I know this question is quite maths-focused; however, it relates closely to numerical methods that are used to solve Physics problems (for example in Fluid Dynamics/CFD). I asked the same question here on Mathematics SE a few days ago, but no-one there has answered it and it was suggested that I might get a better answer here, since PDEs are core to many physical theories.
I am currently reading through the book 'Computational Techniques for Fluid Dynamics', by C.A.J. Fletcher. Chapter 2 discusses classification of PDEs by finding the number and nature of their characteristics. However, there is a section about finding characteristics of second-order PDEs (2.1.3), which I am a little confused about.
They give a generalized second-order PDE as:
$$A\frac{\partial^2u}{\partial x^2}+B\frac{\partial^2u}{\partial x\partial y}+C\frac{\partial^2u}{\partial y^2}+H=0\tag{1}$$
where $A$, $B$ and $C$ are functions of $x,y$ and $H$ contains all the first-derivative terms. They then introduce some new variables to simplify the notation:
$$P=\frac{\partial u}{\partial x}, Q=\frac{\partial u}{\partial y}, R=\frac{\partial^2 u}{\partial x^2}, S=\frac{\partial^2 u}{\partial x\partial y}, T=\frac{\partial^2 u}{\partial y^2}$$
They then state that a curve K is introduced and along a tangent to K, the differentials for $P$ and $Q$ satisfy:
$$dP=Rdx+Sdy\tag{2}$$
$$dQ=Sdx+Tdy\tag{3}$$
Using the substitutions above, the original PDE can be written:
$$AR+BS+CT+H=0$$
Equations (2) and (3) are then used to eliminate $R$ and $T$ in the above equation, which is then re-arranged to give the following:
$$S\Bigl[A\Bigl(\frac{dy}{dx}\Bigr)^2-B\Bigl(\frac{dy}{dx}\Bigr)+C\Bigr]-\Bigl\{\Bigl[A\Bigl(\frac{dP}{dx}\Bigr)+H\Bigr]\frac{dy}{dx}+C\frac{dQ}{dx}\Bigr\}=0\tag{4}$$
It then states that if:
$$A\Bigl(\frac{dy}{dx}\Bigr)^2-B\Bigl(\frac{dy}{dx}\Bigr)+C=0\tag{5}$$
then that eliminates the left-hand term in equation (4), which yields a simpler relationship between $\frac{dP}{dx}$ and $\frac{dQ}{dx}$. The solutions of equation (5) define the characteristic curves for the PDE.
So, here is where I am confused: why did they choose to split up equation (4) in that manner? As I understand it, the goal of finding characteristic curves is to reduce a PDE to a total differential, so that it can be more easily solved. However, how has this goal been met, when the terms $P$, $Q$ and $H$ on the right-hand side of equation (4) still contain partial differentials? Given that it still contains partial differentials, how is the reduced form of the equation more useful and why does equation (5) provide the characteristic curves?
Thanks in advance!