[This is a brute-force way to compute the Laplacian in spherical coordinates.]

Refs:

• “Advanced Engineering Mathematics” by Kreyszig.

• Youtube video: “Converting the Laplacian to Spherical Coords” by Todd Yilk. Link to the video: Link.

[Spherical Coordinates]

Recall Spherical Coordinates

Note that spherical to cartesian conversion is

\[{ x = r \cos(\theta) \sin(\phi), \quad y = r \sin(\theta) \sin(\phi), \quad z = r \cos(\phi) . }\]

Note that cartesian to spherical “conversion” is

\[{ r = \sqrt{x ^2 + y ^2 + z ^2}, \quad \cos(\theta) = \frac{x}{\sqrt{x ^2 + y ^2}} , \quad \cos(\phi) = \frac{z}{\sqrt{x ^2 + y ^2 + z ^2}} . }\]

[Laplacian in spherical coordinates]

Consider a scalar function ${ u . }$

Q) Can we express the Laplacian

\[{ \nabla ^2 u = u _{xx} + u _{yy} + u _{zz} }\]

in terms of the derivatives of ${ u }$ wrt ${ r, \theta, \phi }$?

Note that

\[{ {\begin{aligned} &\, u _x \\ = & u _r r _x + u _{\theta} \theta _{x} + u _{\phi} \phi _{x} . \end{aligned}} }\]

Hence

\[{ {\begin{aligned} &\, u _{xx} \\ = &\, (u _r r _x) _x + (u _{\theta} \theta _{x}) _{x} + (u _{\phi} \phi _x) _{x} \\ = &\, u _{rx} r _{x} + u _r r _{xx} + u _{\theta x} \theta _{x} + u _{\theta} \theta _{xx} + u _{\phi x} \phi _{x} + u _{\phi} \phi _{xx} \\ = &\, (u _{rr} r _x + u _{r\theta} \theta _{x} + u _{r \phi} \phi _{x}) r _x \\ + &\, u _{r} r _{xx} \\ + &\, (u _{\theta r} r _x + u _{\theta \theta} \theta _x + u _{\theta \phi} \phi _{x} ) \theta _x \\ + &\, u _{\theta} \theta _{xx} \\ + &\, (u _{\phi r} r _x + u _{\phi \theta} \theta _x + u _{\phi \phi} \phi _x) \phi _x \\ + &\, u _{\phi} \phi _{xx} \\ = &\, { \begin{pmatrix} r _{xx} &\theta _{xx} &\phi _{xx} \end{pmatrix} } {\begin{pmatrix} u _{r} \\ u _{\theta} \\ u _{\phi} \end{pmatrix}} + {\begin{pmatrix} r _{x} &\theta _{x} &\phi _{x} \end{pmatrix}} {\begin{pmatrix} u _{rr} &u _{r \theta} &u _{r \phi} \\ u _{\theta r} &u _{\theta \theta} &u _{\theta \phi} \\ u _{\phi r} &u _{\phi \theta} &u _{\phi \phi} \end{pmatrix}} {\begin{pmatrix} r _{x} \\ \theta _{x} \\ \phi _{x} \end{pmatrix}} . \end{aligned}} }\]

Likewise for ${ u _{yy}, u _{zz}. }$

Hence the Laplacian

\[{ {\begin{aligned} &\, \nabla ^2 u \\ = &\, \sum _{\nu = x, y, z} u _{\nu \nu} \\ = &\, \sum _{\nu = x, y, z} { \begin{pmatrix} r _{\nu \nu} &\theta _{\nu \nu} &\phi _{\nu \nu} \end{pmatrix} } {\begin{pmatrix} u _{r} \\ u _{\theta} \\ u _{\phi} \end{pmatrix}} + \sum _{\nu = x, y, z} {\begin{pmatrix} r _{\nu} &\theta _{\nu} &\phi _{\nu} \end{pmatrix}} {\begin{pmatrix} u _{rr} &u _{r \theta} &u _{r \phi} \\ u _{\theta r} &u _{\theta \theta} &u _{\theta \phi} \\ u _{\phi r} &u _{\phi \theta} &u _{\phi \phi} \end{pmatrix}} {\begin{pmatrix} r _{\nu} \\ \theta _{\nu} \\ \phi _{\nu} \end{pmatrix}}. \end{aligned}} }\]

Note that atleast in the positive octant

\[{ {\begin{aligned} &\, r _x = \frac{x}{\sqrt{x ^2 + y ^2 + z ^2}}, \\ &\, \left( \frac{-y}{\sqrt{x ^2 + y ^2}} \right) \theta _x = \frac{\sqrt{x ^2 + y ^2} - x \left( \frac{x}{\sqrt{x ^2 + y ^2}} \right)}{x ^2 + y ^2}, \\ &\, \left( \frac{- \sqrt{x ^2 + y ^2}}{\sqrt{x ^2 + y ^2 + z ^2}} \right) \phi _x = \frac{-z \frac{x}{\sqrt{x ^2 + y ^2 + z ^2}}}{x ^2 + y ^2 + z ^2} \end{aligned}} }\]

Hence

\[{ {\begin{aligned} &\, r _x = \frac{x}{\sqrt{x ^2 + y ^2 + z ^2}}, \\ &\, \theta _x = \frac{-y}{x ^2 + y ^2} , \\ &\, \phi _{x} = \frac{zx}{(x ^2 + y ^2 + z ^2) \sqrt{x ^2 + y ^2}} . \end{aligned}} }\]

That is

\[{ {\begin{aligned} &\, r _x = \frac{r \cos(\theta) \sin(\phi)}{r} , \\ &\, \theta _{x} = \frac{- r \sin(\theta) \sin(\phi) }{r ^2 \sin(\phi) ^2}, \\ &\, \phi _{x} = \frac{(r \cos(\phi)) (r \cos(\theta) \sin(\phi))}{r ^2 (r \sin(\phi)) }. \end{aligned}} }\]

That is

\[{ \boxed{ {\begin{aligned} &\, r _x = \cos(\theta) \sin(\phi) , \\ &\, \theta _{x} = \frac{- \sin(\theta)}{r \sin(\phi)} , \\ &\, \phi _{x} = \frac{\cos(\phi) \cos(\theta)}{r } . \end{aligned}} } }\]

Hence

\[{ {\begin{aligned} &\, r _{xx} = - \sin(\theta) \sin(\phi) \theta _{x} + \cos(\theta) \cos(\phi) \phi _{x}, \\ &\, \theta _{xx} = \frac{- \cos(\theta) \theta _{x} r \sin(\phi) + \sin(\theta) (r _{x} \sin(\phi) + r \cos(\phi) \phi _{x} ) }{r ^2 \sin(\phi) ^2} , \\ &\, \phi _{xx} = \frac{(- \sin(\phi) \phi _{x} \cos(\theta) - \cos(\phi) \sin(\theta) \theta _{x} )r - \cos(\phi) \cos(\theta) r _{x}}{r ^2} . \end{aligned}} }\]

That is

\[{ {\begin{aligned} &\, r _{xx} = \frac{(\sin(\theta)) ^2}{r} + \frac{(\cos(\theta)) ^2 (\cos(\phi)) ^2}{r} , \\ &\, \theta _{xx} = \frac{\sin(\theta) \cos(\theta)}{r ^2 \sin ^2 (\phi) } + \frac{\sin(\theta) \cos(\theta)}{r ^2} + \frac{1}{r ^2} \cot ^{2} (\phi) \sin(\theta) \cos(\theta), \\ &\, \phi _{xx} = \frac{- \sin(\phi) \cos(\phi) \cos ^2 (\theta)}{r ^2} + \frac{\cot (\phi) \sin ^2 (\theta)}{r ^2} - \frac{ \sin(\phi) \cos(\phi) \cos ^2 (\theta)}{r ^2} . \end{aligned}} }\]

That is

\[{ \boxed{ {\begin{aligned} &\, r _{xx} = \frac{(\sin(\theta)) ^2}{r} + \frac{(\cos(\theta)) ^2 (\cos(\phi)) ^2}{r} , \\ &\, \theta _{xx} = \frac{\sin(\theta) \cos(\theta)}{r ^2 \sin ^2 (\phi) } + \frac{\sin(\theta) \cos(\theta)}{r ^2} + \frac{1}{r ^2} \cot ^{2} (\phi) \sin(\theta) \cos(\theta), \\ &\, \phi _{xx} = \frac{- 2 \sin(\phi) \cos(\phi) \cos ^2 (\theta)}{r ^2} + \frac{\cot (\phi) \sin ^2 (\theta)}{r ^2} . \end{aligned}} } }\]

Note that atleast in the positive octant

\[{ {\begin{aligned} &\, r _y = \frac{y}{\sqrt{x ^2 + y ^2 + z ^2}}, \\ &\, \left( \frac{-y}{\sqrt{x ^2 + y ^2}} \right) \theta _{y} = \frac{-xy}{(x ^2 + y ^2) ^{3/2}}, \\ &\, \left( \frac{- \sqrt{x ^2 + y ^2}}{\sqrt{x ^2 + y ^2 + z ^2}} \right) \phi _{y} = \frac{-yz}{(x ^2 + y ^2 + z ^2) ^{3/2}} . \end{aligned}} }\]

Hence

\[{ {\begin{aligned} &\, r _y = \frac{y}{\sqrt{x ^2 + y ^2 + z ^2}}, \\ &\, \theta _{y} = \frac{x}{x ^2 + y ^2}, \\ &\, \phi _{y} = \frac{yz}{(x ^2 + y ^2 + z ^2) \sqrt{x ^2 + y ^2}} . \end{aligned}} }\]

That is

\[{ {\begin{aligned} &\, r _y = \frac{r \sin(\theta) \sin(\phi)}{r}, \\ &\, \theta _{y} = \frac{r \cos(\theta) \sin(\phi)}{r ^2 \sin ^2 (\phi)}, \\ &\, \phi _{y} = \frac{(r \sin(\theta) \sin(\phi)) (r \cos(\phi))}{r ^2 (r \sin(\phi)) } . \end{aligned}} }\]

That is

\[{ \boxed{ {\begin{aligned} &\, r _y = \sin(\theta) \sin(\phi), \\ &\, \theta _{y} = \frac{\cos(\theta)}{r \sin (\phi)}, \\ &\, \phi _{y} = \frac{\sin(\theta) \cos(\phi)}{r } . \end{aligned}} } }\]

Hence

\[{ {\begin{aligned} &\, r _{yy} = \cos(\theta) \theta _{y} \sin(\phi) + \sin(\theta) \cos(\phi) \phi _{y} , \\ &\, \theta _{yy} = \frac{- \sin(\theta) \theta _{y} r \sin(\phi) - \cos(\theta) (r _{y} \sin(\phi) + r \cos(\phi) \phi _{y}) }{r ^2 \sin ^2 (\phi) }, \\ &\, \phi _{yy} = \frac{(\cos(\theta) \theta _{y} \cos(\phi) - \sin(\theta) \sin(\phi) \phi _{y} ) r - (\sin(\theta) \cos(\phi)) r _{y} }{r ^2} . \end{aligned}} }\]

That is

\[{ {\begin{aligned} &\, r _{yy} = \frac{\cos ^2 (\theta)}{r} + \frac{1}{r} \sin ^2 (\theta) \cos ^2 (\phi) , \\ &\, \theta _{yy} = \frac{- \sin(\theta) \cos(\theta)}{r ^2 \sin ^2 (\phi)} + \frac{- \sin(\theta) \cos(\theta)}{r ^2} + \frac{- \sin(\theta) \cos(\theta) \cot ^2 (\phi)}{r ^2} , \\ &\, \phi _{yy} = \frac{\cos ^2 (\theta) \cot (\phi)}{r ^2} + \frac{- \sin ^2 (\theta) \sin(\phi) \cos(\phi)}{r ^2} + \frac{- \sin ^2 (\theta) \sin(\phi) \cos(\phi)}{r ^2} . \end{aligned}} }\]

That is

\[{ \boxed{ {\begin{aligned} &\, r _{yy} = \frac{\cos ^2 (\theta)}{r} + \frac{1}{r} \sin ^2 (\theta) \cos ^2 (\phi) , \\ &\, \theta _{yy} = \frac{- \sin(\theta) \cos(\theta)}{r ^2 \sin ^2 (\phi)} + \frac{- \sin(\theta) \cos(\theta)}{r ^2} + \frac{- \sin(\theta) \cos(\theta) \cot ^2 (\phi)}{r ^2} , \\ &\, \phi _{yy} = \frac{\cos ^2 (\theta) \cot (\phi)}{r ^2} + \frac{- 2 \sin ^2 (\theta) \sin(\phi) \cos(\phi)}{r ^2} . \end{aligned}} } }\]

Note that atleast in the positive octant

\[{ {\begin{aligned} &\, r _z = \frac{z}{\sqrt{x ^2 + y ^2 + z ^2}}, \\ &\, \theta _{z} = 0, \\ &\, \left( \frac{- \sqrt{x ^2 + y ^2}}{\sqrt{x ^2 + y ^2 + z ^2}} \right) \phi _{z} = \frac{x ^2 + y ^2}{(x ^2 + y ^2 + z ^2) ^{3/2}} . \end{aligned}} }\]

Hence

\[{ {\begin{aligned} &\, r _z = \frac{z}{\sqrt{x ^2 + y ^2 + z ^2}}, \\ &\, \theta _{z} = 0, \\ &\, \phi _{z} = \frac{- \sqrt{x ^2 + y ^2}}{(x ^2 + y ^2 + z ^2)} . \end{aligned}} }\]

That is

\[{ {\begin{aligned} &\, r _z = \frac{r \cos(\phi)}{r}, \\ &\, \theta _{z} = 0, \\ &\, \phi _{z} = \frac{- r \sin(\phi)}{r ^2} . \end{aligned}} }\]

That is

\[{ \boxed{ {\begin{aligned} &\, r _z = \cos(\phi), \\ &\, \theta _{z} = 0, \\ &\, \phi _{z} = \frac{- \sin(\phi)}{r} . \end{aligned}} } }\]

Hence

\[{ {\begin{aligned} &\, r _{zz} = - \sin(\phi) \phi _{z}, \\ &\, \theta _{zz} = 0, \\ &\, \phi _{zz} = \frac{- \cos(\phi) \phi _{z} r + \sin(\phi) r _{z}}{r ^2} . \end{aligned}} }\]

That is

\[{ \boxed{ {\begin{aligned} &\, r _{zz} = \frac{\sin ^2 (\phi)}{r}, \\ &\, \theta _{zz} = 0, \\ &\, \phi _{zz} = \frac{2 \sin(\phi) \cos(\phi)}{r ^2} . \end{aligned}} } }\]

We are now ready to express ${ \nabla ^2 u }$ in spherical coordinates.

Note that

\[{ {\begin{aligned} &\, \nabla ^2 u \\ = &\, \sum _{\nu = x, y, z} u _{\nu \nu} \\ = &\, \sum _{\nu = x, y, z} { \begin{pmatrix} r _{\nu \nu} &\theta _{\nu \nu} &\phi _{\nu \nu} \end{pmatrix} } {\begin{pmatrix} u _{r} \\ u _{\theta} \\ u _{\phi} \end{pmatrix}} + \sum _{\nu = x, y, z} {\begin{pmatrix} r _{\nu} &\theta _{\nu} &\phi _{\nu} \end{pmatrix}} {\begin{pmatrix} u _{rr} &u _{r \theta} &u _{r \phi} \\ u _{\theta r} &u _{\theta \theta} &u _{\theta \phi} \\ u _{\phi r} &u _{\phi \theta} &u _{\phi \phi} \end{pmatrix}} {\begin{pmatrix} r _{\nu} \\ \theta _{\nu} \\ \phi _{\nu} \end{pmatrix}}. \end{aligned}} }\]

We will look at each term separately.

Note that

\[{ {\begin{aligned} &\, u _{r} \text{ term} \\ = &\, u _{r} \sum _{\nu = x, y, z} r _{\nu \nu} \\ = &\, \frac{2}{r} u _{r} . \end{aligned}} }\]

Note that

\[{ {\begin{aligned} &\, u _{\theta} \text{ term} \\ = &\, u _{\theta} \sum _{\nu = x, y, z} \theta _{\nu \nu} \\ = &\, 0 . \end{aligned}} }\]

Note that

\[{ {\begin{aligned} &\, u _{\phi} \text{ term} \\ = &\, u _{\phi} \sum _{\nu = x, y, z} \phi _{\nu \nu} \\ = &\, \frac{\cot(\phi)}{r ^2} u _{\phi} . . \end{aligned}} }\]

Note that

\[{ {\begin{aligned} &\, u _{rr} \text{ term} \\ = &\, u _{rr} \sum _{\nu = x, y, z} r _{\nu} ^{2} \\ = &\, u _{rr} . \end{aligned}} }\]

Note that

\[{ {\begin{aligned} &\, u _{r\theta} \text{ term} \\ = &\, 2 u _{r \theta} \sum _{\nu = x, y, z} r _{\nu} \theta _{\nu} \\ = &\, 0 . \end{aligned}} }\]

Note that

\[{ {\begin{aligned} &\, u _{r\phi} \text{ term} \\ = &\, 2 u _{r \phi} \sum _{\nu = x, y, z} r _{\nu} \phi _{\nu} \\ = &\, 0 . \end{aligned}} }\]

Note that

\[{ {\begin{aligned} &\, u _{\theta \theta} \text{ term} \\ = &\, u _{\theta \theta} \sum _{\nu = x, y, z} \theta _{\nu} ^{2} \\ = &\, \frac{1}{r ^2 \sin ^{2} (\phi)} u _{\theta \theta} . \end{aligned}} }\]

Note that

\[{ {\begin{aligned} &\, u _{\theta \phi} \text{ term} \\ = &\, 2 u _{\theta \phi} \sum _{\nu = x, y, z} \theta _{\nu} \phi _{\nu} \\ = &\, 0 . \end{aligned}} }\]

Note that

\[{ {\begin{aligned} &\, u _{\phi \phi} \text{ term} \\ = &\, u _{\phi \phi} \sum _{\nu = x, y, z} \phi _{\nu} ^{2} \\ = &\, \frac{1}{r ^2} u _{\phi \phi} . \end{aligned}} }\]

Hence

\[{ \boxed{ {\begin{aligned} &\, \nabla ^{2} u \\ = &\, \frac{2}{r} u _{r} + \frac{\cot(\phi)}{r ^2} u _{\phi} + u _{rr} + \frac{1}{r ^2 \sin ^{2} (\phi)} u _{\theta \theta} + \frac{1}{r ^2} u _{\phi \phi} \end{aligned}} } }\]

Equivalently, clubbing “${ u _{r} }$ and ${ u _{rr} }$ terms” and “${ u _{\phi} }$ and ${ u _{\phi \phi} }$ terms”, we have

\[{ \boxed{ {\begin{aligned} &\, \nabla ^{2} u \\ = &\, \frac{1}{r ^2} \frac{\partial}{\partial r} \left(r ^2 \frac{\partial u}{\partial r} \right) + \frac{1}{r ^2 \sin(\phi)} \frac{\partial}{\partial \phi} \left( \sin (\phi) \frac{\partial u}{\partial \phi} \right) + \frac{1}{r ^2 \sin ^2 (\phi)} \frac{\partial ^{2} u}{\partial \theta ^{2}} \end{aligned}} } }\]

as needed. ${ \blacksquare }$