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PDE question

  1. Apr 29, 2009 #1
    Given this equation:

    [tex]
    u_t-u_{xx}=0
    [/tex]
    [tex]
    u(x,0)=x(2-x)
    x\in[0,2]
    [/tex]
    [tex]
    u(0,t)=u(2,t)=0
    t\in[0,2]
    [/tex]

    Verify that u(x,t)=u(2-x,t) is a solution.

    To do this would I just show that:

    u(0,t)=u(2,t)
    u(2-x,0)=(2-x) but off by a scalar.
     
  2. jcsd
  3. Apr 30, 2009 #2

    tiny-tim

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    Hi chaotixmonjuish! :wink:
    I'm confused :confused:

    u(2-x,0) = (2-x)(2-(2-x)) = … ? :smile:
     
  4. Apr 30, 2009 #3

    HallsofIvy

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    What do you mean by "u(x,t)= u(2-x,t)"? You certainly do NOT know that u(2-x,0)= 2-x, only that u(2-x,0)= u(0,x)
     
  5. Apr 30, 2009 #4
    Oh geeze, so how do I verify this?
     
  6. Apr 30, 2009 #5
    I got u(2-x(2-x),0)=...=u(x^2,0)
     
  7. Apr 30, 2009 #6

    tiny-tim

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    uhh? :confused:

    2 - (2 - x) = x.
     
  8. Apr 30, 2009 #7
    Whoops, I meant this:

    [tex]
    u(x,t)=u(2-x,t)
    [/tex]
    [tex]
    u(x,0)=u(2-x)
    [/tex]
    [tex]
    u(x,0)=(2-x)(2-(2-x))
    [/tex]
    [tex]
    u(x,0)=x(2-x)
    [/tex]
    This satisifies the initial condition.
     
    Last edited: Apr 30, 2009
  9. Apr 30, 2009 #8

    tiny-tim

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    Hi chaotixmonjuish! :smile:

    (erm :redface: … why bother with LaTeX when you could just have typed it as text? :wink:)
    Yup … that nails the middle condition …

    and the third one is easy …

    now how about ut - uxx = 0 ? :smile:
     
  10. Apr 30, 2009 #9
    I'm having problems just getting why we only used 2-x in the substitution
     
  11. Apr 30, 2009 #10

    tiny-tim

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    Because we were seeing what happens if u(x,t) = u(2-x,t),

    and for that we needed to know what u(2-x,t) is for t = 0, ie u(2-x,0). :wink:
     
  12. Apr 30, 2009 #11
    So it that why we can pop a 2-x in for u?
     
  13. Apr 30, 2009 #12
    hi chaotixmonjuish ,, can you tell me what is the chapter name of these questions ? they look cool :)
     
  14. May 1, 2009 #13

    HallsofIvy

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    You aren't putting 2-x in for u, you are replacing one of the variables in u with 2-x. And, of course, you can replace a variable in a function with anything!
     
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