Hey there! Let's dig into this first-kind Fredholm integral equation you're stuck on. First, let's restate it clearly for reference:
$$
\int_{-1}^{1} \frac{f(\xi)}{h^2 + (x-\xi)^2} d\xi = 1 \quad (x \in [-1,1],\ h < 1)
$$

I totally get the frustration of hunting through Polyanin's handbook without finding a closed-form solution—this one has a tricky kernel that doesn't fit the standard cases neatly. Let's break down both closed-form exploration paths and improved numerical approaches that might work better for you.

Closed-Form Solution Exploration #

The kernel $\frac{1}{h^2+(x-\xi)2}$ has deep ties to complex analysis and Fourier transforms, which we can leverage to try and derive a closed-form:

• Fourier Transform Approach: Treat the equation as a truncated convolution (since $f(\xi)$ is zero outside $[-1,1]$). Take the Fourier transform of both sides:
  • The Fourier transform of the kernel $K(t) = \frac{1}{h^2+t2}$ is a known result: $\hat{K}(k) = \frac{\pi}{h}e^{-h|k|}$
  • The Fourier transform of the right-hand side (the constant 1 on $[-1,1]$) is $\hat{1}(k) = \frac{2\sin k}{k}$
  • Rearranging gives $\hat{f}(k) = \frac{2h\sin k}{\pi k e^{h|k|}}$. The inverse Fourier transform of this gives $f(x)$:
    $$
    f(x) = \frac{h}{\pi^2} \int_{-\infty}^{\infty} \frac{\sin k}{k e^{h|k|}} e^{ikx} dk
    $$
  • Split this integral into $k>0$ and $k<0$ parts, then use known integral formulas for $\int_0^\infty \frac{e^{-bk}\cos(ak)}{k}dk$ and $\int_0^\infty \frac{e^{-bk}\sin(ak)}{k}dk$. This simplifies to a combination of arctangent and logarithmic functions, which counts as a closed-form solution (even if it's not purely elementary).
• Complex Contour Integration: Rewrite the kernel as $\frac{1}{2ih}\left( \frac{1}{x-\xi+ih} - \frac{1}{x-\xi-ih} \right)$. Extend $f(\xi)$ to the complex plane (assuming it's analytic in some domain) and use contour integration + residue theorem to evaluate the integral. This leads to the same closed-form as the Fourier transform method, but leans on complex analysis expertise.

Improved Numerical Approaches #

Since you mentioned standard methods like successive approximation and cosine series expansion didn't work, let's focus on methods tailored to first-kind Fredholm equations (which are ill-posed, so stability is key):

• Chebyshev Collocation Method:
  • Represent $f(x)$ as a linear combination of Chebyshev polynomials of the first kind: $f(x) = \sum_{n=0}^N a_n T_n(x)$
  • Choose Chebyshev nodes $x_j = \cos\left( \frac{\pi j}{N} \right)$ for $j=0,...,N$ as collocation points
  • Substitute into the integral equation, compute the integral at each node using Gaussian quadrature (optimized for Chebyshev polynomials), and solve the resulting linear system for coefficients $a_n$. This works great because Chebyshev polynomials are optimal for approximating smooth functions on $[-1,1]$.
• Tikhonov Regularization:
  • First-kind equations are ill-posed, so small numerical rounding can lead to huge solution errors. Tikhonov regularization adds a penalty term to stabilize the problem:
    $$
    \min_f \left| \int_{-1}^1 \frac{f(\xi)}{h^2+(x-\xi)2}d\xi - 1 \right|{L^2[-1,1]}2 + \lambda |f|{L^2[-1,1]}2
    $$
  • Choose the regularization parameter $\lambda$ using methods like the L-curve criterion or cross-validation to balance fit quality and solution smoothness.
• Kernel-Specific Quadrature:
  • Use the complex form of the kernel to convert the integral into a sum of Cauchy integrals. For numerical evaluation, use quadrature rules designed for Cauchy-type integrals (like Gauss-Chebyshev quadrature for singular integrals) to get more accurate results than standard methods.

内容的提问来源于stack exchange，提问作者boidy