# Dynamic Optimization Basic Concepts, Necessary and Sufficient conditions Lecture 30

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Dynamic Optimization Basic Concepts, Necessary and Sufficient conditions Lecture 30
```Dynamic Optimization
Basic Concepts, Necessary and Sufficient conditions
Lecture 30
Concept of function and functional …
• Functional: It is a function of other functions
• Examples:
• (1  , 2  …   )
• (  ,   , )

•    =     It represents distance traveled, v(t) is velocity
• In general
•    , =
•
0

(
0
, )
Concept of function and functional …
• Increment of functional:
• Consider following functional    ,  =
• ∆   ,  =    + (),  −    ,

(
0
---(1)
Perturbation
• Do Taylor’s series expansion of 2nd term
• ∆   ,  =
1 2  .
2!  2   = ∗
∗

1  .
, +
1!    = ∗
2
∗
() + ⋯ −    ,
, )
+
Concept of function and functional …
• ∆   ,  =
1  .
1!    = ∗

1st variation of functional
1 2  .
+
2!  2   = ∗
2nd
()2
2
variation of functional
Neglecting higher order terms
•
•
•
•
•
∆ =  + 2
Necessary condition for a functional to be extremum (min or max)
= 0 and
If  2  > 0
functional value is minimum
If  2  < 0
functional value is maximum
Two point boundary
value problem
(TPBVP)
• Problem 1: Our problem is to find out the optimal trajectory x*(t) for

which the functional    ,  =  (  ,   , )
---(1)
0
• is optimized (Extrimum -> maximize or minimize)
• Assumptions:
• V(.) has 1st and 2nd partial derivatives w.r.t all its arguments
Necessary Conditions:
•
.

.
−

•  . −
•
.

.

=
= 0×1

=0
()
---(1)
=0
=
---(2) if  is free, ⇒  ≠ 0
---(3) if ( ) is free, ⇒  ≠ 0
• (1) Euler’s Lagrange equation(E.L equation).
• (2) & (3) Transversality conditions
Necessary Conditions:…
• Case 1:
• when both end are fixed i.e  and ( ) are fixed
• We need to solve equation (1) to get optimum trajectory  ∗ ()
Necessary Conditions:…
()
This is suboptimal path very close to x*(t)
=  ∗  + ()
B(tf , x(tf))
x(tf)=xf
()
()
x(t0)=x0
Say this is optimal path
∗
A(t0, x(t0))
t0 t
tf
t
Necessary Conditions:…
• Case 2:
• when  and ( ) are free
()
D
C
∗ ()
A
t0
tf  +  t
• We need to solve equations (1), (2) & (3) to get optimum trajectory  ∗ ()
Necessary Conditions:…
• Case 3:
• when  is fixed but ( ) is free
()
( )
∗ ()
t0
tf
+  t
• We need to solve equations (1) & (2) to get optimum trajectory  ∗ ()
Necessary Conditions:…
• Case 4:
• when  is free but ( ) is fixed
()
∗ ()
t0

t
• We need to solve equations (1) & (3) to get optimum trajectory  ∗ ()
Necessary Conditions:…
• Case 5:
• When end point B restricted to a curve g(t)
()
D C
∗ ()
B
g(t)
= ( +  )
A

t0
t
+
Necessary Conditions:…
•  . −
.

− ()
=0
---(4)
=
• Equ (4) is also known as Transversality conditions for the case when
the end point B is restricted on a curve g(t).
• We need to solve equations (1) & (4) to get optimum trajectory  ∗ ()
Sufficient condition
• Condition to check whether the given functional is maximum or
minimum
• Similar to sufficient condition of static optimization problem
Sufficient condition…
2
•⇒ =
1
2! 0
()
()
2  .
2
.

.

2  .
2
()

()
∗
Say P which is a 2nx2n matrix
as each 2nd derivative of V(.) is nxn matrix
Sufficient condition…
• If  2  > 0 ⇒  > 0 (+)
• => functional value will be minimum
• =>J*(.) will be the minimum value along with the optimum trajectory x*(t)
• If  2  < 0 ⇒  < 0 (−)
• => functional value will be maximum
• =>J*(.) will be the maximum value along with the optimum trajectory x*(t)
Example: (Using calculus of variation)
• Example 1: Find the extremal for the functional
•=

0 =0
2 2  + 24
• Where the left hand point (A) is fixed i.e. t0 =0 & x(t0) =0
• And in right hand point (B) tf is free but x(tf) = 2
Example:…
• Solution: This is following problem
()
( ) = 2
∗ ()
t0
•  . = 2 2  + 24
tf
+  t
Example:…
• E.L equation
.

•
−
01×1
• ⇒ 24
=0

.
=

−
4()

(x is scalar variable)
• ⇒ 24 − 4() = 0
• ⇒ () = 6
---(1)
• Solving above equation, Integrating equ(1)
•    = 6⇒   = 3 2 + 1 ---(2)
• ⇒    = 3 2 + 1
• ⇒   =  3 + 1  + 2 ---(3)
again integrating
Example:…
• Put t=t0=0 in equation (3)
•   = 0 = 03 + 1 × 0 + 2
• ⇒ 2 = 0 put in equation (3)
• ⇒   =  3 + 1
---(4)
• Put t=tf &   = 2 in equation (4)
•   = 2 =  3 + 1
• ⇒ 1 =
2− 3

---(5)
Example:…
• Transversality condition: Here  is free, so
•  . −
.

()
=0
=
• 2 2  + 24  − (4())()
• ⇒  2  + 12   − 2 2 ( )
• ⇒  2  − 12   = 0
•⇒
•⇒
2
2
3 + 1 − 12
94 + 12 + 62 1 − 24
=0
=0
=
=0
dividing by 2
put value of () from equ(2)
put   = 2
Example:…
• Put the value of 1 =
• ⇒
94
+
2− 3

2
2− 3

from equ (5) in previous equ
3
2 2−
+ 6

• ⇒ 6 −43 + 1 = 0
• Put 3 =
•  2 − 4 + 1 = 0
• ⇒  = 2 ± 3 = 3.732, 0.2679
• ⇒  =
3
3
3.732 & 0.2679
• ⇒  = 1.551 & 0.6446
− 24 = 0
Example:…
• For  = 1.551 , 1 =
2− 3

= −1.117 so
•  ∗  =  3 + 1  =  3 − 1.117
• For  = 0.6446 , 1 =
2− 3

---(6)
= 2.6871so
•  ∗  =  3 + 1  =  3 + 2.6871 --(7)
• There are two optimal trajectory given by (6) & (7), one will give
minimum value other will give maximum value. For this apply
sufficient condition
Example:…
• sufficient condition
•=
2  .
2
.

.

2  .
2
∗
• If  > 0 (+) => functional J*(.) value will be minimum
• If  < 0 (−) => functional J*(.) value will be maximum
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