Solving Burgers' equations with the method of characteristics I am difficulty with this assignment:
This project  deals with the partial differential  equation  (PDE)
$$
u_t(x, t) + u(x, t)u_x(x, t) = 0, \quad x \in \mathbb{R} \text{ and } t \geq 0 \tag{1}
$$
together  with the initial  condition
$$
u(x, 0) = u_0(x), \quad x \in \mathbb{R} \tag{2}
$$
where $u_0$  is a differentiable  function on $\mathbb{R}$ with  continuous  derivative  $u_0^\prime$.  The  PDE  (1) together
with the initial  condition  (2) is called an initial  value problem.
The initial value problem  (1–2) is used, among other  applications, as a rudimentary model for the motion  of a fluid in a thin  tube.   The  unknown  function  u gives the  velocity  of the  fluid.  That is, $u(x, t)$ is the  velocity  of the  fluid at  the  point $x$ in the  tube  at  time  $t$.  The  initial  value  $u_0$ prescribes  the velocity of the fluid at time $t = 0$.
This project  deals with issues related  to the solutions  of (1–2).
Definition 1.  A solution  of the initial  value problem (1–2)  is a function
$$
  u \colon \mathbb{R} \times \left[0,\infty \right) \to \mathbb{R}
$$
that satisfies the following:


*

*$u$ is continuous  on $\mathbb{R} \times \left[0,\infty \right)$.

*$u$ is differentiable  with continuous  first order partial derivatives  on $\mathbb{R} \times \left[0,\infty \right)$. 

*$u$ satisfies the PDE (1)  at every point $(x, t) \in \mathbb{R} \times \left[0,\infty \right)$.

*$u(x, 0) = u_0(x)$ for every $x \in \mathbb{R}$, that is, $u$ satisfies the initial  condition  (2).
Concerning  the solutions  of (1–2), there  are two basic questions. (1)  Does a solution  exist?
(2)  If a solution  exists, what  does it look like?
The aim of this project is to address these two questions.  The analysis will be based on the method of characteristics.
How do I solve this problem?
 A: There are infinitely many solutions to this problem.  But your supplemental conditions, particularly 1. and 2., eliminate many of them.  As I understand it, you want $u$ to remain continuous and differentiable for all $t$.  (If I have misunderstood, you can stop reading now.)  Only very specific initial conditions will produce this behavior.  Namely, given any $x_0$, and $x_1$ such that $x_1-x_0>0$ and $u(x_0,t)$, it must be the case that
$u(x_1,0)\ge u(x_0,0)$
I'm not going to be able to give you a formal solution via characteristics, but you can see this in a qualitative way.  This equation simply "transports" $u(x,0)$ to some other $x$ at time $t$, and the rate at which it transports $u$ is simply $u$.  So what happens if the above condition is violated, that is, what if we have $u(x_0,0)$, but at some $x<x_0$, $x_{-1}$, say, we have $u(x_{-1},0)>u(x_0,0)$?  Then, eventually, $u(x_{-1},0)$ will "catch up with" $u(x_0,0)$.  When that happens, our function will be multivalued -- both discontinuous and undifferentiable.
A: Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
$\dfrac{dx}{ds}=u=u_0$ , letting $x(0)=f(u_0)$ , we have $x=u_0s+f(u_0)=ut+f(u)$ , i.e. $u=F(x-ut)$
$u(x,0)=u_0(x)$ :
$F(x)=u_0(x)$
$\therefore u=u_0(x-ut)$
