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Consisder Inviscid Burgers' equation $$u_t+uu_x=0$$ Assume we are given a smooth solution $u:\mathbb R\times [0,T]\to\mathbb R$ that is periodic in $x$. meaning that for some $K>0$ we have $u(x+K,t)=u(x,t)$ for all $x\in\mathbb R$, $t\in[0,T]$. Then prove that $$\max_{x\in\mathbb R} u(x,0)-\min_{x\in\mathbb R} u(x,0)\leq \frac{K}{T}$$

Attempt. To be honest I really don't know how to prove this. Let $x$ be the point where the maximum is attained and $y$ be a point where the minimum is attained. Then we know that we can choose $x$ and $y$ such that $0\leq x-y\leq L$. Note that $$u(x,0)-u(y,0)=\int^x_y u_x(z,0)\,dz\stackrel{?}{=}\int^x_y -\frac{u_t(z,0)}{u(z,0)}\,dz $$ I wrote the last integrand as $-\frac d {dt}\log(u(z,t))|_{t=0}$ if it exists at least. But then maybe try to apply Jensen, but that doesn't give much either. I also think it can be related to the shockwave time, but I'm not sure how.

I appreciate any help/hints. Thanks in advance. $$$$

EditPiAf
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Shashi
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1 Answers1

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This property is deduced from the intersection of characteristic curves at some $T^*\ge T$, see this post. Consider two characteristic lines passing through $(x_m, 0)$ and $(x_M, 0)$ such that $x_M\le x_m$, which correspond to the min and the max of $u(\cdot,0)$, respectively. In other words, those abscissas are chosen such that $u(x_m, 0) = m$ and $u(x_M, 0) = M$ where $m = \min_x u(x, 0)$ and $M = \max_x u(x, 0)$. The curves intersect at $$ T^* = -\frac{x_M - x_m}{M- m} \ge T \, , $$ where $T$ denotes the breaking time. If $x_m$ and $x_M$ belong to the same period, we have $0\le x_m-x_M \leq K$. Thus, we find $M-m \leq K/T$, which ends the proof.

EditPiAf
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  • So the solution here is the same one that we can obtain with the method of characteristics? Wow I didn't realize that, I was thinking that method of characteristics provides a solution but that it doesn't necessarily provide all of them. – Shashi May 03 '20 at 02:04
  • @Shashi Yes, it's from the method of characteristics. The general solution $u = F(x-ut)$ in implicit form provides local existence and uniqueness as long as the solution is smooth. – EditPiAf May 03 '20 at 09:43
  • Thank you very much I get it now!!! – Shashi May 03 '20 at 13:11