha-1)}(x) $$ By induction with the fact $L_0^{(\alpha)}(x) = 0$, we therefore get $$L_n^{(\alpha)}(x) = {(\alpha+1 -x)L_{n-1}^{(\alpha+1)}(x) + (L_n^{(\alpha+1)})'(x) \over n} $$ is a degree $n$ polynomial. We further have that the leading coefficient is \begin{align*} L_n^{(\alpha)}(x) &= -{x \over n} L_{n-1}^{(\alpha+1)}(x) +O(x^{n-1}) = {x^2 \over n(n-1)} L_{n-2}^{(\alpha+2)}(x) +O(x^{n-1}) = \cdots = {(-1)^n x^n \over n!} L_0^{(\alpha+n)}(x) +O(x^{n-1}) \\ &= {(-1)^n x^n \over n!} +O(x^{n-1}) \end{align*}

We now show orthogonality with lower degree polynomials using integration by parts: $$ \int_0^\infty L_n^{(\alpha)}(x) p_m(x) x^\alpha \E^{-x} \dx = \int_0^\infty {1 \over n!} {\D^n \over \dx^n}\br[x^{\alpha+n}\E^{-x}] p_m(x) \dx = (-1)^n \int_0^\infty {1 \over n!} \br[x^{\alpha+n}\E^{-x}] p_m^{(n)}(x) \dx = 0 $$ since $p_m^{(n)}(x) = 0$. Note we use the fact that $$ {\D^k \over \dx^k}\br[x^{\alpha+n}\E^{-x}] = O(x^{\alpha+n-k}) $$ hence vanishes at zero to ignore the boundary terms in integration by parts.

Problem 1.2

We showed the second property as part of 1.1. For the first part, it is clear that we have the correct constant. Now we show orthogonality with all degree $m < n-1$ polynomials (using the fact that $x^{\alpha+1} \E^{-x}$ is zero at $x = 0$): $$ \int_0^\infty {\D L_n^{(\alpha)}(x) \over \dx} p_m(x) x^{\alpha+1} \E^{-x} \dx = -\int_0^\infty L_n^{(\alpha)}(x) (x p_m'(x) + (\alpha+1) p_m - x p_m) x^{\alpha} \E^{-x} \dx = 0 $$ since $(x p_m'(x) + (\alpha+1) p_m - x p_m) $ is degree $m+1 < n$.

For the third part, use the product rule on the last derivative: \begin{align*} (n+1) L_{n+1}^{(\alpha)}(x) &= {x^{-\alpha}\E^x \over n!} {\D^{n} \over \dx^{n}} {\D \over \dx} \br[x^{\alpha+n+1} \E^{-x}] \\ &= {x^{-\alpha}\E^x \over n!} {\D^{n} \over \dx^{n}} \br[(\alpha+n+1)x^{\alpha+n} \E^{-x}-x^{\alpha+n+1} \E^{-x}] \\ &= (\alpha+n+1)L_n^{(\alpha)}(x) - xL_n^{(\alpha+1)}(x) \end{align*} For the last result, we apply the product rule $n$ times: \begin{align*} L_{n}^{(\alpha+1)}(x) &= {x^{-1-\alpha}\E^x \over n!} {\D^{n} \over \dx^{n}} {\D \over \dx} \br[x x^{\alpha+n} \E^{-x}] \\ &= {x^{-1-\alpha}\E^x \over n!} {\D^{n-1} \over \dx^{n-1}} \br[ x^{\alpha+n} \E^{-x}] + {x^{-1-\alpha}\E^x \over n!} {\D^{n-1} \over \dx^{n-1}} x {\D \over \dx} \br[ x^{\alpha+n} \E^{-x}] \\ &= {2 \over n}L_{n-1}^{(\alpha+1)}(x) + {x^{-1-\alpha}\E^x \over n!} {\D^{n-2} \over \dx^{n-2}} x {\D^2 \over \dx^2} \br[ x^{\alpha+n} \E^{-x}] \\ &= {3 \over n}L_{n-1}^{(\alpha+1)}(x) + {x^{-1-\alpha}\E^x \over n!} {\D^{n-3} \over \dx^{n-3}} x {\D^3 \over \dx^3} \br[ x^{\alpha+n} \E^{-x}] \\ &\vdots\\ &={n \over n}L_{n-1}^{(\alpha+1)}(x) + {x^{-\alpha}\E^x \over n!} {\D^{n} \over \dx^{n}} \br[ x^{\alpha+n} \E^{-x}] \\ &=L_{n-1}^{(\alpha+1)}(x) + L_n^{(\alpha)}(x) \end{align*}

Problem 1.3

Note that relationship 3 above did not depend on $\alpha >-1$. We therefore have from 1.2, comibing property (3) and (4), \begin{align*} x L_n^{(\alpha)}(x) &= -(n+1)L_{n+1}^{(\alpha-1)}(x) +(n+\alpha)L_n^{(\alpha-1)}(x) \\ &= -(n+1)L_{n+1}^{(\alpha)}(x) + (n+1)L_{n}^{(\alpha)}(x) +(n+\alpha)L_{n}^{(\alpha)}(x) - (n+\alpha)L_{n-1}^{(\alpha)}(x) \\ &= - (n+\alpha)L_{n-1}^{(\alpha)}(x) + (2n+\alpha+1) L_n^{(\alpha)}(x) -(n+1)L_{n+1}^{(\alpha)}(x) \end{align*}

The Jacobi operator therefore has the form $$ x \begin{pmatrix} L_0^{(\alpha)}(x) \\ L_1^{(\alpha)}(x) \\ \vdots \end{pmatrix} = \begin{pmatrix} \alpha+1 &-1 \\ -1-\alpha & \alpha+3 & -2 \\ & -2-\alpha & \alpha+5 &-3 \\ & & -3-\alpha & \alpha+7 &-4 \\ && & -4-\alpha & \alpha+9 &\ddots \\ &&&&\ddots & \ddots \end{pmatrix} \begin{pmatrix} L_0^{(\alpha)}(x) \\ L_1^{(\alpha)}(x) \\ \vdots \end{pmatrix} $$

Demonstration The following command creates the Jacobi operators for a Laguerre polynomial with α = 1:

Problem 2.1

Note that $$ {\D \over \dx} \E^{-x/2} u(x) = \E^{-x/2} (-{u(x) \over 2} +u'(x)) $$ Thus $$ {\D \over \dx} \E^{-x/2} u(x) = \E^{-x/2} (u'(x) -{u(x) \over 2} - x u(x)) $$

We have the derivative operator from $L_k(x)$ to $L_k^{(1)}(x)$ as: $$ D = \begin{pmatrix} 0 & -1 \\ &&-1 \\ &&&\ddots \end{pmatrix} $$ and the Multiplication operator for $\alpha = 0$ (from Problem 1.3) $$ J^\top = \begin{pmatrix} 1 &-1\\ -1 & 3 &-2\\ &\ddots & \ddots & \ddots \end{pmatrix} $$ and the conversion operator (from Problem 1.2 property 4) $$ S = \begin{pmatrix} 1 & -1 \\ & 1 & -1 \\&&\ddots & \ddots \end{pmatrix} $$ We thus have multiplication by $x$ from basis to the other as $$ S J^\top = \begin{pmatrix} 2 & -4 & 2 \\ -1 & 5 & -7 & 3 \\ & -2 & 8 & -10 & 4\\ &&\ddots&\ddots&\ddots&\ddots \end{pmatrix} $$ Putting everything together, we get the operator

Problem 2.2

We have \begin{align*} {\E^x \over x^\alpha} {\D \over \dx} \br[x^{\alpha+1} \E^{-x} {\D L_n^{(\alpha)} \over \dx}] &= -{\E^x \over x^\alpha} {\D \over \dx} \br[x^{\alpha+1} \E^{-x} { L_{n-1}^{(\alpha+1)} \over \dx}] \\ &= -n L_{n}^{(\alpha)}(x) \end{align*} Therefore $\lambda_n = -n$. This can be expanded in the form: $$ x {\D^2 L_n^{(\alpha)} \over \dx^2} + (\alpha+1 - x) {\D L_n^{(\alpha)} \over \dx} = -n L_n^{(\alpha)}(x) $$

Problem 3.1

Define $C_k^{(\alpha)}(z) = \CC[L_k^{(\alpha)} \diamond^\alpha \E^{-\diamond}](z)$ and recall that \begin{align*} C_1(z) &= {{1 \over 2 \pi \I} \int_0^\infty \E^{-x} \dx + (z-a_0) C_0(z) \over b_0} = -{1 \over 2 \pi \I} - (z-1) C_0(z) \\ &= {(z-1) \E^{-z} \Ei z -1 \over 2 \pi \I} \end{align*}

Here we double check the formula, noting that $L_1(x) = \E^x {\D \over \dx} x \E^{-x} = 1 - x$:

We now use these to determine the results with $\alpha = 1$. Note that: $$ C_0^{(1)}(z) = \CC[\diamond \E^{-\diamond}](z) =C_0(z) - C_1(z) = { \E^{-z} \Ei z - (z-1) \E^{-z} \Ei z +1 \over 2 \pi \I} $$

Therefore, we have \begin{align*} C_1^{(1)}(z) &= {{1 \over 2 \pi \I} \int_0^\infty x \E^{-x} \dx + (z-a_0^{(1)}) C_0^{(1)}(z) \over b_0^{(1)}} \\ &= {{1 \over 2 \pi \I} + (z-2) C_0^{(1)}(z) \over -1}\\ &= {1 + (z-2) (\E^{-z} \Ei z - (z-1) \E^{-z} \Ei z +1) \over -2 \pi \I} \end{align*}

Let's check the result using $$ L_1^{(1)}(x) = x^{-1} \E^x {\D \over \dx} x^2 \E^{-x} = 2 - x $$

Problem 3.2

We have $$ \int_x^\infty L_2(x) \E^{-x} \dx = {1 \over 2} x \E^{-x} L_1^{(1)}(x) \E^{-x} $$ Thus from lectures we have $$ {1 \over 2 \pi\I} \int_0^\infty L_2(x) \E^{-x} \log(z-x) \dx = {1 \over 2}\CC[ \diamond \E^{-\diamond} L_1^{(1)}](z) $$ and therefore $$ {1 \over \pi} \int_0^\infty L_2(x) \E^{-x} \log|z-x| \dx = -\Im \CC[ \diamond \E^{-\diamond} L_1^{(1)}](z) = \Re{1 + (z-2) (\E^{-z} \Ei z - (z-1) \E^{-z} \Ei z +1) \over -2 \pi } $$ Let's check the result:

Problem 4.1

Consider integration contours $\gamma_{+x}$ and $\gamma_{-x}$ that avoid $0$ above and below:

So that $$ \Gamma_\pm(\alpha, x) = \int_{\gamma_{\pm x}} \zeta^{\alpha-1} \E^{-\zeta} \D \zeta $$ Note that $$ \int_x^{-r} (\zeta_+^{\alpha - 1} - \zeta_-^{\alpha -1}) \E^{-\zeta} \D\zeta = 0 $$ since $\zeta_+^{\alpha-1} = \E^{\pi \I (\alpha-1)}|\zeta|^{\alpha-1} = \E^{2 \I \pi \alpha} \zeta_-^{\alpha-1}$. Furthermore, the integrals over the arcs tend to zero as $r \rightarrow 0$: $$ |\I r^\alpha \int_0^\pi \E^{- r \E^{\I \theta}} \E^{\I \theta \alpha} \D \theta | \leq r^\alpha \pi \E^r \rightarrow 0 $$ and similarly on the lower arc. Thus we have $$ \Gamma_+(\alpha, x)-\E^{2 \I \pi \alpha} \Gamma_-(\alpha, x) = \lim_{r \rightarrow 0 } \left(\int_{\gamma_{+x}} - \E^{2 \I \pi \alpha} \int_{\gamma_{-x}}\right) \zeta^{\alpha-1} \E^{-\zeta} \D \zeta = (1 - \E^{2 \I \pi \alpha})\int_0^\infty x^{\alpha-1} \E^{-x} \dx = (1 - \E^{2 \I \pi \alpha})\Gamma(\alpha) $$

Problem 4.2

Note that, for $0 < \alpha < 1$, $$ \psi(z) = z^{-\alpha} \E^z \Gamma(\alpha, z) $$ has the following properties:

1. $\psi(z)$ decays as $z \rightarrow \infty$, via integration by parts: $$ z^{-\alpha} \E^z \int_z^\infty \zeta^{\alpha-1} \E^{-\zeta} \D \zeta = z^{-1} \E^z + z^{-\alpha} \int_z^\infty \zeta^{\alpha-2} \E^{z-\zeta} \D \zeta $$ and we have assuming $z$ is bounded away from the negative real axis: $$ |\int_z^\infty \zeta^{\alpha-2} \E^{z-\zeta} \D \zeta| \leq \int_z^\infty |\zeta|^{\alpha-2} \D \zeta = \int_0^\infty |x+z|^{\alpha-2} \D x < \infty $$ (otherwise one would use a deformed contour).
2. We have the subtractive jump: $$\psi_+(x) - \psi_-(x) = \E^x(x_+^{-\alpha} \Gamma_+(\alpha, x) - \Gamma_-(\alpha, x)) = \E^x |x|^\alpha (\E^{-\I \pi \alpha} \Gamma_+(\alpha, z) - \E^{\I \pi \alpha} \Gamma_-(\alpha,x)) = \E^x |x|^\alpha \E^{-\I \pi \alpha} (1-\E^{2 \I \pi \alpha}) $$

We use these properties to verify that $$ \CC[\diamond^\alpha \E^{-\diamond}](z) = {1 \over \Gamma(-\alpha)} {(-z)^\alpha \E^{-z} \Gamma(-\alpha, - z) \over   \E^{-\I\pi\alpha} - \E^{\I\pi\alpha}} $$ via Plemelj.