JUST THE MATHS SLIDES NUMBER 14.7 PARTIAL DIFFERENTIATION 7 - - PDF document

“JUST THE MATHS” SLIDES NUMBER 14.7 PARTIAL DIFFERENTIATION 7 (Change of independent variable) by A.J.Hobson

14.7.1 Illustrations of the method

UNIT 14.7 PARTIAL DIFFERENTIATON 7 CHANGE OF INDEPENDENT VARIABLE 14.7.1 ILLUSTRATIONS OF THE METHOD The following technique would be necessary, for example, in changing from one geometrical reference system to an-

• ther, especially with

“partial differential equations”. The method is an application of the chain rule for partial derivatives and is illustrated with examples. EXAMPLES

• 1. Express, in plane polar co-ordinates, r and θ, the fol-

lowing partial differential equations: (a) ∂V ∂x + 5∂V ∂y = 1; (b) ∂2V ∂x2 + ∂2V ∂y2 = 0.

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Solution Both differential equations involve a function, V (x, y), where x = r cos θ and y = r sin θ. Hence, ∂V ∂r = ∂V ∂x .∂x ∂r + ∂V ∂y .∂y ∂r,

• r

∂V ∂r = ∂V ∂x cos θ + ∂V ∂y sin θ and ∂V ∂θ = ∂V ∂x .∂x ∂θ + ∂V ∂y .∂y ∂θ,

• r

∂V ∂θ = −∂V ∂x r sin θ + ∂V ∂y r cos θ. We eliminate, first ∂V

∂y , and then ∂V ∂x to obtain

∂V ∂x = cos θ∂V ∂r − sin θ r ∂V ∂θ and ∂V ∂y = sin θ∂V ∂r + cos θ r ∂V ∂θ .

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Hence, the differential equation, (a), becomes (cos θ + 5 sin θ)∂V ∂r +

  5 cos θ

r − sin θ

   ∂V

∂θ = 1. To find the second-order derivatives of V with respect to x and y, we write the formulae for the first-order derivatives in the form ∂ ∂x[V ] =

  cos θ ∂

∂r − sin θ r ∂ ∂θ

   [V ]

and ∂ ∂y[V ] =

  sin θ ∂

∂r + cos θ r ∂ ∂θ

   [V ].

From these, ∂2V ∂x2 =

  cos θ ∂

∂r − sin θ r ∂ ∂θ

     cos θ∂V

∂r − sin θ r ∂V ∂θ

   ,

which gives ∂2V ∂x2 = cos2θ∂2V ∂r2 + 2 sin θ cos θ r2 ∂V ∂θ −2 sin θ cos θ r ∂2V ∂r∂θ + sin2θ r ∂V ∂r + sin2θ r2 ∂2V ∂θ2 .

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Similarly, ∂2V ∂y2 = sin2θ∂2V ∂r2 − 2 sin θ cos θ r2 ∂V ∂θ +2 sin θ cos θ r ∂2V ∂r∂θ + cos2θ r ∂V ∂r + cos2θ r2 ∂2V ∂θ2 . Adding these together gives the differential equation, (b), in the form ∂2V ∂r2 + 1 r ∂V ∂r + 1 r2 ∂2V ∂θ2 = 0.

• 2. Express the differential equation,

∂2V ∂x2 + ∂2V ∂y2 + ∂2V ∂z2 = 0, (a) in cylindrical polar co-ordinates and (b) in spherical polar co-ordinates. Solution (a) Using x = r cos θ, y = r sin θ and z = z, we may use the results of the previous example to give ∂2V ∂r2 + 1 r ∂V ∂r + 1 r2 ∂2V ∂θ2 + ∂2V ∂z2 = 0.

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(b) Using x = u sin φ cos θ, y = u sin φ sin θ, and z = u cos φ, we could write out three formulae for ∂V

∂u, ∂V ∂θ and ∂V ∂φ

and then solve for ∂V

∂x, ∂V ∂y and ∂V ∂z ; but this is compli-

cated. However, the result in part (a) provides a shorter method as follows: Cylindrical polar co-ordinates are expressible in terms

• f spherical polar co-ordinates by the formulae

z = u cos φ, r = u sin φ, θ = θ. Hence, by using the previous example with z, r, θ in place of x, y, z, respectively, and u, φ in place of r, θ, respectively, we obtain ∂2V ∂z2 + ∂2V ∂r2 = ∂2V ∂u2 + 1 u ∂V ∂u + 1 u2 ∂2V ∂φ2 . Therefore, to complete the conversion we need only to consider ∂V

∂r .

By using r, u, φ in place of y, r, θ, respectively, the previous formula for ∂V

∂y gives

∂V ∂r = sin φ∂V ∂u + cos φ u ∂V ∂φ . The given differential equation thus becomes

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∂2V ∂u2 + 1 u ∂V ∂u + 1 u2 ∂2V ∂φ2 + 1 u2sin2φ ∂2V ∂θ2 + 1 u sin φ

   sin φ∂V

∂u + cos φ u ∂V ∂φ

    = 0.

That is, ∂2V ∂u2 + 2 u ∂V ∂u + 1 u2 ∂2V ∂φ2 +cot φ u2 ∂V ∂φ + 1 u2sin2φ ∂2V ∂θ2 = 0.

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