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PDE derivation

  1. Nov 21, 2008 #1
    the book gives

    [tex] u_{xx} - u_{tt} - au_{t} - bu = 0; 0<x<L, t>0 [/tex]

    says if you multiply it by

    [tex]2u_{t} [/tex]

    you can get

    [tex]\left( 2u_{t}u_{x}\right)_{x} - \left( u^{2}_{x} + u^{2}_{t} + bu^{2}\right)_{t} -2au^{2}_{t} = 0 [/tex]

    or

    [tex]\frac{\partial}{\partial x} \left( 2 \frac{\partial u}{\partial t}\frac{\partial u}{\partial x} \right) - \frac{\partial}{\partial t} \left[ \left( \frac{\partial u}{\partial x} \right) ^{2} + \left( \frac{\partial u}{\partial t} \right) ^{2} + bu^{2} \right] - 2a\left(\frac{\partial u}{\partial t} \right)^{2} =0; [/tex]

    So far I have:

    [tex] 2 \frac{\partial u}{\partial t} \left( \frac{\partial^{2} u}{\partial x^{2}} - \frac{\partial^{2}u}{\partial t^{2}} - a\frac{\partial u}{\partial t} - b\frac{\partial u}{\partial t} \right) = 0; [/tex]

    [tex] 2\frac{\partial u}{\partial t} \ \frac{\partial^{2}u}{\partial x^{2}} \ - \ 2 \frac{\partial u}{\partial t} \ \frac{\partial^{2}u}{\partial t^{2}} \ - \ 2a\left( \frac{\partial u}{\partial t} \right)^{2} - 2b\left(\frac{\partial u}{\partial t} \right)^2 = 0; [/tex]

    I can pull a d/dx out of the first term, to get [tex] \frac{\partial}{\partial x}\left(2\frac{\partial u}{\partial t} \ \frac{\partial u }{\partial x}\right) [/tex], and the [tex]- \ 2a\left( \frac{\partial u}{\partial t} \right)^{2} [/tex] is already there. How can I get the rest of it?
     
  2. jcsd
  3. Nov 21, 2008 #2

    tiny-tim

    User Avatar
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    Hi Somefantastik! :smile:

    I'm not swearing, but …

    b u! :eek:
     
  4. Nov 21, 2008 #3
    [tex]
    2 \frac{\partial u}{\partial t} \left( \frac{\partial^{2} u}{\partial x^{2}} - \frac{\partial^{2}u}{\partial t^{2}} - a\frac{\partial u}{\partial t} - bu\right) = 0;
    [/tex]

    so now I need to show somehow that

    [tex]
    - \frac{\partial}{\partial t} \left[ \left( \frac{\partial u}{\partial x} \right) ^{2} + \left( \frac{\partial u}{\partial t} \right) ^{2} + bu^{2} \right] \ = \ -2\frac{\partial u}{\partial t} \frac{\partial^{2}u}{\partial t^{2}} \ - \ 2bu\frac{\partial u}{\partial t}
    [/tex]

    [tex]
    -2\frac{\partial u}{\partial t} \frac{\partial^{2}u}{\partial t^{2}} \ - \ 2bu\frac{\partial u}{\partial t} \ = \ -\frac{\partial}{\partial t} \left(2\frac{\partial u}{\partial t}\frac{\partial u}{\partial t} + 2bu^{2} \right) \ = \ -\frac{\partial}{\partial t} \left[2\left(\frac{\partial u}{\partial t}\right)^{2} + 2bu^{2} \right] [/tex]

    I'm still missing the (du/dx)^2 term? And that 2 hanging out in there is wrong too.
     
  5. Nov 21, 2008 #4
    You should use chain rule for composite differentials, you can't pull out that d/dx term like that in the first post.The best way is, write out all the terms of the end result. Simplify and try to collect to get the first expression.Its pretty straightforward.You know chain rule right?

    [tex]
    (AB)_{x}=A(B)_{x}+B(A)_{x}
    [/tex]

    Write out the squares explicitly, like (A*A),and use the chain rule.For example first term

    [tex]
    \left( 2u_{t}u_{x}\right)_{x}=2(u_{t}u_{xx}+u_{xt}u_{x})
    [/tex]

    Second term

    [tex]
    (u_{x}^2)_{t}=2(u_{xt}u_{x})
    [/tex]

    etc, etc
    As you see, some of the terms will cancel out.you can do the rest.
     
  6. Nov 22, 2008 #5
    DUH, of course. I should know better. Thank you.
     
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