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8.12: Appendix III- General Linear Autonomous Inhomogeneous ODEs

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    18746
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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

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    +\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

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    {\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 \[C\ns_{jj'}(t,t')=\sum_{\sigma,\sigma'} {\nu_{\!\sigma\ns}^{j-1}\over P'(\nu\nd_{\!\sigma\ns})}\,{\nu_{\!\sigma'}^{j'-1}\over P'(\nu\nd_{\!\sigma'})} \int\limits_{-\infty}^\infty\!\!{d\omega\over 2\pi}\> { {\hat\phi} (\omega)\, e^{-i\omega (t-t')} \over (\omega - i\nu\ns_{\!\sigma}) (\omega + i\nu\nd_{\!\sigma'}) }\ .\]

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