Let us consider $$f(x) + \lambda \int_0^4 {K(s,x)f(s)ds=0} ,{\text{ x}} \in {\text{(0}}{\text{,1)}}$$ Observe that the kernel is not defined on a square. My question: Can I apply the classical Fredholm theory in a rectangle? Is the theory works? Thanks.

  • 1
    $\begingroup$ Which results are available to you depends more on (a) the properties of your kernel $K$ and (b) the function space in which you're looking for solutions $f$. Could you provide some context about the setting in which you're working? If you're not sure, maybe provide a few details about what you know about your kernel and maybe we can work the rest out from that. I'd imagine that whether your kernel is defined over a square as opposed to a rectangle makes no difference, but cannot say this with confidence without more details about your kernel or the setting in which you intend to work. $\endgroup$ – DCM Dec 24 '18 at 18:19
  • 1
    $\begingroup$ I would expect different behaviour in this "rectangular" case. The problem you wrote cannot be written as $(I+\lambda K)f=0$, because here $K$ maps $L^2(0, 4)$ to $L^2(0,1)$. Thus, I expect the standard Fredholm results to break down. But this is only an educated guess. What result exactly do you need? $\endgroup$ – Giuseppe Negro Dec 25 '18 at 12:10
  • 1
    $\begingroup$ To make a linear algebra analogue; the standard Fredholm equation corresponds to the system of $n\times n$ linear equations $$\lambda K_{n\times n} x_n + x_n = 0, \qquad K\in \mathbb C^{n\times n}, x\in \mathbb C^n.$$ What you have here is more like $$\lambda K_{m\times n} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n\end{bmatrix} + \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_m\end{bmatrix}=0, $$ where $m<n$. $\endgroup$ – Giuseppe Negro Dec 25 '18 at 12:15
  • $\begingroup$ Excuse me for the delay, I was a little occupied. I understood thank you. I have another question: for the first kind fredholm equation, if I have $g(x) = \int\limits_0^1 {k(s,x)f(s)ds} ,{\text{ x}} \in {\text{(0}}{\text{,1/2)}}$ . If I make the substitution $s=t/2$. Am I in the frame of first kind fredholm integral equation? $\endgroup$ – Gustave Dec 30 '18 at 14:06

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.