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An exact solution for linear and nonlinear systems of equations with gaussian and exponential distributions. It includes the determination of the number of times a level is crossed in an interval, the study of limit cycle oscillations, and the integration of terms by parts using taylor's expansion. The document also discusses the general formulation of the problem and the steady state pdf.
Typology: Slides
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Markov Vector Approach-
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2
^
0
,^1
,^
;^ 0;
0
,^ ,
~^1 ,^
~ ~^
(^1) 0;^
2
,^ ;^
1, 2,^
,
2
; ,^
1, 2,^
, 1 2 i^
j^
ij
j^ j
t ij^
ij n
j n j j dX^ t^
f^ t X
t^ dt
G t X
t^ dB t
t^
X^
X
X^ t^
f^ t X
t^
n
G t X
t^
n^ m dB t^
m dB t^
B^ t^
B^ t^
D
f^ t x
j^
n
GDG
x
i j^
m
p^
p t^
x
^
^
^
^
^
^
^
^
^
^
^
^ ^
^
^ ^
^
^
^
^
^
^
^
^
^ ^
^
^ ^
^
Ge n e
r a l :
e n s i o
n a l I t o SD^
E
0
0 1 ;0 |^
;^
BCS m^ m
jk j^ k^
j^ k
n
i^ i i
p x^ x
p x^
x^
x^ x
^
^
^
^
^
^
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4
Further
questions
-^ How
-^ A^ few
selected
examples
for^ which
exact
solutions
are^ possible
-^ Can
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5
^
^
^
^
^
^
Lagrange's method for solving linear PDE-s. Consider the PDE of the form,^
,^
,^ ,^
,^ ,^
1
To obtain an integral of the above equation we considerthe auxiliary equationLe
z^
z
P x y z
Q x y z
R x y z
x^
y
dx^
dy^
dz P^ Q^ ^ R
^
^
^
Recall :
^
^
^
t two independent solutions of this equation be written as,^ ,^ ^
&^
,^ ,^
where
and
are constants.
Then ,
0 is a solution of (1). Alternatively,
( ) is also a solution.
u x y z
a^
v x y z
b^
a^
b
u v
u^ f v
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0
0
1
1
1
1
1
1 ;^ 0;
0
;^0
~^
1;^ ~
;^
~^ 1
0;^
2 t
ij^ m m
I II I II II
II^
I
MX^
CX^
KX^
W^ t^
t^
X^
X^ X^
X
X^ t^
N^
n^ m^
W^ t^
m
W^ t^
W^ t W
t^
D
X^ M
CX^
M^ KX
M^
W^ t
Y^
X Y^ Y
X dY^
Y dt dY^
M^ CY
M
KY^
M^
dB t
dY^ t^
PYdt
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^ ^
^ ^
^ ^
^ ^ ^ ^
^
^
Example : Linear MDOF systems^ ^
^
^
^
0
1
1
0;^01
0
0 QdB t t^ ;
Y^
Y
I
P^
Q
M^ K
M^
C^
M
^
^
^
^
^
^
^
^
^
^
^
^
^
^
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8
^
^
^
^
1
1
1
1
2 2
2
2 1
2 1
0
Consider the eigenvalue problemLet^
be the 2
matrix of
I^
I
II^
II^
m
N^ N^
N^ m
N^
N
dY^ t
Y^ t I^
dt^
dB t
dY^
t^
Y^ t
M^ K
dY^ t
PYdt QdB t t
^
^
^
^
1
eigenvectors and
be the 2
diagonal matrix of complete set of
eigenvalues of
Introduce the transformation
Y^ t^
Z t
^
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2
2 1
1 1
1 2
2
(^02) 0
1 2
2
1
0
Define the conditional characteristic function
;0 exp ; |^
;0 exp N^
m^ m j^ j^
jk
j^
j^ k j^
j^ k
N
N j^
j^
N
j j j
p^
p
z p t^
z^
z^ z
t^ Z^
z^
t
p z t
z^
i^
z^ dz dz
dz
p z t
z^
i^
z
^ ^
^
^
^ ^
^
^
2 1
2 0
1
N j ;0 exp
N
k^
j^ j j
k
dz
t^
z p z t
z^
i^
z^ dz
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^
^
^
^
^
^
^
^
^
(^2) 0
1
2
0
1 2 0
1
2
0
1
;^
; |^
;0 exp 1
; |^
;^
;^ exp 2 ;^
; |^
;0 exp
; 1 ; |^
;^
exp
2
N j^
j j
N j^
j j N
k^
j^ j j
k
N
k^
j^ j j
k
M^
t^
p z t
z^
i^
z^ dz
p z t
z^
M^
t^
i^
z^ d
M^
t^
iz p z t
z^
i^
z^ dz M^
t
z p z t
z^
i^
z^ d
i
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^ ^ ^
^
^
^
^
2
0
1
; 1 ; |^
;^
exp
2
N
k^
k^
j^ j j
k^
M^ t k
z p z t
z^
i^
i^
z^ d
z
^
^
^ ^
^
^
^
^
^
^
^
^
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^
^ 2
2 1
1 1 2
2 2 1
(^1 )
1
2
2 2
1 1
2 2
2 2
(^1 )
1 2 1 2
1
1
2
;^ 1, 2,
, 2^
exp^
;^ 1, 2,
, 2
1
N^
m^ m j^ j^
jk
j^
j^ k j^
j^ k
N^
N^ N j^ j^
jk^ j^ k
j^
j^ k j
N
N^ N N^ N^
jk^ j^ k j^ k
i
i^
i^
i
i^ i p^
p
z p t^
z^
z^ z
M^
M^
M
t
d
d^
d dt^
dM
M
d dt^
i^
N^
t^
t^ i^
N
^
^
^
^
^
^
^ ^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^ ^
^
(^0) i (^) diagonal matrix with entries exp
i
i
t
t
^
^
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^ ^
^
^ ^
^ ^
^
^
2 2
2 2
1 1
1 1 2 2 1 1
2 2 1 1
Consider
&
1
1
2
(^212)
1 2 i^
l
N^ N^
N^ N
i^ i^
l^ l
jk^ j^ k^
jk^ j^ k
j^ k^
j^ k i^ l^
l^ i
i^ i^ l^
l^ i^ l^
l^ i^
i^ l^
N^ N
jk^ j^ k j^ k
l^ i^
l^ i N N
i^ l^
jk j^ k d^
d
dM^
dM
a^
b
M^
M dM
a^
b^
d^
d
M
d^
dM
^ ^
^
^
^ ^
^ ^
^
^ ^
^
^ ^
^ ^
^
^ ^
^
^ ^
^
^
^
^
^
^ ^ ^ ^
^
^
^
2 2 1 1 Multiply both sides by
and sum over
and 2 1 2
j^ k il
N^ N
il
l^ i
i^ l^
i^ l
M
i^ l
dM^
d
M
^
^ ^ ^
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^
^
^
0
0 0
0 0
0
0 0
(^0 00 ) (^1) exp 0 0
exp 2 1 exp^
2 1
1
,^ exp
2
2
1 exp^
2 This is the characteristic function of a multivariateGaussian PDF.
t^
t
t^
t t t^
t^
t
t^
t^
t
M^
i^ z
M^
i^ z M^
t^
i^ z i^ z
^
^
^
^
^
^
^
^
^
^ ^ ^
^
^
^
^ ^
^
^
^
^
^
^
^
^
^
^
^
^
^ ^
^
^
^
^
^
The mean vector and covariance matrix can be evaluated fromthe charateristic function.The PDF in the original coordinate system can be obtainedby using the transformation
. Y^ t^
Z t
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17
RemarksFor linear systems, the exact solution can also be obtained using convolution integral approach discussed earlier in this course.The Markov vector approach does not offer any special advantage
2 2
0
0 0
0
here.The above formulation is also valid when excitations are modeled asfiltered white noise excitations.^2
;^ 0;
0
;^0
2
;^ 0;
0
;^0
0;^
2
x^
x^
x^ f^
t^ t^
x^
x^ x^
x
f^
f^
f^ w t
t^
f^
f^ f^
f
w t^
w t w t
D
X^ t^
x t^
x t^
f^ t
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
is Markov^0 0;^
t f t 0
dX^ t^
PXdt^
QdB t t
X
X
^
^
^
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^ ^
^
^ ^ ^
^ ^
2 2 0 0; lim^32
0
1 exp^
;
Select C such that
1
Example:
1 exp^
; 2
4
0
pdf is Gaussian, as it should be.
x x
d^
x^ p^
d^ pD
p x
dx^
dx dpD
x^ p dx p x^
C^
s ds^
x
D p x dx x^ ax
bx ax^
bx
p x^
C^
x
D b
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^
^ ^
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^ ^
^ ^
^
^ ^
^
^
^
^
^
^
^
^ ^
^
0
0
(^20121 )
2
1
sdof system with nonlinear damping and
nonlinear stiffness
;^ 0;
0
;^0
.
0;^
2 Total energy x 2 x^ xf
H^
g^ x^
w t^
t^
x^
x^ x^
x
w t^
w t w t
D
xH
g u du X^ t^
x t X^ t^
x t dX^ t
X
t dt dX^
t^
X g^
H^ g^ X
^
^
^
^
^
^
^
^
^
^ ^
^
^ ^
^
^ ^
^
^ ^
^
^ ^
^ ^
Example :^ ^ ^
^
1 ^ X 2 2 20
dt^ dB t
XH
g u du
^
^
^
^
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