Question #105380
Find the value of n for which the equation (n - 1)^2 uxx - y^2n uyy = ny^2n-1 uy , is
(1) parabolic
(2) hyperbolic
1
Expert's answer
2020-03-13T12:31:01-0400

Find the value of nn for which the equation (n1)2uxxy2nuyy=ny2n1uy(n-1)^2u_{xx}-y^{2n}u_{yy}=ny^{2n-1}u_y 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)A(x,y)u_{xx}+B(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 (x0,y0)(x_0,y_0) depends on the sign of the discriminant defined as


Δ(x0,y0)=B2A2CB\Delta(x_0,y_0)=\begin{vmatrix} B & 2A \\ 2C & B \end{vmatrix}

Δ(x0,y0)=02(n1)22(y2n)0=4(n1)2y2n\Delta(x_0,y_0)=\begin{vmatrix} 0 & 2(n-1)^2 \\ 2(-y^{2n}) & 0 \end{vmatrix}=4(n-1)^2y^{2n}

(1) parabolic: Δ(x0,y0)=0=>4(n1)2y2n=0=>n=1 or y=0\Delta(x_0,y_0)=0=>4(n-1)^2y^{2n}=0=>n=1 \ or\ y=0

n=1 or y=0n=1 \ or\ y=0


(2) hyperbolic: Δ(x0,y0)>0=>4(n1)2y2n>0=>n1,y0\Delta(x_0,y_0)>0=>4(n-1)^2y^{2n}>0=>n\not=1, y\not=0

n1,y0n\not=1, y\not=0



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