• Support PF! Buy your school textbooks, materials and every day products Here!

General solution of the spherical wave equation

  • #1

Homework Statement:

The radial oscillations of an ideal gas in a spherical cavity of radius r' are governed by the spherical wave equation subject to the boundary condition ψ(r',t)=0. ψ(r,t) is the radial displacement and v is the speed of sound. Show that the general solution of this equation is written
$$ \psi(r,t) = \Sigma_{i=1,\infty} sinc(i\frac{ir\pi}{r'})cos(\omega_{i}t-\phi_{i})$$
where
$$ sinc(x) = \frac{sin(x)}{x} $$
$$ \omega_{i} = i \pi \frac{v}{r'} $$

Homework Equations:

$$ \frac{\partial^2 \psi}{\partial t^2} = v^{2}(\frac{\partial^2 \psi}{\partial r^2} + \frac{2}{r} \frac{\partial \psi}{\partial r}) $$
Since the spherical wave equation is linear, the general solution is a summation of all normal modes.

To find the particular solution for a given value of i, we can try using the method of separation of variables.

$$ ψ(r,t)=R(r)T(t)ψ(r,t)=R(r)T(t) $$​

Plug this separable solution into the spherical wave equation.

$$ \frac{1}{T}\frac{\partial^{2}T}{\partial t^{2}} = \frac{v^{2}}{R} \frac{\partial^{2}R}{\partial r^{2}} + \frac{2 v^{2}}{rR} \frac{\partial R}{\partial r} $$
$$ R(r)=Acos(k_{i}r)+Bsin(k_{i}r) $$
$$ T(t)=Ccos(ω_{i}t)+Dsin(ω_{i}t) $$​

Given the initial condition ψ(r',t)=0,
$$ 0=Acos(kir′)+Bsin(kir′)=Acos(kir′)+Bsin(kir′) $$
Only one boundary condition is given, but I wonder if there is some other boundary condition for spherical waves that would allow us to ascertain more information, or if there is a superior method. I am trying to figure out how to proceed form here.​
 
Last edited:

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
2019 Award
12,736
2,905
According to the general solution, R should be a ##\operatorname {sinc}##. How did you find your ##R## ?

And: do you use ##r'## for ##a## with a specific reason ?
 
  • #3
phyzguy
Science Advisor
4,370
1,346
There is a second boundary condition, not stated explicitly, which is that the solution should remain finite at r=0.
 
  • #4
135
22
Please check your solutions for radial equation. Their solution will provide you with another physical boundary condition.
 
  • #5
Okay, I see that, as @phyzguy indicated, since the power of the wave is spread over a larger surface area as one moves farther from r=0, so the displacement will tend to decrease as one moves farther from r=0. Thus ψ(0,t) is finite and greater than zero provided the displacement is greater than zero at some r. @BvU pointed out that R(r) should contain a sinc term which makes sense because sin(x)/x decreases with increasing x. So, I will try a different solution for R(r)
$$ 0 = Acos(k_{i} r) + Bsinc(k_{i} r) $$ $$ 0 < R(0) = Acos(0) + sinc(0) = A + sinc(0) $$
Although, I don't immediately see how this is useful, and sinc(0) is undefined.
 
  • #6
Dr Transport
Science Advisor
Gold Member
2,273
413
Okay, I see that, as @phyzguy indicated, since the power of the wave is spread over a larger surface area as one moves farther from r=0, so the displacement at is the maximum. Thus ψ(0,t) is finite and greater than zero provided the displacement is greater than zero at some r. @BvU pointed out that R(r) should contain a sinc term which makes sense because sin(x)/x decreases with increasing x. So, I will try a different solution for R(r)
$$ 0 = Acos(k_{i} r) + Bsinc(k_{i} r) $$ $$ 0 < R(0) = Acos(0) + sinc(0) = A + sinc(0) $$
Although, I don't immediately see how this is useful, and sinc(0) is undefined.
[itex] sinc(0) = 1[/itex] not undefined. I don't think your solution is correct.
 
  • #7
[itex] sinc(0) = 1[/itex] not undefined. I don't think your solution is correct.
sinc(x) = sin(x)/x. Why isn't sinc(0) udnefined if we're dividing by zero? Or are you saying that sinc(x) approaches one as x approaches zero? Could you elaborate on why the solution is wrong?
 
Last edited:
  • #8
Dr Transport
Science Advisor
Gold Member
2,273
413
expand [itex] \sin(x)[/itex] in a Taylor series and divide by [itex] x[/itex], it is [itex] 1 [/itex] not undefined.
 
  • #9
expand [itex] \sin(x)[/itex] in a Taylor series and divide by [itex] x[/itex], it is [itex] 1 [/itex] not undefined.
That makes sense.
$$ R(0) = A + B $$
Which just tells us that R(0) is some constant. But you suggested that my solution was incorrect. COuld you elaborate on this? I agree that my solution doesn't seem to be getting me closer to an answer, but I don't know what I should do differently.
 
  • #10
Dr Transport
Science Advisor
Gold Member
2,273
413
[itex] R(r)[/itex] is of the form [itex] \frac{1}{r} \cos(\omega t - kr - \phi)[/itex] not a combination of [itex] \cos(x)[/itex] and [itex] sinc(x)[/itex] as you state, you're missing a factor of [itex] r[/itex].
 
  • #11
@Dr Transport @BvU @phyzguy @Abhishek11235
$$ 0 = \frac{A}{r}cos(\omega t - kr' - \phi) $$ $$ R(0) = \frac{A}{r}cos(\omega t + \phi) $$
Plugging
$$ \frac{A}{r} \cos(\omega t - kr - \phi) $$ into the wave equation, we can verify that it is a solution and that
$$ \omega = kv $$
Presumably, I can somehow use the initial conditions to solve for k for a given value of i and verify that
$$ \omega_{i} = i \pi \frac{v}{r'} $$
If someone could point me in the direction of how to do this, it would be appreciated.
 
Last edited:
  • #12
135
22
@Dr Transport @BvU @phyzguy @Abhishek11235
$$ 0 = \frac{A}{r}cos(\omega t - kr' - \phi) $$ $$ R(0) = \frac{A}{r}cos(\omega t + \phi) $$
Is that solution for R or ##\psi##?
If it is for R,then your solution for R contains time. But remember while using separation of variables,your assumptions were that R and T are independent of each other.
 
  • #13
Is that solution for R or ##\psi##?
If it is for R,then your solution for R contains time. But remember while using separation of variables,your assumptions were that R and T are independent of each other.
I see why that is confusing. I was responding to @Dr Transport who said:
[itex] R(r)[/itex] is of the form [itex] \frac{1}{r} \cos(\omega t - kr - \phi)[/itex] not a combination of [itex] \cos(x)[/itex] and [itex] sinc(x)[/itex] as you state, you're missing a factor of [itex] r[/itex].
I think Dr Transport meant that the solution to the wave equation takes the form
$$ \frac{1}{r} \cos(\omega t - kr - \phi) $$
As I said, I have tried plugging
$$ \frac{A}{r} \cos(\omega t - kr - \phi) = \psi $$
into the wave equation to confirm that it is a solution, so I suppose my previous reply contains solutions for ψ. Regardless, I am trying to figure out how to apply the boundary conditions to solve for ω and ultimately find the general solution.
 
  • #14
Dr Transport
Science Advisor
Gold Member
2,273
413
rewrite the differential for the [itex] R(r)[/itex] as [itex] \frac{d^2(rR(r))}{d^2r}[/itex]. The complete wave function [itex] \psi(r,t) = T(t)R(r)[/itex] which can be rewritten like I said before with a phase factor of [itex]\phi[/itex].
 
  • #15
@Dr Transport I apologize if I'm slow, but I'm not what you're asking me to do. What do you mean by
rewrite the differential for the [itex] R(r)[/itex] as [itex] \frac{d^2(rR(r))}{d^2r}[/itex].
 
Last edited:
  • #16
135
22
@Dr Transport I apologize if I'm slow, but I'm not what you're asking me to do. What do you mean by
He has given you a hint. That is one of the standard trick to solve DE. If you expand that you will see that it reproduces your DE with missing factor. The resultant equation is easy to solve. Try it!
 
  • #17
He has given you a hint. That is one of the standard trick to solve DE. If you expand that you will see that it reproduces your DE with missing factor. The resultant equation is easy to solve. Try it!
I'm still trying to figure out what you're asking me to do. What am I supposed to plug in for R(r)?
 
  • #18
135
22
I'm still trying to figure out what you're asking me to do. What am I supposed to plug in for R(r)?
You are confused. So,I will partially do it for you.

$$\frac{1}{R}\frac{\partial^2 R}{\partial r^2}+ \frac{2}{rR} \frac{\partial R}{\partial r}=k^2$$

Now ,the hint given to you was:
$$\frac{\partial^2 (rR)}{\partial r^2}= r\frac{\partial^2 R}{\partial r^2} + 2\frac{\partial R}{\partial r}$$

Now,there should be no problem in joining the above 2 equations and get a solution for ##R##!
 
  • #19
@Abhishek11235
$$ \frac{\partial^2 (rR)}{\partial r^2} = rRk^{2} $$ $$ Ak^{2}cos(\omega t - kr - \phi) = r[\frac{A}{r}cos(\omega t - kr - \phi)]k^{2} $$ $$ R(r) = \frac{A}{r}cos(\omega t - kr - \phi) $$
 
  • #20
135
22
@Abhishek11235
$$ \frac{\partial^2 (rR)}{\partial r^2} = rR(k^{2}) $$ $$ Ak^{2}cos(\omega t - kr - \phi) = r[\frac{A}{r}cos(\omega t - kr - \phi)]k^{2} $$ $$ R(r) = \frac{A}{r}cos(\omega t - kr - \phi) $$
See post #12. You did same mistake. You should re-read any book on Differential equation if you can't solve this easy one. And I can't post solution or go ahead as it will violate community rules.
 
  • #21
See post #12. You did same mistake. You should re-read any book on Differential equation if you can't solve this easy one. And I can't post solution or go ahead as it will violate community rules.
$$ R(r) = A\frac{e^{-ikr}}{r} $$
 

Related Threads for: General solution of the spherical wave equation

Replies
1
Views
8K
Replies
4
Views
5K
Replies
3
Views
3K
  • Last Post
Replies
3
Views
3K
Replies
9
Views
5K
Replies
0
Views
2K
Replies
3
Views
4K
Top