Volterra series

Remarks

The operator $x \mapsto y$ is causal, since $y(t)$ depends only on $x(s)$ for $s \le t$, and time-invariant, since shifting $x$ in time shifts $y$ by the same amount. The kernels can be taken symmetric in their variables without changing $y$. If only $h_1$ is nonzero, $y = h_0 + h_1 * x$ is a convolution, the output of a linear system with impulse response $h_1$. Some authors also allow kernels with delta functions, or integrals over all of $\R^n$ for noncausal systems.

Example: let $z(t) = \int_0^\infty e^{-\tau} x(t - \tau)\, d\tau$ and $y = z^2$. Then $y(t) = \int_0^\infty \int_0^\infty e^{-\tau_1 - \tau_2}\, x(t - \tau_1)\, x(t - \tau_2)\, d\tau_1\, d\tau_2$ by Fubini's theorem. So $y$ is a Volterra series with $h_2(\tau_1, \tau_2) = e^{-\tau_1 - \tau_2}$, of $L^1$ norm $1$, and all other kernels $0$.

Non-example: the system $y(t) = t\, x(t)$ is not given by any Volterra series, because it is not time-invariant. The input $x = 1$ gives $y(t) = t$, which is not constant, while a Volterra series maps a constant input to a constant output.