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PDE integral help

  1. Mar 16, 2008 #1
    My PDE book does the following:

    [tex] \int \phi_x^2 dx [/tex]

    Where,

    [tex] \phi_x = b-\frac{b}{a} |x| [/tex]

    for [tex]|x|> a [/tex] and x=0 otherwise.

    Strauss claims:

    [tex]\int \phi_x^2 dx = ( \frac{b}{a} ) ^2 2a [/tex]

    However, I think there is a mistake. It can be shown that:

    [tex] \frac{-3a}{b}(b- \frac{b|x|}{a})^3[/tex] is a Soln. Evaluate this between 0<x<a and you get:

    [tex]\frac{b^2 a}{3}[/tex]

    Because the absolute value function is symmetric, its twice this value:

    [tex]\frac{2b^2 a}{3}[/tex]

    Unless I goofed, I think the book is in error.

    *Note: Intergration is over the whole real line.
     
    Last edited: Mar 17, 2008
  2. jcsd
  3. Mar 16, 2008 #2

    Dick

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    I think you must mean |x|<a, right? Otherwise the integral isn't defined. I tried integrating (b-bx)^2 from 0 to a and I don't get anything close to either answer. Can you clarify?
     
  4. Mar 17, 2008 #3
    O crap, its b-b/a|x| sorry. See above I fixed it.
     
  5. Mar 17, 2008 #4

    Dick

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    Then I'm getting the same result as you.
     
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