Results 1 to 4 of 4

Math Help - tough DE theory problem

  1. #1
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1

    tough DE theory problem

    Hey Gang:

    I am going to ask a question this time. I got a hold of the book

    Ordinary DE and Stability Theory by Sanchez I picked up for a song. It has some tough problems. I thought it would be a nice book, but the notation is killer.

    I have been wanting to strengthen my DE skills.

    Here is one if anyone has a good idea.

    Let's take the equation x'=f(x), where x=x(t) is an unknown

    scalar function and f and df/dx are defined and continuous on the strip

    B=[(t,x)|-\infty<t<{\infty}, \;\ a<x<b].

    We can also assume f(x) \neq 0, \;\ a<x<b,

    and let F(x)=\int_{x_{0}}^{x}\frac{1}{f(s)}ds,

    where a<x_{0}<b.

    Show all solutions are described by the family

    x(t)={\phi}(t-c), where {\phi} is the inverse

    of F, and c is a constant determined by initial conditions.


    I tried using the second fundamental theorem of calculus and thought

    about \frac{d}{dx}[f(x)]=\frac{d}{dt}\int_{x_{0}}^{x(t)}\frac{1}{f(s)}ds

    Just a little hung up.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Wingless sent me the following. He gets the credit for this post

    We know these:

    \frac{dx}{dt} = x'(t) = f[x(t)]

    F[x(t)] = \int_{x_0}^{x} \frac{1}{f(s)}~ds

    Now, on the RHS of the second line, substitute s = x(t) to get,

    F[x(t)] = \int_{x^{-1}(x_0)}^{x^{-1}[x(t)]} \frac{x'(t)}{f[x(t])}~dt

    Using the first line, replace x'(t) to get,

    F[x(t)] = \int_{x^{-1}(x_0)}^{x^{-1}[x(t)]} \frac{f[x(t)]}{f[x(t])}~dt = \int_{x^{-1}(x_0)}^{x^{-1}[x(t)]}dt = \underbrace{x^{-1}[x(t)]}_{\text{this is t}}-\underbrace{x^{-1}(x_0)}_{\text{some constant}}+C = t- C

    x(t) = \phi (t-C)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Thank you much fellas. Since the wing isn't here, tell him thanks as well.

    I started out right and then went down the primrose path
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2

    Cool

    My solution is also similar to that by wingless's, however I would like to give it here for showing some technical details.
    Since x^{\prime}(t)=f(x(t))\neq0 for all x\in(a,b), we see that x is either increasing or decreasing, i.e. it has a inverse function x^{-1}.
    Now, consider the followings.
    F[x(t)]=\int\limits_{x_{0}}^{x(t)}\frac{1}{f(s)}ds=\int\l  imits_{x_{0}}^{x(t)}\frac{1}{x^{\prime}(x^{-1}(s))}ds=\int\limits_{x^{-1}(x_{0})}^{t}dv=t-x^{-1}(x_{0}),\qquad(\ast)
    where
    v=x^{-1}(s)\text{ and so }dv=(x^{-1}(s))^{\prime}ds=\frac{1}{x^{\prime}(x^{-1}(s))}ds (see Inverse functions and differentiation - Wikipedia).
    Applying \phi on both sides of (*), we get the desired result.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Tough ODE problem
    Posted in the Differential Equations Forum
    Replies: 12
    Last Post: September 22nd 2010, 03:02 PM
  2. tough problem 1
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: August 10th 2010, 05:01 AM
  3. Group Theory Tough Problems
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: December 2nd 2009, 08:47 PM
  4. Tough Problem from Rotman's Group Theory, 3rd Ed
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 1st 2009, 06:12 AM
  5. Tough Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 14th 2008, 03:08 AM

Search Tags


/mathhelpforum @mathhelpforum