0
$\begingroup$

How can we show that a smooth solution of the problem $\begin{cases} u_t +uu_x = 0 \\ u(x, 0) = \cos(\pi x) \end{cases}$ satisfies the equation $u = \cos \pi(x − ut)$ and that $u$ ceases to exist (as a single-valued continuous function) when $t = 1/\pi$? The only thing I can think of is maybe graphically doing it, but I don't see how.

Can someonpe please edit Robert's answer below for the situtation at hand? I made a mistake before typing it. Thanks

  • 0
    That's not quite right...2012-10-17

1 Answers 1

0

For any real constants $b$ and $c$, on the line $x = c t + b$ we have $\dfrac{d}{dt} (u(ct+b,t) - c)= c u_x(ct+b,t) + u_t(ct+b,t) = (c - u(ct+b,t)) u_x(ct+b,t)$. If $u$ is smooth, this implies that if $u(ct+b,t) - c = 0$ somewhere on that line then it is $0$ everywhere on the line. In particular take $t=0$, $c = u(b,0) = \cos(b)$, to conclude that $u(\cos(b) t+b,t) = \cos(b)$ for all $t$, i.e. $u(x,t) = \cos(b)$ where $\cos(b) t + b = x$. But as soon as $t > 1$ the function $f(b) = \cos(b) t + b$ is not one-to-one, since $f'(b) = - t \sin(b) + 1$ changes sign, so there will be two conflicting values for some $x$.

  • 0
    By l'Hôpital's rule, $\lim_{b \to 1/2} \frac{1/2 - b}{\cos(\pi b)} = \frac{1}{\pi}$ so the solution can't be defined past $t = 1/\pi$.2012-10-22