To solve an integral equation numerically, we can apply Nyström's method. Therefor we approximate the integral operator
$(A\varphi)(x) := \int \limits_G K(x,y)\varphi(y)dy,\; x \in G$
by a sequence of numerical integration operators.
$(A_n\varphi)(x) := \sum \limits_{k=1}^n \alpha_k^{(n)}K(x,x_k^{(n)})\varphi(x_k^{(n)}),\; x\in G$
Hence, the solution to the integral equation
$\varphi - A\varphi=f$
is approximated by the solution of
$\varphi_n - A_n \varphi_n = f$
as for exampled described in Rainer Kress' Linear Integral Equations on p. 201.
Apparently, this reduces to solving a finite-dimensional linear system.
However, I fail to see, where this takes place.
Assuming the Kernel
$K(x,y) = (x+1)e^{-x*y}/2$
and the function
$f(x) = e^{-x}+1/2*(e^{-x-1}-1)$
can someone please show me, step by step, how this is done, given the IEQ
$\varphi(x)-\int \limits_0^1K(x,y)\varphi(y)dy=f(x)$, with exact solution $\varphi(x)=e^{-x}$
Therefore let's discretize the interval [0,1] with $x_i$ in {0, 1/3, 2/3, 1} (so $n=4$).
I then get:
$\varphi_4 - A_4 \varphi_4 = f$, that is
$\varphi_4 - \sum \limits_{k=1}^{4}\alpha_k (x+1)e^{-x*x_k^{(4)}}/2\varphi(x_k^{(4)})=e^{-x}+1/2*(e^{-x-1}-1)$
But here, I don't know how to continue.
UPDATE So this yields
$\varphi_4(x) = f(x) + \sum \limits_{k=1}^4 \alpha_k K(x, x_k) \varphi(x_k)$
$\varphi_4(x) = e^{-x}+1/2*(e^{-x-1}-1) + \alpha_1 K(x, 0) * \varphi(0) + \alpha_2 K(x, 1/3) * \varphi(1/3) + \alpha_3 K(x, 2/3) \varphi(2/3) + \alpha_4 K(x, 1) * \varphi(1)$
I'm certain this is false, as I can't see how to find my solution. Where did I do my mistake?