• Feb 28th 2010, 05:05 PM
yungman
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:
$\frac{\partial^2 u}{\partial t^2} \;=\; c^2 \frac{\partial^2 u}{\partial x^2} \;\;,\;\; u(0,t)=u(L,t)=0 \;\;,\;\; u(x,0)=f(x) \;\;,\;\; \frac{\partial u}{\partial t}(x,0)=g(x)$
Given: f(x)=0 & g(x)=x for 0<x<1. c=1, L=1.

D'Alembert:
$u(x,t)=\frac{1}{2}[f^*(x-ct)+f^*(x+ct)]+\frac{1}{2c}\int_{x-ct}^{x+ct} g^*(s)ds$

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

Where $g^*$ is the odd 2-period extension of g and G is the antiderivative of $g^*.$
Let . $G(x)=\int_{-1}^x g^*(z)dz$ (2)
To complete the solution, we must determine G. We know G on any interval of length 2. Since $g^*(x)=x$ on the interval (-1,1), we obtain
. $G(x)=\int_{-1}^x g^*(z)dz \;=\; \frac{1}{2}x^2 \;-\; \frac{1}{2}$. for x in (-1,1) (3) . Hence
$G(x) \;=\; \frac{1}{2}x^2 \;-\; \frac{1}{2} \;\;\;if\;-1
$G(x) \;=\; G(x+2) \;\;otherwise.$

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
• Mar 2nd 2010, 01:42 AM
yungman