PDE question - boards.ie
Boards.ie uses cookies. By continuing to browse this site you are agreeing to our use of cookies. Click here to find out more x
Post Reply  
 
Thread Tools Search this Thread
08-03-2012, 16:48   #1
Smythe
Registered User
 
Join Date: Mar 2010
Posts: 334
PDE question

I'm just trying to get an understanding of answering PDEs, so wanted to ask what you thought of my answer to this question.

The one-dimensional wave equation is given by the first equation shown in this link;

http://mathworld.wolfram.com/WaveEqu...mensional.html

where Ψ = f(x, t)

Is f(x, t) = cos(x-vt) + (x-vt) a possible solution?


∂^2 f/∂x^2 = -cos(x-vt)

and

∂^2 f/∂t^2 = c^2 cos(x-ct)

So, since -cos(x-vt) ≠ cos(x-vt) I assume this is not a solution.

Last edited by Smythe; 08-03-2012 at 18:50.
Smythe is offline  
Advertisement
08-03-2012, 19:41   #2
Iderown
Registered User
 
Iderown's Avatar
 
Join Date: Mar 2010
Location: Near Belfast
Posts: 110
There is a very wide range of solutions for the 1-D wave equation. Any expression of the form f(x - nt) + g(x + nt) is a solution. Obviously, not all are waves.
The further requirement for a wavelike solution is to choose initial/boundary conditions which are appropriate to a wavelike solution.
Iderown is offline  
08-03-2012, 19:51   #3
Smythe
Registered User
 
Join Date: Mar 2010
Posts: 334
Thanks Iderown.

So do you mean I am correct that f(x, t) = cos(x-vt) + (x-vt) is not a solution?
Smythe is offline  
09-03-2012, 08:42   #4
Iderown
Registered User
 
Iderown's Avatar
 
Join Date: Mar 2010
Location: Near Belfast
Posts: 110
It is a solution. Just not a fully wavelike solution. The h(x,t) = x - vt part is not wavelike. You can easily check that this h(x,t) satisfies the wave equation.
The wave equation is linear - that means that sums of solutions are also solutions. The initial/boundary conditions are an important feature in the solution of partial differential equations.
Iderown is offline  
09-03-2012, 14:53   #5
Michael Collins
Moderator
 
Join Date: Jul 2001
Location: Dublin
Posts: 1,715
Quote:
Originally Posted by Smythe View Post
...
∂^2 f/∂t^2 = v^2 cos(x-vt) [edited to replace 'c' with 'v' like other equations]

So, since -cos(x-vt) ≠ cos(x-vt) I assume this is not a solution.
There's a minus missing here for the second derivative with respect to time. You obviously know the chain rule since you realised that the 'v' must come outside the cos( ) function, so it was probably just a sign slip up.

So you should have



which when divided by then will equal the second derivative with respect to x.

What Iderown says is quite useful too, and in fact it's easy to show that any function of the form



is a solution. Specifically



and the left hand side is



Dividing the first result above by , according to the 1-D wave equation, gives the last result above. So any function of the given form must be a solution (provided all the relevent derivatives exist, of course). Also as Iderown says, initial and boundary conditions must be met for a particular solution to a given problem.
Michael Collins is offline  
Thanks from:
Advertisement
10-03-2012, 21:11   #6
Smythe
Registered User
 
Join Date: Mar 2010
Posts: 334
Quote:
Originally Posted by Michael Collins View Post
There's a minus missing here for the second derivative with respect to time. You obviously know the chain rule since you realised that the 'v' must come outside the cos( ) function, so it was probably just a sign slip up.
Thanks. Don't know how I made that mistake.

Quote:
Originally Posted by Michael Collins View Post



What Iderown says is quite useful too, and in fact it's easy to show that any function of the form



is a solution. Specifically



and the left hand side is



Dividing the first result above by , according to the 1-D wave equation, gives the last result above. So any function of the given form must be a solution (provided all the relevent derivatives exist, of course). Also as Iderown says, initial and boundary conditions must be met for a particular solution to a given problem.
That's interesting. Thank you for that.
Smythe is offline  
Post Reply

Quick Reply
Message:
Remove Text Formatting
Bold
Italic
Underline

Insert Image
Wrap [QUOTE] tags around selected text
 
Decrease Size
Increase Size
Please sign up or log in to join the discussion

Thread Tools Search this Thread
Search this Thread:

Advanced Search



Share Tweet