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Suppose we are given an integral operator $ g(x)=f(x)+ \lambda \int_{0}^{\infty}K(x,t)f(t)dt $

with the kernel $ K(x,t)=K(t,x)$. According Hilbert-Schmidt theory then, the function can be obtained as

$ f(x)= \sum_{n=0}^{\infty}\frac{c_{n}}{\lambda _{n} -\lambda}\phi_{n} (x)$

with $ \phi_{n} (x) $ being the eigenfucnctions of the integral kernel $ K(x,t) = K(t,x) $. How can I get these eigenfunctions? Thanks.

rschwieb
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Jose Garcia
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  • It depends on what $K$ is! The eigenvalue equation is an equation and you have to solve it. – Qiaochu Yuan May 30 '12 at 18:43
  • ok but how about the eigenfunctions ?? – Jose Garcia May 30 '12 at 18:48
  • You simply have to find those too. I don't really understand what you're asking. Of course in the finite-dimensional case you can compute a determinant to find the eigenvalues and then row reduce to find their eigenvectors but in the infinite-dimensional case you really just have to solve the equation! Is there a particular $K$ you're struggling with? – Qiaochu Yuan May 30 '12 at 18:51
  • isn't there a general method to get eigenfunctions ??, i have searched many books of integral equation and found little info about it – Jose Garcia May 30 '12 at 20:05
  • Again, it depends on what $K$ is. I don't think there's anything reasonable that can be said about the general problem. For an arbitrary $K$ the eigenfunctions should not be expressible in terms of elementary functions, so what does it even mean to know what they are? – Qiaochu Yuan May 30 '12 at 20:06
  • what if $k(x,t)=e^{-(x-y)^2/\sigma^{2}}$ ? – Mirar Apr 06 '22 at 13:30

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