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# 8.12: Appendix III- General Linear Autonomous Inhomogeneous ODEs

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We can also solve general autonomous linear inhomogeneous ODEs of the form ${d^n\! x\over dt^n} + a\nd_{n-1}\,{d^{n-1} x\over dt^{n-1}}+ \ldots + a\nd_1\,{dx\over dt} + a\nd_0\,x=\xi(t)\ .$ We can write this as ${\cal L}\nd_t\,x(t)=\xi(t)\ , \label{Leqn}$ where $${\cal L}\nd_t$$ is the $$n^\ssr{th}$$ order differential operator ${\cal L}\nd_t={d^n\over dt^n} + a\nd_{n-1}\,{d^{n-1}\over dt^{n-1}} + \ldots + a\nd_1\,{d\over dt}+a\nd_0\ .$ The general solution to the inhomogeneous equation is given by $x(t)=x\nd_{h}(t)+\!\impi dt'\>G(t,t')\,\xi(t')\ ,\label{time}$ where $$G(t,t')$$ is the Green’s function. Note that $${\cal L}\nd_t\,x\nd_{h}(t)=0$$. Thus, in order for eqns. [Leqn] and [time] to be true, we must have ${\cal L}\nd_t\, x(t)=\stackrel ParseError: EOF expected (click for details) Callstack: at (Bookshelves/Thermodynamics_and_Statistical_Mechanics/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/08:_Nonequilibrium_Phenomena/8.12:_Appendix_III-_General_Linear_Autonomous_Inhomogeneous_ODEs), /content/body/p[2]/span[1], line 1, column 5  {\overbrace ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Thermodynamics_and_Statistical_Mechanics/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/08:_Nonequilibrium_Phenomena/8.12:_Appendix_III-_General_Linear_Autonomous_Inhomogeneous_ODEs), /content/body/p[2]/span[2], line 1, column 1  +\impi dt'\>{\cal L}\nd_t\,G(t,t')\,\xi(t')=\xi(t)\ ,$ which means that ${\cal L}\nd_t\,G(t,t')=\delta(t-t')\ ,$ where $$\delta(t-t')$$ is the Dirac $$\delta$$-function.

If the differential equation $${\cal L}\nd_t\,x(t)=\xi(t)$$ is defined over some finite or semi-infinite $$t$$ interval with prescribed boundary conditions on $$x(t)$$ at the endpoints, then $$G(t,t')$$ will depend on $$t$$ and $$t'$$ separately. For the case we are now considering, let the interval be the entire real line $$t\in (-\infty,\infty)$$. Then $$G(t,t')=G(t-t')$$ is a function of the single variable $$t-t'$$.

Note that $${\cal L}\nd_t={\cal L}\big(\frac{d}{dt}\big)$$ may be considered a function of the differential operator $$\frac{d}{dt}$$. If we now Fourier transform the equation $${\cal L}\nd_t\,x(t)=\xi(t)$$, we obtain $\begin{split} \impi\!dt\,e^{i\omega t}\,\xi(t)&=\impi dt\,e^{i\omega t}\,\bigg\{{d^n\over dt^n} + a\nd_{n-1}\,{d^{n-1}\over dt^{n-1}} +\ldots + a\nd_1\,{d\over dt}+a\nd_0\bigg\}\,x(t)\\ &=\impi dt\,e^{i\omega t}\,\Bigg\{(-i\omega)^n + a\nd_{n-1}\,(-i\omega)^{n-1} + \ldots + a\nd_1\,(-i\omega)+a\nd_0\bigg\}\,x(t)\ . \end{split}$ Thus, if we define ${\hat{\cal L}}(\omega)=\sum_{k=0}^n a\nd_k\,(-i\omega)^k\ ,$ then we have ${\hat{\cal L}}(\omega)\,{\hat x}(\omega)={\hat\xi}(\omega)\ ,$ where $$a\nd_n\equiv 1$$. According to the Fundamental Theorem of Algebra, the $$n^\ssr{th}$$ degree polynomial $${\hat{\cal L}}(\omega)$$ may be uniquely factored over the complex $$\omega$$ plane into a product over $$n$$ roots: ${\hat{\cal L}}(\omega)=(-i)^n\, (\omega-\omega_1)(\omega-\omega_2) \cdots(\omega-\omega_n)\ .$ If the $$\{a\nd_k\}$$ are all real, then $$\big[{\hat{\cal L}}(\omega)\big]^*={\hat{\cal L}}(-\omega^*)$$, hence if $$\Omega$$ is a root then so is $$-\Omega^*$$. Thus, the roots appear in pairs which are symmetric about the imaginary axis. if $$\Omega=a+ib$$ is a root, then so is $$-\Omega^*=-a+ib$$.

The general solution to the homogeneous equation is $x\nd_{h}(t)=\sum_{\sigma=1}^n A\ns_\sigma\,e^{-i\omega_\sigma t}\ ,$ which involves $$n$$ arbitrary complex constants $$A_i$$. The susceptibility, or Green’s function in Fourier space, $${\hat G}(\omega)$$ is then ${\hat G}(\omega)={1\over{\hat{\cal L}}(\omega)}={i^n\over (\omega-\omega_1)(\omega-\omega_2)\cdots(\omega-\omega_n)}\ ,$ Note that $$\big[{\hat G}(\omega)\big]^*={\hat G}(-\omega)$$, which is equivalent to the statement that $$G(t-t')$$ is a real function of its argument. The general solution to the inhomogeneous equation is then $x(t)=x\nd_{h}(t)+\impi \! dt'\>G(t-t')\,\xi(t')\ ,$ where $$x\nd_{h}(t)$$ is the solution to the homogeneous equation, with zero forcing, and where $\begin{split} G(t-t')&=\int\limits_{-\infty}^\infty\!\! {d\omega\over 2\pi}\,e^{-i\omega (t-t')}\,{\hat G}(\omega)\\ &=i^n\!\impi \, {d\omega\over 2\pi}\,{e^{-i\omega (t-t')}\over (\omega-\omega_1) (\omega-\omega_2)\cdots(\omega-\omega_n)}\\ &=\sum_{\sigma=1}^n {e^{-i\omega\nd_\sigma (t-t')}\over i\,{\cal L}'(\omega\nd_\sigma)}\,{\RTheta}(t-t')\ , \label{gfun} \end{split}$ where we assume that $${Im}\,\omega\nd_\sigma <0$$ for all $$\sigma$$. This guarantees causality – the response $$x(t)$$ to the influence $$\xi(t')$$ is nonzero only for $$t > t'$$.

As an example, consider the familiar case \begin{aligned} {\hat{\cal L}}(\omega)&=-\omega^2-i\gamma\omega+\omega_0^2\nonumber\\ &=-(\omega-\omega\nd_+)\,(\omega-\omega\nd_-)\ ,\end{aligned} with $$\omega\nd_\pm=-\frac{i}{2}\gamma\pm \beta$$, and $$\beta=\sqrt{\omega_0^2-\fourth\gamma^2\,}$$. This yields ${\cal L}'(\omega\nd_\pm)=\mp(\omega_+-\omega_-)=\mp 2\beta\ .$ Then according to equation [gfun], $\begin{split} G(s)&=\Bigg\{ {e^{-i\omega\nd_+ s}\over i {\cal L}'(\omega\nd_+)} + {e^{-i\omega\nd_- s}\over i {\cal L}'(\omega\nd_-)} \Bigg\}\,{\RTheta}(s)\\ &=\bigg\{ {e^{-\gamma s/2}\,e^{-i\beta s}\over -2i\beta} + {e^{-\gamma s/2}\,e^{i\beta s}\over 2i\beta} \bigg\}\bvph\,{\RTheta}(s)\\ &=\beta^{-1}\,e^{-\gamma s/2}\,\sin(\beta s)\,{\RTheta}(s)\ . \end{split}$

Now let us evaluate the two-point correlation function $$\big\langle x(t)\,x(t') \big\rangle$$, assuming the noise is correlated according to $$\big\langle \xi(s)\,\xi(s') \big\rangle=\phi(s-s')$$. We assume $$t,t'\to\infty$$ so the transient contribution $$x\ns_{h}$$ is negligible. We then have $\begin{split} \big\langle x(t)\,x(t')\big\rangle & = \impi ds\!\impi ds'\>G(t-s)\,G(t'-s')\,\big\langle \xi(s)\,\xi(s') \big\rangle\\ &=\impi{d\omega\over 2\pi}\>{\hat\phi}(\omega)\, \big|{\hat G}(\omega)\big|^2 \,e^{i\omega(t-t')}\ . \end{split}$

## Higher order ODEs

Note that any $$n^\ssr{th}$$ order ODE, of the general form ${d^n\! x\over dt^n}=F\bigg(x\,,\,{dx\over dt}\,,\,\ldots\,,\,{d^{n-1}\!x\over dt^{n-1}}\bigg)\ ,$ may be represented by the first order system $${\dot\Bvphi}=\BV(\Bvphi)$$. To see this, define $$\varphi\nd_k=d^{k-1}\!x/dt^{k-1}$$, with $$k=1,\ldots,n$$. Thus, for $$k<n$$ we have $${\dot\varphi}\nd_k=\varphi\nd_{k+1}$$, and $${\dot\varphi}\nd_n=F$$. In other words, $\stackrel ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Thermodynamics_and_Statistical_Mechanics/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/08:_Nonequilibrium_Phenomena/8.12:_Appendix_III-_General_Linear_Autonomous_Inhomogeneous_ODEs), /content/body/div/p[1]/span, line 1, column 1  {\overbrace{ {d\over dt}\begin{pmatrix} \varphi\nd_1\\ \vdots \\ \varphi\nd_{n-1}\\ \varphi\nd_n \end{pmatrix}}}= \stackrel{\BV(\Bvphi)}{\overbrace{\begin{pmatrix}\varphi\nd_2\\ \vdots \\ \varphi\nd_n \\F\big(\varphi\nd_1,\ldots,\varphi\nd_p\big)\end{pmatrix}}}\ .$

An inhomogeneous linear $$n^\ssr{th}$$ order ODE, ${d^n\!x\over dt^n} + a\nd_{n-1}\,{d^{n-1}\!x\over dt^{n-1}} + \ldots + a\nd_1\, {dx\over dt} + a\nd_0\,x=\xi(t) \label{ILODE}$ may be written in matrix form, as ${d\over dt}\begin{pmatrix} \varphi\nd_1\\ \varphi\nd_2 \\ \vdots \\ \varphi\nd_n \end{pmatrix}= \stackrel{Q}{\overbrace{\begin{pmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ -a\nd_0 & -a\nd_1 & -a\nd_2 & \cdots & -a\nd_{n-1} \end{pmatrix} }} \begin{pmatrix} \varphi\nd_1\\ \varphi\nd_2 \\ \vdots \\ \varphi\nd_n \end{pmatrix} + \stackrel{\Bxi}{\overbrace{\begin{pmatrix} 0\\ 0\\ \vdots \\ \xi(t) \end{pmatrix} }} \ . \label{Qxieqn}$ Thus, ${\dot\Bvphi}=Q\,\Bvphi + \Bxi\ , \label{phiQeqn}$ and if the coefficients $$c\nd_k$$ are time-independent, the ODE is autonomous.

For the homogeneous case where $$\xi(t)=0$$, the solution is obtained by exponentiating the constant matrix $$Qt$$: $\Bvphi(t)=\exp(Qt)\,\Bvphi(0)\ ;$ the exponential of a matrix may be given meaning by its Taylor series expansion. If the ODE is not autonomous, then $$Q=Q(t)$$ is time-dependent, and the solution is given by the path-ordered exponential, $\Bvphi(t)={\textsf P}\exp\Bigg\{\int\limits_0^t\!\!dt'\,Q(t')\Bigg\}\,\Bvphi(0)\ ,$ where $${\textsf P}$$ is the path ordering operator which places earlier times to the right. As defined, the equation $${\dot\Bvphi}=\BV(\Bvphi)$$ is autonomous, since the $$t$$-advance mapping $$g\nd_t$$ depends only on $$t$$ and on no other time variable. However, by extending the phase space $${\mathbb M}\ni\Bvphi$$ from $${\mathbb M} \to {\mathbb M}\times{\mathbb R}$$, which is of dimension $$n+1$$, one can describe arbitrary time-dependent ODEs.

In general, path ordered exponentials are difficult to compute analytically. We will henceforth consider the autonomous case where $$Q$$ is a constant matrix in time. We will assume the matrix $$Q$$ is real, but other than that it has no helpful symmetries. We can however decompose it into left and right eigenvectors: $Q\ns_{ij}=\sum_{\sigma=1}^n\,\nu\ns_\sigma\, R\ns_{\sigma,i}\,L\ns_{\sigma,j}\ .$ Or, in bra-ket notation, $$Q=\sum_\sigma \nu\ns_\sigma \,\tket{R_\sigma}\tbra{L_\sigma}$$. The normalization condition we use is $\braket{L\ns_\sigma}{R\ns_{\sigma'}}=\delta\ns_{\sigma\sigma'},$ where $$\big\{\nu\ns_\sigma\big\}$$ are the eigenvalues of $$Q$$. The eigenvalues may be real or imaginary. Since the characteristic polynomial $$P(\nu)=\textsf{det}\,(\nu\,{\mathbb I}-Q)$$ has real coefficients, we know that the eigenvalues of $$Q$$ are either real or come in complex conjugate pairs.

Consider, for example, the $$n=2$$ system we studied earlier. Then $Q=\begin{pmatrix} 0 & 1 \\ -\omega_0^2 & -\gamma \end{pmatrix}\ .$ The eigenvalues are as before: $$\nu\ns_\pm=-\half\gamma\pm\sqrt{\fourth\gamma^2-\omega_0^2\,}$$. The left and right eigenvectors are $L\ns_\pm={\pm 1\over\nu\ns_+-\nu\ns_-}\,\begin{pmatrix} -\nu\ns_\mp & 1 \end{pmatrix} \qquad,\qquad R\ns_\pm=\begin{pmatrix} 1 \\ \nu\ns_\pm \end{pmatrix}\ .$

The utility of working in a left-right eigenbasis is apparent once we reflect upon the result $f(Q)=\sum_{\sigma=1}^n f(\nu\ns_\sigma)\,\,\ket{R_\sigma}\,\bra{L_\sigma}$ for any function $$f$$. Thus, the solution to the general autonomous homogeneous case is \begin{aligned} \ket{\Bvphi(t)}&=\sum_{\sigma=1}^n e^{\nu\ns_\sigma t}\,\ket{R\ns_\sigma}\, \braket{L\ns_\sigma}{\Bvphi(0)}\\ \varphi\ns_i(t)&=\sum_{\sigma=1}^n e^{\nu\ns_\sigma t}\, R\ns_{\sigma,i}\,\sum_{j=1}^n L\ns_{\sigma,j}\,\varphi\ns_j(0)\ .\end{aligned} If $$\textsf{Re}\,(\nu\ns_\sigma)\le 0$$ for all $$\sigma$$, then the initial conditions $$\Bvphi(0)$$ are forgotten on time scales $$\tau\ns_\sigma=\nu^{-1}_\sigma$$. Physicality demands that this is the case.

Now let’s consider the inhomogeneous case where $$\xi(t)\ne 0$$. We begin by recasting Equation [phiQeqn] in the form ${d\over dt}\big(e^{-Qt}\,\Bvphi\big)=e^{-Qt}\,\Bxi(t)\ .$ We can integrate this directly: $\Bvphi(t)=e^{Qt}\,\Bvphi(0) + \int\limits_0^t\!\!ds\,e^{Q(t-s)}\,\Bxi(s)\ .$ In component notation, $\varphi\ns_i(t)=\sum_{\sigma=1}^n e^{\nu\ns_\sigma t}\, R\ns_{\sigma,i}\,\braket{L\ns_\sigma}{\Bvphi(0)} +\sum_{\sigma=1}^n R\ns_{\sigma,i} \int\limits_0^t\!\!ds\>e^{\nu\ns_\sigma(t-s)}\, \braket{L\ns_\sigma}{\Bxi(s)} . \label{CNsoln}$ Note that the first term on the RHS is the solution to the homogeneous equation, as must be the case when $$\Bxi(s)=0$$.

The solution in Equation [CNsoln] holds for general $$Q$$ and $$\Bxi(s)$$. For the particular form of $$Q$$ and $$\xi(s)$$ in Equation [Qxieqn], we can proceed further. For starters, $$\tbraket{L\ns_\sigma}{\Bxi(s)}=L\ns_{\sigma,n}\,\xi(s)$$. We can further exploit a special feature of the $$Q$$ matrix to analytically determine all its left and right eigenvectors. Applying $$Q$$ to the right eigenvector $$\tket{R\ns_\sigma}$$, we obtain $R\ns_{\sigma,j}=\nu\ns_\sigma \, R\ns_{\sigma,j-1} \qquad (j > 1)\ .$ We are free to choose $$R\ns_{\sigma,1}=1$$ for all $$\sigma$$ and defer the issue of normalization to the derivation of the left eigenvectors. Thus, we obtain the pleasingly simple result, $R\ns_{\sigma,k}=\nu_\sigma^{k-1}\ .$ Applying $$Q$$ to the left eigenvector $$\tbra{L\ns_\sigma}$$, we obtain \begin{aligned} -a\ns_0\,L\ns_{\sigma,n}&=\nu\ns_\sigma\,L\ns_{\sigma,1}\\ L\ns_{\sigma,j-1} - a\ns_{j-1}\,L\ns_{\sigma,n}&=\nu\ns_\sigma\,L\ns_{\sigma,j}\qquad (j>1)\ .\end{aligned} From these equations we may derive $L\ns_{\sigma,k}=-{L\ns_{\sigma,n}\over\nu\ns_\sigma}\sum_{j=0}^{k-1} a\ns_j\,\nu_\sigma^{j-k-1} ={L\ns_{\sigma,n}\over\nu\ns_\sigma}\sum_{j=k}^n a\ns_j\,\nu^{j-k-1}_\sigma\ .$ The equality in the above equation is derived using the result $$P(\nu\ns_\sigma)=\sum_{j=0}^n a\ns_j\,\nu_\sigma^j=0$$. Recall also that $$a\ns_n\equiv 1$$. We now impose the normalization condition, $\sum_{k=1}^n L\ns_{\sigma,k}\,R\ns_{\sigma,k}=1\ .$ This condition determines our last remaining unknown quantity (for a given $$\sigma$$), $$L\ns_{\sigma,p}\>$$: $\braket{L\ns_\sigma}{R\ns_\sigma}=L\ns_{\sigma,n}\sum_{k=1}^n k\,a\ns_k\,\nu_\sigma^{k-1}=P'(\nu\ns_\sigma)\, L\ns_{\sigma,n}\ ,$ where $$P'(\nu)$$ is the first derivative of the characteristic polynomial. Thus, we obtain another neat result, $L\ns_{\sigma,n}={1\over P'(\nu\ns_\sigma)}\ .$

Now let us evaluate the general two-point correlation function, $C\ns_{jj'}(t,t')\equiv \big\langle \varphi\ns_j(t)\,\varphi\ns_{j'}(t') \big\rangle - \big\langle \varphi\ns_j(t)\big\rangle\, \big\langle \varphi\ns_{j'}(t')\big\rangle\ .$ We write $\big\langle \xi(s)\,\xi(s') \big\rangle = \phi(s-s')=\!\!\int\limits_{-\infty}^\infty \!\!{d\omega\over 2\pi}\>{\hat\phi}(\omega)\,e^{-i\omega(s-s')}\ .$ When $${\hat\phi}(\omega)$$ is constant, we have $$\big\langle \xi(s)\,\xi(s') \big\rangle = {\hat\phi}(t)\,\delta(s-s')$$. This is the case of so-called white noise, when all frequencies contribute equally. The more general case when $${\hat\phi}(\omega)$$ is frequency-dependent is known as colored noise. Appealing to Equation [CNsoln], we have \begin{aligned} C\ns_{jj'}(t,t')&=\sum_{\sigma,\sigma'} {\nu_{\!\sigma\ns}^{j-1}\over P'(\nu\nd_{\!\sigma\ns})}\,{\nu_{\!\sigma'}^{j'-1}\over P'(\nu\nd_{\!\sigma'})} \int\limits_0^t \!\! ds \> e^{\nu\ns_\sigma(t-s)} \!\!\int\limits_0^{t'} \!\! ds' \> e^{\nu_{\!\sigma'}(t'-s')}\,\phi(s-s')\\ &=\sum_{\sigma,\sigma'} {\nu_\sigma^{j-1}\over P'(\nu\nd_{\!\sigma\ns})}\,{\nu_{\sigma'}^{j'-1}\over P'(\nu\nd_{\!\sigma'})} \int\limits_{-\infty}^\infty\!\!{d\omega\over 2\pi}\> { {\hat\phi} (\omega)\, ( e^{-i\omega t} - e^{\nu\ns_\sigma t} ) ( e^{i\omega t'} - e^{\nu_{\!\sigma'} t'} ) \over (\omega - i\nu\ns_{\!\sigma}) (\omega + i\nu\nd_{\!\sigma'}) }\ .\end{aligned} In the limit $$t,t'\to \infty$$, assuming $$\textsf{Re}\,(\nu\ns_\sigma)<0$$ for all $$\sigma$$ ( no diffusion), the exponentials $$e^{\nu\ns_\sigma t}$$ and $$e^{\nu_{\!\sigma'} t'}$$ may be neglected, and we then have $C\ns_{jj'}(t,t')=\sum_{\sigma,\sigma'} {\nu_{\!\sigma\ns}^{j-1}\over P'(\nu\nd_{\!\sigma\ns})}\,{\nu_{\!\sigma'}^{j'-1}\over P'(\nu\nd_{\!\sigma'})} \int\limits_{-\infty}^\infty\!\!{d\omega\over 2\pi}\> { {\hat\phi} (\omega)\, e^{-i\omega (t-t')} \over (\omega - i\nu\ns_{\!\sigma}) (\omega + i\nu\nd_{\!\sigma'}) }\ .$