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Lecture 1
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 19th century and at the beginning of 20th 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 20th
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 Ek is as follows
1 u u
(1)
Ek    i i dV ,
2 V t t
(where ui = 1,2,3 are the components of the displacement vector; Einstein’s summing
3
symbolic is used, i.e.
u u
i 1
i i
 ui ui etc.,  denotes the mass density). In our case one obtains
(back to Lecture 9-10)
l
2
1  v 
Ek      qdx ,
2 0  t 
(2)
v
of a mass element of the string with the mass qdx is defined by the
t
partial derivation of its displacement v (e.g. u1= 0, u2=v, u3=0) with respect to time. The
general expression for the potential energy of an elastic body follows as
(3)
   W (eij )dV   fiui dV   piui dS ,
where the velocity



(W stands for the strain energy density, fi, i =1,2,3 are the components of the vector of body
forces and pi, 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 ds of a line element of deformed string
we obtain (compare with the Problem 2 in Appendix 1) (back to Lecture 4)
2
2
 v    v   2
2
2
2
2
(4)
ds  dx  dv  dx   dx   1     dx .
 x    x  
The change of the length of considered line element due to deflection reads
2


 v 
ds  dx   1     1 dx.
(5)
 x 


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
2
l


 v 
   N  1     1dx .
(6)
x 



0


2
v
<< 1. The square root in Eq. (6) can then
x
be expanded into the truncated binomial series leaving only first two terms of the expansion.
Obtain
Under assumption of small deformation, it holds
1  v 
 v 
1   1    .
(7)
2  x 
 x 
Substituting the last expression into Eq. (6), the following relation for the potential energy is
received
2
2
1
 v 
(8)
  N    dx ,
2 0  x 
The expressions in Eqs. (2) and (8) for Ek and  are further substituted in the Hamilton
principle
t1
t1 l
 1  v 2 1 N  v  2 
  ( Ek   )dt         
(9)
   qdxdt  0 ,
2

t
2
q

x





t0
t0 0 


According to the Hamilton principle formulation, the following conditions must hold
(10)
 v  0 for t  t0 and t  t1 .
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
2
l
t1 l
2
t l
l t
1
1
1  v 
v  v 
v 
q
d
x
d
t



q
d
x
d
t


(v)qdxdt 

 




2

t

t

t

t

t




t0 0
t0 0
0 t0
   
(12)
t1
l t1
 2v
 v 
    v  qdx     2 vqdxdt.
t  t0
t
0
0 t0
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  t0 and t  t1 .
As a consequence, the first term on the right-hand side of the last equation of (12) disappears
and we obtain
l
t1 l
2
t l
1
1  v 
 2v
      qdxdt      2 vqdxdt .
2  t 
t
t0 0
t0 0
(13)
The variation of the integral of the potential energy is treated in a similar fashion:
l
1
1
1
1  v 
v  v 
 2v
 v 
   N   dxdt    N    dxdt    N  v  dt    N 2  vdxdt.
(14)
2  x 
x  x 
x  0
x
t0 0
t0 0
t0 
t0 0
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 .
t1 l
2
t l
t
Thus, Hamilton’s principle leads to the equation
t1 l

 2v N  2v 


   t 2  q x2   vqdxdt  0 .
t0 0 
3
t l
(15)
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
 2v N  2v
,
(16)
 2 
t
q x 2
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
 2T  2T  2T
(17)
T  2  2  2  0,
x
y
z
where  is Laplace‘s differential operator. If volume heat sources exist, then Laplace‘s
equation switches to Poisson’s equation
1
T   rV  x, y, z ,
(18)

where the function rV 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
 2T  2T
 2T  2T
r ( x, y)
,
(18b)


0
,
resp.
 2  V
2
2
2
x
y
x
y
h
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, P   ,
or, more briefly
4
(19)
T   f P .
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
 k (T  T0 )  0, on  ,
(20)
n
where n is the outward unit normal to the surface  and k is a coefficient. Substituting kT0 =
g(P), the boundary condition can be put into the form
 T

(21)
 n  kT   g P  .

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
(22)
 g P .
n 
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  2
(23)

0
x 2 y 2
and on the boundary of this domain the boundary condition

(24)
 ynx  xny ,
n 



where nx = cos( n ,x), ny = cos( n ,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 f
f 
(25)
  x1  y2  z3 d    f1nx  f 2ny  f3nz dS
5
Gauss’ theorem is very useful in dealing with variational principles and performing the
integration by parts in 2D and 3D problems. If f1, f2, f3 are interpreted as components of a
vector f , Gauss’ integral theorem can be written as
 div f d    f n d S

(26)

Substitute f2 =f3 = 0, f1 = 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:
v
u
(27)
 u xd   v x d    uvnxdS .
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:
3
u v
u
u
  vu d    
d   v d    grad u grad vd   v d S ,
xi xi
n
n

 i 1



2
 u 
u
u
 d   v d    grad u 2 d   u d S ,
  uu d     
xi 
n
n

 i 1 



v 
 u
 vu  uv  d     v n  u n  d S .
3
 (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
  d    V ndS   r  d ,
(28)

dt  

 
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   d  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

(29)
  qndS   rd  0 ,


or, using Gauss’s theorem
 (divq  r )d  0 .
(30)

Applying Fourier’s law for the heat conductivity, q = -gradT, where grad is differential



operator, grad  i  j  k , i, j, k are unit base vectors , we arrive at
x
y
z
 (div  gradT  r )d  0
(31)

6
from where, by denoting r = rV, 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  w( x, y ), or z  w( x, y )  0 .
(32)
An elemental rectangle, whose area in the undeformed membrane is dxdy , moves after
deformation into a surface element of area dS ' , and it holds
dxdy
,
(33)
dS ' 
cos( n, z )
where cos( n, z ) is the direction cosine of the unit normal vector of the surface element dS ' , i.
e. the cosine of an angle which contains the unit normal vector of a membrane element with zaxis (note that z-axis coincides with the normal vector direction in undeformed configuration
of the membrane) . It follows from (32) for direction cosines
w w
(34)
cos( n, x) : cos( n, y ) : cos( n, z )  
:
:1,
x
y
Thus, for cos( n, z ) we get
1
.
(35)
cos( n, z ) 
2
2
 w   w 
1      
 x   y 
Substituting (35) in (33) and integrating over the total area  , the area of the deflected
membrane is obtained as
2
2
 w   w 
(36)
 1   x    y  dxdy .
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
2
2


 w 

w


     1        1 hdxdy   pwdxdy .
(37)

 x   y 
 



2
2
 w 
 w 
 are small
where h is the thickness of the membrane. Because the quantities 
 a 
 x 
 y 
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
 w 2  w 2 
1
1
p 

(38)
         hdxdy   pwdxdy  h  (grad w)2  2 wd ,
2   x   y  
2

h






where the definition of the differential operator grad in two dimensions,


grad w  i w x  j w y , 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 


 2w 2w 
 w w w w p 
  0  h  

 wdxdy    h  2  2   p wdxdy 
x x
y y h 
y 
 x
 
 

(39)
 w
w 
  
nx 
n y wds.
x
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)
2w 2w
p
(40)
 2  ,
2
x
y
h
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 and equation of balance (28):

d
qn
r
sd    dS   d   s d ,
(41)

dt 
T
T



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.)
(42)
s   J  , X     J  X  , for =1,2…

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 
,
(43)
X 
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
L
(44)
  0   K X     X  X   ...
2

 
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
(45)
    X  X   0 .
2
 
The constitutive relation (43) then becomes linear
J   L X 
(46)

and, for the case of heat conductivity, it follows
grad T
q   Lqq
.
(47)
T2
Comparing (47) with the relation q = -gradT, one concludes that Lqq =T2. Consider now Eq.
(43), multiply it by the variation of the generalized forces, X, sum over  and integrate over
the whole body. Obtain:

X  d   d .
(48)
 J  X  d    
 
X 

 

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
2
 qn
 r
 grad T   

(50)
  dS  

 d  0 ,
T
 T   
  T

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
  gradT 2 r 
1
1
2
2
    T   T d   T 2   gradT   rT d  Tmin2    gradT   rT d, (51)


where Tmin 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 rV=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 zaxis 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

1  u w    
1  v w    

  
 xz   
 y ,  yz   
 x ,
 
2  z x  2  x
2  z y  2  y
(52)


 xx   yy   zz   xy  0.
From Hooke’s law the stress components are obtained as
 

 

 xz  2G xz  G
 y ,  yz  2G yz  G
 x ,
 x

 y

 xx   yy   zz   xy  0,
(53)
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
2
2

 
1
1
1
  
2
2
2  
W   ij  ij   xz xz   yz yz 
 xz   yz   2 G  x  y    y  x   
2
2G


 
(54)
2
2


 
1
  2
     
 G 2 
x
y   x  y2  .
  2
 
2
x 
 y
 x   y 

Because the volume forces are neglected and the loading on the circumferential surface of the
bar is zero, the potential energy reduces to
10
   2   2

 
1
  
2 
(55)
  G   
x
y  dxdy    x 2  y 2  dxdy  .
  2
 
2
x  
 y

   x   y 

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
  2   2
 
1
  
2
  G         2 
x
y  dxdy 
2

x

y

y

x





 
 

  2   2
  ( x)  ( y )  
1
2
 G   


2



 dxdy 
 
2

x

y

y

x





 
 

(56)
  2   2 
1
2
2
 G   
 dxdy  G    xn y   ynx ds  0,
 
2

x

y
 
 
 


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
  2  2 
 


2
(57)
  G   2  2  dxdy  G 2   nx 
ny  xny  ynx  ds  0

x

y

x

y


 

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  2



 2  0 in the domain  and
nx 
ny 
 ynx  xny on boundary  , (58)
2
x
y
x
y
n
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 u0(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
arbitrary real number and v(P) is an arbitrary function twice differentiable in  fulfilling Dirichlet’condition
prescribed on . Obviously, it holds
F u0  tv  F u0  .
For every fixed function v is the functional F(u0 + 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 u0  tv t  0  0
dt
Let us demonstrate the technique e.g. at Eq. (38). It can be easily shown
2
p 

w0  tv  w0   th  grad w0 grad v  vd  t 2  grad v  d  .
h 


Differenciate the preceding expression with respect to t, substitute t=0 and the result put equal zero. Obtain
p 

h  grad w0 grad v  vd  0 .
h 

According to Green’s formula (see Remark 1) it holds
11
w0
d S,
n



where the second integral on the right-hand side disappears because v = 0 on . Finally obtain
p

h  w0  vd  0.
h 

Since the function v is arbitrary, it must hold w0 = -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
 grad w
0
grad vd    vw0 d    v
function w0 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 2  w 2 
(59)
  x    y  dxdy


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  1, y1  1
(60)
find such a function, for which the integral
1
J  y    x 2 y ' 2 dx
(61)
1
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
arctan  x  
y 
.
(62)
arctan 1  
It is obvious that the function y is continuous including its first derivative in -1  x  1 and
satisfies the conditions (60). Further,
1
1
1
2
dx
2
J  y    x 2 y '2dx   ( x 2   2 ) y '2 dx 

.
(63)
2
2
2

arctan 1   1 x  
arctan 1  
1
1
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 20th 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 un of
this space has a limit u in this space. Here, a Cauchy sequence is a sequence of functions un(x)
from P such that
lim   um , un   0
m 
n 
((un, um) is so called distance of functions un, um)
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:
  u, v   0, while   u, v   0  u  v,
  u , v     v, u  ,
  u, z     u, v     v, z  for all u , v, z  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  M , v  M , a, b 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
u, v   v, u ,
where overbar stands for complex conjugate. In the case of real elements, the preceding condition is replaced by:
u, v  v, u  .(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
1u1  2u2 , v  1u1, v  2 u2 , v
where u1, u2, v  M and 1, 2 are arbitrary constants.
C. It holds
u, u   0 .
D. If u, u   0 , then u = 0 and vice versa. The number
u 
u, u 
is called the norm of element u, and the number u, v   u  v is the distance of elements u, v.
13
E. Every Cauchy sequence {un} 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 un  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, L2(), 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. L2() means the set of all functions u(x) square integrable in  ( thus the integral


u 2  x  d x exists
and is finite) on which there are defined:
the scalar product
 u, v   u  x v  x  d x ,


the norm
 u, u    u 2  x  d x ,
u 

the distance
  u, v   u  v 
 u  x   v  x 
2
dx .
(*)

A sequence of function unL2() is said to converge in this space (or to converge in the mean) to the function
uL2(), and we write
un  u in L2   ,
if
lim   u, un   0 ,
n
or
 u  x   u  x 
lim
n
n 
2
dx 0,

or (what is the same)
lim  u  x   un  x  d x  0
2
n

For two functions u(x), v(x) which need not be equal at all points x , we may have
2
 u  x   v  x  d x  0 .

Thus, their distance in the space L2() 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
yn  arctan( nx) / arctan( n) converges to the function
 1  x  0,
1 for

(64)
y  x    0 for
x  0,
 1 for
0  x  1,

which is the limit of the sequence of functions yn in the space L2(-1,1), i.e. lim  2  yn , y   0
n 
Both, the function (64) and the functions yn are from the space L2(-1,1). It is sufficient to show
that
14
lim   yn , y   0, or, what is the same lim  2  yn , y   0 .
n
n
(65)
Fig.2
According to the definition of the distance in L2(-1,1), and because the functions yn and y are
odd functions, we get

2 arctan 1 n
 arctan nx 

 arctan nx  
  yn , y    
 y  dx  2  
 1 dx  2 


arctan
n
arctan
n
arctan
n
arctan
n




1
0
1
2
2
1
2

ln( 2(1  n 2 )
i
i
1 in
1 i n 
 1  i n 


dilog 
ln 2
.
  dilog 
 
2
2

n arctan n
2n arctan n 
1 in
 2 
 2  4n arctan n
Obviously,
lim  2  yn , y   0 ,
(66)
n 
hence, really
lim yn  x   y  x  in L2  -1,1
n
.
Observe, that while the functions yn(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 L2(). Define the scalar product in this set in the same manner as in
the space L2(), hence
(67)
u, v   uxvx d x, u  N , v  N .

Using this scalar product, define the norm and the distance in N. Thus, a certain metric space
is obtained. Denote it as N L2   . In contrast to the space L2(), the space N L2   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 L2 1,1 is not complete. The
sequence of the functions yn(x) from the previous example is apparently from the space
N L2 1,1 , because these functions are continuous. The sequence yn(x) is a Cauchy sequence in
the space N L2 1,1 , because it is convergent, see Fig.2 and Eq. (66), and thus, it is also the
Cauchy sequence in the L2(-1,1) , since the spaces L2(-1,1) and N L2 1,1 have the same metric.
Yet the sequence yn(x) is not convergent in the space N L2 1,1 . Presume by contraries, that it
does converge to a certain function v  N L2 1,1 , 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
yn(x) in the space L2(-1,1) would also converge to the function v, since L2(-1,1) has the same
metric as the space N L2 1,1 . We have seen in the example that the sequence yn(x) converges in
L2(-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 L2(-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 L2(-1,1),
so that basing on the presumption, that the continuous function v(x) is the limit of the
sequence yn(x) in N L2 1,1 , we arrive to conclusion, that this sequence has two different limits
in L2(-1,1). That is, however, in contradiction with the theorem, that a sequence of functions
in L2(-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
 2u  2u

 f  x, y  in domain ,
(68)
x 2 y 2
u  0 on boundary ,
leads to the variational problem of finding a minimum of the functional
 u 2  u 2 
(69)
J u         dxdy  2 fudxdy
x   y  

 



on a set M, whose elements are sufficiently smooth function satisfying the boundary condition
u = 0. Choose n linearly independent functions in M
(70)
1  x, y  ,..., n  x, y 
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
(71)
b11  x, y   ...  bnn  x, y 
with arbitrary coefficients b1...,bn (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(b1…,bn) of these coefficients. If the functional (69) is required to attain a minimum for a
function
(72)
un  x, y   c11  x, y   ...  cnn  x, y 
in the subset P, following equations must hold
h
c1 ,..., cn   0,..., h c1 ,..., cn   0.
(73)
b1
bn
The form of considered functional (69) suggests that the function h(b1…,bn) is a quadratic
functions of variables b1...,bn, hence the system (73) for unknowns c1...,cn 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  x  , 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 {n} from any system of linearly independent
functions {n} using Schmidt’s orthogonalization process. Note that an orthonormal system of functions {n}
fulfils the relations
1, m  n,
0, m  n.
 m , n   
We proceed according the following scheme
1 
1
,
1
n 1
n 
n   n , k  k
k 1
n 1
n   n , k  k
, n 1
k 1
The completeness of orthonormal system of functions { n} in a Hilbert space H can also be defined as follows:
orthonormal system of functions {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 {n} is
incomplete. If the system {n} is complete, then it follows from
n ,   0, n  1,2,...
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
u u
1
1
 w 
 i i dV    h   dxdy , substitute both expressions into left
2 V t t
2   t 
hand side of Eq. (9) and perfom variation..
 2w 2w 
2w
Answer:  2    2  2 
t
y 
 x
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.
2
L
1
 u  
Hint: The potential energy is given by     EA    d x , see Example 1 of
 x  
2
0 
Appendix 1, where the axial loading px is set equal to zero. The kinetic energy is given
2
form Ek 
1
 u 
by Ek    A   dx . From the Hamilton principle it follows
20
 t 
2
L
2
2
1
 u  1
 u  
  ( Ek   )dt       A    EA    dxdt  0 .
 t  2
 x  
2
t0
t0 0 
Observe
t1
t1 L
17
1
1
 2u
 u 
    A   dxdt   A   2  udxdt ,
2
t
 t 
t0 0
t0 0
2
t1 L
t L
where  u  0 for t  t0 and t  t1 was applied. Fot the variation of  obtain
t1 L
1
 u 
2
t1 L
u  u 
t1
 u

L
t1 L
 2u
   EA   dxdt  AE      dxdt AE    u  dt AE   2  udxdt
2
x  x 
x  0
x
 x 
t 0
t 0
t 
t 0
Thus, Hamilton principle leads to
0
0
L
0
0
1 L
  2u
 2u 
 u 
 AE    u  dt  A   E 2   2   udxdt  0 .
x  0
x
t 
t0 
t0 0 
For the essential boundary conditions (u=0, for x =0, L) the first term disappears.
t1
t
u
 0 (free ends) follow. The second
x 0
L
Otherwise, the natural boundary conditions
 2u
 2u


 0.
x 2
t 2
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.
term provides the searched differential equation E
2
1  2w 
Hint: The potential energy is given by    EI  2  d x , see Example 2 of
2  x 
0
Appendix 1, where the axial loading pz is set equal to zero. The decisive contribution
w
to the kinetic energy is due to the velocity of translation
of the cross section of the
t
L
1
 w 
bar. Thus, Ek    A 
 dx . Hamilton principle gives
20
 t 
L
2
2
2
2
1
 w  1   w  
  ( Ek  )dt       A    EI  2   dxdt  0
2
 t  2  x  
t0
t0 0 

t1
t1 L
with
2
 2w 
 w  0 for t  t0 and t  t1 . Integrate by parts the term   2  d x according to
x 
0
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
 2 w EI  4 w
transversal oscillations as 2 
 0.
t
 A x 4
L
18