Proving the existence of a solution for a autonomous diff.equation

[mathboy20]

Homework Statement

Let $$\mathcal{V} \subset \mathbb{R}^n$$ be open and $$f: \mathcal{V} \rightarrow \mathbb{R}^n$$ be continous. Assume that f has partial derivates which are continous.

Then the autonomous differential equation

$$\frac{dx}{dt}(t) = f(x(t))$$

on the region $$D = \mathbb{R} \times \mathcal{V}$$

Let (I,x) be a solution to the differential equation.

(a) Let $$\tau \in \mathbb{R}$$ and $$I_{\tau} =\{t-\tau|t\in I \}$$.

Define $$x_{\tau} \in C^{1}(I_{\tau}; \mathbb{R}^n)$$ and $$x_{\tau}(t) = x(t + \tau)$$.

Then show $$(I_{\tau},x_{\tau})$$ also is a solution to the differential equation.

(b) Assume that $$t_0 \in I$$ such that $$\frac{dx}{dt}(t_0) = 0$$ show that x is a constant function on I.

(c) Assume that $$t_1,t_2 \in I$$ where $$t_1 < t_2$$ and $$x(t_1) = x(t_2)$$

show that x is still a constant function on I.

(d) Assume that $$t_1, t_2 \in I$$ and that $$x(t_1) = x(t_2)$$. Show that exist a global defined $$(t_2 - t_1)$$ periodic solution $$(\mathbb{R},\hat{x})$$ to the equation such that $$x = \hat{x}$$ on I.

Homework Equations

The Attempt at a Solution

(solution a) In order to show that $$(I_{\tau},y_{\tau})$$ its important to emphize that since f's partial derviates is continuous on D, then according the existence part of the EaU theorem, there is at most one solution of the original equation. Then as I understand this since $$I_\tau$$ where $$\tau \in \mathbb{R} \subset \mathcal{V}$$ is also continoues on D.
Thus $$x_\tau$$ is also a solution for the original equation since this is defined on the same interval as the original solution x(t) for the original equation.

Thus $$(x_\tau, I_{\tau})$$ is a solution for the original equation.

(solution b) For the solution x to the differential equation is going to be constant then the corresponding interval I can only be bounded from the below and thusly making the corresponding solution being constant?

(solution c) By Rolle's theorem then the function f is continuous on the interval $$[t1,t2] \in I$$ and is differentiable on I, then f is constant.

(solution d) we know from (a) that the x is continuous on I and thusly according to the existence and uniqueness theorem then a periodical solution defined on thus interval I exists.

Could somebody please look at this and see if I have covered this correctly??

Sincerely Mathboy

 

[mighty2000]

Hello Mathboy,

Thank you for your post and for sharing your attempt at solving the given problem. Overall, I think you have a good understanding of the concepts and have provided a solid attempt at solving the problem. However, there are a few areas that could be improved or clarified. I will go through each part of your solution and provide some feedback.

Solution a) You have correctly identified that (I_{\tau},x_{\tau}) is a solution to the differential equation. However, your explanation is a bit unclear. I would suggest emphasizing that since x(t) is a solution to the original differential equation, it satisfies \frac{dx}{dt}(t) = f(x(t)) for all t \in I. Then, since x_{\tau}(t) = x(t + \tau) for all t \in I_{\tau}, we can substitute t + \tau for t in the original equation to get \frac{dx}{dt}(t + \tau) = f(x(t + \tau)). This shows that x_{\tau} is also a solution to the differential equation on the interval I_{\tau}.

Solution b) Your explanation is a bit unclear. I would suggest starting by stating that since \frac{dx}{dt}(t_0) = 0, we know that x(t) is constant on some interval containing t_0. Then, using the uniqueness part of the existence and uniqueness theorem, we can conclude that x(t) is constant on the entire interval I.

Solution c) Your explanation is correct, but it could be made more clear. I would suggest starting by stating that since x(t_1) = x(t_2), we know that x(t) is constant on the interval [t_1, t_2]. Then, by the continuity of x(t) and the differentiability of f(x(t)), we can apply Rolle's theorem to show that x(t) is constant on the entire interval I.

Solution d) Your explanation is correct, but it could be made more clear. I would suggest starting by stating that since x(t_1) = x(t_2), we know that x(t) is constant on the interval [t_1, t_2]. Then, by the continuity of x(t) and the periodicity of the solution, we can extend the solution to be defined on the entire interval I by repeating the same periodic pattern.