Question

Verify that the Laplacian has the form in 2=2r2+1rr+1r22θ2+2z2\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

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Cylindrical coordinates are defined by the relations:

x=rcosθy=rsinθz=zr2=x2+y2tanθ=yxz=z\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 5.35.3 we know that in cylindrical coordinates the Laplacian has the form:

2=2r2+1rr+1r22θ2+2z2\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 r2=x2+y2r^2=x^2+y^2, with respect to xx and yy, and we get

2rrx=2x2rry=2y    rx=xr=cosθry=yr=sinθ\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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