Linear Operators

**Elite** · August 1st 2009, 08:59 AM

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

**Opalg** · August 1st 2009, 09:49 AM

Originally Posted by Elite

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).

**Krahl** · August 1st 2009, 11:22 AM

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.

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