1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Contraction, integral, maximum norm

  1. Apr 29, 2016 #1
    1. The problem statement, all variables and given/known data

    Hi everybody! Here is another problem about contraction and Banach fixed-point theorem that I don't get:

    The function ƒ: C([0,½]) → C([0,½]) is defined by:
    [tex]
    [f(x)](t) := 1 + \int_{0}^{t} x(s) ds ∀ t∈[0,\frac{1}{2}].
    [/tex]

    Is ƒ a contraction with respect to the norm || ⋅ ||? If yes, which function is the fixed point of ƒ?

    2. Relevant equations

    Contraction mapping theorem, Banach fixed-point theorem

    3. The attempt at a solution

    Well I could not really get anywhere, because I don't understand really the definition of the function. Here is what I would do anyway:

    [tex]
    || [f(x)](t) - [f(y)](t) ||_{\infty} = \mbox{max } | 1 + \int_{0}^{t} x(s) ds - 1 - \int_{0}^{t} y(s) ds | \\
    = \mbox{max } | \int_{0}^{t} x(s) - y(s) ds |
    [/tex]

    Then I have no idea how to integrate that since x is a function of s... What is the primitive of x(s)? Also not sure if I did the right thing in the first place as well! Some help would be very appreciated. :)

    Thank you in advance for your answers.


    Julien.
     
  2. jcsd
  3. Apr 29, 2016 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Looks OK so far. Do some overestimating. How big can that integral be?
     
  4. Apr 29, 2016 #3
    @LCKurtz Hi and thx for your answer. I have no idea, really. How should I do some estimation of this integral?


    Julien.
     
  5. Apr 29, 2016 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Well, you know you can't actually perform the integration. But remember from calculus$$
    \left | \int_a^b f(x)~dx \right | \le \int_a^b |f(x)|~dx$$and other properties of integration. And that integrand looks like it might be useful calculating ##\| \cdot \|_\infty##.
     
  6. Apr 29, 2016 #5
    @LCKurtz Thank you again for your answer. So I would have:

    [tex]
    || [f(x)](t) - [f(y)](t) ||_{\infty} ≤ max \cdot \int_{0}^{t} | x(s) - y(s) | ds
    [/tex]

    and

    [tex]
    || x - y ||_{\infty} = max | x(t) - y(t) |
    [/tex]

    Right? Mm I still don't see how to compare them, the only thing I know is that t is between 0 and 1 and ƒ is continuous between 0 and 1. I'm also not sure about my || x - y ||. Shouldn't it be || x(t) - y(t) || or something?

    Maybe I just don't know the inequalities of calculus well enough, I'm gonna take a close look at them during the next days.


    Julien.
     
  7. Apr 29, 2016 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You know more about ##t## than it is in ##[0,1]##, and it matters.
     
  8. Apr 29, 2016 #7
    @LCKurtz Yes indeed it is between 0 and ½. Is that what you meant?

    PS: I edited my last post while you were writing I believe, sorry for that.
     
  9. Apr 29, 2016 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes. Now you are close. You just need to continue a couple more inequalities to get your overestimate. It might be easier to see you way through these inequalities if you start with$$
    | [f(x)](t) - [f(y)](t)| = | 1 + \int_{0}^{t} x(s) ds - 1 - \int_{0}^{t} y(s) ds |
    = | \int_{0}^{t} x(s) - y(s) ds |$$and don't worry about taking the max until you are done overestimating.
     
    Last edited: Apr 29, 2016
  10. Apr 29, 2016 #9
    @LCKurtz What about that:

    [tex]
    || [f(x)](t) - [f(y)](t) ||_{\infty} ≤ max \int_{0}^{t} | x(s) - y(s) | ds ≤ (t - 0) \cdot max |x(s) - y(s)| \\
    [/tex]

    Since t is between 0 and ½, ƒ is a contraction. The inequality comes from: if ƒ is bounded on [0,t] and integrable then | ƒ | is integrable and:

    [tex]
    \frac{1}{t - 0} \int_{0}^{t} | f(x) | dx ≤ max | f(x) |
    [/tex]

    Is that a valid proof? Thx a lot for your help.


    Julien.
     
  11. Apr 29, 2016 #10

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I think that is a bit sketchy, especially with ##t##'s in your answer. I had edited my previous post while you posted this. You might want to look at my previous post again.
     
  12. Apr 29, 2016 #11
    @LCKurtz Okay I'm not sure that's exactly what you had in mind but I give it another go:

    [tex]
    | [f(x)](t) - [f(y)](t) | = | 1 + \int_{0}^{t} x(s) ds - 1 - \int_{0}^{t} y(s) ds | = | \int_{0}^{t} | x(s) - y(s) ds | \\
    ≤ \int_{0}^{t} | x(s) - y(s) | ds \\
    ≤ \int_{0}^{\frac{1}{2}} | x(s) - y(s) | ds \\
    ≤ \frac{1}{2} \cdot max | x(s) - y(s) | \\
    \implies || [f(x)](t) - [f(y)](t) ||_{\infty} ≤ \frac{1}{2} \cdot || x(s) - y(s) ||_{\infty} \\
    \implies \mbox{f is a contraction}
    [/tex]

    Very similar to what I've done last post though..


    Julien.
     
  13. Apr 29, 2016 #12

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, it is similar, but much clearer, leaving no doubt each step is correct. Good. I have to leave for a few hours now. When I return I will check and see how you are doing in finding the fixed point.
     
  14. Apr 29, 2016 #13
    @LCKurtz Thanks a lot for all this help, I appreciate it. I might search for the fixed point tomorrow as it is close to midnight here in Germany :)


    Julien.
     
  15. Apr 29, 2016 #14
    Okay i couldn't wait so I gave it a go, but then I get an integral equation, which is something I have never done before:

    [tex]
    1 + \int_{0}^{t} x(s) ds = x(t)
    [/tex]

    The only thing I got out of it is:

    [tex] x(0) = 1 [/tex]

    That's not much I know :D I'll sleep over it and go further tomorrow.


    Julien.
     
  16. Apr 29, 2016 #15
    Could it be that the fixed point function is et? I don't really know how to find it, but it seems to work:

    [tex]
    1 + \int_{0}^{t} e^s ds = 1 + e^t - e^0 = e^t
    [/tex]

    Maybe I've misunderstood how to find the fixed point though, I'm not really sure.


    Julien.
     
  17. Apr 29, 2016 #16

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, you have found the fixed point by inspection, but your equation in post #14$$
    x(t)=1 + \int_{0}^{t} x(s) ds$$will lead you directly to it. Try differentiating both sides with respect to ##t## to change it from an integral equation to a differential equation and see if you can get it from there.
     
  18. Apr 30, 2016 #17
    [tex]
    1 + \int_{0}^{t} x(s) ds = x(t) \\
    \frac{d}{dt} (1 + \int_{0}^{t} x(s) ds) = \frac{d}{dt} (x(t)) \\
    0 + \frac{d}{dt} \int_{0}^{t} x(s) ds = \dot{x}(t) \\
    \mbox{According to the fundamental theorem of calculus: } \\
    x(t) = \dot{x}(t) \\
    \implies x_*(t) = e^t
    [/tex]

    Pretty neat! Thx a lot @LCKurtz, I've learned a lot through this post! Note that x* is the notation my teacher uses for fixed point.

    Julien.
     
  19. Apr 30, 2016 #18

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You're welcome. One tiny oversight, you would get ##x_*(t) = Ce^t##, but as you observed earlier, ##x_*(0) = 1##. You need that.
     
  20. May 1, 2016 #19
    @LCKurtz Thanks a lot, that makes sense!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Contraction, integral, maximum norm
  1. Is this a norm? (Replies: 13)

  2. Integral norm proving (Replies: 4)

Loading...