How Does Symmetry Affect the Solution to D'Alembert's Problem?

[yungman]

Homework Statement

\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)

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

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

Homework Equations

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)]

u(-x,t)=-u(x,t) \;\;,\;\; u(x+2L,t)=u(x,t) \;\;,\;\; u(x-L,t)=u(x+L,t)

The Attempt at a Solution

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

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}) )]

\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)]

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)]


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

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

Thanks
Alan

 

[yungman]

Anyone?

 

[yungman]

Can anyone at least give me some opinion even you might not have the answer?

 

[vela]

Try expressing u(x,t) in terms of the normal modes.

 

[yungman]

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