Monge's method solution

**theta** · October 23rd 2009, 05:57 AM

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

**Jester** · October 25th 2009, 07:26 AM

Originally Posted by theta

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

$ R dy^2 - S dx dy + T dx^2 + U(dx dp + dy dq) =0,$
$ R dy dp + T dx dq + U dp dq - V dx dy = 0 $

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

$ U^2 \lambda^2 + S U \lambda + TR + UV = 0 $ and further, the factors are

$ \left(U dp + T dx + \lambda_1 U dy\right)\left(U dq + \lambda_1 U dx + R dy \right) = 0 $

$ \left(U dp + T dx + \lambda_2 U dy\right)\left(U dq + \lambda_2 U dx + R dy \right) = 0 $

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

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

leading to the first integrals

$ p + x - y = f(q - 2x + y) $ 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

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

giving $U_{XY} = 0.$ With the general solution as $U = F(X) + G(Y)$ , gives the general solution of your PDE via (1).

**ripafi** · May 25th 2010, 03:10 AM

(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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