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Math Help - Linear Operators

  1. #1
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    Linear Operators

    Find the non-zero e-values and corresponding e-functions of a linear integral operator A with kernel
    K(x,y)=(x^2+\frac{1}{x})y, 1 \leq x,y \leq 2.This operator acts on the space of continuous functions
    L={f(x),0 \leq x \leq 1} in the usual way;
    (Af)(x)=(int0..1)K(x,y)f(y)dy.

    No one knows how to do it even at the tutorials, and the lecturer don't wanna know lol. Help please
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  2. #2
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    Quote Originally Posted by Elite View Post
    Find the non-zero e-values and corresponding e-functions of a linear integral operator A with kernel
    K(x,y)=(x^2+\frac{1}{x})y, 1 \leq x,y \leq 2.This operator acts on the space of continuous functions
    L={f(x),0 \leq x \leq 1} in the usual way;
    (Af)(x)=(int0..1)K(x,y)f(y)dy.

    No one knows how to do it even at the tutorials, and the lecturer don't wanna know lol. Help please
    If the kernel is defined for x and y lying between 1 and 2, then the integral must go from 1 to 2 (not from 0 to 1), and the function space L must also consist of functions defined on [1,2], not [0.1].

    If \lambda is an eigenvalue then Af(x) = \lambda f(x), so \int_1^2(x^2+x^{-1})yf(y)\,dy = \lambda f(x). But as far as the integral is concerned, x is a constant, so we can take out the factor (x^2+x^{-1}) and get (x^2+x^{-1})\int_1^2yf(y)\,dy = \lambda f(x). But then the integral is a constant, and we are left with two possibilities. The first is that \lambda and the integral are both 0. The second is that f(x) is a nonzero constant times x^2+x^{-1}, in which case you can plug that formula for f into the integral, and you should then find that \lambda = 19/4 (if I've done the integration correctly).
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  3. #3
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    Yep i agree, either there was a mistake in the actual question about the limits or you made a typo while copying.

    This question is actually much simpler than the similar types of questions your probably able to do.

    the problem here was the 1/x im guessing. but noticing that the function is seperable
    you can do it in no time.
    Last edited by Krahl; August 1st 2009 at 02:08 PM.
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