MHB Why is the solution single-valued ?

[evinda]

Hello! (Wave)

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)

 

[I like Serena]

Hello! (Wave)

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)

Hey evinda!

Isn't it the other way around?
Doesn't $u(x,\frac 1\pi)$ need to be a single-valued continuous function? (Wondering)
It's not if for the same value of $x$ we get at least 2 distinct values of $u$.

Note that if we pick $t=0$ then $u=\cos(\pi x)$ is indeed a single-valued continuous function. (Thinking)

 

[S.G. Janssens]

With single-valued function is it meant that the function is 1-1 ?

To add a minor remark to the previous post: The answer to your above question is: no.

It is part of the definition of "function" that a function yields one and precisely one value for each input. So, in fact, the phrase "single-valued function" is a pleonasm.

For example, the function $g : \mathbb{R}^2 \to \mathbb{R}$ defined by $g(a,b) = a^2 + b$ is single-valued (it is therefore a correctly defined function), but not 1 to 1.

I think that what you are asked to show, is that for $t_0 = \frac{1}{\pi}$ and for at least one value of $x$, say $x = x_0$, the equation $u=\cos{[\pi (x-ut)]}$ has at least two solutions, say $u_1(t_0,x_0)$ and $u_2(t_0,x_0)$. Namely, in that case $u$ is no longer single-valued for $(x,t) = (x_0,t_0)$ or, in other words, it is no longer a function or, in yet other words, your solution ceases to exist (or better: splits into two).

P.S. There are certain areas where it is not unusual to encounter multi-valued "functions" (sometimes these are called correspondences), but I don't think that these need to be considered here.

 

[I like Serena]

I think that what you are asked to show, is that for $t_0 = \frac{1}{\pi}$ and for at least one value of $x$, say $x = x_0$, the equation $u=\cos{[\pi (x-ut)]}$ has at least two solutions, say $u_1(t_0,x_0)$ and $u_2(t_0,x_0)$. Namely, in that case $u$ is no longer single-valued for $(x,t) = (x_0,t_0)$ or, in other words, it is no longer a function or, in yet other words, your solution ceases to exist (or better: splits into two).

It's a bit strange.
The domain of $u$ has not been specified, and it would be a perfectly valid question to ask what the domain is such that $u$ is well-defined.
Either way, I believe for $t=\frac 1\pi$, that $u(x,\frac 1\pi)$ is still a single-valued continuous function that is defined on all of $\mathbb R$.

evinda, can it be that we need to show that for $t>\frac 1\pi$ the function $u$ is not defined on all of $\mathbb R$ any more?
And that for $0\le t\le \frac 1\pi$ is well-defined and continuous on all of $\mathbb R$? (Wondering)

 

[evinda]

I think that what you are asked to show, is that for $t_0 = \frac{1}{\pi}$ and for at least one value of $x$, say $x = x_0$, the equation $u=\cos{[\pi (x-ut)]}$ has at least two solutions, say $u_1(t_0,x_0)$ and $u_2(t_0,x_0)$. Namely, in that case $u$ is no longer single-valued for $(x,t) = (x_0,t_0)$ or, in other words, it is no longer a function or, in yet other words, your solution ceases to exist (or better: splits into two).

So if $u_1=\cos{(\pi x-\pi u_1 t)}$, then we are looking for a function $u_2$ such that $\cos{(\pi x-\pi u_1 t)}=\cos{(\pi x-\pi u_2 t)}$ ?

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It's a bit strange.
The domain of $u$ has not been specified, and it would be a perfectly valid question to ask what the domain is such that $u$ is well-defined.
Either way, I believe for $t=\frac 1\pi$, that $u(x,\frac 1\pi)$ is still a single-valued continuous function that is defined on all of $\mathbb R$.

evinda, can it be that we need to show that for $t>\frac 1\pi$ the function $u$ is not defined on all of $\mathbb R$ any more?
And that for $0\le t\le \frac 1\pi$ is well-defined and continuous on all of $\mathbb R$? (Wondering)

No, it is the exact problem statement. So is it wrong? (Thinking)

 

[I like Serena]

So if $u_1=\cos{(\pi x-\pi u_1 t)}$, then we are looking for a function $u_2$ such that $\cos{(\pi x-\pi u_1 t)}=\cos{(\pi x-\pi u_2 t)}$ ?

A distinct function value $u_1(x)\ne u_2(x)$. (Thinking)

No, it is the exact problem statement. So is it wrong?

I believe so yes.
As I said, for $t=\frac 1\pi$ we can 'just' find a single-valued continuous solution on all of $\mathbb R$. (Worried)