FredholmIntegralEquationToSturmLiouvilleform

The process of converting an integral equation to a differential equation, particularly in the case of a Sturm-Liouville problem, involves a few steps. I'll describe it in a generalized form and then touch on the specifics of a translation-invariant kernel.

General Process

Given an integral equation of the form:

$$ \int_a^b K(s, t) f(t) dt = \lambda f(s) $$

where $K(s, t)$ is the kernel, $f(t)$ is the eigenfunction, and $\lambda$ is the eigenvalue.

1. Differentiate Both Sides: If the kernel $K(s, t)$ depends on $s$ and $t$ in a way that allows for differentiation with respect to $s$, you differentiate both sides of the equation with respect to $s$, potentially multiple times if needed, to obtain derivatives of $f$ with respect to $s$.

2. Apply Leibniz's Rule: When differentiating under the integral sign, you apply Leibniz's rule, which can simplify the kernel into a form involving delta functions or other simplifications, especially if the kernel is made up of absolute value functions or other piecewise definitions.

3. Obtain a Differential Operator: The goal is to express the left-hand side as a differential operator acting on $f$. This often involves the kernel transforming into a differential expression involving the function $f$ and its derivatives $f', f'', \ldots$, along with potentially new functions $p(t), q(t), \ldots$ that are derived from the properties of the kernel.

4. Formulate Sturm-Liouville Problem: The result is a differential equation of the form:

$$ \frac{d}{dt}\left[p(t)\frac{df}{dt}\right] + q(t)f = \lambda w(t)f $$

where $p(t)$, $q(t)$, and $w(t)$ are coefficients that depend on the original kernel and the differential operator is self-adjoint.

Translation-Invariant Kernel

For a translation-invariant kernel, which depends only on the difference $s - t$, denoted as $K(s - t)$, and is absolutely continuous and infinitely differentiable, the simplification is as follows:

1. Kernel Form: The kernel can be expressed as $K(s - t)$, which simplifies the analysis since it only depends on one variable $u = s - t$.

2. Fourier Transform: Such kernels are amenable to analysis via the Fourier transform. The Fourier transform of the kernel $\hat{K}(\omega)$ and the eigenfunction $\hat{f}(\omega)$ can be used to convert the integral equation into an algebraic equation.

3. Simplified Differential Operator: Since the kernel is infinitely differentiable, its derivatives with respect to $s$ under the integral sign will not encounter discontinuities, allowing for a straightforward application of Leibniz's rule.

4. Application of Fourier Transform: By taking the Fourier transform of both sides, the convolution theorem can be applied, which states that the Fourier transform of a convolution is the product of the Fourier transforms:

$$ \hat{K}(\omega) \cdot \hat{f}(\omega) = \lambda \hat{f}(\omega) $$

5. Eigenvalue Problem: This yields a simple eigenvalue problem in the frequency domain, where the eigenvalues are related to the Fourier transform of the kernel:

$$ \hat{f}(\omega) = \frac{1}{\lambda} \hat{K}(\omega) \cdot \hat{f}(\omega) $$

The simplifications provided by the translation-invariant, absolutely continuous, and infinitely differentiable properties of the kernel allow for a more straightforward analysis and solution to the problem, often enabling solutions in closed form, especially when leveraging the properties of the Fourier transform.