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Let $p \ge 1$ be an integer and let $\omega >0$, $\gamma>0$, $\bar{t}_c > \omega$ and $\mu_0 >0$ be real numbers. The question is to solve the following inhomogeneous differential equation. \begin{equation} \frac{d^{p+1} r_t}{d t^{p+1}} + \sum\limits_{i=1}^p \frac{d^{p-i}}{d t^{p-i}}\left(\frac{\omega^{2 i}}{t^{2 i}} r_t\right) = (-1)^{p+1} \mu_0 \gamma^{(p)} \bar{t}_c^{\gamma} \cdot t^{-\gamma-p} \end{equation} In case $p=1$ the solution reads: \begin{eqnarray} &&r_t = \frac{1}{\bar{\omega}} \left[-r_0 \sqrt{\frac{t}{t_0}} \omega \sin\left[\bar{\omega} \log(\frac{t}{t_0}) - \phi\right] + \mu_0 \sqrt{t_0 t}\sin\left[\bar{\omega} \log(\frac{t}{t_0})\right]\right] +\\ &&\frac{\mu_0 \gamma \bar{t}_c^{\gamma}}{\omega^2+\gamma^2-\gamma}\left[t^{1-\gamma} - \frac{t_0^{1/2-\gamma} \sqrt{t}}{2 \bar{\omega}} \sin\left(\bar{\omega} \log\left(\frac{t}{t_0}\right)+\theta\right)\right] \end{eqnarray} subject to $r(t_0)=r_0$ and $r^{'}(t_0) = \mu_0$. Here $\bar{\omega} := \sqrt{\omega^2 - 1/4}$ and $\cos(\phi) = 1/(2 \omega)$ and $\sin(\phi) = \bar{\omega}/\omega$ and $\cos(\theta) = (1/2-\gamma)/\sqrt{\omega^2+\gamma^2-\gamma}$. I have obtained this solution in a usual way, ie by solving the homogeneous equation first and then by constructing a special solution to the inhomogeneous equation via Greens functions. The solution above is a waveform with a ``power-law term'' proportional to $t^{1-\gamma}$ and a log-periodic oscillation term proportional to $\sqrt{t} \sin(\bar{\omega} \log(t/t_0) + \phi_1)$ -- see below. Solution to the ODE for following parameter values. $\omega = 13.$, $t_c = 1000.$, $\gamma=1.1$ and $(r_0,\mu_0) = (0.02,0.03)/256.$

Now, the question is how do I go about solving my ODE for $p>1$. In particular I am interested what happens in the limit $p\rightarrow \infty$.

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  • $\begingroup$ What is $\gamma^{(p)}$? $\endgroup$
    – Ron Gordon
    Commented Jan 20, 2015 at 17:01
  • $\begingroup$ This is the Pochammer symbol. $\gamma^{(p)} := \gamma (\gamma+1) \cdots \dots \cdot (\gamma+p-1)$. $\endgroup$
    – Przemo
    Commented Jan 20, 2015 at 18:19

1 Answer 1

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Here we solve the case $p=2$ . We are using the power-series method to solve the homogeneous equation first. We assume that
\begin{equation} r_t = \sum\limits_{n=0}^\infty a_n t^{-n+\alpha} \end{equation} Inserting this into the ODE we get: \begin{equation} a_n (-n+\alpha)_{(3)} t^{-n+\alpha-3} + a_n \omega^2 (-n+\alpha-2)_{(1)} t^{-n+\alpha-3} + \omega^4 a_n t^{-n+\alpha-4} = 0 \end{equation} for $n=0,1,2,\cdots$. We obtain the indicial equation by equating the coefficient at the highest possible power (namely at the power $\alpha-3$) to zero. We have: \begin{equation} (\alpha)_{(3)} + \omega^2 (\alpha-2)_{(1)}=0 \quad\Leftrightarrow \quad \alpha \in \left\{2, \alpha_\pm\right\} \end{equation} where \begin{equation} \alpha_\pm := \frac{1}{2}\left(1 \pm \sqrt{1 - 4 \omega^2}\right) \end{equation} The recursion relations for the coefficients read: \begin{eqnarray} a_n &=& (-\omega^4) \frac{a_{n-1}}{(-n+\alpha)_{(3)} + \omega^2 (-n+\alpha-2)_{(1)}} \\ &=& (\omega^4) \frac{a_{n-1}}{(n-(\alpha-2))(n-(\alpha-\alpha_+))(n-(\alpha-\alpha_-))} \\ &=& (\omega^4)^n \frac{a_0}{[1-(\alpha-2)]^{(n)} [1-(\alpha-\alpha_+)]^{(n)} [1-(\alpha-\alpha_-)]^{(n)}} \end{eqnarray} for $n=1,2,\cdots,\infty$. The fundamental solutions to the inhomogeneous ODE therefore read: \begin{equation} f_{0,\pm}(t) := t^{\alpha} F_{1,3} \left[ \begin{array}{rrr} 1 \\ 3-\alpha & 1-\alpha+\alpha_+ & 1- \alpha+ \alpha_- \end{array}; \frac{\omega^4}{t} \right] \end{equation} where $\alpha \in \left\{2,\alpha_\pm\right\}$. Before we proceed any further a sanity check. In the limit $t \rightarrow \infty$ we retrieve the solutions to the $p=1$ equation (as it should be). There is also an additional solution $t^2$ which stems from the fact that the equation is now of third order. Now we construct a special solution to the inhomogeneous equation. It of course reads: \begin{equation} r_t = \int\limits_{t_0}^\infty G(t,\xi) rhs(\xi) d \xi \end{equation} where $rhs(t)$ is the right hand side of our ODE and $G(t,\xi)$ is the Greens function of our ODE. The Greens function reads: \begin{equation} G(t,\xi) = \left\{ \begin{array}{rr} C^{0}_+ f_0(t) + C^{+}_+ f_+(t) + C^{-}_+ f_-(t) & \mbox{for $t > \xi$} \\ C^{0}_- f_0(t) + C^{+}_- f_+(t) + C^{-}_- f_-(t) & \mbox{for $t < \xi$} \\ \end{array} \right. \end{equation} where the constants depend on $\xi$ only. We determine the constants from the continuity of the Green's function and its first derivative and from the jump in the second derivative all at $t=\xi$. In other words we have: \begin{equation} \left(\begin{array}{c} C_+^{0} - C_-^{0} \\ C_+^{+} - C_-^{+} \\ C_+^{-} - C_-^{-} \end{array} \right)= \left( \begin{array}{rrr} f_0(\xi) & f_+(\xi) & f_-(\xi) \\ f_0^{'}(\xi) & f_+^{'}(\xi) & f_-^{'}(\xi) \\ f_0^{''}(\xi) & f_+^{''}(\xi) & f_-^{''}(\xi) \end{array} \right)^{-1} \cdot \left(\begin{array}{r} 0 \\ 0 \\ 1\end{array}\right) = \frac{1}{{\mathcal W}[f_0,f_\pm](\xi)} \left( \begin{array}{r} f_+(\xi) f_-^{'}(\xi) - f_-(\xi) f_+^{'}(\xi) \\ -f_0(\xi) f_-^{'}(\xi) + f_-(\xi) f_0^{'}(\xi) \\ f_0(\xi) f_+^{'}(\xi) - f_+(\xi) f_0^{'}(\xi) \end{array} \right) \end{equation} where ${\mathcal W}[f_0,f_\pm](\xi)$ is the Wronskian of the system of fundamental solutions. Out of the six constants we have eliminated three. The remaining three constants are eliminated from the boundary conditions $r(t_0) = r_0$, $r^{'}(t_0) = v_0$ and $r^{''}(t_0) = a_0$. Putting everything together we get: \begin{eqnarray} &&r_t = \left(\begin{array}{r} f_0(t) & f_+(t) & f_-(t)\end{array}\right) \cdot \left( \begin{array}{rrr} f_0(t_0) & f_+(t_0) & f_-(t_0) \\ f_0^{'}(t_0) & f_+^{'}(t_0) & f_-^{'}(t_0) \\ f_0^{''}(t_0) & f_+^{''}(t_0) & f_-^{''}(t_0) \\ \end{array} \right)^{-1} \cdot \left(\begin{array}{r} r_0 \\ v_0 \\ a_0\end{array} \right) +\\ && \int\limits_{t_0}^t \left(\begin{array}{r} f_0(t) & f_+(t) & f_-(t)\end{array}\right) \cdot \left( \begin{array}{r} f_+(\xi) f_-^{'}(\xi) - f_-(\xi) f_+^{'}(\xi) \\ -f_0(\xi) f_-^{'}(\xi) + f_-(\xi) f_0^{'}(\xi) \\ f_0(\xi) f_+^{'}(\xi) - f_+(\xi) f_0^{'}(\xi) \end{array} \right) \cdot \frac{rhs(\xi)}{{\mathcal W}[f_0,f_\pm](\xi)} d \xi \end{eqnarray} Now, all what remains to be done is to calculate the Wronskian and to simplify the solution. Firstly we notice that the Wronskian ${\mathcal W}[f_0,f_\pm](\xi)$ does not depend on $\xi$. Indeed by the generalized Abel's identity we have that the derivative of the logarithm of Wronskian is equal to the coefficient at the $p$th derivative in our ODE. However since that coefficient is zero the said derivative is zero. Since the Wronskian is constant we can evaluate is at any point, in particular at $\xi \rightarrow \infty$ Therefore we have:

\begin{eqnarray} &&{\mathcal W}[f_0,f_\pm](\xi) = \lim\limits_{\xi \rightarrow \infty} {\mathcal W}[f_0,f_\pm](\xi) = \\ &&\left| \begin{array}{r} \xi^\alpha \\ \sum\limits_{q=0}^1 \binom{1}{q} (\alpha)_{(1-q)} \xi^{\alpha-(1-q)} \frac{q!}{(3-\alpha)^{(q)} (1-\alpha+\alpha_+)^{(q)} (1-\alpha+\alpha_-)^{(q)}} \frac{(-\omega^4)^q}{\xi^{2q}} \\ \sum\limits_{q=0}^2 \binom{2}{q} (\alpha)_{(2-q)} \xi^{\alpha-(2-q)} \frac{q!}{(3-\alpha)^{(q)} (1-\alpha+\alpha_+)^{(q)} (1-\alpha+\alpha_-)^{(q)}} \frac{(-\omega^4)^q}{\xi^{2q}} \end{array} \right| = -\sqrt{1-4 \omega^2} (2+\omega^2) + O(\frac{1}{\xi}) \end{eqnarray} Here $\alpha = \left\{2,\alpha_+,\alpha_-\right\}$. In the calculation above we used the fact that \begin{equation} \frac{d}{d t} F_{p,q}\left[ \begin{array}{rrr} a_1 & \cdots & a_p \\ b_1 & \cdots & b_q \end{array}; t \right] = \frac{a_1\cdot \dots \cdot a_p}{b_1 \cdot \dots \cdot b_q} F_{p,q} \left[ \begin{array}{rrr} 1+a_1 & \cdots & 1+a_p \\ 1+b_1 & \cdots & 1+b_q \end{array}; t \right] \end{equation}

Now, as another sanity check we will analyze the integral term in the limit $t\rightarrow \infty$. The term reads: \begin{eqnarray} &&\frac{-\mu_0 \gamma^{(2)} \bar{t}_c^\gamma}{-\sqrt{1-4 \omega^2} (2+\omega^2)} \cdot \left( \begin{array}{r} t^2 & t^{\alpha_+} & t^{\alpha_-} \end{array} \right) \cdot \int\limits_{t_0}^t \left( \begin{array}{r} (\alpha_- -\alpha_+) \xi^0 \\ (-\alpha_-+2) \xi^{1+\alpha_-} \\ (\alpha_+-2) \xi^{1+\alpha_+}\end{array} \right) \cdot \xi^{-\gamma-2} d\xi = \\ &&\frac{-\mu_0 \gamma^{(2)} \bar{t}_c^\gamma}{-\sqrt{1-4 \omega^2} (2+\omega^2)} \cdot \left( \begin{array}{r} t^2 & t^{\alpha_+} & t^{\alpha_-} \end{array} \right) \cdot \left. \left( \begin{array}{r} (\alpha_- -\alpha_+) \frac{\xi^{-\gamma-1}}{(-\gamma-1)} \\ (-\alpha_-+2) \frac{\xi^{-\gamma+\alpha_-}}{(-\gamma+\alpha_-)} \\ (\alpha_+-2) \frac{\xi^{-\gamma+\alpha_+}}{(-\gamma+\alpha_+)}\end{array} \right) \right|_{t_0}^t = \\ &&\frac{\mu_0 \gamma \bar{t}_c^{\gamma}}{\omega^2+\gamma^2-\gamma} \cdot t^{1-\gamma} - \frac{\mu_0 \gamma \bar{t}_c^{\gamma}}{\omega^2+2} \cdot t_0^{-1-\gamma} t^2 +\\ &&\frac{\mu_0 \gamma^{(2)} \bar{t}_c^\gamma}{\bar{\omega} (2+\omega^2)} \cdot t^{1/2} t_0^{1/2-\gamma} \cdot Im\left[\left(\frac{t}{t_0}\right)^{\imath \bar{\omega}} \frac{3+2 \imath \bar{\omega}}{-1+2 \gamma + 2 \imath \bar{\omega}}\right] \end{eqnarray} We clearly retrieve both the drift term $t^{1-\gamma}$ and the oscillation term $t^{1/2} \sin(\bar{\omega} \log(t/t_0) + \phi_1)$. These are the terms that appeared in the $p=1$ solution. However there is now another drift term $t^2$ which didn't appear before.

Now, we will evaluate the integral on the right hand side of our solution. Firstly note that this term can be written as follows. \begin{equation} {\mathcal I} = \frac{\mu_0 \gamma^{(2)} \bar{t}_c^{\gamma}}{\sqrt{1-4 \omega^2} (2+\omega^2)} {\mathcal J}_{-\gamma-2} \end{equation} where \begin{equation} {\mathcal J}_{-\gamma-2} := \int\limits_{t_0}^t \left| \begin{array}{c} f_{0,\pm}(\xi) \\ f_{0,\pm}^{'}(\xi) \\ f_{0,\pm}(t) \end{array} \right| \xi^{-\gamma-2} d\xi \end{equation}

where the integrand is a determinant of a three by three matrix times a power law. We integrate by parts. \begin{eqnarray} {\mathcal J}_{-\gamma-2} &=& \left.\left|\begin{array}{c} f_{0,\pm}(\xi) \\ f_{0,\pm}^{'}(\xi) \\ f_{0,\pm}(t) \end{array} \right| \frac{\xi^{-\gamma-1}}{\gamma+1}\right|_t^{t_0} + \int\limits_{t_0}^t \left| \begin{array}{c} f_{0,\pm}(\xi) \\ f_{0,\pm}^{''}(\xi) \\ f_{0,\pm}(t) \end{array} \right| \frac{\xi^{-\gamma-1}}{\gamma+1} d\xi \\ &=& \left.\left|\begin{array}{c} f_{0,\pm}(\xi) \\ f_{0,\pm}^{'}(\xi) \\ f_{0,\pm}(t) \end{array} \right| \frac{\xi^{-\gamma-1}}{\gamma+1}\right|_t^{t_0} + \left.\left|\begin{array}{c} f_{0,\pm}(\xi) \\ f_{0,\pm}^{''}(\xi) \\ f_{0,\pm}(t) \end{array} \right| \frac{\xi^{-\gamma}}{\gamma(\gamma+1)}\right|_t^{t_0} + \int\limits_{t_0}^t \left\{ \left| \begin{array}{c} f_{0,\pm}^{'}(\xi) \\ f_{0,\pm}^{''}(\xi) \\ f_{0,\pm}(t) \end{array} \right| + \left| \begin{array}{c} f_{0,\pm}(\xi) \\ f_{0,\pm}^{'''}(\xi) \\ f_{0,\pm}(t) \end{array} \right| \right\} \frac{\xi^{-\gamma}}{\gamma(\gamma+1)} d\xi \end{eqnarray} Since the functions $f_{0,\pm}$ satisfy the homogeneous ODE the second term in the curly brackets is simply equal to $-\omega^2/(\gamma(\gamma+1)) {\mathcal J}_{-\gamma-2}$. In order to evaluate the first term in the curly brackets we integrate by parts again. This produces another surface term and -- again on the grounds that our functions satisfy the homogeneous ODE -- a term $(\omega^4 {\mathcal J}_{-\gamma-3} - 2 \omega^2 {\mathcal J}_{-\gamma-2})/((\gamma+1)_{(3)})$.

Bringing everything together we get a following recursion relation: \begin{eqnarray} &&{\mathcal J}_{-\gamma-2} \left[1 + \frac{\omega^2}{(\gamma+1)_{(2)}} + \frac{2 \omega^2}{(\gamma+1)_{(3)}}\right] - {\mathcal J}_{-\gamma-3} \frac{\omega^4}{(\gamma+1)_{(3)}} =\\ && \left.\left|\begin{array}{c} f_{0,\pm}(\xi) \\ f_{0,\pm}^{' }(\xi) \\ f_{0,\pm}(t) \end{array} \right| \frac{\xi^{-\gamma-1}}{\gamma+1}\right|_t^{t_0} + \left.\left|\begin{array}{c} f_{0,\pm}(\xi) \\ f_{0,\pm}^{''}(\xi) \\ f_{0,\pm}(t) \end{array} \right| \frac{\xi^{-\gamma }}{(\gamma+1)_{(2)}}\right|_t^{t_0} + \left.\left|\begin{array}{c} f_{0,\pm}^{'}(\xi) \\ f_{0,\pm}^{''}(\xi) \\ f_{0,\pm}(t) \end{array} \right| \frac{\xi^{1-\gamma}}{(\gamma+1)_{(3)}}\right|_t^{t_0} + \end{eqnarray}

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