# Integral equation with kernel defined in a rectangle

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.

• 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. – DCM Dec 24 '18 at 18:19
• 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? – Giuseppe Negro Dec 25 '18 at 12:10
• 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 $m<n$. – Giuseppe Negro Dec 25 '18 at 12:15
• 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? – Gustave Dec 30 '18 at 14:06