Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
That's not quite right... –  Robert Israel Oct 17 '12 at 1:54
add comment

1 Answer

up vote 0 down vote accepted
+50

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 $x$.

share|improve this answer
    
Can you show how it incorporates u(x,t)? –  Buddy Holly Oct 21 '12 at 2:08
    
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
    
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 u –  Buddy Holly Oct 21 '12 at 6:20
    
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! –  Buddy Holly Oct 21 '12 at 6:23
    
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
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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