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authorcanuel <cattaneo.manuel@hotmail.com>2022-08-23 16:19:51 +0200
committercanuel <cattaneo.manuel@hotmail.com>2022-08-23 16:19:51 +0200
commit3aba88ca005ce951e8052d41d1cb4b448971c3ad (patch)
treebc14c856a0dc1dce64c8dd0550a385ecf24f7f4a /buch/papers
parentkugel: Corrections and normalizations (diff)
downloadSeminarSpezielleFunktionen-3aba88ca005ce951e8052d41d1cb4b448971c3ad.tar.gz
SeminarSpezielleFunktionen-3aba88ca005ce951e8052d41d1cb4b448971c3ad.zip
chapter about recurrence relation of Legendre Associated Functions and Spherical Harmonics
Diffstat (limited to 'buch/papers')
-rw-r--r--buch/papers/kugel/packages.tex4
-rw-r--r--buch/papers/kugel/proofs.tex2
-rw-r--r--buch/papers/kugel/references.bib9
-rw-r--r--buch/papers/kugel/spherical-harmonics.tex166
4 files changed, 169 insertions, 12 deletions
diff --git a/buch/papers/kugel/packages.tex b/buch/papers/kugel/packages.tex
index ead7653..c02589f 100644
--- a/buch/papers/kugel/packages.tex
+++ b/buch/papers/kugel/packages.tex
@@ -16,5 +16,5 @@
\node[gray, anchor = center] at ({#1 / 2}, {#2 / 2}) {\Huge \ttfamily \bfseries TODO};
\end{tikzpicture}}
-\DeclareMathOperator{\sphlaplacian}{\nabla^2_{\mathit{S}}}
-\DeclareMathOperator{\surflaplacian}{\nabla^2_{\partial \mathit{S}}}
+\DeclareMathOperator{\sphlaplacian}{\nabla^2_{S}}
+\DeclareMathOperator{\surflaplacian}{\nabla^2_{\partial S}}
diff --git a/buch/papers/kugel/proofs.tex b/buch/papers/kugel/proofs.tex
index 143caa8..4fbef26 100644
--- a/buch/papers/kugel/proofs.tex
+++ b/buch/papers/kugel/proofs.tex
@@ -166,7 +166,7 @@
\end{proof}
-\begin{lemma}
+\begin{lemma}\label{kugel:lemma:sol_associated_leg_eq}
If $Z_n(z)$ is a solution of the Legendre equation \eqref{kugel:eqn:legendre},
then
\begin{equation*}
diff --git a/buch/papers/kugel/references.bib b/buch/papers/kugel/references.bib
index e5d6452..e3c0f85 100644
--- a/buch/papers/kugel/references.bib
+++ b/buch/papers/kugel/references.bib
@@ -17,6 +17,15 @@
file = {Submitted Version:/Users/npross/Zotero/storage/SN4YUNQC/Carvalhaes and de Barros - 2015 - The surface Laplacian technique in EEG Theory and.pdf:application/pdf},
}
+@article{implementation,
+ title = {New Implementation of Legendre Polynomials for Solving Partial Differential Equations},
+ issn = {272767969},
+ url = {https://www.researchgate.net/publication/272767969_New_Implementation_of_Legendre_Polynomials_for_Solving_Partial_Differential_Equations},
+ shorttitle = {Implementation og Legendre Polynom},
+ date = {2013-12},
+ author = {Ali Davari, Abozar Ahmadi}
+}
+
@video{minutephysics_better_2021,
title = {A Better Way To Picture Atoms},
url = {https://www.youtube.com/watch?v=W2Xb2GFK2yc},
diff --git a/buch/papers/kugel/spherical-harmonics.tex b/buch/papers/kugel/spherical-harmonics.tex
index 72f7402..7dcb461 100644
--- a/buch/papers/kugel/spherical-harmonics.tex
+++ b/buch/papers/kugel/spherical-harmonics.tex
@@ -313,22 +313,20 @@ obtain the \emph{associated Legendre functions}.
The functions
\begin{equation}
P^m_n (z) = (1-z^2)^{\frac{m}{2}}\frac{d^{m}}{dz^{m}} P_n(z)
- = \frac{1}{2^n n!}(1-z^2)^{\frac{m}{2}}\frac{d^{m+n}}{dz^{m+n}}(1-z^2)^n
+ = \frac{1}{2^n n!}(1-z^2)^{\frac{m}{2}}\frac{d^{m+n}}{dz^{m+n}}(1-z^2)^n, \quad |m|<n
\end{equation}
are known as Ferrers or associated Legendre functions.
\end{definition}
+The constraint $|m|<n$, can be justified by considering Eq.\eqref{kugel:eq:associated_leg_func}, in which the derivative of degree $m+n$ is present. A derivative to be well defined must have an order that is greater than zero. Furthermore, it can be seen that this derivative is applied on a polynomial of degree $2n$. As is known from Calculus 1, if you derive a polynomial of degree $2n$ more than $2n$ times, you get zero, which is a trivial solution in which we are not interested.
-\kugeltodo{Discuss $|m| \leq n$.}
-
-\if 0
-The constraint $|m|<n$, can be justified by considering Eq.\eqref{kugel:eq:associated_leg_func}, in which the derivative of degree $m+n$ is present. A derivative to be well defined must have an order that is greater than zero. Furthermore, it can be seen that this derivative is applied on a polynomial of degree $2n$. As is known from Calculus 1, if you derive a polynomial of degree $2n$ more than $2n$ times, you get zero, which is a trivial solution in which we are not interested.\newline
We can thus summarize these two conditions by writing:
\begin{equation*}
\begin{rcases}
m+n \leq 2n &\implies m \leq n \\
m+n \geq 0 &\implies m \geq -n
- \end{rcases} |m| \leq n.
+ \end{rcases} \; |m| \leq n.
\end{equation*}
+\if 0
The set of functions in Eq.\eqref{kugel:eq:sph_harm_0} is named \emph{Spherical Harmonics}, which are the eigenfunctions of the Laplace operator on the \emph{spherical surface domain}, which is exactly what we were looking for at the beginning of this section.
\fi
@@ -345,7 +343,7 @@ domain and combine it with $\Phi(\varphi)$ to get the full result:
\begin{equation*}
f(\vartheta, \varphi)
= \Theta(\vartheta)\Phi(\varphi)
- = P^m_n (\cos \vartheta) e^{im\varphi}.
+ = P^m_n (\cos \vartheta) e^{im\varphi}, \quad |m|<n.
\end{equation*}
This family of functions, which recall are the solutions of the eigenvalue
problem of the surface spherical Laplacian, are the long anticipated
@@ -356,9 +354,9 @@ $Y^m_n(\vartheta, \varphi)$.
\label{kugel:def:spherical-harmonics}
The functions
\begin{equation*}
- Y^m_n (\vartheta, \varphi) = P^m_n(\cos \vartheta) e^{im\varphi},
+ Y^m_n (\vartheta, \varphi) = P^m_n(\cos \vartheta) e^{im\varphi}, \quad |m|<n
\end{equation*}
- where $m, n \in \mathbb{Z}$ and $|m| < n$ are called (unnormalized) spherical
+ where $m, n \in \mathbb{Z}$ are called (unnormalized) spherical
harmonics.
\end{definition}
@@ -645,6 +643,156 @@ harmonics or write $Y^m_n$, we mean the orthonormal spherical harmonics of
definition \ref{kugel:def:spherical-harmonics-orthonormal}.
\subsection{Recurrence Relations}
+The idea of this subsection is to introduce first some recursive relations regarding the Associated Legendre Functions, defined in eq.\eqref{kugel:def:ferrers-functions}. Subsequently we will extend them, in order to derive recurrence formulas for the case of Spherical Harmonic functions as well.
+\subsubsection{Associated Legendre Functions}
+To start this journey, we can first write the following equations, which relate the Associated Legendre functions of different indeces $m$ and $n$ recursively:
+\begin{enumerate}[(i)]
+ \item $(2n+1) x P^m_n(z)= (m+n) P^m_{n-1}(z) + (n-m+1) P^m_{n+1}(z)$, \label{kugel:eq:rec_rel_1}
+ \item $\dfrac{2mz}{\sqrt{1-z^2}} P^m_n(z) = P^{m+1}_n(z) + [n(n+1)-m(m-1)] P^{m-1}_n(z)$, \label{kugel:eq:rec_rel_2}
+ \item $\sqrt{1-z^2} P^m_n(z) = \dfrac{1}{2n+1} \left[ P^{m+1}_{n+1}(z) - P^{m+1}_{n-1}(z) \right]$, \label{kugel:eq:rec_rel_3}
+ \item $\sqrt{1-z^2} P^m_n(z) = \dfrac{1}{2n+1} \left[ (n+m)(n+m-1)P^{m-1}_{n-1}(z) - (n-m+1)(n-m+2)P^{m-1}_{n+1}(z) \right]$. \label{kugel:eq:rec_rel_4}
+\end{enumerate}
+Much of the effort will be proving this bunch of equalities. Then, in the second part, where we will derive the recursion equations for $Y^m_n(\vartheta,\varphi)$, we will basically reuse the ones presented above.
+
+Maybe it is worth mentioning at least one use case for these relations: They are widely used in some software implementations, as they lead to better numerical accuracy and computational cost lower by a factor of six\cite{usecase_recursion}.
+\begin{enumerate}[(i)]
+ \item
+ \begin{proof}
+ This is the relation that links the associated Legendre functions with the same $m$ index but different $n$. Using \ref{} \kugeltodo{ref alla recurrence dei polinomi di legendre (รจ da qualche parte nel libro)}, we have
+ \begin{equation*}
+ (n+1)P_{n+1}(z)-(2n+1)xP_n(z)+nP_{n-1}(z)=0,
+ \end{equation*}
+ that can be differentiated $m$ times, obtaining
+ \begin{equation}\label{kugel:eq:rec_1}
+ (n+1)\frac{d^mP_{n+1}}{dz^m}-(2n+1) \left[z \frac{d^m P_n}{dz^m}+ m\frac{d^{m-1}P_{n-1}}{dz^{m-1}} \right] + n\frac{d^m P_{n-1}}{dz^m}=0.
+ \end{equation}
+ To continue this derivation, we need the following relation:
+ \begin{equation}\label{kugel:eq:rec_2}
+ \frac{dP_{n+1}}{dz} - \frac{dP_{n-1}}{dz} = (2n+1)P_n.
+ \end{equation}
+ The latter will not be derived, because it suffices to use the definition of the Legendre Polynomials $P_n(x)$ to check it.
+
+ We can now differentiate the just presented eq.\eqref{kugel:eq:rec_2} $m-1$ times, that will become
+ \begin{equation}\label{kugel:eq:rec_3}
+ \frac{d^mP_{n+1}}{dx^m} - \frac{d^mP_{n-1}}{dx^m} = (2n+1)\frac{d^{m-1}P_n}{dx^{m-1}}.
+ \end{equation}
+ Then, using eq.\eqref{kugel:eq:rec_3} in eq.\eqref{kugel:eq:rec_1}, we will have
+ \begin{equation}\label{kugel:eq:rec_4}
+ (n+1)\frac{d^mP_{n+1}}{dx^m}- (2n+1)\frac{d^mP_{n+1}}{dx^m} -m\left[\frac{d^m P_{n+1}}{dx^m}+ \frac{d^{m}P_{n-1}}{dx^m}\right] + n\frac{d^m P_{n-1}}{dx^m}=0.
+ \end{equation}
+ Finally, multiplying both sides by $(1-x^2)^{\frac{m}{2}}$ and simplifying the expression, we can rewrite eq.\eqref{kugel:eq:rec_4} in terms of $P^m_n(x)$, namely
+ \begin{equation*}
+ (n+1-m)P^m_{n+1}(x)-(2n+1)xP^m_n(x)+(m+n)P^m_{n-1}(x)=0,
+ \end{equation*}
+ that rearranged, will be
+ \begin{equation*}
+ (2n+1) x P^m_n(x)= (m+n) P^m_{n-1}(x) + (n-m+1) P^m_{n+1}(x).
+ \end{equation*}
+ \end{proof}
+
+ \item
+ \begin{proof}
+ This relation, unlike the previous one, link three expression with the same $n$ index but different $m$.
+
+ In the proof of Lemma \ref{kugel:lemma:sol_associated_leg_eq}, at some point we ran into this expression.
+ \begin{equation*}
+ (1-x^2)\frac{d^{m+2}P_n}{dx^{m+2}} - 2(m+1)x \frac{d^{m+1}P_n}{dx^{m+1}} + [n(n+1)-m(m+1)]\frac{d^mP_n}{dx^m} = 0,
+ \end{equation*}
+ that, if multiplied by $(1-x^2)^{\frac{m}{2}}$, will be
+ \begin{equation*}
+ (1-x^2)^{\frac{m}{2}+1}\frac{d^{m+2}P_n}{dx^{m+2}} - 2(m+1)x (1-x^2)^{\frac{m}{2}}\frac{d^{m+1}P_n}{dx^{m+1}} + [n(n+1)-m(m+1)](1-x^2)^{\frac{m}{2}}\frac{d^mP_n}{dx^m} = 0.
+ \end{equation*}
+ Therefore, as before, expressing it in terms of $P^m_n(x)$:
+ \begin{equation*}
+ P^{m+2}_n(x) - \frac{2(m+1)x}{\sqrt{1-x^2}}P^{m+1}_n(x) + [n(n+1)-m(m+1)]P^m_n(x)=0.
+ \end{equation*}
+ Furthermore, we can adjust the indeces and terms, obtaining
+ \begin{equation*}
+ \frac{2mx}{\sqrt{(1-x^2)}} P^m_n(x) = P^{m+1}_n(x) + [n(n+1)-m(m-1)] P^{m-1}_n(x)
+ \end{equation*}
+
+ \end{proof}
+
+ \item
+ \begin{proof}
+ To derive this expression, we can multiply eq.\eqref{kugel:eq:rec_3} by $(1-x^2)^{\frac{m}{2}}$ and, as always, we could express it in terms of $P^m_n(x)$:
+ \begin{equation*}
+ P^m_{n+1}(x) - P^m_{n-1}(x) = (2n+1)\sqrt{1-x^2}P^{m-1}_n(x).
+ \end{equation*}
+ Afer that we can divide by $2n+1$ resulting in
+ \begin{equation}\label{kugel:eq:helper}
+ \frac{1}{2n+1}[P^m_{n+1}(x) - P^m_{n-1}(x)] = \sqrt{1-x^2}P^{m-1}_n(x).
+ \end{equation}
+ To conclude, we arrange the indeces differently:
+ \begin{equation*}
+ \sqrt{1-x^2}P^{m}_n(x)=\frac{1}{2n+1}[P^{m+1}_{n+1}(x) - P^{m+1}_{n-1}(x)].
+ \end{equation*}
+ \end{proof}
+
+ \item
+ \begin{proof}
+ For this proof we can rely on (\ref{kugel:eq:rec_rel_1}), and therefore rewrite (\ref{kugel:eq:rec_rel_2}) as
+ \begin{equation*}
+ \frac{2m}{(2n+1)\sqrt{1-x^2}} \left[ (m+n)P^m_{n-1}(x) + (n-m+1)P^m_{n+1}(x) \right] = P^{m+1}_n(x) + [ n(n+1)-m(m-1) ]P^{m-1}_n(x).
+ \end{equation*}
+ Rewriting then $P^{m-1}_n(x)$ using eq.\eqref{kugel:eq:helper}, we will have
+ \begin{align*}
+ \frac{2m}{(2n+1)\sqrt{1-x^2}} &\left[ (m+n)P^m_{n-1}(x) + (n-m+1)P^m_{n+1}(x) \right] = P^{m+1}_n(x) \\
+ &+ \frac{n(n+1)-m(m-1)}{(2n+1)\sqrt{1-x^2}} \left[ P^m_{n+1}(x)-P^m_{n-1}(x) \right].
+ \end{align*}
+ The last equation, after some algebric rearrangements, it is easy to show that it is equivalent to
+ \begin{equation*}
+ \sqrt{1-x^2} P^m_n(x) = \dfrac{1}{2n+1} \left[ (n+m)(n+m-1)P^{m-1}_{n-1}(x) - (n-m+1)(n-m+2)P^{m-1}_{n+1}(x) \right]
+ \end{equation*}
+ \end{proof}
+
+\end{enumerate}
+
+\subsubsection{Spherical Harmonics}
+The goal of this subsection's part is to apply the recurrence relations of the $P_n(z)$ functions to the Spherical Harmonics.
+
+With some little adjustments we will be able to have recursion equations for them too. As previously written the most of the work is already done. Now it is only a matter of minor mathematical operations/rearrangements.
+
+We can start by listing all of them:
+\begin{enumerate}[(i)]
+ \item $Y^m_n(\vartheta, \varphi) = \dfrac{1}{(2n+1)\cos \vartheta} \left[ (m+n)Y^m_{n-1}(\vartheta, \varphi) + (m-n+1)Y^m_{n+1}(\vartheta, \varphi) \right]$
+ \begin{proof}
+ We can multiply both sides of equality in eq.\eqref{} by $e^{im \varphi}$ and perform the substitution $z=\cos \vartheta$. After a few simple algebraic steps, we will obtain the relation we are looking for
+ \end{proof}
+ \item $Y^m_n(\vartheta, \varphi) = \dfrac{\tan \vartheta}{2m}\left[ Y^{m+1}_n(\vartheta, \varphi)e^{-i\varphi} + [n(n+1)-m(m-1)]Y^{m-1}_n(\vartheta, \varphi)e^{i\varphi} \right]$
+ \begin{proof}
+ In this proof, as before, we can perform the substitution $z=\cos \vartheta$, and notice that $\sqrt{1-z^2}=\sin \vartheta$, hence, the relation in eq.\eqref{} will be
+ \begin{equation*}
+ \frac{2m \cos \vartheta}{\sin \vartheta} P^m_n(\cos \vartheta) = P^{m+1}_n(\cos \vartheta) + [n(n+1)-m(m-1)]P^{m-1}_n P^m_n(\cos \vartheta).
+ \end{equation*}
+ The latter, multiplied by $e^{im\varphi}$, becomes
+ \begin{align*}
+ \frac{2m \cos \vartheta}{\sin \vartheta} P^m_n(\cos \vartheta)e^{im\varphi} &= P^{m+1}_n(\cos \vartheta)e^{im\varphi} + [n(n+1)-m(m-1)]P^{m-1}_n P^m_n(\cos \vartheta)e^{im\varphi} \\
+ &= P^{m+1}_n(\cos \vartheta)e^{i(m+1)\varphi}e^{-i\varphi} + [n(n+1)-m(m-1)]P^{m-1}_n (\cos \vartheta)e^{i(m-1)\varphi}e^{i\varphi} \\
+ &= Y^{m+1}_n(\vartheta, \varphi)e^{-i\varphi} + [n(n+1)-m(m-1)]Y^{m-1}_n(\vartheta, \varphi)e^{i\varphi} \\
+ \end{align*}
+ Finally, after some ``cleaning''
+ \begin{equation*}
+ Y^m_n(\vartheta, \varphi) = \frac{\tan \vartheta}{2m} \left[ Y^{m+1}_n(\vartheta, \varphi)e^{-i\varphi} + [n(n+1)-m(m-1)]Y^{m-1}_n(\vartheta, \varphi)e^{i\varphi} \right]
+ \end{equation*}
+ \end{proof}
+ \item $Y^m_n(\vartheta, \varphi) = \dfrac{e^{-i\varphi}}{ (2n+1)\sin \vartheta } \left[ Y^{m+1}_{n+1}(\vartheta, \varphi) - Y^{m+1}_{n-1}(\vartheta, \varphi) \right]$
+ \begin{proof}
+ Now we can consider eq.\eqref{}, and multiply it by $e^{im\varphi}$. After the usual substitution $z=\cos \vartheta$, we have
+ \begin{align*}
+ \sin \vartheta P^m_n(\cos \vartheta)e^{im\varphi} &= \dfrac{e^{im\varphi}}{2n+1}\left[ P^{m+1}_{n+1}(\cos \vartheta) - P^{m+1}_{n-1}(\cos \vartheta)\right] \\
+ &= \dfrac{e^{-i\varphi}}{2n+1}\left[ P^{m+1}_{n+1}(\cos \vartheta)e^{i(m+1)\varphi} - P^{m+1}_{n-1}(\cos \vartheta)e^{i(m+1)\varphi}\right] \\
+ \end{align*}
+ A few manipulations later, we will obtain
+ \begin{equation*}
+ Y^m_n(\vartheta, \varphi) = \frac{e^{-i\varphi}}{(2n+1)\sin \vartheta} \left[ Y^{m+1}_{n+1}(\vartheta, \varphi)-Y^{m+1}_{n-1}(\vartheta, \varphi) \right]
+ \end{equation*}
+ \end{proof}
+ \item $Y^m_n(\vartheta, \varphi) = \dfrac{e^{i\varphi}}{(2n+1)\sin \vartheta} \left[ (n+m)(n+m-1)Y^{m-1}_{n-1}(\vartheta, \varphi) - (n-m+1)(n-m+2)Y^{m-1}_{n+1}(\vartheta, \varphi) \right]$
+ \begin{proof}
+ This proof is very similar to the previous one. We just have to perform the substitution $z = \cos \vartheta$, as always. Secondly we can multiply the right side by $e^{im\varphi}$ and the left one too but in a different form, namely $e^{im\varphi}=e^{i(m-1)\varphi}e^{i\varphi}$. Then it is only a question of recalling the definition of $Y^m_n(\vartheta, \varphi)$.
+ \end{proof}
+\end{enumerate}
\section{Series Expansions in $C(S^2)$}