a proposal of testing method on the exact steady state probability density function of non-linear...

Journal of Sound and <ibration (2000) 230(5), 1165}1176doi:10.1006/jsvi.1999.2715, available online at http://www.idealibrary.com on

a

A PROPOSAL OF TESTING METHOD ON THE EXACTSTEADY STATE PROBABILITY DENSITY FUNCTION

OF NON-LINEAR STOCHASTIC SYSTEM

R. WANG AND K. YASUDA

Department of Electronic-Mechanical Engineering, Nagoya ;niversity, Furo-Cho,Chikusa-Ku, Nagoya 464-8603, Japan

AND

Z. ZHANG

Shanghai Post and ¹elecommunication ¹raining Centre, Shanghai, People1s Republic of Chin

(Received 8 July 1999, and in ,nal form 4 October 1999)

This paper is concerned with the exact analysis method of the response processof second order non-linear stochastic systems excited by Gaussian white noise.A main feature is that a testing method is presented in this paper for the exactsteady state probability density of two dimensions to demonstrate the e!ectivenessof the exact results. Examples are given to illustrate the applications of the analysismethod.

( 2000 Academic Press

1. INTRODUCTION

In the past 40 years the response of non-linear dynamic systems to stochasticexcitations has been extensively studied, and both exact and approximate methodshave been developed. When excitation is Gaussian white noise, the response of thesystem is a Markovian process, and the probability density of the response processis governed by the Fokker}Planck}Kolmogrov equation (the FPK equation).

In many areas of random mechanics, sometimes we need to analytically obtainthe exact probability densities of response processes for non-linear stochasticsystems or random oscillators. Recent results on the method of the exact FPKequation may be found in references [1}5, 6]. However, in a general case, no exactsolution can be obtained and numerical methods must be used. Unfortunately, thenumerical methods for solving the FPK equation in higher dimensions are verydi$cult to perform [7, 8], and compared with the exact analytic results, there stillexist many di$culties in testing the accuracy of numerical results [5]. Thiscomparison is especially impossible for the joint probability density made up ofvector random processes of two or more dimension. Hence, this paper presentsgeneral steps to overcome the di$culties. First, the exact probability density of theresponse process of the non-linear stochastic system is solved in accordance with

0022-460X/00/101165#12 $35.00/0 ( 2000 Academic Press

1166 R. WANG E¹ A¸.

the method proposed in section 3. The main point of this method is to determine anundetermined function appearing in the reduced FPK equation by solving theRiccati equation. By using this method, an exact steady state solution of non-linearstochastic system is obtainable. Second, a non-linear stochastic di!erentialequation is constructed in accordance with the analysis technique presented the insection 4. If a known non-linear stochastic di!erential equation is the same with theconstructed non-linear stochastic di!erential equation, then the probability densitycorresponding to the known non-linear stochastic system is exact. Examples aregiven to illustrate the application of the analysis technique.

2. THE STEADY STATE FPK EQUATION

The following general non-linear system is considered:

xK#g(x, xR )"w(t), (1)

where w(t) represents a zero-mean Gaussian white noise with delta-type correlationfunctions E[w(t)w(t#q)]"2n/d(q).

The stationary probability density p(y1, y

2) of the system response is governed by

the reduced Fokker}Planck equation

!y2

LpLy

1

#

L[g (y1, y

2)p]

Ly2

#n/L2pLy2

2

"0, (2)

where y1"x(t) and y

2"xR (t).

Let

p (y1, y

2)"c expC!

1n/

f (y1, y

2)D, (3)

where c is a normalization constant, and f (y1, y

2) is a stationary potential function.

Of course, expression (3) must be non-negative and normalizable or p(y1, y

2) to be

a valid probability densitySince

!y2

LpLy

1

#

L[(g (y1, y

2)#g

1(y

1)/p(y

1, y

2))p (y

1, y

2)]

Ly2

#n/L2pLy2

2

"!y2

LpLy

1

#

L[g (y1, y

2)p (y

1, y

2)]

Ly2

#n/L2pLy2

2

, (4)

g(y1, y

2) and g(y

1, y

2)#g

1(y

1)/p(y

1, y

2) have the same probability density.

Substituting equation (3) into equation (2) yields

p Ay2fy

1

n/!g

fy2

n/#

f 2y2

n/#gy

2!fy

2y2B"0, (5)

NON-LINEAR STOCHASTIC SYSTEM 1167

where

fy1"

LfLy

1

, fy2"

LfLy

2

,

f 2y2"A

LfLy

2B2, fy

2y2"

L2fLy2

2

, gy2"

LgLy

2

Since p(y1, y

2)O0, equation (5) is reduced to

LgLy

2

!

fy2g

n/#y

2

fy1

n/#

f 2y2

n/!fy

2y2"0. (6)

3. THE EXACT ANALYSIS METHOD OF THE FPK EQUATION

If f (y1, y

2) satis"es the following equation set:

fy2y2!gy

2"

1n/

h(y1, y

2), (7)

y2

fy1!g fy

2#f 2y

2"h (y

1, y

2), (8)

then f (y1, y

2) can satisfy the FPK equation (6) where h (y

1, y

2) is an undetermined

function. The above equations show balance of the probability potential function.We can obtain from equation (7),

fy2"g(y

1, y

2)!g(y

1, 0)#fy

2(y

1, 0)#

1n/ P

y2

0

h(y1, y

2) dy

2, (9)

f"Py2

0

g(y1, y

2) dy

2!y

2g(y

1, 0)#y

2fy

2(y

1, 0)#f (y

1, 0)

#

1n/ P

y2

0P

y2

0

h (y1, y

2) dy

2dy

2, (10)

fy1"P

y2

0

gy1dy

2!y

2gy

1(y

1, 0)#y

2fy

2y1(y

1, 0)#fy

1(y

1, 0)

#

1n/ P

y2

0P

y2

0

hy1(y

1, y

2) dy

2dy

2. (11)

Substituting equations (9) and (11) into equation (8) yields

y2CP

y2

0

gy1dy

2!y

2gy

1(y

1, 0)#y

2fy

2y1(y

1, 0)#fy

1(y

1, 0)#

1n/ P

y2

0P

y2

0

hy1(y

1, y

2) dy

2dy

2D!Cg(y

1, 0)!fy

2(y

1, 0)!

1n/ P

y2

0

h (y1, y

2) dy

2D Cg(y1, y

2)!g(y

1, 0)#fy

2(y

1, 0)

#

1n/ P

y2

0

h (y1, y

2) dy

2D"h(y1, y

2). (12)

1168 R. WANG E¹ A¸.

Assuming fy2(y

1, 0)"0, equation (12) may be expressed as

fy1(y

1, 0)"

g(y1, 0)

y2

[g (y1, y

2)!g (y

1, 0)]#y

2gy

1(y

1, 0)

!Py2

0

gy1(y

1, y

2) dy

2#h

1(y

1, y

2), (13)

where

h1(y

1, y

2)"

h(y1, y

2)

y2

!

1n/ P

y2

0P

y2

0

hy1(y

1, y

2) dy

2dy

2!

1n/y

2P

y2

0

h(y1, y

2) dy

2

]Cg(y1, y

2)!2g(y

1, 0)#

1n/ P

y2

0

h (y1, y

2) dy

2D. (14)

The function h (y1, y

2) is selected in order to make the right-hand side of equation

(13) to be a function of y1. Hence, equation (13) may be further expressed as

fy1(y

1, 0)"k

1(y

1)#k

2(y

1, y

2)#h

1(y

1, y

2) (15)

where

k1(y

1)#k

2(y

1, y

2)"

g(y1, 0)

y2

[g(y1, y

2)!g (y

1, 0)]#y

2gy

1(y

1, 0)!P

y2

0

gy1(y

1, y

2) dy

2.

Let

k2(y

1, y

2)#h

1(y

1, y

2)"h

2(y

1), (16)

The above results are substituted into equation (15) yielding

fy1(y

1, 0)"k

1(y

1)#h

2(y

1). (17)

Substituting equation (17) into equation (10) and combining equation (14) yield

f (y1, y

2)"P

y2

0

g (y1, y

2) dy

2!y

2g (y

1, 0)#P

y1

0

[k1(y

1)#h

2(y

1)] dy

1#

1n/

C(y1, y

2),

(18)

where h2(y

1) satis"es (the combination in equation (14))

Cy2y2

y2

!

Cy1

n/!

Cy2

n/y2Cg!2g(y

1, 0)#

Cy2

n/D"h2(y

1)!k

2(y

1, y

2), (19)

with

C(y1, y

2)"P

y2

0P

y2

0

h (y1, y

2) dy

2dy

2.

Solving h2(y

1) is generally di$cult from equation (19). An alternative method is

given below.


By the assumption of the equation set

Cy2y2

y2

"h2(y

2), (20)

Cy1

n/#

Cy2

n/y2Cg!2g (y

1, 0)#

Cy2

n/D"k2(y

1, y

2), (21)

the solution of equations (20) and (21) satisfy equation (19). Integrating equation(20) yields

Cy2"

y222

h2(y

1), C"

y326

h2(y

1), Cy

1"

y326

h@2(y

1), (22}24)

Substituting equations (22) and (24) into equation (21), we obtain the followingRiccati equation:

h@2(y

1)#

32n/

h22(y

1)#

3y22

[g(y1, y

2)!2g(y

1, 0)]h

2(y

1)"

6n/y32

k2(y

1, y

2) . (25)

In general cases, equation (25) cannot be used to express elementary functions.But, if h

2(y

1) is considered to be a function of y

1, then the remaining terms of the

above equation, namely,

3y22

[g(y1, y

2)!2g(y

1, 0)]h

2(y

1)!

6n/y32

k2(y

1, y

2) (26)

must be a function of y1. If function h

2(y

1) exists in formula (26), we can solve

function h2(y

1) from (25). The result is then substituted into equation (25).

If h2(y

1) cannot be found, or substituting h

2(y

1) into equation (25) will not be

established, then it is shown that the conditions assumed in equations (20) and (21)are not valid. In this case, other conditions must be assumed.

If equation (23) and h2(y

1) satisfy the Riccati equation (25), the results of h

2(y

1)

are substituted into equation (18) and yield the following exact result.

f (y1, y

2)"P

y2

0

g (y1, y

2) dy

2!y

2g(y

1, 0)#P

y1

0

[k1(y

1)#h

2(y

1)] dy

1#

y32

6n/h2(y

1).

(27)

Example 1. Consider the following non-linear stochastic system:

xK#2bxR

x2n~2!(n!1)

xR 2x#

anb

x4n~3!n/ (n!1)

bx2n~3"w(t). (28)

Then g(y1, y

2)"2b (y

2/y2n~2

1)!(n!1) y2

2/y

1#((an/b) y4n~3

1)!(n/(n!1)/b)

y2n~31

, where a and b are two positive constants.

1170 R. WANG E¹ A¸.

Substituting the above equations into equation (15), we get

k1(y

1)#k

2(y

1, y

2)"2 Cany2n~1

1!

n/ (n!1)y1

D!(n!1)

b

]y2[any4n~4

1!n/(n!1)y2n~4

1]

#2(n!1)by22

y2n~11

!(n!1)y32

3y21

,

thus k1(y

1)"2 Cany2n~1

1!n/

n!1y1D (29)

and

k2(y

1, y

2)"!

(n!1)b

y2[any4n~4

1!n/ (n!1)y2n~4

1]

#2(n!1)by22

y2n~11

!(n!1)y32

3y21

. (30)

Furthermore, it is desirable for h2(y

1) to be derived from equation (26):

3y22GC2b

y2

y2n~21

!(n!1)y22

y1

!

anb

y4n~31

#

n/(n!1)b

y2n~31 D h

2(y

1)

#2n/(n!1)y22

3y21

!4n/(n!1) by2

y2n~11

#

2n/(n!1)b

[any4n~41

!n/(n!1)y2n~41

]H"!

3(n!1)y1

Ch2(y1)!

2n/3y

1D

#

3y22Ch2 (y

1)!2n/

(n!1)y1D

]Cn/(n!1)

by2n~31

!

anb

y4n~31

#2by2

y2n~21

D .

When

h2(y

2)"2n/

(n!1)y1

, (31)

equation (26) becomes ![6n//y21](n!1)(n!4/3) to be only the function of y

1.

The results obtained by substituting equations (30) and (31) into equation (25)satisfy the Riccati equation (25).

Substituting equations (29) and (31) into equation (27), we get

f (y1, y

2)"ay2n

1#b

y22

y2n~21

. (32)


By making use of equation (3), the exact steady state probability density of thenon-linear system de"ned by equation (28) is given by

p (y1, y

2)"c expC!

1n/Aax2n#b

xR 2x2n~2BD , (33)

where the normalization constant c can be solved as follows:

c"Jabnn2/

. (34)

When n"1, equation (28) becomes the following linear form:

xK#2bxR #abx"w(t). (35)

The exact probability density of the above equation is

p(x, xR )"c expC!1

n/(ax2#bxR 2)], (36)

the result is well-known.It should be pointed out that Zhu [9] has obtained an equation of motion of the

exact stationary solutions of the relative general single-degree-of-freedom (s.d.o.f )non-linear stochastic systems on the basis of results studied by Caughey and Ma[10] as well as Young and Lin [11]. Here, we "rst show that equation (28) does notbelong to the stochastic di!erential equation of this type presented by Zhu [9].Next, we show that some concrete non-linear stochastic systems are found to bestill very di$cult by means of this equation of motion presented by Zhu [9]. Theextent of the di$culties is no less than trying to "nd a solution for the FPKequation [4]. The method provided by the paper can overcome these di$culties.

Example 2. Consider the following non-linear system:

xK#g0(x)#(1#x2)xR #

xxR 21#x2

"w(t). (37)

Then g (y1, y

2)"g

0(y

1)#(1#y2

1)y

2#y

1y22/(1#y2

1).

By combining the above equation and equation (15), we get

k1(y

1)#k

2(y

1, y

2)"g

0(1#y2

1)#g

0

y1y2

1#y21

!y1y22!

y32

3(1#y21)#

2y21y32

3(1#y21)2

.

Thus

k1(y

1)"g

0(1#y2

1), (38)

k2(y

1, y

2)"g

0

y1y2

1#y21

!y1y22!

y32

3(1#y21)#

2y21y32

3(1#y21)2

. (39)

1172 R. WANG E¹ A¸.

By substituting equation (39) into equation (26), one can "nd h2(y

1),

3y32Ch2(y

1)#

2n/y1

1#y21D [(1#y2

1)y

2!g

0(y

1)]#3y

1

h2(y

1)

1#y21

#

2n/1#y2

1

!

4n/y21

(1#y21)2

.

When

h2(y

1)"!2n/

y1

1#y21

, (40)

equation (26) becomes 2n//(1#y21)!10n/ y2

1/(1#y2

1)2 to be a function of y

1.

The results obtained by substituting equations (38) and (40) into equation (25)satisfy the Riccati equation (25).

Substituting equations (38) and (40) into equation (27) yields

f (y1, y

2)"!n/ ln(1#y2

1)#

1#y21

2y22#P

y1

0

g0(y

1)(1#y2

1) dy

1. (41)

When g0(x) in equation (37) possesses certain form, this equation may not belong

to the equation of motion of the exact stationary solutions of the relative generals.d.o.f. non-linear stochastic system presented by Zhu [9].

Example 3. Consider the following non-linear oscillator [10]:

xK#bxR

1#xR 2/2#

1#xR 2/21#x2/2

"w(t). (42)

Then

g (y1, y

2)"

by2

1#y22/2#

1#y22/2

1#y21/2

By combining the above equation and equation (15), we get

k1(y

1)#k

2(y

1, y

2)"

by1

1#y21/2!

by1y22

(2#y21)(1#y2

2/2)

#

y21y2/2!(y3

2/6)(1!y2

1/2)

(1#y21/2)2

.

Then

k1(y

1)"

by1

1#y21/2

, (43)

k2(y

1, y

2)"!b

y1y22

(2#y21) (1#y2

2/2)

#Cy21y2/2!y3

2/6 (1!y2

1/2)

(1#y21/2)2 D . (44)

Under the above condition, equation (26) becomes

3y1h2(y

1)

1#y21/2

#n/ )1!y2

1/2

(1#y21/2)

#

3y22Ch2(y

1)#n/

y1

1#y21/2DC

by2

1#y22/2!

y1

1#y21/2D .

(45)


When

h2(y

1)"!n/

y1

1#y21/2

, (46)

equation (45) becomes n/ (1!2y21)/(1#y2

1/2)2 to be a function of y

1.

The results obtained by substituting equations (43) and (46) into equation (25)satisfy the Riccati equation.

By making use of equations (27), (43) and (46), we get

f (y1, y

2)"lnCA1#

y212 B

b~n(A1#

y222 B

bD , (47)

where b'2n/.The above result is in complete accordance with the solution obtained by

Caughey and Ma [9].

4. A TESTING TECHNIQUE OF THE EXACT STEADY STATE PROBABILITYDENSITY

We shall now turn our attention to the testing method of the above exact steadystate probability density, so that we recall equation (6) in section 2 as follows:

LgLy

2

!

fy2g

n/#y

2

fy1

n/#

f 2y2

n/!fy

2y2"0. (48)

Since g(y1, y

2) is a function of f (y

1, y

2) in equation (48), equation (48) is a "rst

order linear di!erential equation with respect to g(y1, y

2). Then an exact general

solution of equation (48) is

g (y1, y

2)"fy

2(y

1, y

2)!

1n/

expC1

n/f (y

1, y

2)DP

y2

0

y2fy

1(y

1, y

2)

]expC!1

n/f (y

1, y

2)Ddy

2. (49)

Example 1. The exact result of equation (28), which is given in Equation (32) as

f (y1, y

2)"ay2n

1#b

y22

y2n~21

, (50)

where a and b are two positive constants, will be treated here.By substituting the above equation into equation (49), we get

g (y1, y

2)"2b

y2

y2n~21

!(n!1)y22

y1

#

anb

y4n~31

!

n/(n!1)b

y2n~31

. (51)

Namely, the non-linear stochastic system is

xK#2bxR

x2n~2!(n!1)

xR 2x#

anb

x4n~3!n/ (n!1)

bx2n~3"w(t). (52)

1174 R. WANG E¹ A¸.

It is shown that equation (52) is in complete accordance with equation (28) insection 3. Hence, the steady state probability density de"ned by equation (28) isexact.

Example 2. The following exact results obtained by Caughey and Ma [10] aretested, here the exact results are given by

f (y1, y

2)"n/AP

H

0

h (u) du!lnHYB, (53)

where >"y22/2 presents the kinetic energy of the system, and H"H(y

1, y

2)

presents the total energy of the system. It is a "rst integration of the equationyK1#Hy

1/H

Y"0.

By combining equations (49) and (53) yields

g (y1, y

2)"n/Ch (H)H

Y!

HYY

HYD y

2!

1H

Y

expCPH

0

h(u) duD]P y

2Ch (H)Hy1!

H>y1

HYD expC!P

H

0

h(u) duDHYdy

2

"n/Ch (H)HY!

HYY

HYD y

2!

expCPH

0

h(u) duDH

YP [H

Yh (H)Hy

1!H>y

1]

]expC!PH

0

h(u) duD d> (54)

because

PH>y1expC!P

H

0

h (u) duDd>"Hy1expC!P

H

0

h (u) duD#P Hyh(H)Hy

1

]expC!PH

0

h(u) duD d> (55)

Substituting equation (55) into equation (54) yields

g(y1, y

2)"n/y

2Ch (H)HY!

HYY

HYD#

H>1

HY

. (56)

The resulting non-linear system now becomes

xK#n/Ch(H)HY!

HYY

HYDxR #

Hx

HY

"w(t). (57)


It is shown that equation (57) is in complete accordance with equation (4) byCaughey and Ma [10]. Hence, the result obtained by reference [10] is exact,namely, the exact steady state probabilistic density of the above equation is shownin the resulting equation

p(x, xR )"AHYexpC!P

H

0

h (u) duD . (58)

ACKNOWLEDGMENTS

The work is supported by the Japan Society for the Promotion of Science.

5. CONCLUDING REMARKS

This paper is aimed at developing an analysis method of non-linear randommechanics. In a general case, exact solutions for randomly excited non-linearsystems are di$cult to obtain mathematically, and there has been no uniformanalysis method for the response of a non-linear stochastic system. For this reason,di!erent exact analysis methods have been developed for di!erent non-linearstochastic systems. In this paper the main points are the following:

1. The most important aspect of this paper is to present a testing method for theexact steady state probability density of two dimensions in order to demonstratethe e!ectiveness of the exact results. Examples are given to show that this testingmethod is e!ective. Moreover, this testing method can be generalized to thenon-linear stochastic system of higher dimensions.

2. By the proposed method, non-linear damping of various types can be treated.3. This paper presents the exact analysis method, which when applied to a system

treated by Caughey and Ma, yields the same solution obtained by them.4. If an undetermined function in a non-linear stochastic system can satisfy the

Riccati equation (25), then exact steady state solutions are obtainable.5. When some concrete non-linear stochastic systems cannot satisfy the relative

general motion equation of the s.d.o.f. non-linear stochastic system presented byZhu [9], obtaining an exact solution is still possible. Of course Zhu's method canalso solve the equation that the method of this paper cannot solve.

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a proposal of testing method on the exact steady state probability density function of non-linear...

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