Confusion in college book on the introduction of PDE (method of characteristics) In my course book, the method of characteristics for solving 1st order PDE of the form $P \frac{\partial u}{\partial x} + Q\frac{\partial u}{\partial y} = R$ where $u$ is a function of $x$ and $y$ is presented as follows : 
Suppose the value of $u$ in known on a parameterized curve $\Gamma \equiv (x(s), y(s))$ with $s$ being the parameter used to describe this curve. 
First, the book says that if we knew the partials $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ on $\Gamma$, we could, in practice, find the values of $u$ near $\Gamma$ and then near those points and so on because of the following relation : 
$u(x+dx, y+dy) = u(x,y) + dx\frac{\partial u}{\partial x}(x,y) + dy\frac{\partial u}{\partial y}(x,y)$ 
Indeed, because we know $u(s)$, we can write that $\frac{du}{ds} = \frac{\partial u}{\partial x}\frac{dx}{ds} + \frac{\partial u}{\partial y}\frac{dy}{ds}$ and then form with the PDE the following system : 
$$
        \left(\begin{matrix}
        P & Q  \\
        \frac{dx}{ds} & \frac{dy}{ds} \\
        \end{matrix}\right) \left(\begin{matrix}
        \frac{\partial u}{\partial x} \\
        \frac{\partial u}{\partial y} \\
        \end{matrix}\right) = \left(\begin{matrix}
        R \\
        \frac{du}{ds} \\
        \end{matrix}\right)
$$
The Cauchy problem is then well-posed if the determinant of this 2 by 2 matrix is non zero for every point of $\Gamma$. This method to find $u(x,y)$ by knowing its partials is then discarded because it is not a very convenient way to find $u$.
Second, the book says that we now suppose that the problem is well-posed. Let's consider this time directions $(dx,dy)$ that are non parallel to $\Gamma$. The variation of $du$ associated with these directions is given by : 
$du = \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy$
We can then form the following system with the PDE : 
$$
        \left(\begin{matrix}
        P & Q  \\
        dx & dy \\
        \end{matrix}\right) \left(\begin{matrix}
        \frac{\partial u}{\partial x} \\
        \frac{\partial u}{\partial y} \\
        \end{matrix}\right) = \left(\begin{matrix}
        R \\
        du \\
        \end{matrix}\right)
$$
And then the book says (that's what really bugs me) : "Because the problem is well-posed, there exist for every point of $\Gamma$ a particular direction $(dx,dy)$ such that the determinant of this 2 by 2 matrix is zero, leading to the characteristic equation : $Pdy = Qdx$"
My question : Why there must exists a unique direction $(dx,dy)$ for which this determinant is zero? I don't see how the assumption of a well-posed problem leads to conclude that.
 A: It seems to me that, modulo issues with differentials, this is purely a linear algebra point. You have the nonzero row vector $(P,Q)$. Introducing an additional row to this matrix fails to result in an invertible matrix iff the new row vector is a multiple of $(P,Q)$. There is a single line of such vectors. Strictly speaking I would say there are actually two directions, though (one "positive" one and one "negative" one).
A: Passing from
$\quad\left(\begin{matrix}
        P & Q  \\
        \frac{dx}{ds} & \frac{dy}{ds} \\
        \end{matrix}\right) \left(\begin{matrix}
        \frac{\partial u}{\partial x} \\
        \frac{\partial u}{\partial y} \\
        \end{matrix}\right) = \left(\begin{matrix}
        R \\
        \frac{du}{ds} \\
        \end{matrix}\right) \quad
$ to
$\quad \left(\begin{matrix}
        P & Q  \\
        dx & dy \\
        \end{matrix}\right) \left(\begin{matrix}
        \frac{\partial u}{\partial x} \\
        \frac{\partial u}{\partial y} \\
        \end{matrix}\right) = \left(\begin{matrix}
        R \\
        du \\
        \end{matrix}\right)$
you just multiply the last row of the matrix by $ds$ so that the determinant does not change.
Now, the resulting matrix having determinant null, means that the vector $(P,Q)$ is normal to the vector $(dx,dy)$, and, provided that $P$ and $Q$ are not both null, there is only one normal vector $(dx,dy)$, apart from a multiplication constant.
Hence the direction individuated by $(dx,dy)$ is unique. 
