On the Application of the Generating Series for Nonlinear Systems with Polynomial Stiffness 147 For the asymmetric Duffing oscillator under investigation here, performing the ensemble average for all the terms in the generating series, the following result is obtained: g =−4ε1 σ 2 2 ! x0 x0 x0 x0 x0 −a1 −a2 −2a1 −a1 −a2 −2a2" +48ε1ε2 σ 2 2 2 ! x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 −a1 −a2 −2a1 −a1 −a2 −4a1 −3a1 −a2 −2a1 −2a2 2a1 −a1 −a2 −2a2" +48ε1ε2 σ 2 2 2 ! x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 −a1 −a2 −2a1 −a1 −a2 −2a1 −3a1 −a2 −2a1 −2a2 2a1 −a1 −a2 −2a2" +· · · (52) As discussed above, because of the third condition in the recursion (50), many terms are zero and do not contribute. For example, the second significant term in the expansion above is actually the 39th term from the total of 360 in g2. To move the computation forward, it is necessary to carry out the partial fraction calculations implicit in the terms in Eq. (52). At this point, it is useful to specify numerical values for a1 and a2, as the partial fraction calculations can be cumbersome when carried out algebraically. In the case of the asymmetric Duffing equation, the system parameters are m,c,k1,k2 and k3. The relevant parameters in the generating series can then be calculated via the scaled version of the Duffing equation in Eq. (12); starting with values here of m=1kg, c =15Nsm−1, k1 =25Nm−1, k2 =625Nm−1 and k3 =7500Nm−1, one obtains a1 =− 1 2(3+ √5), a2 =−1 2(3+ √5), ε1 =1 and ε2 =0.5. By decomposing the terms in Eq. (52) into partial fractions and applying the inverse Laplace–Borel transforms as given in Table 1, the mean response of the systemE[y(t)] can be computed; the result is y(t) =−4ε1 σ 2 2 !0.08332− 0.0005835 0.1910 e− t 0.1910 + 0.02288 0.3333 e− t 0.3333 − 0.6315 2.618 e− t 2.618 − 0.03514 0.3820 e− t 0.3820 + 0.2409 1.309 e− t 1.309 "+48ε1ε2 σ 2 2 !0.0001508+ 0.0005778 0.01667 e− t 0.01667 − 0.001053 1.3092 (1− t 1.309 )e− t 1.309 ) − 0.0002746 0.1910 e− t 0.1910 + 0.000006524 0.09551 e− t 0.09551 + 0.0005542 0.3333 e− t 0.3333 − 0.003057 2.618 e− t 2.618 − 0.00002157 0.19102 (1− t 0.1910 )e− t 0.1910 − 0.00005400 0.1214 e− t 0.1214 − 0.0007346 0.3820 e− t 0.3820 − 0.0007298 0.33332 (1− t 0.3333 )e− t 0.3333 + 0.001947 1.309 e− t 1.309 "+. . . (53) Choosing the somewhat arbitrary value σ = 1, the result in Fig. 2 is obtained. One observes a transient that occurs from ‘switching on’ the excitation at t = 0. In fact, because the asymmetric Duffing oscillator has stationary response if the input is stationary, the expectation will tend to a constant value as t −→∞; because the restoring force is asymmetric, that constant value will be non-zero. Table 1 Laplace–Borel transforms of common functions [7,11] x(t) g[x(t)] Unit step 1 tn n! xn 0 n−1 i=0 i n−1 ai ti i! eat (1−ax0)−n cos(ωt) (1+ω 2 x 2 0) −1
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