Myint Tyn U., Debnath L. Linear Partial Differential Equations for Scientists and Engineers

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8.11 Green’s Functions for Ordinary Diﬀerential Equations 311

where A and B are arbitrary constants, y

= y

(x)andy

= y

(x)aretwo

linearly independent solutions of the corresponding homogeneous equation

L [y]=0,andy

(x) is a particular integral of (8.11.1).

Using the method of variation of parameters, the particular integral will

be sought by replacing the constants A and B by arbitrary fu nctions of x,

(x)andu

(x), to get

(x)=u

(x) y

(x)+u

(x) y

(x) . (8.11.5)

These arbitrary functions are to be determined so that (8.11.5) satisﬁes

equation (8.11.1). The substitution of the above trial form for y

(x)into

(8.11.1) imposes one condition that (8.11.5) must satisfy. However, there are

two arbitrary functions, and hence, two conditions are needed to determine

them. If follows that another condition is available in solving the problem.

The ﬁrst task is to substitute the trial solution into equation (8.11.1).

Diﬀerentiating y

(x), we obtain

′

= u

′

+ u

′

+ u

′

+ u

′

It is convenient to set the second condition noted above to require that

′

+ u

′

=0, (8.11.6)

leaving

′

= u

′

+ u

′

. (8.11.7)

A second diﬀerentiation of y

gives

′′

= u

′′

+ u

′′

+ u

′

+ u

′

Putting these results into equation (8.11.1) and grouping yields

[p (x) y

′′

+ p

′

+ qy

]+u

[p (x) y

′′

+ p

′

+ qy

]

+p (x) {u

′

+ u

′

} = −f (x) .

Since y

(x)andy

(x) are solutions of the homogeneous equations, both

the square brackets in the above expression vanish. The result is t hat

′

+ u

′

= −

f (x)

p (x)

. (8.11.8)

Thus, two equations ((8.11.6) and (8.11.8)) determine u

(x)andu

(x).

Solving these equations algebraically for u

′

and u

′

produces

′

(x)=

f (x) y

(x)

p (x) W (x)

′

(x)=−

f (x) y

(x)

p (x) W (x)

312 8 Eigenvalue Problems and Special Functions

where W (x)=(y

′

− y

′

) is the non-zero Wronskian of the solutions y

and y

. Integration of these results yields

(x)=+



f (x) y

(x) dx

p (x) W (x)

(x)=−



f (x) y

(x) dx

p (x) W (x)

The substitution of these results into (8.11.5) gives the solution in (a, b)

(x)=y

(x)



f (ξ) y

(ξ)

p (ξ) W (ξ)

dξ − y

(x)



f (ξ) y

(ξ)

p (ξ) W (ξ)

dξ

= −



(x) y

(ξ)

p (ξ) W (ξ)

f (ξ) dξ −



(x) y

(ξ)

p (ξ) W (ξ)

f (ξ) dξ. (8.11.9)

This form suggests the deﬁnition

G (x, ξ)=

⎧

⎪

⎨

⎪

⎩

−

(x)y

(ξ)

p(ξ)W (ξ)

,a≤ ξ<x,

−

(x)y

(ξ)

p(ξ)W (ξ)

,x<ξ≤ b.

(8.11.10)

This is called Green’s function. Thus, the solution becomes

(x)=



G (x, ξ) f (ξ) dξ (8.11.11)

provided G (x, ξ) is continuous and f (ξ) is at least piecewise continuous in

(a, b). The existence of Green’s function is evident from equations (8.11.10)

provided W (ξ) =0.

In order to obtain a deeper insight into the role of Green’s function,

certain properties are important to note. These are

(i) G (x, ξ) is continuous at x = ξ, and consequently, throughout the inter-

val (a, b). This follows from the fact that G (x, ξ) is constructed from

the solutions of the homogeneous equation which are continuous in the

intervals a ≤ ξ<xand x<ξ≤ b, W (ξ) =0.

(ii) Its ﬁrst and second derivatives are continuous for all x = ξ in a ≤ x,

ξ ≤ b.

(iii) G (x, ξ) is symmetric in x and ξ,thatis,ifx and ξ are interchanged

in (8.11.10), the deﬁnition is not changed.

(iv) The ﬁrst derivative of G (x, ξ) has a ﬁnite discontinuity at x = ξ

lim

x→ξ+



∂G

∂x



− lim

x→ξ−



∂G

∂x



= −

p (ξ)

. (8.11.12)

(v) G (x, ξ) is a solution of the homogeneous equation throughout the inter-

val except at x = ξ. This point can be pressed further by substitution

of (8.11.11) and (8.11.1), note that y

(x) is a solution of (8.11.1). This

leads to

8.11 Green’s Functions for Ordinary Diﬀerential Equations 313







p (x)

dG (x, ξ)



+ q (x) G (x, ξ)



f (ξ) dξ = −f (x) ,

where the quantity in the square bracket is zero except at x = ξ.If

follows that, in order for this result to hol d,



p (x)

dG (x, ξ)



+ q (x) G (x, ξ)=−δ (ξ − x) , (8.11.13)

where δ (ξ − x) is the Dirac function which has the following properties:

δ (ξ − x)=0, except at x = ξ,



δ (ξ − x) dξ =1, a<x<b.

For any continuous function f (ξ)in(a, b) and for a ≤ x ≤ b,



f (ξ) δ (ξ − x) dξ = f (x) .

(vi) For ﬁxed ξ, G (x, ξ) satisﬁes the given boundary conditions (8.11.2)–

(8.11.3).

Diﬀerentiating (8.11.11) with respect to x by Leibniz’s rule, we obtain

′

(x)=



′

(x, ξ) f (ξ) dξ + G (x, x−) f (x)



′

(x, ξ) f (ξ) dξ − G (x, x+) f (x)



′

(x, ξ) f (ξ) dξ,

since G (x, ξ) is continuous in ξ,thatis,G (x, x−)=G (x, x+).

Because G (x, ξ) satisﬁes the boundary conditions, it follows that

(a)+a

′

(a)=



G (a, ξ)+a

′

(a, ξ)] f (ξ) dξ =0.

Similarly,

(b)+b

′

(b)=0.

The result embodied in equation (8.11.13) permits a meaningful physi-

cal interpretation of the r ole of Green’s function. If we assume that the

diﬀerential equation (8.11.1) represents a vibrating mechanical system

driven by a uniformly distributed force −f (x) in the interval (a, b),

where the system is deﬁned, then G (x, ξ) governed by (8.11.13) repre-

sents the displacement of the system at a point x resulting from a unit

314 8 Eigenvalue Problems and Special Functions

impulse of force at x = ξ. The displacement at the point x due to uni-

formly distributed force f (ξ) per unit length over an elementary interval

(ξ,ξ + dξ)isgivenbyf (ξ) G (x, ξ) dξ. Finally, the total displacement

of the system at the point x results from superposition (addition) of

these contributions so that the total solution over the entire interval

(a, b) is given by (8.11.1), that is,

(x)=



G (x, ξ) f (ξ) dξ.

Combining all above results, we state the fundamental theorem for

Green’s function:

Theorem 8.11.1. If f (x) is continuous on [a, b], then the function

y (x)=



G (x, ξ) f (ξ) dξ

is a solution of the boundary-value problem

L [y]=−f (x) ,

y (a)+a

′

(a)=0,b

(b)+b

′

(b)=0.

Example 8.11.1. Consider the problem

′′

= −x, y (0) = 0,y(1) = 0. (8.11.14)

For a ﬁxed value of ξ, Green’s function G (x, ξ) satisﬁes the associated

homogeneous equation

′′

in 0 <x<ξ, ξ<x<1, and the boundary conditions

G (0,ξ)=0,G(1,ξ)=0.

In addition, it satisﬁes

(x, ξ)



x=ξ+

x=ξ−

= −

p (ξ)

Now choose G (x, ξ) such that

G (x, ξ)=

⎧

⎨

⎩

(x, ξ)=(1− ξ) x, for 0 ≤ x ≤ ξ

(x, ξ)=(1− x) ξ, for ξ ≤ x ≤ 1.

It can be seen that G

′′

= 0 over the intervals 0 <x<ξ, ξ<x<1. Also

8.12 Construction of Green’s Functions 315

(0,ξ)=0,G

(1,ξ)=0.

Moreover,

′

(x, ξ) − G

′

(x, ξ)=−ξ − (1 − ξ)=−1

which is the value of the jump −1/p (ξ), because in this case p = 1. Hence,

by Theorem 8.11.1, keeping in mind that ξ is the variable in G (x, ξ), the

solution of (8.11.14) is

y (x)=



G (x, ξ) f (ξ) dξ +



G (x, ξ) f (ξ) dξ



(1 − x) ξ

dξ +



x (1 − ξ) ξdξ =



1 − x



8.12 Construction of Green’s Functions

In the above example, we see that the solution was obtained immediately as

soon as Green’s function was obtained properly. Thus, the real problem is

not that of ﬁnding the solution but that of determining Green’s function for

the problem. We will now show by construction that there exists a Green’s

function for L [y] satisfying the prescribed boundary conditions.

We ﬁrst assume that the associated homogeneous equation satisfying the

conditions (8.11.2) and (8.11.3) has the trivial solution only, as in Example

8.11.1. We construct the solution y

(x) of the equation

L [y]=0

satisfying a

y (a)+a

′

(a) = 0. We see that c

(x) is the most general

such solution, where c

is an arbitrary constant.

In a similar manner, we let c

(x), with c

is an arbit rary constant,

be th e most general solution of

L [y]=0

satisfying b

y (b)+b

′

(b) = 0. Thus, y

and y

exist in the interval (a, b)

and are linearly independent. For, if they were linearly dependent, then

= cy

, which shows that y

would satisfy both the boundary conditions

at x = a and x = b. This contradicts our assumption about the trivial

solution. Consequently, Green’s function can take the form

G (x, ξ)=

⎧

⎨

⎩

(ξ) y

(x) , for x<ξ

(ξ) y

(x) , for x>ξ

. (8.12.1)

Since G (x, ξ) is continuous at x = ξ,wehave

316 8 Eigenvalue Problems and Special Functions

(ξ) y

(ξ) − c

(ξ) y

(ξ)=0. (8.12.2)

The discontinuity in the derivative of G at the point requires that

(x, ξ)



x=ξ+

x=ξ−

= c

(ξ) y

′

(ξ) − c

(ξ) y

′

(ξ)=−

p (ξ)

. (8.12.3)

Solving equations (8.12.2) and (8.12.3) for c

and c

, we ﬁnd

(ξ)=

−y

(ξ)

p (ξ) W (y

; ξ)

(ξ)=

−y

(ξ)

p (ξ) W (y

; ξ)

, (8.12.4)

where W (y

; ξ) is the Wronskian given by W (y

; ξ)=y

(ξ) y

′

(ξ) −

(ξ) y

′

(ξ). Since the two solutions are linearly ind ependent, the Wronskian

diﬀers from zero.

From Theorem 8.2.6, with λ =0,wehave

pW = constant = C. (8.12.5)

Hence, Green’s function is given by

G (x, ξ)=

⎧

⎨

⎩

−y

(x) y

(ξ) /C, for x ≤ ξ

−y

(x) y

(ξ) /C, for x ≥ ξ.

(8.12.6)

Thus, we state the following theorem:

Theorem 8.12.1. If the associated homogeneous boundary-value problem

of (8.11.1)–(8.11.3) has the trivial solution only, then Green’s function ex-

ists and is unique.

Proof. The proof for uniqueness of Green’s function is left as an exercise

for the reader.

Example 8.12.1. Consider the problem

′′

+ y = −1,y(0) = 0,y





=0. (8.12.7)

The solution of L [y]=(dy

′

/dx)+y = 0 satisfying y (0) = 0 is

(x)=sinx, 0 ≤ x<ξ

and the solution of L [y] = 0 satisfying y (π/2) = 0 is

(x)=cosx, ξ < x ≤

The Wronskian of y

and y

is then given by

W (ξ)=y

(ξ) y

′

(ξ) − y

(ξ) y

′

(ξ)=−1.

8.13 The Schr¨odinger Equation and Linear Harmonic Oscillator 317

Since in this case p = 1, (8.12.6) becomes

G (x, ξ)=

⎧

⎨

⎩

sin x cos ξ, for x ≤ ξ

cos x sin ξ, for x ≥ ξ.

Therefore, the solution of (8.12.7) is

y (x)=



G (x, ξ) f (ξ) dξ +



π/2

G (x, ξ) f (ξ) dξ



cos x sin ξdξ+



π/2

sin x cos ξdξ

= −1+sinx +cosx.

It can be seen in the formula (8.12.6) that Green’s function is symmetric

in x and ξ.

Example 8.12.2. Construct the Green’s function for the two-point boundary-

value problem

′′

(x)+ω

y = f (x) ,y(a)=y (b)=0.

This describes the forced oscillation of an elastic string with ﬁxed ends at

x = a and x = b.

It is easy to check that sin ωx and cos ωx are two functions which satisfy

the homogeneous equation y

′′

+ ω

y = 0. These are used to construct two

functions y

(x)andy

(x) which satisfy the boundary conditions y

(a)=

(b) = 0. Accordingly, y

(x)=A sin ωx +B cos ωx and y

(x)=C sin ωx+

D cos ωx, and the resulting functions are

(x)=sinω (x − a) ,y

(x)=sinω (x − b) .

The corresponding Wronskian is W = −ω sin ω (a − b). Substituting these

results into (8.11.10) yields

G (x, ξ)=

⎧

⎪

⎨

⎪

⎩

sin ω(ξ−a)sinω(x−b)

−ω sin ω(a−b)

,a≤ ξ<x

sin ω(x−a)sinω(ξ−b)

−ω sin ω(a−b)

,x≤ ξ ≤ b

provided sin ω (a − b) =0.

8.13 The Schr¨oding er Equation and Linear Harmonic

Oscillator

The quantum mechanical motion of the harmonic oscillator is described by

the one-dimensional Schr¨odinger equation

318 8 Eigenvalue Problems and Special Functions

Hψ(x)=Eψ (x) , (8.13.1)

where the Hamiltonian H is given by

H = −







+ V (x) ,V(x)=

Mω

, (8.13.2)

and E is the energy, V (x) is the potential, h =2π is t he Planck con-

stant, M is the mass of the particle, and ω is the classical frequency of the

oscillator.

We solve equation (8.13.1) subject to the requirement that the solution

be bounded as |x|→∞. The solution of (8.13.1) is facilitated by the ﬁrst

solving the equation for large x. In terms of the constants

β =

2ME



,α=

Mω



> 0,

the equation (8.13.1) takes the form



β − α



ψ =0. (8.13.3)

For small β and large x, β − α

∼−α

so that the equation becomes

− α

ψ =0.

As |x|→∞, ψ (x)=x

exp



αx



satisﬁes (8.13.3) for a ﬁnite n so far

as leadi ng terms



∼−α



are concerned. The positive exponential fac-

tor is unacceptable because of the boundary conditions, so the asymptotic

solution ψ (x)=x

exp



−

αx



suggests the possibility of th e exact solu-

tion in the form ψ (x)=v (x)exp



−

αx



where v (x)istobedetermined.

Substituting this result into (8.13.3), we obtain

− 2αx

+(β − α) v =0. (8.13.4)

In terms of a new independent variable ζ = x

√

α, this equation reduces

to the form

dζ

− 2ζ

dζ



− 1



v =0. (8.13.5)

We seek a power series solution

v (ζ)=

∞



n=0

. (8.13.6)

8.13 The Schr¨odinger Equation and Linear Harmonic Oscillator 319

Substituting this series into equation (8.13.5) and equating the coeﬃcients

of ζ

to zero, we obtain the recurrence relation

n+2

(2n +1− β/α)

(n +1)(n +2)

(8.13.7)

which gives

n+2

∼

as n →∞. (8.13.8)

This ratio is the same as that of the series for ζ

exp







∼ x

αx



with

ﬁnite n. This leads to the fact that ψ (x)=v (x) e

−αx

∼ x

αx

which

does not satisfy the basic requirement for |x|→∞. This unacceptable

result can only be avoided if n is an integer and the series terminates so

that it becomes a polynomial of degree n. This means that a

n+2

= 0 but

= 0 so that

2n +1−

=0, (8.13.9)

=(2n +1).

Substituting the values for α and β, it turns out that

E ≡ E



n +



ω,n=0, 1, 2,.... (8.13.10)

This represents a discrete set of energies. Thus, in quantum mechanics,

a stationary state of the harmonic oscillator can assume only one of the

values from the set E

. The energy is thus quantized, and forms a discrete

spectrum. According to the classical theory, the energy forms a continuous

spectrum, that is, all non-negative numb ers are allowed for the energy of a

harmonic oscillator. This shows a remarkable contrast between the results

of the classical and quantum theory.

The number n which characterizes the energy eigenvalues and eigen-

functions is called the quantum number.Thevalueofn = 0 corresponds to

the minimum value of the quantum numb er with the energy

ω. (8.13.11)

This is called the lowest (or ground) state energy which never vanishes as

the lowest possible classical energy would. E

is proportional to ,repre-

senting a quantum phenomenon. The discrete energy spectrum is in perfect

agreement with the quantization rules of the quantum theory.

320 8 Eigenvalue Problems and Special Functions

To determine the eigenfunctions for the harmonic oscillator associated

with the eigenvalues E

, we obtain the solution of equation (8.13.5) which

has the form

dζ

− 2ζ

dζ

+2nv =0. (8.13.12)

This is a well-known diﬀerential equation for the Hermite polynomials

(ζ)ofdegreen. Thus, the complete eigenfunctions can be expressed

in terms of H

(ζ)as

(x)=A



√



exp



−

αx



, (8.13.13)

where A

are arbitrary constants.

The Hermite polynomials H

(x) are usually deﬁned by

(x)=(−1)



−x



,D≡

. (8.13.14)

They form an orthogonal system in (−∞, ∞) with the weight function

exp



−x



The orthogonal relation for these polynomials is



∞

−∞

−x

(x) H

(x) dx =

⎧

⎨

⎩

0,n= m

√

π, n = m.

The Hermite polynomials H

(x)forn =0, 1, 2, 3, 4are

(x)=1

(x)=2x

(x)=−2+4x

(x)=−12x +8x

(x)=12− 48x

+16x

Finally, the eigenf unction s ψ

of the linear harmoni c oscillator for the quan-

tum number n =0, 1, 2, 3 are given in Figure 8.13.1.