(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

This is an example I copy from the book. The book showed the steps of solving and provide the answer. I don't understand the book at all. Below I show the question and the solution from the book. Then I am going to ask my question at the bottom.

Question

Use D'Alembert method to solve the wave equation with boundary and initial value:

[tex]\frac{\partial^2 u}{\partial t^2} \;=\; c^2 \frac{\partial^2 u}{\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]

Given: f(x)=0 & g(x)=x for 0<x<1. c=1, L=1.

2. Relevant equations

D'Alembert:

[tex] u(x,t)=\frac{1}{2}[f^*(x-ct)+f^*(x+ct)]+\frac{1}{2c}\int_{x-ct}^{x+ct} g^*(s)ds[/tex]

3. The attempt at a solution

f(x) = 0 [tex].\;\;\Rightarrow\; u(x,t)\; =\; \frac{1}{2c}\int_{x-ct}^{x+ct} g^*(s)ds \;=\; \frac{1}{2c}[G(x+t) \;-\; G(x-t)][/tex](1)

Where [tex]g^*[/tex] is the odd 2-period extension of g and G is the antiderivative of [tex]g^*.[/tex]

Let .[tex]G(x)=\int_{-1}^x g^*(z)dz [/tex] (2)

To complete the solution, we must determine G. We know G on any interval of length 2. Since [tex]g^*(x)=x[/tex] on the interval (-1,1), we obtain

.[tex]G(x)=\int_{-1}^x g^*(z)dz \;=\; \frac{1}{2}x^2 \;-\; \frac{1}{2} [/tex]. for x in (-1,1) (3). Hence

[tex]G(x) \;=\; \frac{1}{2}x^2 \;-\; \frac{1}{2} \;\;\;if\;-1<x<1[/tex]

[tex]G(x) \;=\; G(x+2) \;\;otherwise.[/tex]

My question:

1) How does it jump from (1) to (3)??

2) Is (2) just the first Fundamental Theorem of Calculus where -1 is the lower limit of x on [-1,1]?

3) Where is (3) come from? How are (x+t) and (x-t) change to x and -1 respectively?

I don't even understand the answer of the book!! Can anyone explain to me?

Thanks

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# Homework Help: D'Alembert solution with f(x)=0, g(x)=x

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