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Solve D'Alembert problem.

  • Thread starter yungman
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  • #1
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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
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Anyone?
 
  • #3
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Can anyone at least give me some opinion even you might not have the answer?
 
  • #4
vela
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Try expressing u(x,t) in terms of the normal modes.
 
  • #5
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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
 

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