Lecture annotation
Historical introduction. Definition of variational problems, examples of the equivalence of differential equation integration and searching for the minimum of a suitable functional.
A number of problems of the mechanics, hydromechanics, mathematical physics etc., usually lead to partial differential equations – less often to ordinary differential equations, which are to be integrated under specified initial and boundary conditions . As regards to applications, there is important to obtain even approximate numerical values of the solution to these equations. In some cases, when both differential equation and a considered domain are very simple, the solution of a problem can be obtained in closed form, most often in the form of infinite series. Such results are usually obtained by Fourier’s method or by the method of
Laplace’s integral transform. In 19 th
century and at the beginning of 20 th
century mathematicians were trying to modify these methods and/or to find new methods such that they would have allowed obtaining the closed form of solution also in more general cases.
However, this effort failed, and new diverse methods started to develop in order to find a suitable approximate solution. Particularly, the method of finite differences and variational methods acquitted well. As to the variational methods, Ritz method and Galerkin method became most popular, and, with advent of powerful computers, their modification - the finite element method turned to be the most used one. Note that the finite difference method often gives unreliable values of approximate solution near the boundary of a considered domain.
In theory and in practical applications as well, we encounter a fact that has a fundamental meaning. Namely, in many cases it turns out that the integration of a differential equation can be replaced by an equivalent problem of obtaining a function which minimizes (extremizes) a suitably constructed functional. Such problems are called variational problems ; the integration of a differential equation is replaced by an equivalent variational problem. Within this context observe that differential equations are usually derived from corresponding physical principles (conservation of the momentum and the moment of momentum, conservation of the energy, mass, electric charge, and etc.), constitutive equations, and complementary relations (kinematical relations between displacements and deformations in the solid mechanics, or fluid flow assumptions in the hydromechanics etc.) put down for an element whose size converges to zero. These differential equations hold in every point of a domain, i.e. they have a local character. However, the corresponding physical principles and complementary relations can be expressed also in a different way, which is often not only more useful, but also possesses a deeper physical meaning. Already at the beginning of 20 th century the famous mathematician David Hilbert pushed forward a proposition that physical principles should not be primary formulated in terms of differential equations, but instead, in terms of integral relations. In doing so, either the energy conservation is used, or the variational principles for the energy – i.e. general thermodynamic conditions, which postulate that a system in equilibrium reaches the state exhibiting lowest value of the energy.
Hereafter, derivation of differential equations describing some physical phenomena directly from the mentioned integral relations will be demonstrated in this lecture.
Partial differential equations (PDE) can be classified into two large groups. The equations of the first group describe processes during which the searched quantities are markedly changing
1

in course of time. These unknown quantities are functions of coordinates and time. The simplest but important representative of this group is the wave equation, whose special case is the equation describing string oscillations. The wave equation belongs to so–called hyperbolic
PDEs. The derivation of the wave equation from the variational principle for the energy will now be displayed.
Assume that all points of the string lie in the same plane, i.e. the string oscillates in the plane, e.g. x-y plane. We need to express the kinetic and the potential energy of the string. The general relation for the kinetic energy E k
is as follows
E k

1
2

V


 u t i

 u t i d
V , (1)
(where u i
= 1,2,3 are the components of the displacement vector; Einstein’s summing symbolic is used, i.e. i
3 

1 u i u i
 u i u i
etc.,

denotes the mass density). In our case one obtains
( back to Lecture 9-10 )
E k

1
2 l
0
 
 v
 t


2 q d x
,
 v where the velocity
 t
(2)
of a mass element of the string with the mass
 q d x is defined by the partial derivation of its displacement v (e.g. u
1
= 0, u
2
= v , u
3
=0) with respect to time. The general expression for the potential energy of an elastic body follows as
  

( )d ij
V


 i i d


 i i d , (3)
( W stands for the strain energy density, f i
, i =1,2,3 are the components of the vector of body forces and p i
, i =1,2,3 denote the components of the vector of surface loading acting on the boundary of a body). In the investigated case, the second and the third term on the right-hand side of Eq. (3) are zero because body forces are neglected and the displacements at the ends of the string are zero (both string ends are fixed). Thus, only inner forces in the string contribute to the potential energy. We need to determine the strain energy, i.e. the work of deformation.
This work is exerted by the force N , which stretches the string, during an extension of the string. We assume that the work is proportional to the stretching force N and to the change of string length caused by its deflection. For the length d s of a line element of deformed string we obtain (compare with the Problem 2 in Appendix 1) ( back to Lecture 4 ) d s 2  d x 2  d v 2  d x 2 
 v
 x d x
2
 



1 

 v x
2 
 d x 2
. (4)
The change of the length of considered line element due to deflection reads d s  d x 



1 
 v
 x

2
 1


 d x .
(5)
Multiply the change of the element length by the force N acting across the whole cross section q of the string and obtain the work exerted by the force N on the element extension. Integrate over the total string length and finally obtain the strain energy and, thus in the given case, also the potential energy
 of the string
  l
0

N


1

 
 v x
 2


1 d x . (6)
2

Under assumption of small deformation, it holds
 v
 x
<< 1. The square root in Eq. (6) can then be expanded into the truncated binomial series leaving only first two terms of the expansion.
Obtain
1

 v
 x

2

1

1
2
 v
 x
2
. (7)
Substituting the last expression into Eq. (6), the following relation for the potential energy is received
 
1
N
 l  
 v x
 2 d x , (8)
2
0
The expressions in Eqs. (2) and (8) for E k
and

are further substituted in the Hamilton principle
 t
1  t
0
( E k
 
)d t
  t
0

0


1
2

  v
 t
 2

1
2
N q
 
 v x
 2

 d d

0 ,
According to the Hamilton principle formulation, the following conditions must hold
 v

0 for t
 t
0
and t
 t
1
.
(9)
(10)
The string ends are fixed, hence
 v  0 pro x  0 a x  l . (11)
Let us first follow the variation of the integral of the kinetic energy. According to the rules of variational calculus, the order of variation and integration is changed and the variation of the integrand is carried out. In the result obtained, the commutative property of the variation and derivative is invoked which makes possible to integrate by parts. We get by sequel
 t
1 t
0
 l
0
1
2

 v
 t
2 q d x d t  t
1 t
0
 l
0

 v
 t

 v
 t q d x d t  
 v
 t

 t
(  v ) q d x d t 
 l

0

 v
 t
 v t
1 t
0 q d x  l t
1
0
 t
0

 2 v
 t 2
 vq d x d t .
l t
1
0
 t
0
(12)
According to the Hamilton principle presumption, the variation of initial and final position of the string coincides with the actual position of the string, hence
 v  0
for t  t
0
and t  t
1
.
As a consequence, the first term on the right-hand side of the last equation of (12) disappears and we obtain
 t
1 t
0

0 l 1
2

 v
 t

2 q d x d t   t
1 t
0

0 l

 2 v
 t 2
 vq d x d t . (13)
The variation of the integral of the potential energy is treated in a similar fashion:
 t
1 t
0
 l
0
1
2
N
 
 v x
 2
 t
1 t
0
 l
0
N
 v
 x

  v
 x

 t
1  t
0

N
 v
 x l
 v d t

0
 t
1 t
0
 l
0
N
 2 v
 x
2

(14)
Here, the first term on the right-hand side of the last equation of (14) disappears because
 v  0
for x  0
and x  l
.
Thus, Hamilton’s principle leads to the equation t
0

0


 


2 t
2 v

N q

 x
2
2 v


 d d

0 . (15)
3

The preceding equations holds for an arbitrary continuous functions
 v ( x ) for 0 <x<l . Thus, basing on fundamental lemma of the variational calculus
(see also this) , it follows
immediately from Eq. (15) that



2 t
2

 2 v N v q
 x
2
, (16) which is the differential equation describing the transverse string oscillations, and in the sence of the variational calculus,
Euler’s equation
(in dynamics often called
Lagrange’s equation
) associated with the functional (9), (Hamilton principle in the given case). ( back to Lecture 11 )
Second important example of the PDE of the first group is the equation describing heat conduction, which belongs to so–called parabolic PDEs.
Equations of the second group describe stationary phenomena, where the searched quantities are not changing in course of time, or are changing only very little. These equations usually belong to so–called elliptic PDEs. The simplest representative of the elliptic PDEs is
Laplace’s equation, which describes a wide range of physical phenomena, e.g. temperature distribution in an unevenly heated body without heat sources, particle velocity distribution in potential flow, axial displacement distribution during torsion of a shaft etc. Let us follow e.g. the temperature distribution T ( x , y,z ) in a body. T ( x , y,z ) must satisfy Laplace’s equation containing three independent coordinates x , y,z
 T 


2 x
T
2



2 y
T
2

 2 T
 z 2
 0 ,
(17) where

is Laplace‘s differential operator. If volume heat sources exist, then Laplace‘s equation switches to Poisson’s equation
 T  
1
 r
V
 x , y , z

,
(18) where the function r
V
describes the volume density of heat sources and

is the coefficient of internal heat conduction. If the domain is a plate and the temperature is constant along its thickness, then the problem simplifies and can be treated as two-dimensional


2 x
T
2



2 y
T
2

0 , resp.


2
T x
2

 2
T
 y
2
  r
V
(
 x , h y )
, (18b) where h denotes the thickness of the plate. For to obtain the temperature distribution in the domain, we have to integrate Eq. (17) or (18), respectively. Let us now focus on the case when internal heat sources do not exist. Apparently, infinite number of functions satisfies
Laplace’s equation (a solution to Laplace’s equation is a harmonic function ( back to Lecture
7) ). Physically, there is however clear that in a given body subjected to a specified conditions, the temperature distribution must be definite. It means that a solution to the given problem cannot be based only on Laplace’s equation. Apparently, we have to append some additional conditions. These conditions can be found e.g. by measuring the temperature at arbitrary points on the surface of the body. Assume that the temperature was measured in a great number of points sufficiently closely distributed on the surface of the body. Then, from practical point of view, we can assume that the temperature at an arbitrary point on the surface is known. This information can be considered as an additional condition for the integration of
Laplace’s equation. Denote by 
the domain, which includes the heated body and by

its surface. Further, let P is arbitrary point on the surface

and f ( P ) is the known value of temperature at the point P . The additional condition ( boundary condition ) can be put down in the following form ( back to Lecture 3 )
T  f
 
, P  
, or, more briefly
4

T

 f . (19)
The problem of the temperature distribution can be stated mathematically as follows: find a function T ( x , y,z ) such that it satisfies Laplace’s equation in interior points of the domain
 and on its boundary

takes prescribed values. In mathematics, such problem is known as
Dirichlet’s boundary-value problem.
Boundary conditions can also exhibit a form different from the Dirichlet’s problem. Assume that the temperature of the surroundings is known and let there is a heat flow from the surroundings through the surface

into the body. According to Newton’s law, the heat flow density through the boundary is proportional to the difference of temperatures of surroundings and the surface of body. Newton’s law allows obtaining the boundary condition in the form

T
 n

(

T
0
)

0, on

, (20) where n is the outward unit normal to the surface

and k is a coefficient. Substituting kT
0
= g ( P ), the boundary condition can be put into the form



T
 n
 kT



 g . (21)
The problem of Laplace‘s equation integration along with the boundary condition (21) is often called the mixed boundary-value problem. If on the left-hand side of (21) we put k = 0, we obtain new boundary condition

T
 n

 g
 
.
(22)
The problem of integration of Laplace‘s equation along with the boundary condition (22) is called Neumann’s boundary-value problem.
Neumann’s problem is of a great importance not only in the theory of heat conduction, but also in many other cases. E. g. the problem of torsion of a shaft leads to searching a function

( x , y ), which is referred to as the warping function, describing the axial deformation od each cross section along the bar. The warping function

( x , y ) satisfies 2D Laplace’s equation in the domain formed by the cross section of the bar:
 2 
 x 2

 2 
 y 2
 0
(23) and on the boundary of this domain the boundary condition
 
 n

 yn x
 xn y
, where n x
= cos( n
, x ), n y
= cos( n
(24)
, x ) are the components of unit outward normal vector n
.
Many other physical problems lead to Neumann’s problem, such as bending of a bar loaded by a force, circumfluence of a rigid body by an ideal fluid etc.
Laplace’s equation will now be derived, similarly as the wave equation for oscillating string, from integral relations for several physical phenomena. The purpose of performing this derivation is twofold – partly the equivalence beween the differential equation and corresponding integral relation will be pointed out, and partly we get acquainted with
Gauss’ integral theorem which is essential for transforming certain integrals over a domain into integrals over the boundary and vice versa:




 f
 x
1 
 f
 y
2 
 f
 z
3

 d
 


 f
1 n x
 f
2 n y
 f
3 n z
 d
S
(25)
5

Gauss’ theorem is very useful in dealing with variational principles and performing the integration by parts in 2D and 3D problems. If f
1
, f
2
, f
3
are interpreted as components of a vector f
, Gauss’ integral theorem can be written as

 div f d
  

(26)
Substitute f
2
= f
3
= 0, f
1
= uv . Calculate the derivation in the volume integral and move one of the terms obtained on the right-hand side of the equation. Then the formula for integration by parts emerges:

 u
 v
 x d
   
 v
 u
 x d  

 uvn x d
S
. (27)
Remark 1. Formula (27) leads to Green’s relations, which are very important in mathematical physics. For a 3D domain Green’s relations for Laplace’s operator

read:
 
 v
 u d
 
3

1
 u
 x i

 v x i d
  
 v
 u
 n d
  
 grad u grad v d
  
 v
 u
 n d S
,


 u
 u d
 
3
 i 1


 u
 x i


2 d
 


 u v
 n d
 

  grad u

2 d
 

 u
 u
 n d S ,

  v
 u
 u
 v
 d
 

 v
 u
 n
 u
 v
 n d S .

( back to Lecture 2 ) ( back to Lecture 4 ) ( back to Lecture 5-6 ) ( back to Lecture 7 ) ( back to Lecture 8 ) ( back to
Lecture 9-10)
Begin with the derivation of Eq. (18). The starting point is the equation of balance for the energy. Generally, an equation of balance for a physical quantity represented by a tensor field

, has the form d dt




  d



 

V d
S 

 r
   d

, (28) where

is the mass density, V (

), a tensor of 1 order higher than

, is called an efflux of

, while r (

) is called a source density of

. An equation of balance expresses the rate of growth of
 
as the sum of two parts: a rate of flow V (

) inward through the boundary


and a creation r (

) in the interior points of domain. Equations of this form occur frequently in mathematical physics. For now assume that

stands for the internal energy density u and r stands for heat source density (e.g. by induction heating) in the domain

. Efflux V (

) has a meaning of heat flux - q through the boundary
 into the domain. If the total internal energy does not change in course of time (we are interested only in a stationary state), then Eq. (28) leads to the condition


   d S 

 r  d
  0 , (29) or, using Gauss’s theorem
 (div q  r  ) d
  0
. (30)

Applying Fourier’s law for the heat conductivity, q = -
 grad T , where grad is differential operator, grad
 i

 x
 j

 y
 k

 z
, we arrive at

 (div  grad T  r  ) d
  0
(31)
6

from where, by denoting r

= r
V
, already Eq. (18) follows. A different approach to deriving the equation for string oscillation and Laplace’s equation for temperature distribution is immediately obvious. Namely, while the former was derived from the variational principle as
Euler’s equation
, the latter followed from the equation of balance, which expresses equality between the rate of the internal energy and the energy supply from surroundings (It cannot be understood as Euler’s equation at this moment!) From the physical point of view is not quite clear, what a physical meaning of considered functional is, whose extremization would provide the differential equation (17) or (18), respectively. We will address this question later on, after deriving the equation for stationary deflection of a membrane caused by transverse loading p , which is formally identical with Eq. (18).
The relation (3) for the potential energy of an elastic body is a starting point. The strain energy density W is considered, by analogy with the case of string, to be proportional to an increase of membrane area due to deflection, and to the uniform tension per unit length in the membrane,

. Let the surface taken by the membrane after deflection is given by the following equation z

( , ), or z

( , )

0 . (32)
An elemental rectangle, whose area in the undeformed membrane is d x d y deformation into a surface element of area d S '
, and it holds d
S '  d x d y
, cos( n , z )
, moves after
(33) where cos( n , z ) is the direction cosine of the unit normal vector of the surface element d
S ' , i. e. the cosine of an angle which contains the unit normal vector of a membrane element with z axis (note that z -axis coincides with the normal vector direction in undeformed configuration of the membrane) . It follows from (32) for direction cosines cos( n , x ) : cos( n , y ) : cos( n , z )  
 w
 x
: 
 w
 y
: 1
, (34)
Thus, for cos( n , z ) we get cos( n , z )

1

1
. (35)
 w
 x

2



 w
 y


2
Substituting (35) in (33) and integrating over the total area

, the area of the deflected membrane is obtained as


1 

 w x
2



 w
 y


2 d x d y
. (36)
Bearing in mind the remarks concerning the considered form of the strain energy, and Eq. (3), the potential energy of the membrane

can be written as
   




1 

 w x
2




 w y


2
 1


 h d x d y  
 pw d x d y . (37) where h is the thickness of the membrane. Because the quantities
 w
 x

2
a


 w
 y


2
are small compared to 1 under the assumption of small deformation, the square root in Eq. (37) can be expanded into the truncated binomial series leaving only first two terms of the expansion.
After this simplification we arrive at the expression
7

 
1
2
 








 w
 x
2



 w
 y


2


 h d x d y
 
 pw d x d y

1
2
 h



 (grad w )
2 
2 p
 h w

 d

, (38) where grad w  i

 w  definition x  j  w  y of the differential operator grad in two dimensions,
, was applied. The minimum of the potential energy is searched on the set of all admissible membrane deflections w ( x , y ), i.e. such that they satisfy the boundary condition w ( x , y ) = 0 on the flange of the membrane

  
0
  h




 w
 x
  w
 x

 w
 y
  w
 y
 p
 h
 w 

 d x d y
 




  h


 2 w
 x 2

 2 w
 y 2


 p 


 w d x d y

 w
 x n x
 w
 y


 w d s .
(39)





 n y
To derive (39), Gauss’ theorem (27) for a 2D domain was used, choosing u=
 w /
 x and u=
 w /
 y respectively, and v =
 w . The countour integral in Eq. (39) is equal to zero because
 w = 0 on

.
 w is an arbitrary continuous functions in the domain

, hence the necessary and sufficient condition for the intergral to be equal zero is ( fundamental lemma , see also this )


2 x w
2

 2 w
 y 2
  p
 h
, (40) which is the equation type of Eq. (18) for a 2D domain and Euler’s equation associated with the functional

(38).( back to Lecture 2 ) ( back to Lecture 3 )
Let us return to the question of the physical origin of a functional whose extremization would lead to the differential equation for temperature distribution in a body. First of all, it should be pointed out from the thermodynamical point of view, that a body inside which heat flows exist, is not in the state of thermodynamical equilibrium and irreversible processes take place.
According to the irreversible thermodynamics principles , the entropy production is associated with irreversible dissipative processes. Let s be the entropy density, then it follows from II. law of thermodynamics d dt


 s d
   
 q n
T d S

and equation of balance (28):


 r
T d
  

   d

, (41) where

( s )

0 is the production entropy density and T >0 is the absolute temperature . The production entropy density is generally expressed by the product of not yet determined thermodynamic forces , X  , and thermodynamic fluxes , J  (thermodynamic forces are generalized forces, which work on associated thermodynamic fluxes, e.g. mechanical stress works on an increment of inelastic strain, electrical gradient works on electrical current, temperature gradient works on heat flow etc.)

  
J

, X


 

J

X

, for
 =1,2…
(42)
With respect to Eq.(42) we write in the case of heat flow J = q and X =grad(1/ T ). So called
Gyarmati principle of minimal energy dissipation states that the most likely evolution of a system in course of time is that one, for which the the energy dissipation reaches a minimum.
Energy dissipation depends on a material through the dissipation potential

, which defines a specific constitutive relation. (There is a certain analogy to the strain energy of elastic material W ). It has the same physical dimension as the entropy production, but it is interpreted as the energy dissipated for a specific functionality between thermodynamic fluxes and forces.
The dissipation potential

depends on thermodynamic forces and, in the space of thermodynamic forces, constitutes a closed convex surface, which encompasses the origin
X  =0, see Fig.1.
8

J


 

X

0
X 

Fig.1.
The flux associated with a corresponding thermodynamic force is then determined by differentiation
J


 
 X

, (43) which states that the flux is normal to the equipotential surface

= const. in the space of thermodynamic forces. The product of thermodynamic fluxes and forces in Eq. (42) has a simple geometric visualization of the scalar product of vectors, which is always positive or zero provided the equipotential surface

= const is a convex surface encompassing the origin.
Thus, it is automatically guaranteed that the entropy production is positive. A typical application is e.g. the theory of plastic flow of materials, where the plastic strain rate is obtained by differentiation of plastic potential with respect to stress.
Consider for now the potential

in the case of small deviations from the equilibrium (i.e. the absolute value of forces X  is small). In the vicinity of thermodynamic equilibrium ( X  =0) the potential

can then be expanded in Taylor’s series
    

K

X

 
 
L

2
X

X

 ...
(44)
In the thermodynamic equilibrium, the dissipation is zero, thus

(0)=0, and with respect to the property of non-negativity of the potential

, it must hold that coefficients K  = 0. For these reasons, the dissipation potential for systems near to thermodynamic equilibrium obtains the positive definite form
  
 
L

2
X

X

 0
. (45)
The constitutive relation (43) then becomes linear
J

 

L

X

(46) and, for the case of heat conductivity, it follows q   L qq grad
T 2
T
.
Comparing (47) with the relation q = -
 grad T , one concludes that L
(47) qq
=

T
2
. Consider now Eq.
(43), multiply it by the variation of the generalized forces,

X  , sum over

and integrate over the whole body. Obtain:




J

 X
 d
 




 
 X

 X
 d
 

   d

. (48)
During variation of thermodynamic forces, the fluxes are considered to be constant, and with respect to the definition of entropy production, we put Eq.(48) into the form
9



    
 d
  0
. (49)
Eq. (49) expresses the principle of minimal energy dissipation. Consider the problem of heat conductivity and a stationary state. In the stationary state, the rate of the entropy change is equal to zero. Express the total entropy production from Eq. (41), substitute it in Eq.(49) and substitute for the dissipation potential the expression

=

[grad( T )/ T ]
2
. Get







T
 d S 



T r
  grad
T
T

2 d




 0 , (50) where the variation is taken with respect to T . In case of Dirichlet’s condition, T = f ( P ), P

, the variation of the surface integral in (50) is equal to zero. The volume integral in (50) can be estimated as follows



  grad
T
T
 2


T r


 d
  

T
1
2

  grad T

2   rT
 d
 
1
2
T min



  grad T

2   rT
 d , (51) where T min
denotes the lowest temperature in the domain

. To find a minimum of energy dissipation, it is sufficient to minimize the last integral (51). The mathematical structure of the last integrand is identical with the structure of integrand on the right-hand side of Eq. (38), and in both cases Dirichlet’s condition is considered. Aparently, by minimizing the last integral in (51) we obtain Poisson’s equation (18), or for r
V
= 0, Laplace’s equation (17), describing temperature distribution in the domain

, emerges. Thus, Eqs. (18), or (17) can be understood as Euler’s equation associated with the functional (51).
The last example is concerned with torsion of a straight bar. Let the bar axis coincides with z axis and one of the end cross sections lies in the plane z =0. Introduce the angle of twist per unit length of bar,

. In the cross section with a general position z , the displacement components u and v are given by u = -
 zy a v =
 zx.
Cross section warping is described by an unknown function w=

( x , y ). The strain components follow as
 xz

1
2

 u z


 w x


2


 x
 y ,
 yz

1
2



 v z


 w y




2




 y
 x


,
(52)
 xx
  yy
  zz
  xy

0 .
From Hooke’s law the stress components are obtained as
 xz

2 G
 xz

G



 x
 y ,
 yz

2 G
 yz

G





 y
 x


,
(53)
 xx
  yy
  zz
  xy

0 , where G is the shear modulus. ( back to Lecture 5-6 ) The strain energy density W is defined as
W = 1/2
 ij
 ij
(Einstein’s symbolic is used). Using (52) and (53) obtain
W

1
2

G
1
2

  ij ij
   xz xz
   yz yz
2






 x


2





 y


2


2


1
2 G
 
 y
 x

2
 xz


 
 x
2 yz


1
2
G
 y


 x
2  y
2
2


.





 
 x
2
 y




 
 y
 x


2

 
(54)
Because the volume forces are neglected and the loading on the circumferential surface of the bar is zero, the potential energy reduces to
10

 
1
2
G

2 









 x


2





 y


2

2


 
 y x

 
 x y


 
 d d
 

 x
2  y
2
 d d 

. (55)
The second integral is the polar moment of inertia of the cross section, which is constant.
Hence, the variational problem is reduced to minimizing a simpler functional
  
1
2
G




  
 x
 2



 
 y


2

2




 y x



 x y


 
 d d


1
2
G




 

 x
 2



 
 y


2

2



(

 y x )


(

 x y )






(56)

1
2
G




 

 x
 2

   

 y

2

  

 xn y
  yn ds x


0, where Gauss’ theorem was applied. The minimum of the potential energy is searched on the set of functions continuous including the first derivative and no additional restrictions are imposed. In analogy with Eq. (39) we derive
   
G

2




 2
 x

2

 2
 y

2


  
2






 x n x



 y n y
 xn y
 yn x


 ds

0 (57)
In the cross section domain

, and on its boundary

as well,

is an arbitrary function, thus the necessary and sufficient condition for

=0 is that both integrands are zero, i.e.
 2
 x

2

 2
 y

2

 
 x n x

 
 y n y



 n
 yn x
 xn y
on boundary

, (58) which are actually the equations (23) and (24). While the first equation (58) is Euler’s equation associated with the functional

, the second equation (58) is the boundary condition.
Observe that the boundary condition (24) is a dicrect consequence of minimalization of the potential energy (on the contrary to Dirichlet’s boundary condition). It is referred to as natural boundary condition . Natural and essential boundary conditions
(Dirichlet’s conditions in preceding examples) will be mentioned also in Lecture 2 . ( back to Lecture 4 )
Remark 2. Matematicians use slightly different approach to the determination of functional extremum, which is based upon the conception of Gâteaux differential . Assume that a Dirichlet boundary value problem for
Laplace’s equation is solved in the domain

. The minimum of corresponding functional F is searched on the set of all function twice differentiable in

, which satisfy Dirichlet’s boundary condition on

. Let u
0
( P ) ( P is a general point in

) is a function for which the functional F attains its minimum on the given set. Let t is an

fulfilling Dirichlet’condition arbitrary real number and v ( P ) is an arbitrary function twice differentiable in prescribed on

. Obviously, it holds
F
 u
0
 tv


F
 
0
.
For every fixed function v is the functional F ( u
0
+ tv ) only a function of the variable t . Last inequality shows that this function exhibits a minimum for t = 0. Then it must hold ( also see this ) d
F
 u
0
 tv
 t

0

0 dt
Let us demonstrate the technique e.g. at Eq. (38). It can be easily shown

 w
0
 tv

 
 
0
 t
 h



 grad w
0 grad v
 p
 h v

 d
  t
2

  grad v
 2 d
 .
Differenciate the preceding expression with respect to t , substitute t= 0 and the result put equal zero. Obtain
 h



 grad w
0 grad v

 p h v

 d
 
0
.
According to Green’s formula (see Remark 1) it holds
11


 grad w
0 grad v d
   
 v
 w
0 d
  
 v

 w n
0 d S , where the second integral on the right-hand side disappears because v = 0 on

. Finally obtain
 h



 w
0

 p h

 v d
 
0 .

Since the function v is arbitrary, it must hold
 w
0
= p /
 h in

, which is, however, the equation (40). The function v is sometimes called a testing function and if its norm v

1 , it coincides with the variation of the function w
0 frequently used in technical and physical literature, and in this and subsequent lectures.

( back to
Lecture 4 )
We now focus to purely mathematical questions. Previous examples demonstrated the equivalence of the integration of a differential equation with a variational problem of obtaining a function, which minimalizes (extremizes) a specific integral - functional. In the process of extremization, the differential equation emerged as Euler’s equation of corresponding functional. Instead of deriving Euler’s equation (which has to be integrated for to obtain the solution), we can attempt to obtain the solution directly by an a p p r o x i m a t e extremization of the functional. For the first time in history, the variational problem was presented in the form of “Dirichlet’s principle” for a 2D domain. This principle states that among all functions, which take a prescribed value on the boundary of considered domain

, it is a harmonic function in

(satisfies Laplace’s equation in

, see the first equation in
(18b), (23), or the first equation in(58)), for which “Dirichlet’s integral”









 w
 x
2



 w
 y


2


 d x d y (59) attains a smallest value, see e.g. the first integral in Eqs. (38), or (55). The mathematician
Riemann considered as evident, that a function, which mimimalizes Dirichlet’s integral, exists. It inspired critical comments of another mathematician Weierstrass, who showed, using a simple one-dimensional example that a smallest value of the integral does not have to be reached. To gain insight into mathematical way of thinking, it is desirable to show
Weierstrass’s example.
Weierstrass investigated following problem: among all continuously differentiable functions in the interval -1

x

1, satisfying the boundary conditions y
 
  1 , y
 
 1
(60) find such a function, for which the integral
J 
1

 1 x 2 y ' 2 d x
(61) attains a smallest value. The value of the integral J ( y ) depends on the choice of function y ( x ).
Clearly, J ( y )

0, hence the integral (61) is lower bounded and thus, it has an infimum. We prove that this infimum is equal to zero; it is sufficient to assure ourselves that it is possible to find a function y ( x ), which satisfies the preceding boundary conditions, and which assigns a lower value to the integral J ( y ) than an arbitrary, predefined positive number. Select an arbitrary number

0 and substitute y

 arctan arctan
 
  . (62)
It is obvious that the function y  is continuous including its first derivative in -1

x

1 and satisfies the conditions (60). Further,
J  
1
 1 x 2 y ' 2
 d x  
1
 1
( x 2   2 ) y ' 2
 d x 
 2 arctan 2

1
 1 x 2 d x
  2

2  arctan
. (63)
12

Eq. (63) shows that J ( y  ) attains an arbitrary small value for arbitrary small

. From the definition of infimum, it follows that the infimum of the integral J ( y ) is equal to zero.
However, there exists no such function, continuous including its first derivative in -1

x

1, which satisfies the conditions (60) and renders the integral J ( y ) equal to zero. Really, let J ( y )
=0. Since the integrand in (61) is non-negative, it must equal to zero. Then y '  0 and y = const, which contradicts to the conditons (60). Thus, it may happen that the integral does not attain its infimum and the corresponding variational problem is not solvable. This circumstance raised doubts about Dirichlet’s principle.
Weierstrass’s objections resulted in that Dirichlet’s principle was missed out for a long time.
An interest in this principle and variational methods started up again in the early 20 th century due to works of D. Hilbert, who showed that problems, relating to correct application of
“Dirichlet’s principle”, were not caused by a “small error” in formulation, but they had much deeper origin connected with what we call now the completeness of metric space . Remind the theorem about the completeness of metric space:
Theorem 1 : A metric space P is called complete , if every Cauchy sequence of functions u n
of this space has a limit u in this space. Here, a Cauchy sequence is a sequence of functions u n
( x ) from P such that lim m
 n

  u u m n
 
0
(

( u n
, u m
) is so called distance of functions u n
, u m
)
Remark 3 . A set M is called metric space, if there is defined the distance for every pair of its elements
(functions) u , v ,

( u,v ), which satisfies following requirements:

,

0, while
   u v ,
      
,
          
for all , ,

M .

Remark 4 . A set M is called linear if for its elements the operations of the addition of two elements (functions) and of the multiplication of an element by a real number are defined, with properties familiar from vector algebra
( of the form au +bu = ( a + b) u , etc), while u

,

, , real numbers
 au bv M .

( back to Lecture 2 ) ( back to Lecture 3 ) ( back to Lecture 4 )
( back to Lecture 5-6 )
Remark 5. A linear set M is so called Hilbert space , if to every pair of its elements (functions) u , v , there is assigned a number ( scalar product ), denoted by ( u , v ), satisfying following conditions:
A. It holds
 

 
, where overbar stands for complex conjugate. In the case of real elements, the preceding condition is replaced by:
   
.( back to Lecture 11 )
We say that two functions u , v

H are orthogonal in the space H if ( u , v ) = 0.

B. It holds
 u
  u , v

 
 
 
 u , v

1 1 2 2 where u
1
, u
2
1 1
, v

M and

1
,

2 2
2
are arbitrary constants.
C. It holds
 

0 .
D. If u
 

 u ,
0
 
, then u = 0 and vice versa. The number is called the norm of element u , and the number

 
 u
 v is the distance of elements u , v.
13

E. Every Cauchy sequence { u n
} from M converges in M to an element u

M.

Remark 6.
A set M of functions from Hilbert space H is called dense in this space, if every function u

H can be approximated to arbitrary accuracy by an appropriate function from M. In other words, every function u

H can be obtained as the limit of sequence u n

M.
From the point of applications, the most important Hilbert spaces are those, which have a dense set. If there exists such a subset M of a Hilbert space H , which is dense and countable , the Hilbert space is separable . Note that the most important space for applications, L
2
(

), is separable. The concept of density is of fundamental importance in numerical methods because it concerns the possibility of the approximation of functions (e.g. of the solution of a problem to be found) by functions of a certain type (e.g. by polynomials)

( back to Lecture 2 ) ( back to Lecture 3 ) ( back to Lecture 5-6 ) ( back to Lecture
11 )
Remark 7 . L
2
(

) means the set of all functions u ( x ) square integrable in

( thus the integral

 u 2
  d x exists and is finite) on which there are defined: the scalar product
       d x

 , the norm u
    
 u
2
  d x
, the distance
      

2 d x
. (*)
A sequence of function u u

L
2
(

), and we write u n
 u in L
2
 
, n

L
2
(

) is said to converge in this space (or to converge in the mean ) to the function if lim n

 
, n
 
0 , or lim n



   u n
  2 d x

0
, or (what is the same) lim
    u
  2 d x

0 n
 n

For two functions u ( x ), v ( x ) which need not be equal at all points x

, we may have
 2 d x

0
.

Thus, their distance in the space L
2
(

) is equal to zero. Such functions are called equivalent . They differ in

on a set of (Lebesque) measure zero , or as we say, they are equal almost everywhere in

.

( back to Lecture 2 )
( back to Lecture 3 ) ( back to Lecture 5-6 ) ( back to Lecture 9-10 )
Fig. 2 shows courses of the function y  for various

=1/ n , where n is arbitrary natural number.
It can be intuitively deduced from Fig. 2 that for n

, the sequence of functions y n
 arctan( nx ) / arctan( n )
converges to the function
 




1 for 1
0 for x

0,
1 for 0 1,
0,
(64) which is the limit of the sequence of functions y n
in the space L
2
(-1,1), i.e. n lim
 
 2
 y n
, y

 0
Both, the function (64) and the functions y n
are from the space L
2
(-1,1). It is sufficient to show that
14

n lim

  n
 
0, or, what is the same lim n


2
 n
 
0 . (65)
Fig.2
According to the definition of the distance in L
2
(-1,1), and because the functions y n and y are odd functions, we get


2
 y n
, y


1
 1

 ln( 2 ( 1  n 2 ) n arctan n
 arctan
 arctan n
 y


2 dx
 i
2 n arctan 2 n

 dilog
2

1
0

 arctan arctan n
2 i n
 dilog
 1


2 d x 
2 i n
2 



 arctan n

2 arctan 1 arctan n i
4 n arctan 2 n ln 2
1 
1  i i n n
.
n

(66)
Obviously, n lim
 
 2
 y n
, y

 0
, hence, really n lim
 y n
    
in L
2
 
.
Observe, that while the functions y n
( x ) are continuous in the interval

-1,1

, the limit function y is not continuous in this interval.
Not every metric space is complete. Consider a simple example: let N is a linear set of functions from the space L
2
(

). Define the scalar product in this set in the same manner as in the space L
2
(

), hence
 
 u
    d x , u

N , v

N . (67)


Using this scalar product, define the norm and the distance in N.
Thus, a certain metric space is obtained. Denote it as N
L
2
  . In contrast to the space L
2
(

), the space N
L
2 need not be complete. For example, consider a linear set of all functions continuous in the closed interval

-1,1

. It can be easily shown that the corresponding space N
L
2
 
1 , 1
 is not complete. The sequence of the functions y n
( x ) from the previous example is apparently from the space
N
L
2
 
1 , 1
 , because these functions are continuous. The sequence y n
( x ) is a Cauchy sequence in the space N
L
2


1 , 1
 , because it is convergent, see Fig.2 and Eq. (66), and thus, it is also the
Cauchy sequence in the L
2
(-1,1) , since the spaces L
2
(-1,1) and N
L
2
 
1 , 1
 have the same metric.
Yet the sequence y n
( x ) is not convergent in the space N
L
2


1 , 1
 does converge to a certain function v

N
L
2
 
1 , 1

. Presume by contraries, that it
, which is, thus, continuous (since it is from
15

this space) in the interval

-1,1

and we come to contradiction. Actually, then the sequence y n
( x ) in the space L
2
(-1,1) would also converge to the function v , since L
2
(-1,1) has the same metric as the space N
L
2
 
1 , 1
 . We have seen in the example that the sequence y n
( x ) converges in
L
2
(-1,1) to the function y ( x ) given by (64), which is not continuous in the interval

-1,1

. y ( x ) is not even equivalent in L
2
(-1,1) with a continuous function, because y ( x ) cannot be rendered continuous in

-1,1

by changing it values on the set of measure zero . Thus, v

y in L
2
(-1,1), so that basing on the presumption, that the continuous function v ( x ) is the limit of the sequence y n
( x ) in N
L
2
 
1 , 1
 , we arrive to conclusion, that this sequence has two different limits in L
2
(-1,1). That is, however, in contradiction with the theorem, that a sequence of functions in L
2
(-1,1) can have one limit at most.
Another significant factor that drew attention of theoreticians and practicians to the exploitation of Dirichlet’s principle, was a method suggested by German engineer W. Ritz in
1913. We outline the essential idea of his method on the solution of Dirichlet’s problem for
Poisson’s equation (one of the first problems, their solution Ritz attempted using his method).
We have seen in foregoing examples that Dirichlet’s problem for the Poisson’s equation


2 x u
2



2 y u
2
   
(68) u
 leads to the variational problem of finding a minimum of the functional
J
 









 u x
2



 u
 y


2


 d x d y

2

 fu d x d y (69) on a set M , whose elements are sufficiently smooth function satisfying the boundary condition u = 0. Choose n linearly independent functions in M

1
 
,...,
 n
 
(70) and seek an approximate solution of the considered problem as a function which minimalizes the functional J ( u ), however not in the set M , but in a „ n -dimensional“ subset P , whose elements are all functions exhibiting the form b
     b n
 n
 
(71) with arbitrary coefficients b
1
..., b n
(so called a p p r o x i m a t e m i n i m i z a t i o n ). If Eq. (71) is substituted into (69) in place of function u ( x,y ), the functional (69) becomes a function h ( b
1
…, b n
) of these coefficients. If the functional (69) is required to attain a minimum for a function u n
   c
  
...
c n
 n
 
(72) in the subset P , following equations must hold
 h
 b
1
 c
1
,..., c n


0 ,...,
 h
 b n
 c
1
,..., c n


0 .
(73)
The form of considered functional (69) suggests that the function h ( b
1
…, b n
) is a quadratic functions of variables b
1
..., b n
, hence the system (73) for unknowns c
1
..., c n
is linear. If the solution of (73) is found, then the function (72) is, in a certain sense, the approximation of searched solution of the problem (68). ( back to Lecture 3 )
Remark 8 . A system of functions {
 n
}
 n
 
, n

1, 2...,
(74)
16

from a Hilbert space H (

), (not orthogonal, in general), is called complete in this space if the set of all linear combinations of these functions is a dense set in H (

). If, moreover, the system (74) is linearly independent in
H (

) (none of the functions (74) can be expressed as a linear combination of the others) it is called a base in this space. In particular, every complete orthonormal system in H (

) is a base in this space (in this case we speak of an orthonormal base.
) We can construct an orthonormal system { functions {

 n
} from any system of linearly independent n
} using Schmidt’s orthogonalization process . Note that an orthonormal system of functions {
 n
} fulfils the relations
 m n
  

1, m
 n ,
0, m
 n .
We proceed according the following scheme

1



1
1
,
 n

 n
 n

 k n

1 

1 k n

1 

1
 n k

 n k
 k k
, n

1
The completeness of orthonormal system of functions { orthonormal system of functions {
 

 n
} in a Hilbert space H can also be defined as follows: n
} is complete in H , if there does not exist an element in H different from zero, which would be orthogonal to all functions of the system. If it were to the contrary, then the system { incomplete . If the system {
 n
} is complete, then it follows from
 n
,
 
0 , n

1 , 2 ,...
 n
} is that

= 0. Necessary and sufficient condition for the existence of a complete, countable or finite, orthonormal system in a Hilbert space H is that this space is separable, cf. Remark 6. For applications, the following lemma is of great importance:
A function orthogonal to all functions of a dense set is equal to zero.

( back to Lecture 2 ) ( back to Lecture 3 )
( back to Lecture 5-6 ) ( back to Lecture 11 )
Problems
( back to Lecture 11 )
1.
Derive the equation describing membrane oscillations using Hamilton principle.
Hint: Use Eq. (38) for the potential energy with p =0, Eq. (1) in the form E k

1
2

V

  i u i t t d V

1
2


 h
 
 w t
 2
, substitute both expressions into lefthand side of Eq. (9) and perfom variation..
Answer:



2 t w
2
 


 2
 x w w
2



2 y
2


2.
Derive the equation describing axial oscillations of a simple bar (length L ,
Young’s modulus E , density

, cross-sectional area A , axial displacement u ) using the Hamilton principle.
Hint: The potential energy is given by
L
0


1
2
EA
  u
 x
 2

 d x , see Example 1 of
Appendix 1, where the axial loading p x
is set equal to zero. The kinetic energy is given by E k
 t
1  t
0
( E k

1
2

L
0

A
 
 u t
 2 d x . From the Hamilton principle it follows
 
)d t
  t
0

0


1
2

A
  u
 t
 2

1
2
EA
  u
 x
 2 
 d d

0 .
Observe
17

 t
1 t
0

L
0
1
2

A
  u
 t
 2 where
 u
 d d
 
A t
1 t
0

L
0


 t
2 u
2

0 for t
 t
0
and t
 t
1
, was applied. Fot the variation of

obtain
 t
1 t
0

L
0
1
2
EA
  u


 x

2 d d

AE t
1 t
0

L
0
 u
 x

  u


 x

Thus, Hamilton principle leads to

AE t
1  t
0

 u
 x
 u

L
0 d
 t
0

0


E
 2 u
 x
2
 
 2 u
 t
2



 t
1  t
0

 u
 x
L
 u d


0 d d

0 . t
1 t
0

L
0
 2 u
 x
2

For the essential boundary conditions ( u =0, for x =0, L ) the first term disappears.
Otherwise, the natural boundary conditions
 u
 x
L

0 (free ends) follow. The second
0 term provides the searched differential equation E
 2 u
 x
2
 
 2 u
 t
2

0 .
3.
Derive the equation describing transversal oscillations of a simple bar (length L ,
Young’s modulus E , density

, cross-sectional area A , tranverse displacement w ) using the Hamilton principle.
Hint: The potential energy is given by
  
L
0
1
2
EI


 2 w
 x
2


2 d x , see Example 2 of
Appendix 1, where the axial loading p z
is set equal to zero. The decisive contribution
 w to the kinetic energy is due to the velocity of translation
 t of the cross section of the bar. Thus, E k

1
2

L
0

A
 
 w t
 2 d x . Hamilton principle gives
 t
1  t
0
( E k
 
)d t
  t
0

0


1
2

A
  w
 t
 2

1
2
EI


 2
 x w
2


2 
 d d

0 with
 w

0 for t
 t
0
and t
 t
1
. Integrate by parts the term

L
0


 2 w
 x
2


2 d x according to
Eq.(2f) in Appendix 1, apply the essential and/or natural boundary conditions at x
=0, L according to Eq. (2g) in Appendix 1 and get the differential equation describing transversal oscillations as

 t
2 w EI w
2


 4
4

0 .
18