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 November 3rd, 2015, 07:56 PM #1 Newbie   Joined: Nov 2015 From: Chicago Posts: 2 Thanks: 0 Power series on ODE solution Dear all, I am a bit rusty on power series expansion, and I am stucked with the following problem. I have the following ODE in terms of two continuous differentiable generic functions $\displaystyle u(t)$ and $\displaystyle v(t)$: $\displaystyle \dot{u} = -g_L u + g_D (v(t)-u(t)) + g(t) (E-u)$ where $\displaystyle g_L,g_D,E$ real and >0 and $\displaystyle g(t)$ is continuous and non-negative. The solution outline in my book suggests that under the assumption that $\displaystyle g_D \gg g_L$ it is possible to expand in power series of $\displaystyle 1/g_D$ and obtain up to the leading order: $\displaystyle u = (1-\lambda)E + \lambda v* + O(1/g_D)$ where: $\displaystyle \lambda = (g_D+g_L)/g_T$ $\displaystyle g_T = g_L+g_D+g(t)$ $\displaystyle v* = g_D/(g_L+g_D)$ I cannot figure out the practical steps to obtain such result. Any help would be very welcome! Thanks in advance! M November 4th, 2015, 04:09 AM #2 Newbie   Joined: Nov 2015 From: Chicago Posts: 2 Thanks: 0 The way is by integration by parts There is a straight forward solution to my question, after a bit of playing around. The solution of the ODE is in fact: $\displaystyle u(t) = e^{-(g_L+g_D)t}\int_0^t dz\, g_D v(z) e^{(g_L+g_D)z}$. Then one notes that the integral on the rhs of the above equation may be solved by integration by parts whereby $\displaystyle \int_0^t d\left(v(z) e^{(g_L+g_D)z}\right) = \int_0^t dz\, v'(z)e^{(g_L+g_D)z} + \frac{g_D}{g_D+g_L}\int_0^t v(z)d\left(e^{(g_D+g_L)z}\right)$ $\displaystyle \Rightarrow \int_0^t v(z)d\left(e^{(g_D+g_L)z}\right) = \frac{g_D}{g_D+g_L} \left( v(t)e^{(g_D+g_L)t} + \int_0^t dz\, v'(t)e^{(g_D+g_L)z}\right)$. By replacing the latter in the solution, and ordering terms by powers of $\displaystyle 1/g_D$ it is straight forward to see that, up to the leading order, it is $\displaystyle u(t) = e^{-(g_L+g_D)t} \frac{g_D}{g_D+g_L} \left( v(t)e^{(g_D+g_L)t} + \int_0^t dz\, v'(t)e^{(g_D+g_L)z}\right)$ $\displaystyle \Rightarrow u(t) = \frac{g_D}{g_D+g_L} v(t) + O\left(\frac{1}{g_D}\right)$. Last edited by skipjack; November 4th, 2015 at 06:33 AM. Tags ode, power, series, solution Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post leo255 Calculus 3 December 10th, 2014 08:54 PM g0bearmon Real Analysis 2 May 22nd, 2012 12:10 PM aaron-math Calculus 4 November 28th, 2011 10:53 AM kEnji Complex Analysis 2 May 7th, 2007 03:03 AM g0bearmon Calculus 1 December 31st, 1969 04:00 PM

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