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Math Help - 2nd PDE for Exam

  1. #1
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    2nd PDE for Exam

    Hi again,

    x^2u_x - y^2u_y = 0 subject to u(1,y) = F(y)

    Solution

    x = 1, y = s, u(s) = F(s)

    \frac{dx}{dt} = x^2

    \frac{dx}{x^2} = dt

    \frac{-1}{x} = t +c. Sub in x = 1 we have - 1 = 0 + c so

    \frac{-1}{x} = t - 1

    x = \frac{1}{1 - t}

    Now \frac{dy}{dt} = -y^2

    \frac{dy}{y^2} = -dt

    \frac{-1}{y^2} = -t + c

    but I don't understand this next line....

    \frac{-1}{y} = -t - \frac{1}{s} I don't know how c = \frac{-1}{s}

    From there, y = \frac{1}{t+ \frac{1}{s}}

    From x and y it can be shown that

    s = \frac{1}{\frac{1}{x} + \frac{1}{y} - 1}

    Then
    \frac{du}{dt} = 0, \ \ u = F(s) Why is u = F(s)?

    so

    u(x,y) = F \Big( \frac{1}{\frac{1}{x} + \frac{1}{y} - 1} \Big) <br />
    = F \Big(\frac{xy}{x+y - xy} \Big) Again I'm not sure where these values are coming from.

    Is it just a simple substitution somewhere?

    Any help would be really appreciated.
    Thanks
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  2. #2
    MHF Contributor
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    Let me put in some details that might help. In your problem you are actually transforming coordinates from (x,y) \to (t,s). So by the chain rule

    u_t = u_x x_t + u_y y_t

    Choosing x_t = x^2, and y_t = -y^2 gives

    u_t = u_x x_t + u_y y_t = x^2 u_x - y^2 u_y = 0 (from your PDE).

    Thus, we are required to solve

    x_t = x^2,\;\;y_t = -y^2,\;\;\;u_t = 0 (these are PDEs but independent of s so really ODEs).

    What about the IC x = 1 where u = F(y). We need some kind of IC to go with this in the (t,s) plane. Here we choose simple and let t = 0 and connect the two boundary curves via y = s (as you've done).

    Thus, solve

    x_t = x^2,\;\;y_t = -y^2,\;\;\;u_t = 0,\;\;(*)

    subject to when t = 0 then

    x = 1,\;\;y = s\;\;\;u = F(y) = F(s),\;\;(**).

    Remember when interating (*) we get functions of integration and not constants of integration. Use (**) to find these.
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