How to obtain the characteristics equation for the inviscid Burgers equation?

Question

I am trying to plot the characteristic lines for the inviscid Burgers equation which is $$u_t +uu_x=0.$$

From what I understand, with the initial condition $u(x,0)=f(x)$ and using the method of characteristics, the solution can be obtained implicitly as

$$u(x,t)=f(\xi)$$

where $\xi=x-ut$ and $\frac{dx}{dt}=f(\xi)$ describes the characteristics curve.

Then to plot the characteristics onto the $(x,t)$ plane, one could rearrange to get

$$t=\frac{1}{u}(x-\xi)=\frac{1}{f(\xi)}(x-\xi) $$

Assuming the above is correct, then consider the following initial condition:

\begin{equation} u(x,0)=f(x)= \left \{ \begin{aligned} &a^2-x^2, && \text{if}\ |x|)\leq a \\ &0, && \text{if}\ |x|)\geq a \\ \end{aligned} \right. \end{equation}

which gives the solution \begin{equation} u(x,t)=f(\xi)= \left \{ \begin{aligned} &a^2-\xi^2, && \text{if}\ |x|)\leq a \\ &0, && \text{if}\ |x|)\geq a \\ \end{aligned} \right. \end{equation}

where $\xi=x-ut=x-(a^2-\xi^2)t.$

The characteristics equation would then be $$t=\frac{1}{a^2-\xi^2}(x-\xi)$$

and substituting $\xi$

$$t=\frac{1}{a^2-(x-ut)^2}(ut) \tag{1}$$

which shouldn't be correct since we are expecting the characteristics to be straight lines.

Furthermore, I think I have started to enter into a "loop" with the substitutions. May I ask what have I done wrong?

An example of the characteristic lines onto the $x-t$ plane can be seen in Inviscid Burgers equation

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This example was obtained from Debnath, Example 5.2.1.

Answer 1

Indeed, we have $u=f(x-ut)$. It follows from the method of characteristics which gives the set of lines $$ x(t) = x_0 + f(x_0) t , \quad\quad x_0\in \Bbb R , $$ where the function $f = u(\cdot ,0)$ represents the initial data $x \mapsto (a^2-{x}^2)\,\Bbb I_{|x|\leq a}$ and $\Bbb I$ is the indicator function (see e.g. this post for the case $a=1$). Along these curves, $u(t) = f(x_0)$ is constant and equal to its initial value. The equation for $x(t)$ above can be rewritten as $x_0 = x - u t$, and thus, we get $u=f(x-ut)$. However, starting from this equation to recover the equation of the characteristic lines may not be a good idea.

To solve for $u(x,t)$, we must distinguish two cases here:

• The first case is $|x-ut|\geq a$ for which $u=0$. Therefore, if $|x|\geq a$, we have $u=0$.

• The second case is $|x-ut|\leq a$ for which $u=a^2 - (x-ut)^2$. Solving the quadratic equation, we find the expression of $ u $ for $|x|\leq a$.

Finally, $$ u(x,t) = \frac{2 t x - 1 + \sqrt{1 - 4 t x + 4 a^2 t^2} }{2 t^2} \Bbb I_{|x|\leq a} . $$ Note that characteristics intersect at the breaking time $t_b = 1/(2a)$, where the classical solution becomes multi-valued: a shock wave is generated.