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9

LINEAR OPERATOR THEORY IN MECHANICS

One of the most useful concepts in the study of mechanics is the linear operator. Finite dimensionallinear operators, namely matrices, have been studied in Chapters 1 to 3. Many of the techniques and results developed for matrices also apply in the study of more general linear operators. In infmite dimensional function spaces, however, technical complexities arise. Fortunately, most work can be done effectively in~. which is a Hilbert space. In this setting, there is a strong mathematical relation between force and displacement functions, which also provides physical insight into problems of mechanics. Finally, boundary-value problems that arise naturally in applications are studied and completeness of eigenfunctions is established.

9.1 LINEAR FUNCTIONALS

Bounded Functionals

Let H be a Hilbert space and S be a subset of H. If to each u in S there corresponds a real number T(u), then Tis called a functional on S. The subsetS is called the domain of the functional T and is denoted Dr· A functional T is said to be a bounded functional if there exists a finite constant c such that

I T(u) I ::;; c II u II (9.1.1)

for all u in DT. The smallest c for which this inequality holds is called the norm of the functional T, denoted as II T 11. and is given by

U T I = sup I T(u) I ue Dr· u:;t:O If U I

(9.1.2)

Thus, IT( u) I ::;; II T lilt u II, for all u in ~· A functional T is a continuous functional at

u0 if, for any£> 0, there exists a o > 0 such that if II u0 - u II< o, then I T(u0)- T(u) I< e. Equivalently, Tis continuous at u0 if lim un = u0 implies that lim T(un) = T(u0) [15, 18].

n~oo

383

384 Chap. 9 Unear Operator Theory in Mechanics

Example 9.1.1

On a Hilbert space H. the scalar product is real valued and defines the functional

T(u) = ( u, u ) = II u 112 (9.1.3)

To see that this functional is not bounded, note that for any v o~: 0 in H, u = av is in H for any scalar a ~ 0. For any finite constant c, choosing a > c/11 v II,

T(u) = II u 1111 u II = a II v 1111 u II > c II u II

so T is not bounded. Even though T is not bounded, it is continuous at all u0 in H. To see this, let

u0 o~: 0 be in H. Then, for any u in H,

I T(u0)- T(u) I = I ( u0, u0 ) - ( u, u) I

= I ( u0, u0 ) - ( U, u0 ) + ( u, u0 ) - ( u, u) I

= I ( u0 - U, u0 ) - ( u, u - u0 ) I

~ I ( u0 - U, u0 ) I + I ( U, u - u0 ) I

(9.1.4)

For any e > o, choose S < ( e/4) II u0 tt small enough so that for II u- u0 II< B, 11 u II< 211 u0 11. This is possible since

II u II - II u0 II ~ If u - u0 II ~ B

implies that

Now, Eq. 9.1.4 yields

o e e 3 I T(u ) - T(u) I ~ 4 + 2 = 4 e < e

if II u- u0 II< B. For the remaining possibility, u0 = 0,

I T(u0) - T(u) I = I T(u) I = II u 112

For any e>O, letS<{£. Then, llu- u0 11 = lluii<S implies that IT(u0)-T(u) l<e. Thus, even though T is not bounded, it is continuous at all u0 in H. •

Sec. 9.1 Linear Functionals 385

Linear Functionals

Definition 9.1.1. A functional T is said to be a linear functional if (1) its domain Dr is a linear space and (2)

T( u1 + u2 ) = T(u1) + T(u2)

T( au ) = a T(u) (9.1.5)

for all u 1 and u2 in DT and all real a. • Note that Eq. 9.1.5 is equivalent toT( au1 + ~u2 ) = aT(u1) + ~T(u2) and, ifT is

linear, then T(O) = 0.

Example 9.1.2

For any f in a Hilbert space H, the functional

T(u) = ( f, u) (9.1.6)

with domain DT = H is a linear functional, since its domain is a linear space and it satisfies Eqs. 9.1.5. By the Schwartz inequality,

I T(u) I = I ( f, u ) I s II f 1111 u II

Since II f 11 is finite, Tis bounded and II T II s II f 11. In fact, II Ttl =II f 11. To show this, for purposes of contradiction, assume that II T n <II f u. Evaluating Eq. 9.1.6 at f,

T(t) = ( f, f) = II f 1111 f II > U T 1111 f II

which is a contradiction. Thus, indeed II T II = II f 11. For any u0 in H,

I T(u0) - T(u) I = I ( f, u0 - u ) I

s II f II II u0 - u II

If II f II ¢ 0, choose o <£/If f IJ. Thus, II u0 - u II < o implies I T(u0) - T(u) I < £, so T is continuous. If II f II = 0, T(u) = 0 is trivially continuous. •

Example 9.1.3

LetT be a linear functional on Rn and e1, e2, ..• , en be the usual basis vectors for R n; i.e., ei =: [ oij lnxl· Then, any u in Rn can be written as u = uiei and T(u) = ui T(e1). Let f = [ T(e1), .•. , T(en) ]T. Since any vector u can be represented as u = uiei and Tis linear, T(u) = ui T(ei) = rTu = ( f, u ). That is, for any linear functional Ton Rn, there exists a vector fin Rn such that T(u) = ( f, u ). As shown in Example 9 .1.2, this is a bounded linear functional. •


The theorems that follow demonstrate some of the remarkable regularity properties of linear functionals.

Theorem 9.1.1. If a linear functional T is continuous at u = 0, it is continuous on its entire domain 0,.. •

To prove Theorem 9 .1.1, let un be a sequence in 0,. with lim un = u "# 0 in DT. It is n-+oo

to be shown that lim T(un) = T(u). By linearity ofT, T(u) - T(un) = T( u- un ). Since DT n-+oo

is a linear space, the sequence ~ = u - un is in 0,. and ~ approaches 0. Since T is continuous at 0, T(~) = T( u- un) = T(u)- T(un) approaches T(O) = 0, as required. •

As a result of Theorem 9 .1.1, it is not necessary to speak of a linear functional as continuous at a particular u0 in its domain. A linear functional is either continuous or not continuous everywhere in its domain.

Theorem 9.1.2. A linear functional Tis continuous if and only if it is bounded .

• To prove Theorem 9 .1.2, suppose first that T is bounded. Then,

Thus, lim T(un) = T(u), which implies that Tis continuous. n-+oo

Suppose now that T is continuous. Assume, for purposes of proof by contradiction, that T is not bounded. Then, for each n there must exist a vector un such that

un I T(un) I~ n II un 11. The sequence vn = approaches zero and has the property

n I un I

Thus, lim T(vn) ;?:: 1 "# 0, which violates the continuity assumption. Thus, T is bounded.• n-+oo

It is interesting to note that the functional of Example 9.1.1 was shown to be continuous, but not bounded. This does not contradict Theorem 9.1.2, since the functional of Example 9 .1.1 is not linear. The reader is cautioned that linearity is a very special property of a functional and that characteristics of linear functionals are not shared by nonlinear functionals. As will be seen later, some properties of functionals on finite dimensional spaces do not carry over to infinite dimensional spaces. The very important property of Rn shown in Example 9.1.3 is, however, valid on all Hilbert spaces, even if they are infinite dimensional.

Sec. 9.1 Linear Functionals 387

Riesz Representation Theorem

Theorem 9.1.3 (Riesz Representation Theorem). Every continuous linear functional Ton a Hilbert space H can be expressed in the form T(u) = ( u, f), where f is in H. Furthermore, f is unique. •

To prove this, let N be the null space of T; i.e., N = { u: T(u) = 0 } . It is easily shown that N is a closed subspace of H. If N = H, then select f = 0. If N "# H, then write H = N + Nl., where Nl. is the set of all vectors that are orthogonal to each vector inN. Since N "# H, N.l contains a nonzero element, say f0. By normalization, set II f0 II = 1. Let v = T(u) f0 - u T(f0), where u is arbitrary. Clearly vis inN, since T(v) = 0. Thus, with f0 in Nl.,

( v, f0 ) = ( T(u) f0 - u T(f0), f0 ) = 0

or,

T(u) - T(f0) ( u, f0 ) = 0

and

T(u) = T(f0) ( u, f0 ) == ( u, T(f0) f0 )

Thus, f = T(f0) f0. To prove uniqueness, suppose that T(u) == ( u, f) = ( u, g ), for all u in H. Then, ( u, f - g ) = 0 for all u, in particular for u = f - g, so II f- g 112 = 0, which implies f = g. •

Consider n generalized coordinates q1, ••. , ~ of a mechanical system. The linear space Rn is the space of generalized coordinates. The generalized force Q, for any q in Rn, is a vector, such that the virtual work oW= ( Q, oq) due to a differential oq, called a virtual displacement, is a continuous linear functional. The component Qi of Q in the scalar product ( Q, oq) is called the component of generalized force corresponding to the generalized coordinate qi. If qi has the physical dimension of length, then Qi will have the physical dimension of force. If qi is dimensionless, then Qi will have the dimension of energy.

Example 9.1.4

The configuration of the double pendulum in Fig. 9.1.1 is uniquely determined by the angles q1 and q2. The moments M1 and M2 are the corresponding generalized forces, since the work performed by the moments as they act through differential rotations Oq1 and oq2 is

oW= M·Oq· 1 1

= ( QM, Oq)

where QM = [ M1, M2 ]T.


F

Figure 9.1.1 Double Pendulum

For the force F that acts at the end of the second bar, the work done is

f,W = By F

where By is a small vertical displacement of the end of the second bar. If the bars have unit length,

y = cos ql + cos ~

Taking the total differential of y,

Thus,

f,y = - f,q1 sin ql - &J.2 sin ~

oW = ( - F sin q1 ) Oq1 + ( - F sin ~ ) f,q2

= < QF, oq)

and the generalized force corresponding to F is QF = [-F sin q1,-F sin~ ]T. •

Example 9.1.5

Consider the beam shown~ under the action of a distributed force f(x). Let ou(x) be a differential displacement of the beam, in the interval 0 s; x s; .R. This generalized displacement is in the Hilbert space L2(0, l ), so the virtual work, which is a

Sec. 9.2 Linear Operator Equations 389

bounded linear functional, must be of the form

'OW = L~ 'Ou(x) f(x) dx = ( f, 'Ou )

where, by the Riesz Representation Theorem, f is also in L2(0, .t ). Thus, the generalized force for a flexible beam is the distributed load f(x).

u

Figure 9.1.2 Beam

EXERCISES 9.1

1. Prove that every linear functional on Rn is continuous.

2. Show that iff is in ~(0, l), then the linear functional

T(u) = ( u, f)

for u in L2(0, 1), is bounded.

3. Show that a linear functional

T(u) = ( u, f)

•x

•

where u is in L2(0, 1 ), need not be bounded if f is not in L2(0, 1 ). Hint: Let f(x) = 1/x.

9.2 LINEAR OPERATOR EQUATIONS

Linear algebra in Rn and linearity properties of matrix operations played a key role in the theory and solution methods for matrix equations in Chapters 1 to 3. Linear operator concepts are introduced in this section that form the foundation for treating elliptic boundaryvalue problems as extensions of matrix equations. The reader is encouraged to note both the similarities and differences between matrix and differential operators. A goal is to use knowledge gained in matrix theory to exploit similarities between the two. A second goal is to identify fundamental differences, so that pitfalls due to inappropriate use of matrix ideas in manipulating differential operators can be avoided.


Linear Operators

Definition 9.2.1. Let V be a function space and A an operator that assigns to each element u in a linear subspace D A of V a vector g =Au in V, such that for any u and v in D A and for any real a and ~.

A (au + ~v) = aAu + ~Av (9.2.1)

The operator is called a linear operator with domain D A and range RA, where

RA = { g: Au = g, for some u in D A }

Example 9.2.1

(9.2.2)

• Let the subspace D A= C2(0, ..t) of the function space ~(0, ..t) be the domain of the second order differential operator

d2u Au=-

dx2 (9.2.3)

for 0 < x < ..t. Application of the second derivative operator to a function u in C2(0, ..t) yields a continuous function. Thus,

RA = C0(0, ..t) (9.2.4)

Linearity properties of differentiation imply that A satisfies Eq. 9.2.1, so it is a linear operator. •

Example 9.2.2

Let the spaceD A= ~(0, ..t) be the domain of the integral operator

(9.2.5)

for 0 < x < ..t. Note that Au is a function of x. Since integration yields a continuous function,

RA = { g e c0(0, ..t): Au = g, u e ~(0, ..t) } (9.2.6)

Linearity properties of integration imply that A satisfies Eq. 9.2.1, so it is a linear operator. •

Bounded Linear Operators

A linear operator A with domain D A is said to be a bounded linear operator if there exists a real number c such that

II Au If ~ c II u II (9.2.7)


for all u in D A- The smallest number c that satisfies the above inequality is called the norm of the linear operator A, denoted II A 11. and given by

IJAul IJAI = sup -

UE DA> Uii:O u u I (9.2.8)

Thus, II Au II~ U A 1111 u 11. for all u in D A- The linear operator A is said to be a continuous

linear operator if lim I u - un I = 0 implies that lim I Au - Aun I = 0· n~- n~-

Theorem 9.2.1. A linear operator is continuous if and only if it is bounded. Furthermore, if it is contiauous at 0, then it is continuO\ls on its entire domain. •

The proof of Theorem 9.2.1 follows the same arguments used in the proofs of Theo-rems 9.1.1 and 9.1.2. •

Example 9.2.3

Consider the linear operator Au = - d2u/dx2 defined on D A = C2(0, 2x), which is a subspace of Lz(O, 2x). To see that A is unbounded, as an operator from DA to L2(0, 21t), consider the sequence of functions

n.n = cos nx, "' _ ~ n = 1, 2, ...

n"'V1t

Since

AP cos nx A"' = n-........

~

the norm of Acpn is

[ i 27t ( cos nx ) 2 ]l/2 U Af I = n {i dx = n

0 1t

whereas II cpn II ~ 0 as n ~ oo, so

IIAfl sup = + oo

n II <If I

which means that there is no real constant c such that II A<Pn II ~ c U q,n II for all n. Thus, the operator A is unbounded. Moreover, since II q,n II ~ 0 and ll Aq,n If ~ oo

as n --7 oo, A is not a continuous operator. This result is consistent with Theorem 9.2.1. . •


Example 9.2.4

Consider next the integral operator of Example 9.2.2, whose domain DAis all of L2(0, .t). Note that

I Auf= JJ J: u(!;ldl;J dx

= JJ I J: I X u(!;) dl; I r dx

~ fol (fox 12 d~) (fox u2(~)d~) dx

,; J.~ u.l 12 d~ ) u.l u2(!;) d~ ) dx

Thus,

II Au II ~ .t II u II

so A is a bounded linear operator and II A II ~ .t. • Symmetric Operators

Definition 9.2.2. A linear operator A, with domain D A and range RA that are subspaces of a Hilbert space H, is said to be a symmetric linear operator if

(Au, v) ;:: ( u, Av)

for all u and v in DA-

Example 9.2.5

(9.2.9)

• Since D A and RA for the second order differential operator A of Example 9.2.1 are subspaces of the Hilbert space ~(0, .t), the operator may be checked for symmetry. For u and v in D A~ C2(0, .t), using integration by parts,

JJ d2u ( Au, v ) = - 2 v dx

o dx

= [ - du v J J + r J du ~ dx dx 0 J0 dx dx

(9.2.10)

Sec. 9.2 Linear Operator Equat~ons 393

Whereas the second tenn on the right of Eq. 9 .2.10 is symmetric in u and v, the first term is not. Thus, the linear differential operator of Example 9 .2.1, whose domain is all of C2(0, -t), is not symmetric. •

Example 9.2.6

Consider the linear differential operator A of Example 9 .2.1, but with its domain a proper subset of C2(0, 1) that satisfies homogeneous boundary conditions; i.e.,

d2u Au = - 2 , 0 < x < 1

dx (9.2.11)

DA = { u in C2(0, 1): u(O)= u(l)= 0}

Even though the differential formula in the definition of the operators here and in Example 9.2.1 is the same, the operators have different domains. Thus, they are different operators.

To see that the operator of Eq. 9.2.11 is symmetric, note that Eq. 9.2.10 is valid, but for v in DA, v(O) = v(l) = 0. Thus,

r1 du dv (Au, v) = Jo dx dx dx (9.2.12)

for all u and v in DA. Since the right side of Eq. 9.2.12 is symmetric in u and v, (Au, v) = ( Av, u) = ( u, Av ). Thus the operator of Eq. 9.2.11 is symmetric. •

Positive Operators

Definition 9.2.3. Let A be a symmetric linear operator that is defined on a dense subspace D A of a Hilbert space H, with range in H. Then, A is said to be a positive semidefinite linear operator if

(Au, u) i?: 0 (9.2.13)

for all u and if ( Au, u ) = 0 for some u * 0 in D A. It is said to be a positive definite linear operator if

(Au, u) > 0 (9.2.14)

for all u * 0 in D A- In other words, if A is positive definite, ( Au, u ) = 0 implies u = 0. Finally, A is said to be a positive bounded below linear operator if there exists a constant c > 0 such that

( Au, u ) ~ c ( u, u ) = c II u 11 2

for all u in DA. Clearly, if A is positive bounded below, it is positive definite.

(9.2.15)

•


In many mechanics problems, u is a deflection and Au is a force, so ( Au, u ) is proportional to the energy that is required to produce the deflection u. The property ( Au, u ) > 0, for all admissible u -:¢:. 0, is a statement that a positive amount of energy is required to produce a nonzero deflection. The property ( Au, u ) ~ c II u 112, for c > 0, is a statement that a lower bound exists for the amount of energy that must be expended in achieving a displacement of fixed norm, which implies stability of the system. That is, large deflections can only be produced by large expenditures of energy. If A is positive definite, but not positive bounded below, then for any integer n there is a 11n in DA such that ( AUn, Un ) S: II Un 112/n. Putting lin = Un/11 Un II, II lin II = 1 and ( Alin, lin ) S: ( 1/n) ~ 0 as n ~ oo, This means that a nonzero deflection lin can be produced with a very small expenditure of energy ( Alin, lin ) .

Example 9.2. 7

For the operator A of Example 9.2.6, Eq. 9.2.12 yields

r.e (du)2 (Au, u) = Jo di" dx ~ 0

In fact, (Au, u) = 0 implies that du/dx = 0, or u =c. Due to the boundary conditions ofEq. 9.2.11, c = 0, so u(x) = 0. Thus, A is positive definite. The fact that it is positive bounded below follows as a special case of the following example. •

Example 9.2.8

Consider the Laplace operator

Au= -V2u

for functions in the domain

DA _= { u in C2(0): u(x) = 0, on r }

(9.2.16)

(9.2.17)

where r is the boundary of 0, which is a bounded subset of Rn. The second Green's formula in Exercise 3 (b) of Section 6.1 shows that the operator is symmetric and that

for all u and v in D A· Setting v = u in this relation,

( Au, u ) = J J 0 Vu Tvu dO ~ 0

(9.2.18)

(9.2.19)

for all u in D A· It remains to show that if ( Au, u ) = 0, then u = 0. This is true, since if (Au, u) = 0, then by Eq. 9.2.19, Vu = 0, which implies that u is constant on 0. Since u = 0 on r, the constant is zero and u == 0. Thus, the Laplace operator, with domain DA ofEq. 9.2.17, is positive definite.


In fact, the Laplace operator ofEq. 9.2.16 with domain ofEq. 9.2.17 is positive bounded below. To simplify the proof of this result, consider n in R2• Then, by Eq. 9.2.19,

(9.2.20)

Since n is assumed to be bounded, it can be enclosed in some rectangle 0 1 with the coordinate axes along two of its sides, as shown in Fig. 9 .2.1.

y

b ~------------------------------,

u(x) = 0

~----------------------------------~X a

Figure 9.2.1 Rectangular Enclosure of 0

Extending the physical domain from 0 to 0 1 and setting admissible functions equal to zero on 01 - n,

f xt d u(x, y1)

0 ax dx = u(xl, Y1) - u(O, Y1) = u(xl, Y1)

for any point (x1, y1) in 0. Applying the Schwartz inequality,

2 ( rxl au(x,yl) )2 u (x1, y1) = Jo 1 X ax dx

rxt 'f rxt [ au(x, YI) ]2 s Jo (1 dx Jo ax dx

rxt [ d u(x, Y1) ] 2 = xl Jo ax dx

i a [ a u(x, y1) ] 2 sa a dx

0 X


Integrating this inequality over the rectangle 0 S: x1 S: a, 0 S y1 S b,

21b Ja[ au(x, Yt.) ]2 = a a dx dyt

0 0 X

= .z s: JJ il u~ y) r dx dy

Letting a2 = 1/a and noting that the integrands are zero outside Q, this inequality becomes,

(9.2.21)

which is known as Friedrichs' inequality. From Eq. 9.2.20, this inequality is

(9.2.22)

which shows that the Laplace operator of Eq. 9.2.16, with the domain of Eq. 9.2.17, is positive bounded below. •

Matrices as Linear Operators

While the power of linear operator theory is best exploited in dealing with boundary-value problems, it is instructive to see that the concepts introduced thus far in this section are related to properties of matrices. Let H = Rn and V = Rm, with m S: n, be Hilbert spaces with the norm of Eq. 2.4.4. Consider the m x n matrix A, written both in terms of its elements and its columns as

A = [ ~j lmxn = [ a 1, · • · , ~ ] (9.2.23)

where aj are m x 1 column matrices. With the domain D A = H, for any u in D A• the matrix A defmes the operator

Au = [ a .. y.] 1 = U·a· "'lJ J mx J J (9.2.24)

The range RA of this operator is thus a subspace of V.


If the matrix A has full row rank, then there are m linearly independent columns aj in Eq. 9.2.23 that form a basis for V. Thus, for any v in V, there exist constants uj corresponding to the aj that form a basis of V, such that

v = ujaj = Au* (9.2.25)

where u* is defined as ann-vector that contains the parameter uj in row j. However, u* may not be unique if m < n. Thus, if the matrix A has full row rank, RA = V.

Since D A = H is a vector space and

A ( au + ~w ) = o:Au + ~A w (9.2.26)

follows from properties of matrix multiplication and addition, for all real a and ~ and vectors u and w in D A• A is a linear operator.

To see that any matrix operator A is bounded, using the Schwartz inequality, note that

[ m ( n )2 ]1/2

HAu I= ~ ~~juj 1=1 J=1

= [ t { t ~ } { t uJ} ] 112

1=1 j=1 J=1

= F(A)Iu I (9.2.27)

where

[ m n ]1/2

F(A} = ~~afj (9.2.28)

is called the Frobenius norm of A. Thus, A is a bounded linear operator from H to V, with its operator norm II A II bounded by

II A II s; F(A) (9.2.29)

To determine the operator norm of A, the theorem that follows can be used.


Theorem 9.2.2. The operator nonn of A is

IIAI= max IAI A.ecr(ATA)

(9.2.30)

where a( AT A) is the spectrum of AT A, which is the set of all eigenvalues of AT A. •

For proof, note that the eigenvalues of AT A are nonnegative. To see this, let

where u =1: 0. Then,

( u, AT Au ) = ( u, AU ) = A II u 112

and

Thus 'A ~ 0. Let the eigenvalues of AT A be arranged in ascending order,

and let cp 1, cp2, ... , q,n, be the corresponding orthonormalized eigenvectors. For any u e Rn,

II Au If ~ ( Au, Au ) = ( u, AT Au )

Since { q,i } forms a basis for Rn, u can be expressed as

and

Then,

i=l i=l

and

Sec. 9.2 Unear Operator Equations

n n

= An L l: ~cfiTaj~ = An U u f i=l j=l

Thus,

IIAIISA.n

However, if u = .n. then II u II = 1 and

Thus,

II Au If = ( u, AT Au) = ( .n. AT Aq,n) = An

HAl=~= max IA.I A.eo{ATA)

399

• Consider next the case in which m = n; i.e., A is a square matrix. Using properties

of matrix transpose and the scalar product on Rn,

(Au, v) = (Au )Tv = UTA Tv (9.2.31)

If the operator A is symmetric, then

(Au,v) • uTATv = uTAv • (u,Av) (9.2.32)

for all u and v in Rn. In particular, for u = ek • [ Oki] and v = ej • [ Oji ], Eq. 9.2.32 becomes

ajk = akj (9.2.33)

Thus, the matrix A assoclated with a symmetric operator is a symmetric matrix. Conversely, if A is a symmetric matrix, then for all u and v in Rn,

(Au,v) = uTATv = uTAv = (u,Av) (9.2.34)

So the operator associated with a symmetric matrix A is a symmetric operator. For m = n and A a symmetric matrix, if the operator A is positive semidefmite,

( u, Au) = UT Au ~ 0 (9.2.35)

for all u in Rn and there is au*'# 0 such that ( u*, Au*)= 0. Thus, the matrix A is positive semidefinite.


If the symmetric operator A is positive definite,

( u, Au ) = u 1 Au > 0 (9.2.36)

for all u:;:. 0 in Rn. Thus, the matrix A that defines the operator is positive definite. Let +i be orthonormal eigenvectors of the n x n positive defmite matrix A, with cor

responding eigenvalues "-i > 0, i = 1, ... , n. For any u in Rn,

Thus,

Finally,

n

= b·b-~- = ~ b:Z 1 j'ij """' 1

i=l

n

n

= ~ b~A,. """' 1 1 i=l

~ ~ Aj L bf = ~n Aj a u 12

J i=l J

(9.2.37)

(9.2.38)

(9.2.39)

This shows that the linear operator associated with a positive definite matrix is also positive bounded below. This result is valid only for finite dimensional vector spaces and operators that can be defined by matrices. It is not in general true for infinite dimensional, differential operators on L,z(O).

Operator Eigenvalue Problems

As in use of separation of variables in Section 8.5, eigenvalue problems associated with operator equations are often encountered. For symmetric linear operators A and B whose common domain is a subspace D of a Hilbert space H and whose ranges are in H, define the eigenvalue operator equation

Au= ABu (9.2.40)

where u ¢0 is in D. The function u is called an eigenfunction for the operators and A, is the associated eigenvalue.


Let u 1 and u2 be eigenfunctions that correspond to different eigenvalues, A1 :F- A2;

i.e.,

Au1 = A1Bu1

Au2 = ~Bu2

Taking the scalar product of both sides ofEq. 9.2.41 with u2 and u1,

( u2, Au1 ) = A1 ( u2, Bu1 )

(u1,Au2 ) = A2 (u1, Bu2 )

Since u1 and u2 are both in D and A and B are symmetric,

(u1,Au2 ) = (Au1,u2 ) = (u2,Au1 )

(u1,Bu2 ) = (Bu1,u2 ) = (u2,Bu1 )

(9.2.41)

(9.2.42)

These results can be substituted into the second of Eqs. 9.2.42 and the result subtracted from the first, to obtain

(9.2.43)

Here, ( u1, Bu2 ) is said to be the scalar product with respect to the operator B. This proves the theorem that follows.

Theorem 9.2.3. Eigenfunctions of the operator equation of Eq. 9.2.40, with symmetric operators A and B, that correspond to different eigenvalues are orthogonal with respect to the operator B. Further, if the operator B is positive defmite, the eigenvalues are real. •

The proof of Theorem 3.4.2 is applicable here, with a modification of the definition of the scalar product. •

EXERCISES 9.2

1. Find the null space of the operator

d2u Au=--

dx2

from DA = c2(0, 21t) to C0(0, 21t); i.e., NA = { u in DA: u" = 0 }, and determine its dimension.

2. Let A be the operator in Exercise 1, with domain restricted to those functions that satisfy u(O) = 0. Determine the null space of this operator and its dimension. Answer the same question for the operator A, with domain restricted to functions that satisfy u(O) = u(21t) = 0.


3. Let A be the operator defmed on 4(0, 1) into 4(0, 1) by

Au = fo1 K(s, t) u(t) dt

where

Is this operator bounded? If so, what is its norm?

4. Show that for a symmetric, positive definite, linear operator A, if the equation Au = f has a solution, it is unique.

5. Let A be a symmetric, positive definite operator that is defined on a dense subspace D A of a Hilbert space H Verify that

[ u, v ]A = ( Au, v )

is a scalar product on D A. This scalar product is referred to as the energy scalar product with respect to A. Also show that II u IIA = [ u, u lA 112 is a norm.

6. Let 0 be a region in R3, with boundary f, and let A denote the Laplace operator -V2, defined on the domain

DA = { u in C(Q): ~=Oon r} Detennine whether the operator is symmetric.

9.3 STURM·LIOUVILLE PROBLEMS

In this section, results concerning sets of orthogonal functions that are generated as solutions of certain types of boundary-value problems are summarized.

Eigenfunctions and Eigenvalues

A problem that arises often in mechanics consists of a homogeneous linear differential equation of the form

d [ du J - dx p(x) di' + q(x) u = A. r(x) u (9.3.1)

together with homogeneous boundary conditions that are prescribed at the end points of an interval [a, b]. This problem has a nontrivial solution only if the parameter A. has a certain value, say A. = A.k. The permissible values of A. are known as eigenvalues and the corre-

Sec. 9.3 Sturm-liouville Problems 403

sponding functions u = q,k(x), which satisfy Eq. 9.3.1 with A.= A.k, are known as eigenfunctions.

In most cases that occur in practice, the functions p(x) and r(x) are positive in the interval [a, b], except possibly at one or both of the end points. Defining the linear differential operator

and the scalar operator

Bu = ru

the differential equation ofEq. 9.3.1 takes the operator form

Au = A.Bu

Note that

b

( Au, v ) = { [- ( pu' )' + qu ] v dx

(9.3.2)

(9.3.3)

(9.3.4)

= {b u [- ( pv' )' + qv] dx - [ p ( u'v - uv') ]~ (9.3.5)

so the operator A is symmetric if its domain is limited to functions that satisfy

[ p ( u'v -- uv' ) ]~ = 0 (9.3.6)

Note also that

( Bu, v ) := Lb ruv dx = ( u, Bv ) (9.3.7)

so B is symmetric. Furthermore, if r(x) ~ a. > 0, for a ~ x ~ b, then

soB is positive bounded below, hence positive definite. Assuming that Aj '#~are eigenvalues of Eq. 9.3.4, Theorem 9.2.2 shows that if tht:

specified boundary conditions imply Eq. 9.3.6, then the corresponding eigenfunctions <j>1

and q,i are orthogonal relative to the weighting function r(x); i.e.,

(9.3.8)


Boundary conditions that give rise to this situation include the following:

( 1) At each endpoint of the interval, u, du/dx, or au + ~ ( du/dx) vanish. (2) If it happens that p(x) vanishes at x = a or at x = b, u and du/dx remain finite at

that point and impose one of the conditions in (1) at the other endpoint. (3) If it happens that p(b) = p(a), u(b) = u(a) and u'(b) = u'(a).

A collection of functions u in C2(a, b) that satisfy boundary conditions (1), (2), or (3) is defined to be the domain D of the operators A and B of Eqs. 9.3.2 and 9.3.3.

In most practical cases; in particular if p, q., and rare in C1(a, b) and both p and rare positive throughout the interval (a, b) that is of finite length, it will be shown in Section 9.7 that there exists an infinite set of distinct eigenvalues A.1, "-2 •.... If q(x) :2! 0 in (a, b) and if

(9.3.9)

then the eigenvalues are all nonnegative (see Theorems 9.2.2 and 9.7.1). Further, except in the case of the periodicity condition of (3) above, to each eigenvalue there corresponds one and only one eigenfunction. In case (3), two linearly independent eigenfunctions generally correspond to each eigenvalue (see Example 9.3.1). Such pairs of functions can be orthogonalized by the Gram-Schmidt procedure.

The boundary-value problem considered here is known as a Sturm-Liouville Problem. The importance of such problems stems from the fact that the sets of orthogonal eigenfunctions that are generated by these problems are complete in L 2(a, b). Further, a positive statement can be made about pointwise convergence of the series representation of a sufficiently well-behaved function f(x) in terms of the eigenfunctions.

In practice, it is often inconvenient to normalize the eigenfunctions. In such cases, the coefficients in a series representation

00

f(x) = "L ci~(x) (9.3.10) i=l

are given by the formula

ci Iab r ( 4i )2 dx = Lb rf<ji dx (9.3.11)

or,

. 2 . cd 4f I r = (f, 4f h (9.3.12)

This result is obtained formally by multiplying both sides ofEq. 9.3.10 by the product rq,i, integrating the result term-by-term over (a, b), and taking into account the orthogonality of the eigenfunctions relative to the weighting function r(x).

Sec. 9.3 Sturm-Uouvme Problems 405

Convergence theorems for series expansion of functions in the eigenfunctions of the Sturm-Liouville problem play a central role in engineering mathematics. Two theorems on convergence and completeness are stated here, one of which is proved in Section 9.7. The first theorem concerns completeness, in the L;z sense, of the eigenfunctions of the SturmLiouville problem.

Completeness of Eigenfunctions In l 2

Theorem 9.3.1. Let the functions p(x), q(x), and r(x) be in C1(a, b) with p(x)>O and r(x) > 0 in a S x S b. Then, the eigenfunctions of Eq. 9.3.4, with boundary conditions of type (1), are complete in l.;z(a, b). That is, for any fin L2(a, b),

as n ~ oo, where the ci are given by Eq. 9.3.12. • The proof of this theorem requires mathematical tools of Section 9.7, so it will be

postponed. The value of this theorem is that it establishes the L;z convergence of the generalized Fourier series of Eq. 9.3.10, for sequences of eigenfunctions { q,i(x)} of SturmLiouville problems that arise in solution of specific problems of mechanics. These eigenfunctions are called special functions and were introduced in Section 4.4.

Example 9.3.1

Consider the differential equation

d2u ---=AU

dx2 (9.3.13)

which is a special case of Eq. 9.3.1, in which p(x) = r(x) = 1 and q(x) = 0. Consider the interval (0, -f) and impose the boundary conditions

u(O) = 0, u(1) = 0 (9.3.14)

Note that the left side of Eq. 9.3.13, with the boundary conditions of Eq. 9.3.14, is just the operator of Eq. 9.2.11, which was shown in Examples 9.2.6 and 9.2.7 to be symmetric and positive definite. The eigenvalues are of the form A.k = k27t2/12, where k is any positive integer. The corresponding eigenfunctions are

q,k :::: sin~. k = 1, 2, ... 1

Thus, Eq. 9.3.10 is just the Fourier sine series representation

"" f(x) = I ck sin k1tx

k::::l 1


for 0 < x < .t, where, with r(x) = 1, Eq. 9.3.11 yields

2J.e . kn:x ck = - f(x)sm- dx

,t 0 ,t

In a similar way, the conditions u'(O) = u'(..t) = 0 give rise to a Fourier cosine series representation, while the periodicity conditions u(-.t) = u(.t) and u'(-.t) = u'(.t) on the interval (-.t, .t) lead to the general Fourier series representation over that interval, involving both sines and cosines of period 2.t. This establishes the 4 completeness of the Fourier sine-cosine series that was stated in Theorem 5.3.4. II

In addition to the important 4 completeness result for eigenfunctions of SturmLiouville problems, a theorem is stated here that guarantees pointwise convergence results similar to those proved for sine-cosine series in Section 5.5. For proof see Ref. 20.

Theorem 9.3.2. Let the Sturm-Liouville problem of Eq. 9.3.4 satisfy the hypotheses of Theorem 9.3.1. Then,

(1) lff(x) is in D1(a, b), the generalized Fourier series ofEq. 9.3.10 converges to

00

I [ f( x + 0 ) + f( x - 0 ) ] = ,L ckf(x) k=l

for a< x <b. (2) lff(x) is in C2(a, b) and satisfies the boundary condition of type (1), then the

generalized Fourier series of Eq. 9.3.10 converges uniformly to f(x) in a~x~~ II

EXERCISES 9.3

1. Find the eigenvalues and eigenfunctions of the following Sturm-Liouville problems:

(a) u" + A.u = 0

u(O) = u'(n:) = 0

(b) u" + A.u = 0

u(O) = u(2x). u'(O) = u'(2n:)

2. Expand the function f(x) =sin x, 0 ~ x ~ 11:, in the eigenfunctions of Exercise 1 (a).

Sec. 9.4 Separation of Variables and Eigenfunction Expansions 407

9.4 SEPARATION OF VARIABLES AND EIGENFUNCTION EXPANSIONS

Thus far, the study of dynamics problems has been limited to linear second order equations in two independent variables. While the study of this special class of problems is illuminating, extensions to equations with more independent variables and higher order equations that arise in applications should be considered. In this section, the separation of variables method is extended to higher order and higher dimensional problems. For these problems, the separation method and the formal theory of eigenfunction expansions is found to be quite broadly applicable.

Example 9.4.1 (Vibration of a Cantilever Beam)

Natural vibration of a beam is governed by the differential equation

(9.4.1)

where a2 = EI/pS, E is Young's modulus, I is the second moment of the crosssectional areaS, pis mass density, and A is a fourth order differential operator. For the cantilever beam of Fig. 9.4.1, the boundary conditions at the fixed end x = 0 are

u(O, t) = Ux(O, t) = 0 (9.4.2)

At the free end, the bending moment and the shear force must be equal to zero, so

(9.4.3)

Figure 9.4.1 Cantilever Beam

In order to completely define the motion of the beam, initial conditions that prescribe the displacement and velocity of the beam at t = 0 must be given, in the form

u(x, 0) = f(x)

ut(x, 0) = g(x) (9.4.4)


The problem considered therefore leads to the solution ofEq. 9.4.1, with boundary conditions of Eqs. 9 .4.2 and 9 .4.3 and initial conditions of Eqs. 9 .4.4.

Following the idea of separation of variables introduced in Section 8.2, a solution of the following form is sought:

u(x, t) = X(x) T(t) (9.4.5)

Substituting from Eq. 9.4.5 into Eq. 9.4.1,

T(t) AX(x) 4

a2T(t) = X(x) = a (9.4.6)

where AX= d4X/dx4 and the positive separation constant is selected to avoid exponential growth of the solution in time. The second of Eqs. 9 .4.6 yields the eigenvalue problem

X(O) = 0, X'(O) = 0, X"(..t) = 0, X"'(..t) = 0

(9.4.7)

(9.4.8)

for X(x). To see that the constant in Eq. 9.4.6 must be positive, from another point of view, observe that A is a positive definite operator. By taking the scalar product on both sides ofEq. 9.4.7 with X, (AX, X)= a4 ( X, X)> 0, since X¢ 0. Thus, a4 > 0 is justified.

The general solution of Eq. 9 .4. 7 is

X(x) = A cosh ax + B sinh ax + C cos ax + D sin ax

From the conditions X(O) = 0 and X'(O) = 0 ofEq. 9.4.8, C =-A and D =-B. It follows that

X(x) = A [cosh ax- cos ax] + B [sinh ax- sin ax] (9.4.9)

From the conditions X "(..t) = 0 and X "'(..t) = 0 of Eq. 9.4.8,

A [ cosh al + cos a..t ] + B [ sinh a..t + sin al ] = 0

A [ sinh a..t - sin a..t ] + B [ cosh a..t + cos a..t ] = 0

This system of homogeneous linear equations has a nontrivial solution for A and B only if the determinant of the coefficient matrix is zero. Setting the determinant equal to zero, eigenvalues are obtained from the transcendental equation


Because cosh2aJ - sinh2aJ = 1 and sin2aJ + cos2aJ = 1, this equation can be reduced to

( cosh aJ ) ( cos aJ ) = - 1

The roots ofEq. 9.4.10 can be calculated numerically as [21]

a 1J = 1.875

a2J = 4.694

a3J = 7.854

1t an l = 2 ( 2n - 1 ), for n > 3

(9.4.10)

(9.4.11)

The last formula gives a value of au for n = 3 that is accurate to the third decimal place and for n = 7 that is accurate to the sixth place.

The time dependence in this problem is governed by the first equality of Eq. 9.4.6; i.e.,

(9.4.12)

which is satisfied by the trigonometric functions

(9.4.13)

with the natural frequencies of vibration

(9.4.14)

i.e., the natural frequencies Vn of vibration behave as the square of au. Because

the second natural frequency is more than two and one-half octaves higher than the base tone, while the third frequency is more than four octaves higher than the base tone.

If the cantilever beam is a tuning fork with a basic frequency of 440 cycles per second, then the second frequency of the tuning fork is 27 57.5 cycles per second, while the third frequency of 7721.1 cycles per second is already higher than fre-

410 Chap. 9 Linear Operator Theory in Mechanics

quencies that are used in music. If the tuning fork is set in vibration by a blow, higher frequencies appear in addition to the first, which explains the initial metallic sound. The higher harmonics are quickly damped, so the tuning fork soon rings out with the pure basic tone.

Substituting the ai of gq. 9.4.11 into Eq. 9.4.9, with the associated normalized eigenvectors [ Ai, Bi ]T, eigenfunctions Xi(x) are obtained that satisfy the homogeneous boundary conditions of the problem. Solutions of the vibration problem may thus be written in the form

00

u (x, t) = l Xi(x) Ti(t) i=l

00

= l Xi(x) [ ai cos 21tvit + bi sin 21tVit ] i::::l

(9.4.15)

where the constants~ and bi must be chosen to satisfy the initial conditions of Eq. 9.4.4. More specifically,

co

u(x, 0) = f(x) = l ~ Xi(x) i=l

(9.4.16) 00

u1(x, 0) = g(x) = l bi21tVi Xi(x) i== 1

so ~and 21tvibi are Fourier coefficients of functions f(x) and g(x), respectively, with respect to coordinate functions Xi(x).

If there had been a lateral force pS h(x, t) applied to the beam of Fig. 9 .4.1, the differential equation of motion would have been

(9.4.17)

To solve this forced vibration problem, with homogeneous initial conditions, assume that the eigenfunctions Xn(x) are complete in L:z(O, ..e), so

00

h(x, t) = l hn(t) Xn(x) (9.4.18) n=l

where

1 hn(t) = fo h(x, t) Xn(x) dx (9.4.19)


Assuming a particular solution of the form

00

up(x, t) = L TPn (t) Xn(x) (9.4.20) n=l

and substituting Eqs. 9.4.18 and 9.4.20 into Eq. 9.4.17,

(9.4.21)

A particular solution TPn ofEq. 9.4.21 is

1 Jt TP (t) = - 2 1tn(t) sin {~a (t-t)} dt n ana 0

(9.4.22)

and, since

it follows that

00

up(x, t) = L TPn (t) Xn(x) n=l

is a particular solution of Eq. 9 .4.17 that satisfies homogeneous initial and boundary conditions. The solution for forced vibration of a cantilever beam is obtained by superposing the solutions of Eqs. 9.4.15 and 9.4.22. That is, the forced solution u(x, t) is given by

u(x, t) = u(x, t) + up(x, t) (9.4.23)

The completeness assumption for the eigenfunctions Xn(x) in the foregoing analysis is motivated by the fact that the eigenfunctions of the Sturm-Liouville problem are complete. It will be shown later in this chapter that these eigenfunctions are indeed complete in ~(0, .l). •

Example 9.4.2 (Vibrating Membrane)

From Section 6.3, the equation of a vibrating membrane on a rectangular domain 0 = { X = ( X1, X2 }: 0 S X1 Sa, 0 S X2 S b }, with boundary r, is

for x in 0,

2 1 Au = - V u = - - Uu

m2 (9.4.24)


u(x, t) = 0

for X on r, and

u(x, 0) = f(x)

ut(x, 0) = g(x)

(9.4.25)

(9.4.26)

(9.4.27)

for X inn, where m2 = T/p. This is similar to the form of problem treated in Example 8.6.2. The problem separates with u(x, t) = T(t) X(x), and the solution is

00

u(x, t) = I, (~cos m~t + bi sin mait) ~(x) i=l

where Xi(x) is the normalized solution of the eigenvalue problem

AX=- V2X = a2X, xinO

X= 0, xonr

(9.4.28)

(9.4.29)

On the rectangular domain 0, the solution X(x) of Eq. 9.4.29 may be attempted by separation of variables; i.e.,

X(x) = U(x1) V(x2) (9.4.30)

The boundary condition ofEq. 9.4.29 implies that

U(O) = U(a) = V(O) = V(b) = 0

Substitution from Eq. 9.4.30 into the frrst ofEqs. 9.4.29 yields

U"(x1) V(x2) + U(xt) V"(x2) + a2U(xl) V(x2) = 0 (9.4.31)

Dividing by-U(x1) V(x~ yields

U "(x1) V "(x2) -~-= U(x1) V(x2)

(9.4.32)

In order for this equation to hold for all x1 and x2, the left and right sides must be equal to a constan!. Since - U" is a positive definite operator with the boundary conditions treated here, the constant must be positive, say p2. Thus,

- U"(x1) = P2 U(x1)

- V"(x2) = ( a 2 - P2 ) V(x2)

(9.4.33)

(9.4.34)

Since the o~rator - V " with these boundary conditions is also positive definite, a2 - p2 = "( > 0.

Sec. 9.4 Separation of Variables and Eigenfunction Expansions

Equations 9.4.33 and 9.4.34 have solutions

U = C cos J3x1 + D sin J3x1

V = E cos yx2 + F sin yx2

413

(9.4.35)

(9.4.36)

Since U(O) = V(O) = 0, C = E = 0. Further, since U(a) = V(b) = 0, sin J3a == 0 and sin yb == 0. Thus,

in l3i = -a· i = 1, 2, ...

jn "(j = b ' j = 1 ' 2, . . .

(9.4.37)

but since y = ci -132, a double sequence of eigenvalues aij is obtained,

(lij = ( 'Yj2 + 13t )1/2 = 1t[ (~i)2 + (bj )2]1/2 (9.4.38)

with eigenfunctions

(9.4.39)

Thus, the general solution is

Loo LQO . . i1tXt . j1tx2 u(x t) = [ H· · cos ma··t + b· · sm mt:l··t ] sm - sm -' ,J lJ IJ lJ a b

i=l j::::l

The initial conditions may now be written as

Loo L"" inxl . j1tXz u(x, 0) = f(x) = a·· sm -- sm -b-

IJ a i=l j=l

Thus, coefficients ~j and bij are Fourier coefficients of the functions f(x) and g(x) in double sine series expansions; i.e.,

4 Ja fb . i1tX1 . j1tX2 aij = ab 0 Jo f(xl> x2) sm -a- sm b dx1 dx2

4 J a Jb . inxl jnxz biJ. :::= b g(x1, x2) sm- sin -b- dx1 dx2

maija 0 0 a •


EXERCISES 9.4

1. Obtain equations for the frequency of vibration of a clamped-clamped beam; i.e., for boundary conditions u(O, t) = ux(O, t) = u(.t, t) = ux(.t, t) = 0.

2. Show that the selection of a nonnegative constant on the right of Eq. 9.4.6 is justified by the fact that the operator

AX= x<4>

with domain

DA = {X e ct'(O, 1): X(O) = X'(O) = X"(.t) = X"'(.t) = 0}

is positive defmite.

3. Derive formulas for forced vibration of the membrane of Example 9.4.2; i.e., replace Eq. 9.4.24 by

1 - Uu - V2u = h(x, t) m2

9.5 A FORMAL TREATMENT OF THE EIGENFUNCTION EXPANSION METHOD

As seen in Chapter 8 and Section 9.4, the Fourier expansion of a function f(x) in terms of the eigenfunctions ui(x) of a symmetric, positive defmite operator is often possible. Applications studied in Section 9.4 show that when this happens, the separation of variables technique can be employed to construct solutions of linear dynamics problems that arise in mechanics. In light of this apparent success, it is desirable to formalize and carefully justify the eigenfunction expansion method, which is the purpose of this section. It is noted here, and further developed later, that existence of a Green function for an operator is key to proving completeness of the eigenfunctions.

Let n be the domain of the independent variables x; i.e., an inteiVal on the x-axis, an area in the x1-xrplane, or a volume in x1-x2-xrspace, that has a piecewise smooth boundary r. Suppose the state of a continuum that occupies 0 is characterized by a function u(x, t) that vanishes identically if the system is in stable equilibrium under zero load. Let A be a differential operator in the independent variables X, defined on n, let p(x) represent the mass density of the material at any point in n, and let Q(x, t) represent a given external distributed force. A solution of the differential equation

putt+ Au = Q (9.5.1)

is sought that satisfies homogeneous time-independent boundary conditions on the bound-

Sec. 9.5 A Formal Treatment of the Eigenfunction Expansion Method

ary r of 0 and the initial conditions

u(x, 0) = f(x)

u1(x, 0) = g(x)

415

(9.5.2)

All derivatives that occur are assumed to be continuous. For applications with homogeneous boundary conditions considered here, the operator A has been shown to be positive defmite.

Separation of Variables

Consider first free vibration that is characterized by solutions of the homogeneous differential equation

putt + Au == 0 (9.5.3)

that satisfy prescribed homogeneous boundary conditions. Specifically, consider solutions of the form u(x, t) = X(x) T(t). Each such solution is associated with an eigenvalue a 2, with the properties

T =- a1' (9.5.4)

(9.5.5)

where X(x) must satisfy the boundary conditions that are imposed on u, so the eigenvalue a 2 of the positive definite operator A is positive. Thus,

T(t) == a cos at + b sin at

The problem of Eq. 9.5.5 is to determine values of the eigenvalues a 2 for which the homogeneous differential equation of Eq. 9.5.5 has eigenfunctions that satisfy the prescribed boundary conditions. The state of vibration that satisfies Eq. 9.5.3 is then represented by

u(x, t) = T(t) X(x) = ( a cos at + b sin at ) X(x)

Completeness of Eigenfunctions

(9.5.6)

In the case of a bounded domain n, the following statements are true, but remain to be proved:

(1) The eigenvalues a 2 form a countably infinite ascending sequence 2 -r 2

at s; a2 s; a3 s; · · · (2) There exists a sequence of associated eigenfunctions X1 (x), X2(x), ... that is

complete in the Lz sense and satisfy the orthonormality relations


J 0 pXiXk dx = 0, i ¢ k

J 0 pXf dx = 1, i = 1, 2, ...

(9.5.7)

(3) Every function w(x) that satisfies the prescribed homogeneous boundary conditions, and for which Aw is continuous, may be expanded in an absolutely and uniformly convergent series in the eigenfunctions ~(x); i.e.,

00

w(x) = .2, ci Xi(x) i=l

(9.5.8)

This pseudo-theorem will be justified in subsequent sections. On the basis of these properties, which must be proved for each problem, an infinite

sequence of functions ( ai cos exit+ bi sin ait ) Xi(x) is obtained. From these functions, solutions of the initial-boundary-value problem for the differential equation of Eq. 9.5.3 are obtained by superposition,

00

u(x, t) = .2, Ti(t)Xi(x) i=l

if the constants llj and bi are chosen as

ai = Jo pfXi dx

(9.5.9)

For the nonhomogeneous equation of Eq. 9.5.1, with homogeneous boundary conditions, u(x, t) is found by determining its time dependent expansion coefficients Ti(t). To this end, use

and

00

uP(x, t) = .2, TPi(t)Xi(x) i=l

00

Q(x, t) = p .2, Qi(t)Xi(x) i=l

Sec. 9.5 A Formal Treatment of the Eigenfunction Expansion Method 417

in Eq. 9.5.1, to obtain

00 00 00

~ pTp. + ~ T AX· = p ~ QX· ~ J ~ Pj J ~ J J j=l j=l j=l

Multiplying both sides of this equation by Xi, integrating over 0, and using Eqs. 9.5.5 and 9.5.7,

(9.5.10)

where Qi(t) is the expansion coefficient of Q(x, t) p-1 with respect to the orthonormalized eigenfunction Xi(x). A particular solution Tpi(t) of this equation is given by

1 1t Tp. (t) = - Qi(t) sin ai ( t - 't ) dt 1 (li 0

The function formed using these expansion coefficients is a particular solution of Eq. 9.5.1. Other solutions are obtained by adding solutions of Eq. 9.5.3. Thus, the initial value problem under consideration is reduced to the problem of solving the homogeneous equation of Eq. 9.5.3.

Application of the separation of variables method to the heat conduction problem also leads to eigenvalue problems. If the units of time and length are suitably chosen, the differential equation of heat conduction takes the fonn

ut +Au== 0 (9.5.11)

where u denotes temperature, which is a function of position x and time t. The convection of heat from a homogeneous body 0 with surface r into an infinite medium of zero temperature is characterized at the surface r by a boundary condition of the form

au dn +ou=O (9.5.12)

where 0" is a positive constant. This condition states that the rate of change of temperature in the direction of the inner nonnal is proportional to the difference in temperature between the exterior and the interior of the body. A solution of the equation of heat conduction is sought that satisfies this boundary condition and a prescribed initial temperature condition at timet= 0.

Writing u(x, t) in the form u(x, t) = X(x) T(t) yields

(9.5.13)


since A is positive defmite. This yields the following eigenvalue problem for X(x):

AX = a2x, in o ax Tn + ax = 0, on r

(9.5.14)

For a given eigenvalue a2 and its eigenfunction X(x), the corresponding solution of the differential equation has the form

2 u(x, t) = c X(x) e-a. t (9.5.15)

By the eigenfunction exp.ansion method, the solution can be made to satisfy a given initial condition u(x, 0) = f(x), where f(x) is continuous in 0, together with its derivatives of ftrst and second order, and satisfies the boundary conditions. If the normalized eigenfunctions X1 (x), X2(x), ... , form a complete sequence in ~(0), the desired solution is given by

00

u(x, t) = L ci Xi(x) e-«2t

i=l (9.5.16)

Note that the positive character of the eigenvalues a2 implies that the solution u(x, t) approaches zero asymptotically as t increases, as is expected from the physical meaning of the problem.

Green's Function

Eigenfunctions can also be used to solve the equilibrium problem; i.e., the boundary-value problem for the differential equation Au = Q(x). Let

'"' Q(x) == L Xi(x) J Q{;) Xi(~)d~

i::l 0

co

u(x) = L ci Xi(x) i=l

To find ci(t), substitute Q and u into Au = Q, to obtain

00 00 00

L ci AXi(x) = L cif1.j2 Xi(x) = L Xi(x) J Q(~) Xi(~) d~ i=l i=l i=l n

Sec. 9.6 Green's Function for Ordinary Boundary-Value Problems 419

Equating coefficients of Xi(x),

(9.5.17)

Hence, the solution is given formally as

(9.5.18)

Formally interchanging summation and integration in this expression, a function

(9.5.19)

is defined, so that the solution of the boundary-value problem can formally be written in the form ·

u(x) = fo Q~) K(x, ~) d~ (9.5.20)

The function K(x, ~)is called Green's function for the operator. It was characterized in Section 8.4 in a quite different manner. It forms the basis for a more detailed investigation, which reaches beyond the formal structure of the present treatment

9.6 GREEN'S FUNCTION FOR ORDINARY BOUNDARY-VALUE PROBLEMS

Consider the linear ordinary differential operator A of second order,

Au = - ( pu' )' + qu (9.6.1)

where u(x) is defined in the interval aS: x S: b, p(x) is in C1(a, b), q(x) is in c0(a, b), and p(x) > 0. The associated nonhomogeneous differential equation is of the form

Au = h(x) (9.6.2)

where h(x) is in D0(a, b). The goal in solving the boundary-value problem is to find a solution ofEq. 9.6.2 that satisfies homogeneous boundary conditions at a and b; i.e.,

u(a) = u(b) = 0 (9.6.3)


Influence Function

It is natural to start with physical considerations. Consider Eq. 9.6.2 as the condition for static equilibrium of a string, under the influence of a time-independent force that is distributed with density h(x) over the string. A limiting process may be used to transform the continuously distributed force to a point force; i.e., to a force that acts at a single point x =~with a given intensity. Let K(x, ~)denote the deflection of the string at point x, as a result of the action of a point force of unit intensity that is applied at point ~. The effect at x of the distributed force h(x) can then be considered as superposition of the effects of differential forces h(~) d;. On physical grounds, it is expected that the desired solution is of the form

u(x) = Lb K(x, ~) h(~) d~ (9.6.4)

Note that if h(~) is a point load at ;, then h(~) = 3(;) and u(x) = K(x, ~). The function K(x, ;), which satisfies the prescribed boundary conditions at a and b, is called the influence function or Green's function for the differential operator A. It follows that the function u(x) that is represented in Eq. 9.6.4 by an integral in terms of the kernel K(x, ;) and the load density h(x) also satisfies these boundary conditions.

The Green function K(x, ~). for ~ flXed, satisfies the differential equation

AK = 0 (9.6.5)

everywhere except at the point x = ~. since K corresponds to a zero force when x ¥: ;. At the point x = ~. the function K(x, ~) must have a singularity, which can be found in the following way. Consider the point force as the limiting case of a force h£(x) that vanishes in (a, b) if I x- ~I>£, but for which the total intensity is

(9.6.6)

Denote the associated deflection of the string as ~(x, ~). Thus,

(9.6.7)

Integrating this equation between the limits ~ - o and ~ + o, where o ;;::: e may be chosen arbitrarily provided the interval of integration remains in (a, b),

f ~+a [ d ( dK ) J ~-S - (h p dx £ + qK£ dx = 1

Taking the limit as e -+ 0, assuming that K£ converges to a continuous function K(x, ~) that is continuously differentiable except at x = ;,


f ~+3 [ d ( dK ) ] ~-a - dx p dx. - qK dx = 1

Now, as~~ 0, integration of the second term qK becomes zero and

. d K(x, ~) x=~+li 1 lim - --5--+0 ax x~-5 - p(~)

(9.6.8)

which characterizes the singularity of the Green function at x = ~. Properties of Green's Functions

Summarizing the foregoing heuristic discussion, a rigorous mathematical theory can be obtained.

Theorem 9.6.1. Let K(x, ~) be a Green function of the differential operator A and homogeneous boundary conditions, with the following properties:

(1) For fixed~. K(x, ~)is a continuous function of x that satisfies the prescribed boundary conditions.

(2) Except at the point x = ~. the first and second derivatives of K with respect to x are continuous. At the point x = ~. the first derivative has a jump discontinuity that is given by

d K(x, ~) I x=~+O = - ...!._ dX x=~-0 p(~)

(9.6.9)

(3) Considered as a function of x, K(x, ~) satisfies the differential equation AK = 0, except at the point x = ~.

If h(x) is a continu<;>us or piecewise continuous function of x, then the function u(x) in Eq. 9.6.4 is a solution of the differential equation of Eq. 9.6.2 that satisfies the boundary conditions. Conversely, if the function u(x) satisfies Eq. 9.6.2 and the boundary conditions, it can be represented by Eq. 9.6.4. •

To prove the first conclusion in Theorem 9.6.1, Leibniz's rule for differentiation of an integral with respect to a parameter of Eq. 6.1.8 is needed. Differentiating Eq. 9 .6.4,

u'(x) = Ib K'(x, ~) h(~) d~

= Ix K'(x, ~) h(~) d~ + Lb K'(x, ~) h(~) d~ (9.6.10)


Multiplying both sides of Eq. 9.6.10 by p(x) and differentiating again,

( pu' )'(x) = fax ( pK' Y(x, ~) h(x, ~) d~ + Lb ( pK' )'(x, ~) h(x, ~) d~ + p K'( X, X - 0 ) h(x) - p K'( X, X + 0 ) h(x)

= lb ( pK' Y(x, ~) h(~) d~ + p [ K'( x + 0, x) - K'( x- 0, x)] h(x)

= lb ( pK' Xx, ~) h(~) d~ - h(x)

where K'( x, x - 0 ) = K'( x + 0, x ), K'( x, x + 0 ) = K'( x- 0, x ), and Eq. 9.6.9 have been used. Therefore,

Au = - ( pK' Y + qu

= lb (- ( pK' )' + qK ) h(~) d~ + h(x)

= lb (AK) h(~) ~ + h(x)

This proves that u is a solution, since AK = 0, except at ~ = x. To prove the converse, form the integral identity

f ~ ,~ ( vAu - uAv) dx = - p ( u'v - v'u )

xl xl (9.6.11)

With v = K(x, ~) in the intervals a S x S l; and ~ S x S b,

Jab K(x, ~) Au(x) dx = -pel;)[ u'(~)( K( ~- 0, ~) - K( ~ + 0, ~))

- u~)( K'( ~- 0, ~ ) - K'( ~ + 0, ~ ) ) ]

Since K(x, ~)is continuous, the first term on the right vanishes. By Eq. 9,6.9, the second term on the right becomes u(~). By interchanging ~ and x, this is just

u(x) = Ib K(~, x) Au~) d~

Substituting for Au from Eq. 9.6.2, the desired formula of Eq. 9.6.4 is obtained for the ~~~ .


Green's function for a symmetric differential operator is a symmetric function of the arguments ~ and x; i.e.,

K(x, ~) = K(~, x) (9.6.12)

This follows almost immediately from Eq. 9.6.11 by substituting v = K(x, 11) and u = K(x, ~), dividing the domain of integration into the intervals a :s; x :s; ~ and ~ :s; x s; 11, and treating each interval separately. The proof is completed by taking into account both the discontinuity relation of Eq. 9.6.10 at the points x = ~ and x = 11 and the boundary conditions. Symmetry of Green's function expresses a reciprocity that frequently occurs in mechanics; i.e., a unit force that is applied at point ~ produces the displacement K(x, ~) at point x and a unit force that acts at x produces the same result at ~.

To construct Green's function for Au, with boundary conditions u(a) = u(b) = 0, consider any solution Uo(x) of the differential equation Au = 0 that satisfies the given boundary condition at x =a. Then, CoUo(X) is the most general such solution. Similarly, let c1 u1 (x) be the family of solutions of Au = 0 that satisfy the boundary condition at x = b. There are two possible cases. Either the two families of solutions are distinct, which is the general case, or they are identical.

In the first case, the functions Uo(X) and u1 (x) are linearly independent. In this case, the constants c0 and c1 can be chosen such that the point of intersection of the solutions is x = ~ and such that the discontinuity of the derivative at this point has precisely the value -1/p(~). In this way, Green's function K(x, ~) is obtained explicitly as

u 1 (~) u0(x)

K(x, ~) = (9.6.13)

In the second case, Uo(x) and u1(x) differ only by a constant factor. Every solution that belongs to one family also belongs to the other. The function Uo(X) satisfies not only the boundary condition at x = a, but also the boundary condition at x = b. Thus, the equation Au = 0 has a nontrivial solution Uo(x) that satisfies the boundary conditions. This can also be expressed by stating that A. = 0 is an eigenvalue of Au = A.u. Hence, the above construction fails and no Green's function exists.

The existence of Green's function is equivalent to the existence of a unique solution of the homogeneous boundary-value problem for the differential equation Au = h(x). Therefore, the following alternative exists: Under given hoiil.Qgeneous boundary conditions, either the equation Au = h(x) has a unique solution u(x) for any given h(x), or the homogeneous equation J\.u = 0 has a nontrivial solution.

Ordinary differential equations of higher order are not essentially different. The discussion here is restricted to a typical example associated with the differential equation u"" = f, the uniform beam of Example 9.4.1. As before, the influence function, or


1\

Green's function K(x, ;>is introduced as the displacement of the beam under the influence of a unit point force that acts at the point x = ; and satisfies the prescribed homogeneous boundary conditions. In the same manner as above, the following typical conditions for the Green function are obtained:

1\

(1) For every value of the parameters. the function K(x, ;>and its first and second 1'1.

derivatives are continuous and K(x, ;> satisfies the prescribed homogeneous boundary conditions.

(2) At any point x '# ;, the third and fourth derivatives with respect to x are continuous. However, at x = ; the following discontinuity condition holds:

1\ A lim [ K" I ( ; + £, ; ) - K" I ( ; - £, ; )] = 1 £-+0

(9.6.14)

(3) Except at the point x =;,the differential equation

1\

K""(x, ;) = 0 (9.6.15)

is satisfied.

The fundamental property of Green's function for the fourth order operator Au = u"" can be stated as follows: Let u(x) be a continuous function that satisfies the boundary conditions and has three piecewise continuous derivatives. Let h(x) be a piecewise continuous function. If u(x) and h(x) are connected by the relation

1\

Au = u"" = h(x) (9.6.16)

then

Jxl " u(x) = K(x, ;> h(;) d;

xo (9.6.17)

and conversely.

EXERCISES 9.6

1. Carry out the calculations outlined in the text to obtain Eq. 9.6.8.

2. Derive Eq. 9.6.12.

3. Carry out the calculations outlined in the text, using properties (1), (2), and (3) of K(x, ~)in Theorem 9.6.1, to derive Eq. 9.6.13.

4. Show that u(x) ofEq. 9.6.17 is a solution ofEq. 9.6.16.

Sec. 9.7 Completeness of Eigenfunctions 425

9.7 COMPLETENESS OF EIGENFUNCTIONS

With the aid of the Green function, completeness relations for eigenfunctions of positive bounded below ordinary differential operators are proved for the Sturm-Liouville operator. Literature is cited for more general results and an outline of the proof for general ordinary differential operators is given.

Green's Function for the Sturm-Liouville Operator

Consider first the Sturm-Liouville operator

Au = -( p(x) u' )' + q{x)u

u(O) = u(l) = 0 (9.7.1)

where q(x) is in C'(O, 1), p(x) is in C1(0, 1), and h(x) is in Lz(O, 1). Suppose that p(x) is positive and that q(x) is nonnegative, for 0 s; x s;; 1. Since the operator Au =- ( pu' )' + qu is positive bounded below on DA = { u E C2(0, 1): u(O) = u(1) = 0 }, it follows that the boundary-value problem

- ( pu' Y + qu = h(x)

u(O) = u(l) = 0 (9.7.2)

has a unique solution and, as shown in Section 9 .6, the Green function G(x, ;) exists. That is, the solution of this problem is

u(x) = L1 G(x, ~) h(~) d;

Setting h(~) = A.p u(~) in Eq. 9. 7 .2, the relation

u(x) = A. Ll G{x, ~) p(~) u(;) d~ (9.7.3)

is obtained for the eigenfunctions of Au= A.p(x)u, where p(x) > 0 is in C'(O, 1). Conversely, if u(x) satisfies Eq. 9.7.3, it is an eigenfunction and A. is an eigenvalue of Au= A.pu.

Given the eigenfunctions Un(x) of A, the equation

(9. 7.4)

can be thought of as providing Fourier coefficients ofG(x, ;) f'i}'©, in terms of normal-


{()© [ t ]t/2 ized functions p ~:n(~), where Kn = fo p(~) ~(~) d~ . The Bessel inequality

ofEq. 5.4.14, applied to the Fourier coefficients ofEq. 9.7.4, yields

Multiplying both sides by p(x) and integrating from 0 to 1 yields the relation

(9.7.5)

Since G(x, ~) is continuous in x and ~. the right side of Eq. 9.7.5 is finite. Since At, ... , A.m are eigenvalues, it follows that if there are an infinite number of eigenvalues,

00

the series :L, ~ converges. In particular, n=t An

(9.7.6)

It is shown in Ref. 20 that there are always an infinite number of eigenvalues for the Sturm-Liouville operator.

The following special case of a more general minimization principle for eigenvalues of positive bounded below differential operators is required for theoretical arguments. This result is a special case of the theory developed in Chapter 10.

Theorem 9.7.1. The smallest eigenvalue At of the Sturm-Liouville operator A is given by

Jot { pqi2 + qqt } dx

J ol pqt dx

(9.7.7)

where minimization is over functions in D A that are continuous and have piecewise continuous derivatives. The normalized function u(x) for which the minimum is achieved is the corresponding eigenfunction. Successive eigenvalues Ak are given by the relation

Sec. 9.7 Completeness of Eigenfunctions

Jot { pq#2 + qqt} dx

Jot pf dx

The principle theorem of this section is now stated and proved.

Completeness of Sturm-Liouville Equations

427

(9.7.8)

•

Theorem 9.7.2. The eigenfunctions Un(X) of Au= A.pu for the Sturm-Liouville operator A are complete in the space of functions ~(0, 1); i.e., functions that satisfy

• To prove Theorem 9.7.2, let f be any continuous, piecewise continuously differen

tiable function with f(O) = f(l) = 0. The Fourier series for f(x) in terms of the Un(X) is

00

f(x) = l: Cn Un(x) n=l

where the coefficients are

By definition of en,

1 ( k-1 ) J p f - l: CnUn ui dx = 0, i = 1, ... , k- 1 0 n=1

( k-1 )

Putting cp = f - l: CnUn in Eq. 9.7.8, n=1


Integration by parts yields

fol ( pf'un' + qfUn) dx = Jol f { - (PUn' )' + qlln } dx

= An fol pfUn dx

= AnCn fol P~ dx

Also,

= { An fol p~ dx,

0,

m=n

m:;t:n

Thus,

s: f.- { J.' ( pf'2 + qf2 ) dx }


Since A.k ~ oo, it follows that any function f(x) that is continuously differentiable and satisfies f(O) = f(l) = 0 can be approximated in the mean by a finite linear combination of the

ui(x). Since any function for which J 01 pf2 dx is finite can be approximated in the mean

by a continuously differentiable function [14], it follows that the eigenfunctions { ui(x)}

are complete in 11. This completes the proof of the theorem. •

Completeness of Eigenfunctions of General Operators

Consider now the general linear boundary-value problem with the nth order ordinary differential equation

Au = p0(x) u<n) + p1(x) u<n-l) + ... + Pn(x) u = h(x) (9.7.9)

in a~ X~ b, where Pi(x) is in cn-i(a, b) and Po(x) ~ 0 on a~ X~ b, with the boundary conditions

n

Uju == L, ( Mjk u(k-l}(a) + Njk u(k-l)(b)) = 0, j = 1, ... , n (9.7.10) k=l

where Mjk and Njk are constants. Three theorems that are proved in Ref. 20 (see Section 7.4) are now stated. The first

theorem provides existence of an infinite set of distinct eigenvalues. The second theorem is an extension of Theorem 9.7.2, which is proved in the same way. The third theorem provides results on pointwise convergence.

Theorem 9.7.3. If the operator A of the boundary-value problem of Eq. 9.7.9, with the boundary conditions of Eq. 9.7.10, is symmetric, then it has an infinite set of distinct eigenvalues with no finite cluster point. •

Theorem 9.7.4. Let the boundary-value problem be as in Theorem 9.7.3. Then, its normalized eigenfunctions { Un(X) } are orthogonal and complete in ~(a, b). •

Theorem 9.7.5. Let the boundary-value problem be as in Theorem 9.7.3. If f(x) is in cn(a, b) and satisfies the boundary conditions ofEq. 9.7.10, then

co

f(x) = L, ( f, uk ) uk(x), a ~ x ~ b (9.7.11) k=O

and convergence is uniform in a ~ x ~ b. • The foregoing results are next generalized to higher dimensions. Let A be a positive

bounded below differential operator of order 2n, defined on a dense subspace D A of


L2(0). If there exists a Green function G(x, ~)for Au= h(x); i.e., a function G(x, ~)in

J.1(0) such that u(x) = J0 G(x, ~) h(~) d~, and if there are an infinite number of distinct

eigenvalues "-n of A, then An-+ co as n-+ co,

To show that this is true, let Un(x) be eigenfunctions of Au = A.pu, with the corresponding eigenvalues An· Then,

J J 0 G(x, ~) p(~) un(~) d~

[ JJ./ ~~) d~ r (9.7.12) ---------------- =

This equation may be thought of as providing the Fourier coefficients of G(x, ~)...J p(~), in terms of functions ...J p(~) Un(~) that are L2-orthogonal. Bessel's inequality yields

m 2( ) I, Jf un x , Jf oltx, ~l p~l ~ n=1 An p(~) ~(~) d; 0

0

Multiplying both sides of this inequality by p(x) and integrating,

m

I.~ s; J J J J G2(x, ~) p(x) p(~) d~ dx n=1 ~ 0 0

(9.7.13)

from which it follows that An -+ co as n -+ oo,

The foregoing and results proved in Chapter 10 yield the result that follows.

Theorem 9.7.6. Let A be an operator with the above properties. Then the smallest eigenvalue of A is given by

where U c1> I~ = J J 0 p f(~) d~. The successive eigenvalues are given by

A,= +••~""' { (cjl, ui)p=O

i= 1, ... 'k-1

(9.7.14)

(9.7.15)

•


Theorem 9.7.7. Let the operator A be as in Theorem 9.7.6. Then the eigenfunc-tions of A are complete in J.1(0). •

To prove Theorem 9.7.7, let f(x) be in DA, which is a dense subspace of ~(Q). 00

Expanding f(x) in a Fourier Series f(x) - L cn un(x), where the Un(X) are normalized by

IIUn lip = 1, the coefficients are

Cn = ( f, lln )p

n=1

where ( u, v )p = J J 0 puv dQ. By definition of cn,

([ f - ~ c,u,]. ut )P = 0, i = I, ... , k- I

By Eq. 9.7.15,

{ k-1

= i; ( Af, f) - ~ A..cn ( u,, flp

( k-1 ) k-1 }

- Af, ~ Cnlln + ~ "-nc~

{ ~1 }

= r ( Af, f) - L "-nc~ k n=1

(9.7.16)

Further, Af is in ~. so its Fourier coefficients are

b· = ( U· Af) = A·C· 1 1' p 1 1

Thus, both sequences { ci} and { A.ici} are in the space .t 2 defined in Section 5.3, so 00 00

L AnC~ < oo. Therefore, the series L A.nc~ converges and the right side of Eq. 9.7.16 n=1 n=l


can be made arbitrarily small fork sufficiently large, which proves that any f(x) in D A can

be approximated by a finite linear combination of the ui(x). Since D A is dense in ~(!l), it

follows that the sequence { ~(x)} is complete in ~(!l). •

The proof in Section 9.2 that the Laplace operator with reasonable boundary condi· tions is positive bounded below, the construction of Green's function for this operator for simple sets !lin Section 8.4, and Theorem 9.7.7 shows that its eigenfunctions are com· plete. While the hypotheses of Theorem 9.7.7 have not been verified for all conceivable linear operators, it has been shown to apply to a broad variety of problems.

It is worthy of note that physical intuition and reasoning indicates that Theorem 9.7.7 is generally applicable to linear conservative mechanics problems. Such problems are generally governed by variational principles, so the resulting operators should be symmetric [18] and those associated with elastic deformation (strain energy) are generally positive definite. The positive bounded below property was seen in Section 9.2 to be as· sociated with stable equilibrium, hence it is also expected to be valid.

Finally, while Green functions have been constructed only for a few problems, the Green function has been shown to be the response at point ~ due to a unit input at point x, which certainly exists for wide classes of mechanics problems. Indeed, there is an extensive literature that presents Green's functions for a variety of problet¥• The physical mo. tivation for validity of the hypotheses of Theorem 9. 7. 7 and the exttM\lsive literature that rigorously verifies these hypotheses in specific cases provide confidence that the theorem is widely applicable. This theoretical support for the formal eigenfunction method pre· sented in Section 9.5 provides one of the most useful mathematical tools for the study of dynamics of linear continua.

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