using [[Separation of Variables for Heat and Waves in 1D]], we found Bessel's equation, read more in [[Bessel's Equation]]. In this file, we introduce the *Legendre Equation*, similar to the Bessel's equation, it's pretty complicated. --- ### **Intro** Consider PDE: $\nabla^2 \cdot u = -\lambda u$, We are in a spherical coordinate, the boundary condition will be discussed during the solving process. There are 3 parameters describing the spherical coordinate, they helps with separation of variables: * $r$ is the radial distance, which is $0 < r < a$ * $\psi$ is the longitude, $-\pi \le \psi \le \pi$, and this is the same convention as the argument of a complex number. * $\theta$ is the co-latitude, $0 \le \theta \le \pi$, where the north pole is $0$ and the south pole is $\pi$ **Avoiding solutions that blows up**: it's likely to blow up at the center and the poles of the sphere, and we want it to be $2\pi$ periodic at the longitude, meaning that function wrt to $\psi$ will be $2\pi$ periodic. **Spherical Laplacian Operator** $ \nabla^2 \cdot u = \frac{1}{r^2}\partial_r[r^2\partial_r[u]] + \frac{1}{r^2\sin(\theta)}\partial_\theta[\sin\theta\partial_\theta [u]] + \frac{1}{r^2\sin^2\theta} \partial_\psi^2[u] $ And we are going to take this for granted, and this crucial for solving in alternative coordinate. **Objectives** * Use the separations of variables. * Derive the Spherical Bessel's equation * Derive the Legendre Equation * Reveal the relations of the eigenvalues. --- ### **Separation of Variable** Let: $u(r, \theta, \psi) = R(r)Y(\theta, \psi)$, recall $\nabla^2 \cdot u = -\lambda u$, then using the PDE: $ \begin{aligned} u(r, \theta, \psi) &= R(r)Y(\theta, \psi) \\ \underset{(1)}{\implies} \frac{\nabla\cdot [R(r)Y(\theta, \psi)] } {R(r)Y(\theta, \psi)} &= \frac{\frac{1}{r^2}\partial_r[r^2\partial_r[R]]} {R} + \frac{\frac{1}{r^2\sin\theta}\partial_\theta[\sin\theta \partial_\theta[Y]] + \frac{1}{r^2\sin^2\theta}\partial_\psi^2[Y]}{Y} \\ \underset{(2)}{\implies} -\lambda^2 &= \frac{\frac{1}{r^2}\partial_r[r^2\partial_r[R]]} {R} + \frac{\frac{1}{r^2\sin\theta}\partial_\theta[\sin\theta \partial_\theta [Y]] + \frac{1}{r^2\sin^2\theta}\partial_\psi^2[Y]}{Y} \\ \underset{(3)}{\implies} -\lambda^2r^2 &= \frac{\partial_r[r^2\partial_r[R]]} {R} + \frac{\frac{1}{\sin\theta}\partial_\theta[\sin\theta \partial_\theta[Y]] + \frac{1}{\sin^2\theta}\partial_\psi^2[Y]}{Y} \end{aligned} \tag{1} $ The separation of variable results are $ \underset{(4)}{\implies} \begin{cases} Y^{-1} ((\sin\theta)^{-1} \partial_\theta[\sin\theta \partial_\theta[Y]] + (\sin^2\theta)^{-2} \partial_\psi^2[Y]) = -\eta & \\ R^{-1}\partial_r[r^2\partial_r[R]] = -\eta & \\ \eta = \lambda r^2 \end{cases} $ **Explainations:** 1. Expanding the Spherical Laplacian Operator. 2. Substitute the expanded expression back to the original PDE, and put $-\lambda^2$ on the left hande side of the expression. 3. Multipy $r$ on both side of the expression. 4. Functions of $r, y$ are on the RHS and they equals to a constant, using the old trick from separations of variable, we get both equation equals to the same constant respectively. The sign for $\eta$ is not assumed. Focusing on the part involving $R$, firstly, divide by $r$, this will be: $ \begin{aligned} -\lambda^2 - \frac{\partial_r[r^2\partial_r[R]]}{Rr^2} &= \frac{-\eta}{r^2} \\ -R\lambda^2 - \frac{\partial_r[r^2\partial_r[R]]}{r^2} &= \frac{-\eta R}{r^2} \\ \text{Substitute: } \partial_r[r^2\partial_r[R]] &= 2r\partial_r[R] + r^2\partial_r^2[R] \\ \implies -R\lambda^2 - \left( \frac{2}{r}\partial_r[R] +\partial_r^2[R] \right) &= \frac{-\eta R}{r^2} \end{aligned} $ So that mean: > $ > R'' + \frac{2}{r}R' - > \left( > \frac{\eta}{r^2} - \lambda^2 > \right) R = 0 > \tag{1.1} > $ > This is the *Sphereical Bessel's equation*. Focusing $Y$, $ \begin{aligned} \frac{\frac{1}{\sin\theta}\partial_\theta[\sin\theta \partial_\theta[Y]] + \frac{1}{\sin^2\theta}\partial_\psi^2[Y]}{Y} + \eta &= 0 \\ \frac{1}{\sin\theta}\partial_\theta[\sin\theta \partial_\theta[Y]] + \frac{1}{\sin^2\theta}\partial_\psi^2[Y] + \eta Y &= 0 \end{aligned} $ And, take notice that, this is still a partial differential equation, which means that we are going to apply the SOV again here. This is call the *Spherical Harmonic Equation*. using the same tricks, we begin with $Y(\theta, \psi) = \Theta (\theta) \Psi(\psi)$, then $ \begin{aligned} \frac{\partial_\theta[\sin\theta \Theta'(\theta)\Psi(\psi)]}{\sin\theta} + \frac{1}{\sin^2\theta}\Psi''(\psi)\Theta(\theta) + (\sin\theta)^2\eta Y &= 0 \\ (\sin\theta) \partial_\theta[\sin\theta \Theta'(\theta)\Psi(\psi)] + \Psi''(\psi)\Theta(\theta) + \eta \Theta(\theta)\Psi(\psi)(\sin\theta)^2 &= 0 \\ \frac{ (\sin\theta) \partial_\theta[\sin\theta \Theta'(\theta)] }{\Theta(\theta)} + \frac{\Psi''(\psi)}{\Psi(\psi)} + \frac{\eta(\sin\theta)^2}{\Theta(\theta)} &= 0 \\ \text{ introduce new constant}\quad \frac{ (\sin\theta) \partial_\theta[\sin\theta \Theta'(\theta)] + \eta (\sin\theta)^2 }{\Theta} &= \frac{-\Psi''}{\Psi} = \alpha^2 \end{aligned}\tag{1.2} $ And we introduced another constant $\alpha^2$ for the separation of variables. For $\Psi(\psi)$, we have the ODE, $\Psi'' + \alpha^2 \Psi = 0, -\pi \le \psi \le \pi$. And the boundary condition is periodic, and notice that this is the same as the $\Theta(\theta)$ function in the Cylinder case, see [Helmholtz Equation in a Cylinder](Helmholtz%20Equation%20in%20a%20Cylinder.md) for more information. Recall from there that we had $\alpha = m, \; m\in \mathbb{Z} = 0, \pm 1, \pm 2\cdots$, with basis the basis functions $\Psi_m = \exp \left(im\psi\right)$. Finally, for $\Theta(\theta)$, and we devide $\sin^2(\theta)$ on expression (1.2) to yield the ODE $ \frac{1}{\sin\theta} \partial_\theta[\sin\theta \partial_\theta[\Theta]] + \left( \eta - \frac{m^2}{\sin^2\theta} \right)\Theta = 0 \tag{1.3} $ We also multiplied by $\Theta$ and at the same time, substitute the $\alpha= m$ for $\Psi''/\Psi$. Now, we know that $\theta\in[0, \pi]$, and we want out solution to be bounded at the poles (CLARIFICATION NEEDED), and this will mean that $\Theta(\pm \pi) \ne \infty$. To get insight, we make a *Change of Variable* with $x = \cos\theta$ so $dx = -\sin\theta d\theta$ then $\theta\in[0, \pi]$ means $x\in [-1, 1]$, let $y(x) = \Theta(\theta)$ For expression (1.3) we use the above *Change of Variable*, obtaining a new ODE: > $ > \partial_x[(1 - x^2)\partial_x[y]] > + > \left[\eta - \frac{m^2}{1 - x^2} > \right]y = 0 > \tag{1.4} > $ > Imposing condition: $y(\pm 1) \ne \infty$, and this is: *Associated Legendre Equation* a 2nd order ODE that is hard to solve. --- ### **Solving the Associated Bessel's Equation** For more information, please see [Legendre Polynomials](Legendre%20Polynomials.md), [Bessel's Equation](Bessel's%20Equation.md) for more information on how to solve these 2 complicated ODEs.