Ambiguities on Solving a First Order PDE by Method of Characteristics I am having difficulties to understand that how the method of characteristics works. I will clarify my question with an example. Consider a real valued function of two real variables $u(x,y)$ which satisfies the following first order partial differential equation (PDE)
$$\frac{{\partial u}}{{\partial x}}(x,y) + a\frac{{\partial u}}{{\partial y}}(x,y) = 0\tag{1}$$
where $a$ is some real constant. We know that 
$$\left\{ \begin{array}{l}
x = x\\
y = y\\
z = u(x,y)
\end{array} \right.\tag{2}$$
is a surface with its parameters being $x$ and $y$. The PDE is telling us that the normal to the surface,$\left( {{{\partial u} \over {\partial x}},{{\partial u} \over {\partial y}}, - 1} \right)$, is perpendicular to the vector $\left( {1,a,0} \right)$. In other language, the vector $\left( {1,a,0} \right)$ is tangent to the surface.
Next, we define a curve on the surface as follows
$$\left\{ \matrix{
  x = x(s) \hfill \cr 
  y = y(s) \hfill \cr 
  z = z(s) = u(x(s),y(s)) \hfill \cr}  \right.,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \matrix{
  {{dx} \over {ds}} = 1,\,\,\,\,\,\,\,x({s_0}) = {x_0} \hfill \cr 
  {{dy} \over {ds}} = a,\,\,\,\,\,\,\,y({s_0}) = {y_0} \hfill \cr}  \right.\tag{3}$$
then the PDE tells us that
$${{dz} \over {ds}} = 0,\,\,\,\,\,\,\,z({s_0}) = u({x_0},{y_0})\tag{4}$$
Now, I solve the first order ordinary differential equations (ODEs) in $(3)$ and $(4)$ to get
$$\left\{ \matrix{
  x = (s - {s_0}) + {x_0} \hfill \cr 
  y = a(s - {s_0}) + {y_0} \hfill \cr 
  z = u({x_0},{y_0}) \hfill \cr}  \right.\tag{5}$$
hence, I found my special curve on the surface $u(x,y)$. Here is where I got stuck! :)

Question
How can I proceed to get the surface $u(x,y)$ itself? I need some explanation that describes a systematic procedure to achieve this in conjunction with its geometric interpretation. Any helps will be appreciated! :)
 A: Let $u_0(x) = u(x,0)$. Now, if we started with some $(x_0,y_0)$ such that $y_0=0$, then by what you have above we have
$$ u(x,y) = u(x(s),y(s)) = z(s) = u(x_0,y_0) = u_0(x_0) $$
where $x = x_0 + s$ and $y = y_0 + as = as$ (take $s_0 = 0$ for convenience). This means that $s = \frac{y}{a}$, and hence $x_0 = x - \frac{y}{a}$, so
$$ u(x,y) = u_0\left(x-\frac{y}{a}\right).$$

The above is how you would solve it provided boundary conditions, which are necessary to get a unique solution (very much like how, in solving first-order ODEs, you need some initial condition to specify a unique solution). But what if you wanted all solutions? Well, just notice that in the above case, we didn't put any conditions on the "boundary values" at $y=0$ (aside from it being differentiable so that we can have a differentiable solution). In other words, the boundary value $u_0$ is arbitrary, i.e. any differentiable $f$ can serve as the boundary value, and we would get a solution. So the solution to this PDE is
$$ u(x,y) = f\left(x - \frac{y}{a}\right) $$
where $f$ is an arbitrary differentiable function (akin to the arbitrary constant in 1-dimensional indefinite integration).

Of course, in solving the boundary value problem, we didn't have to specify the boundary values at $y=0$. What if we looked at the values at $x=0$ instead, and set $u_0(y) = u(0,y)$? Then we would want to solve the equations $x = x_0 + s = s$ and $y = y_0 + as$, with $u(x,y) = u_0(y_0)$. We then have $y_0 = y-ax$, and hence
$$ u(x,y) = u_0(y-ax) $$
for a specified initial data $u_0$, so the general solution is
$$ u(x,y) = f(y-ax) $$
for an arbitrary differentiable $f$. I'll leave it to you to check that the two answers I've given here are actually the same. :)
A: Here is a slightly different way of looking at the method of characteristics that tries to avoid tangents and normals to surfaces which is in the standard formulation. The aim here is not to be general, but to present the method in a simpler way to make it more clear what is going on.

We are going to find curves $(x(s),y(s))$ in the plane with the property that $u(x(s),y(s))$ stays constant as we move along these curves. Then given any point $(x_1,y_1)$ in the plane then we will follow such a curve backwards until we hit a point $(x_0,y_0)$ in the initial/boundary set. Since $u$ stays constant along these curves we know that $u(x_1,y_1) = u(x_0,y_0)$ and since $u(x_0,y_0)$ is known by the initial/boundary conditions we will then know what $u(x_1,y_1)$ is.
Finding such curves is not hard. The condition $u(x(s),y(s)) = $ constant is just the statement that 
$$0 = \frac{du(x(s),y(s))}{ds} = u_x \frac{dx}{ds} + u_y\frac{dy}{ds}$$
and since the PDE reads $u_x + u_ya = 0$ we see that taking $\frac{dx}{ds}=1$ and $\frac{dy}{ds}=a$ works. The curve, starting at $(x_0,y_0)$ when $s=0$, defined by these equations is the straight line
$$(x(s),~y(s)) = (x_0 + s,~y_0 + as) \implies y(s) = y_0 + a (x(s)-x_0)$$
Now lets assume the inital conditions are set at the $x$-axis, i.e. $u(x,0) = f(x)$. The characteristic curve through the point $(x_1,y_1)$, namely $y(x) = y_1 + a(x-x_1)$, intersects the $x$-axis at $(x_0,y_0) = (x_1 - \frac{y_1}{a},0)$. Since $u$ is constant along the curve we have $u(x_1,y_1) = u(x_0,y_0)$ and the general solution to the PDE can therefore be written
$$u(x_1, y_1) = u\left(x_1 - \frac{y_1}{a}, 0\right) =  f\left(x_1-\frac{y_1}{a}\right)$$
Note that $(x_1,y_1)$ was an arbitrary point in the plane so the formula above holds for all $x_1,y_1$.
A: I think a reconsideration should be made from Eq.$(3)$ above and what is after that. Our surface can be parameterized  by any other two parameters (I don't know that why notes and books on the subject insist on writing a parametrized curve instead of a paratmetrized surface even though it causes lots of ambiguities)
$$\left\{ \matrix{
  x = x(s,t) \hfill \cr 
  y = y(s,t) \hfill \cr 
  z = z(s,t) = u(x(s,t),y(s,t)) \hfill \cr}  \right.\tag{1}$$
In fact, this is a change of variables from $(x,y)$ to $(s,t)$. If we want our change of variable to be useful then we may require that
$$\left\{ \matrix{
  {{\partial x} \over {\partial s}} = 1 \hfill \cr 
  {{\partial y} \over {\partial s}} = a \hfill \cr}  \right.\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\left\{ \matrix{
  x(s,t) = s + p(t) \hfill \cr 
  y(s,t) = as + q(t) \hfill \cr}  \right.\,\tag{2}$$
because, this will turn our PDE into a simple one by just using chain-rule
$$\eqalign{
  & {{\partial z} \over {\partial s}}\left( {s,t} \right) = {{\partial u} \over {\partial x}}\left( {x(s,t),y(s,t)} \right){{\partial x} \over {\partial s}}\left( {s,t} \right) + {{\partial u} \over {\partial y}}\left( {x(s,t),y(s,t)} \right){{\partial y} \over {\partial s}}\left( {s,t} \right)  \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{\partial u} \over {\partial x}}\left( {x(s,t),y(s,t)} \right) + a{{\partial u} \over {\partial y}}\left( {x(s,t),y(s,t)} \right) = 0 \cr} \tag{3}$$
and hence we can say that
$$z(s,t) = h(t)\tag{4}$$
where $h(t)$ is some arbitrary function of $t$. According to $(1)$, we can say that
$$u\left( {x(s,t),y(s,t)} \right) = h(t)\tag{5}$$
Now, if we could write $h(t)$ in terms of $x(s,t)$ and $y(s,t)$ then we were able to derive $u(x,y)$ from $(5)$. Hence, by using $(2)$ we can conclude that
$$\eqalign{
  & ax(s,t) - y(s,t) = ap(t) - q(t)  \cr 
  & f\left( {ax(s,t) - y(s,t)} \right) = f\left( {ap(t) - q(t)} \right)  \cr 
  & h(t) = f\left( {ap(t) - q(t)} \right)  \cr 
  & h(t) = f\left( {ax(s,t) - y(s,t)} \right) \cr}\tag{6}$$
and then combining $(5)$ with $(6)$ we can get
$$\eqalign{
  & u\left( {x(s,t),y(s,t)} \right) = f\left( {ax(s,t) - y(s,t)} \right)  \cr 
  & u(x,y) = f(ax - y) \cr} \tag{7}$$
A: I have read just your solution of your question which always is the best chosen solution by yourself :) I think first argument (mentioned in question) is good and tells u is a "cylindrical surface". Actually it is a complete proof :) it tells solution is any cylindrical surface in a certain direction. Your solution make sense too but I cant understand what the hell is f?! Use this change of variables:
$$\frac{{\partial u}}{{\partial x}}(x,y) + a\frac{{\partial u}}{{\partial y}}(x,y) = 0\tag{1}\\x(s,t)=s \,\,,\,\,y(s,t)=as+t \,\, \\ \,\,(s(x,y)=x\,\,,\,\,t(x,y)=y-ax)\\u=u(s(x,y),t(x,y))\\ {{\partial u}\over{\partial x}}={{\partial u}\over{\partial s}}-a{{\partial u}\over{\partial t}}\,\,,\,\,{{\partial u}\over{\partial y}}={{\partial u}\over{\partial t}}\\ \Rightarrow (1) \,\, \text{holds}\,\,\text{iff} \,\,{{\partial u}\over{\partial s}}=0\\ \text{or}\,\,u(s,t)=h(t)\,\,\text{for some differentiable} \,\,h\\ \text{or}\,\,u(x,y)=h(y-ax)$$
