Verify that the Laplacian has the form in $\nabla^{2}=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r} \frac{\partial}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2}}{\partial \theta^{2}}+\frac{\partial^{2}}{\partial z^{2}}$ in cylindrical coordinates.

Solution

VerifiedCylindrical coordinates are defined by the relations:

$\begin{align*} x&=r\cos \theta\\ y&=r\sin \theta\\ z&=z\\\\ r^2&=x^2+y^2\\ \tan \theta &=\frac{y}{x}\\ z&=z\\ \end{align*}$

From equation $5.3$ we know that in cylindrical coordinates the Laplacian has the form:

$\bigtriangledown^2=\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r} +\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}+\frac{\partial^2}{\partial z^2}$

We differentiating the equation $r^2=x^2+y^2$, with respect to $x$ and $y$, and we get

$\begin{align*} 2r\frac{\partial r}{\partial x}&=2x\\ 2r\frac{\partial r}{\partial y}&=2y\\\\ \implies \frac{\partial r}{\partial x}&=\frac{x}{r}\\ &=\cos \theta\\\\ \frac{\partial r}{\partial y}&=\frac{y}{r}\\ &=\sin \theta\\ \end{align*}$

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