1. Not finding help here? Sign up for a free 30min 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!

Finding a continous solution to an integral

  1. Jul 3, 2011 #1
    1. The problem statement, all variables and given/known data
    find a continous solution to the integral equation:
    f(x)=(x^3)+(1/2)*integral from 0 to 1 of ((x*y)/(y+1))*f(y)dy, by finding a fixed point of the function v:(C([0,1]), d)-->((C([0,1]), d) defined by
    v(f)(x)=(x^3)+(1/2)*integral from 0 to 1 of ((x*y)/(y+1))*f(y)dy
    2. Relevant equations
    d is the supremum norm.
    d(v(f1),v(f2))=sup(x in [0,1]) |(v(f1)(x)-v(f2)(x)|<=(1/4)*sup(y in [0,1]) |f1(y)-f2(x)|

    3. The attempt at a solution
    it says to use maple to do calculatins. tried and failed
     
  2. jcsd
  3. Jul 4, 2011 #2
    Am I right this is complex analysis?
     
  4. Jul 4, 2011 #3

    Gib Z

    User Avatar
    Homework Helper

    Susanne217- This would be categorized as the theory of Integral equations. Here, specifically, the theory borrows tools from the theory of Metric spaces, in particular the Banach Contraction Mapping theorem.

    OP - I don't know how to use Maple, so I'm not sure how this question was intended to be solved. Though I will complain, the question could have been better worded, since saying "Find the fixed point of v(f)(x) " is exactly the same as saying "solve the integral equation".

    An idea though: We know that the fixed point of v(f)(x) is the limit of the recursion [itex] f_{n+1} = v(f_n) [/itex] with any initial [itex] f_0 \in C([0,1], \mathbb{C} ) [/itex]. I started with [itex] f_0(x) = 1 [/itex]. Then [itex] f_1(x) = x^3 + x(1-\log 2)/2 [/itex]. Repeat this a few more times, and you'll notice that the same integrals appear over and over, with only some constants changing, and we always end up having [itex] f_n (x) = x^3 + C_n x [/itex] where [itex] C_n [/itex] is a sequence of real constants (it's an n-th degree polynomial in log 2, but we don't need that). So then one could well conjecture that our fixed point has the form [itex] f(x) = x^3 + Cx[/itex] where C is some constant. So plug this into our fixed point relation v(f) = f, and solve for the value of C which satisfies this. This gives you a fixed point, which you know is also unique, and completes the problem.
     
  5. Jul 4, 2011 #4
    f(x)=x^3+Cx where C is some constant. So plug this into our fixed point relation v(f) = f, and solve for the value of C which satisfies this. This gives you a fixed point, which you know is also unique, and completes the problem.

    Is this what you mean?...

    1/2*integral (xy/y+1)(y^3 +Cy)dy +x^3=x^3+CX
     
  6. Jul 4, 2011 #5

    Gib Z

    User Avatar
    Homework Helper

    Yes, it is.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding a continous solution to an integral
  1. Continous or not? (Replies: 2)

  2. Finding a solution (Replies: 5)

  3. Integral solution (Replies: 3)

  4. Find the solution? (Replies: 13)

Loading...