=== Method of characteristics ===
First, it is crucial for you to understand the method of characteristics for solving simple quasilinear equations without shock waves. Let us focus on **Problem 7** from the 1st list, i.e. semi-linear equations. Suppose we would like to solve something similar, say 

$$
   \begin{cases}
      u_t + x t \, u_x = t(1+u^2), & t>0, \quad x\in\mathbb{R} \\
      u(x,0)=\frac{x^2}{1+x^2}, & x\in\mathbb{R}.
   \end{cases}
$$ 
So that $c(x,t) = x t$, that is the speed of wave propagation increases with $x$ and $t$. Note that it is negative for $x<0$. Moreover, the source term $t(1+u^2)$ increases both with time and the concentration. In the method of characteristics you have to form a system of ODEs (and understand how it arises)

$$
   \begin{cases}
      X' = t X, & X(0) = \xi, \\
      U' = t(1+U^2), & U(0) \stackrel{df}{=} u(X(0),0) = u(\xi, 0) = \frac{\xi^2}{1+\xi^2}. \\ 
   \end{cases}
$$

Remember that $\xi$ denotes the point at which the characteristic intersects the $t=0$ axis. It //enumerates// every member from the characteristic family. These equations usually have to be solved simultaneously however, in this simple example we can start with finding $X$. Since this is a simple ODE we immediately find out that

$$
   X(t) = \xi e^{\frac{1}{2}t^2},
$$

and from here it is useful to find $\xi$, i.e.

$$
   \xi(x,t) = x e^{-\frac{1}{2}t^2}.
$$

Note that $\xi(x,t)$ is a $x$-coordinate of a point of intersection with $t = 0$ of a unique characteristic passing through $(x,t)$ (this is very important). A simple drawing of characteristics looks like this:

{{ :characteristics7.png?nolink |}}

Knowing that we can proceed to solving for $U$, that is

$$
   \frac{U'}{1+U^2} = t \quad \rightarrow \quad U(t) = \tan \left( \frac{1}{2} t^2 + C\right),
$$

where the integration constant can be determined from the initial condition, that is

$$
   \frac{\xi^2}{1+\xi^2} = U(0) = \tan C \quad \rightarrow \quad C = \arctan \frac{\xi^2}{1+\xi^2}.
$$

Now, we know everything. We can combing the two above solutions with the definition of $U$ and write

$$
   u(x,t) = U(t, \xi(x,t)) = \tan \left( \frac{1}{2} t^2 + \arctan \frac{x^2 e^{-t^2}}{1+x^2 e^{-t^2}}\right),
$$

where in the first equality we have clearly indicated that $U$ depends also on $\xi$. This is a highly artificial example, however, it outlines all the necessary steps used in solving semi-linear equations. \\

Now, in **Problem 9** you proceed similarly. However, there is an important difference: you solve your problem on a quarter-plane $x>0$ and $t>0$. So that, you have two sets on which the conditions are given: $t = 0$ (initial) and $x = 0$ (boundary). The most important point is to remember that if you know characteristics, you know everything. Now, since your data is given on a different axes, you have to **separate** the characteristics according which axis it **intersects** first. This is clear from the following picture

{{ ::characteristics9.png?nolink |}}

We can see that the thick line is a characteristic that separates two families of them: one intersecting the $t=0$ while the other $x = 0$. Solving the characteristic equation for $X$ will help you to determine which characteristics belong to which family. Remember that for each family you have **different** integration constants (in equation for $X$). The red family is the one you know very well - it is an initial value problem and you solve it as usual with determining $\xi$. However, the green family does not intersect $t = 0$ but rather $x = 0$. And it is very useful to determine the integration constant for which $ x = 0 $ when $t = \tau$ (since there the condition is given). Call it $\tau$. For the simple example above, the characteristics have the linear form

$$
   X(t) = x-\tau.
$$

Compare this with the initial value problem for which you had $ t = 0 $ for $x = \xi$. Finally, the whole Problem 9 can be solved by separating the characteristics as above and integrating the equation for $U$ using **different** condition for each **family**.\\

In **Problem 10** we have quasilinear equations, however, the overall method stays the same. For example, the first problem gives us the following characteristic equations

$$
   \begin{cases}
      X' = \ln U, & X(0) = \xi, \\
      U' = 0, & U(0) = e^{\xi}.
   \end{cases}
$$

Note that now we cannot first solve for $X$ since we do not know $U$. But from the second equation we obtain $U(t) = \text{const.} = \exp(\xi)$. Then, $X' = \ln \exp(\xi) = \xi$ and hence

$$
   X(t) = (1+t)\xi,
$$

and from here $\xi(x,t) = x/(1+t)$ whence, $u(x,t) = U(t,\xi(x,t)) = \exp(x/(1+t))$. \\

Finally, **Problem 11** combines everything of the above into one equation. You should not have any problems in solving it. Remember, it is very useful to **draw** everything.