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

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

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

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

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

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

$\displaystyle
R dy^2 - S dx dy + T dx^2 + U(dx dp + dy dq) =0,$
$\displaystyle
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 $\displaystyle \lambda$ are two real distinct solutions of

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

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

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

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

which gives rise to

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

or

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

leading to the first integrals

$\displaystyle
p + x - y = f(q - 2x + y)
$ and
$\displaystyle p + x - 2y = g(q -x + y)$

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

$\displaystyle
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
$ $\displaystyle \;\;\;(1)$

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

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