A trick on PDE?

  • Thread starter hanson
  • Start date
  • #1
319
0
A trick on PDE??

Hi all.
I am reading a text in mathematical wave theory.
I saw and am confused by a manipulation of a PDE, as shown in the attached figure.

I don't really undertand how the equation (1.9) is transformed by "introducing the charcteristic variables). (as indicated by the red arrow and the question mark)

I guess there are some missing steps? Could somone fill in the missing links so that I could know what's going on there.
 

Attachments

Answers and Replies

  • #2
cristo
Staff Emeritus
Science Advisor
8,107
73
New parameters are introduced such that [tex]\xi=x-t ,\hspace{1cm} \zeta=x+t[/tex]. You need to work out utt and uxx in terms of the new parameters. For example, [tex]u_t=\frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial t}+\frac{\partial u}{\partial \zeta}\frac{\partial \zeta}{\partial t}=-u_{\xi}+u_{\zeta}[/tex], using the chain rule. Use a similar method to find ux, and then to find utt and uxx.

When you've worked out utt and uxx sub them into the LHS of (1.9) and you should obtain the result.
 
Last edited:
  • #3
319
0
Thanks! I think I could work it out.
But what is the motivation of doing this?
I mean, is there any phyical meaning attached? Or What benefits can be obtained after this transformation?
 
  • #4
cristo
Staff Emeritus
Science Advisor
8,107
73
Thanks! I think I could work it out.
But what is the motivation of doing this?
I mean, is there any phyical meaning attached? Or What benefits can be obtained after this transformation?
I'm not sure of the physical meaning, but the advantage of performing the transformation is that you can integrate the equation! Note that f is now a function of only [itex]\xi[/itex], so the primes denote differentiation wrt [itex]\xi[/itex]. Thus you can integrate once wrt [itex]\xi[/itex] and once wrt [itex]\zeta[/itex] and obtain the result given (noting that this is a PDE and so the constant of integration is not just a constant in the normal sense, but is a function of the coordinate not integrated with repect to.)

IF you don't make the substitution, then the equation is a fair bit harder to solve.
 
  • #6
319
0
Oops...
I still find it difficult to implement the two initial conditions to determine functions A and B...
how to relate the conditions in terms of the characteristic variables?
 

Related Threads on A trick on PDE?

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
3K
Replies
6
Views
2K
  • Last Post
Replies
3
Views
2K
Replies
2
Views
766
  • Last Post
Replies
1
Views
4K
Top