I am studying from old exams and there is a problem which is traffic flow with a ramp.
I have never seen this type of problem in class, so for the simple case, how would I solve
$$u_t + uu_x =\delta(x)\\
u(0,x) = 1$$
with method of characteristics.
I think we have
$$\frac{dt}{ds} = 1 \quad \text{ with } t(0) = 0$$
$$\frac{dx}{ds} = u(s) \quad \text{ with } x(0) = x_0$$
$$\frac{du}{ds} = \delta(x(s)) \quad \text{ with } u(0) = 1$$
From these, I can get
$$\frac{d^2x}{ds^2} = \delta(x(s))$$
and how would I continue?
Further attempt:
Now we pick some $x(0) = x_0 \neq 0$ as the initial condition for
$$\frac{d^2x}{ds^2} = \delta(x(s))$$
Now we assume that $x$ is continuous, so in a short time, $x(s) \neq 0$ for $0<s<S$, this means we have
$$\frac{du}{ds} = 0\quad\quad \frac{dx}{ds} = u(s)=u(0) = 1$$
so we get the solution $x(s) = 1\cdot s + x_0$. Next we see that when $x_0<0$, after time $s=|x_0|$, $x$ will become zero, this means there will be a jump in the value of $u$, and after the jump, $u$ will take the constant value $1+1 = 2$, $1$ from the initial value, $1$ from the dirac. Then the characteristics we have $x(s) = 2\cdot s – 2\cdot|x_0|$ for $s>|x_0|$.
Is this approach okay?
Best Answer
From your characteristic equations you can swap $s$ for $t$ (i.e. find $x(t)$ and $u(t)$ such that $u(x(t),t)=u(t)$) and you are left with having to solve $$ u'(t)=\delta(x(t)) \\x'(t)=u(t) $$ Where $x(t)\neq 0$, you get $u=\text{const}=u_0$, and so $x(t)=x_0+u_0t$. If $x_0 > 0$, these lines are diagonal, $u_0=1$ due to the initial condition and they continue for every like this unperturbed. If $x_0<0$, these lines are initially also diagonal, i.e. $u_0=1$ until they cross the $x=0$ line at $t=t_0=-x_0>0$. After the characteristics have crossed the $x=0$ line, they will continue with a constant speed again (because away from $x=0$, $u=\text{const}$ again), but this speed will have a distinct value, say $u_+ \neq 1$, because there will be a jump in $u$ at $x=0$ induced by the Dirac term. (Make a drawing, it will be clear. I think that was your intuition as well.)
To determine the size of the jump in $u$ at $x=0$, we integrate the equations above over $s\in(t_0-, t_0+)$, where $t_0\pm=t_0\pm \epsilon$ for $\epsilon > 0, \epsilon\to 0$, and $t_0=-x_0$ is the time at which the characteristic encounters the ramp ($x=0$). Assuming that $u(s)$ is bounded by $M>0$, integrating the second equation and taking the absolute value gives $|x(t_0+)-x(t_0-)|=|\int_{t_0-}^{t_0+}dt\,u(t)|\leq \int_{t_0-}^{t_0+}dt\,|u(t)|\leq 2 M \epsilon\to 0$ by Lebesgue theorem of dominated convergence, which shows that $x(t)$ is continuous at $t=t_0$ as expected.
Integrating the first equation should give the jump in $u(t)$ at $t=t_0$. The problem is that $x(t)$ is not smooth at $t=t_0$ (slope change), but let's go on for a bit just to see: for a function $x(t)$ with a single zero at $t=t_0$, one has $\delta(x(t)) = \frac{\delta(t-t_0)}{|x'(t_0)|}$, so that $$ \int_{t_0-}^{t_0+}dt\,u'(t) = u(t_0+)-u(t_0-) = \int_{t_0-}^{t_0+}dt\,\delta(x(t)) = \int_{t_0-}^{t_0+}dt\frac{\delta(t-t_0)}{|x'(t_0)|}=\frac{1}{|u(t_0)|} $$ The problem here is clear: the size of the jump in $u(t)$ at $t=t_0$ is ill-defined in this formula: $u(t_0)$ doesn't have a value. I think that the trick to get rid of this problem goes as follows. Assuming $u(t)>0$, rewrite $$ u'(t)=\delta(x(t))=\frac{\delta(t-t_0)}{|x'(t_0)|}=\frac{\delta(t-t_0)}{|x'(t)|} \implies u(t)u'(t) = \delta(t-t_0) \\\implies \frac{d}{dt}u^2(t) = 2\delta(t-t_0) $$ Now integrating this equation instead gives $u^2(t_0+)-u^2(t_0-)=2$ and so $u(t_0+)=u_+=\sqrt{3}$. The characteristics that cross the $x=0$ line will instantaneously change their slope from 1 to $u_+ = \sqrt{3}>1$, and will therefore collide with the characteristics that left with $x_0>0$ (shock formation). It's probably worth checking that this procedure will indeed give you a solution that solves the initial problem...