Question

x^2(y-1)uxx-x(y^2-1)uxy+y(y-1)uyy+(xy)ux-uy =0
Reduce into canonical form and solve it

Answer 1

Transforming the PDE into a canonical form.

Comparing the given partial differential equation with

$A(x,y)u_{xx}+B(x,y)u_{xy}+C(x,y)u_{yy}+D(x,y)u_x+E(x,y)u_y+F(x,y)u+G(x,y)= 0,$

We have

$$A=x^2(y-1);\ B=-x(y^2-1);\ C=y(y-1)$$

The roots of the equations $Aα^2+Bα+C= 0$ i.e., $x^2(y-1)\alpha-x(y^2-1)\alpha+y(y-1)=0$ are given by $λi=±\frac1x$.The differential equations for the family of characteristics curves are

$$\frac{dy}{dx}\pm \frac1x=0$$

Whose solutions are $y+ln\ x=c_1$ , and $y-ln\ x=c_2$. Choose

$$\xi=y+ln\ x,\ \eta=y-ln\ x$$

An easy computation shows that

$u_x=u_{\xi}\xi_x+u_{\eta}\eta_x,\\ u_{xx}=u_{\xi \xi}\xi^2_x-2u_{\xi \eta}x^2+u_{\eta \eta}x^2+u_{\xi}-u_{\eta},\\ u_{yy}=u_{\xi \xi}+2u_{\xi \eta}+u_{\eta \eta}$

Substituting these expression in the equation $x^2(y-1)u_{xx}-x(y^2-1)u_{xy}+y(y-1)u_{yy}+(xy)u_x-u_y =0$ yields the conical form and the solution to the PDE.