Consider a stochastic differential equation: $$ dx(t) = a x(t)dt + b y(t)dt \quad (1) $$ where $y(t)$ is a stochastic process satisfying $\langle y(t)y(t')\rangle = \delta(t-t')$. We will assume that the first statistical moments are zero and that $a<0$. I am interested in the variance $\langle x(t)^2 \rangle$. When I solve this equation in the frequency domain, I obtain the value $\langle x(t)^2 \rangle = \frac{-b^2}{2a}$. This is the steady state variance,($\lim_{x\rightarrow \infty} \langle x(t)^2 \rangle$) but does not give any information about the transient variance $\langle x(t)^2 \rangle$ at $t\neq \infty$.

My question is: why does solving this equation in the frequency domain not capture the transient behavior of the variance?

More mathematical detail are below:

My question is: why does solving in the frequency domain give the steady state variance, and not a function which depends on $t$?

We can be solve in the time domain to give: $$ \langle x(t)^2 \rangle = e^{2at}\langle x(0)^2 \rangle - \frac{b^2}{2a}(1-e^{2at}) $$ The steady state can then be found by taking $t \rightarrow \infty$.

We can also solve in the frequency domain. Fourier transforming $(1)$ gives: $$ x(\omega) = \frac{b}{-a-i\omega} y(\omega) \quad (2) $$ where $y(\omega)$ now obeys $\langle y(\omega)y(\omega')\rangle = \delta(\omega + \omega')$. We then invert the Fourier transform to find the variance:

$$ \langle x(t)^2 \rangle = \frac{1}{2 \pi} \int^{\infty}_{-\infty}\int^{\infty}_{-\infty}\langle x(\omega)x(\omega')\rangle^{-it(\omega + \omega')} d\omega d\omega' \quad (3) $$ Plugging $(2)$ into $(3)$ and integrating over the delta function gives a value of $\frac{-b^2}{2a}$, which is the steady state variance obtained earlier, but does not contain any information about the transient behavior. Why is this?

  • $\begingroup$ Can you show your time-domain analysis? When Fourier transforming the original differential equation, I dealt with $e^{-i\omega t}dx(t)$ by integrating $d(e^{-i\omega t}x(t)) + i\omega e^{-i\omega}x(t)dt$ and assuming that the boundary terms arising from integrating the first term, are zero. Is that an error? Perhaps $\hat{x}(\omega) = b\hat{y}(\omega)/(i\omega - a) + c$ for some nonzero constant $c$. $\endgroup$
    – Joe Mack
    Commented Jun 18, 2020 at 17:53

1 Answer 1


BOTTOM LINE UP FRONT: I think the exponential decay growth in $\left<|x(t)|^2\right>$ can be shown in the frequency domain only if the "boundary terms" are nonzero when we compute the Fourier transform of $dx(t)$ from the original SDE.

I provide only a start in the work below.

Since these processes seem like they could possibly be complex-valued, I will consider $\left<x(t)\overline{x(t')}\right>$ and $\left<y(t)\overline{y(t')}\right>$, where the $\texttt{overline}$ indicates complex conjugation.
I will write out some steps, because I need to see some things that you did not write explicitly.
\begin{equation} dx(t) = ax(t)dt + by(t)dt. \end{equation} Fourier transform: \begin{equation} \begin{split} \int e^{-i\omega t}dx(t) &=~ \int e^{-i\omega t}ax(t)dt + \int e^{-i\omega t} by(t)dt\\ \int\left[d\left(e^{-i\omega t}x(t)\right) - x(t)d(e^{-i\omega t})\right] &=~ a\widehat{x}(\omega) + b\widehat{y}(\omega)\\ \underbrace{\int d\left(e^{-i\omega t}x(t)\right)}_{\textrm{Boundary terms: $-c$, a constant}} - \int x(t)(-i\omega)e^{-i\omega t}dt &=~ a\widehat{x}(\omega) + b\widehat{y}(\omega)\\ i\omega\widehat{x}(\omega) &=~ a\widehat{x}(\omega) + b\widehat{y}(\omega) + c \end{split} \end{equation} Result: \begin{equation} \widehat{x}(\omega) = \frac{b\widehat{y}(\omega) + c}{i\omega - a} \end{equation}
\begin{equation} \begin{split} x(t) &=~ \frac{1}{2\pi}\int \widehat{x}(\omega) e^{it\omega}d\omega\\ &=~ \frac{1}{2\pi}\int \frac{b\widehat{y}(\omega) + c}{i\omega - a} e^{it\omega}d\omega\\ & \\ x(t') &=~ \frac{1}{2\pi}\int \widehat{x}(\omega') e^{it'\omega'}d\omega'\\ &=~ \frac{1}{2\pi}\int \frac{b\widehat{y}(\omega') + c}{i\omega' - a} e^{it'\omega'}d\omega' \end{split} \end{equation} \begin{equation} \end{equation} \begin{equation} \begin{split} \left<x(t)\overline{x(t')}\right> &=~ \left<\frac{1}{2\pi}\int \frac{b\widehat{y}(\omega) + c}{i\omega - a} e^{it\omega}d\omega\frac{1}{2\pi}\int \frac{\overline{b\widehat{y}(\omega') + c}}{-i\omega' - a} e^{-it'\omega'}d\omega'\right>\\ &=~ \frac{1}{4\pi^2}\int\int\frac{b^2\left<\widehat{y}(\omega)\overline{\widehat{y}(\omega')}\right> + bc\left<\overline{\widehat{y}(\omega')}\right> + b\overline{c}\left<\widehat{y}(\omega)\right> + |c|^2}{(i\omega - a)(-i\omega' - a)}e^{it\omega}e^{-it'\omega'}d\omega d\omega' \end{split} \end{equation}
\begin{equation} \begin{split} \left<\widehat{y}(\omega)\overline{\widehat{y}(\omega')}\right> &=~ \left<\int y(\tau)e^{-i\omega\tau}d\tau\overline{\int y(\tau')e^{-i\omega'\tau'}d\tau'}\right>\\ &=~ \int\int\left<y(\tau)\overline{y(\tau')}\right>e^{-i\omega\tau}e^{i\omega'\tau'}d\tau d\tau'\\ &=~ \int\int\delta(\tau-\tau')e^{-i\omega\tau}e^{i\omega'\tau'}d\tau d\tau'\\ &=~ \int e^{-i(\omega - \omega')\tau}d\tau, \end{split} \end{equation} after integrating in $\tau'$. This integral does not converge, but in the sense of distributions, \begin{equation} \int e^{-i(\omega - \omega')\tau}d\tau = 2\pi\delta(\omega - \omega'). \end{equation}
What to do about the cross-terms? We must consider the integrals in $\omega$ and $\omega'$ separately for those. Consider the one for $\omega$: \begin{equation} \int\frac{\left<\widehat{y}(\omega)\right>}{i\omega - a}e^{i t\omega}d\omega \end{equation} If we assume analyticity of $\left<\widehat{y}(\omega)\right>$ (as a function of complex-valued $\omega$) and assume that it decays rapidly enough for large $|\omega|$ in the complex-$\omega$ plane, then we can appeal to contour integration and Cauchy's integral formula. The only pole in the complex-$\omega$ plane is found at $\omega = -ia$, where I assume $a$ is real. The integral is \begin{equation} \int\frac{\left<\widehat{y}(\omega)\right>}{i\omega - a}e^{i t\omega}d\omega ~=~ \lim_{N\to\infty}\oint_{\gamma_N}\frac{\left<\widehat{y}(\omega)\right>}{i\omega - a}e^{i t\omega}d\omega ~=~ 2\pi i\left<\widehat{y}(-ia)\right>e^{at}, \end{equation} where $\gamma_N$ is a path that includes the real interval $[-N,N]$ and a semi-circular arc connecting $N$ and $-N$. You can find examples of this path in almost any undergraduate complex analysis book. We assume that the integral on this arc decays to zero as $N\to\infty$.
I think this is how we pick up the exponentially decaying growing behavior in $\left<|x(t)|^2\right>$.

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