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Math Help - Partial differential equation-wave equation(2)

  1. #1
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    Exclamation Partial differential equation-wave equation(2)

    Can you help me to solve equation....
    ((∂u)/(∂x))(x,t)=(1/(c))((∂u)/(∂x))(x,t)
    Use the equation above to determine the dimension of the constant c
    But why ((∂u)/(∂x))=LL⁻and ((∂u)/(∂x))=LT⁻........
    How can it evolution?
    Please help me to solve
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  2. #2
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    Quote Originally Posted by Steve Lam View Post
    Can you help me to solve equation....
    ((∂u)/(∂x))(x,t)=(1/(c))((∂u)/(∂x))(x,t)
    Use the equation above to determine the dimension of the constant c
    But why ((∂u)/(∂x))=LL⁻and ((∂u)/(∂x))=LT⁻........
    How can it evolution?
    Please help me to solve
    The wave in equation in general is

     <br />
\frac{\partial ^2 u}{\partial t^2} = c^2 \frac{\partial ^2 u}{\partial x^2}<br />

    If the dimensions of t is T and x L, the due to the equality and the presence of u on both sides, the dimensions of c^2 is

     <br />
c^2 = \frac{L^2}{T^2}<br />
, the dimension of c is like m/s and often c is called the wave speed.

    As for the solution, if you introduce new coordinates

    r = x + c \,t and s = x - c \,t, then the PDE transforms to

     <br />
\frac{\partial ^2 u}{\partial r \partial s} = 0<br />

    which has as it's olution

     <br />
u = f(r) + g(s)<br />
    or

     <br />
u = f(x+c \,t) + g(x-c \,t)<br />
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  3. #3
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    But why ((∂u)/(∂x))=LL⁻.....
    Please explain more details
    Thanks!
    At the same time, why (∂u)=L and (∂x)=L^2
    Thanks!
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  4. #4
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    Quote Originally Posted by Steve Lam View Post
    Can you help me to solve equation....
    ((∂u)/(∂x))(x,t)=(1/(c))((∂u)/(∂x))(x,t)
    Use the equation above to determine the dimension of the constant c
    But why ((∂u)/(∂x))=LL⁻and ((∂u)/(∂x))=LT⁻........
    How can it evolution?
    Please help me to solve
    You have written an error in your PDE. Anyway the correct equation is:
    \frac{\partial^2u}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2u}{\partial t^2}
    The first term can be written as:
    \frac{\partial^2u}{\partial x^2}=\frac{1}{\partial x}\left( \frac{\partial u}{\partial x} \right)
    The dimensions of these are now:
    \frac{1}{L}\left( \frac{V}{L} \right)
    In which I used V as the dimension of the dependent variable. This can be anything the PDE stands for, a pressure, a voltage, a current, ....
    So using this on the complete PDE you get:
    \frac{1}{L}\left(\frac{V}{L}\right) =\frac{1}{C^2} \frac{1}{T}\left( \frac{V}{T} \right)
    C is here the dimension of c. From which we can extract the dimension of c, i.e. C as:
    C=\frac{L}{T}

    coomast
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  5. #5
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    Thank you! I am so happy!
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