Nair S. Advanced Topics in Applied Mathematics: For Engineering and the Physical Sciences

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Green’s Functions 9

(b)

δ(e

αx

−β) =

αβ



x −

lnβ



. (1.31)

We may solve these examples using the basic properties of the delta

function as follows:

(a)



∞

−∞

φ(x)

∂

∂α

δ(αx)dx =

∂

∂α



∞

−∞

φ(x)δ(αx) dx

∂

∂α



∞

−∞

φ(x/α)δ(x) dx/α

∂

∂α

φ(0)

=−

φ(0). (1.32)

Comparing

∂

∂α

δ(αx) =−

δ(x). (1.33)

(b)



∞

−∞

φ(x)δ(e

αx

−β)dx =



∞

−∞



ln y



δ(y −β)

αy



∞

−∞

αy



ln y



δ(y −β)dy

αβ



ln β



. (1.34)

Thus,

δ(e

αx

−β) =

αβ



x −

ln β



. (1.35)

Other well-known examples of delta sequences are

, limit a →0, (1.36)

sinλx

, limit λ →∞. (1.37)

10 Advanced Topics in Applied Mathematics

1.3 LINEAR DIFFERENTIAL OPERATORS

Consider the differential equation

Lu(x) =f (x); a < x < b. (1.38)

Here, u(x) is the unknown, f (x) is a given forcing function, and L is a

differential operator. For a differential equation of order n, we need n

boundary conditions. For the time being, let us assume all the needed

boundary conditions are homogeneous. The differential operator L

has the form

L =a

(x)

n−1

(x)

n−1

+···+a

(x). (1.39)

A linear operator satisﬁes the properties

L(u

) =Lu

+Lu

, (1.40)

L(cu) =cLu, (1.41)

where u

,andu are functions in C

,andc is a constant. Recall C

indicates the set of differentiable functions with all derivatives up to

and including the nth continuous.

1.3.1 Example: Boundary Conditions

For the system

+u =sinx; u(1) =1, u(2) = 3, (1.42)

with nonhomogeneous boundary conditions, we introduce a new

dependent variable v,as

u =v +Ax +B. (1.43)

The boundary conditions become

u(1) =1 =v(1) +A +B, (1.44)

u(2) =3 =v (2) +2A +B. (1.45)

Green’s Functions 11

We get homogeneous boundary conditions for v, if we choose

A +B = 1, (1.46)

2A +B =3, (1.47)

with the solutions, A =2andB =−1. The original differential equation

becomes

+v = sinx −2x +1; v(1) =0, v(2) =0. (1.48)

1.4 INNER PRODUCT AND NORM

Given two real functions u(x) and v(x) on x ∈ (a, b), we deﬁne their

inner product as

u,v =



u(x)v(x) dx. (1.49)

The two functions u and v are orthogonal if

u,v =0. (1.50)

The Euclidian norm of a function u associated with the above inner

product is deﬁned as

u=



u,u=







1/2

. (1.51)

If the norm of a function is unity, we call it a normalized function.

For any function, by dividing it by its norm, we obtain a new nor-

malized version of the function. Generalized inner product and norms

are deﬁned by inserting a weight function w, which is positive, in the

deﬁnitions, as

u,v =



uvw dx; w(x)>0. (1.52)

A sequence of functions, u

,i = 1,2,...,n, is called an ortho-normal

sequence if

u

=δ

, (1.53)

12 Advanced Topics in Applied Mathematics

where δ

is the Kronecker delta, deﬁned by



1, i =j,

0, i =j.

(1.54)

1.5 GREEN’S OPERATOR AND GREEN’S FUNCTION

For linear systems of equations in the matrix form,

Ax = y, (1.55)

if A is non-singular, we can write the solution x as

x = By; B = A

−1

. (1.56)

Here B is the inverse of the operator A. For the differential equation

Lu(x) =f (x), (1.57)

formally, we can write the solution u as

u =L

−1

f . (1.58)

Since L is a differential operator, we expect its inverse to be an integral

operator. Thus,

u =Gf , G =L

−1

, (1.59)

where G is called the Green’s operator. Explicitly, we have

u(x) =



g(x,ξ)f (ξ )dξ . (1.60)

The kernel inside the integral, g(x,ξ), is called the Green’s function

for the operator L.

The Green’s function depends on the differential operator and the

boundary conditions. Once g(x,ξ)is obtained, solutions can be gener-

ated by entering the function f (x) inside the integral. Thus, the task

Green’s Functions 13

of obtaining complementary and particular solutions of differential

equations for speciﬁc forcing functions becomes much simpler. Even

discontinuous forcing functions can be accommodated inside the

integral.

1.5.1 Examples: Direct Integrations

For the ﬁrst-order differential equation

=f (x); u(a) =0, (1.61)

by integrating we ﬁnd

u(x) =



f (ξ )dξ. (1.62)

For this case,

g(x,ξ)= h(x −ξ)=



1, ξ<x,

0, ξ>x.

(1.63)

The second-order differential equation describing the deﬂection of a

taut string,

=f (x); u(a) =0; u(b) =0, (1.64)

after one integration gives

=C +



f (ξ )dξ, (1.65)

where C has to be found. Integrating again,

u(x) =Cx +D +







ξ=x



ξ=a

f (ξ )dξdx



. (1.66)

Using u(a) = 0 we get D =−Ca. As shown in Fig. 1.6, the double

integration has to be performed on the upper triangular area. By

interchanging x



-andξ-integrations, we have

u(x) =C(x −a) +



ξ=x

ξ=a





=ξ

f (ξ )dx



dξ. (1.67)

14 Advanced Topics in Applied Mathematics

x⬘= ξ

x⬘

Figure 1.6. Area of integration in the x



,ξ–plane.

This interchange of integrations with appropriate changes in the limits

is called Fubini’s theorem. After completing the x



-integration, we get

u(x) =C(x −a) +



(x −ξ)f (ξ )dξ . (1.68)

Using u(b) =0, we solve for C,

C =−

b −a



(b −ξ)f (ξ )dξ. (1.69)

Now u has the form

u =

b −a





(b −a)(x −ξ)f (ξ ) dξ −



(x −a)(b −ξ)f (ξ ) dξ



In the second integral the range of integration, a to b, can be split into

a to x and x to b.

u =−

b −a





(b −x)(ξ −a)f (ξ) dξ +



(x −a)(b −ξ)f (ξ ) dξ



We can extract the Green’s function in the form

g(x,ξ)=

b −a



(x −b)(ξ −a), ξ<x,

(ξ −b)(x −a), ξ>x.

(1.70)

Green’s Functions 15

0.2 0.4 0.6 0.8

–0.2

–0.1

0.1

0.2

Figure 1.7. Green’s function for the point-loaded string when ξ = 0.25.

Observe the symmetry, g(x,ξ) =g(ξ , x), for this case. The differential

equation represents the deﬂection of a string under tension subjected

to a distributed vertical load, f (x). With a = 0andb = 1, we have

plotted the Green’s function in Fig. 1.7 for ξ =0.25.

Figure1.7showsthedeﬂectionofthestringundera concentra ted

unit load at x =ξ . The Green’s function g(x,ξ) satisﬁes

g(x,ξ)

=δ(x −ξ), g(0,ξ)= g(1,ξ)= 0. (1.71)

With ξ as a parameter, and the delta-function zero for x <ξ and for

x>ξ,wecansolveEq.(1.71)intwoparts.Ifu

and u

represent

the left and right solutions which satisfy homogeneous equations, we

can satisfy the left boundary condition by choosing the constants of

integration to have u

(0) = 0. Similarly, we can choose u

to have

(1) =0.

g(x,ξ)= Au

(x) =Ax, x <ξ, (1.72)

g(x,ξ)= Bu

(x) =B(1 −x), x >ξ. (1.73)

Integrating Eq. (1.71) from ξ − to ξ +,weﬁnd

lim

→0



(ξ +,ξ)−

(ξ −,ξ)



=1. (1.74)

16 Advanced Topics in Applied Mathematics

This can be written as





x=ξ

=1. (1.75)

The slope of g is discontinuous at x =ξ (the quantity inside the double

bracket denotes a jump), but g itself is continuous. We can enforce the

continuity by choosing the constants A and B as

A =Cu

(ξ), B =Cu

(ξ). (1.76)

So far we have

g(x, ξ)=C



(x)u

(ξ), x <ξ,

(ξ)u

(x), x >ξ.

(1.77)

Using the jump condition (1.75), we ﬁnd

C{u

(ξ)u



(ξ) −u

(ξ)u



(ξ)}=1, C{ξ(−1) −(1−ξ)(1)}=1, C =−1.

(1.78)

Finally,

g(x, ξ)=



x(ξ −1), x <ξ,

ξ(x −1), x >ξ,

(1.79)

which is the same as the Green’s function we found by direct

integration.

1.6 ADJOINT OPERATORS

Before we extend the idea of constructing Green’s functions for arbi-

trary linear operators using the δ function as a forcing function, it is

useful to extend the notion of symmetry in matrices to differential

operators. With n-vectors x and y and an n ×n matrix A, we ﬁnd, if

Ax = x

By, (1.80)

then, as these are scalars, taking the transpose of one of them,

B = A

, (1.81)

Green’s Functions 17

and, if A is symmetric, B = A. Using the inner product notation, we

have

y,Ax=x,By⇒ B = A

. (1.82)

The term “transpose" of a matrix operator translates into the “adjoint"

of a differential operator. If we follow the terminology of matrix alge-

bra, it is more reasonable to speak of the transpose of an operator.

However, it is the convention to use the word “adjoint" for differen-

tial operators. Of course, the adjoint of a matrix is a totally different

quantity. With L and L

∗

denoting two nth order linear differential

operators and u and v functions in C

,wecallL

∗

the adjoint of L,if

v,Lu=u,L

∗

v, (1.83)

with u and v satisfying appropriate homogeneous boundary conditions.

1.6.1 Example: Adjoint Operator

Find the adjoint operator and adjoint boundary conditions of the

system,

Lu =x





+2u, u(1) =0, u



(2) +u(2) =0. (1.84)

For this system, we use integration-by-parts to apply the differential

operator on v,

v,Lu=



v[x





+2u]dx





−(x



u +vu







u[(x



−v



+2v ]dx.

From the quantity inside the integral, we see

∗

v = (x



−v



+2v = x



+(4x −1)v



+4v,

∗

+(4x −1)

+4. (1.85)

18 Advanced Topics in Applied Mathematics

The quantity that has to be evaluated at the boundaries is called the bi-

linear concomitant, P(x), as it is linear in u and in v. The homogeneous

boundary conditions have to be such that P should vanish at each

boundary.

P(2) =4v(2)u



(2) −4v(2)u(2) −4v



(2)u(2) +v(2)u(2)

=4v(2)u



(2) −[3v (2) +4v



(2)]u(2).

From the given condition, u



(2) =−u(2),wehave

P(2) =−[7v(2) +4v



(2)]u(2).

Since u(2) is arbitrary,



(2) +7v(2) =0. (1.86)

At the other boundary, we have

P(1) =v(1)u



(1) −2v(1) +v



(1)u(1) +v(1)u(1)

=v(1)u



(1) −[v(1) −v



(1)]u(1).

Using, u(1) =0, we get

v(1) =0. (1.87)

In general, L

∗

and the boundary conditions associated with it are dif-

ferent from L and its boundary conditions. When L

∗

is identical to L,

we call L a self-adjoint operator. This case is analogous to a symmetric

matrix operator.

1.7 GREEN’S FUNCTION AND ADJOINT

GREEN’S FUNCTION

Let L and L

∗

be a linear operator and its adjoint with independent

variable x. We assume there are associated homogeneous bound-

ary conditions that render the bi-linear concomitant P = 0 at the

boundaries. Consider

Lg(x, x

) =δ(x −x

); L

∗

(x,x

) =δ(x −x

), (1.88)