Wave equation, general solution, cylindrical symmetry

AI Thread Summary
The discussion focuses on the general solutions to the wave equation in different spatial coordinates, particularly cylindrical symmetry. It highlights that while the general solutions for linear and radial spherical coordinates are established, the proposed solution for cylindrical coordinates involving a logarithmic term does not satisfy the wave equation, indicating limitations in its applicability. The conversation also notes that the wave equation in cylindrical coordinates is more complex, requiring Bessel's functions for solutions, which are not closed forms but rather represented by Taylor series. The distinction between steady-state solutions and general solutions is emphasized, with references to established literature on the topic. Overall, the complexities of solving the wave equation in cylindrical coordinates are underscored, particularly in relation to initial conditions and separability of solutions.
lalbatros
Messages
1,247
Reaction score
2
I was interrested in the general solutions to the wave equation depending on only one spatial coordinate.

For one linear coordinate, the general solution is:
a f(x-ct) + b g(x+ct)​

For one radial spherical coordinate, the general solution is:
a f(r-ct)/r + b g(r+ct)/r​

I thought that for a radial cylindrical coordinate, the solution would be:
a f(r-ct)*Log(r) + b g(r+ct)*Log(r)​

Yet I found this is not a solution since I got this residual for the wave equation:
D²(f(r-ct)/r) = a (2+Log(r))*f'(r-ct)/r + b (2+Log(r))*g'(r+ct)/r​

This would indicate that the cylindrical solution applies only to the static case.

Any comment on this surprise?
Did I do a mistake in the calculations, of does that instead mean something?
Thanks.
 
Last edited:
Physics news on Phys.org
The complete wave equation in cylindrical coordinates is:

\frac{\partial^2 u}{\partial t^2}=c^2\left[\frac{1}{\rho}\frac{\partial}{\partial \rho}\left(\rho\frac{\partial u}{\partial \rho}\right)+\frac{1}{\rho^2}\frac{\partial^2 u}{\partial \theta^2}+\frac{\partial^2 u}{\partial z^2}\right]

But if you want a symmetry around the z-axis, you can suppose that the function is constant for \theta and z, so you get the equation:

\frac{\partial^2 u}{\partial t^2}=\frac{c^2}{\rho}\frac{\partial}{\partial \rho}\left(\rho\frac{\partial u}{\partial \rho}\right)

Which look like the spherical coordinates version but is much more difficult to solve. In fact, this is the wave equation in 2D polar coordinates.

The solution to this equation uses Bessel's functions, which are represented by Taylor series, not a closed form. It's strange that the solution in 2D is much more complicated that the solution in 3D.
 
(warning: my post does not really contribute anything meaningful to the discussion)
GPPaille said:
The solution to this equation uses Bessel's functions.
I was under the impression that the solution to this equation was what defined bessel functions. I could be wrong, though.
 
siyphsc: Yes, but I forgot to mention that this not gives a general solution, but only the steady state solution. So this says that the solution u(\rho,t) is separable:

u(\rho,t)=u_s(\rho)u_t(t)

Bessel's functions are modes in a cylindrical space for the radial component, like sinusoidal functions are modes in a 1D space.
 
In Landau and Lifgarbagez, Fluid Mechanics book, it gives as a general solution to the wave equation with cylindrical symmetry (Section 71):

\begin{equation}
\psi = \int_{ct-\rho}^{ct+\rho}\frac{F(\xi)}{\sqrt{\rho^2-(\xi-ct)^2}}
\end{equation}

The way he derives this form is by integrating out the z-dependence of the spherically symmetric solutions in 3D. Note that F is a general function that is found by satisfying the initial conditions of the wave.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...

Similar threads

Back
Top