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| -rw-r--r-- | buch/papers/kugel/packages.tex | 4 | ||||
| -rw-r--r-- | buch/papers/kugel/proofs.tex | 2 | ||||
| -rw-r--r-- | buch/papers/kugel/references.bib | 9 | ||||
| -rw-r--r-- | buch/papers/kugel/spherical-harmonics.tex | 166 |
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 $C(S^2)$} |