Answer to Question #105583 in Differential Equations for Gayatri Yadav

Question

Find the values of n for which the equation (n-1)²uxx - y^2n uyy = ny^(2n-1)uy is
i) parabolic ii) hyperbolic.

Answer 1

$(n-1)²u_{xx} - y^{2n} u_{yy} = ny^{2n-1}u_y\\ A(x,y)u_{xx}+2B(x,y)u_{xy}+C(x,y)u_{yy}=\\ =F(x,y,u,u_x,u_y)\\$

The type of second-order PDE at a point 

$(x_0,y_0)$  depends on the sign of the discriminant defined as

$\Delta(x_0,y_0)=B^2-AC$

In our case

$A=(n-1)^2, B=0, C=-y^{2n}\\ \Delta =0^2 -(n-1)^2(-y^{2n})=(n-1)^2y^{2n}$

i) parabolic

$\Delta(x_0,y_0)=0\implies\\ (n-1)^2y^{2n}=0\implies\\ n=1 \quad or \quad y=0.$

ii) hyperbolic

$\Delta(x_0,y_0)>0\implies\\ (n-1)^2y^{2n}>0\implies\\ n\neq1 \quad \quad y\neq 0.$