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Homework Help: Really easy PDE, confused about how to put in side conditions

  1. Jan 30, 2012 #1
    Hey, before you read this over I'll mention that I've checked the general solution and it works. So if you don't feel like following my steps to get the general solution just jump down to the end of my attempt, because the real problem for me is figuring out what to do with the side conditions.

    1. The problem statement, all variables and given/known data
    get the particular solution for this PDE, for side conditions u(x,1) = 0 ; u(0,y) = 0

    yuxy + 2ux = x

    2. Relevant equations
    nothing really...

    3. The attempt at a solution

    first I integrate w.r.t x, which is suggested in the text:

    yuy + 2u = x2/2 + h(y) ; where y is some function of y

    divide through by y...

    uy + 2u/y = x2/2y + h(y)/y

    now take integrating factor y2 (I think this next line follows obviously, if it is confusing please let me know and I'll show you how I got it):

    y2u = (x2y2)/4 + ∫h(y)ydy + k(x)

    at this point I have done the integral with respect to y on each side, basically. so some function k(x) shows up. Now I replace ∫h(y)ydy with just H(y), some new function of y because the whole expression ∫h(y)ydy will end up being only dependent on y so we just rename the thing:

    y2u = (x2y2)/4 + H(y) + k(x)

    and divide through by y2...

    u = (x2)/4 + H(y)/(y2)+ k(x)/(y2)

    so here's where I get confused: if I put u(x,1) into the general solution I get:
    k(x) = -x2/4 - H(1)

    and for u(0,y) I get:

    H(y) = -k(0)

    But I see no way of getting solutions for k(x) and H(y) on their own. Is there something I'm missing here? Some sort of "trick" or other thing I should be aware of? Thanks for any advice.
  2. jcsd
  3. Jan 30, 2012 #2
    aaaand just got it.
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