Wiener–Hopf equation

Remarks

The unknown is usually sought in $L^p(0, \infty)$ with $1 \le p \le \infty$, or in a space of bounded or of polynomially growing functions. Kernels such as $\frac{1}{2} E_1(\abs{x})$, with $E_1$ the exponential integral, are also used; this one is in $L^1(\R)$.

Example: $k(x) = \frac{1}{2} e^{-\abs{x}}$ and $f(x) = \frac{1}{2} e^{-x}$. The constant $\varphi = 1$ is a solution. Check: $\int_0^\infty \frac{1}{2} e^{-\abs{x - t}}\, dt = \frac{1}{2}(1 - e^{-x}) + \frac{1}{2} = 1 - \frac{1}{2} e^{-x}$, so $\varphi - \int_0^\infty k(x - t) \varphi(t)\, dt = \frac{1}{2} e^{-x}$. The function $1 + x$ solves the homogeneous equation with the same kernel, so this equation has more than one solution among functions of linear growth.

Non-example: $\varphi(x) - \int_{-\infty}^{\infty} k(x - t)\, \varphi(t)\, dt = f(x)$ for all $x \in \R$ is a convolution equation on the whole line. It is not a Wiener–Hopf equation, and the Fourier transform solves it directly when $1 - \hat k$ has no zeros.

The integral operator $\varphi \mapsto \int_0^\infty k(x - t) \varphi(t)\, dt$ is the continuous analogue of a Toeplitz matrix, and it is called a Wiener–Hopf operator.