# Is it implied that the problem is well-defined?

• MHB
• evinda
In summary: Yes, when we have a different $g$ we get a different solution, so $u$ depends continuously on the data of the problem, right? (Thinking)Of course $u$ depends on $g$, but continuous dependence is another matter. Think of the map that sends $g$ to $u$:$g \mapsto u(\cdot,\cdot, g) \qquad (*)$where the dots are place-holders for $x$ and $t$. So, this map sends an initial condition $g$ to the solution $u(\cdot,\cdot,g)$, which is a function
evinda
Gold Member
MHB
Hello! (Wave)

We have the Cauchy problem of the equation

$u_t+xu_x=xu, x \in \mathbb{R}, 0<t<\infty$

with some given smooth ($C^1$) function $g$ as initial value.

I want to check if the problem is well defined for each time. We know that a problem is well defined if the solution exists, is unique and depends continuously on the data of the problem.

I have computed that the solution of the problem is $u(x,t)=g(xe^{-t}) e^{x(1-e^{-t})}$.

So we have that the problem is well-defined if $g$ does not take two different values for some specific $t$, right?

But is this implied from the fact that $g$ is smooth? (Thinking)

It it implied from the fact that $g$ is a function: For each argument of $g$ (i.e. $x e^{-t}$) there exists precisely one function value of $g$.

So, you have found a (global, classical) solution $u \in C^1(\mathbb{R} \times (0,\infty))$.
Now you would need to check that it is unique, and you would need to think what it means for $u$ to depend continuously on the initial datum $g$.

Krylov said:
It it implied from the fact that $g$ is a function: For each argument of $g$ (i.e. $x e^{-t}$) there exists precisely one function value of $g$.

So, you have found a (global, classical) solution $u \in C^1(\mathbb{R} \times (0,\infty))$.
Now you would need to check that it is unique, and you would need to think what it means for $u$ to depend continuously on the initial datum $g$.

How could we show the uniqueness of the solution? Do we suppose that there is an other solution? (Thinking)

Isn't it implied directly that $u$ depends on the initial data, since it contains $g$ ?

evinda said:
How could we show the uniqueness of the solution? Do we suppose that there is an other solution? (Thinking)

Isn't it implied directly that $u$ depends on the initial data, since it contains $g$ ?

As $g$ is smooth of class $C^1$, that means that the solution is well-defined for any $x$ and for any $t$.

And since the solution contains $g$ as the initial value for (presumably) $t=0$, doesn't that imply that the solution is unique?
After all, a different instance of $g$ means that we have a different solution, don't we? (Wondering)

I like Serena said:
And since the solution contains $g$ as the initial value for (presumably) $t=0$, doesn't that imply that the solution is unique?

So because of the fact that $g$ is well-defined for any $x$ and $t$, and $u$ contains $g$, we get that the solution $u$ is unique? (Thinking)
I like Serena said:
After all, a different instance of $g$ means that we have a different solution, don't we? (Wondering)

Yes, when we have a different $g$ we get a different solution, so $u$ depends continuously on the data of the problem, right? (Thinking)

evinda said:
So because of the fact that $g$ is well-defined for any $x$ and $t$, and $u$ contains $g$, we get that the solution $u$ is unique? (Thinking)

I don't think the uniqueness of the solution follows like that.

It seems to me it is best to go back and consider how you obtained your solution. Was it with the method of characteristics? If so, then what does the method of characteristics tell you about existence and uniqueness of global (that is, defined for all $t > 0$) solutions for this problem?

evinda said:
Yes, when we have a different $g$ we get a different solution, so $u$ depends continuously on the data of the problem, right? (Thinking)

Of course $u$ depends on $g$, but continuous dependence is another matter. Think of the map that sends $g$ to $u$:
$g \mapsto u(\cdot,\cdot, g) \qquad (*)$
where the dots are place-holders for $x$ and $t$. So, this map sends an initial condition $g$ to the solution $u(\cdot,\cdot,g)$, which is a function of the two variables $x$ and $t$. As the domain of $(*)$ we could take $C^1(\mathbb{R})$ and as its co-domain $C^1(\mathbb{R} \times [0,\infty))$.

Now the question of continuous dependence on initial conditions becomes a question of continuity of this map, but continuity with respect to what? We need a way to measure the changes in $u(\cdot,\cdot,g)$ as a result in changes in $g$. (Put more fancily: We need a topology on the domain and co-domain of (*).)

I can think of many ways to do that, typically involving $C^1$ function norms of $g$ and $u$ and compact subsets of $\mathbb{R}$ and $[0,\infty)$, but it may be better to ask you whether in your lectures you were provided with any more precise definition of "continuous dependence" for this kind of problem. It is this definition that you should probably recall, formulate and apply.

(Ultimately, this application is probably not going to be difficult, because you have an explicit solution that depends on $g$ in a nice way.)

## 1. What does it mean for a problem to be well-defined?

A well-defined problem is one that has a clear and specific question or goal, with all necessary information and constraints clearly stated. This allows for a precise and unambiguous solution to be found.

## 2. How can I determine if a problem is well-defined?

To determine if a problem is well-defined, you should check if all necessary information and constraints are clearly stated, if the question or goal is specific and unambiguous, and if there is a clear method or approach to finding a solution.

## 3. Why is it important for a problem to be well-defined?

A well-defined problem is important because it allows for a clear and precise solution to be found. It also helps to avoid confusion and misunderstandings, which can lead to incorrect or incomplete solutions.

## 4. Can a problem be well-defined in one field but not in another?

Yes, a problem can be well-defined in one field but not in another. This is because different fields may have different standards and criteria for what constitutes a well-defined problem. For example, a problem in mathematics may have a clear and precise solution, but a problem in psychology may have more subjective and open-ended solutions.

## 5. What are some common characteristics of well-defined problems?

Some common characteristics of well-defined problems include a clear and specific question or goal, all necessary information and constraints are stated, there is a clear method or approach to finding a solution, and the problem is free from ambiguity and confusion.

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