# Solve D'Alembert problem.

## 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

## Answers and Replies

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Anyone?

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

vela
Staff Emeritus
Science Advisor
Homework Helper
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Try expressing u(x,t) in terms of the normal modes.

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