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diff --git a/buch/papers/kugel/spherical-harmonics.tex b/buch/papers/kugel/spherical-harmonics.tex index 9349b61..b540531 100644 --- a/buch/papers/kugel/spherical-harmonics.tex +++ b/buch/papers/kugel/spherical-harmonics.tex @@ -33,16 +33,6 @@ that mathematics offers us. \subsection{Eigenvalue Problem} \label{kugel:sec:construction:eigenvalue} -\begin{figure} - \centering - \includegraphics{papers/kugel/figures/tikz/spherical-coordinates} - \caption{ - Spherical coordinate system. Space is described with the free variables $r - \in \mathbb{R}_0^+$, $\vartheta \in [0; \pi]$ and $\varphi \in [0; 2\pi)$. - \label{kugel:fig:spherical-coordinates} - } -\end{figure} - From Section \ref{buch:pde:section:kugel}, we know that the spherical Laplacian in the spherical coordinate system (shown in Figure \ref{kugel:fig:spherical-coordinates}) is is defined as @@ -80,16 +70,17 @@ that deserves its own name. is called the surface spherical Laplacian. \end{definition} -In the definition, the subscript ``$\partial S$'' was used to emphasize the -fact that we are on the spherical surface, which can be understood as being the -boundary of the sphere. But what does it actually do? To get an intuition, -first of all, notice the fact that $\surflaplacian$ have second derivatives, -which means that this a measure of \emph{curvature}; But curvature of what? To -get an even stronger intuition we will go into geometry, were curvature can be -grasped very well visually. Consider figure \ref{kugel:fig:curvature} where the -curvature is shown using colors. First we have the curvature of a curve in 1D, -then the curvature of a surface (2D), and finally the curvature of a function on -the surface of the unit sphere. +In the definition, the subscript ``$\partial S$'' was used to emphasize the fact +that we are on the spherical surface, which can be understood as being the +boundary of the sphere. But what does it actually do? To get an intuition, first +of all, notice the fact that $\surflaplacian$ have second derivatives, which +means that this a measure of \emph{curvature}; But curvature of what? To get an +even stronger intuition we will go into geometry, were curvature can be grasped +very well visually. Consider figure \ref{kugel:fig:curvature} where the +curvature is shown using colors: positive curvature in red, and negative +curvature in blue. First we have the curvature of a curve in 1D, then the +curvature of a surface (2D), and finally the curvature of a function on the +surface of the unit sphere. \begin{figure} \centering @@ -111,12 +102,12 @@ that satisfy the equation \surflaplacian f = -\lambda f. \end{equation} Perhaps it may not be obvious at first glance, but we are in fact dealing with a -partial differential equation (PDE)\footnote{ - Considering the fact that we are dealing with a PDE, - you may be wondering what are the boundary conditions. Well, since this eigenvalue problem is been developed on - the spherical surface (boundary of a sphere), the boundary in this case are empty, i.e no boundary condition has to be considered.}. -unpack the notation of the operator $\nabla^2_{\partial S}$ according to -definition +partial differential equation (PDE)\footnote{Considering the fact that we are +dealing with a PDE, you may be wondering what are the boundary conditions. Well, +since this eigenvalue problem is been developed on the spherical surface +(boundary of a sphere), the boundary in this case are empty, i.e no boundary +condition has to be considered.}. If we unpack the notation of the operator +$\nabla^2_{\partial S}$ according to definition \ref{kugel:def:surface-laplacian}, we get: \begin{equation} \label{kugel:eqn:eigen-pde} \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( @@ -189,23 +180,35 @@ require a dedicated section of its own. \begin{figure} \centering - \kugelplaceholderfig{.8\textwidth}{5cm} + \subfigure[Spherical coordinates. \label{kugel:fig:spherical-coordinates}] + {\includegraphics[width=.45\linewidth]{papers/kugel/figures/tikz/spherical-coordinates}} + \qquad + \subfigure[Substitution. \label{kugel:fig:legendre-substitution}] + {\includegraphics[ + width=.45\linewidth + % scale = 1.2, + % trim = 0 40 0 0, clip, + ]{papers/kugel/figures/tikz/legendre-substitution}} \caption{ - \kugeltodo{Why $z = \cos \vartheta$.} + (a) Spherical coordinate system. Space is described with the free variables + $r \in \mathbb{R}_0^+$, $\vartheta \in [0; \pi]$ and $\varphi \in [0; + 2\pi)$. (b) Geometrical intuition for the substitution to obtain the + Legendre equation. } \end{figure} To solve \eqref{kugel:eqn:ode-theta} we start with the substitution $z = \cos -\vartheta$ \kugeltodo{Explain geometric origin with picture}. The operator -$\frac{d}{d \vartheta}$ becomes +\vartheta$, which has a neat geometrical justification. To see it consider +figure \ref{kugel:fig:legendre-substitution}, where we sketched the geometry of +the problem. Algebraically, the operator $\frac{d}{d \vartheta}$ becomes \begin{equation*} \frac{d}{d \vartheta} = \frac{dz}{d \vartheta}\frac{d}{dz} = -\sin \vartheta \frac{d}{dz} = -\sqrt{1-z^2} \frac{d}{dz}, \end{equation*} -since $\sin \vartheta = \sqrt{1 - \cos^2 \vartheta} = \sqrt{1 - z^2}$, and -then \eqref{kugel:eqn:ode-theta} becomes +since $\sin \vartheta = \sqrt{1 - \cos^2 \vartheta} = \sqrt{1 - z^2}$, which +agrees with our sketch. Thus, \eqref{kugel:eqn:ode-theta} becomes \begin{align*} \frac{-\sqrt{1-z^2}}{\sqrt{1-z^2}} \frac{d}{dz} \left[ \left(\sqrt{1-z^2}\right) \left(-\sqrt{1-z^2}\right) \frac{d \Theta}{dz} @@ -236,8 +239,10 @@ can perform the substitution backwards and get back to our eigenvalue problem. However, the associated Legendre equation is not any easier, so to attack the problem we will look for the solutions in the easier special case when $m = 0$. This reduces the problem because it removes the double pole, which is always -tricky to deal with. In fact, the reduced problem when $m = 0$ is known as the -\emph{Legendre equation}: +tricky to deal with\footnote{It however to notice that the differential equation +is still singular because of the factor in front of the second derivative of +$Z$}. In fact, the reduced problem when $m = 0$ is known as the \emph{Legendre +equation}: \begin{equation} \label{kugel:eqn:legendre} (1 - z^2)\frac{d^2 Z}{dz^2} - 2z\frac{d Z}{dz} @@ -316,11 +321,19 @@ 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, \quad |m|<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}, where we differentiate $m+n$ times. We all know that a differentiation, to be well defined, must have an order that is greater than zero \kugeltodo{is that always true?}. 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, that would be a trivial solution. This is the power of zero: It is almost always a (boring) solution. +The constraint $|m|<n$, can be justified by considering equation +\eqref{kugel:eq:associated_leg_func}, where we differentiate $m+n$ times. We all +know that a differentiation, 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, that would be a +trivial solution. This is the power of zero: It is almost always a (boring) +solution. We can thus summarize these two conditions by writing: \begin{equation*} @@ -337,8 +350,8 @@ section \ref{kugel:sec:construction:eigenvalue}. We had left off in the middle of the separation, were we had used the Ansatz $f(\vartheta, \varphi) = \Theta(\vartheta) \Phi(\varphi)$ to find that $\Phi(\varphi) = e^{im\varphi}$, and we were solving for $\Theta(\vartheta)$. As you may recall, previously we -performed the substitution $z = \cos \vartheta$. Now we can finally bring back the -solution to the associated Legendre equation $P^m_n(z)$ into the $\vartheta$ +performed the substitution $z = \cos \vartheta$. Now we can finally bring back +the solution to the associated Legendre equation $P^m_n(z)$ into the $\vartheta$ domain and combine it with $\Phi(\varphi)$ to get the full result: \begin{equation*} f(\vartheta, \varphi) @@ -534,6 +547,15 @@ product: \end{theorem} \begin{proof} We will begin by doing a bit of algebraic maipulaiton: + \footnote{ + Essentially, what we just did was to turn + \eqref{kugel:eq:spherical-harmonics-inner-prod} in this form: + \( + \langle Y^m_n, Y^{m'}_{n'} \rangle_{\partial S} + = \langle P^m_n, P^{m'}_{n'} \rangle_z + \; \langle e^{im\varphi}, e^{-im'\varphi} \rangle_\varphi + \). + } \begin{align*} \int_{0}^\pi \int_0^{2\pi} Y^m_n(\vartheta, \varphi) \overline{Y^{m'}_{n'}(\vartheta, \varphi)} @@ -561,17 +583,12 @@ product: \end{equation*} where in the second step we performed the substitution $z = \cos\vartheta$; $d\vartheta = \frac{d\vartheta}{dz} dz= - dz / \sin \vartheta$, and then we - used lemma \ref{kugel:thm:associated-legendre-ortho}. - We are allowed to use - the lemma because $m = m'$. After the just mentioned substitution we can write eq.\eqref{kugel:eq:spherical-harmonics-inner-prod} in this form - \begin{equation*} - \langle Y^m_n, Y^{m'}_{n'} \rangle_{\partial S} = \langle P^m_n, P^{m'}_{n'} \rangle_z \; \langle e^{im\varphi}, e^{-im'\varphi} \rangle_\varphi. - \end{equation*} - Now we just need look at the case when $m \neq m'$. Fortunately this is - easier: the inner integral is $\int_0^{2\pi} e^{i(m - m')\varphi} d\varphi$, - or in other words we are integrating a complex exponential over the entire - period, which always results in zero. Thus, we do not need to do anything and - the proof is complete. + used lemma \ref{kugel:thm:associated-legendre-ortho}. We are allowed to use + the lemma because $m = m'$. Now we just need look at the case when $m \neq + m'$. Fortunately this is easier: the inner integral is $\int_0^{2\pi} e^{i(m - + m')\varphi} d\varphi$, or in other words we are integrating a complex + exponential over the entire period, which always results in zero. Thus, we do + not need to do anything and the proof is complete. \end{proof} These proofs for the various orthogonality relations were quite long and @@ -619,10 +636,12 @@ regrettably sometimes even ourselves, would write instead: reader. \end{proof} -Lemma \ref{kugel:thm:legendre-poly-ortho} has a very similar proof, while the theorem \ref{kugel:thm:spherical-harmonics-ortho} for the spherical harmonics is proved by the following argument. -The spherical harmonics are the solutions to the eigenvalue problem $\surflaplacian f = -\lambda f$, -which as discussed in the previous section is solved using the separation Ansatz. So to -prove their orthogonality using the Sturm-Liouville theory we argue that +Lemma \ref{kugel:thm:legendre-poly-ortho} has a very similar proof, while the +theorem \ref{kugel:thm:spherical-harmonics-ortho} for the spherical harmonics is +proved by the following argument. The spherical harmonics are the solutions to +the eigenvalue problem $\surflaplacian f = -\lambda f$, which as discussed in +the previous section is solved using the separation Ansatz. So to prove their +orthogonality using the Sturm-Liouville theory we argue that \begin{equation*} \surflaplacian = L_\vartheta L_\varphi \iff \surflaplacian f(\vartheta, \varphi) @@ -639,8 +658,8 @@ At this point we have shown that the spherical harmonics form an orthogonal system, but in many applications we usually also want a normalization of some kind. For example the most obvious desirable property could be for the spherical harmonics to be ortho\emph{normal}, by which we mean that $\langle Y^m_n, -Y^{m'}_{n'} \rangle = \delta_{nn'}$. To obtain orthonormality, we simply add an -ugly normalization factor in front of the previous definition +Y^{m'}_{n'} \rangle = \delta_{nn'} \delta_{mm'}$. To obtain orthonormality, we +simply add an ugly normalization factor in front of the previous definition \ref{kugel:def:spherical-harmonics} as follows. \begin{definition}[Orthonormal spherical harmonics] @@ -677,204 +696,314 @@ is a so called Condon-Shortley phase factor $(-1)^m$ in front of the square root in the definition of the normalized spherical harmonics. It is yet another normalization that is added for physical reasons that are not very relevant to our discussion, but we mention this potential source of confusion since many -numerical packages (such as \texttt{SHTOOLS} \kugeltodo{Reference}) offer an -option to add or remove it from the computation. +numerical packages (such as \texttt{SHTOOLS} +\cite{markwieczorek_shtoolsshtools_2022}) offer an option to add or remove it +from the computation. Though, for our purposes we will mostly only need the orthonormal spherical harmonics, so from now on, unless specified otherwise when we say spherical harmonics or write $Y^m_n$, we mean the orthonormal spherical harmonics of definition \ref{kugel:def:spherical-harmonics-orthonormal}. -\subsection{Recurrence Relations}\kugeltodo{replace x with z} -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. +\subsection{Recurrence Relations} + +The idea of this subsection is to introduce first some recursive relations +regarding the associated Legendre functions, defined in equation +\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: + +To start this journey, we can first write the following equations, which relate +the associated Legendre functions of different indices $m$ and $n$ recursively: \begin{subequations} \begin{align} - P^m_n(z) &= \dfrac{1}{(2n+1)x} \left[ (m+n) P^m_{n-1}(z) + (n-m+1) P^m_{n+1}(z) \right] \label{kugel:eq:rec-leg-1} \\ - P^m_n(z) &= \dfrac{\sqrt{1-z^2}}{2mz} \left[ P^{m+1}_n(z) + [n(n+1)-m(m-1)] P^{m-1}_n(z) \right] \label{kugel:eq:rec-leg-2} \\ - P^m_n(z) &= \dfrac{1}{(2n+1)\sqrt{1-z^2}} \left[ P^{m+1}_{n+1}(z) - P^{m+1}_{n-1}(z) \right] \label{kugel:eq:rec-leg-3} \\ - P^m_n(z) &= \dfrac{1}{(2n+1)\sqrt{1-z^2}} \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-leg-4} + P^m_n(z) &= \dfrac{1}{(2n+1)x} \left[ + (m+n) P^m_{n-1}(z) + (n-m+1) P^m_{n+1}(z) + \right] \label{kugel:eqn:rec-leg-1} \\ + P^m_n(z) &= \dfrac{\sqrt{1-z^2}}{2mz} \left[ + P^{m+1}_n(z) + [n(n+1)-m(m-1)] P^{m-1}_n(z) + \right] \label{kugel:eqn:rec-leg-2} \\ + P^m_n(z) &= \dfrac{1}{(2n+1)\sqrt{1-z^2}} \left[ + P^{m+1}_{n+1}(z) - P^{m+1}_{n-1}(z) + \right] \label{kugel:eqn:rec-leg-3} \\ + P^m_n(z) &= \dfrac{1}{(2n+1)\sqrt{1-z^2}} \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:eqn:rec-leg-4} \end{align} \end{subequations} -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: In some software implementations (that include lighting computations in computer graphics, antenna modelling softwares, 3-D modelling in medical applications, etc.) -they are widely used, as they lead to better numerical accuracy and computational cost lower by a factor of six\cite{usecase_recursion_paper}. -\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{search the general equation of recursion for orthogonal polynomials (is somewhere in the book)}, 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*} - Further, 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 eq.\eqref{kugel:eq:rec-leg-1}, and therefore rewrite eq.\eqref{kugel:eq:rec-leg-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} + +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: In some +software implementations (that include lighting computations in computer +graphics, antenna modelling software, 3D modelling in medical applications, +etc.) they are widely used, as they lead to better numerical accuracy and +computational cost lower by a factor of six \cite{davari_new_2013}. + +\begin{proof}[Proof of \eqref{kugel:eqn:rec-leg-1}] + This is the relation that links the associated Legendre functions with the + same $m$ index but different $n$. Using theorem + \ref{buch:orthogonal:satz:drei-term-rekursion}, we have + \begin{equation*} + (n+1)P_{n+1}(z)-(2n+1)zP_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 $P_n(z)$ to check it. We can now differentiate $m-1$ + times the just presented equation \eqref{kugel:eq:rec_2}, that so that is + becomes + \begin{equation} + \label{kugel:eq:rec_3} + \frac{d^mP_{n+1}}{dz^m} - \frac{d^mP_{n-1}}{dz^m} + = (2n+1)\frac{d^{m-1}P_n}{dz^{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}}{dz^m} + - (2n+1)\frac{d^mP_{n+1}}{dz^m} + - m\left[\frac{d^m P_{n+1}}{dz^m} + + \frac{d^{m}P_{n-1}}{dz^m}\right] + + n\frac{d^m P_{n-1}}{dz^m}=0. + \end{equation} + Finally, multiplying both sides by $(1-z^2)^{\frac{m}{2}}$ and simplifying the + expression, we can rewrite eq. \eqref{kugel:eq:rec_4} in terms of $P^m_n(z)$, + namely + \begin{equation*} + (n+1-m)P^m_{n+1}(z)-(2n+1)zP^m_n(z)+(m+n)P^m_{n-1}(z)=0, + \end{equation*} + that rearranged, will be + \begin{equation*} + (2n+1) z P^m_n(z)= (m+n) P^m_{n-1}(z) + (n-m+1) P^m_{n+1}(z). + \qedhere + \end{equation*} +\end{proof} + +\begin{proof}[Proof of \eqref{kugel:eqn:rec-leg-2}] + This relation, unlike the previous one, link three expression with the same + $n$ index but different $m$. In the proof of Lemma + \ref{kugel:thm:extend-legendre}, at some point we ran into this expression. + \begin{equation*} + (1-z^2)\frac{d^{m+2}P_n}{dz^{m+2}} + - 2(m+1)z \frac{d^{m+1}P_n}{dz^{m+1}} + + [n(n+1)-m(m+1)]\frac{d^mP_n}{dz^m} = 0, + \end{equation*} + which, if multiplied by $(1-z^2)^{\frac{m}{2}}$, becomes + \begin{equation*} + (1-z^2)^{\frac{m}{2}+1}\frac{d^{m+2}P_n}{dz^{m+2}} + - 2(m+1)z (1-z^2)^{\frac{m}{2}}\frac{d^{m+1}P_n}{dz^{m+1}} + + [n(n+1)-m(m+1)](1-z^2)^{\frac{m}{2}}\frac{d^mP_n}{dz^m} = 0. + \end{equation*} + Therefore, as before, expressing it in terms of $P^m_n(z)$: + \begin{equation*} + P^{m+2}_n(z) - \frac{2(m+1)z}{\sqrt{1-z^2}}P^{m+1}_n(z) + + [n(n+1)-m(m+1)]P^m_n(z)=0. + \end{equation*} + Further, we can adjust the indices and terms, obtaining + \begin{equation*} + \frac{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). + \qedhere + \end{equation*} +\end{proof} + +\begin{proof}[Proof of \eqref{kugel:eqn:rec-leg-3}] + To derive this expression, we can multiply eq. \eqref{kugel:eq:rec_3} by + $(1-z^2)^{\frac{m}{2}}$ and, as always, we could express it in terms of + $P^m_n(z)$: + \begin{equation*} + P^m_{n+1}(z) - P^m_{n-1}(z) = (2n+1)\sqrt{1-z^2}P^{m-1}_n(z). + \end{equation*} + After that we can divide by $2n+1$ resulting in + \begin{equation}\label{kugel:eq:helper} + \frac{1}{2n+1}[P^m_{n+1}(z) - P^m_{n-1}(z)] = \sqrt{1-z^2}P^{m-1}_n(z). + \end{equation} + To conclude, we arrange the indices differently: + \begin{equation*} + \sqrt{1-z^2}P^{m}_n(z)=\frac{1}{2n+1}[P^{m+1}_{n+1}(z) - P^{m+1}_{n-1}(z)]. + \qedhere + \end{equation*} +\end{proof} + +\begin{proof}[Proof of \eqref{kugel:eqn:rec-leg-4}] + For this proof we can rely on \eqref{kugel:eqn:rec-leg-1}, and therefore + rewrite \eqref{kugel:eqn:rec-leg-2} as + \begin{equation*} + \frac{2m}{(2n+1)\sqrt{1-z^2}} \left[ + (m+n)P^m_{n-1}(z) + (n-m+1)P^m_{n+1}(z) + \right] = P^{m+1}_n(z) + [ n(n+1)-m(m-1) ]P^{m-1}_n(z). + \end{equation*} + Rewriting then $P^{m-1}_n(z)$ using eq.\eqref{kugel:eq:helper}, we will have + \begin{align*} + \frac{2m}{(2n+1)\sqrt{1-z^2}} + &\left[ (m+n)P^m_{n-1}(z) + (n-m+1)P^m_{n+1}(z) \right] = P^{m+1}_n(z) \\ + &+ \frac{n(n+1)-m(m-1)}{(2n+1)\sqrt{1-z^2}} \left[ + P^m_{n+1}(z)-P^m_{n-1}(z) + \right]. + \end{align*} + The last equation, after some algebraic rearrangements, it is easy to show + that it is equivalent to + \begin{equation*} + \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]. + \qedhere + \end{equation*} +\end{proof} \subsubsection{Spherical Harmonics} -The goal of this subsection's part is to apply the recurrence relations of the $P^m_n(z)$ 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: +The goal of this subsection's part is to apply the recurrence relations of the +$P^m_n(z)$ 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{subequations} \begin{align} - 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] \label{kugel:eq:rec-sph_harm-1} \\ - 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] \label{kugel:eq:rec-sph_harm-2} \\ - 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] \label{kugel:eq:rec-sph_harm-3} \\ - 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] \label{kugel:eq:rec-sph_harm-4} + 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] \label{kugel:eqn:rec-sph-harm-1} \\ + 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] \label{kugel:eqn:rec-sph-harm-2} \\ + 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] \label{kugel:eqn:rec-sph-harm-3} \\ + 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] \label{kugel:eqn:rec-sph-harm-4} \end{align} \end{subequations} -\begin{enumerate}[(i)] - \item - \begin{proof} - We can multiply both sides of equality in eq.\eqref{kugel:eq:rec-leg-1} 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 - \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{kugel:eq:rec-leg-2} 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 - \begin{proof} - Now we can consider eq.\eqref{kugel:eq:rec-leg-3}, 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 - \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} +\begin{proof}[Proof of \eqref{kugel:eqn:rec-sph-harm-1}] + We can multiply both sides of equality in \eqref{kugel:eqn:rec-leg-1} 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} + +\begin{proof}[Proof of \eqref{kugel:eqn:rec-sph-harm-2}] + 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 + \eqref{kugel:eqn:rec-leg-2} 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] + \qedhere + \end{equation*} +\end{proof} + +\begin{proof}[Proof of \eqref{kugel:eqn:rec-sph-harm-3}] + Now we can consider \eqref{kugel:eqn:rec-leg-3}, 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]. + \qedhere + \end{equation*} +\end{proof} + +\begin{proof}[Proof of \eqref{kugel:eqn:rec-sph-harm-4}] + 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} \section{Series Expansions in $L^2(S^2)$} -We have now reach a point where we have all the tools that are necessary to build something truly amazing: a general series expansion formula for -function on the surface of the sphere. -Before starting we want to recall the definition of the inner product on the spherical surface of definition \ref{kugel:def:inner-product-s2} + +We have now reach a point where we have all the tools that are necessary to +build something truly amazing: a general series expansion formula for function +on the surface of the sphere. Before starting we want to recall the definition +of the inner product on the spherical surface of definition +\ref{kugel:def:inner-product-s2} \begin{equation*} \langle f, g \rangle = \int_{0}^\pi \int_0^{2\pi} f(\vartheta, \varphi) \overline{g(\vartheta, \varphi)} \sin \vartheta \, d\varphi \, d\vartheta. \end{equation*} -To be a bit technical we can say that the set of spherical harmonic functions, toghether with the inner product just showed, -form something that we call Hilbert Space\footnote{For more details about Hilber space you can take a look in section \ref{kugel:sec:preliminaries}}. -This function space is defined over the space of ``well-behaved'' \footnote{The definitions of ``well-behaved'' is pretty ambigous, even for mathematicians. -It depends basically on the context. -You can sumarize it by saying: functions for which the theory we are considering (Fourier theorem) is always true. In our case we can say that well-behaved functions -are functions that follow some convergence contraints (pointwise, uniform, absolute, ...) that we don't want to consider further anyway.} functions. -We can say that the theory we are about to show can be applied on all twice differentiable complex valued functions, -to be more concise: complex valued $L^2$ functions $S^2 \to \mathbb{C}$. - -All these jargons are not really necessary for the practical applications of us mere mortals, namely physicists and engineers. -From now on we will therefore assume that the functions we will dealing with fulfill these ``minor'' conditions. - -The insiders could turn up their nose, but we don't want to dwell too much on the concept of Hilbert space, convergence, metric, well-behaved functions etc. -We simply think that this rigorousness could be at the expense of the possibility to appreciate the beauty and elegance of this theory. -Furthermore, the risk of writing 300+ pages to prove that $1+1=2$\cite{principia-mathematica} is just around the corner (we apologize in advance to Mr. Whitehead and Mr. Russel for using their effort with a negative connotation). - -Despite all, if you desire having definitions a bit more rigorous (as rigorous as two engineers can be), you could take a look at the chapter \ref{}. +To be a bit technical we can say that the set of spherical harmonic functions, +together with the inner product just showed, form something that we call Hilbert +Space\footnote{For more details about Hilbert space you can take a look in +section \ref{kugel:sec:preliminaries}}. This function space is defined over the +space of ``well-behaved'' \footnote{The definitions of ``well-behaved'' is +pretty ambiguous, even for mathematicians. It depends basically on the context. +You can summarize it by saying: functions for which the theory we are considering +(Fourier theorem) is always true. In our case we can say that well-behaved +functions are functions that follow some convergence constraints (pointwise, +uniform, absolute, ...) that we don't want to consider further anyway.} +functions. We can say that the theory we are about to show can be applied on +all twice differentiable complex valued functions, to be more concise: complex +valued $L^2$ functions $S^2 \to \mathbb{C}$. + +All this jargon is not really necessary for the practical applications of us +mere mortals, namely physicists and engineers. From now on we will therefore +assume that the functions we will dealing with fulfill these ``minor'' +conditions. + +The insiders could turn up their nose, but we don't want to dwell too much on +the concept of Hilbert space, convergence, metric, well-behaved functions etc. +We simply think that this rigorousness could be at the expense of the +possibility to appreciate the beauty and elegance of this theory. Furthermore, +the risk of writing 300+ pages to prove that $1+1=2$\cite{principia-mathematica} +is just around the corner (we apologize in advance to Mr. Whitehead and Mr. +Russel for using their effort with a negative connotation). + +Despite all, if you desire having definitions a bit more rigorous (as rigorous +as two engineers can be), you could take a look at the chapter \ref{}. \subsection{Spherical Harmonics Series} @@ -882,54 +1011,72 @@ To talk about a \emph{series expansion} we first need a series, so we shall build one using the spherical harmonics. \begin{definition}[Spherical harmonic series] - \label{kugel:definition:spherical-harmonics-series} + \label{kugel:def:spherical-harmonics-series} \begin{equation} f(\vartheta, \varphi) = \sum_{n=0}^\infty \sum_{m =-n}^n - c_{m,n} Y^m_n(\vartheta, \varphi). \label{kugel:definition:spherical-harmonics-series} + c_{m,n} Y^m_n(\vartheta, \varphi). + \label{kugel:eqn:spherical-harmonics-series} \end{equation} \end{definition} -With this definition we are basically saying that any function defined on the spherical surface can be represented as a linear combination of spherical harmonics. -Does eq.\eqref{kugel:definition:spherical-harmonics-series} sound familiar? Well that is prefectly normal, since this is analog to the classical Fourier theory. -In the latter is stated that ``any'' $T$-periodic function $f(x)$, on any interval $[x_0-T/2,x_0+T/2]$, can be represented as a linear combination of complex exponentials. More compactly: +With this definition we are basically saying that any function defined on the +spherical surface can be represented as a linear combination of spherical +harmonics. Does equation \eqref{kugel:definition:spherical-harmonics-series} +sound familiar? Well that is perfectly normal, since this is analog to the +classical Fourier theory. In the latter is stated that ``any'' $T$-periodic +function $f(x)$, on any interval $[x_0-T/2,x_0+T/2]$, can be represented as a +linear combination of complex exponentials. More compactly: \begin{equation*} - f(x) = \sum_{n \in \mathbb{Z}} c_n e^{i \omega_0 x}, \quad \omega_0=\frac{2\pi}{T} + f(x) = \sum_{n \in \mathbb{Z}} c_n e^{i \omega_0 x}, + \quad \omega_0=\frac{2\pi}{T} \end{equation*} -In the case of definition \ref{kugel:definition:spherical-harmonics-series} the kernels, instead of $e^{i\omega_0x}$, have become $Y^m_n$. In addition, the sum is now over the two indices $m$ and $n$. +In the case of definition \ref{kugel:def:spherical-harmonics-series} the +kernels, instead of $e^{i\omega_0x}$, have become $Y^m_n$. In addition, the sum +is now over the two indices $m$ and $n$. \begin{lemma}[Spherical harmonic coefficients] - \label{kugel:lemma:spherical-harmonic-coefficient} - \begin{align*} + \label{kugel:thm:spherical-harmonic-coefficient} + \begin{equation*} c_{m,n} - &= \langle f, Y^m_n \rangle_{\partial S} \\ - &= \int_0^\pi \int_0^{2\pi} f(\vartheta,\varphi) \overline{Y^m_n(\vartheta,\varphi)} \sin\vartheta \,d\varphi\,d\vartheta - \end{align*} + = \langle f, Y^m_n \rangle_{\partial S} + = \int_0^\pi \int_0^{2\pi} + f(\vartheta,\varphi) \overline{Y^m_n(\vartheta,\varphi)} + \sin\vartheta \,d\varphi\,d\vartheta + \end{equation*} + \kugeltodo{Would be better as a definition? I don't get what is being proved.} \end{lemma} + \begin{proof} - To develop this proof we will take advantage of the orthogonality property of the Spherical Harmonics. We can start and finish by applying the inner product on both sides of eq.\eqref{kugel:definition:spherical-harmonics-series}: + To develop this proof we will take advantage of the orthogonality property of + the Spherical Harmonics. We can start and finish by applying the inner product + on both sides of \eqref{kugel:eqn:spherical-harmonics-series}: \begin{align*} \langle f, Y^{m}_{n} \rangle_{\partial S} &= \left\langle \sum_{n'=0}^\infty \sum_{m' =-n'}^{n'} - c_{m',n'} Y^{m'}_{n'}(\vartheta, \varphi) \right\rangle_{\partial S} \\ - &= \sum_{n'=0}^\infty \sum_{m' =-n'}^{n'} + c_{m',n'} Y^{m'}_{n'} \right\rangle_{\partial S} \\ + &= \sum_{n'=0}^\infty \sum_{m' =-n'}^{n'} \langle c_{m',n'} Y^{m'}_{n'}, Y^{m}_{n} \rangle_{\partial S} \\ - &= \sum_{n'=0}^\infty \sum_{m' =-n'}^{n'} c_{m',n'} \langle Y^{m'}_{n'}, Y^{m}_{n} \rangle_{\partial S} = c_{m,n} + &= \sum_{n'=0}^\infty \sum_{m' =-n'}^{n'} c_{m',n'} + \langle Y^{m'}_{n'}, Y^{m}_{n} \rangle_{\partial S} = c_{m,n} \end{align*} - We omitted the $\vartheta, \varphi$ dependency to avoid overloading the notation. \end{proof} -Thanks to Lemma \ref{kugel:lemma:spherical-harmonic-coefficient} we can now calculate the series expansion defined in \ref{kugel:definition:spherical-harmonics-series}. -It can be shown that, for the famous ``well-behaved functions'' $f(\vartheta, \varphi)$ mentioned before, theorem \ref{fourier-theorem-spherical-surface} is true +Thanks to Lemma \ref{kugel:thm:spherical-harmonic-coefficient} we can now +calculate the series expansion defined in +\ref{kugel:def:spherical-harmonics-series}. It can be shown that, for the famous +``well-behaved functions'' $f(\vartheta, \varphi)$ mentioned before, the +following theorem is true. The connection to Theorem +\ref{kugel:thm:fourier-theorem} is pretty obvious. + \begin{theorem}[Fourier Theorem on $\partial S$] \label{fourier-theorem-spherical-surface} \begin{equation*} - \lim_{N \to \infty} + \lim_{N \to \infty} \int_0^\pi \int_0^{2\pi} \left\| f(\vartheta,\varphi) - \sum_{n=0}^N\sum_{m=-n}^n c_{m,n} Y^m_n(\vartheta,\varphi) \right\|_2 \sin\vartheta \,d\varphi\,d\vartheta = 0 \end{equation*} \end{theorem} -The connection to Theorem \ref{fourier-theorem-1D} is pretty obvious. \subsection{Spectrum} @@ -974,4 +1121,22 @@ but with a spherical function $f(\vartheta, \varphi)$. \begin{proof} \end{proof} -\subsection{Visualization}
\ No newline at end of file + +\if 0 +\begin{theorem}[Spherical harmonic series expansion] + A complex valued piecewise continuous function on the surface of the sphere + $f: S^2 \to \mathbb{C}$, $f \in L^2$ has a series expansion + \begin{equation*} + \hat{f}(\vartheta, \varphi) + = \sum_{n \in \mathbb{Z}} \sum_{m \in \mathbb{Z}} + c_{m,n} Y^m_n(\vartheta, \varphi), + \end{equation*} + where $c_{m,n} = \langle f, Y^m_n \rangle$, that converges everywhere uniformly + on $f$, i.e. $\|f(\vartheta, \varphi) - \hat{f}(\vartheta, \varphi)\|_2 = 0$. +\end{theorem} +\begin{proof} + Sadly, this proof is beyond the scope of this text. +\end{proof} +\fi + +\subsection{Visualization} |