Question about characteristics and classification of second-order PDEs 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!
 A: If $A\neq0$ then the equation (5) is a quadratic equation, and unless your PDE is elliptic ($B^2-4AC<0$), locally it has two solutions $y_1(x)$ and $y_2(x)$ which satisfy Vieta's formula,
$$
A\left(\frac{dy_1}{dx}+\frac{dy_2}{dx}\right)=B,\quad A\frac{dy_1}{dx}\frac{dy_2}{dx}=C.
$$
If your PDE is parabolic ($B^2-4AC=0$) then $y_1=y_2$. It follows that
$$
A\partial_x^2+B\partial_x\partial_y+C\partial_y^2=A\left(\partial_x^2+\left(\frac{dy_1}{dx}+\frac{dy_2}{dx}\right)\partial_x\partial_y+\frac{dy_1}{dx}\frac{dy_2}{dx}\partial_y^2\right)
$$
$$
=A\left(\partial_x+\frac{dy_1}{dx}\partial_y\right)\left(\partial_x+\frac{dy_2}{dx}\partial_y\right)+\mbox{first order terms}.
$$
Therefore (1) now becomes
$$
A\left(\partial_x+\frac{dy_1}{dx}\partial_y\right)\left(\partial_x+\frac{dy_2}{dx}\partial_y\right)u+\mbox{lower order terms}=0,
$$
which is the expected simplification.
This is the high school level answer to the question why characteristics matter. The scientific answer is much longer and involves microlocal analysis (propagation of singularities), quantum field theory (trajectories of classical point particles corresponding to a linear quantum field), etc.
This is, however, not a complete answer to the question posted, for I have no suggestion as to why (4) is relevant.
Edit: By no means could I read the author's mind when they wrote equation (4), but let me elaborate on what could be done along those lines (at OP's request).
(5) is a first order ODE, of which the solutions depend on one parameter (initial data). For every $(x_0,y_0)$ in the domain let $y_1(x;x_0,y_0)$ and $y_2(x;x_0,y_0)$ be the two solutions of (5) satisfying
$$
y_1(x_0;x_0,y_0)=y_0,\quad y_2(x_0;x_0,y_0)=y_0.
$$
In other words, $y_1(x;x_0,y_0)$ and $y_2(x;x_0,y_0)$ are the two characteristics through the point $(x_0,y_0)$. Fix $(x_0,y_0)$ for a moment, and omit their mention for brevity. Let us do with the characteristic $y_1(x)$ first. Consider $P$ and $Q$ as independent functions, then their restriction to the characteristic $P(x,y_1(x))$ and $Q(x,y_1(x))$ as unknown functions of $x$. If $H$ contains no zero order terms then $H(P,Q,x,y_1(x))$ is a function of $P$,$Q$ and $x$. This renders (4) a first order ODE relating two unknown functions $P$ and $Q$ of one variable $x$,
$$
\left[A(x,y_1(x))\frac{dP}{dx}+H(P,Q,x,y_1(x))\right]y_1'(x)+C\frac{dQ}{dx}=0.
$$
If you can solve this equation in one way or another then you obtain
$$
F_1(P,Q,x)=C_1,
$$
where $F_1$ is a function and $C_1$ is a number, both depending on the choice of the characteristics, i.e., $(x_0,y_0)$. Now if you do the same thing with the characteristic $y_2(x)$ you end up with
$$
F_2(P,Q,x)=C_2.
$$
On the other hand, if your $H$ contains zero ordet terms, $H=H(P,Q,x,y,u)$, then
$$
u(x,y_1(x))=\int^x\left[P+y_1'(t)Q\right]dt,
$$
and (4) becomes an integral-differential equation. Good luck solving it.
Now the final solution of (1) depends on the specification of the boundary (Cauchy) data. Suppose that you have a Cauchy curve $\Sigma$ that intersects every characteristic exactly once, and suppose that $P$, $Q$, $R$, $S$, $T$ and $u$ are known along $\Sigma$. Consider only those $(x_0,y_0)\in\Sigma$, and let $p(x_0,y_0)$, $q(x_0,y_0)$ be the values of $P$ and $Q$ at $(x_0,y_0)$. We now recall all dependences upon $(x_0,y_0)$ above. If $(x,y)$ is any point in the domain then let $(x_1,y_1)$ and $(x_2,y_2)$ be the intersections of $y_1(x)$ and $y_2(x)$ with $\Sigma$, respectively,
$$
y=y_1(x;x_1,y_1),\quad y=y_2(x;x_2,y_2).
$$
This brings us to the system
$$
\begin{cases}
y=y_1(x;x_1,y_1)\\
y=y_2(x;x_2,y_2)\\
F_1(P,Q,x;x_1,y_1)=F_1(p(x_1,y_1),q(x_1,y_1),x_1;x_1,y_1)\\
F_2(P,Q,x;x_2,y_2)=F_2(p(x_2,y_2),q(x_2,y_2),x_2;x_2,y_2)
\end{cases}
$$
This is the parametric solution of the problem (1) with given boundary data $p$ and $q$.
If we want an explicit solution, we make a few more steps. The curve $\Sigma$ can be locally parameterized by one variable $s$, so that $(x_1,y_1)=\eta(s_1)$ and $(x_2,y_2)=\eta(s_2)$. Therefore the above is a system of 4 equation for 4 variables $P$, $Q$, $s_1$ and $s_2$. Eliminating $s_1$ and $s_2$ we find the desired solution
$$
P(x,y),\quad Q(x,y).
$$
Finally,
$$
u(x,y)=u(x_0,y_0)+\int_{(x_0,y_0)}^{(x,y)}\left[Pdx+Qdy\right]
$$
along your favourite curve joining $(x_0,y_0)$ and $(x,y)$. The point $(x_0,y_0)\in\Sigma$ therefore we know $u(x_0,y_0)$ from the boundary data.
This is how, I think, you would solve (1) if you want to start from (4). Generally, the method of characteristics always ends with a parametric solution, and for an explicit solution you need to exliminate the dependence upon the characteristic.
