The Laplacian in polar coordinates

Branko Ćurgus

Mathematical facts

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• In two Cartesian coordinates $x$ and $y$ for a differentiable real valued function $u$ we have \[ ({\rm grad}\, u)(x,y) = \vec{i} \, \frac{\partial u}{\partial x}(x,y) + \vec{j} \, \frac{\partial u}{\partial y}(x,y) \]
• In two Cartesian coordinates $x$ and $y$ for a differentiable vector function $\vec{F}(x,y) = \vec{i} \, f(x,y) + \vec{j} \, g(x,y)$ we have \[ \bigl({\rm div} \, \vec{F}\bigr)(x,y) = \frac{\partial f}{\partial x}(x,y) + \frac{\partial g}{\partial y}(x,y) \]
• It is useful to introduce the vector differential operator, called del and denoted by nabla. In Cartesian coordinates it is defined as \[ \vec{\nabla} = \vec{i} \, \frac{\partial}{\partial x} + \vec{j} \, \frac{\partial}{\partial y}. \] Then \[ {\rm grad}\, u = \vec{\nabla}\, u, \qquad {\rm div} \, \vec{F} = \vec{\nabla} \cdot \vec{F}. \]
• If $u$ is a twice-differentiable real-valued function, then the Laplacian of $u$ is defined as the divergence of the gradient of $u$: \[ {\rm div} \, ({\rm grad}\, u) = \vec{\nabla} \cdot \vec{\nabla} u = \nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \]
• The chain rule for functions of two variables: (to be added)
• The Cartesian coordinates and polar coordinates in the plane are related by the following formulas: \begin{equation*} x = r \cos \theta, \qquad y = r \sin \theta. \end{equation*}

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Derivation

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Let $U:{\mathbb R}^2 \to {\mathbb R}$ be a twice-differentiable function. Here we think of $U$ as a function $U(x,y)$ of the Cartesian coordinates. Let \begin{equation} \label{eq1} W(r,\theta) = U(r \cos \theta, r \sin \theta) \end{equation} Notice that $W:[0,+\infty)\times [0,2\pi) \to {\mathbb R}$. We want to derive the expression for

$\displaystyle \frac{\partial^2 U}{\partial x^2}(r \cos \theta, r \sin \theta) + \frac{\partial^2 U}{\partial y^2}(r \cos \theta, r \sin \theta) $

in terms of the function

$W(r,\theta)$

First differentiate \eqref{eq1} with respect to $r$ and $\theta$ using the chain rule for multivariable functions: \begin{equation} \label{eq2} \frac{\partial W}{\partial r} = ( \cos \theta ) \, \frac{\partial U}{\partial x} + ( \sin \theta ) \, \frac{\partial U}{\partial y} \end{equation} \begin{equation} \label{eq3} \frac{\partial W}{\partial \theta} = (- r \sin \theta ) \, \frac{\partial U}{\partial x} + (r \cos \theta ) \, \frac{\partial U}{\partial y} \end{equation} Next differentiate \eqref{eq2} with respect to $r$: \begin{equation} \label{eq4} \frac{\partial^2 W}{\partial r^2} = ( \cos \theta )^2 \, \frac{\partial^2 U}{\partial x^2} + ( \cos \theta )( \sin \theta ) \, \frac{\partial^2 U}{\partial y \partial x} + ( \sin \theta ) ( \cos \theta ) \, \frac{\partial^2 U}{\partial x \partial y} + ( \sin \theta )^2 \, \frac{\partial^2 U}{\partial y^2}; \end{equation} and differentiate \eqref{eq3} with respect to $\theta$ \begin{align} \label{eq5} \frac{\partial^2 W}{\partial \theta^2} & = (- r \cos \theta ) \, \frac{\partial U}{\partial x} - (r \sin \theta ) \, \frac{\partial U}{\partial y} \\ \nonumber & \phantom{++} + (r \sin \theta )^2 \, \frac{\partial^2 U}{\partial x^2} - r^2 (\sin \theta ) (\cos \theta ) \, \frac{\partial U}{\partial y \partial x} \\ \nonumber & \phantom{++++} - r^2 (\cos \theta ) (\sin \theta ) \, \frac{\partial U}{\partial x \partial y} + (r \cos \theta )^2 \, \frac{\partial^2 U}{\partial y^2} \end{align} Observe that by \eqref{eq2} the expression on the right-hand side of \eqref{eq5} equals $\displaystyle -r \frac{\partial W}{\partial r}$. Using this observation the last equality can be rewritten as \begin{equation} \label{eq6} \frac{1}{r^2} \left(\frac{\partial^2 W}{\partial \theta^2} + r \, \frac{\partial W}{\partial r} \right) = ( \sin \theta )^2 \, \frac{\partial^2 U}{\partial x^2} - ( \sin \theta )( \cos \theta ) \, \frac{\partial^2 U}{\partial y \partial x} - ( \cos \theta ) ( \sin \theta ) \, \frac{\partial^2 U}{\partial x \partial y} + ( \cos \theta )^2 \, \frac{\partial^2 U}{\partial y^2} \end{equation} Adding \eqref{eq4} and \eqref{eq6} yields \[ \frac{1}{r^2} \frac{\partial^2 W}{\partial \theta^2} + \frac{1}{r} \frac{\partial W}{\partial r} + \frac{\partial^2 W}{\partial r^2} = \frac{\partial^2 U}{\partial x^2} + \frac{\partial^2 U}{\partial y^2}. \] This is the desired expression: \[ \frac{\partial^2 U}{\partial x^2}(r\cos\theta,r\sin\theta) + \frac{\partial^2 U}{\partial y^2}(r\cos\theta,r\sin\theta) = \frac{1}{r^2} \frac{\partial^2 W}{\partial \theta^2}(r,\theta) + \frac{1}{r} \frac{\partial W}{\partial r}(r,\theta) + \frac{\partial^2 W}{\partial r^2}(r,\theta). \] Thus the Laplacian in polar coordinates of a function $W(r,\theta)$ is \[ \frac{1}{r^2} \frac{\partial^2 W}{\partial \theta^2}(r,\theta) + \frac{1}{r} \frac{\partial W}{\partial r}(r,\theta) + \frac{\partial^2 W}{\partial r^2}(r,\theta). \] An equivalent form for the Laplacian in polar coordinates of a function $W(r,\theta)$ is \[ \frac{1}{r^2} \frac{\partial^2 W}{\partial \theta^2}(r,\theta) + \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial W}{\partial r}(r,\theta) \right). \]

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Examples

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Example 1. Consider the following function in the orthogonal coordinate system $xy$, \[ u(x,y) = x y\bigl(x^2 + y^2\bigr) \quad x,y \in \mathbb{R}. \]

One can think of this function as giving the temperature at each point in the $xy$-plane. So, at the point $P$, whose coordinates in $xy$ orthogonal coordinate system are $(2,2),$ the temperature is $32.$ The same temperature in polar coordinates is given by the function \[ w(r,\theta) = r^4 (\cos \theta)(\sin \theta) \quad r \in [0,+\infty), \quad \theta \in [0, 2 \pi). \] The same point $P$ from above has coordinates $r = 2\sqrt{2},$ $\theta = \pi/4.$ So, the temperature at the point $P$ is \[ w\bigl(2\sqrt{2}, \pi/4 \bigr) = 2^4 2^2 (\sqrt{2}/2) (\sqrt{2}/2) = 32. \]

The Laplacian of the function $u(x,y)$ is \[ (\nabla^2 u)(x,u) = \frac{\partial^2 u}{\partial x^2}(x,y) + \frac{\partial^2 u}{\partial y^2}(x,y) = 12 x y. \]

The Laplacian of the function $w(r,\theta)$ is \[ (\nabla^2 w)(r,\theta) = \frac{1}{r^2} \frac{\partial^2 w}{\partial \theta^2}(r,\theta) + \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial w}{\partial r}(r,\theta) \right) = 12 r^2 (\cos \theta)(\sin \theta). \]

As expected, we have \[ 12xy \quad \text{in polar coordinates is} \quad 12 r^2 (\cos \theta)(\sin \theta). \]

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Example 2. Let \[ U(x,y) = \frac{x^2-y^2}{\bigl(x^2+y^2\bigr)^2}. \] A lengthy calculation (or the fact that $U$ is the real part of $(x+iy)^{-2}$), yields that $U$ is a harmonic function. That is $\nabla^2 U =0$.

In polar coordinates we have \[ W(r,\theta) = U(r \cos\theta,r\sin\theta) = \frac{\cos(2\theta)}{r^2}. \] Let us verify that $\nabla^2 W =0$ in polar coordinates. We calculate: \begin{align*} \frac{\partial W}{\partial r} & = -2 \frac{\cos(2\theta)}{r^3}, \\ \frac{\partial^2 W}{\partial r^2} & = 6 \frac{\cos(2\theta)}{r^4}, \\ \frac{\partial W}{\partial \theta} & = -2 \frac{\sin(2\theta)}{r^2}, \\ \frac{\partial^2 W}{\partial \theta^2} & = -4 \frac{\cos(2\theta)}{r^2}, \end{align*} and substitute in the expression for the Laplacian in polar coordinates to get: \[ \frac{1}{r^2}( -4) \frac{\cos(2\theta)}{r^2} + \frac{1}{r} (-2) \frac{\cos(2\theta)}{r^3} + 6 \frac{\cos(2\theta)}{r^4} = 0. \]

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Example 3. Let \[ U(x,y) = \frac{x^2-y^2}{x^2+y^2}. \] Then \[ W(r,\theta) = \cos(2\theta). \] A lengthy calculation shows that \[ \bigl(\nabla^2 U\bigr)(x,y) = - 4 \frac{x^2-y^2}{\bigl(x^2+y^2\bigr)^2}. \] Let us verify this in polar coordinates by calculating: \begin{align*} \frac{\partial W}{\partial r} & = 0, \\ \frac{\partial^2 W}{\partial r^2} & = 0, \\ \frac{\partial W}{\partial \theta} & = -2 \sin(2\theta), \\ \frac{\partial^2 W}{\partial \theta^2} & = -4 \cos(2\theta), \end{align*} and substituting in the expression for the Laplacian in polar coordinates: \[ \frac{1}{r^2}( -4) \cos(2\theta) \] which is exactly $\bigl(\nabla^2 U\bigr)(r \cos\theta,r \sin \theta)$.

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