
## Lecture 22. Separation of variables: Misc equations

#### Laplace operator in polar coordinates

In the next several lectures we are going to consider Laplace equation in the disk and similar domains and separate variables there but for this purpose we need to express Laplace operator in polar coordinates. Recall that (from 1st year Calculis) polar coordinates are $(r,\theta)$ connected with Cartesian coordinates by $x=r\cos (\theta)$, $y=\sin(\theta)$ and inversely \begin{equation*} \left\{\begin{aligned} &r=\sqrt{x^2+y^2},\\ &\theta = \arctan \bigl(\frac{y}{x}\bigr); \end{aligned}\right. \end{equation*} surely the second formula is not exactly correct as changing $(x,y)\to (-x,-y)$ does not change it ratio but replaces $\theta$ by $\theta +\pi$ (or $\theta-\pi)$ as $\theta$ is defined modulo $2\pi n$ with $n\in \mathbb{Z}$. It does not really maater as we are interested only in derivatives: $$r_x=\cos(\theta),\ r_y=\sin(\theta),\ \theta_x = -r^{-1}\sin(\theta),\ \theta_y = r^{-1}\cos(\theta). \label{eq-1}$$ Exercise. Prove (\ref{eq-1}).

Then by chain rule \left\{\begin{aligned} &\partial_x = \cos(\theta)\partial_r - r^{-1}\sin(\theta)\partial_\theta,\\ &\partial_x = \sin(\theta)\partial_r + r^{-1}\cos(\theta)\partial_\theta \end{aligned}\right. \label{eq-2} and therefore \begin{equation*} \Delta = \partial_x^2+ \partial_y^2= \bigl(\cos(\theta)\partial_r - r^{-1}\sin(\theta)\partial_\theta\bigr)^2+ \bigl(\sin(\theta)\partial_r + r^{-1}\cos(\theta)\partial_\theta\bigr)^2 \end{equation*} and after tedious calculations one can get $$\Delta = \partial_r^2 +\frac{1}{r}\partial_r +\frac{1}{r^2}\partial_\theta^2. \label{eq-3}$$ Exercise. Do it.

Instead I want to use different method requiring much less error prone calculations but more delicate arguments (useful in more complicated cases).

Note first the identity $$\iint\Delta u\cdot v\,dxdy = -\iint \nabla u \cdot \nabla v\,dxdy \label{eq-4}$$ provided $v=0$ near $\Gamma$ (boundary of $\mathcal{D}$) and integrals are taken over $\mathcal{D}$.

Now let us express the left- and right-hand expression in polar coordinates. Recall that polar coordinates are orthogonal (i.e. level lines of $r$ (circles) and level lines of $\theta$ (rays from origin) are orthogonal in the points where they intersect) and the distance between two close points $ds$ can be calculated as $$ds^2 =dx^2+dy^2 =dr^2+r^2d\theta^2 \label{eq-5}$$ and therefore area element is $dA=dx dy= rdrd\theta$.

But what about $\nabla u\cdot \nabla v$? We claim that $$\nabla u\cdot \nabla v = u_r v_r +\frac{1}{r^2}u_\theta v_\theta. \label{eq-6}$$ Really $\nabla u$ is a vector of the different nature than $d\mathbf{s}=(dx,dy)$. They are connected by $du=\nabla u\cdot d\mathbf{s}$ and when we change coordinates $d\mathbf{s}=Qd\mathbf{s}$ with some matrix $Q$ and since \begin{equation*} du=\nabla u\cdot d\mathbf{s}=\nabla u'\cdot d\mathbf{s}'= \nabla u'\cdot Qd\mathbf{s}= Q^T \nabla u'\cdot d\mathbf{s} \end{equation*} we conclude that $d\mathbf{s}'=Q^{T\,-1}\nabla u$ where $^T$ means transposed matrix. Such dual vectors mathematicians call covectors.

Therefore (\ref{eq-4}) becomes \begin{multline*} \iint \Delta u\cdot v\,r drd\theta = -\iint \bigr(u_r v_r +\frac{1}{r^2}u_\theta v_\theta\bigr)r\,drd\theta=\\ -\iint \bigr(r u_r v_r +\frac{1}{r}u_\theta v_\theta\bigr)\,drd\theta= \iint \bigr(\bigr(r u_r\bigl)_r +\bigr(\frac{1}{r}u_\theta\bigl)_\theta \bigr)v\,drd\theta \end{multline*} where we integrated by parts. This identity \begin{equation*} \iint r \Delta u\cdot v\,drd\theta = \iint \bigr(\bigr(r u_r\bigl)_r +\bigr(\frac{1}{r}u_\theta\bigl)_\theta \bigr)v\,drd\theta. \end{equation*} holds for any $v$ vanishing near $\Gamma$ and therefore we can nix integration and $v$: \begin{equation*} r \Delta u = \bigr(r u_r\bigl)_r +\bigr(\frac{1}{r}u_\theta\bigl)_\theta . \end{equation*} Exercise. Think about this. Finally we get \begin{equation*} \Delta u = r^{-1}\bigr(r u_r\bigl)_r +\bigr(\frac{1}{r}u_\theta\bigl)_\theta \end{equation*} which is exactly (\ref{eq-3}).

It may look too complicated for polar coordinates but in more general cases this approach is highly fenefitial.

#### Laplace operator in spherical coordinates

Spherical coordinates are $\rho$ (radius), $\phi$ (latitude) and $\theta$ (longitude): \begin{equation*} \left\{\begin{aligned} &x=\rho \sin(\phi)\cos(\theta),\\ &y=\rho \sin(\phi)\sin(\theta)\\ &z=\rho\cos(\phi). \end{aligned}\right. \end{equation*} Conversely \begin{equation*} \left\{\begin{aligned} &\rho=\sqrt{x^2+y^2+z^2},\\ &\phi =\arctan\bigl(\frac{\sqrt{x^2+y^2}}{z}\bigr),\\ &\theta = \arctan \bigl(\frac{y}{x}\bigr); \end{aligned}\right. \end{equation*} and using chain rule and "simple" calculations becomes rather challenging.

Instead we recall that these coordinates are also orthogonal: if we fix $\phi$ and $\theta$ we get rays from origin, which are orthogonal to the speres which we get if we fix $r$. On the spheres if we fix $\theta$ we get meridians and if we fix $\phi$ we get parallels and those are also orthogonal. Then $$ds^2 =dx^2+dy^2+dy^2+dz^2 =d\rho^2+\rho^2d\phi^2+\rho^2 \sin^2(\phi)d\theta^2 \tag*{(5)'}$$ where $d\rho$, $\rho d\phi$ and $\rho\sin(\phi)d\theta$ are distances along rays, meridians and parallels and therefore volume element is $dV=dx dy dz= \rho^2\sin(\theta)d\rho d\phi d\theta$.

Therefore $$\nabla u\cdot \nabla v = u_\rho v_\rho +\frac{1}{\rho^2}u_\phi v_\phi+ \frac{1}{\rho^2\sin(\phi)} u_\theta v_\theta. \tag*{(6)'}$$ Plugging this into $$\iiint\Delta u\cdot v\,dxdydz = -\iiint \nabla u \cdot \nabla v\,dxdydz \tag*{(4)'}$$ we get \begin{multline*} \iiint \Delta u\cdot v \rho^2\sin(\phi)\,d\rho d\phi d\theta = \\ -\iiint \bigr(u_\rho v_\rho +\frac{1}{\rho^2} u_\phi v_\phi + \frac{1}{\rho^2\sin(\phi)} u_\theta v_\theta\bigr)\rho^2\sin(\phi)\,d\rho d\phi d\theta=\\ \iiint \Bigr(\bigr(\rho^2\sin(\phi) u_\rho\bigl)_\rho + \bigr(\sin(\phi) u_\phi\bigl)_\phi+ \bigr(\frac{1}{\sin(\phi)}u_\theta\bigl)_\theta \Bigr)v\,d\rho d\phi d\theta. \end{multline*}

Then we can nix integration and factor $v$: \begin{equation*} \Delta u\cdot \rho^2\sin(\phi)= \bigr(\rho^2\sin(\phi) u_\rho\bigl)_\rho + \bigr(\sin(\phi) u_\phi\bigl)_\phi+ \bigr(\frac{1}{\sin(\phi)}u_\theta\bigl)_\theta \end{equation*} and then \begin{equation*} \Delta u= \frac{1}{\rho^2\sin(\phi)}\Bigl(\bigr(\rho^2\sin(\phi) u_\rho\bigl)_\rho + \bigr(\sin(\phi) u_\phi\bigl)_\phi+ \bigr(\frac{1}{\sin(\phi)}u_\theta\bigl)_\theta \Bigr) \end{equation*} and finally $$\Delta =\partial_\rho^2 + \frac{2}{\rho}\partial_\rho + \frac{1}{\rho^2}\bigl(\partial_{\phi}^2 +\cot(\phi)\partial_\phi \bigr)+\frac{1}{\rho^2\sin^2(\phi)}\partial_{\theta}^2. \label{eq-7}$$ (compare with (\ref{eq-3}))

Definition. $$\Lambda:= \partial_{\phi}^2 +\cot(\phi)\partial_\phi +\frac{1}{\sin^2(\phi)}\partial_{\theta}^2 \label{eq-8}$$ is a spherical Laplacian (aka Laplace-Beltrami operator omn the sphere).

#### Special knowledge: Generalization

If the length element is $$ds^2 =\sum_{j,k} g_{jk}d q^j dq^k \tag*{(5)''}$$ where $q=q^1,\ldots,q^n)$ are new coordinates and we prefer to write $dqi^j$ rather than $dq_j$ (to half-follow Einstein notations), $g_{jk}$ is symmetric matrix, then $$\nabla u\cdot \nabla v = \sum_{j,k} g^{jk}u_{q^j}u_{q^k} \tag*{(6)''}$$ where $(g^{jk})$ is an inverse matrix to $(g_{jk})$: $\sum_{k} g^{jk}g_{kl}= \sum_{k} g_{lk}g^{kj}= \delta^j_l$ .

Then volume element is $|\det (g_{jk})|^{\frac{1}{2}}\,dq_1\cdots dq_n$ and $$\Delta u = |\det (g_{jk})|^{-\frac{1}{2}}\sum_{j,k} \bigl( |\det (g_{jk})|^{\frac{1}{2}} g^{jk}u_{q^k} \bigr)_{q^j}. \label{eq-9}$$

#### Secret knowledge: elliptical coordinates

Elliptical coordinates on plane are $(\mu, \nu)$: \left\{\begin{aligned} &x=c\cosh (\mu) \cos(\nu),\\ &y=c\sinh (\mu) \sin(\nu). \end{aligned}\right. \label{eq-10} Level lines $\mu=\const$ are ellipses with focai at $(-c,0)$ and $(c,0)$ and level lines $\nu=\const$ are hyperbolas with the same focai; so we have confocal ellipses and hyperbolas.

These coordinates are not only orthogonal but they are conformal ($ds^2$ is proportional to $d\mu^2+d\nu^2$) $$ds^2= \bigl(\sinh^2(\mu)+\sin^2(\nu) \bigr)(d\mu^2+d\nu^2) \label{eq-11}$$ and therefore $$\Delta = \frac{1}{c^2\bigl(\sinh^2(\mu)+\sin^2(\nu) \bigr)}(\partial_\mu^2 +\partial_\nu^2 ). \label{eq-12}$$