# PDE on smoothness conditions and existence values

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

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That's not quite right... – Robert Israel Oct 17 '12 at 1:54

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 $x$.
How what incorporates $u(x,t)$? $u(x,t) = \cos(b)$ where $\cos(b)t + b = x$. This is also $u(x,t) = \cos(x - \cos(b) t) = \cos(x - u(x,t) t)$, if you prefer to write it that way. Note that there is no $\pi$ involved. – Robert Israel Oct 21 '12 at 5:47
Either you stated the problem wrong, or the solution you were given is wrong. Take $t=0$ in $u = \cos \pi(x - ut)$ and you get $u = \cos \pi x$, not $u = \cos x$. – Robert Israel Oct 21 '12 at 16:22