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Following is a recurrence relation written by Robert Israel from Poisson arrivals followed by locking

$u_n(t) = \int_0^{t-nT} \lambda \exp(-\lambda y) u_{n-1}(t-y-T)\, dy$.

The solution he wrote is:

$u_n(t) = 1 - q_n(t-nT) \exp(-\lambda (t-nT))$ where $q_n(s) = \sum_{j=0}^{n-1} (\lambda s)^j/j!$.

I wonder how the recurrence is solved and what kind of method it is?

Thanks in advance!

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The integral is a convolution. You can solve it using Laplace transforms (convolutions in the time domain correspond to products in the Laplace domain). The boundary condition $u_0(t)$ can be derived separately. If it helps consider $v_n(t) \equiv u_n(t+nT)$ –  Emre Apr 25 '11 at 6:43

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up vote 4 down vote accepted

Following Emre's hint I write $v_n(t):=u_n(t+nT)$ and arbitrarily assume $v_0(t)\equiv 1$. The $v_n$ satisfy the simpler looking recursion $$v_n(t)=\int_0^t \lambda e^{-\lambda y} v_{n-1}(t-y) dy\ .\qquad(*)$$ Computing the first few $v_n$'s by hand one gets the idea that the $v_n$ might be of the form $v_n(t)=1-q_n(t)e^{-\lambda t}$ with polynomials $q_n(t)$. Entering this "Ansatz" into the recursion $(*)$ one obtains after some calculation the following recursion for the $q_n$: $$q_n(t)=1+\lambda \int_0^t q_{n-1}(y) dy\qquad (n\geq1)$$ with $q_0(t)\equiv0$. Now it is easy to see that the $q_n(t$) are the partial sums $s_{n-1}$ of the series for $e^{\lambda t}$.

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