# Why is the solution single-valued ?

• MHB
• evinda
In summary, the conversation discusses the problem of showing that a smooth solution to the equation $u_t+uu_x=0$ with initial condition $u(x,0)=\cos{(\pi x)}$ satisfies the equation $u=\cos{[\pi (x-ut)]}$ and that this solution ceases to exist as a single-valued continuous function when $t=\frac{1}{\pi}$. The conversation also delves into the definition of a single-valued function and the need to show that for $t>\frac{1}{\pi}$, the function $u$ is not defined on all of $\mathbb{R}$ anymore.
evinda
Gold Member
MHB
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)

evinda said:
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)

evinda said:
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.

Last edited:
Krylov said:
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)

Krylov said:
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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I like Serena said:
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)

evinda said:
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)

evinda said:
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)

## 1. Why is it important for a solution to be single-valued?

A single-valued solution ensures that there is only one correct answer to a problem, providing a clear and unambiguous solution. This is crucial in scientific research as it allows for reproducibility and accuracy in experiments.

## 2. What does it mean for a solution to be single-valued?

A single-valued solution means that there is only one unique value that satisfies the given conditions or equations. This value is the only correct answer to the problem, and any other values would not be valid solutions.

## 3. How is a single-valued solution different from a multi-valued solution?

A single-valued solution has only one unique value that satisfies the given conditions, while a multi-valued solution can have multiple values that fulfill the requirements. In other words, a single-valued solution is specific and unique, while a multi-valued solution is more general and can have different possible answers.

## 4. What types of problems have single-valued solutions?

Problems that have well-defined parameters and conditions usually have single-valued solutions. These can include mathematical equations, scientific experiments, and engineering designs. In contrast, problems that involve subjective or open-ended criteria may have multi-valued solutions.

## 5. Are there any exceptions to the rule of a single-valued solution?

In some cases, a problem may have no solution or an infinite number of solutions, making it impossible to have a single-valued solution. This can occur when the given conditions contradict each other or when the problem is ill-defined. However, in most scientific research, a single-valued solution is the desired outcome.

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