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Math Help - Monge's method solution

  1. #1
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    Monge's method solution

    solve the p.d.e-

    r+3s+t-(s^2-rt)=1

    by monge's method

    where p,q,r,s,t have their usual meanings.

    thnx
    Last edited by CaptainBlack; October 23rd 2009 at 01:18 PM.
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  2. #2
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    Quote Originally Posted by theta View Post
    solve the p.d.e-

    r+3s+t-(s^2-rt)=1

    by monge's method

    where p,q,r,s,t have their usual meanings.

    thnx
    The Monge-Ampere equation

    R r+S s+T t+U\left(r t-s^2\right) = V

    was studied extensively in the 1800's most notably by Lie. He
    showed that it was possible to find first integrals to this
    equation. If one solves

    dp = r dx + s dy,\;\;\; dq = s dx + t dy

    for r and t and substitutes into our PDE and separate wrt s then
    we obtain the Monge equations

     <br />
R dy^2 - S dx dy + T dx^2 + U(dx dp + dy dq) =0,
     <br />
R dy dp + T dx dq + U dp dq - V dx dy = 0<br />

    Considering a linear combination of these two shows that they can
    be factored if \lambda are two real distinct solutions of

     <br />
U^2 \lambda^2 + S U \lambda + TR + UV = 0<br />
and further, the factors are

     <br />
\left(U dp + T dx + \lambda_1 U dy\right)\left(U dq + \lambda_1 U dx + R dy \right) = 0<br />

     <br />
\left(U dp + T dx + \lambda_2 U dy\right)\left(U dq + \lambda_2 U dx + R dy \right) = 0<br />

    In your case where R = 1, S = 3, T = 1, U = 1 \; \text{and}\; V = 1 then we have \lambda^2 + 3 \lambda + 2 = 0 which has factors \lambda = -1, -2. Thus, we have

    \left(dp + dx - dy\right)\left( dq - dx + dy \right) = 0
    \left( dp + dx - 2 dy\right)\left( dq - 2 dx + dy \right) = 0

    which gives rise to

    dp + dx - dy = 0, \;\; dq - 2 dx + dy = 0

    or

     <br />
dp + dx - 2 dy = 0 ,\;\;\; dq - dx + dy =0<br />

    leading to the first integrals

     <br />
p + x - y = f(q - 2x + y)<br />
and
    p + x - 2y = g(q -x + y)

    Sophus Lie (1877) showed that via a contact transformation that it
    possible to transform PDE like yours to the wave equation. In
    your case the transformation is

     <br />
x = U_X + U_Y,\;\;\; y = X - Y,\;\;\;u=(2X-Y)(U_X+U_Y) -<br />
\frac{1}{2}(U_X+U_Y)^2-\frac{1}{2}(X-Y)^2-U<br />
\;\;\;(1)

    giving U_{XY} = 0. With the general solution as U = F(X) + G(Y), gives the general solution of your PDE via (1).
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  3. #3
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    Solve by Monge's Method

    (1) (q^2)r-2pqs+(p^2)t=(p^2)qz
    (2) x[r+2xr+(x^2)t]=p+2(x^3)
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