3D Laplace solution in Cylindrical Coordinates For a Hollow Cylindrical Tube

  • #1
jkthejetplane
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4
Homework Statement:
Find the general series solution for laplace in cylindrical coordinates
Relevant Equations:
for this i have always used (s,phi,z)
Here is the initial problem and my attempt at getting Laplace solution. I get lost near the end and after some research, ended up with the Bessel equation and function. I don't completely understand what this is or even if this i the direction I go in.
This is a supplemental thing that I want to nail down for review to get my brain up to speed again for this semester

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Answers and Replies

  • #2
pasmith
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You are missing a term from your final equation.

The solution must be periodic in [itex]\phi[/itex], so the dependence will be [itex]\Phi'' = -n^2\Phi[/itex]. You do then have [itex]Z'' = CZ[/itex], but at this point there's no reason to believe that [itex]C \geq 0[/itex]. So your radial dependence satisfies [tex]
s^2 S'' + s S' + (Cs^2 - n^2)S = 0.[/tex] Setting [itex]k = |C|^{1/2}[/itex] and [itex]x = ks[/itex] turns this into [tex]
x^2 \frac{d^2S}{dx^2} + x \frac{dS}{dx} + (\operatorname{sgn}(C) x^2 - n^2)S = 0[/tex] which is the Bessel equation if [itex]C > 0[/itex] and the modified Bessel equation if [itex]C < 0[/itex]. If [itex]C = 0[/itex] then the dependence on [itex]z[/itex] is [itex]Az + B[/itex] and the radial dependence is [itex]s^\alpha[/itex] where [itex]\alpha[/itex] depends on [itex]n[/itex].

The Bessel functions are oscillatory with amplitude decaying to zero as [itex]x \to \infty[/itex]. The Bessel function of the first kind ,[itex]J_n[/itex], is bounded at the origin with [itex]J_0(0) = 1[/itex] and [itex]J_n(0) = 0[/itex] for [itex]n \geq 1[/itex]. The Bessel function of the second kind, [itex]Y_n[/itex], blows up at the origin. The modified Bessel functions are monotonic and positive with the modified function of the first kind, [itex]I_n[/itex], being bounded at the origin with [itex]I_0(0) = 1[/itex] and [itex]I_n(0) = 0[/itex] for [itex]n \geq 1[/itex] and increasing without limit as [itex]x \to \infty[/itex], and the modified function of the second kind, [itex]K_n[/itex], blowing up at the origin and decaying to 0 as [itex]x \to \infty[/itex].
 
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