The goal of this page is to derive the formula for the Laplacian in polar coordinates step by step. The Laplacian operator in two dimensions is often expressed in Cartesian coordinates: \[ (\nabla^2 u)(x,y) = \frac{\partial^2 u}{\partial x^2} (x,y) + \frac{\partial^2 u}{\partial y^2} (x,y). \] However, for problems with circular symmetry, it is often more convenient to work in polar coordinates.
The expression for the Laplacian depends on the coordinate system in which a function of two variables is given. This derivation will show that for a function \(w(r,\theta)\) given in the polar coordinates the Laplacian is given by the following expression: \[ (\nabla^2 w)(r,\theta) = \frac{1}{r^2} \frac{\partial^2 w}{\partial \theta^2}(r,\theta) + \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial w}{\partial r}(r,\theta) \right). \] The above expression is known as the Laplacian in polar coordinates
By the end of this explanation, you will understand the step-by-step process behind this expression.
For example, given \[ u(x,y) = x^2 + y^2. \] The Laplacian of this function is \[ (\nabla^2 u)(x,y) = 4. \] The same function given in polar coordinates is \[ w(r,\theta) = r^2. \] When we calculate its Laplacian in polar coordinates we should be getting the same value at the corresponding points. Calculating \begin{align*} \frac{1}{r^2} \frac{\partial^2 w}{\partial \theta^2} + \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial w}{\partial r} \right) & = 0 + \frac{1}{r} \frac{\partial}{\partial r} \left( r \mkern 2mu 2 r \right) \\ & = \frac{1}{r} 4 r \\ & = 4. \end{align*}
In this derivation, I use the capital letters \(U(x,y)\) and \(W(r,\theta)\) for functions. Typically, lowercase letters \(u(x,y)\) and \(w(r,\theta)\) are used, but I chose capitals because I believe they improve readability in this context.
(I) 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 | $W(r,\theta)$ |
(II) 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}
(III) 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}
(IV) 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: \[ \frac{\partial^2 U}{\partial x^2}(r\cos\theta,r\sin\theta) + \frac{\partial^2 U}{\partial y^2}(r\cos\theta,r\sin\theta) = \frac{1}{r^2} \frac{\partial^2 W}{\partial \theta^2}(r,\theta) + \frac{1}{r} \frac{\partial W}{\partial r}(r,\theta) + \frac{\partial^2 W}{\partial r^2}(r,\theta). \]
(V) 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 \[ \frac{1}{r^2} \frac{\partial^2 W}{\partial \theta^2}(r,\theta) + \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial W}{\partial r}(r,\theta) \right). \]
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 \[ 12xy \quad \text{in polar coordinates is} \quad 12 r^2 (\cos \theta)(\sin \theta). \]
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: \[ \frac{1}{r^2}( -4) \frac{\cos(2\theta)}{r^2} + \frac{1}{r} (-2) \frac{\cos(2\theta)}{r^3} + 6 \frac{\cos(2\theta)}{r^4} = 0. \]
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 $\bigl(\nabla^2 U\bigr)(r \cos\theta,r \sin \theta)$.
(I) Recall the definition of the gradient in two Cartesian coordinates $x$ and $y$. For a differentiable real valued function $U(x,y)$ we have \begin{align*} (\operatorname{grad} U)(x,y) & = \vec{i} \, \frac{\partial U}{\partial x}(x,y) + \vec{j} \, \frac{\partial U}{\partial y}(x,y) \\ & = \left\langle \frac{\partial U}{\partial x}(x,y), \frac{\partial U}{\partial y}(x,y) \right\rangle \end{align*}
(II) In Section Derivation part (II) we derived the following equalities \begin{equation} \label{eq7} \frac{\partial W}{\partial r} = ( \cos \theta ) \, \frac{\partial U}{\partial x} + ( \sin \theta ) \, \frac{\partial U}{\partial y} \end{equation} \begin{equation} \label{eq8} \frac{\partial W}{\partial \theta} = (- r \sin \theta ) \, \frac{\partial U}{\partial x} + (r \cos \theta ) \, \frac{\partial U}{\partial y} \end{equation} Divide equation \eqref{eq8} by \(r\) to get \begin{equation} \label{eq9} \frac{1}{r}\mkern 1mu \frac{\partial W}{\partial \theta} = (- \sin \theta ) \, \frac{\partial U}{\partial x} + (cos \theta ) \, \frac{\partial U}{\partial y} \end{equation}
Next we look at equations \eqref{eq7} and \eqref{eq9} as a system to be solved for \(\frac{\partial U}{\partial x}\) and \(\frac{\partial U}{\partial y}\). First, multiply equation \eqref{eq7} by \(\cos \theta\) and equation \eqref{eq9} by \(-\sin\theta\) and add the resulting equations to get \[ \frac{\partial U}{\partial x} = (\cos\theta) \frac{\partial W}{\partial r} - (\sin\theta) \frac{1}{r} \frac{\partial W}{\partial \theta} \] Second, multiply equation \eqref{eq7} by \(\sin \theta\) and equation \eqref{eq9} by \(\cos\theta\) and add the resulting equations to get \[ \frac{\partial U}{\partial y} = (\sin\theta) \frac{\partial W}{\partial r} + (\cos\theta) \frac{1}{r} \frac{\partial W}{\partial \theta} \]
(III) Recall that the gradient $(\nabla U)(x,y)$ is given by \[ (\nabla U)(x,y) = \left\langle \frac{\partial U}{\partial x}(x,y),\; \frac{\partial U}{\partial y}(x,y) \right\rangle. \] Since we calculated in (II) that \begin{align*} \frac{\partial U}{\partial x}(r \cos\theta, r\sin\theta) & = (\cos\theta) \frac{\partial W}{\partial r}(r,\theta) - (\sin\theta) \frac{1}{r} \frac{\partial W}{\partial \theta}{\partial r}(r,\theta), \\ \frac{\partial U}{\partial y}(r \cos\theta, r\sin\theta) & = (\sin\theta) \frac{\partial W}{\partial r}{\partial r}(r,\theta) + (\cos\theta) \frac{1}{r} \frac{\partial W}{\partial \theta}{\partial r}(r,\theta), \end{align*} we have \begin{align*} \nabla W & = \left\langle (\cos\theta) \frac{\partial W}{\partial r} - (\sin\theta) \frac{1}{r} \frac{\partial W}{\partial \theta},\; (\sin\theta) \frac{\partial W}{\partial r} + (\cos\theta) \frac{1}{r} \frac{\partial W}{\partial \theta} \right\rangle \\ & = \left\langle (\cos\theta) \frac{\partial W}{\partial r},\; (\sin\theta) \frac{\partial W}{\partial r} \right\rangle + \left\langle - (\sin\theta) \frac{1}{r} \frac{\partial W}{\partial \theta},\; (\cos\theta) \frac{1}{r} \frac{\partial W}{\partial \theta} \right\rangle \\ & = \frac{\partial W}{\partial r} \bigl\langle \cos\theta , \sin\theta\bigr\rangle + \frac{1}{r} \frac{\partial W}{\partial \theta} \bigl\langle - \sin\theta , \cos\theta \bigr\rangle \\ \end{align*} We often rewrite the gradient vector in polar coordinates in terms of the orthonormal basis $\mathbf{a}_\theta$ and $\mathbf{b}_\theta$, defined by \[ \mathbf{a}_\theta = \bigl\langle \cos\theta , \sin\theta\bigr\rangle, \quad \mathbf{b}_\theta = \bigl\langle - \sin\theta , \cos\theta \bigr\rangle. \] This basis is called the polar basis.