I have several hints for problems with shock waves.

• First, determine characteristics and check whether they intersect. Points of intersections are the origins of shock waves.
• In order to deal with a shock waves you have to use the Rankine-Hugoniot's condition. For this, you have to know what are the values of the solution on both sides of the discontinuity.
• R-H conditions will give you an ODE

$$ \frac{ds}{dt} = \frac{[q]}{[u]}, $$ where the square bracket is the change of respective quantity over the discontinuity. The initial condition for the ODE can be determined from the spacetime diagram, so that it is very important for you to draw everything.

• The above ODE can sometimes be solved explicitly and sometimes not. Do not be afraid of the latter.
• If after drawing the characteristics on a spacetime diagram you see an empty region - a void - then it is straightforward to fill it with the fan. The solution is called the rarefaction wave. The appropriate formula is given in the notes.
• Sometimes there are both shock and rarefaction waves. And they can intersect.
• Finally, consult the lecture notes as I have included there several examples. The book by Logan has also a lot of them.
• If you have any specific questions please fell free to email me. I can post solutions here.

The next topic is heat equation and it should be easier

This problem can pose some difficulties however, it is vert important to understand what is going on here. Here, you will see how to merge all different phenomena happening for one dimensional waves described by a scalar quasilinear PDE. It is completely normal that you may have difficulties with it. Be patient and work hard.

We start with 13a since it is probably most demanding. We want to solve

$$ u_t + u u_x = 0, $$ with condition $$ u(x,0) = \Phi(x) = \begin{cases} 1, & x < 0, \\ -1, & 0 < x < 1, \\ 0, & x > 1. \\ \end{cases} $$ Note that we are not taking care of using $\leq$ since, after all, we are dealing with weak (integral) solutions that are not very sensitive on changing the function on a measure zero set. Note that the wave speed $c(u) = u$, therefore, it depends on the solution $u$ in an increasing way. Hence, larger values of $u$ will travel faster than lower.

First, as always, we start with drawing characteristics since they will tell us everything. They are

$$ X(t) = \Phi(\xi) t + \xi = \begin{cases} t + \xi, & \xi < 0, \\ -t + \xi, & 0 < \xi < 1, \\ \xi, & \xi > 1,\\ \end{cases} $$ with $u(x,t) = \Phi(\xi(x,t))$ since $U' = 0$ (solution is constant on characteristics). Whence, the solution has a particularly simple form - it is piecewise constant apart from voids. The spacetime diagram looks like the following.

Note that we make the drawing only for some initial times since we do not know what is happening in the future. We immediately see that slanted characteristics intersect at $(0,0)$ and there we must fit a shock wave. Moreover, there is a region without any characteristics - void - in which we will introduce the rarefaction wave.

1st shock wave. The shock wave (discontinuous solution) is fitted from the Rankine-Hugoniot's condition

$$ \frac{ds}{dt} = \frac{[q]}{[u]}, \quad s(0) = 0, $$ where $[f] = f_+ - f_-$ is a change of some $f$ over the discontinuity. The initial condition $s(0) = 0$ is the point at which the characteristics intersect for the first time (see Figure). For our problems, the flux is always $q(u) = \int c(u) du$ and hence

$$ q(u) = \frac{1}{2} u^2, $$ which gives $$ \frac{ds}{dt} = \frac{1}{2} \frac{u_+^2-u_-^2}{u_+-u_-} = \frac{1}{2} (u_+ + u_-). $$

Since, in this shock, the value of $u$ on characteristics that intersect is equal to $u_+ = -1$ (right) and $u_- = 1$ (left) we have

$$ \frac{ds}{dt} = \frac{1}{2} (-1+1) = 0, $$

hence, $s$ is constant and by the initial condition we have

$$ s(t) = 0. $$

The shock wave is thus stationary.

Rarefaction wave. Fans are always fitted using the formula (3.15) from the lecture

$$ u(x,t) = c^{-1} \left( \frac{x-\xi_0}{t} \right), $$

where $\xi_0$ is the $x$-coordinate of the discontinuity. Here, $c^{-1}(u) = u$ and $\xi_0 = 1$ (see Figure!). Therefore,

$$ u(x,t) = \frac{x-1}{t}, \quad t+1 < x < 1. $$

And the rarefaction wave is fitted with continuously joining adjacent characteristics.

The new situation is depicted on the next diagram. The shock wave is depicted on red while the fan with green. Notice that now the rarefaction wave intersects with characteristics carrying $u=1$. We thus have to introduce another shock wave.

2nd shock wave. As always we start from the Rankine-Hugoniot's condition

$$ \frac{ds}{dt} = \frac{1}{2} (u_+ + u_-). $$ Notice that this equation has exactly the same form as before. However, the values of $u_\pm$ are now different. On the right we have the rarefaction wave, and hence $u_+ = (s-1)/t$, $u_- = 1$. Important when plugging into R-H condition we have to take the limit $x\rightarrow s(t)$, viz. the definition. Many students forget about this making a mistake. The initial condition is now $s(1) = 0$ since the first point of intersection is $(0,1)$. We have

$$ \frac{ds}{dt} = \frac{1}{2} \left( \frac{s-1}{t} +1\right), $$

which is a linear ODE. I will skip the details about solving it (you should know this very well!) and just write

$$ s(t) = 1 + t - 2\sqrt{t} = \left( \sqrt{t} - 1 \right)^2. $$

The next picture shows the present situation (sorry for the quality - I am doing it really fast).

Notice that eventually the second shock wave terminates at the end of rarefaction. Here, at $(1,4)$ the it meets with left and right characteristics. In order to find that point notice that $(\sqrt{4}-1)^2 = 1$. We thus have to fit the last wave.

3rd shock wave. This is simple: $u_+ = 0$ and $u_- = 1$, hence with $s(4) = 1$ we have

$$ \frac{ds}{dt} = \frac{1}{2} \left(0 + 1\right) = \frac{1}{2}, $$ therefore,

$$ s(t) = \frac{1}{2}t-1. $$ And the final picture is the following.

We can see that the rest can be continued indefinitely.

Lastly, we would like to see how does the solution behave for various times. This can be done very easily by collecting the values of $u$ for a fixed time $t>0$ (it is a horizontal line).

We can summarize this as follows:

• For $0\leq t<1$

$$ u(x,t) = \begin{cases} 1, & x <0, \\ -1, & 0 < x < 1-t, \\ \frac{x-1}{t}, & 1-t < x < 1, \\ 0, & x >1.\\ \end{cases} $$

• For $1\leq t < 4$

$$ u(x,t) = \begin{cases} 1, & x < (\sqrt{t}-1)^2, \\ \frac{x-1}{t}, & (\sqrt{t}-1)^2 < x < 1, \\ 0, & x >1.\\ \end{cases} $$

• For $t \geq 4$

$$ u(x,t) = \begin{cases} 1, & x < \frac{1}{2}t-1, \\ 0, & x > \frac{1}{2}t-1.\\ \end{cases} $$

You are encouraged to make a plot of this! If you understand this example completely then the rest should be much easier.

The next problems in 13 are somewhat easier. There are certain points that you should keep in mind.

• They exhibit a “triangle” in which characteristics meet at single point. They terminate there and, hence, do not intersect.
• What does intersect are pure characteristics or rarefactions.
• In Problem 13d we do not have a void but $u$ is not piecewise constant. Therefore, for the R-H condition we take the value of $u$ evaluated as a one-sided limit on the shock wave.
• Keep in mind that the flux can always be calculated from the speed, i.e. $q(u) = \int c(u) du$.