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Math Help - PDE, Second Order Differential

  1. #1
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    PDE, Second Order Differential

    Sorry for the poor lay-out to follow (i'm useless at latex) but I think I'm missing a trick here, I've got the answer but I can't get to it.

    The function u(x, y) satisfies the partial differential equation;

    6d^2u/dx^2 + y(d^2/dxdy) - y^2 (d^2u/dy^2) + 2(du/dx) = 0.

    Using the method of characteristics (told to) I got

    p = x - 2ln (y) r =x + 3ln(y) (just use those letters as I don't have the greek letters on my computer that are used.)

    Using the chain rule show that this equation transforms to 5(d^2u/d!d?) + du/di = 0.

    I've got it down to

    d^2u/dx^2 = d^2u/dp^2 + 2(d^2u/drdp) + d^2u/dr^2

    d^2u/dy^2 = 9/y^2(d^2u/dr) - 12/y^2(du/drdp) + 4/y^2(d^2u/dp^2) - 3/y^2 (du/dr) + 2/y^2 (du/dp)


    d^2u/dxdy = 3/y(d^2u/dr^2) + 1/y(d^2u/dpdr) - 2/y(d^2u/dp)

    When substituting these values back into the original equation the terms I want to cancel out do and as its an hyperbolic I want to be left with the du/dxdy terms.

    But instead of getting 5 as the co-efficient like the answer, I keep getting 25, I can't for the life of me see where I'm dropping the minuses. Spent about an hour and a half on this question so far and driving me crazy.
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  2. #2
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    Quote Originally Posted by Broadhallian View Post
    Sorry for the poor lay-out to follow (i'm useless at latex) but I think I'm missing a trick here, I've got the answer but I can't get to it.

    The function u(x, y) satisfies the partial differential equation;

    6d^2u/dx^2 + y(d^2/dxdy) - y^2 (d^2u/dy^2) + 2(du/dx) = 0.

    Using the method of characteristics (told to) I got

    p = x - 2ln (y) r =x + 3ln(y) (just use those letters as I don't have the greek letters on my computer that are used.)

    Using the chain rule show that this equation transforms to 5(d^2u/d!d?) + du/di = 0.

    I've got it down to

    d^2u/dx^2 = d^2u/dp^2 + 2(d^2u/drdp) + d^2u/dr^2

    d^2u/dy^2 = 9/y^2(d^2u/dr) - 12/y^2(du/drdp) + 4/y^2(d^2u/dp^2) - 3/y^2 (du/dr) + 2/y^2 (du/dp)


    d^2u/dxdy = 3/y(d^2u/dr^2) + 1/y(d^2u/dpdr) - 2/y(d^2u/dp)

    When substituting these values back into the original equation the terms I want to cancel out do and as its an hyperbolic I want to be left with the du/dxdy terms.

    But instead of getting 5 as the co-efficient like the answer, I keep getting 25, I can't for the life of me see where I'm dropping the minuses. Spent about an hour and a half on this question so far and driving me crazy.
    When you substitute your three transformed derivatives (I checked - they are correct) into the second order derivatives of your PDE, you should get

    25 u_{rp} + 3 u_r - 2 u_p.

    This, with the first order derivative gives the entire PDE as

    25 u_{rp} + 5 u_r = 0,

    or u_{rp} + \frac{1}{5}\,u_r = 0.

    BTW - this PDE is totally solvable.
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  3. #3
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    I see where I went wrong now, it was all in the first order terms with what I was being cocky about and ignoring .

    All I needed to do was pay more attention to the signs there and realised I had to divide it by 5!!

    BTW - this PDE is totally solvable.
    Aye, thats the next part, I've solved it as it gives you the answer in the question but it was driving me crazy as I couldn't see where I was going wrong to get the pde.

    Cheers.
    Last edited by mr fantastic; April 7th 2009 at 10:49 PM.
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  4. #4
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    Quote Originally Posted by Broadhallian View Post
    Aye, thats the next part, I've solved it as it gives you the answer in the question but it was driving me crazy as I couldn't see where I was going wrong to get the pde.
    You want to solve: u_{rp} + \tfrac{1}{5}\,u_r = 0.
    Let v = u_r and that gives: v_p + \tfrac{1}{5}v = 0.
    Now multiply by e^{p/5} so: v_p e^{p/5} + \tfrac{1}{5}e^{p/5} v = 0 \implies \left( v e^{p/5} \right)_p = 0

    Can you continue now?
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