Solution to the Inviscid Burgers' Equation
Definition
The following quasilinear partial differential equation is called the Burgers’ equation.
$$ \begin{cases} u_{t} + u u_{x} = 0 & , t>0 \\ u(t,x) = f(x) & , t=0 \end{cases} $$
Here, $t$ represents time, $x$ represents position, and $u(t,x)$ represents the waveform at position $x$ at time $t$. $f$ represents the initial condition, specifically the waveform at $t=0$.
Explanation
Burgers’ equation represents the case where the diffusion coefficient $\nu$ is $0$ in $\displaystyle u_{t} + u u_{x} = \nu u_{xx}$.
If $f ' (x)>0$, it means that the speed increases with increasing $x$, hence the characteristic curves form a rarefaction wave as shown above. The slope at $(t,x)$ represents the speed of the wave, implying that in a rarefaction, the speed either consistently increases or decreases. In such cases, the characteristic curves never intersect.
Drawing the waveform shows that as time progresses, the fast points move faster and the slow points move slower, resulting in an increasing difference.
On the other hand, in cases where the speed decreases with increasing $x$, characteristic curves intersect as shown.
In this case, as shown in the figure, points at the back overtake the points in front. This is particularly noticeable from the time $t=1$, where $u$ starts to have multiple values for a given $x$. A natural phenomenon that could be imagined to exemplify this is the formation of a wave; the water at the bottom slows down due to friction with the sand or gravel, while the top moves over it. This is called a blow-up, which mathematically means the function ceases to exist, and physically, it signifies the superposition of several states at the same time.
The case where characteristic curves form a rarefaction is too straightforward, hence the main interest lies elsewhere. Although finding the solution is not complicated, converting it into an explicit function form is challenging, and often impossible. In solving Burgers’ equation, determining the blow-up time and blow-up location is also a crucial problem. The solution to the inviscid Burgers’ equation, if it exists, is as follows.
Solution
Step 1. Let $\xi = x - tu$.
Step 2. Let $u = f(x - tu)$.
Then, since $\xi = x - t f( x - tu) = x - t f (\xi )$, the characteristic lines become $x = f(\xi) t + \xi$. If $f ' (x) > 0$ for all $x \in \mathbb{R}$, the characteristic lines form a rarefaction and do not meet. If there exists a $x$ such that $f ' (x) < 0$, the characteristic lines will meet at some point and blow-up at that point.
Step 3. Convert the initial condition to the form of $f(x-tu) = f(\xi)$.
If possible, convert $u = f(\xi) = f(x - tu)$ into an explicit function form.
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If the solution to the inviscid Burgers’ equation blows up within a finite time, the time and location are as follows:
Step 1. If $u$ has been found in the form of an explicit function, find $t = t_{\ast}$ where $u$ diverges.
That $t_{\ast}$ is the blow-up time.
Step 2. If it has not been possible to find the explicit function form, find $f ' (x)$.
$$ t_{\ast} : = \inf \left\{ \left. - {{1} \over {f ' (x) }} \ \right| \ f '(x) < 0 \right\} $$
That is the blow-up time.
Step 3. Find the point $x_{0}$ where the characteristic line $x = f(\xi) t + \xi$ intersects with the $x$ axis.
$x_{*} = x_{0} + f(x_{0}) t_{\ast}$, obtained by substituting $\xi=x_{0}$ and $t = t_{\ast}$ into $x = f(\xi) t + \xi$, is the blow-up location.
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Examples
1
- Find the blow-up time of $\displaystyle \begin{cases} u_{t} + u u_{x} = 0 & , t>0 \\ u(t,x) = \alpha (x - tu) + \beta & , t=0 \end{cases}$.
This is the type where $u$ is neatly represented as an explicit function.
Organizing $u = \alpha ( x - tu ) + \beta$ in terms of $u$ yields $\displaystyle u(t,x) = {{\alpha x + \beta} \over {1 + \alpha t}}$ and the blow-up time is $\displaystyle t_{\ast} = - {{1} \over {\alpha}} $.
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2
- Find the blow-up time of $\displaystyle \begin{cases} u_{t} + u u_{x} = 0 & , t>0 \\ u(t,x) = {{1} \over {2}} \pi - \tan^{-1} x & , t=0 \end{cases}$.
This is the type where representing $u$ as an explicit function is cumbersome.
$$ f ' (x) = \left( {{1} \over {2}} \pi - \tan^{-1} x \right) = - {{1} \over {1 + x^2}} $$
and for $x \in \mathbb{R}$, $f ' (x) < 0$ holds. Therefore,
$$ t_{\ast} = \inf \left\{ \left. - {{1} \over {f ' (x) }} \ \right| \ f '(x) < 0 \right\} = \inf \left\{ \left. (1+ x^2) \ \right| \ f '(x) < 0 \right\} = 1 $$
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