Given the equation
$$u_t + uu_x = 0, x \in \mathbb{R}, t > 0$$
with the initial conditions:
$$u(x,0)=u_0(x)=
\begin{cases}
1 & \text{if} \ x < -1 \\
x^2 & \text{if} \ -1 < x < 0 \\
0 & \text{if} \ x > 0
\end{cases}$$
Using the method of characteristics I obtain the following solutions in a parametrized form:
$$(x,t,u) = (tu_0(s)+s,t,u_0(s))$$
and hence
$$u(x,t) =
\begin{cases}
1 & \text{if} \ x – t < 1 \\
\frac{x}{t} & \ \text{if} \ 0 < \frac{x}{t} < 1 \\
0 & \text{if} \ x > 0
\end{cases}$$
Thus the shocks occur "everywhere" and I do not know how to use Rankine-Hogoniot condition to handle the shocks. Even if I just formally use Rankine condition, does it lose many informations of $u(x,t)$ in the region $0 < x/t < 1$?
Burgers equations with piecewise initial data
characteristicsinitial-value-problemspartial differential equations
Related Solutions
The variable $u$ is constant along the characteristic curves, which satisfy \begin{aligned} x'(t) & = u(x(t),t) \, , \\ & = u(x(0),0) \, . \end{aligned} Thus, the latter are straight lines in the $x$-$t$ plane, determined by the initial data. Here, the initial data is piecewise constant, i.e. we solve Riemann problems. As displayed in the figure below,
- characteristics separate in the vicinity of $x=0$, and a rarefaction wave occurs;
- characteristics cross in the vicinity of $x=1$, and a shock-wave occurs.
The rarefaction wave is a continuous self-similar solution, deduced from the self-similarity Ansatz $u (x,t)=v (\xi)$ with $\xi = x/t$. Indeed, $$ \partial_t u(x,t) + u(x,t)\, \partial_x u(x,t) = \left(v(\xi) - \xi\right) \frac{v'(\xi)}{t} \, , $$ and thus, $v(\xi) = \xi$ or $u(x,t) = x/t$. The shock speed $s$ is given by the Rankine-Hugoniot jump condition: $$ s = (1+0)/2\, . $$ As long as the rarefaction and the shock don't interact, the solution is therefore $$ u(x,t) = \left\lbrace \begin{aligned} & 0 &&\text{if }\; x\leq 0 \, , \\ & x/t &&\text{if }\; 0\leq x\leq t \, , \\ & 1 &&\text{if }\; t \leq x < 1+ t/2 \, ,\\ & 0 &&\text{if }\; 1+t/2 < x \, , \end{aligned} \right. $$ valid for times $t<t^*$ such that $t^* = 1 + t^*/2 = 2$. At the time $t^*$, both waves interact. The new shock speed is determined from the Rankine-Hugoniot condition $$ x'(t) = (x(t)/t+0)/2 \, , $$ with initial shock speed $x'(t^*) = s$. Hence, the solution for $t\geq t^*$ is $$ u(x,t) = \left\lbrace \begin{aligned} & 0 &&\text{if }\; x\leq 0 \, , \\ & x/t &&\text{if }\; 0\leq x< \sqrt{2t} \, , \\ & 0 &&\text{if }\; \sqrt{2t} < x \, . \end{aligned}\right. $$
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 $u=n(x-ut) \quad\to\quad u=\frac{nx}{1+nt}$
$$\begin{cases} u=0, & \text{if } x < 0\\ u(x,t)=\frac{nx}{1+nt},& \text{if } 0 \leq \frac{x}{1+nt} \leq \frac{1}{n} \quad \text{ i.e. }\quad 0 \leq x \leq \frac{1}{n}+t\\ u=1, & \text{if } \frac{x}{1+nt} > \frac{1}{n} \quad \text{ i.e. }\quad x> \frac{1}{n}+t \end{cases}$$
Best Answer
The shock wave starts at the minimal $t$ where for a given point $(x,t)$ there are two solutions to $x=x_0+tu_0(x_0)$, which happens on the line $x=-1+t$ when it is equal for infinitesimal $h$ to $-1+h+t(-1+h)^2=-1+t+h(1-2t+h)$, thus at $t=\frac12$, $x=-\frac12$.
The shock wave follows a curve $x=v(t)$, where the characteristic lines with $v(t)=x_{01}+t$, $x_{01}<-1$ and $v(t)=x_{02}+tx_{02}^2$, $x_{02}\in(-1,0)$ meet, that is,
The Rankine-Hogoniot equation for the speed of the shock now gives \begin{align} v'(t)&=\frac12(1+x_{02}(t)^2)\\ &=(1+2tx_{02})x_{02}'(t)+x_{02}(t)^2 \end{align} Solve this (numerically) for the inverse function $t_s(x)$ with $$ t_s'(x)=1/x_{02}'(t_s(x))=\frac{2(1+2t_s(x)x)}{1-x^2} $$ with initial conditions $t_s(-1)=0.5$ over the interval $[-1,0]$. This now is a linear DE with integrating factor $(1-x^2)^{2}$. Carrying out the integration ends with \begin{align} (1-x^2)^2t_s(x)&=\int 2(1-x^2)\,dx=2x-\frac23x^3+C\\ 0&=-2+\frac23+C\implies C=\frac43\\ t_s(x) &=\frac23\frac{3(x+1)-(x+1)(x^2-x+1)}{(1-x^2)^2} \\ &= \frac{2(2-x)}{3 (1-x)^2} \end{align}
At time $T=t_s(0)=\frac43$ the middle phase has collapsed and the outer phases meet in a shock wave with speed $0.5$.
Constructing the solution picture from the characteristic curves ending at the shock curves gives the plot