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Homework Help: C vs C1 continuous

  1. Sep 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Let F be a function defined on a open set E where [tex]E \in V[/tex]. F is said to be continuous at each point [tex]x \in V[/tex]. Which according to my textbook at hand can be written as [tex]F \in \mathcal{C}(E)[/tex]

    But the expression [tex]F \in \mathcal{C}^{1}(E)[/tex] is that equal to saying that F have first order derivatives defined on the set E?
     
  2. jcsd
  3. Sep 13, 2010 #2

    HallsofIvy

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    More than that. [tex]F \in \mathcal{C}^{1}(E)[/tex] says that F has a continuous first derivative. For example, f(x)= |x| is continuous in any neighborhood of x= 0 but not differentiable at x= 0 so [tex]f \in \mathcal{C}^{1}(E)[/tex] but f is not in [tex]\mathcal{C}^{1}(E)[/tex]. A slightly more complicated example is [tex]f(x)= x^2sin(1/x)[/tex] if x is not 0, f(0)= 0. It is fairly easy to show that f is continuous in any neighborhood of x= 0 and, in fact, that it has a derivative for all x, including x= 0, but the derivative function is not continuous at x= 0.
     
  4. Sep 13, 2010 #3
    Thank you Mr. Hall,

    I have a follow-up question.

    Let say we have autonomous differential equation

    [tex]x' = f(x)[/tex] and that we have an open set [tex]E \subset \mathbb{R}[/tex] and that [tex]f \in \mathcal{C}^1(E)[/tex]. We then assume that (I,x) is a solution to the diff-eqn also that [tex]x(t_1) = x(t_2)[/tex] where [tex]t_1, t_2 \in I[/tex] and that [tex]t_1 < t_2[/tex]

    Show for n = 1 that y is a constant solution.

    I know from the definition of f that it has first derivatives on E and that E is open. And since from the definition of the solution to a diff.eqn that [tex]I \subset \mathbb{R}^n[/tex]. Since I is considered to be a subset of [tex]R[/tex] and that E is also a subset of [tex]\mathbb{R}[/tex] for n = 1.

    Then here is my questions:

    1) Do I need to show that f also has derivatives on I?

    2) I can't well assume that since f has derivatives on a subset of R then if f's domain consistets of both of R and E. Can this domain be seen a as closed set? And I then assume by Rollo's theorem that there exists some value on f's domain, [tex] t_3 [/tex]. Where if x is a solution of f, and then [tex]x'(t_1) = f(x(t_1)) = x'(t_2) = 0[/tex] and then [tex]x'(t_3) = f(x(t_3))= 0[/tex] and than the solution x is constant?
     
    Last edited: Sep 14, 2010
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