Calculus of Variations
Functional
Euler’s 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
Euler’s equation
V-5
Functional
Approximate method
1. Method of Weighted Residuals

Galerkin method
2. Variational Method


Kantorovich Method
Raleigh-Ritz Method
V-1 Maximum and Minimum of Functions
Part A Functional
Euler’s equation
Maximum and Minimum of functions
(a) If f(x) is twice continuously differentiable on [x0 , x1]
Nec. Condition for a max. (min.) of f(x) at x [ x0 , x1 ]
i.e.
is that F   x   0
Suff. Condition for a max (min.) of f(x) at x [ x0 , x1 ] are that F   x   0
also F   x   0 ( F   x   0 )
(b) If f(x) over closed domain D. Then nec. and suff. Condition for a max. (min.)
of f(x) at x0
2 f
that
xi x j
 D  D
x  x0
are that
f
xi
x  x0
 0 i = 1,2…n
and also
is a negative infinite .
2
(c) If f(x) on closed domain D
If we want to extremize f(x) subject to the constraints
gi ( x1 , , xn )  0 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

dg  g x dx  g y dy  0
gx
dy   dx
gy
To extremize f

df  f x dx  f y dy  0
gx
( f x  f y )dx  0
gy
3
We have
fx g y  f y gx  0
and
g 0
to find (x0,y0) which is to extremize f subject to g = 0
(ii) 2nd method : (Lagrange Multiplier)
Let

v( x, y,  )  f ( x, y )   g ( x, y)
extrema of v without any constraint
extrema of f subject to g = 0
To extremize v

v
 fx   gx  0
x
v
 fy  gy  0
y

fx g y  f y gx  0
v
g 0

We obtain the same equations to extrimizing. Where  is called
The Lagrange Multiplier.
4
V-2 Maximum and Minimum of Functionals
2.1 What are functionals.
y ( x)
y
Functional is function’s function.
y1
x1
y ( x)   ( x)
 ( x)
y2
x2
2.2 The simplest problem in calculus of variations.
Determine y ( x)  c 2 [ x , x ] such that the functional：
0
1
I  y  x     F  x, y  x  , y  x  dx
x1
x0
as an extrema
2
where F  c over its entire domain , subject to y(x0) = y0 , y(x1) = y1at the end
points.
5
x
On integrating by parts of the 2nd term
[ Fy ( x, y, y) ]xx
1
since

0
d
  [ Fy ( x, y, y)  Fy ( x, y, y)]dx  0 ------- (1)
x0 dx
x1
 ( x0 )   ( x1 )  0 and since  ( x) is arbitrary.
d
 Fy ( x, y, y)   Fy ( x, y, y)  0
dx
-------- (2)
Euler’s Equation
Natural B.C’s
 F 
 y   0 or /and
  x0
 F 
0
 y 
  x  x1
The above requirements are called national b.c’s.
6
V-3
The Variational Notation
Variations
Imbed u(x) in a “parameter family” of function
variation of u(x) is defined as
 ( x,  )  u ( x)   ( x)
the
 u   ( x)
The corresponding variation of F , δF to the order in ε is ,
since
 F  F ( x  y   , y   )  F ( x, y, y)
F
F
y
 y

y
y

and
x1
I (u   )   F ( x, u   , u   )dx  G ( )
x0
Then
x1
 I    F ( x, y, y)dx
x0
x1
   F ( x, y, y)dx
x0
 F

F
 
 y   y dx
x0
y
 y

x1
7
1
 F d  F  
 F 
 

  ydx     y 

 y
 x0
 y dx  y  
x

x1
x0
Thus a stationary function for a functional is one for which the first variation = 0.
F
y
For the more general cases
(a) Several dependent variables.
Ex :
I   F ( x, y, z ; y , z )dx
x1
x0
Euler’s Eq. 
F d F
 ( )0
y dx y
,
F d F
 ( )0
z dx z 
(b) Several Independent variables.
Ex :
I   F ( x, y, u, ux , u y )dxdy
Euler’s Eq. 
R
F  F
 F
 (
) (
)0
u x ux y u y
8
(C) High Order.
x1
Ex :
I   F ( x, y, y, y)dx
Euler’s Eq. 
x0
F d F
d 2 F
 ( ) 2 ( ) 0
y dx y dx y
Variables
Causing more equation.
Order
Causing longer equation.
9
V-4 Constraints and Lagrange Multiplier
Lagrange multiplier
Lagrange multiplier can be used to find the extreme value of a multivariate
function f subjected to the constraints.
Ex :
(a) Find the extreme value of I 
where u ( x1 )  u1
v( x1 )  v1

x1
x0
F ( x, u , v, u x , vx )dx
u ( x2 )  u2
v( x2 )  v2
and subject to the constraints
G ( x, u , v )  0
------------(1)
 F d  F   
x   F
d  F  

From  I     

u

 
v  dx  0





x
2
1

  v
dx  ux  
 v
dx  vx  


-----------(2)
10
Because of the constraints, we don’t get two Euler’s equations.
From
G 


G
G
u 
v  0
u
v
Gv
v  u
Gu
So (2)


 Gv  F d  F    F d  F  
 I      
    
   vdx  0
x0
 Gu  u dx  ux    v dx  vx  
G  F d  F   G  F d  F  
 
 


 
  0
v  u dx  u x   u  v dx  vx  
x1
The above equations together with (1) are to be solved for u , v.
11
(b) Simple Isoparametric Problem
I 
To extremize

x2
x1
F  x, y , y  dx , subject to the constraint：
(1) J  x2 G x, y, y dx  const.

 
x1
(2) y (x1) = y1 , y (x2) = y2
Take the variation of two-parameter family： y   y  y  11  x    22  x 
(where 1  x  and 2  x  are some equations which satisfy
  x    x    x    x   0 )
1
1
2
1

)
Then , I (1 ,  2 ) 
J (1 ,  2
1
x2
x1
x2
x1
2
2
2
F ( x, y  11   22 , y  11   2 2 )dx
G ( x, y  11   22 , y  11   22 )dx
To base on Lagrange Multiplier Method we can get :
12


 I   J  1  2 0  0
1

 I   J  1   2  0  0
 2
x2 
 G d  G   
  F d  F  





x1  y dx  y   y dx  y  i dx  0


i = 1,2
So the Euler equation is：


d  
 F  G      F  G    0
y
dx  y

G d  G 
 
 0 ,  is arbitrary numbers.
when

y dx  y 
 The constraint is trivial, we can ignore 
.
13
Examples
Euler’s equation
Functional
Helmholtz Equation
Ex : Force vibration of a membrane.
 2u 2  2u  2u
 c ( 2  2 )  f ( x, y, t )
2
t
x y
-----(1)
if the forcing function f is of the form
f ( x, y, t )  P( x, y ) sin(t   )
we may write the steady state disp u in the form
(1)

u  v( x, y ) sin(t   )
 2v  2v
c ( 2  2 )   2v  p  0
x
y
2
14
 2  2v  2v

2
c
(

)


v

p
 vdxdy  0 -----(2)
R  x2 y 2

Consider
c2  vxx vdxdy
c2  vyy vdxdy
 c2  [(vx v) x  vx vx ]dxdy
 c2  [(vy v) y  vy vy ]dxdy
R
R
R
Note that (vx v) x  vxx v  vx vx
V  Vx i  Vy j
R
(v y v) y  v yy v  v y v y
Vx  vx v
Vy  v y v
n  cos i  sin  j
Vx Vy ( x v) ( y v)
V 

=

x
y
x
y
15


  V nds
  (v  v cos   v  v sin  )ds
( V )da 
x
y
c2  vxx vdxdy   c2vyy vdxdy
R
R
1 2
c  (vx ) 2 dxdy

R 2
1 2
2
c  v y v sin  ds   c  (v y ) 2 dxdy

R 2
 c 2  vx v cos  ds  
 
1
2 (v 2 )dxdy  
R 2
R
(2)  
1 2
c  [(vx ) 2  (v y ) 2 ]dxdy
R 2
P vdxdy  0
c 2 (vx cos   v y sin  ) vds  
  c 2 v vds   R [ 1 c 2 (v)2  1  2v 2  Pv]dxdy  0
n
2
2
16
Hence :
(i) if v  f ( x, y ) is given on
i.e.  v  0
on 

then the variational problem

c2
1
  [ (v)2   2v 2  pv]dxdy  0
R 2
2
-----(3)
v
(ii) if n  0 is given on 
the variation problem is same as (3)
(iii) if
v
  ( s ) is given on
n
1
2
1
2



  [ R  c 2 (v)2   2v 2  pv dxdy   c 2 vdx]  0
17
Diffusion Equation
Ex :
Steady state Heat condition
  (k T )  f ( x, T )
B.C’s :
in
T  T1
kn T  q2
kn T  h(T  T3 )
on
on
on
D
B1
B2
B3
multiply the equation by  T , and integrate over the domain D. After integrating
by parts, we find the variational problem as follow.
T
1

2
 [   k (T )   f ( x, T )dT d
D 2
T0

1 
  q2Td 
B2
with
T = T1 on
2 B3
h(T  T3 ) 2 d ]  0
B1
18
Poison Equation
Ex : Torsion of a primatri Bar
 2  2
 0
in
R
on

where  is the Prandtl stress function and
  z  G

y
,
 zy   

x
The variation problem becomes

 [(D )  4 ]dxdy  0
2
R
with   0 on 
19
V-5-1 Approximate Methods
(I) Method of Weighted Residuals (MWR)
L[u ]  0 in
D
+homo. b.c’s in B
Assume approx. solution
n
u  un   Cii
i 1
where each trial function i satisfies the b.c’s
The residual
Rn  L[un ]
In this
method (MWR), Ci are 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..
20
(II) Galerkin Method
wj are chosen to be the trial functions  j 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.
Ex: Torsion of a square shaft
 2  2
 0
on
x  a , y  a
(i) one – term approximation
 1  c1 ( x 2  a 2 )( y 2  a 2 )
Ri   2 1  2  2c1[( x  a) 2  ( y  a) 2 ]  2
1  ( x 2  a 2 )( y 2  a 2 )
21
a
 
From
a
a a
R11dxdy  0
5 1
 c1 
8 a2
therefore
1 
5
2
2
2
2
(
x

a
)(
y

a
)
2
8a
the torsional rigidity
D1  2G   dxdy  0.1388G(2a)4
R
the exact value of D is
Da  0.1406G (2a) 4
the error is only -1.2%
22
(ii) two – term approximation
 2  ( x 2  a 2 )( y 2  a 2 )[c1  c2 ( x 2  y 2 )]
By symmetry → even functions
 R2   2  2
1  ( x 2  a 2 )( y 2  a 2 )
2  ( x 2  a 2 )( y 2  a 2 )( x 2  y 2 )


From
and
R
R21dxdy  0
R
R22 dxdy  0
we obtain
c1 
1295 1
525 1
,
c

2
2216 a 2
4432 a 2
therefore
D2  2G   2 dxdy  0.1404G(2a)4
R
the error is only -0.14%
23
V-5-2 Variational Methods
(I) Kantorovich Method
n
Assuming the approximate solution as :
u   Ci ( xn )U i
i 1
where Ui is a known function decided by b.c. condition.
Ci is a unknown function decided by minimal “I”.

Euler Equation of Ci
Ex : The torsional problem with a functional “I”.
u 2 u 2
I (u )    [( )  ( )  4u ]dxdy
 a  b x
y
a
b
24
Assuming the one-term approximate solution as :
u ( x, y)  (b2  y 2 )C ( x)
Then,
I (C)  
a

b
 a b
{(b2  y 2 )2 [C( x)]2  4 y 2C2 ( x)  4(b2  y 2 )C( x)}dxdy
Integrate by y
16
8
16
I (C)   [ b5C2  b3C2  b3C]dx
 a 15
3
3
a
Euler’s equation is
C( x) 
5
5
C(
x
)


2b 2
2b 2
where b.c. condition is C(  a )  0
General solution is
C( x)  A1 cosh(
5x
5x
)  A2 sinh(
) 1
2b
2b
25
where
A1  
1
cosh(
5a
)
2b
, A2  0
and

 cosh(
C( x)  1 
 cosh(

5
2
5
2
x 
)
b 

a 
)
b 
So, the one-term approximate solution is

 cosh(
u  1 
 cosh(

5
2
5
2
x 
)
b  (b 2  y 2 )

a 
)
b 
26
(II) Raleigh-Ritz Method
This is used when the exact solution is impossible or difficult to obtain.
n
First, we assume the approximate solution as :
u   CiU i
i 1
Where, Ui are some approximate function which satisfy the b.c’s.
Then, we can calculate extreme I .
I  I (c1 ,
Ex:
Sol :
From

1
0
, cn )
y  xy   x
choose c1 ~ cn i.e.
,
I

c1

I
0
cn
y (0)  y (1)  0
1  2 1 2



y

xy

x

ydx

0
I

y

xy

xy




0 
dx
1
2
2

27
Assuming that
y  x 1  x   c1  c2 x  c3 x 2
(1)One-term approx
y  c1 x 1  x   c1  x  x 2 

y  c1 1  2 x 
x 2 2
1 2
2
3
4
2 
I
c

c
1

4
x

4
x

c
x

2
x

x

c
x
x

x
dx



Then,  1  0  1 
1 
1 

2
2

c12 
4  c12  1 2 1 
 1 1   19 c 2  c1
 1  2         c1   
1
120
12
2 
3 2 4 5 6
3 4
I
19
1
0 
c1   0  c1  0.263  y (1)  0.263x(1  x)
c1
60
12
1
(2)Two-term approx
y  x(1  x)(c1  c2 x)  c1 ( x  x 2 )  c2 ( x 2  x3 )
28
Then
y  c1 (1  2 x)  c2 (2 x  3 x 2 )

1 2
I (c1 , c2 )   [ c1 1  4 x  4 x 2   2c1c2  2 x  7 x 2  6 x 3 
0 2
1 2 3
2
2
3
4
c3 (4 x  12 x  9 x )  c1 x  2 x 4  x5  2c1c2 x 4  2 x5  x 6
2
1
 




c2 2 ( x5  2 x6  x 7 )  c1  x 2  x3   c2  x3  x 4  ]dx
c12 
4 1 2 1
 7 3 1 1 1
 1  2       c1c2 1      
2 
3 4 5 6
 3 2 5 3 7
c12  4
9 1 2 9  c1 c2
  3      
2 3
5 6 7 8  12 20
19 2 11
107 2 c1 c2

c1  c1c2 
c2  
120
70
1680
12 20
29

I
0
c1

I
0
c2

19
11
1
c1  c2 
60
70
12
11
109
1
c1 
c2 
70
840
20
0.317 c1 + 0.127 c2 = 0.05

c1 = 0.177 , c2 = 0.173
 y(2)  (0.177 x  0.173x 2 )(1  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) )
30