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Higher order partial derivatives

  1. Jun 8, 2010 #1
    Consider the partial di erential equation, (y4-x2)uxx - 2xyuxy - y2uyy = 1. We will make the substitution x = s2 - t2 and y = s - t, to simplify

    (a) Solve for s and t as functions of x and y

    the farthest point i got to was
    x = s^2 - t^2 = (s+t)(s-t) = y(s+t)
    y = s - t
    s+t = x/y

    i don't know what to do after that.. i have the solution, but i have no idea how to get to it.
    the solution is
    s = x + y^2 / 2y
    t = x - y^2 / 2x
     
  2. jcsd
  3. Jun 8, 2010 #2

    tiny-tim

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    Hi magnifik! :smile:

    No, you're mssing the point …

    that isn't the solution, it's just writing s and t in terms of x and y …

    s+t = x/y, s-t = y, so 2s = x/y + y = (x + y2)/y, 2t = (x - y2)/y.

    To get the solution, you need to find uss ust and utt (and I expect ust will be zero, which will make the solution fairly easy :wink:).
     
  4. Jun 8, 2010 #3
    errr.. i meant the solution for part a, which was putting x and y in terms of s and t. thanks!
     
  5. Jun 8, 2010 #4
    but i don't understand how you got from s+t = x/y, s-t = y to 2s and 2t...how did you combine the previous two equations to get that?
     
  6. Jun 8, 2010 #5

    HallsofIvy

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    Excuse me, tiny-tim, but the only question asked was "(a) Solve for s and t as functions of x and y".

    Presumably, there is a "b" but so far all we want to do is solve for s and t.

    magnifik, you have [itex]x= s^2- t^2[/itex] and [itex]y= s- t[/itex].

    From [itex]y= s- t[/itex] you can say that [itex]s= y+ t[/itex] and, putting that into [itex]x=s^2- t^2[/itex], [itex]x= y^2+2yt+ t^2- t^2= y^2+ 2yt[/itex]. Solve that equation for t, then replace t by that expression in [itex]s= y+ t[/itex].
     
  7. Jun 8, 2010 #6

    THANK YOU!! makes so much sense now :)
     
  8. Jun 8, 2010 #7

    tiny-tim

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    erm :redface: … I added, and subtracted!

    Add s+t to s-t, you get 2s; subtract, you get 2t …

    this is a common transformation, and you should be familiar with it. :wink:
     
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