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November 3rd, 2015, 07:56 PM   #1
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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
chrlbrwn is offline  
 
November 4th, 2015, 04:09 AM   #2
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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.
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