# Solving conservation law IVP when characteristics intersect at $t=0$

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, &13 \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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• What text are you working from? i.e. what tools do you have to solve with? – RobertTheTutor Feb 23 at 17:03
• @RobertTheTutor we're working from Evans (AMS) and Renardy (Springer), both of which I have access to. – bosco98 Feb 23 at 17:39

• 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 $$v_t + (\tfrac12 v^2)_x = 0\, ,$$ $$v(x,0) = \begin{cases} 0, & x\leq 1\\ 4, & 13 \end{cases}$$