Let $a,b,c \,$ be continuous functions defined on $\Bbb R^2$.

Question

I am stuck on the following problem:

Let $a,b,c \,$ be continuous functions defined on $\Bbb R^2$.Let $V_1,V_2,V_3$ be non-empty subsets of $\Bbb R^2$ such that $V_1 \cup V_2 \cup V_3 =\Bbb 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

1. $V_1,V_2,V_3$ are open sets in $\Bbb R^2$

2. $V_1,V_3$ are open sets in $\Bbb R^2$

3. $V_1,V_2$ are open sets in $\Bbb R^2$

4. $V_2,V_3$ are open sets in $\Bbb R^2$

What I know that $$a(x,y)u_{xx}+b(x,y)u_{xy}+c(x,y)u_{yy}=0$$ is elliptic if $\,\,b^2-4ac <0$, parabolic if $\,\,b^2-4ac =0$ and hyperbolic if $\,\,b^2-4ac >0$. So, looking at this I think option 2 is the right choice. Since we are talking about open sets ,the case where $b^2-4ac=0$ will not arise. Am I right ? Is there any better way to look at the problem?

Answer 1

Since $\,a,b,c\,$ are continuous, the non-empty subset $V_2$ is closed. Hence the cases $1,3,4$ are to be excluded. The only case left satisfies all the requirements. So you are quite right, and no better way could be found.

Answer 2

Adding more details to the above answer: $b^2-4ac$ is continuous since $a,b,c$ are continuous, and $(b^2-4ac)^{-1}[\{0\}]=V_2$, assured by the facts that

1. $V_1\cup V_2\cup V_3=\Bbb R^2$,
2. $(b^2-4ac)<0$ in $V_1$ and
3. $(b^2-4ac)>0$ in $V_3$

and since inverse image of a closed set under continuous map is closed, $V_2$ is closed, ruling out options 1,3 and 4.