---
tag: 199B14MWGC
kind: definition
name: Fourier transform method for convolution equations
msc: ["45E10", 42A38]
defines: NRG03YP2PW
uses: [EY75YD988Z, DPVM34ADZV, SASP14MA1Z]
sources:
  - url: https://en.wikipedia.org/wiki/Fourier_transform
---

## Definition

Let $k \in L^1(\R)$, let $f \in L^2(\R)$, and let $\lambda \in \C$.
Write $\hat g(\xi) = \int_{-\infty}^{\infty} g(x) e^{-i x \xi} \, dx$ for the [[Fourier transform|DPVM34ADZV]] of $g$.
The *Fourier transform method* for the [[convolution integral equation|EY75YD988Z]]
$$\varphi(x) - \lambda \int_{-\infty}^{\infty} k(x - t) \, \varphi(t) \, dt = f(x), \quad x \in \R,$$
with unknown $\varphi \in L^2(\R)$ consists of the following steps.

1. Apply the Fourier transform to both sides; by the convolution theorem the equation becomes $(1 - \lambda \hat k(\xi)) \, \hat\varphi(\xi) = \hat f(\xi)$ for almost every $\xi \in \R$.
2. Solve for $\hat\varphi(\xi) = \hat f(\xi) / (1 - \lambda \hat k(\xi))$.
3. Recover $\varphi$ from $\hat\varphi$ with the inverse Fourier transform.

The function $\xi \mapsto 1 - \lambda \hat k(\xi)$ is the *symbol* of the equation.

## Remarks

If $1 - \lambda \hat k(\xi) \ne 0$ for every $\xi \in \R$, the method gives the unique solution in $L^2(\R)$.
Indeed $\hat k$ is continuous and tends to $0$ at infinity by the [[Riemann–Lebesgue lemma|WVKDC011KP]], so $\abs{1 - \lambda \hat k}$ is bounded below by a positive constant.
Then $\hat f / (1 - \lambda \hat k)$ lies in $L^2(\R)$, and the [[Plancherel theorem|HPZHRZ0502]] makes every step reversible.

Example: $k(s) = e^{-\abs{s}}$, $\lambda = \frac14$, $f(x) = \frac12 e^{-\abs{x}}$.
Here $\hat k(\xi) = \frac{2}{1 + \xi^2}$, the symbol $1 - \frac{1}{2(1 + \xi^2)}$ is at least $\frac12$, and step 2 gives $\hat\varphi(\xi) = \frac{1}{1/2 + \xi^2}$, so $\varphi(x) = \frac{1}{\sqrt 2} e^{-\abs{x}/\sqrt 2}$.

Non-example: for $k(s) = \frac12 e^{-\abs{s}}$ and $\lambda = 1$ the symbol $\frac{\xi^2}{1 + \xi^2}$ vanishes at $\xi = 0$.
Step 2 then gives a function $\hat f(\xi) (1 + \xi^2)/\xi^2$ that need not be in $L^2(\R)$, so the method does not produce a solution for every $f \in L^2(\R)$.

The method needs the integral to run over the whole group $\R$.
On a half-line one gets a [[Wiener–Hopf equation|QYY4NMDY5H]], which needs the [[Wiener–Hopf method|H4NXEY6QTZ]] instead.
For equations on $[0, \infty)$ of Volterra type, the [[Laplace transform method|EHGECREYB4]] plays the same role.