Give Laplace’s equation in polar coordinates. In other words, if \(x=r\cos\theta\), \(y=r\sin\theta\), write down a relation equivalent to \[ \frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=0 \] in terms of \(r\), \(\theta\), and the partial derivatives of \(f\) with respect to \(r\) and \(\theta\).
Solution
1. Laplace’s equation in polar coordinates is\[\frac{\partial^2 f}{\partial r^2} + \frac{1}{r}\frac{\partial f}{\partial r} + \frac{1}{r^2}\frac{\partial^2 f}{\partial \theta^2} = 0.\]
change of variables from \((x,y)\) to \((r,\theta)\).
2. Derivation.
2.1. \(r = \sqrt{x^2 + y^2}\) and \(\theta = \arctan(y/x)\), so \(\frac{\partial r}{\partial x} = \cos\theta\), \(\frac{\partial r}{\partial y} = \sin\theta\), \(\frac{\partial \theta}{\partial x} = -\frac{\sin\theta}{r}\), \(\frac{\partial \theta}{\partial y} = \frac{\cos\theta}{r}\).
direct differentiation.
2.2. By the chain rule, \(\frac{\partial f}{\partial x} = \cos\theta\, f_r - \frac{\sin\theta}{r} f_\theta\) and \(\frac{\partial f}{\partial y} = \sin\theta\, f_r + \frac{\cos\theta}{r} f_\theta\).
chain rule with 1.
2.3. Applying the chain rule again and simplifying gives \(\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = f_{rr} + \frac{1}{r} f_r + \frac{1}{r^2} f_{\theta\theta}\).
a standard (but lengthy) computation; the cross terms cancel.
3. Hence the relation equivalent to \(\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0\) is\[f_{rr} + \frac{1}{r} f_r + \frac{1}{r^2} f_{\theta\theta} = 0.\]
1 and 2.
4. Q.E.D.
3.