Find the value of n for which the equation (n−1)2uxx−y2nuyy=ny2n−1uy is
(1) parabolic
(2) hyperbolic
Second-order PDE
A(x,y)uxx+B(x,y)uxy+C(x,y)uyy=F(x,y,u,ux,uy) The type of second-order PDE at a point (x0,y0) depends on the sign of the discriminant defined as
Δ(x0,y0)=∣∣B2C2AB∣∣
Δ(x0,y0)=∣∣02(−y2n)2(n−1)20∣∣=4(n−1)2y2n
(1) parabolic: Δ(x0,y0)=0=>4(n−1)2y2n=0=>n=1 or y=0
n=1 or y=0
(2) hyperbolic: Δ(x0,y0)>0=>4(n−1)2y2n>0=>n=1,y=0
n=1,y=0
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