Find the value of n n n for which the equation ( n − 1 ) 2 u x x − y 2 n u y y = n y 2 n − 1 u y (n-1)^2u_{xx}-y^{2n}u_{yy}=ny^{2n-1}u_y ( n − 1 ) 2 u xx − y 2 n u yy = n y 2 n − 1 u y is
(1) parabolic
(2) hyperbolic
Second-order PDE
A ( x , y ) u x x + B ( x , y ) u x y + C ( x , y ) u y y = F ( x , y , u , u x , u y ) A(x,y)u_{xx}+B(x,y)u_{xy}+C(x,y)u_{yy}=F(x,y,u,u_x,u_y) A ( x , y ) u xx + B ( x , y ) u x y + 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 ) (x_0,y_0) ( x 0 , y 0 ) depends on the sign of the discriminant defined as
Δ ( x 0 , y 0 ) = ∣ B 2 A 2 C B ∣ \Delta(x_0,y_0)=\begin{vmatrix}
B & 2A \\
2C & B
\end{vmatrix} Δ ( x 0 , y 0 ) = ∣ ∣ B 2 C 2 A B ∣ ∣ Δ ( x 0 , y 0 ) = ∣ 0 2 ( n − 1 ) 2 2 ( − y 2 n ) 0 ∣ = 4 ( n − 1 ) 2 y 2 n \Delta(x_0,y_0)=\begin{vmatrix}
0 & 2(n-1)^2 \\
2(-y^{2n}) & 0
\end{vmatrix}=4(n-1)^2y^{2n} Δ ( x 0 , y 0 ) = ∣ ∣ 0 2 ( − y 2 n ) 2 ( n − 1 ) 2 0 ∣ ∣ = 4 ( n − 1 ) 2 y 2 n (1) parabolic: Δ ( x 0 , y 0 ) = 0 = > 4 ( n − 1 ) 2 y 2 n = 0 = > n = 1 o r y = 0 \Delta(x_0,y_0)=0=>4(n-1)^2y^{2n}=0=>n=1 \ or\ y=0 Δ ( x 0 , y 0 ) = 0 => 4 ( n − 1 ) 2 y 2 n = 0 => n = 1 or y = 0
n = 1 o r y = 0 n=1 \ or\ y=0 n = 1 or y = 0
(2) hyperbolic: Δ ( x 0 , y 0 ) > 0 = > 4 ( n − 1 ) 2 y 2 n > 0 = > n ≠ 1 , y ≠ 0 \Delta(x_0,y_0)>0=>4(n-1)^2y^{2n}>0=>n\not=1, y\not=0 Δ ( x 0 , y 0 ) > 0 => 4 ( n − 1 ) 2 y 2 n > 0 => n = 1 , y = 0
n ≠ 1 , y ≠ 0 n\not=1, y\not=0 n = 1 , y = 0
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