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8/11/2019 Chapter5 Calculus of Variations
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Calculus of Variations
Functional Eulers equation
V-1 Maximum and Minimum of Functions
V-2 Maximum and Minimum of Functionals
V-3 The Variational Notaion
V-4 Constraints and Lagrange Multiplier
Eulers equation Functional
V-5 Approximate method
1. Method of Weighted Residuals
Raleigh-Ritz Method
2. Variational Method
Kantorovich Method
Galerkin method
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V-1 Maximum and Minimum of Functions
Max imum and Min imum of func t ions
(a)If f(x) is twice continuously differentiable on [x0, x1] i.e.
Nec. Condition for a max. (min.) of f(x) at is that
Suff. Condition for a max (min.) of f(x) at are that
also ( )
(b)If f(x) over closed domain D. Then nec. and suff. Condition for a max. (min.)
i = 1,2n and also
that is a negative infinite .
0 1[ , ]x x x 0F x
0 1[ , ]x x x
0F x 0F x
0F x
D D 0 0x xi
fx
of f(x) at x0 are that
0
2
x x
i j
f
x x
Part A Functional Eulers equation
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(c)If f(x) on closed domain DIf we want to extremize f(x) subject to the constraints
i=1,2,k ( k < n )
Ex :Find the extrema of f(x,y) subject to g(x,y) =0
(i) 1st method : by direct diff. of g
1( , , ) 0i ng x x
0x ydg g dx g dy x
y
gdy dx
g
0x ydf f dx f dy ( ) 0xx y
y
gf f dx
g
To extremize f
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We have
and
to find (x0,y0) which is to extremize f subject to g = 0
0x y y x
f g f g 0g
(ii) 2nd method : (Lagrange Multiplier)
extrema of v without any constraint
extrema of f subject to g = 0
To extremize v
( , , ) ( , ) ( , )v x y f x y g x y
0x xv
f gx
0y yv
f gy
0x y y xf g f g
0
v
g
Let
We obtain the same equations to extrimizing. Where is called
The Lagrange Multiplier.
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On integrating by parts of the 2nd term
------- (1)
since and since is arbitrary.
-------- (2) Eulers Equation
Natural B.Cs
or /and
The above requirements are called national b.cs.
0)],,(),,([]),,([ 10
1
0
dxyyxFyyxF
dx
dyyxF
x
x yy
x
xy
0 1( ) ( ) 0x x ( )x
( , , ) ( , , ) 0y yd
F x y y F x y ydx
0
0
x
F
y
1
0
x x
F
y
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Variat ions
Imbed u(x) in a parameter family of function the
variation of u(x) is defined as
The corresponding variation of F , F to the order in is ,
since
and
( , ) ( ) ( )x u x x
( )u x
1
0( ) ( , , ) ( )
x
xI u F x u u dx G
( , ) ( , , )F F x y y F x y y
F Fy y
y y
1
0
( , , )x
xI F x y y dx
1
0
( , , )x
xF x y y dx
Then
1
0
x
x
F F
y y dxy y
V-3 The Variational Notation
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Thus a stationary function for a functional is one for which the first variation = 0.
For the more general cases
,
1
1
0
0
xx
xx
F d F Fydx y
y dx y y
1
0( , , z ; , z )
x
xI F x y y dx Ex :
Ex : ( , , , , )x y
RI F x y u u u dxdy
Eulers Eq.
Eulers Eq.
F
y
( ) 0F d F
y dx y
( ) 0
F d F
z dx z
(b) Several Independent variables.
(a) Several dependent variables.
( ) ( ) 0x y
F F F
u x u y u
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Variables Causing more equation.
Order Causing longer equation.
Ex :1
0
( , , , )x
xI F x y y y dx
Eulers Eq.2
2( ) ( ) 0
F d F d F
y dx y dx y
(C) High Order.
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V-4 Constraints and Lagrange Multiplier
Lagrange multiplier can be used to find the extreme value of a multivariatefunction f subjected to the constraints.
Ex :(a) Find the extreme value of
------------(1)
From -----------(2)
1
0
( , , , , )x
x xx
I F x u v u v dx 1 1
( )u x u2 2( )u x u
1 1( )v x v 2 2( )v x v
where
and subject to the constraints
( , , ) 0G x u v
2
1
0x
xx x
F d F F d FI u v dx
v dx u v dx v
Lagrange mu lt ip l ier
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Because of the constraints, we dont get two Eulers equations.
From
The above equations together with (1) are to be solved for u , v.
0G G
G u vu v
vu
Gv u
G
So (2)
1
0
0x
v
xu x x
G F d F F d F I vdx
G u dx u v dx v
0x xG F d F G F d F
v u dx u u v dx v
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(b) Simple Isoparametric Problem
To extremize , subject to the constraint
(1)
(2) y (x1) = y1, y (x2) = y2
Take the variation of two-parameter family
(where and are some equations which satisfy
)
2
1 , ,
x
xI F x y y dx
2
1
, , .x
xJ G x y y dx const
1 1 2 2y y y x x 1 x 2 x 1 1 2 1 1 2 2 2 0x x x x
Then ,
To base on Lagrange Multiplier Method we can get :
2
11 2 1 1 2 2 1 1 2 2( , ) ( , , )
x
x
I F x y y dx
21
1 2 1 1 2 2 1 1 2 2( , ) ( , , )x
xJ G x y y dx
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1 2 0
1
0I J
1 2 02
0I J
2
1
0x
ix
F d F G d Gdx
y dx y y dx y
i = 1,2
So the Euler equation is
when , is arbitrary numbers.
The constraint is trivial, we can ignore .
0d
F G F Gy dx y
0
G d G
y dx y
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( , , ) ( , )sin( )f x y t P x y t
( , )sin( )u v x y t
2 22 2
2 2( ) 0
v vc v p
x y
Eulers equation Functional
Examples
we may write the steady state disp u in the form
if the forcing function f is of the form
(1)
2 2 22
2 2 2( ) ( , , )
u u uc f x y t
t x y
Ex : Force vibration of a membrane.
-----(1)
Helmholtz Equation
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2 22 2
2 2( ) 0
R
v vc v p vdxdy
x y
-----(2)
2
xxR
c v vdxdy2 [( ) ]
x x x xR
c v v v v dxdy
Consider
2
yyR
c v vdxdy2
y y y y[( ) ]
Rc v v v v dxdy
( )x x xx x xv v v v v v ( )y y yy y yv v v v v v
cos sinn i j
x yV V i V j
x xV v v
y yV v v
yx ( v)( v)
=yx
VVV
x y x y
Note that
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2 2
xx yyR R
c v vdxdy c v vdxdy
2 2 21sin ( )
2y y
Rc v v ds c v dxdy
(2)
2 2 2 21
( cos sin ) [( ) ( ) ]2
x y x yR
c v v vds c v v dxdy
2
2
1( ) 0
2R Rv dxdy P vdxdy
2 2 2 2 21 1[ ( ) ] 02 2R
vc vds c v v Pv dxdy
n
( )V da V nds ( cos sin )ds
x y
v v v v
2 2 21
cos ( )2x xRc v v ds c v dxdy
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Hence :
(i) if is given on
i.e. on
then the variational problem
-----(3)
(ii) if is given onthe variation problem is same as (3)
(iii) if is given on
( , )v f x y
0v
2
2 2 21[ ( ) ] 0
2 2R
cv v pv dxdy
0
v
n
( )v
sn
2 2 2 2 21 1
[ ( ) ] 02 2R c v v pv dxdy c vdx
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Ex : Steady state Heat condition
in D( ) ( , )k T f x T
B.Cs :
on B1
on B2on B3
multiply the equation by , and integrate over the domain D. After integrating
by parts, we find the variational problem as follow.
with T = T1 on B1
1T T
2kn T q
3( )kn T h T T T
0
21[ ( ) ( , )
2
T
D T
k T f x T dT d
2 3
2
2 3
1( ) ] 0
2B Bq Td h T T d
Diffus ion Equation
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2 2 0 on
in R
Ex : Torsion of a primatri Bar
where is the Prandtl stress function and
,
The variation problem becomes
with on
z G
y
zy x
2[( ) 4 ] 0R
D dxdy 0
Poison Equat ion
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V-5-1 Approximate Methods
(I) Method o f Weigh ted Residu als (MWR)
in D+homo. b.cs in B
Assume approx. solution
where each trial function satisfies the b.cs
The residual
In thismethod (MWR), Ciare chosen such that Rn is forced to be zero in an average
sense.
i.e. < wj, Rn > = 0, j = 1,2,,n
where wj are the weighting functions..
[ ] 0L u
1
n
n i ii
u u C i
[ ]n n
R L u
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(II) Galerk in Method
wjare chosen to be the trial functions hence the trial functions is chosen
as members of a complete set of functions.
Galerkin method force the residual to be zero w.r.t. an orthogonal complete set.
j
Ex: Torsion of a square shaft
22
0 ,on x a y a
(i) oneterm approximation
2 2 2 2
1 1( )( )c x a y a 2 2 2
1 12 2 [( ) ( ) ] 2iR c x a y a 2 2 2 2
1 ( )( )x a y a
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From 1 1 0a a
a aR dxdy
1 25 18c atherefore
2 2 2 2
1 2
5( )( )
8x a y a
a
the torsional rigidity
4
1 2 0.1388 (2 )R
D G dxdy G a the exact value of D is
40.1406 (2 )
aD G a
the error is only -1.2%
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2 2 2 2 2 2
2 1 2( )( )[ ( )]x a y a c c x y
By symmetry even functions
2 2 2R 2 2 2 2
1 ( )( )x a y a
2 2 2 2 2 2
2 ( )( )( )x a y a x y
From 2 1 0R
R dxdy
and 2 2 0R
R dxdy we obtain
1 22 2
1295 1 525 1,
2216 4432c c
a a
therefore
4
2 22 0.1404 (2 )RD G dxdy G a the error is only -0.14%
(ii) twoterm approximation
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(I) Kanto rovich Method
1
( )n
i n i
i
u C x U
Assuming the approximate solution as :
where Uiis a known function decided by b.c. condition.
Ex : The torsional problem with a functional I.
2 2
( ) [( ) ( ) 4 ]
a b
a b
u uI u u dxdyx y
V-5-2 Variational Methods
Ciis a unknown function decided by minimal I. Euler Equation of Ci
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Assuming the one-term approximate solution as :
2 2( , ) ( ) ( )u x y b y C x
Then,2 2 2 2 2 2 2 2
(C) {( ) [C ( )] 4 C ( ) 4( )C( )}a b
a bI b y x y x b y x dxdy
Integrate by y
5 2 3 2 316 8 16(C) [ C C C]15 3 3
a
aI b b b dx
Eulers equation is
2 2
5 5
C ( ) C( )2 2x xb b where b.c. condition is C( ) 0a
General solution is
1 2
5 5C( ) A cosh( ) A sinh( ) 1
2 2
x xx
b b
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1 2
1A , A 0
5cosh( )
2
a
b
where
and
5cosh( )
2C( ) 1
5cosh( )2
x
bx
a
b
So, the one-term approximate solution is
2 2
5
cosh( )2u 1 ( )
5cosh( )
2
x
bb y
a
b
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(II) Raleigh -Ritz Method
This is used when the exact solution is impossible or difficult to obtain.
First, we assume the approximate solution as :
Where, Uiare some approximate function which satisfy the b.cs.
Then, we can calculate extreme I .
choose c1~ cn i.e.
1
n
i i
i
u C U
1( , , )nI I c c1
0n
I I
c c
y xy x (0) (1) 0y y
1
00y xy x ydx
1 2 2
0
1 1
2 2I y xy xy dx
Ex: ,
Sol :From
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Assuming that
(1)One-term approx
(2)Two-term approx
21 2 31y x x c c x c x
21 11y c x x c x x 1 1 2y c x
1
2 2 2 2 3 4 2
1 1 1 10
11 4 4 2
2 2
xI c c x x c x x x c x x x dx
2 2
1 11
4 1 2 1 1 11 2
2 3 2 4 5 6 3 4
c cc
2 11
19
120 12
cc
1
0I
c
119 1
0
60 12
c
1 2(1 )( )y x x c c x 2 2 3
1 2( ) ( )c x x c x x
(1) 0.263 (1 )y x x 1 0.263c
Then,
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Then2
1 2(1 2 ) (2 3 )y c x c x x
1
2 2 2 3
1 2 1 1 20
1( , ) [ 1 4 4 2 2 7 62I c c c x x c c x x x
2 2 3 4 2 3 4 5 4 5 63 1 1 21
(4 12 9 ) 2 2 22
c x x x c x x x c c x x x
2 5 6 7 2 3 3 42 1 2( 2 ) ]c x x x c x x c x x dx 2
11 2
4 1 2 1 7 3 1 1 11 2 1
2 3 4 5 6 3 2 5 3 7
cc c
2
1 1 24 9 1 2 93
2 3 5 6 7 8 12 20
c c c
2 2 1 21 1 2 2
19 11 107
120 70 1680 12 20
c cc c c c
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0.317 c1+ 0.127 c2= 0.05 c1= 0.177 , c2= 0.173
1
0I
c
1 219 11 1
60 70 12c c
2
(2) (0.177 0.173 )(1 )y x x x ( It is noted that the deviation between the successive approxs. y(1) and y(2) is
found to be smaller in magnitude than 0.009 over (0,1) )
2
0I
c
1 211 109 170 840 20
c c