Given
$$u_t + uu_x = 0$$
and initial value
$$ u(x,t) =
\begin{cases}
0, & \text{if } x < 0\\
nx,& \text{if } 0 \leq x \leq \frac{1}{n}\\
1, & \text{if } x > \frac{1}{n}
\end{cases} $$
What is the solution to this? I came up with
$$ u(x,t) = f(x_0) = f(x – u(x_0,0)t)$$ where $f(x)$ is as defined above.
How would the characteristics look like? I have no idea how to proceed? Also in my solution, how do I find $u(x_0,0)$? Is that a correct term?
What would the solution look like (weak) if $n \to \infty $? I assume $u(x,t)$ would look like
$$ u(x,t) =
\begin{cases}
0, & \text{if } x < 0\\
1, & \text{if } x > 0
\end{cases}$$
Best Answer
The calculus of the general solution of the Burgers PDE $u_t+u_x u=0$ with the method of characteristics was already shown several times on the forum. Only the initial conditions are different from one case to another.
For example :
Inviscid Burgers' Equation and Implicit Solutions
Find the breaking time in IVP for classical Burgers equation
It isn't necessary to repeat again how one comes to the general solution : $$u=F(x-ut)$$ where $F$ is any differentiable function.
In the present case, the initial condition is : $$F(x)=u(x,0) = \begin{cases} 0, & \text{if } x < 0\\ nx,& \text{if } 0 \leq x \leq \frac{1}{n}\\ 1, & \text{if } x > \frac{1}{n} \end{cases}$$ This determines the function $F$ : $$F(X)=\begin{cases} 0, & \text{if } X < 0\\ nX,& \text{if } 0 \leq X \leq \frac{1}{n}\\ 1, & \text{if } X > \frac{1}{n} \end{cases}$$ We use a dummy variable $X$ in order to avoid the confusion between $X=x$ if $t=0$ and $X=(x-ut)$ if $t\neq 0$.
Thus, the solution according to the initial condition is : $$u(x,t) = \begin{cases} 0, & \text{if } (x-ut) < 0\\ n(x-ut),& \text{if } 0 \leq (x-ut) \leq \frac{1}{n}\\ 1, & \text{if } (x-ut) > \frac{1}{n} \end{cases}$$
This case is very simple because $u=n(x-ut) \quad\to\quad u=\frac{nx}{1+nt}$
$$\begin{cases} u=0, & \text{if } x < 0\\ u(x,t)=\frac{nx}{1+nt},& \text{if } 0 \leq \frac{x}{1+nt} \leq \frac{1}{n} \quad \text{ i.e. }\quad 0 \leq x \leq \frac{1}{n}+t\\ u=1, & \text{if } \frac{x}{1+nt} > \frac{1}{n} \quad \text{ i.e. }\quad x> \frac{1}{n}+t \end{cases}$$