          Astron. Astrophys. 335, 329-340 (1998)

## 6. Probabilistic description

The aim of this section is to find a relationship between the mean value of the amplitude of the kinetic energy after a stochastic driving with identical pulses and the eigenfrequencies of the loop. To avoid complicated notations we denote the eigenfrequency by .

We assume the time intervals between the identical pulses to satisfy a well-defined probability distribution. In order to show the effect of the stochastic driving most clearly, we modify the pulse. Instead of the model pulse of the previous section ( ), we now consider pulses with the following time dependence: ### 6.1. Energy after pulses

In particular we are interested in the single contributions of the eigenfunctions to the total kinetic energy which is represented by Eq. (11) by: The time dependence appears in the second factor . After some algebra we obtain the following result for this time derivative of T after driving with pulses: where and R.
R is the random time interval before the ith pulse.

We can rewrite both summations in expression (15) as follows: where Substituting this result in Eqs. (14) and (15), we obtain the following expression for E: Hence the final energy oscillates like the square of a sine, with a specific amplitude dependent on the time intervals between the pulses, R. As this R dependence is included in the factor A, in the following subsection we study the influence of the frequency on A.

### 6.2. Probabilistic approach

To work out our solution in an analytical way, we assume the intervals between the pulses R to satisfy a gamma distribution . We are interested in the probability distribution of : or, more specifically, in the mean value of this amplitude.

We solve this problem in four steps:
At first we need the probability distribution function of S. Since R for j going from 2 to k and since all R are independent, it is easy to check that S (Wonnacott & Wonnacott 1990).

As a second step, we calculate the mean value of the first term of A. It is convenient to rewrite this 'cosine part' of A as Using the definition of mean value and the gamma distribution function, the terms in the first summation can be calculated as  The calculation of the double summation in Eq. (17) is less obvious because X and Y are two dependent stochastic variables and the mean of a product of two variables only equals the product of the two means if the two variables are independent.

We tackle this problem by rewriting the two successive summations over k and l as a sum of 3 summations where respectively k, k and k. Since the function is symmetric in k and l, the first two terms in the sum of summations are equal and so we only have to consider the cases k and k.

If k we can calculate:  and if k we obtain: Following the derivation presented in the second step, we rewrite the 'sine part' of A. The counterpart of Eq. (17) yields where  and  if k and if k.
Finally we put the two previous solutions together to find the mean value of A : Only the last but one term still has to be worked out. Therefore we need the probability distribution function of X. Since in this term k is larger than l we know that X, such that X. So the double summation in Eq. (18) equals where c.
Substituting this result in Eq. (18) we obtain the final result: If we multiply this expression by the factor we obtain, according to Eq. (16), the mean value for the final amplitude of as function of .
In Fig. 9 we have plotted the mean value of this final amplitude as function of for a, , and . A gamma distribution with the given parameters and is a strongly peaked distribution which is quasi symmetric around the value 0.5. Fig. 9. The mean value of the final amplitude of after driving by a succession of five identical pulses with time intervals in between. These intervals satisfy a gamma distribution with parameters and .

We remark that although this figure shows a continuous graph, the results are only relevant for the discrete eigenfrequencies . Nevertheless the graph shows remarkable peaks which we try to explain in the next section.    © European Southern Observatory (ESO) 1998

Online publication: June 12, 1998 