Hi all,

I need the solution of the following exercise:

Show that the integral equation g(s)=$\displaystyle \lambda$$\displaystyle \int^{\pi}_{0} (\sin s \sin2t) g(t) dt$ has no eigenvalues

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- Jul 12th 2010, 10:46 PMraedIntegral equations
Hi all,

I need the solution of the following exercise:

Show that the integral equation g(s)=$\displaystyle \lambda$$\displaystyle \int^{\pi}_{0} (\sin s \sin2t) g(t) dt$ has no eigenvalues - Jul 17th 2010, 07:36 AMInvisibleMan
After one integration wrt s from 0 to pi, you get:

$\displaystyle 2* \lambda \int^{\pi}_{0} ( \sin2t) g(t) dt = \int^{\pi}_{0} g(t)dt$

so you get:

$\displaystyle \int_{0}^{\pi} (2\lambda \sin(2t) -1) g(t) dt=0$

g is different than zero, otherwise it wouldn't be an eigenfunction of this operator.

So 2 lambda *sin(2t)-1 should be zero almost everywhere in the interval [0,pi], there isn't such lambda cause lambda is constant and sin(2t) varies along this interval. - Jul 20th 2010, 04:30 PMJose27Quote:

g is different than zero, otherwise it wouldn't be an eigenfunction of this operator.

Quote:

So 2 lambda *sin(2t)-1 should be zero almost everywhere in the interval [0,pi]

For the problem I only get that if $\displaystyle |\lambda | <\frac{1}{\pi}$ then there is no eigenvector. I don't know how to deal with the other cases. - Jul 20th 2010, 06:17 PMAckbeet
Is the integral zero for all functions g? If so, then the fundamental theorem of the calculus of variations kicks in, and your conclusion is sound. Otherwise not. I'm just not entirely sure of the logic here, whether you get to say the integral is zero for all choices of the function g.

- Jul 21st 2010, 09:05 AMJose27
I don't think this is the case, since he is trying to solve an eigenvalue problem so assuming the identity holds for all your domain is equivalent to asserting that your operator is a multiple of the identity, which clearly has no eigenvalues.

If you're working on $\displaystyle C[0,\pi ]$ (or some interval) then maybe you could appeal to the compactness of your operator (Fredholm alternative comes to mind, but as far as I know it's only valid in Hilbert spaces, I'm not too knowledgeable in Banach algebras though so maybe there you could find something). - Jul 21st 2010, 09:06 AMJose27
Sorry, delete this.