Stochastic modeling of nonlinear oscillators under combined Gaussian and Poisson white noise: a viewpoint based on the energy conservation law

Abstract

A stochastic differential equation model is considered for nonlinear oscillators under excitations of combined Gaussian and Poisson white noise. Since the solutions of stochastic differential equations can be interpreted in terms of several types of stochastic integrals, it is sometimes confusing about which integral is actually appropriate. In order for the energy conservation law to hold under combined Gaussian and Poisson white noise excitations, an appropriate stochastic integral is introduced in this paper. This stochastic integral reduces to the Di Paola–Falsone integral when the multiplicative noise intensity is infinitely differentiable with respect to the state. The stochastic integral introduced in this paper is applicable in more general situations. Numerical examples are presented to illustrate the theoretical conclusion.

Appendix: Proof of the energy–work law (33) from the solution (32)

For convenience, we introduce the following notations: $g(s)=\dfrac{1}{m} g(x(s),\dot{x}(s))$, $f(s)=\dfrac{1}{m} f(x(s), \dot{x}(s))$, and $f_{\dot{x}} (s)=\dfrac{1}{m} f_{\dot{x}(s)}(x(s), \dot{x}(s))$. In this Appendix, all stochastic integrals with respect to Brownian motion are in the sense of Ito (we have dropped $\star$ notation). Then the energy–work law (33) is equivalent to

$$\begin{aligned}&\frac{1}{2} \left[ \dot{x}^2(t) + \omega ^2 x^2(t)\right] - \frac{1}{2} \left[ \ \dot{x}^2(0) + \omega ^2 x^2(0) \right] \nonumber \\&\quad = \int _0^t \left[ \left( g(s) + \dfrac{1}{2} f(s)f_{\dot{x}} (s) \right) \dot{x}(s)+\frac{1}{2 }f^2(s) \right] \,\mathrm{d}s \nonumber \\&\quad + \int _0^t f(s) \dot{x}(s)\, \mathrm{d}B(s). \end{aligned}$$

(48)

Next, we show the solution given in (32) satisfies the energy–work law (48).

Denote the right- and left-hand sides of (48) by RHS and LHS, respectively. Substitute (32) into the left-hand side of (48), we get

$$\begin{aligned} \hbox {LHS}&=\frac{1}{2} \left( \int _0^t \sin \omega (t\!-\!s)\left[ g(s) \!+\! \frac{1}{2}f(s) f_{\dot{x}} (s) \right] \,\mathrm{d}s \right) ^2 \nonumber \\&\quad +\frac{1}{2} \left( \int _0^t \sin (\omega (t\!-\!s)) f(s) \,\mathrm{d}B(s) \right) ^2\nonumber \\&\quad + \frac{1}{2} \left( \int _0^t \cos \omega (t\!-\!s)\left[ g(s) \!+\! \frac{1}{2}f(s) f_{\dot{x}} (s) \right] \,\mathrm{d}s \right) ^2 \nonumber \\&\quad +\frac{1}{2}\left( \int _0^t \cos (\omega (t-s)) f(s) \,\mathrm{d}B(s) \right) ^2 \nonumber \\&\quad + \left( \omega \cos (\omega t) x_0 + \sin (\omega t) \dot{x}_0 \right) \nonumber \\&\quad \times \int _0^t \sin \omega (t-s)\left[ g(s) + \frac{1}{2}f(s) f_{\dot{x}} (s) \right] \,\mathrm{d}s\nonumber \\&\quad + ( \omega \cos (\omega t) x_0 + \sin (\omega t) \dot{x}_0)\int _0^t \sin (\omega (t-s))\nonumber \\&\quad \times f(s)\,\mathrm{d}B(s)\nonumber \\&\quad + \left( \int _0^t \sin \omega (t-s) \left[ g(s) + \frac{1}{2}f(s) f_{\dot{x}} (s) \right] \,\mathrm{d}s \right) \nonumber \\&\quad \times \left( \int _0^t \sin (\omega (t-s)) f(s) \,\mathrm{d}B(s)\right) \nonumber \\&\quad + \left( -\omega \sin (\omega t) x_0 + \cos (\omega t) \dot{x}_0 \right) \nonumber \\&\quad \times \int _0^t \cos \omega (t-s)\left[ g(s) + \frac{1}{2}f(s) f_{\dot{x}} (s) \right] \,\mathrm{d}s\nonumber \\&\quad + (-\omega \sin (\omega t) x_0 + \cos (\omega t) \dot{x}_0)\int _0^t \cos (\omega (t-s))\nonumber \\&\quad \times f(s) \,\mathrm{d}B(s)\nonumber \\&\quad + \left( \int _0^t \cos (\omega (t-s) \left[ g(s) + \frac{1}{2}f(s) f_{\dot{x}} (s) \right] \,\mathrm{d}s\right) \nonumber \\&\quad \times \left( \int _0^t \cos (\omega (t-s)) f(s) \,\mathrm{d}B(s)\right) . \end{aligned}$$

(49)

Substituting (32) into the right-hand side of (48), we get

$$\begin{aligned}&\hbox {RHS} = \int _0^t \left( g(s)+ f(s)f_{\dot{x}}(s) \right) \nonumber \\&\quad \times \left( -\omega \sin (\omega s) x_0 + \cos (\omega s) \dot{x}_0\right) \,\mathrm{d}s\nonumber \\&\quad + \int _0^t\int _0^s \Bigg [\cos (\omega (s-p)) \left( g(s)+ \frac{f(s)f_{\dot{x}}(s)}{2}\right) \nonumber \\&\quad \times \left( g(p)+ \frac{f(p)f_{\dot{x}}(p)}{2}\right) \Bigg ]\,\mathrm{d}p\,\mathrm{d}s\nonumber \\&\quad + \int _0^t\int _0^s \Bigg [\cos (\omega (s-p))\nonumber \\&\quad \times \left. \left( g(s)+ \frac{f(s)f_{\dot{x}}(s)}{2}\right) f(p)\right] \,\mathrm{d}B(p) \,\mathrm{d}s \nonumber \\&\quad + \frac{1}{2}\int _0^t f^2(s) \,\mathrm{d}s + \int _0^t f(s)\nonumber \\&\quad \times \left( -\omega \sin (\omega s) x_0+ \cos (\omega s) \dot{x}_0 \right) \,\mathrm{d}B(s)\nonumber \\&\quad + \int _0^t \int _0^s \cos (\omega (s-p)) f(s)\nonumber \\&\quad \times \left( g(p)+ \frac{f(p)f_{\dot{x}}(p)}{2}\right) \,\mathrm{d}p \,\mathrm{d}B(s)\nonumber \\&\quad + \int _0^t \int _0^s \cos (\omega (s-p)) f(s)f(p) \,\mathrm{d}B(p)\,\mathrm{d}B(s). \end{aligned}$$

(50)

To prove LHS in (49) is equal to RHS in (50), we claim the following facts

$$\begin{aligned}&\int _0^t\int _0^s \Bigg [ \cos (\omega (s-p)) \left( g(s)+ \frac{f(s)f_{\dot{x}}(s)}{2}\right) \nonumber \\&\quad \times \left( g(p)+ \frac{f(p)f_{\dot{x}}(p)}{2}\right) \Bigg ]\,\mathrm{d}p\, \mathrm{d}s\nonumber \\&= \frac{1}{2} \left[ \int _0^t \sin (\omega (t-s)) \left( g(s)+ \frac{f(s)f_{\dot{x}}(s)}{2}\right) \,\mathrm{d}s \right] ^2 \nonumber \\&\quad + \frac{1}{2} \left[ \int _0^t \cos (\omega (t\!-\!s)) \left( g(s)\!+\! \frac{f(s)f_{\dot{x}}(s)}{2}\right) \,\mathrm{d}s \right] ^2, \end{aligned}$$

(51)

$$\begin{aligned}&\int _0^t \int _0^s \cos (\omega (s-p)) f(s)f(p) \,\mathrm{d}B(p) \,\mathrm{d}B(s) \nonumber \\&\quad + \frac{1}{2} \int _0^t f^2(s) \,\mathrm{d}s\nonumber \\&=\frac{1}{2} \left( \int _0^t \sin (\omega (t-s)f(s) \,\mathrm{d}B(s) \right) ^2 \nonumber \\&\quad \quad +\frac{1}{2} \left( \int _0^t \cos (\omega (t-s)f(s) \,\mathrm{d}B(s) \right) ^2, \end{aligned}$$

(52)

$$\begin{aligned}&\int _0^t\int _0^s \left[ \cos (\omega (s-p)\left( g(s)+ \frac{f(s)f_{\dot{x}}(s)}{2}\right) \right] \nonumber \\&\quad \times f(p) \,\mathrm{d}B(p) \,\mathrm{d}s \nonumber \\&\quad + \int _0^t\int _0^s \left[ \cos (\omega (s-p)\left( g(p)+ \frac{f(p)f_{\dot{x}}(p)}{2}\right) \right] \nonumber \\&\quad \times f(s) \,\mathrm{d}p\,\mathrm{d}B(s)\nonumber \\&= \int _0^t \cos (\omega (t-s)) \left( g(s)+ \frac{f(s)f_{\dot{x}}(s)}{2}\right) \,\mathrm{d}s \nonumber \\&\quad \times \int _0^t \cos (\omega (t-s)) f(s) \,\mathrm{d}B(s) \nonumber \\&\quad + \int _0^t \sin (\omega (t-s)) \left( g(s)+ \frac{f(s)f_{\dot{x}}(s)}{2}\right) \,\mathrm{d}s \nonumber \\&\quad \times \int _0^t \sin (\omega (t-s))f(s) \,\mathrm{d}B(s)\, \end{aligned}$$

(53)

$$\begin{aligned}&\int _0^t \left( g(s)+ \frac{f(s)f_{\dot{x}}(s)}{2}\right) \nonumber \\&\quad \times \Bigg (-\omega \sin (\omega s) x_0 + \cos (\omega s) \dot{x}_0\Bigg ) \,\mathrm{d}s\nonumber \\&= (-\omega \sin (\omega t) x_0 + \cos (\omega t) \dot{x}_0) \nonumber \\&\quad \times \int _0^t \cos (\omega (t-s))\left( g(s)+ \frac{f(s)f_{\dot{x}}(s)}{2}\right) \,\mathrm{d}s\nonumber \\&\quad + (\omega \cos (\omega t) x_0 + \sin (\omega t) \dot{x}_0) \nonumber \\&\quad \times \int _0^t \sin (\omega (t-s))\left( g(s)+ \frac{f(s)f_{\dot{x}}(s)}{2}\right) \,\mathrm{d}s, \end{aligned}$$

(54)

and

$$\begin{aligned}&\int _0^t f(s) \left( -\omega \sin (\omega s) x_0+ \cos (\omega s) \dot{x}_0 \right) \,\mathrm{d}B(s)\nonumber \\&\quad = (-\omega \sin (\omega t) x_0 + \cos (\omega t) \dot{x}_0)\nonumber \\&\qquad \times \int _0^t \cos (\omega (t-s)) f(s) \,\mathrm{d}B(s)\nonumber \\&\qquad + (\omega \cos (\omega t) x_0 + \sin (\omega t) \dot{x}_0)\nonumber \\&\qquad \times \int _0^t \sin (\omega (t-s)) f(s) \,\mathrm{d}B(s). \end{aligned}$$

(55)

One can easily see that (54) and (55) are true by using the trignometric identities

$$\begin{aligned}&\cos (\omega s)=\cos (\omega t)\cos (\omega (t-s))\\&\quad +\sin (\omega t)\sin (\omega (t-s)), \end{aligned}$$

and

$$\begin{aligned}&\sin (\omega s)=\sin (\omega t)\cos (\omega (t-s))\\&\quad -\cos (\omega t)\sin (\omega (t-s)), \end{aligned}$$

In the following, we give the proofs for (51) and (52). The proof of (53) is similar to those for (51) and (52) and is not given here.

To prove (51) is true, we rewrite the right-hand side of (51) as double integrals

$$\begin{aligned}&\left( \int _0^t \sin (\omega (t-s)) \left( g(s)+ \frac{f(s)f_{\dot{x}}(s)}{2}\right) \,\mathrm{d}s \right) ^2 \\&\qquad + \left( \int _0^t \cos (\omega (t-s)) \left( g(s)+ \frac{f(s)f_{\dot{x}}(s)}{2}\right) \,\mathrm{d}s \right) ^2\\&= \int _0^t \int _0^t \Bigg [\sin (\omega (t-p))\sin (\omega (t-s))\nonumber \\&\qquad \times \left( g(p)+ \frac{f(p)f_{\dot{x}}(p)}{2}\right) \\&\quad \quad \times \left( g(s)+ \frac{f(s)f_{\dot{x}}(s)}{2}\right) \Bigg ] \mathrm{d}p\,\mathrm{d}s\\&\quad \quad + \int _0^t \int _0^t \Bigg [ \cos (\omega (t-p))\cos (\omega (t-s))\\&\qquad \times \left( g(p)+ \frac{f(p)f_{\dot{x}}(p)}{2}\right) \\&\quad \quad \times \left( g(s)+ \frac{f(s)f_{\dot{x}}(s)}{2}\right) \Bigg ] \,\mathrm{d}p\,\mathrm{d}s\\&\quad =\int _0^t \int _0^t \Bigg [\cos (\omega (s-p))\left( g(p)+ \frac{f(p)f_{\dot{x}}(p)}{2}\right) \\&\quad \quad \times \left( g(s)+ \frac{f(s)f_{\dot{x}}(s)}{2}\right) \Bigg ]\,\mathrm{d}p\,\mathrm{d}s\\&\quad =2\int _0^t \int _0^s \Bigg [ \cos (\omega (s-p))\left( g(p)+ \frac{f(p)f_{\dot{x}}(p)}{2}\right) \\&\quad \quad \times \left( g(s)+ \frac{f(s)f_{\dot{x}}(s)}{2}\right) \Bigg ]\,\mathrm{d}p\,\mathrm{d}s. \end{aligned}$$

Similarly, to prove (52), we rewrite the right-hand side (52) as

$$\begin{aligned}&\left( \int _0^t \sin (\omega (t-s)) f(s) \,\mathrm{d}B(s) \right) ^2 \nonumber \\&\quad \quad \quad + \left( \int _0^t \cos (\omega (t-s)) f(s) \,\mathrm{d}B(s) \right) ^2 \nonumber \\&=\int _0^t \int _0^t \cos (\omega (s-p)) f(s)f(p) \,\mathrm{d}B(p) \,\mathrm{d}B(s). \end{aligned}$$

(56)

The integral domain for the right-hand side of (56) is a square given by $A=\{(s,p) \big | s\in [0,t], p\in [0, t]\}$. Decompose the square into three parts:

$$\begin{aligned} A_1= & {} \{(s,p)~\big | ~0~\le ~s<p~ \le ~t\},\\ A_2= & {} \{(s,p)\big | 0\le p<s\le t\}, \end{aligned}$$

and

$$\begin{aligned} A_3=\{(s,s) \big | 0\le s \le t\}, \end{aligned}$$

then the right-hand side of (56) becomes

$$\begin{aligned}&\int _0^t \int _0^t \cos (\omega (s-p)) f(s)f(p) \,\mathrm{d}B(p) \,\mathrm{d}B(s)\nonumber \\&=\iint \limits _{A_1+A_2+A_3} \cos (\omega (s-p)) f(s) f(p) \,\mathrm{d}B(p) \,\mathrm{d}B(s). \end{aligned}$$

(57)

Note that

$$\begin{aligned}&\iint _{A_1} \cos (\omega (s-p)) f(s) f(p) \,\mathrm{d}B(p) \,\mathrm{d}B(s)\nonumber \\&= \iint _{A_2} \cos (\omega (s-p)) f(s)f(p) \,\mathrm{d}B(p) \,\mathrm{d}B(s) \nonumber \\&= \int _0^t \int _0^s \cos (\omega (s-p)) f(s)f(p) \,\mathrm{d}B(p) \,\mathrm{d}B(s), \end{aligned}$$

(58)

and

$$\begin{aligned}&\iint _{A_3} \cos (\omega (s-p)) f(s)f(p) \,\mathrm{d}B(p) \,\mathrm{d}B(s)\nonumber \\&\quad =\int _0^t f^2(s) \,\mathrm{d}s. \end{aligned}$$

(59)

It follows from (58) and (59) that (52) is true.

Adding Eqs. (51) to (55), we get LHS $=$ RHS, and hence, (33) is proved.