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D'Alembert's Wave. Seems Easy Or Maybe I'm Missing Something?

  1. Jul 13, 2008 #1
    D'Alembert's Wave. Seems Easy Or Maybe I'm Missing Something???

    Hi I have this D'Alembert's question. What I've done seems easy so I wanted to clarify just in case I'm missing something, cus it seems so easy. Second part I don't understand and would appreciate a guidance.

    (PART A)

    D'Alembert's Solution is:
    [itex]z(x,t) = \frac{1}
    {2}\left[ {F(x + ct) + F(x - ct)} \right] + \frac{1}
    {{2c}}\int\limits_{x - ct}^{x + ct} {G(s)ds}[/itex]
    of the one dimensional wave equation:
    [itex]
    \frac{{\partial ^2 z}}
    {{\partial x^2 }} = \frac{1}
    {{c^2 }}\frac{{\partial ^2 z}}
    {{\partial t^2 }},{\text{ }} - \infty < x < \infty ,{\text{ t}} \geqslant {\text{0,}}[/itex]

    with "c" a constant, when the initial conditions are:
    [itex]z(x,0) = F(x),{\text{ }}z_t (x,0) = G(x),{\text{ }} - \infty < x < \infty[/itex]

    Evaluate the solution if F(x)=0 for [itex]-\infty < x < \infty[/itex] and
    [itex]G(x) = \left\{ {\begin{array}{*{20}c}
    {\frac{1}
    {{1 + x}}} & {\left\{ {{\text{x}} \geqslant {\text{0}}} \right\}} \\
    {\text{0}} & {\left\{ {{\text{x}} < {\text{0}}} \right\}} \\

    \end{array} } \right.[/itex]

    (PART B)

    Show the values of z(x,t) for t>0 in different regions of the (x,t)- plane. Show that the value of z(x,t) is continuous for all values of x and t, gives the correct initial value of [itex]z_t[/itex] and [itex]z_t \to 0[/itex] as |x|[itex]\to 0[/itex]

    ------------------------------------------------
    SOLUTION TO PART A:
    ------------------------------------------------


    GRAPH:
    [​IMG]


    [itex]
    z(x,t) = \left\{ {\begin{array}{*{20}c}
    {\int\limits_0^{x + ct} {\frac{1}
    {{1 + x}}} } & {\left\{ {x + ct > 0,x - ct < 0} \right\}} & {{\text{Region I}}} \\
    0 & {\{ x + ct < 0,x - ct < 0\} } & {{\text{Region II}}} \\
    {\int\limits_{x - ct}^{x + ct} {\frac{1}
    {{1 + x}}} } & {\{ x + ct > 0,x - ct > 0\} } & {{\text{Region III}}} \\

    \end{array} } \right.[/itex]


    Working out region 1:
    -----------------------
    [itex]\begin{gathered}
    {\text{Region I}}:{\text{ }}\int\limits_0^{x + ct} {\frac{1}
    {{1 + x}}} = \left[ {\ln (1 + x)} \right]_0^{x + ct} \hfill \\
    = \ln (1 + x + ct) - \ln 1 \hfill \\
    = \ln \left( {\frac{{1 + x + ct}}
    {1}} \right) = \ln (1 + x + ct) \hfill \\
    \end{gathered} [/itex]


    Working out region 2:
    -----------------------
    Region 2 = 0.


    Working out region 3:
    -----------------------
    [itex]\begin{gathered}
    {\text{Region III: }}\int\limits_{x - ct}^{x + ct} {\frac{1}
    {{1 + x}}} = \left[ {\ln (1 + x)} \right]_{x - ct}^{x + ct} \hfill \\
    = \ln (1 + x + ct) - \ln (1 + x - ct) \hfill \\
    = \ln \left( {\frac{{1 + x + ct}}
    {{1 + x - ct}}} \right) \hfill \\
    \end{gathered}[/itex]





    FINAL SOLUTION:
    -----------------------------

    [itex]
    z(x,t) = \left\{ {\begin{array}{*{20}c}
    {\ln (1 + x + ct)} & {\left\{ {x + ct > 0,x - ct < 0} \right\}} & {{\text{Region I}}} \\
    0 & {\{ x + ct < 0,x - ct < 0\} } & {{\text{Region II}}} \\
    {\ln \left( {\frac{{1 + x + ct}}
    {{1 + x - ct}}} \right)} & {\{ x + ct > 0,x - ct > 0\} } & {{\text{Region III}}} \\

    \end{array} } \right.[/itex]



    Thats part a done. As far as I know that's all there is to evaluating the question. Is there anythin else I need to add???


    PART B: any ideas on this please?
     
  2. jcsd
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