6
$\begingroup$

I have shown that a smooth solution of the problem $u_t+uu_x=0$ with $u(x,0)=\cos{(\pi x)}$ must satisfy the equation $u=\cos{[\pi (x-ut)]}$. Now I want to show that $u$ ceases to exist (as a single-valued continuous function) when $t=\frac{1}{\pi}$.

When $t=\frac{1}{\pi}$, then we get that $u=\cos{(\pi x-u)}$.

With single-valued function is it meant that the function is 1-1 ?

If so, then we have that $\cos{(2 \pi-u)}=\cos{(4 \pi -u)}$, i.e. for two different values of $x$, we get the same $u$, and so for $t=\frac{1}{\pi}$, $u$ is not 1-1.

But if this is meant, how are we sure that for $t \neq \frac{1}{\pi}$ the function is single-valued?

$\endgroup$
8
$\begingroup$

The solution $u(x,t)$ is implicitly defined by the equation $F(x,t,u) = u - \cos\Big(\pi(x - ut)\Big) = 0$. The Implicit Function Theorem asserts that $F(x,t,u) = 0$ defines $u$ as a function of $x,t$ if $\dfrac{\partial F}{\partial u}\neq 0$, otherwise we expect characteristics to intersect (well, along a shock curve), i.e. $u(x,t)$ becomes multi-valued.

The partial derivative of $F$ with respect to $u$ is $$ \frac{\partial F}{\partial u}\colon = \partial_u F = 1 - \pi t\sin\Big(\pi(x-ut)\Big) $$ and along the characteristic $x = \cos(\pi x_0)t + x_0 = u(x_0,0)t + x_0$ for an arbitrary but fixed $x_0\in\mathbb{R}$ we have that $$ \partial_u F = 1 - \pi t\sin(\pi x_0). $$ It follows that $\partial_u F = 0$ whenever $$ t = \frac{1}{\pi\sin(\pi x_0)}. \tag{1}$$

Denote by $t^*$ the shortest time where $\partial_u F = 0$. Clearly, $\partial_u F\neq 0$ for all $t\ge 0$ if $$ \pi x_0\in [n\pi, (n+1)\pi], n=\pm 1,\pm 3, \pm 5, \dots, $$ so suppose not. It follows from $(1)$ that $$ t^* = \min_{x_0\in\mathbb{R}}\frac{1}{\pi\sin(\pi x_0)} = \frac{1}{\pi}. $$

$\endgroup$
  • $\begingroup$ First of all, why do we consider $\frac{\partial{F}}{\partial{u}}$ along the characteristic line? Secondly, how do we deduce that $\frac{\partial{F}}{\partial{u}} \neq 0$ for all $t \geq 0$ if $\pi x_0 \in [n \pi, (n+1) \pi], n =\pm 1, \pm 3, \pm 5, \dots$ ? And also, how do we use this fact in order to find the shortest time? $\endgroup$ – Evinda May 28 '18 at 23:44
  • $\begingroup$ 1) We exploit the fact that the solution is constant along the characteristics and so we may rewrite $x-ut$ as $x_0$ where $x_0$ is some initial point at $t=0$. 2) $\partial_u F\neq 0$ (actually, $\ge 1$) if the term $-\pi t\sin(\pi x_0)\ge 0$; since $t\ge 0$, it is enough to requiring $\sin(\pi x_0)\le 0$ which gives the desired ranges for $\pi x_0$. $\endgroup$ – Chee Han May 29 '18 at 5:00
  • $\begingroup$ 3) We look for shortest time because we are considering the worst case scenario, that is, when is the solution starting to become multi-valued? This follows from implicit function theorem, because it says that provided $\partial_u F\neq 0$, then locally we can write $u$ as a function of $x,t$. In particular, for each fixed time $T>0$, $u$ is a function of $x$. Of course, this will fail for $t\ge t^*$ for some critical time $t^*$. $\endgroup$ – Chee Han May 29 '18 at 5:13
  • $\begingroup$ I have understood now how we get the ranges for $\pi x_0$, but I haven't understood how we get the shortest time when the solution is starting to become multi-valued. How do we use the implicit function theorem to deduce that the shortest time is $\frac{1}{\pi}$? Could you explain it further to me? @CheeHan $\endgroup$ – Evinda May 29 '18 at 7:15
  • $\begingroup$ For a fixed time $T>0$, the solution $u(x,T)$ fails to be single-valued when $u(x,T) = u(x)$ is not a function of $x$. But we do not have an explicit expression for the solution $u$ since it is defined implicitly by the equation $F(x,t,u) = 0$. Implicit function theorem says that $u$ is multi-valued if $\partial_u F = 0$, which is the case when $t = \frac{1}{\pi\sin(\pi x_0)}$ for some initial point $x_0$ and clearly the time depends on $x_0$, i.e. for each initial point $x_0$ there is a corresponding critical time, say $t_c$. $\endgroup$ – Chee Han May 29 '18 at 7:27
2
$\begingroup$

Based on the proof in this post, we have the expression for the breaking time $$ t_b = \frac{-1}{\min \partial_x u(x,0)} = \frac{1}{\pi} \, . $$ At $t=t_b$, characteristics intersect and a shock wave occurs. This is illustrated in the figure below, where the characteristic lines are displayed in the $x$-$t$ plane:

characteristics

To see that the solution deduced from the method of characteristics becomes multi-valued for $t>t_b$, one may be interested by the discussion here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.