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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:

2fy2= sinθ sinϕ 2 fr y+ cosθ sinϕr 2fθ y + cosϕrsinθ 2fϕ y=sinθ sinϕ ( r fy)+cosθ sinϕr (fθ fy)+cosϕrsinθ (fϕ fy)=sinθ sinϕ [ r (sinθ sinϕ f r+cosθ sinϕr fθ+cosϕrsinθ fϕ)]......  +cosθ sinϕr [θ (sinθ sinϕ f r+cosθ sinϕr fθ+cosϕrsinθ fϕ)]......  +cosϕrsinθ [ϕ (sinθ sinϕ f r+cosθ sinϕr fθ+cosϕr sinθ fϕ)] .

 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:

2fy2=sinθ sinϕ (sinθ sinϕ 2f r2+cosθ sinϕr 2fr θcosθ sinϕr2 fθ......  +cosϕrsinθ 2fr ϕcosϕr2sinθ fϕ)......  +cosθ sinϕr (sinθ sinϕ 2fr θ+cosθ sinϕ fr+cosθ sinϕr 2fθ2......  sinθ sinϕr fθ+cosϕr sinθ 2fθ ϕcosθ cosϕr sin2θ fϕ)......  +cosϕrsinθ (sinθ sinϕ 2fr ϕ+sinθ cosϕ fr+cosθ sinϕr 2fθ ϕ......  +cosθ cosϕr fθ+cosϕr sinθ 2fϕ2sinϕr sinθ fϕ) .

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

2fy2=sin2θ sin2ϕ 2f r2+sinθ cosθ sin2ϕr 2fr θsinθ cosθ sin2ϕr2 fθ ......  +sinθ sinϕ cosϕr sinθ 2fr ϕsinϕ cosϕr2 fϕ+sinθ cosθ sin2ϕr 2fr θ ......  +cos2θ sin2ϕr fr+cos2θ sin2ϕr2 2fθ2cosθ sinθ sin2ϕr2 fθ ......  +cosθ sinϕ cosϕr2 sinθ 2fθ ϕcos2θ sinϕ cosϕr2 sin2θ fϕ+sinϕ cosϕr 2fr ϕ ......  +cos2ϕr fr+cosθ sinϕ cosϕr2 sinθ 2fθ ϕ+cosθ cos2ϕr2 sinθ fθ ......  +cos2ϕr2 sin2θ 2fϕ2sinϕ cosϕr2 sin2θ fϕ .

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

2fy2=sin2θ sin2ϕ 2f r2+2 sinθ cosθ sin2ϕr 2fr θ2 sinθ cosθ sin2ϕr2 fθ ......  +2 sinϕ cosϕr 2fr ϕsinϕ cosϕr2 fϕ+cos2θ sin2ϕr fr ......  +cos2θ sin2ϕr2 2fθ2+2 cosθ sinϕ cosϕr2 sinθ 2fθ ϕcos2θ sinϕ cosϕr2 sin2θ fϕ ......  +cos2ϕr fr+cosθ cos2ϕr2 sinθ fθ+cos2ϕr2 sin2θ 2fϕ2sinϕ cosϕr2 sin2θ fϕ .

 

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:

2fz2= cosθ 2 fr z sinθr 2fθ z=cosθ ( r fz)sinθr (θ fz)=cosθ [ r (cosθ f r  sinθr fθ)]......  sinθr [θ (cosθ f r  sinθr fθ)] .

After carrying out the differentiations, we obtain:

2fz2= cosθ (cosθ 2f r2sinθr 2fr θ+sinθr2 fθ)......  sinθr (cosθ 2fr θsinθ frsinθr 2fθ2cosθr fθ) .
2fz2=cos2θ 2f r2sinθ cosθr 2fr θ+sinθ cosθr2 fθ......  sinθ cosθr 2fr θ+sin2θr fr+sin2θr2 2fθ2+sinθ cosθr2 fθ .
2fz2=cos2θ 2f r22 sinθ cosθr 2fr θ......  +2 sinθ cosθr2 fθ+sin2θr fr+sin2θr2 2fθ2 .

 

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:

2fx2+2fy2+2fz2==sin2θ cos2ϕ 2f r2+2 sinθ cosθ cos2ϕr 2fr θ2 sinθ cosθ cos2ϕr2 fθ ......  2 sinϕ cosϕr 2fr ϕ+sinϕ cosϕr2 fϕ+cos2θ cos2ϕr fr ......  +cos2θ cos2ϕr2 2fθ22 cosθ sinϕ cosϕr2 sinθ 2fθ ϕ+cos2θ sinϕ cosϕr2 sin2θ fϕ ......  +sin2ϕr fr+cosθ sin2ϕr2 sinθ fθ+sin2ϕr2 sin2θ 2fϕ2+sinϕ cosϕr2 sin2θ fϕ ......  +sin2θ sin2ϕ 2f r2+2 sinθ cosθ sin2ϕr 2fr θ2 sinθ cosθ sin2ϕr2 fθ ......  +2 sinϕ cosϕr 2fr ϕsinϕ cosϕr2 fϕ+cos2θ sin2ϕr fr ......  +cos2θ sin2ϕr2 2fθ2+2 cosθ sinϕ cosϕr2 sinθ 2fθ ϕcos2θ sinϕ cosϕr2 sin2θ fϕ ......  +cos2ϕr fr+cosθ cos2ϕr2 sinθ fθ+cos2ϕr2 sin2θ 2fϕ2sinϕ cosϕr2 sin2θ fϕ......  +cos2θ 2f r22 sinθ cosθr 2fr θ......  +2 sinθ cosθr2 fθ+sin2θr fr+sin2θr2 2fθ2 .

Ten of the terms cancel out immediately. However, in order to reduce all of the remaining terms to (Equation S), trigonometric identities, such as, sin2x+cos2x=1, and algebra must be used after combining like terms:

2fx2+2fy2+2fz2==[sin2θ cos2ϕ+sin2θ sin2ϕ+cos2θ]2f r2......  +[2 sinθ cosθ cos2ϕr+2 sinθ cosθ sin2ϕr2 sinθ cosθr] 2fr θ......  +[2 sinθ cosθ cos2ϕr2+cosθ sin2ϕr2 sinθ2 sinθ cosθ sin2ϕr2......  +cosθ cos2ϕr2 sinθ+2 sinθ cosθr2] fθ......  +[cos2θ cos2ϕr+sin2ϕr+cos2θ sin2ϕr+cos2ϕr+sin2θr]fr......  +[cos2θ cos2ϕr2+cos2θ sin2ϕr2+sin2θr2]2fθ2......  +[sin2ϕr2 sin2θ+cos2ϕr2 sin2θ]2fϕ2 .

Examples of how the terms above can be reduced:

[sin2θ cos2ϕ+sin2θ sin2ϕ+cos2θ] 2f r2=[sin2θ (cos2ϕ+sin2ϕ+cos2θsin2θ)] 2f r2=[sin2θ (1+cos2θsin2θ)] 2f r2=(sin2θ+cos2θ) 2f r2=(1) 2f r2 .[2 sinθ cosθ cos2ϕr2+cosθ sin2ϕr2 sinθ2 sinθ cosθ sin2ϕr2......  +cosθ cos2ϕr2 sinθ+2 sinθ cosθr2] fθ=[2 sinθ cosθ(cos2ϕr2+sin2ϕr21r2)......  +cosθ sin2ϕr2 sinθ+cosθ cos2ϕr2 sinθ] fθ=[0+cosθ (sin2ϕr2 sinθ+cos2ϕr2 sinθ)] fθ=cos θr2 sinθ fθ .

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 sinθ f θ + 1r2 sin2 θ 2fϕ2=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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