A PDE is elliptic in $v_1$, parabolic in $v_2$ and hyperbolic in $v_3$ then which are open sets from $v_1$, $v_2$ and $v_3$

Question

Let $a$, $b$, $c$ be continuous functions defined on $R^{2}$. Let $V_{1}$, $V_{2}$, $V_{3}$ be nonempty subsets of $R^{2}$ such that $ V_{1}\cup V_{2}\cup V_{3} = R^{2}$ and the PDE $$a(x,y) u_{xx} + b(x,y) u_{xy} + c(x,y) u_{yy}=0$$ is elliptic in $V_{1}$, parabolic in $V_{2}$ and hyperbolic in $ V_{3}$ then which one is true?

1. $V_{1}$, $V_{2}$, $V_{3}$ are open sets in $R^{2}$

2. $V_{1}$, $V_{3}$ are open sets in $R^{2}$

3. $V_{1}$, $V_{2}$ are open sets in $R^{2}$

4. $V_{2}$, $V_{3}$ are open sets in $R^{2}$

   I know $$b^{2}-4ac$$

   • $< 0$ is elliptic

   • $> 0$ is hyperbolic

   • $ =0$ is parabolic

   but I am unable to relate this with open set concept

Answer 1

Hint

If $a,b$ and $c$ are continuous then: $$D:\Bbb R^2 \to \Bbb R$$ defined by: $$D=b^2-4ac$$ is also continuous.

But according to your last paragraph: $$V_1=D^{-1}(\{0\})$$ $$V_2=D^{-1}((0,+\infty))$$ $$V_3=D^{-1}((-\infty,0))$$ so the sets $V_2$ and $V_3$ are inverse of open sets by a continuous function.

As for $V_1$ you can show that it is closed so the only case it is open is if $V_1=\emptyset$ or $V_1=\Bbb R^2$.