Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2005 / Differential equations / 9 / Solution

For a nonzero principal part the discriminant $\Delta =b^2-ac$ classifies the equation: it is hyperbolic for $\Delta >0$, parabolic for $\Delta =0$, and elliptic for $\Delta <0$. These are the three definitions requested by the inline roman labels.

In the hyperbolic case the characteristic equation $a(dy{)}^2-2bdxdy+c(dx{)}^2=0$ gives two distinct real tangent directions. Locally, with suitably smooth coefficients, solve the corresponding characteristic ordinary differential equations to obtain two independent families of curves. Label their level sets $\varphi =\text{constant}$ and $\psi =\text{constant}$. Their gradients satisfyand the coordinate Jacobian is nonzero. The chain rule therefore eliminates $u_{\varphi \varphi }$ and $u_{\psi \psi }$. The remaining mixed coefficient $2[a{\varphi }_x{\psi }_x+b({\varphi }_x{\psi }_y+{\varphi }_y{\psi }_x)+c{\varphi }_y{\psi }_y]$ is nonzero: otherwise the invertible coordinate change would send the nondegenerate principal quadratic form to zero. Divide by this coefficient and absorb all first-order terms into $g$, giving $u_{\varphi \psi }=g(\varphi ,\psi ,u,u_{\varphi },u_{\psi })$. This canonical reduction is local; a global chart is not automatic.

For the specific equation $a=y,b=0,c=-1$, so it is hyperbolic exactly on $y>0$, elliptic for $y<0$, and degenerate parabolic on $y=0$. For $\varphi =3x-2y^{3/2}$ and $\psi =3x+2y^{3/2}$, the Jacobian is $18\sqrt{y}\ne 0$. Direct differentiation givesSince $\psi -\varphi =4y^{3/2}$, the canonical form isThe characteristic families are $3x-2y^{3/2}=c$ and $3x+2y^{3/2}=c$, with slopes $dx/dy=\pm \sqrt{y}$. They are distinct for $y>0$ and meet the degeneracy line with coincident vertical tangent directions.