# How to build the solution to a first-order PDE with the method of characteristics?

I'm having a hard time to understand the idea of the initial curve in the method of characteristics. Suppose we have a quasi-linear first order PDE in two variables:

$$a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u).$$

I already know that if we consider $S$ the graph of $u$, that is

$$S = \{(x,y,u(x,y))\in \mathbb{R}^3 : (x,y)\in \mathbb{R}^2\},$$

then the characteristic curves $c: I\subset \mathbb{R}\to \mathbb{R}^3$ are curves which lie on the surface $S$ and satisfies

$$\dfrac{dc_1(t)}{dt}=a(c_1(t)), \quad \dfrac{dc_2(t)}{dt}=b(c_2(t)), \quad \dfrac{dc_3(t)}{dt}=c(c_3(t)),$$

which is the characteristic system. All of that is nice, with this we can get a curve in $S$ passing through each point of the surface.

Now, usually to actually solve a problem we also need to give one initial curve which intercepts the characteristic curves.

The problem is that I don't get the idea of this initial curve. In fact, I'm so confused with this that I didn't even know how to properly explain what those initial curves are above.

In that case what are those initial curves and what is the idea behind them? And mainly, after solving the characteristic system and knowing the parametrization of the characteristic curves, how does one use the initial curve to find the solution to the problem?

How to build the solution to the PDE from the characteristic curves and the initial curve together?

The characteristic system is a first-order ODE for the curve $c$, so (assuming enough regularity on $a,b,c$ to get uniqueness) specifying an initial value $c(0) \in \mathbb R^3$ determines a solution $c : \mathbb R \to \mathbb R^3$.
Thus by specifying the value of $u$ at a single point, we determine the value of $u$ along the curve $(c_1,c_2): \mathbb R \to \mathbb R^2$. But we want to know the value of $u$ everywhere! Since choosing an initial value at a point only gives us a curve, we need to choose initial values at enough points that the resulting curves cover the whole plane.
If it's not immediately intuitive that a curve of such points is what we need, consider the simple case $a = 0$. In this case the characteristic curves have constant $c_1$; i.e. when looking down on the plane they are just vertical lines. Thus specifying an initial value $u(x_0,y_0)$ determines the value of $u$ along the vertical line $x = x_0$; so in order to uniquely determine the solution we need to specify the initial value at one point on each vertical line. The most sensible way to do this is by choosing a curve of points transverse to the vertical lines - e.g. a horizontal line. The general case is just a distorted version of this picture - the characteristic curves are all wobbly, but can we can choose one point on each of them by drawing some other curve transverse to them.
By choosing these points $(x,y)$ to lie on a regular curve $(x(s),y(s))$ and specifying our initial value as a smooth function of $s$, we can ensure that when we "glue" all the resulting characteristic curves back together they will form a smooth function.