Solve D'Alembert problem.

  • Thread starter yungman
  • Start date
  • #1
5,511
195

Homework Statement



[tex] \frac{\partial u^2}{\partial t^2} = c^2 \frac{\partial u^2}{\partial x^2} \;\;,\;\; u(0,t)=u(L,t)=0 \;\;,\;\; u(x,0)=f(x) \;\;,\;\; \frac{\partial u}{\partial t}(x,0)=g(x)[/tex]

[tex]f(x) \;and\; g(x) \;are\; symmetric\; about\;\; x=\frac{L}{2} \;\Rightarrow f(L-x)=f(x) \;\;and\;\; g(L-x)=g(x)[/tex]

Show [tex]u(x,t+\frac{L}{c})=-u(x,t)[/tex]

Homework Equations



[tex]u(x,t)=\frac{1}{2}[f(x+ct)+f(x-ct)]+\frac{1}{2}[G(x+ct)-G(x-ct)] \;\;\;where\;\;\; G(x)=\frac{1}{c}[G(x+ct)-G(x-ct)][/tex]

[tex]u(-x,t)=-u(x,t) \;\;,\;\; u(x+2L,t)=u(x,t) \;\;,\;\; u(x-L,t)=u(x+L,t)[/tex]


The Attempt at a Solution



u(x,t) is periodic with T=2L.

[tex]u(x, t+\frac{L}{c} ) =\frac{1}{2}[f(x+c (t+\frac{L}{c}) )+f(x-c(t+\frac{L}{c}) )]+\frac{1}{2}[G(x+c(t+\frac{L}{c}) )-G(x-c(t+\frac{L}{c}) )][/tex]

[tex]\Rightarrow u(x, t+\frac{L}{c} ) =\frac{1}{2}[ f((x+L)+ct )+f((x-L)-ct)]+\frac{1}{2}[G((x+L)+ct)-G((x-L)-ct)][/tex]

[tex]u(x-L,t)=u(x+L,t) \Rightarrow \; u(x, t+\frac{L}{c} ) =\frac{1}{2}[ f((x+L)+ct )+f((x+L)-ct)]+\frac{1}{2}[G((x+L)+ct)-G((x+L)-ct)][/tex]


I can see odd and even function with symmetric at the middle of the period like sin(x) and cos(x) resp. That [tex]sin(x+\pi)=-sin(x) \;and\; cos(x+\pi)=-cos(x)[/tex]

I just don't know how to express in mathametical terms. Can someone at least get me hints or answer?

Thanks
Alan
 

Answers and Replies

  • #2
5,511
195
Anyone?
 
  • #3
5,511
195
Can anyone at least give me some opinion even you might not have the answer?
 
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,874
1,449
Try expressing u(x,t) in terms of the normal modes.
 
  • #5
5,511
195
Try expressing u(x,t) in terms of the normal modes.
You mean in fouries series expansion? I'll look into this and post back. Thanks
 

Related Threads on Solve D'Alembert problem.

Replies
0
Views
1K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
979
  • Last Post
Replies
1
Views
7K
Replies
3
Views
1K
Replies
1
Views
896
Top