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Math Help - Please help me on understanding the answer of this D'Alembert problem.

  1. #1
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    Please help me on understanding the answer of this D'Alembert problem.

    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<x<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
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  2. #2
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    Anyone please?
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  3. #3
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