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Let's be this PDE:

$\begin{cases}u_t+f(u)u_x=0\\ u(x,0)=\varphi(x) \end{cases}$

and $f\in 1-1$.

  • I have these thoughts:

We can imagine $x'x$ having sticky particles. As we know $\frac{dx}{dt}=f(u)$. So $f(u)$ is the velocity of the particle at the position $x$ at the time $t$. Also, at the time characteristics intersect we have inelastic collision.

So, if we find the velocity of each particle at position $x$ at time $t$ (let's be $v(x,t)$, the weak solution of PDE will be $u=f^{-1}(v)$.

  1. Am I wrong?
  2. If I 'm not wrong, I would like to find a reference for that.

Thanks in advanced!

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  • $\begingroup$ Try method of charcteristics on wikipedia. $\endgroup$
    – username
    Commented Dec 22, 2022 at 18:51
  • $\begingroup$ I know how to use characteristics. I have a problem with that (mathoverflow.net/questions/436604/…) problem. So, I want to find another way to find weak solution. $\endgroup$ Commented Dec 22, 2022 at 19:51
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    $\begingroup$ The solution expressed on implicit form is $$u(x,t)=\varphi\left(x-\int_{\tau=0}^{\tau=t} f\Big(u(x,\tau)\Big)d\tau\right)$$ One cannot expect the explicit form $u(x,t)$ without knowing explicitly the functions $f$ and $\varphi$. $\endgroup$
    – JJacquelin
    Commented May 21, 2023 at 9:18

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