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:

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

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

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

  4. $u(x, 0) = u_0(x)$ 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?

  • $\begingroup$ The method of characteristics gives $u = u_0(x-ut)$, which is valid as long as characteristics do not intersect, i.e. for times $t$ smaller than the breaking time $t_b = -1/\inf u_0'$. $\endgroup$
    – EditPiAf
    Feb 19, 2019 at 14:18

2 Answers 2


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.


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. $u=F(x-ut)$

$u(x,0)=u_0(x)$ :


$\therefore u=u_0(x-ut)$


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