( )z ( ) ( ) ( ) ( ) 0

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Next, we will observe a special category of equation (1) such as. 0 x xx(t)F xx x(t) F xx x n ..... [3] Rogai E. – Exercitii si probleme de ecuatii diferentiale si integrale ...
The Exact Analitical Solutions for Problems Concerning Free Vibrations of the Elastic Systems Ghiorghe CAUTES Facultatea de Inginerie din Braila Calea Calarasilor Nr. 29, 810017 Braila E-Mail [email protected] Gheorghe OPROESCU Facultatea de Inginerie din Braila Calea Calarasilor Nr. 29, 810017 Braila E-Mail [email protected] (Received 22 May 2006; accepted in revised form 6 august 2006)

The work presents some method in order to solve differential equations of the linear or non-linear vibrations. The solutions result on analytical way only and have an exactly form. These solutions can be used later for the exactly research of the dynamics of he mechanical systems.

1. INTRODUCTION

Replacing x& = z, z = z(t) x

Free vibrations of non-linear mechanical systems with variable parameters are described with differential equations as follows

m&x& + Ff ⋅ x& + Fel ⋅ x = 0

(1)

where m is the mass of the system, Ff the damping force, Fel the elastic force, all of them depending on time, x movement and its derivations on relation with the time. After division to m, the mass appears not more in the equation. For the first equation, it doesn’t exist any method for calculating the precise analytical solutions but for the particular ones. Next, we will observe a special category of equation (1) such as x m ⋅ x& n ⋅ &x& p + F1 (t) ⋅ x n ⋅ x& p ⋅ &x& m + + F2 (t) ⋅ x p ⋅ x& m ⋅ &x& n = 0

(2)

homogeneous for x, x& and &x& with m, n, p ∈ N . We will show using particular examples that equation (2) represents the general form of some cases of real systems.

2. THE SOLVING METHODS RJAV vol III no 2/2006

65

(3)

the z(t) derivative function and knowing that

(

x& = xz, &x& = x& z = xz 2 + xz& = x z 2 + z&

)

it results

) n + F2 (t )x p x m z m x n (z 2 + z& ) = 0 (

(

)

p m x m x n z n x p z 2 + z& + F1 (t )x n x p z p x m z 2 + z& +

If

min(m,n,p)=m

and

dividing

with

x m+n +p ⋅ z m (z 2 + z& ) m we obtain z n − m (z 2 + z& ) p − m + F1 (t) ⋅ z p − m + + F2 (t) ⋅ (z 2 + z& ) n − m = 0 ordinar differential eqution of the first order, of Bernoulli type, Riccati or liniar depending of the values of m,n,p. Calculating later z(t), for the replacement, it results

ISSN 1584-7284

() x (t ) = e ∫ z t dt

(4)

after just one integration operation. In the second equation can appear expressions with other combinations x, x& , &x& on the condition that equation remains homogeneous. Next, we prezent other cases that lead to the equation (2) and then to liniar equations (homogeneous or not), Bernoulli or Riccati.

z=e



2t dt 1- t 2

[c1 + ∫

1 1− t2

2 = e −ln(1−t ) [c1 + ∫

⋅e

−∫

2t dt 2 dt] = 1- t

ln(1−t 2 )

1

⋅e 1− t2 1 1 = c +t (c1 + ∫ dt) = 1 1− t2 1− t2

(

dt] =

)

Using (4), it results the general solution of the equation

2.1 FIRST CASE The equation



c1 + t

2 x (t ) = e 1+ t

1 2t 1 &x& - ( ⋅ x& + )x& − x=0 2 x 1− t 1− t 2

dt

=

( )

1 ⎛ t +1 ⎞ 1 2 c1 ln ⎜ ⎟ ln 1− t + k 2 t 1 2 − ⎝ ⎠ =e =

describes an elastically mechanical system with the damping as variable in the time and non-linear dependent from the speed. These gives later an homogeneous equation

c ⎛ t + 1⎞ 1 = ⎜ ⎟ ⎝ t −1⎠

=

(1 − t 2 )x ⋅ &x& − (1 − t 2 )x& 2 − 2t ⋅ x ⋅ x& − x 2 = 0

1 1− t2

(t + 1)c1 (t + 1)c1 1 − t 2

(

)

c2 =

c2

Replacing x& = z, z = z(t) x

where c1 , c 2 , k ∈ R ; c 2 = e k

2.2. SECOND CASE with x& = x ⋅ z, &x& = x(z 2 + z& ) we obtain

The equation

(1 - t 2 ) ⋅ x ⋅ x(z 2 + z& ) − (1 − t 2 )x 2 z 2 −

1 &x& − [ ⋅ x& + f(t)] ⋅ x& − g(t) ⋅ x = 0 x

− 2t ⋅ x ⋅ xz − x 2 = 0 After dividing to x2 and a series of calculations, it results

can be written x ⋅ &x& − x& 2 − f(t) ⋅ xx& − g(t) ⋅ x 2 = 0

(1 - t 2 )z& − 2tz = 1

With (3), after calculations we obtain which means

z& − f(t)z = g(t) , z'−

2t 1− t

2

z=

1 1− t2

a linear non-homogeneous equation with the general solution.

linear equation of first order, non-homogeneous, with the general solution

z = e∫

f(t)dt

[c + ∫ g(t) ⋅ e

− ∫ f(t)dt

dt], c ∈ R

and using the substitution we can write the solution of the equation. RJAV vol III no 2/2006

66

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1 1 z1 = , z 2 = − . t t

2.3. THIRD CASE The equation &x& +

The general solution it is obtained from z − z1

1 2 1 ⋅ x& − ⋅ x& = 0 t x

(z − z )P(t )dt z − z 2 = c1e ∫ 1 2

can be written where c1 ∈ R and P(t) are the coefficient of z 2 in the Riccati equation. We obtain

tx ⋅ &x& + t ⋅ x& 2 − xx& = 0

and using the replacement (3) it becomes 1 z& − z = −2z 2 t

z=

c1t 2 + 1 t(1 − c1 t 2 )

and than

a Bernoulli equation. After dividing with z 2 and writing z −1 = u, u = u(t) , we obtain the a linear 1 equation like u& + ⋅ u = 2 with the solution t

x(t) =

c1 + t 2

&x& +

, c1 ∈ R

1 2 k ⋅ x& − ⋅ x = 0, k ∈ R x t2

can be written

and than

x(t) = e

, c 2 ∈ R.

The non-linear differential equation

and results

t

1 − c1 t 2

2.5. FIFTH CASE

1 u = (c1 + t 2 ) t

z(t) =

c2t

t 2 x&x& + t 2 x& 2 − kx 2 = 0

∫c

tdt

2 1+ t

1 ln(c1 + t 2 ) + k = e2 = c 2 c1 + t 2

and it is homogeneous. With substitution (3), we obtain, after simple calculations, the Riccati equation

where c 2 = e k , k ∈ R .

z& = −2z 2 +

2.4. FOURTH CASE

t2

⋅ k.

for which we look for particular solutions such as

The equation

c z = , c∈R . t

1 1⎞ ⎛2 &x& − ⎜ x& − ⎟ x& + x=0 t⎠ ⎝x t2

From the verifying condition of the equation, it results

becomes 1 1 z& = z 2 − z = − t t2 a Riccati equation with the particular solutions

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2c 2 − c − k = 0 with the real solutions c1 , c 2 67

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REFERENCE

The particular solutions of the equation will be

[1] Cautes, Gh., Oproescu, Gh. – Ecuatii diferentiale. Aplicatii la studiul dinamicii sistemelor elastice. ISBN 973-85449-1-2, Ed. Zedax, Focsani, 2002 [2] Ghermanescu M. – Culegere de probleme de ecuatii diferentiale, Ed. Didactica si Pedagogica , Bucresti , 1962 [3] Rogai E. – Exercitii si probleme de ecuatii diferentiale si integrale, Ed. Tehnica, Bucuresti , 1965.

c c z1 = 1 , z 2 = 2 t t and than from z − z1 (z −z )P(t)dt = c 3e ∫ 1 2 z − z2 with c 3 ∈ R and P(t) = −2 we can obtain z(t ) . From (4) we can calculate x (t ) .

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