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This is my first post here so I apologize if there is anything wrong with it. If so, please let me know and I can edit the post.

I just tried to solve the following question:

Problem 2.

  1. Give the explicit solution for all $t\leqslant 1/2$, to the following equation: $$ \left\lbrace\begin{aligned} &u_t+f(u)_x=0,\\ &u(x,0)=\begin{cases} 1, & x\leqslant 1\\ 1/2, &1<x\leqslant 3\\ 3/2, &x>3 \end{cases} \end{aligned}\right. $$ where $f(u)=4u(2-u)$.
  2. Represent the solution at times $t=0$, $t=1/4$ and $t=1/2$.
  3. Determine the solution beyond $t=1/2$?

What I have so far:

  • $f'(u) = 8 - 8u$, and for a solution $u(x,t), u$ is constant along the characteristic curves defined by $x'(t) = 8 - 8u, x(0) = a$ ($a$ real number), ie $u(x(t), t) = u(x(0), 0) = u(a, 0)$.
  • Then $u$ does not depend on $t$, so the characteristic curves are $x(t) = (8 - 8\cdot u(a,0)) \cdot t + a$.
  • ie $x(t)$ is given by $$\begin{cases} 0t + a & \text{if $a<=1$} \\ 4t + a & \text{if $1 <= a <= 3$} \\ -4t + a & \text{if $a>3$} \end{cases}$$

The question asks for the solution before $t = 1/2$, however I'm struggling with the following:

  1. Based on the above, it seems the characteristic curves already intersect at $t = 0$ (and $x = 3$), long before $t = 1/2$. Am I supposed to construct a shock wave solution using the Rankine-Hugoniot jump condition? I actually think there may be 2 shock wave solutions needed, because the characteristic curves do not cover a space around $(1, 0)$, it seems, since the curve $x = -4t + 3$ intersects the t-axis at $t=4/3$... However if I do this, wouldn't it be a solution for some range of t that isn't $t \leq 1/2$?
  2. On the other hand, is my reasoning above (before bullet #1) correct? This is, I think, analogous to what I have seen in class and in some other resources I looked at, however they all used the function $f(u) = u^{2} / 2$ and while I don't think this would change much, maybe I'm missing something vital.

Thank you in advance for any help.

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    $\begingroup$ What text are you working from? i.e. what tools do you have to solve with? $\endgroup$ – RobertTheTutor Feb 23 at 17:03
  • $\begingroup$ @RobertTheTutor we're working from Evans (AMS) and Renardy (Springer), both of which I have access to. $\endgroup$ – bosco98 Feb 23 at 17:39
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The situation is very similar to this post and related ones, where piecewise constant initial data is considered (i.e, a series of Riemann problems).

  • About $x=1$, we have the characteristic speeds $f'(1) = 0$ on the left side and $f'(1/2) = 4$ on the right side of the discontinuity. Since the right slope $4>0$ is larger than that on the left, characteristic lines separate: a rarefaction wave forms, with a linear fan solution.
  • About $x=3$, we have the characteristic speeds $f'(1/2) = 4$ on the left side and $f'(3/2) = -4$ on the right side of the discontinuity. Since the right slope $-4<4$ is smaller than that on the left, characteristic lines intersect: a shock wave forms, which speed $s=0$ follows from the Rankine-Hugoniot condition (it's a static shock).
  • These two waves interact at the time $t^*$ such that $1+4t^* = 3$, i.e $t^*=1/2$.

To link with the Burgers equation, one may consider the change of variable $v=8(1-u)$ such that $$ v_t + (\tfrac12 v^2)_x = 0\, , $$ $$ v(x,0) = \begin{cases} 0, & x\leq 1\\ 4, & 1<x\leq 3\\ -4, & x>3 \end{cases} $$

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  • $\begingroup$ Hi, thank you, I really appreciate the answer and have accepted it. We were told in class that if there is a region with no characteristic lines, the solution in that region should be a rarefaction wave. This agrees with what you wrote, I believe -- a rarefaction wave solution between x = 1 and x = 1 + 4t (for t <= 1/2). However, isn't there an area within that region where there are characteristic lines, namely 1/4 < t < 1/2 and 3 - 4t < x < 1 + 4t (corresponding to characteristics with slope -4 in the x-t plane)? I'm having an issue reconciling this. $\endgroup$ – bosco98 5 hours ago

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