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Volterar operator is Vf(x) = int (f(s)ds,0,x)

  1. Feb 6, 2009 #1
    Given that the volterar operator is Vf(x) = int (f(s)ds,0,x)

    The adjoint of this operator is

    V*f(x) = int(f(s)ds,x,1)

    How do I show this ?
  2. jcsd
  3. Feb 6, 2009 #2


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    Re: Operator

    Well, I would think by using the definition of "adjoint operator"! Which is?
  4. Feb 6, 2009 #3
    Re: Operator

    If A in B(H,K), then the unique operator B in B(K,H) satisfying u(h,k) = <Ah,k>=<h,Bk> is called the adjoint of A.

    How does this definition tell you that you integrate from x to 1?
  5. Feb 7, 2009 #4


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    Re: Operator

    Well, I guess now the question is "How is <u, v> defined?"
  6. Feb 7, 2009 #5
    Re: Operator

    [tex] <f,g> = \int f(x) \overline{g(x)} dx dy [/tex]

    and given that

    [tex] Vf(x) = \int^{x}_{0} f(s) ds [/tex]

    [tex] <Vf,g> = \int^{x}_{0} f(s) \overline{g(s)}[/tex]

    Here we have to integrate by parts.
    [tex] <Vf,g> = f(s)g(s)|^{x}_{0} - \int^{x}_0 f(s)g(s) ds[/tex]

    I'm still don't see where the integral from 1 to x arises or how I get rid of g
    Last edited: Feb 7, 2009
  7. Feb 7, 2009 #6
    Re: Operator

    Let k:[0,1]X[0,1]-> R be the characteristic function of {(x,y);y<x>}. The corresponding operator V: L^2(0,1)->L^2(0,1) defined by

    Vf(x) = int k(x,y)f(y)dy on [0,1] is called the volterra operator.

    That was the textbook definition. Then they note Vf(x) = int f(y) dy on [0,x]

    as with most online sources.

    What's the later tell you?
  8. Feb 7, 2009 #7
    Re: Operator

    I"m sure I just use the later.

    When integrating by parts when do you switch the order of integration?
  9. Feb 7, 2009 #8
    Re: Operator

    First let's make sure we've answered Halls' questions, Nusc.

    You have: [tex]\langle f, g \rangle = \int_0^1 f(x) \overline{g(x)} \, dx[/tex]
    and: [tex]Vf = \int_0^x f(s) \, ds[/tex]

    Now, you want: [tex]\langle Vf, g \rangle = \langle f, V^\ast g \rangle [/tex]. The task being to find [tex]V^\ast[/tex]. You have actually been given a [tex]V^\ast[/tex], so are only required to check it.

    The left-hand side of this equality translates to
    [tex]\int_0^1 (Vf)(x) \overline{g(x)} dx = \int_0^1 \left( \int_0^x f(s) ds\right) \overline{g(x)} dx[/tex]

    We may write this as a double integral: [tex]\int_0^1 \int_0^x f(s) \overline{g(x)} \, ds \, dx \; \; \; (\ast)[/tex]

    Noting that the required right-hand side is of the form
    [tex]\langle f, V^\ast g \rangle = \int_0^1 f(t) \overline{(V^\ast g)(t)} dt[/tex]
    (where t is an arbitrary parameter: above we used 'x')

    we see from (*) that finding [tex]V^\ast[/tex] just requires one to switch the order of integration: take t=s; we want ds dx to become dx ds.

    There is no "integration by parts" required. This work comes solely under "double integrals".
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