Form of a solution of an equation similiar to the Burgers' Equation I'm fairly new to the whole field of PDE and I tried to work on some examples in order to become more familiar with the method of characteristics.
I've been working on this task:
You have the PDE
$$\partial_t u(x,t)+a(u(x,t)) \partial_x u(x,t)=0~~ \forall x \in \mathbb{R}, t \in [0, T)$$
$$u(x,0)=u_0(x) ~~\forall x \in \mathbb{R}$$
where $a, u_0 \in C^1_b(\mathbb{R})$.
Let now $u \in C^1(\mathbb{R} \times [0,T])$ be a classical solution of the PDE above. Prove that this solution fulfills
$$ u(x,t)=u_0(x-t\cdot a(u(x,t))).$$
My approach so far was to first come up with the characteristic equations:
$$\frac{d}{ds} \left(\begin{array}{c} x(s)\\t(s) \\ y(s) \end{array}\right) = \left(\begin{array}{c} a(y(s)\\1 \\ 0 \end{array}\right)   $$ with the start values 
$$ \left(\begin{array}{c} x(0)\\t(0) \\ y(0) \end{array}\right)= \left(\begin{array}{c} x_0 \\0 \\ u_0(x_0) \end{array}\right).$$
Solving these equations gave me:
$$\left(\begin{array}{c} x(s) \\t(s) \\ y(s) \end{array}\right)= \left(\begin{array}{c} s \cdot a(y(s)) + x_0 \\s \\ u_0(x_0)\end{array}\right)$$
where I'm however not sure about the $x(s)$ part. I assume that there's something missing as my $y$ depends on $s$ which has to be considered when differentiating, but I don't know how to do this. 
Using the substitution $t=s$ and $x_0=x-t \cdot (u_0(x_0))$, I get the solution $$u(x,t)=u_0(x-t \cdot a(u_0(x_0))),$$
which has at least some resemblance to the actual solution...
I would be very grateful if someone could point out my mistakes and maybe even correct them. Thank you very much!
 A: The purpose of the method of characteristics is to reduce the difficult problem of solving your PDE on the domain $\mathbb R \times [0, T)$ to the relatively easy of solving ODEs along curves within your domain. We want  to chose these curves cleverly, so that these ODEs are nice. These cleverly-chosen curves are called the characteristic curves of the PDE.
So suppose that $u(x,t)$ is a solution to your PDE. And suppose that $$s \mapsto (t(s), x(s))$$
is a parametrisation of a curve inside our domain. Then along this curve, we can also think of the independent variable $u$ as a function,
$$ s \mapsto u(s) := u(x(s), t(s)).$$
We're going to focus on a particular, cleverly-chosen family of curves: namely, the family of curves obeying the following set of ODEs:
$$ \frac{dt(s)}{ds} = 1, \ \ \ \ \ \frac{dx(s)}{ds} = a(u(s)).$$
It is a simple exercise, involving nothing more than the chain rule, to verify that the PDE implies that $u(s)$ must itself satisfy the ODE
$$ \frac{du(s)}{ds} = 0$$
along any curve within our family.
Solving these ODEs, we discover that the functions $t(s)$, $x(s)$ and $u(s) := u(t(s),x(s))$ for any curve within our family are given by
$$ t(s) = s+ c_0, \ \ \ \ \ x(s) = a(c_2)s+c_1, \ \ \ \ \ u(s) = c_2,$$
for some constants $c_0, c_1, c_2$, specific to the particular curve that we're looking at.
However, these constants are not all independent. First, we may set $c_0 = 0,$ by redefining the parameter $s$ by a constant shift, $s_{\rm new} = s_{\rm old} + c_0$.
We can also determine the value of $c_2$ in terms of $c_1$, using the initial condition, $u(0,x) = u_0(x)$. Indeed, at $s = 0$, we have $$t = 0,  \ \ \ x =  c_1, \ \ \ u = c_2,$$
and applying the initial condition at $t = 0$ and $x = c_1$, we learn that
$$ c_2 =  u_0(c_1) .$$
So we may summarise as follows. We have a family of "special" curves, whose members are labelled by different choices of $c_1$: each choice of $c_1$ gives a different curve within our family. The points along a given curve within our family are parametrised by a parameter $s$, as follows:
$$ s \mapsto (t(s), x(s)) = (s, a(u_0(c_1))s +c_1),$$
And along this curve, the values of $u$ are given by
$$ u(t(s), x(s)) = u_0(c_1).$$

Now let's make use of this information to solve the PDE. We'll pick an arbitrary point $(t_1, x_1)$ in $\mathbb R \times [0,T)$, and we'll try to determine the value of $u$ at this point.
First, we should ask ourselves: which of the curves within our family contains the point $(t_1, x_1)$? Clearly, from our expressions above, the point $(t_1, x_1)$ lies on the curve with a value of  $c_1$ that obeys
$$ c_1 = x_1 - a(u_0(c_1)) t_1$$
and the value of the paramter $s$ at our point $(t_1, x_1)$ is
$$ s = t_1.$$
The value of $u$ at our point $(t_1, x_1)$ is then given by
$$ u(t_1, x_1) = u_0(c_1), $$
for this value of $c_1$.
It only remains to eliminate $c_1$, to leave an expression for $u(t_1, x_1)$ in terms of $t_1$ and $x_1$ alone. Plugging $u(t_1, x_1) = u_0(c_1)$ into the equation $ c_1 = x_1 - a(u_0(c_1)) t_1$, we learn that $$c_1 = x_1 - a(u(t_1, x_1))t_1,$$ and substituting this back into $u(t_1, x_1) = u_0(c_1)$, we find that
$$ u(t_1, x_1) = u_0 ( x_1 - a(u(t_1, x_1)) t_1),$$
which is the expression we are looking for.
Finally, since our choice of $t_1 $ and $x_1$ was arbitrary, the solution to our PDE is
$$ u(t,x) = u_0(x - a(u(t,x))t).$$

In terms of manipulating algebra, my explanation is identical to your solution (except that you could have finished off by substituting $u_0(x) \mapsto u(x,t)$ at the very end). All I've really done is flesh out some of the logic!
