Difference between revisions of "Helmholtz's Equation"

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[http://en.wikipedia.org/wiki/Bessel_function Bessel functions] of the first kind
 
[http://en.wikipedia.org/wiki/Bessel_function Bessel functions] of the first kind
 
and Hankel functions of order <math>\nu</math>.
 
and Hankel functions of order <math>\nu</math>.
 +
The potential outside the circle can therefore be written as
 +
<center>
 +
<math>
 +
\phi (r,\theta) =  \sum_{\nu = -
 +
\infty}^{\infty} \left[ D_{\nu} J_\nu (k r) + E_{\nu} H^{(1)}_\nu (k
 +
r) \right] \mathrm{e}^{\mathrm{i} \nu \theta},
 +
</math>
 +
</center>
  
 +
We consider the case where we have Neuman boundary condition on the circle. Therefore
 +
we have <math>\partial_n\phi=0</math> at <math>r=a</math>. We can therefore obtain
 +
<center>
 +
<math>
 +
E_{\nu} = - \frac{D_{\nu} J^{\prime}_\nu (k r)}{ H^{(1)}^{\prime}_\nu (kr)}
 +
</math>
 +
</center>
  
 
[http://en.wikipedia.org/wiki/Helmholtz_equation External link]
 
[http://en.wikipedia.org/wiki/Helmholtz_equation External link]
  
 
[[Category:Linear Water-Wave Theory]]
 
[[Category:Linear Water-Wave Theory]]

Revision as of 00:33, 9 June 2010


Indroduction

This is a very well known equation given by

[math]\displaystyle{ \nabla^2 \phi + k^2 \phi = 0 }[/math].

It applies to a wide variety of situations such as electromagnetics and acoustics. In water waves it arises when we Remove The Depth Dependence. Often there is then a cross over from the study of water waves to the study of scattering problems more generally. Also, if we perform a Cylindrical Eigenfunction Expansion we find that the modes all decay rapidly as distance goes to infinity except the solutions which satisfy Helmholtz's equation. This means that many asymptotic results in linear water waves can be derived from results in acoustic or electromagnetic scattering.

Solution for a Circle

We can solve for the scattering by a circle using separation of variables. This is the basis of the method used in Bottom Mounted Cylinder

Helmholtz equation in cylindrical coordinates is

[math]\displaystyle{ \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial \phi}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 \phi}{\partial \theta^2} = -k^2 \phi(r,\theta), }[/math]

we use the separation

[math]\displaystyle{ \phi(r,\theta) =: R(r) \Theta(\theta)\,. }[/math]

Substituting this into Laplace's equation yields

[math]\displaystyle{ \frac{r^2}{R(r)} \left[ \frac{1}{r} \frac{\mathrm{d}}{\mathrm{d}r} \left( r \frac{\mathrm{d} R}{\mathrm{d}r} \right) +k^2 R(r) \right] = - \frac{1}{\Theta (\theta)} \frac{\mathrm{d}^2 \Theta}{\mathrm{d} \theta^2} = \eta^2, }[/math]

where the separation constant [math]\displaystyle{ \eta }[/math] must be an integer, say [math]\displaystyle{ \nu }[/math], in order for the potential to be continuous. [math]\displaystyle{ \Theta (\theta) }[/math] can therefore be expressed as

[math]\displaystyle{ \Theta (\theta) = C \, \mathrm{e}^{\mathrm{i} \nu \theta}, \quad \nu \in \mathbb{Z}. }[/math]

We also obtain the following expression

[math]\displaystyle{ r \frac{\mathrm{d}}{\mathrm{d}r} \left( r \frac{\mathrm{d} R}{\mathrm{d} r} \right) - (\nu^2 - k^2 r^2) R(r) = 0, \quad \nu \in \mathbb{Z}. }[/math]

Substituting [math]\displaystyle{ \tilde{r}:=k r }[/math] and writing [math]\displaystyle{ \tilde{R} (\tilde{r}) := R(\tilde{r}/k) = R(r) }[/math], this can be rewritten as

[math]\displaystyle{ \tilde{r}^2 \frac{\mathrm{d}^2 \tilde{R}}{\mathrm{d} \tilde{r}^2} + \tilde{r} \frac{\mathrm{d} \tilde{R}}{\mathrm{d} \tilde{r}} - (\nu^2 - \tilde{r}^2)\, \tilde{R} = 0, \quad \nu \in \mathbb{Z}, }[/math]

which is Bessel's equation. Substituting back, the general solution is given by

[math]\displaystyle{ R(r) = D_\nu \, J_\nu(k_m r) + E_\nu \, H^{(1)}_\nu(k_m r),\ \nu \in \mathbb{Z}, }[/math]

where Bessel functions of the first kind and Hankel functions of order [math]\displaystyle{ \nu }[/math]. The potential outside the circle can therefore be written as

[math]\displaystyle{ \phi (r,\theta) = \sum_{\nu = - \infty}^{\infty} \left[ D_{\nu} J_\nu (k r) + E_{\nu} H^{(1)}_\nu (k r) \right] \mathrm{e}^{\mathrm{i} \nu \theta}, }[/math]

We consider the case where we have Neuman boundary condition on the circle. Therefore we have [math]\displaystyle{ \partial_n\phi=0 }[/math] at [math]\displaystyle{ r=a }[/math]. We can therefore obtain

[math]\displaystyle{ E_{\nu} = - \frac{D_{\nu} J^{\prime}_\nu (k r)}{ H^{(1)}^{\prime}_\nu (kr)} }[/math]

External link