Bessel vs Modified Bessel Eqn solve PDE

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I'm having trouble understanding the boundary conditions and when you would need to use Bessel vs Modified Bessel to solve simple cylindrical problems (I.e. Heat conduction or heat flow with only two independent variables). When do you use Bessel vs Modified Bessel to solve Strum-Louville problem for a cylinder and what is the intuition behind using either one?
 

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  • #2
pasmith
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I'm having trouble understanding the boundary conditions and when you would need to use Bessel vs Modified Bessel to solve simple cylindrical problems (I.e. Heat conduction or heat flow with only two independent variables). When do you use Bessel vs Modified Bessel to solve Strum-Louville problem for a cylinder and what is the intuition behind using either one?
If you separate variables in Laplace's equation in such a way that the axial dependence is exponential, then the equation for the radial dependence is Bessel's Equation:
[tex]\nabla^2(R(r)e^{\pm kz}e^{\pm in\theta}) = 0 \quad\Rightarrow\quad r^2R'' + rR' + (k^2r^2 - n^2)R = 0[/tex]

If you separate variables in Laplace's equation in such a way that the axial dependence is sinusoidal, then the equation for the radial dependence is the modified Bessel's Equation:
[tex]
\nabla^2(R(r)e^{\pm ikz}e^{\pm in\theta}) = 0 \quad \Rightarrow \quad r^2R'' + rR' - (k^2r^2 + n^2)R = 0
[/tex]

If you're working in a region where [itex]z[/itex] is unbounded then you will only need [itex]e^{\pm kz}[/itex]. If you're working in a region where [itex]z[/itex] is bounded then you will generally need both [itex]e^{\pm kz}[/itex] and [itex]e^{\pm ikz}[/itex].
 
  • #3
jasonRF
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The asymptotic (large argument) forms of the Bessel functions provide some insight. See:
http://en.wikipedia.org/wiki/Bessel_function

Here is a summary (see the link for more of the story).

regular Bessel:
J ~ cos(x-a)/sqrt(x)
Y ~ sin(x-a)/sqrt(x)

Modified bessel:
I ~ exp(x)/sqrt(x)
K ~ exp(-x)/sqrt(x)

So regular bessel functions oscillate, and modified do not.

The hankel functions (H = J +/i Y) thus go like exp(+/- i (x-a))/sqrt(x), which is why they are useful for cylindrical wave problems.

The behavior at zero matters for some applications, especially those for which your solution must be bounded at zero. J and I are bounded, while Y and K (and hence the hankel functions H) are not.

In light of what pasmith wrote, when the Z-dependence oscillates, the R does not, and vice versa.

jason
 

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