Lecture 2
Lecture annotation
Linear operators, symmetric and self adjoint operators. Positive operators and their physical
meaning. Functionals and operators in a Hilbert space. Energy space of a positive definite
operator. Essential and natural boundary condition of a differential equation. Generalized
solution of the problem on minimum of the functional of energy.
Linear operators – bounded and unbounded, symmetric and self-adjoint operators
It can be generally stated that every boundary problem of mechanics, hydromechanics,
mathematical physics etc. is connected with a certain operator, which is called the operator
pertinent to given boundary value problem. An operator operates in a suitable Hilbert space.
A given problem can be put in the form
Au  f ,
(1)
where A is the operator of boundary value problem, u and f are elements of a Hilbert space,
the former is a searched one, while the latter is known. We make this statement clear using
following example. In a domain , we are searching the solution of Poisson’s equation
(2)
 u  f P ,
(P denotes a general point of the domain ), which satisfies the homogeneous Dirichlet
boundary condition
(3)
u   0.
For simplicity assume that the function f(P) is continuous in the closed domain      .
We introduce into consideration the Hilbert space L2(). Apparently, the function f in Eq. (2)
is from this space. We will presume that the searched function u is also from L2(). In the
space L2() we choose a set o functions, DA, which have the following properties: 1. they are
twice continuously differentiable in  ; 2. they are equal to zero on . Such functions will be
called admissible functions. On the set DA we define the operator A by
(4)
Au  u .
It is obvious that the problem (2) and (3) can now be written in the form (1). The set DA, thus
the set of all admissible functions, will be called the domain of definition of the operator A.
Let us point out that the domain of definition of the operator A is not the domain , but a set
of functions.
Analogously one can construct operator pertinent to other boundary problems in case of
homogenous boundary conditions. In case of inhomogeneous boundary conditions, the
problem of operator construction is a bit more complicated. (back to Lecture 3) (back to
Lecture 11)
Remark 1. The operator A is called linear, if its domain of definition is linear and it holds (a1, a2 are arbitrary
constants)
Aa1u1  a2u2   a1 Au1  a2 Au2
If A is a linear operator, then A0 = 0. An operator is called continuous, if
lim Au  Av
u v
and bounded, if it is linear and if it holds
Au  C u , C  const .
(*)
The smallest constant C is called the norm of a bounded operator A and is denoted by A . Obviously,
Au  A u .
1
A typical representative of bounded operators is an integral operator. E.g. in the space L2(0,1), the operator A
defined as
1
Au  vs    K x, s u x dx ,
0
where so-called kernel K(x,s) is integrable for almost all s0,1. On the contrary, a typical property of
differential operators is that they are unbounded. Consider in the space L2(0,1) a linear set DA = C(1)(0,1) (thus
one times continuously differentiable functions in the interval (0,1)) and define on this set the operator A by
du .
Au 
dx
This operator is unbounded on DA, i.e. there does not exist such a constant C, that Eq. (*) would hold for all u
DA. Let us presume by contraries, that such constant exists and we come to contradiction. Let us consider the
function u(x)=sin nx (n natural number). Obviously, u DA. Further
Au  n cos nx .
It holds according to the definition of the norm:
1
u
L2

0 ,1
 sin
2
nx d x 
0
1
,
2
1
Au
L2
n 

0 ,1
2
2
cos 2 nx d x  n
0
1
.
2
If we choose the function u(x) such that n > C, we get the contradiction with the presumption that (*) holds for
every function from DA.
A (linear) operator A is called symmetric on DA, if for every pair of admissible functions u, v the following
equality holds
Au, v  u, Av.
A symmetric operator is called self-adjoint on a Hilbert space H, if the following properties are satisfied: let u,
u* are elements from H, v is an arbitrary element from DA and let it holds
 Av, u   v, u * ,
then u  DA, Au = u*.(back to Lecture 11)
Remark 2. In the next we will presume that the domain of definition DA, which an operator A maps into a
Hilbert space H (DA  H), is dense in H.
Positive and positive definite operators and their physical meaning
A (linear) operator A is positive on DA, if it is symmetric and if for every u  DA it holds
(5)
 Au, u   0 ,
while the equality (Au,u) = 0 holds if and only if u = 0.
Example 1. Let the operator B operates in the space L2(0,1) on a set of functions DB, which are twice
continuously differentiable in the interval 0,1 and satisfy the boundary conditions
(1a)
u0  u1  0 .
Consider that operator B is given by
d2 u
(1b)
Bu   2 .
dx
We show that, on DB this operator is symmetric and positive. It holds
1
1
1
2
Bu , v    v d u2 dx   v d u   d v d u dx
dx
dx 0 0 dx dx
0
(1c)
The function v  DB satisfies the boundary conditions (1a), thus the first term on the right-hand side is equal to
zero and we get
1
Bu , v    d v d u dx .
(1d)
dx dx
Similarly we find
0
2
1
u, Bv    u d
1
2
v
dv du
dx  
dx
2
dx
dx dx
0
0
Formulas (d) and (e) show, that the operator B is symmetric. If we substitute v =u into (d) we obtain
2
u
 du 
dx   
 dx  0 .
2
d
x
d
x


0
0
Assume for now that (Bu,u) = 0. It means that
1
Bu , u    v d
1
2
(1e)
(1f)
2
 du 
0  d x  dx  0 .
1
The integrated function is non-negative and the integral is equal to zero, whence it follows d u  0 and thus, u(x)
dx
= const. The boundary conditions show then that u = 0. Summarizing we conclude that the operator B is positive.
Example 2. Let us consider in the same space the operator C, which operates in the same way as the B according
to the formula
d2 u
(2a)
Cu   2 ,
dx
and its domain of definition DC differs from the domain DB only by boundary conditions. Let functions from DC
satisfy the boundary conditions
(2b)
u0  u0  0, u1  u1  0; ,   const .
The set DC is dense in L2(0,1). Further, let u, v  DC. Then we have
1
1
2
(2c)
Cu, v    v d u2 dx  v0u0  v1u1   d v d u dx .
dx
dx dx
0
0
Eliminating u’(0), u’(1) by using (2b) we get
1
(2d)
Cu, v   v0u0  v1u1   d v d u dx .
d
x
d
x
0
Right-hand side of (2d) is symmetric with respect to u, v , hence (Cu,v)=(Cv,u)=(u,Cv) and thus, the operator C
is symmetric. For arbitrary values of ,  it is difficult to figure out whether the operator C is positive, however,
it is easy to find the sufficient conditions for the operator C being positive. Namely, the operator C will be
positive, if 0, 0, while one of the constants ,  must be different from zero. Actually, let e.g. >0, 0.
Then
Cu, u   u 2 0  u 2 1    d u 
1
0
2
dx
dx  0 .
(2e)
If (Cu,u)=0, it can be seen that it must hold u(0)=0, d u  0 . The last equations results in u(x)=const. However,
dx
u(0)=0 and thus u(x)=0.
If both constants ,  are equal to zero, the operator C is non-positive. The conditions (2b) have then the form
(2f)
u0  u1  0 .
Moreover
2
Cu, u     d u  dx .
dx
0
The function u=1 satisfies the boundary conditions (2f), hence, it belongs to the domain of definition of the
operator C but simultaneously (Cu,u)=0.
1
Example 3. Let us consider Laplace’s operator. Let admissible functions from its domain of definition satisfy
the homogeneous Dirichlet boundary condition
(3a)
u  0,

where  denotes the boundary of domain , in which function u is defined. Note that admissible functions are
required to be twice continuously differentiable in the close domain  . It can be proved that the operator - is
positive on such domain of definition. The Green formula leads to


 u, u     uu d      u 
m

xi 
 i 1 
2
d  u

u
dS ,
n
(3b)
3
where m =2 stands for two-dimensional domain and m =3 for three-dimensional domain. The boundary condition
(3a) indicates the integral over the domain boundary is equal to zero, thus


 u, u      u 
m
 i 1
 xi 
2
d  0.
(3c)
At the same time, if (-u,u) = 0, then necessary u  0 , since the integrated function in the last integral is nonxi
negative. However, it follows then u = const and from (3a) there follows u = 0. Thus, the positiveness of the
operator - has been proved. (back to Lecture 3)
We attempt to express the physical meaning of the notion – “positive operator”. Consider e.g.
a membrane loaded by transverse loading p and clamped at its boundary. Stationary deflection
w of the membrane satisfies the equation (40) of the Lecture 1
2w 2w
p
 2  .
2
x
y
h
Observe that the following expression
 w 2  w 2 
(6)
 w, w        dxdy
x   y  

 


differs only by the constant coefficient h/2 from the strain energy, see the first term in (7) for
the total potential energy of loaded membrane, see also the equation (38) of the Lecture 1.
 w 2  w 2 
1
(7)
         hdxdy   pwdxdy .
2   x   y  



Apparently, in the problem of membrane deflection, the positiveness of the operator -
expresses the physically obvious fact that the potential energy of deflected membrane is
positive. In other words, it is impossible to deflect a membrane without the energy
consumption.
From numerous examples (some of them were presented in the Lecture 1) there is possible to
infer: Let a physical system exhibits due to an external cause characterised by function f(P) a
response u(P). Let these two functions are connected by Eq. (1), where A is a positive
operator. Then, as it can be seen, the quantity (Au,u) is proportional to the energy consumed
in the system for the response u(P) to occur.
Hereafter, if the operator A is positive, we will call the quantity (Au,u) the energy of function
u  DA.
A symmetric operator is called positive definite on DA, if there exists such a constant  > 0,
independent of u, that for every admissible function u  DA it holds
 Au, u    2 u 2 ,
(8)
Obviously, every positive definite operator is positive. The opposite conclusion is false, as it
can be seen from the next example.(back to Lecture 5-6).(back to Lecture 11)
Example 4. Examine in space L2(0,1) the operator L given by
d  3 du 
(4a)
Lu  
x
, 0  x  1 .
dx dx
Let functions u  DL satisfy one boundary condition u(1) = 0. Moreover, these functions are required to twice
continuously differentiable in the closed interval 0,1. The operator L is positive. Actually, it can be easily seen
that it is symmetric
4
Lu, v   u, Lv    u
d  3 dv 
d  3 d u 
x
v x
 dx 
dx dx
d x  d x 
0
1
1
d  3 d v
d u 
 3 d v
d u 
 d x  x  u d x  v d x  d x   x  u d x  v d x 
0
(4b)
1
0
0
since for x = 0 is the term x3 equal to zero and for x = 1 both functions u(x) and v(x) are equal to zero. Thus,
(Lu,v) = (u,Lv) and the operator L is symmetric. Further,
1
Lu, u     u
0
2
d  3 du 
3 d u 
x
dx   x 
 dx  0.
d x d x 
dx
0
1
(4c)
When (Lu,u) = 0, then d u  0 , u = const, and because u(1) = 0, it holds u = 0. Hence, the operator L is positive.
dx
Now it can be shown that L is not positive definite. Inequality (8) can be put into the form
 Au, u    2
u2
Obviously, in case of a positive definite operator, the last expression cannot be lesser than a fixed positive
constant 2. If we show that this expression can achieve an arbitrary small value for the operator L, we thus prove
that operator L is not positive definite.
Let  is a number between 0 and 1. We choose
  x 3 , 0  x  
u  
  x 1
 0,
The function u(x) is equal to zero for x = 1 and it is twice continuously differentiable. Hence, u(x) is from the
domain of definition of the operator L. Set up the quotient
Lu , u  
u2

2
3 d u 
0 x  d x  d x
1
1
u
2

9  x 3   x  d x
2

0

   x  d x
6
dx
0
0
Because u(x) = 0 for x > , the corresponding integrals are dropped. It can be found by simple calculation that
Lu , u   9  ,
40
u2
which can be arbitrary small for a sufficiently small .
Positiveness, or positive definiteness, is a certain “criterion of reasonability” of a given
problem, as seen from the following example:
Example 5. Consider the problem
d2 u
 2 2
dx
u  0  u 1  0
`
(5a)
(5b)
It follows immediately that the operator A, given by
Au  u
(5c)
on the set of admissible functions
DA  u; u  C  2 0,1 , u  0   u 1  0


(5d)
is symmetric on DA because, for u,v  DA, we have
1
1
1
0
0
 Au, v     uv d x   uv 0   uv d x   uv d x 
1
0
1
1
0
0
(5e)
 uv 0   uv d x    uv d x   u , Av  .
1
(C(2)0,1 is the set of functions 2-times continuously differentiable in the closed interval 0,1). However, it is
not positive on DA: Let
5
 Au, u   0
i.e. according to (5e), let
1
 u
2
dx 0.
0
This implies u  x   0 and u  x   const. as is in Example 1, but nothing more, because boundary conditions
of the form (1a) are not prescribed here. Thus (Au,u)=0 implies only u(x)  const., where this constant need not
be equal to zero. Thus the operator (5c) is not on the set (5d) positive.
The loss of positiveness has curious consequences:
From (5a), we get immediately
u  x   2 x  C .
The first of the boundary conditions (5b) yields C=0, while the second implies C=2. These results represent an
evident contradiction; thus the problem (5a) cannot have any solution in spite of the fact that both the differential
equation and the boundary conditions are extremely simple.
This is not all. Replace the right-hand side of the differential equation (5a) by the function 2x-1, i.e. solve the
equation
(5f)
u  2x 1
with same boundary conditions (5b). Integrating, we get
u   x 2  x  C .
Each of the boundary conditions is satisfied for C=0; the solution of the given problem is
x3 x 2
u  K,
3 2
where K is arbitrary. Thus, the problem (5f) has an infinite number of solutions.
Such a thing cannot occur if the operator A is positive. In fact, the following theorem holds:
Theorem 1. (on uniqueness). Let A be a positive operator on its domain of definition DA.
Then the equation
Au  f
(1)
cannot have more than one solution u0 on DA.
Proof: Presume conversely that Eq. (1) has two solution u1, u2, hence Au1 = f, Au2 = f.
Substitute u1  u2  u~ . Subtracting Au2 = f from Au1 = f and making use of the linearity of the
operator A we get Au~  0. Multiplying the last equation by u~ to set up the scalar product, we
get  Au~, u~   0. However, because the operator A is positive, it must hold u~  0 and thus
u1  u2  0 .
Functionals in Hilbert space
We say that a functional lu is defined on a set M belonging to Hilbert space H, if there is
assigned to every element u  M a number lu. The set M is called the domain of definition of
functional l and it is denoted by Dl.
The simplest example of a functional is the norm of an element. Also the scalar product (u,v),
where the second factor v is fixed, can be understood as a functional.
A functional l is linear if its domain of definition Dl is linear and it holds (a1, a2 are arbitrary
constants)
(9)
l a1u1  a2u2   a1lu1  a2lu2 ,
where u1, u2 are elements from its domain of definition.
A linear functional is called bounded if it holds
(10)
lu  N u for every u  Dl ,
where N is a constant. The smallest of all numbers N, which satisfy the inequality (10), is
called the norm of the bounded functional lu and is denoted as l .
The functional lu is called continuous , if
6
lim lu  lv .
(11)
u v
Every linear bounded operator, defined on a whole space, is continuous.
Theorem 2. (Riesz theorem). In Hilbert space H, every linear, bounded functional has the
form of the scalar product
(12)
lu  u, v
where v is a fixed element from the space H. The element v is defined uniquely. Hence, a
linear, bounded functional is uniquely represented by the element v. Besides, l  v . (back
to Lecture 5-6)
Consider a set of all linear, bounded functionals in a Hilbert space H and define the sum of
two functionals, and multiplication by a number as follows:
l1  l2  u  l1u  l2u,  al  u  a lu 
for every u  Dl.. Further, define the norm of functional, see (10). In this way, there is defined
a linear, normed space (whose elements are all linear, bounded functionals on H). This space
is called the adjoint space or the dual space to the space H and is denoted by H*.
Important roles in the next have bilinear and quadratic functionals.
Consider for simplicity a real Hilbert space, in which there exists a dense linear set M. Let to
every pair of functions, u,vM, there is assigned just one number (u,v). The number (u,v)
will be called bilinear functional, if it holds:
1. for fixed v is the functional  linear,
2. the following identity holds
(13)
u, v  v, u  .
If (u,v) is bilinear functional, the functional (u,u) = (u) will be called the homogeneous
quadratic functional or quadratic form. A quadratic form satisfies the identity
(14)
u  v  u   2u, v  v .
If (u) is a quadratic form and l is a linear functional, then the expression
(15)
F u   u   2lu  const
will be called the quadratic functional.
Theorem on the minimum of the quadratic functional and its consequences
Theorem 3. Let A be a positive operator on DA. Then:
1. If the equation
Au  f
(16)
has a (classical) solution u0  DA, then the quadratic functional
(17)
F u    Au, u   2 f , u 
(the so-called functional of energy) assumes, for this function u0(x), its least value on DA.
(This minimum is even strict: If u  DA, u  u0, then F(u)>F(u0)).
2. Conversely, if for some function u0  DA the functional (17) assumes its least value on DA,
then the function u0(x) is the solution of equation (16). (A unique solution according to
Theorem 1)
Proof: First it is obvious that for all u  DA the functional F(u) is defined.
1. Let u0 satisfies Eq. (16) in H, hence f = Au0. If we substitute for f in (17), we obtain for u 
DA
F u    Au, u   2 Au0 , u  .
It is easy to find that
7
F u    Au, u   2 Au0 , u    Au, u    Au0 , u   u, Au0  
 Au, u    Au0 , u    Au, u0    Au, u    Au0 , u    Au, u0    Au0 , u0    Au0 , u0  
 Au  u0 , u  u0    Au0 , u0 .
(18)
(The symmetry of the operator A, and the symmetry of the scalar product was used). If it
holds Au0 = f, then
(19)
F u    Au  u0 , u  u0    Au0 , u0  .
The term (Au0,u0) in (19) does not depend on u and stays constant while u is changing;
further, the operator A is presumed to be positive, so that it holds for the first term on the
right-hand side of (19)
 A  u  u0  , u  u0   0 for every u  DA ,
whereas
 A  u  u0  , u  u0   0
just if and only if u  u0  0 on DA .
From (19) it follows that
F  u   F u0  for every u  DA ,
whereas F(u) = F(u0) just if and only if u = u0 on DA. Thus, if the equation Au0 = f is satisfied,
the functional F(u) assumes its least value on DA just for the function u = u0.
2. Let the functional F(u) assumes on DA its least value for the function u0. Hence, it means
that if an arbitrary element v  DA and an arbitrary real number t are chosen (so that
apparently u0 + tv  DA), it holds
(20)
F u0  tv  F u0 
or
(21)
F u0  tv  F u0   0 .
If the symmetry of the operator A and the symmetry of the scalar product is again used, we get
F u0  tv   Au0  tv, u0  tv  2 f , u0  tv   Au0  tAv, u0  tv  2 f , u0   2t  f , v  
  Au0 , u0   t  Av, u0   t  Au0 , v   t 2  Av, v   2t  f , v   2 f , u0  
  Au0 , u0   2t  Au0 , v   t 2  Av, v   2t  f , v   2 f , u0 .
Hence
F u0  tv   Au0 , u0   2t  Au0 , v   t 2  Av, v   2t  f , v   2 f , u0 .
(22)
Since u0  DA and f  H are fixed functions, it follows from (22) that for a fixed chosen v 
DA, F(u0 + tv) is a quadratic function of variable t. Eq. (21) indicates, that this function shall
attain a local minimum for t = 0, whence its first derivative must equal to zero for t = 0:
d
(23)
F u0  tv  0 ,
dt
t 0
or, according to (22)
2 Au0 , v   2 f , v   0 .
(24)
The condition (24) can be easily modified to the form
 Au0  f , v   0 .
(25)
The function v  DA was chosen fixed but arbitrary, so that we can arrive to the equation (24)
from the condition (21) for every chosen function v  DA. The linear set DA is dense in H
according to presumption (see Remark 2). Hence, it follows from (25) that the function Au0 –
f is orthogonal in the space H to all functions from the linear set DA dense in H and, according
to the basic lemma of the variational calculus, is thus equal to zero, Au0 – f = 0 in H. Hence,
u0 is the solution of the equation Au0 = f, which completes the proof. (back to Lecture 3) (back
to Lecture 5-6) (back to Lecture 11)
8
The functional (17) will be called the functional of energy and the Theorem 3, the Theorem on
the Minimum of the Functional of Energy.
The main sense of Theorem 3 consists in the fact that, on its base, the problem of solving
equation (16) on the set DA of its admissible functions is replaced by the problem of
minimizing a functional on DA. For an approximate solution of the latter problem effective
methods are developed (the Ritz method, etc) However, note that Theorem 3 does not have
the character of an existence theorem. It asserts that if u0  DA is the solution of equation (16),
then u0 minimizes the functional F(u) on DA, or if F(u) attains, on DA. its least value for an u0,
then this u0 is the solution of (16). However, nothing is said about the existence of a solution
of the investigated equation or about the existence of a minimum of the considered functional,
on DA. In fact, neither the equation (16) need have a solution from the set DA, nor the
functional (17) need attain, on DA, its least value, even in very simple cases. Consider the
problem
(26)
u  f  x 
u  0  0, u 1  0
with a positive (even positive definite) operator
Au  u
on the set of its admissible functions
DA  u; u  C  2 0,1 , u  0   0, u 1  0


(27)
(28)
(29)
(see Example 1), and let
0 for 0  x  0.5
(30)
f  x  
1 for 0.5  x  1.
Obviously, f(x)  L2(0,1). The problem (26), (27) cannot have a solution u0  DA: In fact,
u0  DA  u0  C  2 0,1  u0  C 0,1 ,
(31)
so that, for u0  DA, Au0 is a continuous function in the interval 0,1, while the function (30)
does not poses this property (and, at the same time, cannot be made continuous in that interval
even when changing its values on a set of measure zero, see this). Thus, the equation (26)
cannot have a solution u0  DA. From the second assertion of Theorem 3 it then follows that
even the functional (17) cannot attain its minimum on DA since the equation (16) should have
a solution u0  DA in that space.
It follows from the preceding example that it is necessary to generalize the concept of the
solution in an appropriate way, because without doing this we would not be able to include
into our theory even very simple problems from applications (deflection of a bar with a
discontinuous loading, etc). The idea of how to perform this generalization is relatively
simple: To extend the domain DA of definition of the considered operator A by adding some
suitable functions and to extend the functional (17) onto this extended domain. If we succeed
in constructing this extension in such a way that the so extended functional will actually attain
its least value, for a certain function u0, we shall call this function a generalized solution of
given problem (we say also weak solution, because the requirements upon the smoothness of
functions from the domain DA were weakened.)
Such an extension can be constructed in a relatively simple way, if the operator A is positive
definite on DA (thus in the case most frequently encountered in applications). To this end we
introduce so-called energetic space HA, in which the existence and uniqueness of generalized
(weak) solution of given problem can be easily proved.
9
Energetic space HA
Let A be a positive definite operator on DA. Let us define on DA a new scalar product by
u, v A   Au, v  .
(32)
(to distinguish the new scalar product from the old one, the subscript A is introduced) The
operator A being positive definite (positiveness is even sufficient at his stage) all axioms of
the scalar product are actually satisfied, (see axioms A-D in Remark 5 of Lecture 1). Really:
Axiom A. When an operator is positive definite, it is also symmetric. Therefore, if u,v  DA, it
holds
u, v A   Au, v   u, Av   Av, u   v, u A .
(33)
Axiom B. It holds
a1u1  a2u2 , v A   Aa1u1  a2u2 , v , v   a1 Au1  a2 Au2 , v  
(34)
a1  Au1 , v   a2  Au2 , v  a1 u1 , v A  a2 u2 , v A
Axioms C and D. It holds
 Au, u    2 u 2  0
(35)
If (Au,u) = 0, necessary u  0 , hence u =0.
(36)
The norm is denoted by u A (so-called energetic norm), so that
u
2
A
 u, u A   Au, u , u  DA .
(37)
Basing on the inequality (35), it holds
1
u  u A.

The distance is defined as
 A  u, v   u  v A   u  v, u  v  A .
(38)
(39)
In this way, we replace the domain DA by a metric space. Let us denote it by PA. Our new
metric space can be shown to be not complete. It can be completed by adding limit functions
(elements) (so called ideal elements). These ideal elements are functions of the space L2()
which cannot be characterized very intuitively here. Let us note that these functions have, in
, the so called generalized derivatives (see further Remark 4) up to the k-th order inclusive
(2k being the order of the given equation)and fulfil the prescribed boundary conditions in
some generalized sense. From the mathematical point of view, they can be characterized as
such functions from the space L2() which are limit elements – in the metric of the space
L2() – of sequences which are Cauchy sequences in the space PA but which do not have a
limit in that space. On the such extended set of functions, the scalar product (u,v)A, defined
originally by (33) for functions from DA only, can be extended, as well as the norm (37) and
the distance (39). It can be shown that the so constructed space – let us denote it by HA –is
complete in this metric, thus that it is a Hilbert space. The set of functions from DA is dense in
this space.
Remark 3. Convergence in the energetic space is called the convergence according to energy. A system of
functions, complete in the energetic space, is called complete according to energy. (back to Lecture 3)(back to
Lecture 11)
Remark 4. Consider a set of functions   C0      , where C0      denotes the set of those functions from
C       (the set of functions infinitely times continuously differentiable in ), which have in  so called
10
compact support. Here, the compact support of a function (P) (denoted by supp ) is the closure of the set of all
the points P   for which (P)  0. Thus a function (P) belongs to C0      if
  C      and supp   
An example of a function   C0   3,3 is the function defined by
1 4  x 2

 e   for
x 2
  x  
for 2  x <3

0
Since the set supp  is closed and should lie in , every function from C0      - and thus every its derivative–
is equal to zero in a certain neighbourhood of the boundary  of the domain . Let us introduce so-called
multiindex i = (i1,…,in) the component of which are nonnegative integers. Denote i= i1 +…..in and write briefly
i
 .
(40)
D i instead of i1
x1 ...xnin
For example, if n = 2, i = (1,2), we have
 3 .
Di 
x1 x22
Let u be an arbitrary function from C  k     and  an arbitrary function from C0      . Applying i-times
Gauss’- theorem and making use of the mentioned property of functions from C0      , we obtain
 uD  d V   1  D u d V
i
i

(41)
i

for everyi  k. The relation (41) holds for every function from C  k     . Moreover, it may hold even for a
broader class of functions from L2() than there are functions from C  k     , for example, for functions, which
do not have in some points a derivative.
Define: Let u  L2() and let there exists such a function vi  L2(), that for all functions   C0      it holds
 uD  d V   1  v  d V .
i
i
(42)
i


Here, the function vi is said to be the i-th generalized derivative of the function u L2(). It is customary to use
the same notation Diu as in the case of classical derivative (namely, if the function u has the derivative Diu in
classical sense, then the generalized derivative Diu is equal to the classical derivative). The function vi is
uniquely determined by the function u. In particular, for n =1 the first generalized derivative of a function u is
defined as such a function u’  L2(a,b) for which
b
b
a
a
 u d x    u  d x
(43)
holds for every   C0   a, b  .(back to Lecture 4) (back to Lecture 5-6)
The functional (17) is then extended onto the whole space HA by
F  u    u, u  A  2  f , u  .
(44)
This functional then actually attains its minimum on HA for a function u0  HA, which is
uniquely determined by the right-hand side f of the equation (16). As stated earlier, this
function u0(x) is called the generalized (weak) solution of the given problem.
Thus, if the operator in (16) is positive definite and f L2(), then there exists a generalized
solution u0 of the considered problem and it is unique. Actually, Eq. (44) can be put into the
following form
2
2
F u   u  u0 A  u0 A .
(45)
But u A is the norm in the space HA, hence
u  u0
u  u0
A
 0, if u  u0  0 in H A , i.e. if u  u0 in H A and
A
 0 for u  u0 in H A .
11
(46)
From (45) and (46) it follows that the functional F(u) attains its minimum on HA just if u = u0
2
in HA and this minimum is equal to  u0 A . The solution u0 is uniquely determined by the
relation
(47)
u0 , v  A   f , v  for all v  H A ,
which is the extension of the relation (24) onto the space HA. (back to Lecture 3) (back to
Lecture 4) (back to Lecture 5-6)
Remark 5. From (47) and (38) we obtain using Schwarz inequality
f
v A for every v  H A .
 u0 , v  A   f , v   f  v 

Specially, for v = u0 we get
f
2
u0 A  u0 , u0 A 
u0 A ,

from where an important inequality follows
f .
u0 A 

This inequality states the continuous dependence of the solution on the right-hand side of the given equation Au
= f: If f(x) changes “not too much” in the norm of the space L2(), then the solution u0(x) “changes not too
much” in the energetic norm
.
A
We have seen that differential equations, which describe certain physical phenomena, are
often derived from various physical principles (principle of minimum of the potential energy,
etc.) using variational calculus. Mathematically, there is necessary to minimize certain
functionals. Using Euler equations, which express the necessary conditions for the extreme of
these functionals, we arrive to differential equations, (e.g. equations (16), (17), (18), (23) in
Lecture 1). These equations contain, so to speak, “unneeded” higher derivatives in
comparison to those, contained in corresponding functionals, (see e.g. equations (9), (38),
(51), (55), in Lecture 1). Since our well-known functional F corresponds in these cases to the
elastic energy, potential energy etc., there is more natural, from the physical point of view, to
seek a generalized solution of given problem than to seek a classical solution, i.e. the solution
of a differential equation.
Essential and natural boundary conditions
We have mentioned the essential and natural boundary conditions already in Lecture 1.
Consider a domain  of the Euclid space and let  is the boundary of . Let us investigate in
the domain  an inhomogeneous differential equation
(48)
Lu  f  P  ,
where P is a general point of the domain , with homogeneous boundary conditions
B j u  0, j  1,2,....k .

(49)
As usual, this boundary value problem generates a certain operator, which will be denoted by
A. We will presume that A operates in Hilbert space L2() and the domain of definition of the
operator A is a set of all functions, which are k- times continuously differentiable in the closed
domain      , (k is the order of differential equation (48)), and which fulfil the
boundary conditions (48). (back to Lecture 4) (back to Lecture 5-6)
Presume that such constructed operator is positive and consider its energetic space HA. It
should be realized that there are situations, when functions from the energetic space must
12
satisfy the boundary conditions of given problem, and there are situations, when these
functions need not satisfy these conditions. It will make clear using the following example.
Example 6. Find the energetic space HC of the operator C, which was investigated in the example 2. Let u be a
function of the space HC. First it is seen that u is also from the basic space L2(0,1). Further, there exists such a
sequence un  DC, that it holds un  u  0 . Then it also holds un  uk
 0 . To functions un, uk, we can
C
C
n 
n ,k 
applied the formula (2e) from the Example 2, hence
2
du 
 du
2
2
 C u n  u k , u n  u k   u n 0  u k 0  u n 1  u k 1    n  k  dx  0 .
C
n.k 
dx
dx 
0
Assume that >0, 0. Then it holds
1
un  uk
2
(6a)
2
du 
 du
0,
0  d xn  d xk  dx n.
k 
1
(6b)
u n 0   u k 0   0.
n.k 
It is seen from the second relation of (6b) that there exists a limit of numerical sequence {un(0)}. Let
lim un 0  c0 .
n 
(6c)
First of the relations (6b) shows, that the derivative dun/dx converges in the mean to a limit v  L2(0,1). If in
the Newton-Leibnitz formula
x
un x   un 0   un t dt
0
we come to the limit, we get
x
ux   c0   vt dt .
(6d)
0
The functions, which can be expressed in the form of (6d), are called absolutely continuous. Hence, if u  HC,
then the function u(x) is absolutely continuous in the interval 0,1 and its derivative is square integrable. Also
the opposite statement can be proved: An arbitrary function absolutely continuous in 0,1,, having square
integrable derivative, is from the energetic space of the C. Observe, that functions from the space HC need not
fulfil any boundary conditions!
The boundary conditions, which the functions from the domain of the definition must satisfy
and which the functions from the energetic space HA need not satisfy, are called natural for
differential operator A. The boundary conditions, which also the functions from the energetic
space HA must satisfy, are called essential. The boundary conditions in the Example 2 are
natural boundary conditions, while the boundary conditions u(0) = 0, u(1) = 0 in the Example
1 are essential boundary conditions.
Remark 6. In the theory of elasticity, the essential boundary conditions are usually called geometrical or
kinematical conditions and the natural conditions are called force conditions.
The entire just presented theory, based on the theorem on the minimum of the functional of
energy, has the advantage that it is relatively simple and rather intuitive from the technical
point of view. However, it has some drawbacks: Only symmetric problems are considered
(moreover, some smoothness of coefficients of the corresponding operators in needed); the
characterization of the space HA (i.e. of its elements) is not simple; inhomogeneous boundary
conditions make certain difficulties. Therefore, another theory has been built up, based on the
Lax-Milgram theorem, the theory of Sobolev spaces and the concept of so-called traces,
which remove quite a number of these drawback. An interested reader is referred to
specialized literature, e.g. Rektorys, K., Variational methods in Mathematics, Science and
Engineering, Ed. Dordrecht-Boston, Reidel 1979
13
Problems
1. Show that the operator A given by
Au   u  1  cos 2 x  u, u  0   0, u    0
is symmetric, and positive definite on the set of admissible functions DA
DA  u; u  C  2 0,  , u  0   0, u    0 (back to Lecture 3)


Hint. Symmetry: We have to prove that


0
0
 Au, v     u  1  cos2 x  u v d x   u, Av    u  v  1  cos 2 x  v  d x
(P1)
The symmetry of u was proved in Example 1. Further, we obviously have


 1  cos x  uv d x   u 1  cos x  v d x .
2
2
0
(P2)
0
Positiveness: It follows from (1d) and (P2)


0
0
 Au, v     u  1  cos 2 x  u v d x   uv  1  cos 2 x  uv  d x
(P3)
wherefrom



0
0
0
 Au, u    u2  1  cos 2 x  u 2  d x   u2 d x   1  cos 2 x  u 2 d x  0
(P4)
We have to prove that (Au,u) = 0 holds if and only if u = 0. Each of the integral (P4)3 is
nonnegative, consequently (Au,u) = 0 is fulfilled only if both these integral are equal to zero.
2
Now, u  C   0,  , thus u  x  is continuous in 0,. So we get

 u
2
d x  0  u   x   0  u  x   const in 0,
(P5)
0
But u  DA , thus u  0   0 and according to (P5) u  x   0 . Summarizing we have
 Au, u   0  u  x   0 .
Positive definiteness. We have obviously


 Au, u    u2  1  cos 2 x  u 2  d x   u 2 d x  u
0
2
0
Thus it is sufficient to put in (8)  = 1.
2. Show that the operator A given by
Au   u  x 1  x  u, u  0  0, u 1  0
is symmetric, and positive definite on the set of admissible functions DA
DA  u; u  C  2 0,1 , u  0   0, u 1  0


14