Let's say that we have the Tricomi Equation from Tsoumpelis PDE Part 1 (page 130)
\begin{align} u_{yy}-yu_{xx}=0 \label{1}\tag1 \end{align}
We know $a=1, \ b=0,\ c=-y$ And we want to classify this PDE in two ways, so it would be hyperbolic.
- The discriminant is: $$d=b^2-ac=y$$
So the equation has hyperbolic form when $y>0$.
- We could write the above second-order PDE into a first-order system
$$u_x - \frac{1}{\sqrt{y}} u_y =0 \tag2$$ $$u_x+ \frac{1}{\sqrt{y}}u_y=0 \tag3$$
Edited
Let us differentiate the first equation w.r.t. $y$ and the second one w.r.t. $x$: \begin{aligned} a_1 u_{xy} + a_2 v_{yy} &= 0 \\ b_1 v_{xx} + b_2 u_{yx} &= 0 \end{aligned} Using the equality of mixed derivatives for $u$ and the fact that $a_1 = b_2 \neq 0$, we obtain the equations $$ u_{xy} = u_{yx} = -a_2/a_1 v_{yy} = -b_1/b_2 v_{xx} $$ Since $a_2\neq 0$, we are left with the second-order linear PDE $A v_{xx} + 2 B v_{xy} + Cv_{yy} = 0$ satisfied by $v$, with $A = 1$, $B = 0$ and $C = -\frac{a_1 b_1}{a_2 b_2}$.
So in our case $A=1, B=0, C=-y$
Then the discriminant is:
$$D=B^2-AC=y$$ So the equation is in hyperbolic form when $y>0$
Is my consideration right?