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evinda

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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? (Thinking)