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Existence of solution to integral equation

  1. Feb 2, 2008 #1

    quasar987

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    [SOLVED] Existence of solution to integral equation

    1. The problem statement, all variables and given/known data
    There's k:[0,1]²-->R square integrable and the operator T from L²([0,1],R) to L²([0,1],R) defined by

    [tex]T(u)(x)=\int_{0}^{1}k(x,y)u(y)dy[/tex]

    (a) Show that T is linear and continuous.

    (b) If ||k||_2 < 1, show that for any f in L²([0,1],R) , there exists a u in L²([0,1],R) solution of the integral equation

    [tex]u(x) = f(x)+\int_{0}^{1}k(x,y)u(y)dy[/tex]

    (almost everywhere on [0,1])

    3. The attempt at a solution

    I have done (a), and in demonstrating that T is continuous, I have shown that [tex]||T(u)||_2\leq ||k||_2||u||_2[/tex]

    Therefor, the condition ||k||_2 < 1 in (b) + the fact that T is linear is equivalent to saying that T is a contraction. So by Banach's fixed point theorem, there exists a (unique) fixed point to T; call it u*:

    [tex]u^*(x)=\int_{0}^{1}k(x,y)u^*(y)dy[/tex]

    Fine, but what about the equation

    [tex]u(x) = f(x)+\int_{0}^{1}k(x,y)u(y)dy[/tex]

    ??? If we define an operator T_f by

    [tex]T_f(u)(x)=f(x)+\int_{0}^{1}k(x,y)u(y)dy[/tex]

    this is not a contraction.

    How do the dots connect?!?!?
     
  2. jcsd
  3. Feb 3, 2008 #2

    morphism

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    Why is T_f not a contraction?
     
  4. Feb 3, 2008 #3

    quasar987

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    You're right, I spoke too fast!

    I wanted to look at ||T_f(u) - T_f(0)||=||T_(u)||, but I looked at ||T_f(u)|| instead.

    Thank you morphism!
     
  5. Feb 3, 2008 #4

    Defennder

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    Wow, morphism's one line reply was enough to answer your lengthy post.
     
  6. Feb 3, 2008 #5

    HallsofIvy

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    'Tis often so.

    And often, the lengthy (and, hopefully, complete) first post results in the response being brief.
     
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