Is the interval I of an autonomous diff closed?

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Is the interval I of an autonomous diff closed???

Homework Statement



Given this autonomous diff.eqn

Where we have an open set E defined on R^n and [tex]f \in \mathcal{C}^1(E)[/tex]

[tex]x' = f(x)[/tex] where [tex]x(t_a) = x(t_b)[/tex] and [tex]t_a,t_b \in I[/tex] and where [tex]t_a < t_b[/tex].

Show for n = 1, that the solution x is constant.

The Attempt at a Solution



According to the definition of the autonomous differential equation. Then for x to be a solution to the diff.eqn then certain conditions must upheld.

1) x(t) is differentiable on I and [tex]\forall t \in I[/tex], [tex]x(t) \in E[/tex] and

2) [tex]x'(t) = f(x(t))[/tex].

according to how f is defined its continous on E and have first derivatives on E. E is open on subset on [tex]\mathbb{R}^n[/tex] and in our case it must mean that if E is defined as [tex]E \subset \mathbb{R} \times \mathbb{R}[/tex] and then [tex]I \in \mathbb{R}[/tex] which is the interval of all solutions of the original problem. Then if [tex]x(t_a)[/tex] and [tex]x(t_b)[/tex] are both solution of the original equation. Then it must mean (as I see it!) that x(t_a) = x(t_b) = 0. Hence a t_c defined on I will result in [tex]x'(t_c) = f(x(t_c)) = 0[/tex]. Thusly for n = 1 x is a constant solution of of the autonomous diff.eqn x' = f(x).

maybe I have misunderstood something but isn't if E is a subset of R x R and I is subset R doesn't mean that both sets E and I are subset of the same set R?

Have expressed this satisfactory?
 
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Answers and Replies

  • #2
Dick
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i) E isn't a subset of RxR. It's a subset of R. You are given it's a subset of R^n and n=1. ii) f(t_a) isn't a solution to the DE. It's not even a function of t. It's a single point on the solution. I think the point here is that if n=1 then the DE is separable, You can express t as a function of x.
 
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i) E isn't a subset of RxR. It's a subset of R. You are given it's a subset of R^n and n=1. ii) f(t_a) isn't a solution to the DE. It's not even a function of t. It's a single point on the solution. I think the point here is that if n=1 then the DE is separable, You can express t as a function of x.

Hi Dick,

You mean f(t_a) and f(t_b) are points on I ?

and also I consider that that interval I and the set E are both subsets of R? Aren't they?

Don't I need to show here that since we know has first derivatives on E (according to my original post) and is continious on E. Then since E is a subset of R for n = 1.

Then for x to be a solution of f then it has to be differentiable on I?
 
  • #4
Dick
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Hi Dick,

You mean f(t_a) and f(t_b) are points on I ?

and also I consider that that interval I and the set E are both subsets of R? Aren't they?

Don't I need to show here that since we know has first derivatives on E (according to my original post) and is continious on E. Then since E is a subset of R for n = 1.

Then for x to be a solution of f then it has to be differentiable on I?

Hi, Susanne. t_a and t_b are points on I. f(t_a) and f(t_b) are points in the subset E. Yes, they both subsets of R. And yes, x(t) is differentiable on I (as a function of t) and f(x) is differentiable on E as a function of x. But they don't have all that much to do with each other. One is the domain of x and the other is the domain of f and contains the range of x. Look. Solve x'=x (where '=d/dt) on the interval t_a=1 and t_b=2. That is the sort of problem they are talking about.
 
  • #5
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Hi, Susanne. t_a and t_b are points on I. f(t_a) and f(t_b) are points in the subset E. Yes, they both subsets of R. And yes, x(t) is differentiable on I (as a function of t) and f(x) is differentiable on E as a function of x. But they don't have all that much to do with each other. One is the domain of x and the other is the domain of f and contains the range of x. Look. Solve x'=x (where '=d/dt) on the interval t_a=1 and t_b=2. That is the sort of problem they are talking about.

okay thanks :)

is x(t_a) and x(t_b) are points on the subset I?

I have come up with my own idear using Rolle's theorem. So please bare with me :)

Let E defined as in the original post. The solution x is defined on I. Therefore the diff.eqn is differentiable on I. Assume that [tex]x(t_a), x(t_b) \in E[/tex]. Then by Rolle's theorem the diff.eqn has both maximum and minimal solution on I. Thusly none of these are obtained from an interior point of I. Thusly [tex]x(t_a)= x(t_b) [/tex] and x is there constant on I.

Because I need to show the above problem with Rolle's theorem in mind.

How is this Dick?
 
  • #6
Dick
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I already told you x(t_a) and x(t_b) aren't in I. They're in E. And my version of Rolle's theorem doesn't say anything about maxima or minima. The rest of what you are saying makes even less sense. Are you using theorems I don't know about? How do you know the extrema aren't in the interior?

But it is true that if x has a interior maximum on [t_a,t_b] you can see the DE can't be autonomous. Loosely speaking, you can draw a horizontal line a little below the maximum, say at t_max and find t+<t_max<t- such that x(t+)=x(t-)<x(t_max) but x'(t+)>0 (since it's heading up to the max) and x'(t-)<0 since it's heading down from the max. I'm not sure how to make that completely rigorous just now. But do you see why that would lead to a contradiction?
 
  • #7
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I already told you x(t_a) and x(t_b) aren't in I. They're in E. And my version of Rolle's theorem doesn't say anything about maxima or minima. The rest of what you are saying makes even less sense. Are you using theorems I don't know about? How do you know the extrema aren't in the interior?

But it is true that if x has a interior maximum on [t_a,t_b] you can see the DE can't be autonomous. Loosely speaking, you can draw a horizontal line a little below the maximum, say at t_max and find t+<t_max<t- such that x(t+)=x(t-)<x(t_max) but x'(t+)>0 (since it's heading up to the max) and x'(t-)<0 since it's heading down from the max. I'm not sure how to make that completely rigorous just now. But do you see why that would lead to a contradiction?

I don't Dick,

They use Rolle's theorem to show bla bla. But then they don't say. Which part of Rolle's theorem? The whole part or just the idear of it :(
 
  • #8
Dick
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I don't Dick,

They use Rolle's theorem to show bla bla. But then they don't say. Which part of Rolle's theorem? The whole part or just the idear of it :(

Who is 'they'? Are you trying to present the argument given as a solution? If so can you give me the whole thing word for word? Would you also state what you think Rolle's theorem is?
 
  • #9
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Who is 'they'? Are you trying to present the argument given as a solution? If so can you give me the whole thing word for word? Would you also state what you think Rolle's theorem is?

Hi Dick,

The original post is in post number 2 in this thread. "They" are the people who formulated the problem ;)

Rolle's theorem which I'm trying to use comes from Apostol's Mathematical Analysis p. 110.

I know this book pre-dates the Vietnam war but it was the one we used on our Mathematical Analysis course.
 
  • #10
Dick
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Hi Dick,

The original post is in post number 2 in this thread. "They" are the people who formulated the problem ;)

Rolle's theorem which I'm trying to use comes from Apostol's Mathematical Analysis p. 110.

I know this book pre-dates the Vietnam war but it was the one we used on our Mathematical Analysis course.

Ok, so Rolle's theorem says if x(t_a)=x(t_b) then there is a point t_c in between where x'(t_c)=0. How does this help?
 

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