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Integration for a PDE

  1. Oct 16, 2013 #1
    Ok this qusestion has to do with completing the square for a diffusion equation.

    Initial Cond: u(x,0) = e-x

    Now they say plug this into the general formula:

    u(x,t) = 1/(4[itex]\pi[/itex]kt)1/2 ∫ e-(x-y)1/2/4kte-y dy where k is a constant

    now the first step they say is completing the square of:

    -( x2-2xy+y2)+4kty/4kt

    with respect to the y variable, and they get:

    - [(y+2kt-x)2]/4kt + kt - x

    Now I could not get this, I also tried expanding out the final result and reverse engineer the result but in doing so I got stuck with an extra term:

    y2+ 2y(2kt-x) + x2 + (2kt-x)2 - (2kt-x)2

    this step is when I perform the process of completing the square before trying to factorize everything and it is here that I am having trouble. Please help if you can I have the midterm in a couple hrs.

    Thanks
     
  2. jcsd
  3. Oct 16, 2013 #2

    LCKurtz

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    The first thing is to state the problem correctly:
    $$\frac{-(x^2-2xy+y^2)+4kty}{4kt}$$which is not what you have written. Rearrange the numerator to$$
    -(y^2+(2x-4kt)y)-x^2$$Now complete the square, not forgetting the - out in front$$
    -(y^2+(2x-4kt)y+(x-2kt)^2)+(x-2kt)^2-x^2$$ $$
    =-(y+(x-2kt))^2+x^2-4kxt+4t^2-x^2$$Now put the denominator back in and simplify it.
     
  4. Oct 16, 2013 #3

    tiny-tim

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    hi trap101! :smile:
    [(y+2kt-x)2]/4kt

    = -[y2 + 4k2t2 + x2 - 2xy + 4kty - 4ktx]/kt + kt - x :wink:
     
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