## Laplace operator in Spherical Coordinates

Laplace operator plays a vital role in many outstanding problems, such as Helmholtz’s equations, Allen-Cahn’s equations, sine-Gordon equations, and so forth. In Cartesian coordinates, it is denoted by $\Delta_{C}$ and has the following form.

$\displaystyle{\Delta_{C} = \sum_{i=1}^{3}\dfrac{\partial^2}{\partial x_{i}^{2}}}.$

However, its form in spherical coordinates is

$\Delta_{S} = \dfrac{1}{r^2}\dfrac{\partial}{\partial r}\left(r^2 \dfrac{\partial}{\partial r}\right) + \dfrac{1}{r^2 \sin{\theta}}\dfrac{\partial}{\partial \theta}\left(\sin{\theta}\dfrac{\partial}{\partial \theta} \right) + \dfrac{1}{r^2 \sin^2{\theta}}\dfrac{\partial^2}{\partial \phi^2}.$

We aim to prove this formula, means that converting the operator from Cartesian to spherical coordinates is in order.

As usual, we have the following transformation.

$x_1 = r\sin{\theta}\cos{\phi}, \quad x_2 = r\sin{\theta}\sin{\phi}, \quad x_3 = r\cos{\phi}, \quad (1)$

where $r>0, \theta\in [0,\pi], \phi\in [0,2\pi)$ are called radius, inclination and azimuth, respectively.

It follows that

$\displaystyle{r = \sqrt{\sum_{i=1}^{3} {x_{i}^2}}, \quad \cos{\theta} = \dfrac{x_3}{\sqrt{\sum_{i=1}^{3} {x_{i}^2}}}, \quad \sin{\theta} = \dfrac{\sqrt{\sum_{i=1}^{2} {x_{i}^2}}}{\sqrt{\sum_{i=1}^{3} {x_{i}^2}}}},$

$\displaystyle{\cos{\phi} = \dfrac{x_1}{\sqrt{\sum_{i=1}^{2} {x_{i}^2}}}, \quad \sin{\phi} = \dfrac{x_2}{\sqrt{\sum_{i=1}^{2} {x_{i}^2}}}}. \quad (2)$

Combining (1) and (2), we first have

$\dfrac{\partial r}{\partial x_1} = \dfrac{x_1}{r} = \sin{\theta}\cos{\phi},$

$\dfrac{\partial r}{\partial x_2} = \dfrac{x_2}{r} = \sin{\theta}\sin{\phi},$

$\dfrac{\partial r}{\partial x_3} = \dfrac{x_3}{r} = \cos{\theta}. \quad (3)$

Moreover, the derivatives of $\theta$ with respect to $x_1,x_2,x_3$ are

$\dfrac{\partial \theta}{\partial x_1} = \dfrac{\cos{\theta}\cos{\phi}}{r},$

$\dfrac{\partial \theta}{\partial x_2} = \dfrac{\cos{\theta}\sin{\phi}}{r},$

$\dfrac{\partial \theta}{\partial x_3} = -\dfrac{\sin{\theta}}{r}. \quad (4)$

Simultaneously, the derivatives of $\phi$ are simply

$\dfrac{\partial \phi}{\partial x_1} = -\dfrac{\sin{\phi}}{r\sin{\theta}},\quad \dfrac{\partial \phi}{\partial x_2} = \dfrac{\cos{\phi}}{r\sin{\theta}}. \quad (5)$

Observing chain rule is now necessary. For $i = {1;2;3}$,

$\displaystyle{\frac{\partial}{\partial x_i} = \frac{\partial}{\partial r}\frac{\partial r}{\partial x_i} + \frac{\partial}{\partial\theta}\frac{\partial\theta}{\partial x_i} + \frac{\partial}{\partial\phi}\frac{\partial\phi}{\partial x_i}}. \quad (6)$

By substituting (3)-(4)-(5) into (6), we then have

$\displaystyle{\frac{\partial}{\partial x_1} = \sin\theta\cos\phi\frac{\partial}{\partial r} + \frac{\cos\theta\cos\phi}{r}\frac{\partial}{\partial\theta} - \frac{1}{r}\frac{\sin\phi}{\sin\theta}\frac{\partial}{\partial\phi}}, \quad (7)$

$\displaystyle{\frac{\partial}{\partial x_2} = \sin\theta\sin\phi\frac{\partial}{\partial r} + \frac{\cos\theta\sin\phi}{r}\frac{\partial}{\partial\theta} + \frac{1}{r}\frac{\cos\phi}{\sin\theta}\frac{\partial}{\partial\phi}}, \quad (8)$

$\displaystyle{\frac{\partial}{\partial x_3} = \cos\theta\frac{\partial}{\partial r} - \frac{\sin\theta}{r}\frac{\partial}{\partial\theta}}. \quad (9)$

In brief, we shall obtain from (7)-(8)-(9) that

$\displaystyle{\Delta_{S} = \frac{\partial^2}{\partial r^2} + \frac{1}{r^2}\frac{\partial^2}{\partial\theta^2} + \frac{1}{r^2\sin^2{\theta}}\frac{\partial^2}{\partial\phi^2} + \frac{2}{r}\frac{\partial}{\partial r} + \frac{\cos\theta}{r^2\sin\theta}\frac{\partial}{\partial\theta}},$

which gives the desired result.

It is so surprising that the final result is so short after directly having extremely heavy computation. It requires meticulosity, diligence and none of scare when confront with tremendous computations. But do not try to pay more attention to this, the important thing we finally obtain is the formula of Laplace operator in spherical coordinates. From this formula, we also would like to introduce the more general case which is known as the Laplace-Beltrami operator. However, for further posts, we still aim to focus on many issues about Laplace operator in spherical coordinates on account of a wide range of applications.