Convolution integral equation

Remarks

Example: take $k(s) = e^{-\abs{s}}$, $\lambda = \frac{1}{4}$ and $f(x) = \frac{1}{2} e^{-\abs{x}}$. The Fourier transform of $e^{-a\abs{s}}$ is $\frac{2a}{a^2 + \xi^2}$ for $a > 0$, so the transformed equation reads $\hat\varphi(\xi) \left(1 - \frac{1}{2(1 + \xi^2)}\right) = \frac{1}{1 + \xi^2}$. Hence $\hat\varphi(\xi) = \frac{2}{1 + 2\xi^2} = \frac{1}{1/2 + \xi^2}$, and $\varphi(x) = \frac{1}{\sqrt{2}} e^{-\abs{x}/\sqrt{2}}$.

Non-example: the equation $\varphi(x) - \int_{-\infty}^{\infty} e^{-x^2 - t^2} \varphi(t) \, dt = f(x)$ is a linear integral equation, but its kernel $e^{-x^2 - t^2}$ is not a function of $x - t$ alone, so it is not of convolution type.

The equation on the half-line $\varphi(x) - \int_0^\infty k(x - t) \varphi(t) \, dt = f(x)$, $x > 0$, is the Wiener–Hopf equation. The Volterra form $\varphi(x) - \int_0^x k(x - t) \varphi(t) \, dt = f(x)$ is the Volterra equation of convolution type.