Originally Posted by

**Danny** First off, let's determine the type.

If $\displaystyle A = 4y^2,\;\; B = 2(1-y^2),\;\; C = -1$ then

$\displaystyle B^2 - 4AC = 4(1-y^2)^2+ 16y^2 = 4(1+y^2)^2 > 0$ so it's hyperbolic. This means under a change of variable we should be able to transform you PDE to a standard form, i.e.

$\displaystyle u_{rs} + lots = 0$ (I'm using r and s instead of $\displaystyle \xi $ and $\displaystyle \eta$. lots - lower order terms). We now introduce first and secord order transforms. If $\displaystyle r = r(x,y), \;\;s = s(x,y)$ then

$\displaystyle u_x = u_r r_x + u_s s_x,\;(1a)$

$\displaystyle u_y = u_r r_y + u_s s_y,\;(1b)$

$\displaystyle u_{xx} = u_{rr} r_x^2 + 2 u_{rs} r_x s_x + u_{ss} s_x^2 +u_r r_{xx} + u_s s_{xx},\;(1c)$

$\displaystyle u_{xy} = u_{rr} r_x r_y + u_{rs}\left( r_x s_y + r_y s_x\right) + u_{ss} s_x s_y + u_r r_{xy} + u_s s_{xy},\;(1d)$

$\displaystyle u_{yy} = u_{rr} r_y^2 + 2 u_{rs} r_y s_y + u_{ss} s_y^2 +u_r r_{yy} + u_s s_{yy},\;(1e)$.

Substitute these into your PDE and group terms

$\displaystyle \left( 4y^2 r_x^2 +2(1-y^2)r_x r_y - r_y^2\right) u_{rr}$

$\displaystyle + \left( 4y^2 r_x s_x +2(1-y^2)(r_x s_y + r_y s_x) - r_y s_y\right) u_{rs}$$\displaystyle + \left( 4y^2 s_x^2 +2(1-y^2)s_x s_y - s_y^2\right) u_{ss} + lots = 0$.

In order to hit the target we need the coefficient of $\displaystyle u_{rr}$ and $\displaystyle u_{ss}$ to vanish. Thus,

$\displaystyle 4y^2 r_x^2 +2(1-y^2)r_x r_y - r_y^2 = 0,\;\;\;

4y^2 s_x^2 +2(1-y^2)s_x s_y - s_y^2 = 0$,

or

$\displaystyle \left(2r_x - r_y\right)\left(2y^2 r_x + r_y\right) = 0,\;\;\;\left(2s_x - s_y\right)\left(2y^2 s_x + s_y\right) = 0$

They are the same so choose one for $\displaystyle r$ and the other for $\displaystyle s$, i.e.

$\displaystyle 2r_x - r_y = 0,\;\;\;2y^2 s_x + s_y = 0$.

These are first oder PDEs, so by the method of characteristics, we fiind the solutions

$\displaystyle r = R(x + 2y),\;\;\; s = S(x - \frac{2}{3}y^3)$

which we choose the function $\displaystyle R$ and $\displaystyle S$ simple. So

$\displaystyle r = x + 2y,\;\;\; s = x - \frac{2}{3}y^3$ (as you said).

Now put these into your transformations (1) and put all of these into you PDE and simply. See how that goes.