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All the Steps in Transforming the Laplace Equation (Δf=0)
from Rectangular to Cylindrical and to Spherical Coordinates

(Copyright©2023 by Daniel B. Sedory.)

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Obtaining the Second Order Partial Differentials:  ∂2/∂y2

  The second order partial derivative of the function f (with respect to y) is also found by substituting some of the right-sides of the equations already shown there into the first order differentials highlighted below, and then making use of the chain rule, ddx(uw)=udwdx+wdudx, as before:

∂2f∂y2= sin⁡θ sin⁡ϕ ∂2 f∂r ∂y+ cos⁡θ sin⁡ϕr ∂2f∂θ ∂y + cos⁡ϕrsin⁡θ ∂2f∂ϕ ∂y=sin⁡θ sin⁡ϕ (∂∂ r ∂f∂y)+cos⁡θ sin⁡ϕr (∂f∂θ ∂f∂y)+cos⁡ϕrsin⁡θ (∂f∂ϕ ∂f∂y)=sin⁡θ sin⁡ϕ [∂∂ r (sin⁡θ sin⁡ϕ ∂f∂ r+cos⁡θ sin⁡ϕr ∂f∂θ+cos⁡ϕrsin⁡θ ∂f∂ϕ)]......  +cos⁡θ sin⁡ϕr [∂∂θ (sin⁡θ sin⁡ϕ ∂f∂ r+cos⁡θ sin⁡ϕr ∂f∂θ+cos⁡ϕrsin⁡θ ∂f∂ϕ)]......  +cos⁡ϕrsin⁡θ [∂∂ϕ (sin⁡θ sin⁡ϕ ∂f∂ r+cos⁡θ sin⁡ϕr ∂f∂θ+cos⁡ϕr sin⁡θ ∂f∂ϕ)] .

 Upon comparison of what follows to the equations for  ∂2f/∂x2,  they will be found to vary only in the sign (+/-) in front of some terms and some changes from cos to sin or vice versa. All of the differentiations being quite similar to those on the previous page, producing:

∂2f∂y2=sin⁡θ sin⁡ϕ (sin⁡θ sin⁡ϕ ∂2f∂ r2+cos⁡θ sin⁡ϕr ∂2f∂r ∂θ−cos⁡θ sin⁡ϕr2 ∂f∂θ......  +cos⁡ϕrsin⁡θ ∂2f∂r ∂ϕ−cos⁡ϕr2sin⁡θ ∂f∂ϕ)......  +cos⁡θ sin⁡ϕr (sin⁡θ sin⁡ϕ ∂2f∂r ∂θ+cos⁡θ sin⁡ϕ ∂f∂r+cos⁡θ sin⁡ϕr ∂2f∂θ2......  −sin⁡θ sin⁡ϕr ∂f∂θ+cos⁡ϕr sin⁡θ ∂2f∂θ ∂ϕ−cos⁡θ cos⁡ϕr sin2⁡θ ∂f∂ϕ)......  +cos⁡ϕrsin⁡θ (sin⁡θ sin⁡ϕ ∂2f∂r ∂ϕ+sin⁡θ cos⁡ϕ ∂f∂r+cos⁡θ sin⁡ϕr ∂2f∂θ ∂ϕ......  +cos⁡θ cos⁡ϕr ∂f∂θ+cos⁡ϕr sin⁡θ ∂2f∂ϕ2−sin⁡ϕr sin⁡θ ∂f∂ϕ) .

And after multiplying through to eliminate the parentheses, we get:

∂2f∂y2=sin2⁡θ sin2⁡ϕ ∂2f∂ r2+sin⁡θ cos⁡θ sin2⁡ϕr ∂2f∂r ∂θ−sin⁡θ cos⁡θ sin2⁡ϕr2 ∂f∂θ ......  +sin⁡θ sin⁡ϕ cos⁡ϕr sin⁡θ ∂2f∂r ∂ϕ−sin⁡ϕ cos⁡ϕr2 ∂f∂ϕ+sin⁡θ cos⁡θ sin2⁡ϕr ∂2f∂r ∂θ ......  +cos2⁡θ sin2⁡ϕr ∂f∂r+cos2⁡θ sin2⁡ϕr2 ∂2f∂θ2−cos⁡θ sin⁡θ sin2⁡ϕr2 ∂f∂θ ......  +cos⁡θ sin⁡ϕ cos⁡ϕr2 sin⁡θ ∂2f∂θ ∂ϕ−cos2⁡θ sin⁡ϕ cos⁡ϕr2 sin2⁡θ ∂f∂ϕ+sin⁡ϕ cos⁡ϕr ∂2f∂r ∂ϕ ......  +cos2⁡ϕr ∂f∂r+cos⁡θ sin⁡ϕ cos⁡ϕr2 sin⁡θ ∂2f∂θ ∂ϕ+cos⁡θ cos2⁡ϕr2 sin⁡θ ∂f∂θ ......  +cos2⁡ϕr2 sin2⁡θ ∂2f∂ϕ2−sin⁡ϕ cos⁡ϕr2 sin2⁡θ ∂f∂ϕ .

Which after combining some of the terms, leaves us with:

∂2f∂y2=sin2⁡θ sin2⁡ϕ ∂2f∂ r2+2 sin⁡θ cos⁡θ sin2⁡ϕr ∂2f∂r ∂θ−2 sin⁡θ cos⁡θ sin2⁡ϕr2 ∂f∂θ ......  +2 sin⁡ϕ cos⁡ϕr ∂2f∂r ∂ϕ−sin⁡ϕ cos⁡ϕr2 ∂f∂ϕ+cos2⁡θ sin2⁡ϕr ∂f∂r ......  +cos2⁡θ sin2⁡ϕr2 ∂2f∂θ2+2 cos⁡θ sin⁡ϕ cos⁡ϕr2 sin⁡θ ∂2f∂θ ∂ϕ−cos2⁡θ sin⁡ϕ cos⁡ϕr2 sin2⁡θ ∂f∂ϕ ......  +cos2⁡ϕr ∂f∂r+cos⁡θ cos2⁡ϕr2 sin⁡θ ∂f∂θ+cos2⁡ϕr2 sin2⁡θ ∂2f∂ϕ2−sin⁡ϕ cos⁡ϕr2 sin2⁡θ ∂f∂ϕ .

 

Obtaining the Second Order Partial Differentials:  ∂2/∂z2

  Again, as noted above, the second order partial derivative of the function f (this time, with respect to z) is found by substituting some of the right-sides of the equations already shown on the previous page into the first order differentials highlighted below, and then making use of the chain rule as before:

∂2f∂z2= cos⁡θ ∂2 f∂r ∂z− sin⁡θr ∂2f∂θ ∂z=cos⁡θ (∂∂ r ∂f∂z)−sin⁡θr (∂∂θ ∂f∂z)=cos⁡θ [∂∂ r (cos⁡θ ∂f∂ r − sin⁡θr ∂f∂θ)]......  −sin⁡θr [∂∂θ (cos⁡θ ∂f∂ r − sin⁡θr ∂f∂θ)] .

After carrying out the differentiations, we obtain:

∂2f∂z2= cos⁡θ (cos⁡θ ∂2f∂ r2−sin⁡θr ∂2f∂r ∂θ+sin⁡θr2 ∂f∂θ)......  −sin⁡θr (cos⁡θ ∂2f∂r ∂θ−sin⁡θ ∂f∂r−sin⁡θr ∂2f∂θ2−cos⁡θr ∂f∂θ) .
∂2f∂z2=cos2⁡θ ∂2f∂ r2−sin⁡θ cos⁡θr ∂2f∂r ∂θ+sin⁡θ cos⁡θr2 ∂f∂θ......  −sin⁡θ cos⁡θr ∂2f∂r ∂θ+sin2⁡θr ∂f∂r+sin2⁡θr2 ∂2f∂θ2+sin⁡θ cos⁡θr2 ∂f∂θ .
∂2f∂z2=cos2⁡θ ∂2f∂ r2−2 sin⁡θ cos⁡θr ∂2f∂r ∂θ......  +2 sin⁡θ cos⁡θr2 ∂f∂θ+sin2⁡θr ∂f∂r+sin2⁡θr2 ∂2f∂θ2 .

 

The Sum of All Three Second Order Partial Differentials

  Here we sum up all three of the previously derived terms of the second order differentials of f with respect to x, y and z:

∂2f∂x2+∂2f∂y2+∂2f∂z2==sin2⁡θ cos2⁡ϕ ∂2f∂ r2+2 sin⁡θ cos⁡θ cos2⁡ϕr ∂2f∂r ∂θ−2 sin⁡θ cos⁡θ cos2⁡ϕr2 ∂f∂θ ......  −2 sin⁡ϕ cos⁡ϕr ∂2f∂r ∂ϕ+sin⁡ϕ cos⁡ϕr2 ∂f∂ϕ+cos2⁡θ cos2⁡ϕr ∂f∂r ......  +cos2⁡θ cos2⁡ϕr2 ∂2f∂θ2−2 cos⁡θ sin⁡ϕ cos⁡ϕr2 sin⁡θ ∂2f∂θ ∂ϕ+cos2⁡θ sin⁡ϕ cos⁡ϕr2 sin2⁡θ ∂f∂ϕ ......  +sin2⁡ϕr ∂f∂r+cos⁡θ sin2⁡ϕr2 sin⁡θ ∂f∂θ+sin2⁡ϕr2 sin2⁡θ ∂2f∂ϕ2+sin⁡ϕ cos⁡ϕr2 sin2⁡θ ∂f∂ϕ ......  +sin2⁡θ sin2⁡ϕ ∂2f∂ r2+2 sin⁡θ cos⁡θ sin2⁡ϕr ∂2f∂r ∂θ−2 sin⁡θ cos⁡θ sin2⁡ϕr2 ∂f∂θ ......  +2 sin⁡ϕ cos⁡ϕr ∂2f∂r ∂ϕ−sin⁡ϕ cos⁡ϕr2 ∂f∂ϕ+cos2⁡θ sin2⁡ϕr ∂f∂r ......  +cos2⁡θ sin2⁡ϕr2 ∂2f∂θ2+2 cos⁡θ sin⁡ϕ cos⁡ϕr2 sin⁡θ ∂2f∂θ ∂ϕ−cos2⁡θ sin⁡ϕ cos⁡ϕr2 sin2⁡θ ∂f∂ϕ ......  +cos2⁡ϕr ∂f∂r+cos⁡θ cos2⁡ϕr2 sin⁡θ ∂f∂θ+cos2⁡ϕr2 sin2⁡θ ∂2f∂ϕ2−sin⁡ϕ cos⁡ϕr2 sin2⁡θ ∂f∂ϕ......  +cos2⁡θ ∂2f∂ r2−2 sin⁡θ cos⁡θr ∂2f∂r ∂θ......  +2 sin⁡θ cos⁡θr2 ∂f∂θ+sin2⁡θr ∂f∂r+sin2⁡θr2 ∂2f∂θ2 .

Ten of the terms cancel out immediately. However, in order to reduce all of the remaining terms to (Equation S), trigonometric identities, such as, sin2⁡x+cos2⁡x=1, and algebra must be used after combining like terms:

∂2f∂x2+∂2f∂y2+∂2f∂z2==[sin2⁡θ cos2⁡ϕ+sin2⁡θ sin2⁡ϕ+cos2⁡θ]∂2f∂ r2......  +[2 sin⁡θ cos⁡θ cos2⁡ϕr+2 sin⁡θ cos⁡θ sin2⁡ϕr−2 sin⁡θ cos⁡θr] ∂2f∂r ∂θ......  +[−2 sin⁡θ cos⁡θ cos2⁡ϕr2+cos⁡θ sin2⁡ϕr2 sin⁡θ−2 sin⁡θ cos⁡θ sin2⁡ϕr2......  +cos⁡θ cos2⁡ϕr2 sin⁡θ+2 sin⁡θ cos⁡θr2] ∂f∂θ......  +[cos2⁡θ cos2⁡ϕr+sin2⁡ϕr+cos2⁡θ sin2⁡ϕr+cos2⁡ϕr+sin2⁡θr]∂f∂r......  +[cos2⁡θ cos2⁡ϕr2+cos2⁡θ sin2⁡ϕr2+sin2⁡θr2]∂2f∂θ2......  +[sin2⁡ϕr2 sin2⁡θ+cos2⁡ϕr2 sin2⁡θ]∂2f∂ϕ2 .

Examples of how the terms above can be reduced:

[sin2⁡θ cos2⁡ϕ+sin2⁡θ sin2⁡ϕ+cos2⁡θ] ∂2f∂ r2=[sin2⁡θ (cos2⁡ϕ+sin2⁡ϕ+cos2⁡θsin2⁡θ)] ∂2f∂ r2=[sin2⁡θ (1+cos2⁡θsin2⁡θ)] ∂2f∂ r2=(sin2⁡θ+cos2⁡θ) ∂2f∂ r2=(1) ∂2f∂ r2 .[−2 sin⁡θ cos⁡θ cos2⁡ϕr2+cos⁡θ sin2⁡ϕr2 sin⁡θ−2 sin⁡θ cos⁡θ sin2⁡ϕr2......  +cos⁡θ cos2⁡ϕr2 sin⁡θ+2 sin⁡θ cos⁡θr2] ∂f∂θ=[−2 sin⁡θ cos⁡θ(cos2⁡ϕr2+sin2⁡ϕr2−1r2)......  +cos⁡θ sin2⁡ϕr2 sin⁡θ+cos⁡θ cos2⁡ϕr2 sin⁡θ] ∂f∂θ=[0+cos⁡θ (sin2⁡ϕr2 sin⁡θ+cos2⁡ϕr2 sin⁡θ)] ∂f∂θ=cos θr2 sin⁡θ ∂f∂θ .

After applying similar algebraic operations to all the terms, the equation does in fact reduce to:

∇2f=∂2f∂ r2 + 2r ∂f∂ r +1r2 ∂2f∂ θ2 +cos θr2 sin⁡θ ∂f∂ θ + 1r2 sin2⁡ θ ∂2f∂ϕ2=0(Equation S)

 

 

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The Math symbols and equations displayed here use LaTeX under MathJaX.

 

First Posted on: 23 November 2023 (2023.11.23).
Updated on: 24 NOV 2023 (2023.11.24); made a sign correction and finished the page.

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