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

  1. Sep 8, 2008 #1
    1. The problem statement, all variables and given/known data

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

    Then the autonomous differential equation

    [tex]\frac{dx}{dt}(t) = f(x(t))[/tex]

    on the region [tex]D = \mathbb{R} \times \mathcal{V}[/tex]

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

    (a) Let [tex]\tau \in \mathbb{R}[/tex] and [tex]I_{\tau} =\{t-\tau|t\in I \}[/tex].

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

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

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

    (c) Assume that [tex]t_1,t_2 \in I[/tex] where [tex]t_1 < t_2[/tex] and [tex]x(t_1) = x(t_2)[/tex]

    show that x is still a constant function on I.

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



    2. Relevant equations



    3. The attempt at a solution

    (solution a) In order to show that [tex](I_{\tau},y_{\tau})[/tex] its important to emphize that since f's partial derviates is continous 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 [tex]I_\tau[/tex] where [tex]\tau \in \mathbb{R} \subset \mathcal{V}[/tex] is also continoues on D.
    Thus [tex]x_\tau[/tex] 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 [tex](x_\tau, I_{\tau})[/tex] 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 continous on the interval [tex][t1,t2] \in I[/tex] and is differentiable on I, then f is constant.

    (solution d) we know from (a) that the x is continous 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
     
    Last edited: Sep 8, 2008
  2. jcsd
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