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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...2012-10-17

1 Answers 1

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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 $x$.

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    Can you show how it incorporates u(x,t)?2012-10-21
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    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.2012-10-21
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    but the whole problem depends on pi, that is, a hint of the problem said to graph: cos^-1(u) vs. pi(x-u*t) as functions of u2012-10-21
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    The other part i am not getting "show that u ceases to exist". And can you further explain how u is a solution to u = cos(pi(x-ut)). I am having a hard time reading it. Thanks!2012-10-21
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    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 $u = \cos x$.2012-10-21
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    please see the edit above. thanks2012-10-22
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    OK, so now it's $u(x,t) = \cos(\pi b)$ where $\cos(\pi b) t + b = x$, and thus $u(x,t) = \cos(\pi (x - \cos(\pi b) t) = \cos(\pi (x - u(x,t) t)$, where we take $c = \cos(\pi b)$ in the equation $u(ct+b,t) - c = 0$.2012-10-22
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    For example, with $b=1/2$ we get $u(x,t)=0$ on the line $x=1/2$. With some other $b$, say $b=b_1$ (for which $\cos(\pi b_1) \ne 0$) we get $u(x,t) = \cos(\pi b_1)$ on the line $\cos(\pi b_1) t + b_1 = x$. This intersects $x = 1/2$ at $t = (1/2 - b_1)/\cos(\pi b_1)$. That means that $u(x,t)$ can't be well-defined by that point, because it can't be both $0$ and $\cos(\pi b_1)$ at the same point.2012-10-22
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    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