I am numerically solving a Fredholm equation of the second kind: \begin{align} f(x) &= g(x) + \int_0^\infty K(x,t)f(t)\,dt, \end{align} by using a Gaussian quadrature rule to convert it into a system of linear equations. For the particular kernel $K$ in this application, the solutions $f(t)$ are not unique, except at a particular point $x_0$ where all solutions pass through the same point. As it happens I only need $f(x_0)$, so the non-uniqueness for $x\neq x_0$ is in theory not a problem. In practice, the non-uniqueness presents itself numerically by different choices of quadrature leading to different solutions, and in some occasions the system becomes highly ill-conditioned, which is often associated with the solution having a large derivative through $x_0$.
In an effort to improve numerical stability, I would like to be able to pick the solution with the minimal derivative through $x_0$, and have started thinking about using Lagrange multipliers.
I can write the derivative at $x_0$ as \begin{align} f_x(x_0) &= \int_0^\infty \left[ K_x(x_0,t)f(t) + g_x(t)\delta(t-x_0)\right]\,dt \end{align} where the subscript $x$ indicates the derivative w.r.t $x$. So I believe I want to minimise this subject to the constraints (for each $x$): \begin{align} \int_0^\infty \left [K(x,t)-\delta(t-x)\right]f(t)\,dt = -g(x). \end{align} I have derived the following Euler-Lagrange equation for the Lagrange multipliers $\lambda(x)$: \begin{align} \left[K(x,t)-\delta(t-x)\right]\lambda(x) &= -K_x(x_0,t) \\ \lambda(x) &= -\frac{K_x(x_0,t)}{K(x,t)-\delta(t-x)}, \end{align} but then I am really not sure where to go next.
Edit: substituting into the equation for $f_x(x_0)$, I get: \begin{align} f_x(x_0) &= g_x(x_0) - \lambda(x)\int_0^\infty\left[K(x,t)-\delta(t-x)\right]f(t)\,dt \\ &= g_x(x_0) - \lambda(x)g(x), \end{align} and then if I assume a solution can be found with $f_x(x_0)=0$, I have $$\lambda(x) = g_x(x_0)/g(x).$$ Then substituting this back into the Euler-Lagrange equation, I get \begin{align} \left[K(x,t)-\delta(t-x)\right] &= -K_x(x_0,t)\frac{g(x)}{g_x(x_0)} , \end{align} and then substituting back into the original equation: \begin{align} \int_0^\infty \left [K(x,t)-\delta(t-x)\right]f(t)\,dt &= -g(x) \\ \int_0^\infty K_x(x_0,t)\frac{g(x)}{g_x(x_0)}f(t)\,dt &= g(x). \end{align} This would appear to give an equation I can solve for $f(t)$ with the constraint $f_x(x_0)=0$ built in. I'm sure there is some fatal flaw in my reasoning though.