A series somewhat similar to that representingδ(x−t )results when we expand the Green’s function in the eigenfunctions of the corresponding homogeneous equation. In the inhomo- geneous Helmholtz equation we have
∇2ψ (r)+k2ψ (r)= −ρ(r). (10.82) The homogeneous Helmholtz equation is satisfied by its orthonormal eigenfunctionsϕn,
∇2ϕn(r)+kn2ϕn(r)=0. (10.83) As outlined in Section 9.7, the Green’s functionG(r1,r2)satisfies the point source equa- tion
∇2G(r1,r2)+k2G(r1,r2)= −δ(r1−r2) (10.84) and the boundary conditions imposed on the solutions of the homogeneous equation. Be- causeGis real, we expand the Green’s function in a series of real eigenfunctions of the homogeneous equation (10.83); that is,
G(r1,r2)= ∞
n=0
an(r2)ϕn(r1), (10.85) and by substituting into Eq. (10.84) we obtain
− ∞
n=0
an(r2)k2nϕn(r1)+k2 ∞
n=0
an(r2)ϕn(r1)= − ∞
n=0
ϕn(r1)ϕn(r2). (10.86) Hereδ(r1−r2)has been replaced by its eigenfunction expansion, Eq. (1.190). When we employ the orthogonality ofϕn(r1)to isolatean,this yields
∞
m=0
am(r2)
k2−km2 ϕn(r1)ϕm(r1) d3r1= − ∞
m=0
ϕm(r2)
ϕn(r1)ϕm(r1) d3r1, or
an(r2)
k2−k2n
= −ϕn(r2).
Then substituting this into Eq. (10.85), the Green’s function becomes G(r1,r2)=
∞
n=0
ϕn(r1)ϕn(r2)
k2n−k2 , (10.87)
a bilinear expansion, symmetric with respect to r1and r2, as expected. Finally,ψ (r1), the desired solution of the inhomogeneous equation, is given by
ψ (r1)=
G(r1,r2)ρ(r2) dτ2. (10.88)
If we generalize our inhomogeneous differential equation to
Lψ+λψ= −ρ, (10.89)
whereLis a Hermitian operator, we find that G(r1,r2)=
∞
n=0
ϕn(r1)ϕn(r2)
λn−λ , (10.90)
whereλnis thenth eigenvalue andϕn is the corresponding orthonormal eigenfunction of the homogeneous differential equation
Lψ+λψ=0. (10.91)
The eigenfunction expansion of the Green’s function in Eq. (10.90) makes the symmetry propertyG(r1,r2)=G(r2,r1)explicit and is often useful when comparing with solutions obtained by other means.
Green’s Functions — One-Dimensional
The development of the Green’s function for two- and three-dimensional systems was the topic discussed in the preceding material and in Section 9.7. Here, for simplicity, we restrict ourselves to one-dimensional cases and follow a somewhat different approach.
Defining Properties
In our one-dimensional analysis we consider first the inhomogeneous equation
Ly(x)+f (x)=0, (10.92) in whichLis the self-adjoint differential operator
L= d dx
p(x) d
dx
+q(x). (10.93)
As in Section 10.1,y(x)is required to satisfy certain boundary conditions at the endpoints aandbof our interval[a, b].
We now proceed to define a rather strange and arbitrary functionG over the interval [a, b]. At this stage the most that can be said in defense ofGis that the defining prop- erties are legitimate, or mathematically acceptable. Later,Gwill appear as a reasonable tool for obtaining solutions of the inhomogeneous ODE, Eq. (10.92); this role dictates its properties.
1. The intervala≤x≤bis divided by a parametert. We labelG(x)=G1(x)fora≤ x < t andG(x)=G2(x)fort < x≤b.
2. The functionsG1(x)andG2(x)each satisfy the homogeneous15equation; that is, LG1(x)=0, a≤x < t,
LG2(x)=0, t < x≤b. (10.94)
15Homogeneous with respect to the unknown function. The functionf (x)in Eq. (10.92) is set equal to zero.
3. Atx=a, G1(x)satisfies the boundary conditions we impose ony(x),a solution of the inhomogeneous ODE, Eq. (10.92). Atx=b, G2(x)satisfies the boundary condi- tions imposed ony(x)at this endpoint of the interval. For convenience, the boundary conditions are taken to be homogeneous; that is, atx=a,
y(a)=0, or y′(a)=0, or αy(a)+βy′(a)=0 and similarly atx=b.
4. We demand thatG(x)be continuous,16
x→tlim−G1(x)= lim
x→t+G2(x). (10.95)
5. We require thatG′(x)be discontinuous, specifically that15 d
dxG2(x)
t− d dxG1(x)
t= − 1
p(t ), (10.96)
wherep(t )comes from the self-adjoint operator, Eq. (10.93). Note that with the first derivative discontinuous, the second derivative does not exist.
These requirements, in effect, make G a function of two variables,G(x, t ). Also, we note thatG(x, t )depends on both the form of the differential operatorLand the boundary conditions thaty(x)must satisfy. Note that we have described the properties of Green’s functions for second-order differential equations. Note that for Green’s functions for first- order differential equations, the discontinuities arise inGitself.
Now, assuming that we can find a functionG(x, t )that has these properties, we label it a Green’s function and proceed to show that a solution of Eq. (10.92) is
y(x)= b
a
G(x, t )f (t ) dt. (10.97)
To do this we first construct the Green’s functionG(x, t ). Letu(x)be a solution of the homogeneous equation that satisfies the boundary conditions atx=a, and letv(x) be a solution that satisfies the boundary conditions atx=b. Then we may take17
G(x, t )=
+c1u(x), a≤x < t,
c2v(x), t < x≤b. (10.98) Continuity atx=t(Eq. (10.95)) requires
c2v(t )−c1u(t )=0. (10.99) Finally, the discontinuity in the first derivative (Eq. (10.96)) becomes
c2v′(t )−c1u′(t )= − 1
p(t ). (10.100)
16Strictly speaking, this is the limit asx→t.
17The “constants”c1andc2are independent ofx, but they may (and do) depend on the other variable,t.
There will be a unique solution for our unknown coefficientsc1andc2if the Wronskian determinant
u(t ) v(t ) u′(t ) v′(t )
=u(t )v′(t )−v(t )u′(t )
does not vanish. We have seen in Section 9.6 that the nonvanishing of this determinant is a necessary condition for linear independence. Let us assumeu(x)andv(x)to be inde- pendent. (Ifu(x)andv(x)are linearly dependent, the situation becomes more complicated and is not considered here. See Courant and Hilbert in Additional Readings of Chapter 9.) For independentu(x)andv(x)we have the Wronskian (again from Section 9.6 or Exer- cise 10.1.4)
u(t )v′(t )−v(t )u′(t )= A
p(t ), (10.101)
in whichAis a constant. Equation (10.101) is sometimes called Abel’s formula. Numerous examples have appeared in connection with Bessel and Legendre functions. Now, from Eq. (10.100), we identify
c1= −v(t )
A , c2= −u(t )
A . (10.102)
Equation (10.99) is clearly satisfied. Substitution into Eq. (10.98) yields our Green’s func- tion
G(x, t )=
−1
Au(x)v(t ), a≤x < t,
−1
Au(t )v(x), t < x≤b.
(10.103)
Note thatG(x, t )=G(t, x). This is the symmetry property that was proved earlier in Sec- tion 9.7. Its physical interpretation is given by the reciprocity principle (via our propagator function) — a cause att yields the same effect atxas a cause atxproduces att. In terms of our electrostatic analogy this is obvious, the propagator function depending only on the magnitude of the distance between the two points:
|r1−r2| = |r2−r1|.
Green’s Function Integral — Differential Equation
We have constructedG(x, t ), but there still remains the task of showing that the integral (Eq. (10.97)) with our new Green’s function is indeed a solution of the original differential equation (10.92). This we do by direct substitution. WithG(x, t )given by Eq. (10.103),18 Eq. (10.97) becomes
y(x)= −1 A
x a
v(x)u(t )f (t ) dt− 1 A
b x
u(x)v(t )f (t ) dt. (10.104)
18In the first integral,a≤t≤x. HenceG(x, t )=G2(x, t )= −(1/A)u(t )v(x). Similarly, the second integral requiresG=G1.
Differentiating, we obtain y′(x)= −1
A x
a
v′(x)u(t )f (t ) dt− 1 A
b x
u′(x)v(t )f (t ) dt, (10.105) the derivatives of the limits canceling. A second differentiation yields
y′′(x)= −1 A
x a
v′′(x)u(t )f (t ) dt− 1 A
b x
u′′(x)v(t )f (t ) dt
− 1 A
u(x)v′(x)−v(x)u′(x) f (x). (10.106) By Eqs. (10.100) and (10.102) this may be rewritten as
y′′(x)= −v′′(x) A
x a
u(t )f (t ) dtưu′′(x) A
b x
v(t )f (t ) dt−f (x)
p(x). (10.107) Now, by substituting into Eq. (10.93), we have
Ly(x)= −Lv(x) A
x a
u(t )f (t ) dt−Lu(x) A
b x
v(t )f (t ) dt−f (x). (10.108) Sinceu(x)andv(x)were chosen to satisfy the homogeneous equation, theL-factors are zero and the integral terms vanish, and we see that Eq. (10.92) is satisfied.
We must also check thaty(x)satisfies the required boundary conditions. At pointx=a, y(a)= −u(a)
A b
a
v(t )f (t ) dt=cu(a), (10.109) y′(a)= −u′(a)
A b
a
v(t )f (t ) dt=cu′(a), (10.110) since the definite integral is a constant. We choseu(x)to satisfy
αu(a)+βu′(a)=0. (10.111)
Multiplying by the constantc, we verify thaty(x)also satisfies Eq. (10.111). This illus- trates the utility of the homogeneous boundary conditions: The normalization does not matter. In quantum mechanical problems the boundary condition on the wave function is often expressed in terms of the ratio
ψ′(x) ψ (x) = d
dxlnψ (x), compared to d
dxlnu(x)
x=a= −α β, Eq. (10.111). The advantage is that the wave function need not be normalized yet.
Summarizing, we have Eq. (10.97), y(x)=
b a
G(x, t )f (t ) dt, which satisfies the differential equation (Eq. (10.92)),
Ly(x)+f (x)=0,
and the boundary conditions, these boundary conditions having been built into the Green’s function,G(x, t ).
Basically, what we have done is to use the solutions of the homogeneous equation Eq. (10.94) to construct a solution of the inhomogeneous equation. Again, Poisson’s equa- tion is an illustration. The solution (Eq. (9.148)) represents a weighted[ρ(r2)]combination of solutions of the corresponding homogeneous Laplace’s equation. (We followed these same steps early in this section.)
It should be noted that oury(x), Eq. (10.97), is actually the particular solution of the differential equation, Eq. (10.92). Our boundary conditions exclude the addition of so- lutions of the homogeneous equation. In an actual physical problem we may well have both types of solutions. In electrostatics, for instance (compare Section 9.7), the Green’s function solution of Poisson’s equation gives the potential created by the given charge dis- tribution. In addition, there may be external fields superimposed. These would be described by solutions of the homogeneous equation, Laplace’s equation.
Eigenfunction, Eigenvalue Equation
The preceding analysis placed no special restrictions on ourf (x). Let us now assume that f (x)=λρ(x)y(x).19Then we have
y(x)=λ b
a
G(x, t )ρ(t )y(t ) dt (10.112) as a solution of
Ly(x)+λρ(x)y(x)=0 (10.113) and its boundary conditions. Equation (10.112) is a homogeneous Fredholm integral equa- tion of the second kind, and Eq. (10.113) is the homogeneous eigenvalue equation (with the weighting functionw(x)replaced byρ(x)).
There is a change in the interpretation of our Green’s function. It started as a propagator function, a weighting function giving the importance of the chargeρ(r2) in producing the potential ϕ(r1). The charge ρ was the inhomogeneous term in the inhomogeneous differential equation (10.92). Now the differential equation and the integral equation are both homogeneous.G(x, t )has become a link relating the two equations, differential and integral.
To complete the discussion of this differential equation–integral equation equivalence, let us now show that Eq. (10.113) implies Eq. (10.112), that is, that a solution of our differential equation (10.113) with its boundary conditions satisfies the integral equa- tion (10.112). We multiply Eq. (10.113) byG(x, t ), the appropriate Green’s function, and integrate fromx=atox=bto obtain
b a
G(x, t )Ly(x) dx+λ b
a
G(x, t )ρ(x)y(x) dx=0. (10.114)
19The functionρ(x)is some weighting function, not a charge density.
The first integral is split in two(x < t, x > t ), according to the construction of our Green’s function, giving
− t
a
G1(x, t )Ly(x) dx− b
t
G2(x, t )Ly(x) dx=λ b
a
G(x, t )ρ(x)y(x) dx. (10.115) Note thattis the upper limit for theG1integrals and the lower limit for theG2integrals.
We are going to reduce the left-hand side of Eq. (10.115) toy(t ). Then, withG(x, t )= G(t, x), we have Eq. (10.112) (withx andt interchanged).
Applying Green’s theorem to the left-hand side or, equivalently, integrating by parts, we obtain
− t
a
G1(t, x) d
dx
p(x) d dxy(x)
+q(x)y(x)
dx
= −
G1(x, t )p(x)y′(x) x=tx=a+ t
a
∂
∂xG1(x, t )
p(x)y′(x) dx
− t
a
G1(x, t )q(x)y(x) dx, (10.116)
with an equivalent expression for the second integral. A second integration by parts yields
− t
a
G1(x, t )Ly(x) dx= − t
a
y(x)LG1(x, t ) dx
−
G1(x, t )p(x)y′(x) xx==ta +
G′1(x, t )p(x)y(x) xx==ta. (10.117) The integral on the right vanishes becauseLG1=0. By combining the integrated terms with those from integratingG2, we have
−p(t )
G1(t, t )y′(t )−y(t ) ∂
∂xG1(x, t )
x=t−G2(t, t )y′(t )+y(t ) ∂
∂xG2(x, t )
x=t
+p(a)
y′(a)G1(a, t )−y(a) ∂
∂xG1(x, t )
x=a
−p(b)
G2(b, t )y′(b)−y(b) ∂
∂xG2(x, t )x=b
. (10.118)
Each of the last two expressions vanishes, forG(x, t )andy(x)satisfy the same boundary conditions. The first expression, with the help of Eqs. (10.95) and (10.96), reduces toy(t ).
Substituting into Eq. (10.115), we have Eq. (10.112), thus completing the demonstration of the equivalence of the integral equation and the differential equation plus boundary conditions.
Example 10.5.1 LINEAROSCILLATOR
As a simple example, consider the linear oscillator equation (for a vibrating string):
y′′(x)+λy(x)=0. (10.119)
We impose the conditionsy(0)=y(1)=0, which correspond to a string clamped at both ends. Now, to construct our Green’s function, we need solutions of the homogeneous equa- tionLy(x)=0, which isy′′(x)=0. To satisfy the boundary conditions, we must have one solution vanish atx=0, the other atx=1. Such solutions (unnormalized) are
u(x)=x, v(x)=1−x. (10.120)
We find that
uv′−vu′= −1 (10.121)
or, by Eq. (10.101) withp(x)=1, A= −1. Our Green’s function becomes G(x, t )=
+x(1−t ), 0≤x < t, t (1−x), t < x≤1.
(10.122) Hence by Eq. (10.112) our clamped vibrating string satisfies
y(x)=λ 1
0
G(x, t )y(t ) dt. (10.123)
You may show that the known solutions of Eq. (10.119), y=sinnπ x, λ=n2π2,
do indeed satisfy Eq. (10.123). Note that our eigenvalueλis not the wavelength.
Green’s Function and the Dirac Delta Function
One more approach to the Green’s function may shed additional light on our formulation and particularly on its relation to physical problems. Let us refer once more to Poisson’s equation, this time for a point charge:
∇2ϕ(r)= −ρpoint
ε0 . (10.124)
The Green’s function solution of this equation was developed in Section 9.7. This time let us take a one-dimensional analog
Ly(x)+f (x)point=0. (10.125) Heref (x)pointrefers to a unit point “charge,” or a point force. We may represent it by a number of forms, but perhaps the most convenient is
f (x)point=
1
2ε, t−ε < x < t+ε, 0, elsewhere,
(10.126) which is essentially the same as Eq. (1.172). Then, integrating Eq. (10.125), we have
t+ε t−ε
Ly(x) dx= − t+ε
t−ε
f (x)pointdx= −1 (10.127)
from the definition off (x). Let us examineLy(x)more closely. We have t+ε
t−ε
d dx
p(x)y′(x) dx+ t+ε
t−ε
q(x)y(x) dx
=p(x)y′(x)t+ε
t−ε+ t+ε
t−ε
q(x)y(x) dx= −1. (10.128) In the limitε→0 we may satisfy this relation by permittingy′(x)to have a discontinu- ity of−1/p(x)atx=t, y(x)itself remaining continuous.20These, however, are just the properties used to define our Green’s function,G(x, t ). In addition, we note that in the limitε→0,
f (x)point=δ(x−t ), (10.129) in whichδ(x−t )is our Dirac delta function, defined in this manner in Section 1.15. Hence Eq. (10.125) has become
LG(x, t )= −δ(x−t ). (10.130) This is a one-dimensional version of Eq. (9.159), which we exploit for the development of Green’s functions in two and three dimensions — Section 9.7. It will be recalled that we used this relation in Section 9.7 to determine our Green’s functions.
Equation (10.130) could have been expected since it is actually a consequence of our differential equation, Eq. (10.92), and Green’s function integral solution, Eq. (10.97). If we letLx (subscript to emphasize that it operates on thex-dependence) operate on both sides of Eq. (10.97), then
Lxy(x)=Lx b
a
G(x, t )f (t ) dt.
By Eq. (10.92) the left-hand side is just−f (x). On the rightLx, is independent of the variable of integrationt, so we may write
−f (x)= b
a
'LxG(x, t )( f (t ) dt.
By definition of Dirac’s delta function, Eqs. (1.171b) and (1.183), we have Eq. (10.130).
Exercises
10.5.1 Show that
G(x, t )=
+x, 0≤x < t, t, t < x≤1,
is the Green’s function for the operatorL=d2/dx2and the boundary conditions y(0)=0, y′(1)=0.
20The functions p(x) and q(x) appearing in the operator L are continuous functions. With y(x) remaining continuous, q(x)y(x) dxis certainly continuous. Hence this integral over an interval 2ε(Eq. (10.128)) vanishes asεvanishes.
10.5.2 Find the Green’s function for (a) Ly(x)=d2y(x)
dx2 +y(x),
+y(0)=0, y′(1)=0.
(b) Ly(x)=d2y(x)
dx2 −y(x), y(x)finite for−∞< x <∞. 10.5.3 Find the Green’s function for the operators
(a) Ly(x)= d dx
xdy(x)
dx
.
ANS.G(x, t )=
+−lnt, 0≤x < t,
−lnx, t < x≤1.
(b) Ly(x)= d dx
xdy(x)
dx
−n2
x y(x), withy(0)finite andy(1)=0.
ANS.G(x, t )=
1 2n
x t
n
−(xt )n
, 0≤x < t, 1
2n t
x n
−(xt )n
, t < x≤1.
The combination of operator and interval specified in Exercise 10.5.3(a) is pathological, in that one of the endpoints of the interval (zero) is a singular point of the operator. As a consequence, the integrated part (the surface integral of Green’s theorem) does not vanish. The next four exercises explore this situation.
10.5.4 (a) Show that the particular solution of d dx
x d
dxy(x)
= −1 isyP(x)= −x.
(b) Show that
yP(x)= −x= 1
0
G(x, t )(−1) dt, whereG(x, t )is the Green’s function of Exercise 10.5.3(a).
10.5.5 Show that Green’s theorem, Eq. (1.104) in one dimension with a Sturm–Liouville-type operator(d/dt )p(t )(d/dt )replacing∇ã∇, may be rewritten as
b a
u(t )d
dt
p(t )dv(t ) dt
−v(t )d dt
p(t )du(t ) dt
dt
=
u(t )p(t )dv(t )
dt −v(t )p(t )du(t ) dt
b a
.
10.5.6 Using the one-dimensional form of Green’s theorem of Exercise 10.5.5, let v(t )=y(t ) and d
dt
p(t )dy(t ) dt
= −f (t ), u(t )=G(x, t ) and d
dt
p(t )∂G(x, t )
∂t
= −δ(x−t ).
Show that Green’s theorem yields y(x)=
b a
G(x, t )f (t ) dt+
G(x, t )p(t )dy(t )
dt −y(t )p(t )∂
∂tG(x, t )
t=b t=a
. 10.5.7 Forp(t )=t, y(t )= −t,
G(x, t )=
+−lnt, 0≤x < t
−lnx, t < x≤1, verify that the integrated part does not vanish.
10.5.8 Construct the Green’s function for x2d2y
dx2+xdy dx +
k2x2−1 y=0, subject to the boundary conditions
y(0)=0, y(1)=0.
10.5.9 Given that
L=
1−x2 d2
dx2−2x d dx and
G(±1, t )remains finite,
show that no Green’s function can be constructed by the techniques of this section. (u(x) andv(x)are linearly dependent.)
10.5.10 Construct the one-dimensional Green’s function for the Helmholtz equation d2
dx2+k2
ψ (x)=g(x).
The boundary conditions are those for a wave advancing in the positivex-direction — assuming a time dependencee−iwt.
ANS.G(x1, x2)= i 2kexp
ik|x1−x2| . 10.5.11 Construct the one-dimensional Green’s function for the modified Helmholtz equation
d2 dx2−k2
ψ (x)=f (x).
The boundary conditions are that the Green’s function must vanish forx → ∞and x→ −∞.
ANS.G(x1, x2)= 1 2kexp
−k|x1−x2| . 10.5.12 From the eigenfunction expansion of the Green’s function show that
(a) 2 π2
∞
n=1
sinnπ x sinnπ t
n2 =
+x(1−t ), 0≤x < t, t (1−x), t < x≤1.
(b) 2 π2
∞
n=0
sin(n+12)π xsin(n+12)π t (n+12)2 =
+x, 0≤x < t, t, t < x≤1.
Note. In Section 10.4 the Green’s function ofL+λis expanded in eigenfunctions. The λthere is an adjustable parameter, not an eigenvalue.
10.5.13 In the Fredholm equation,
f (x)=λ2 b
a
G(x, t )ϕ(t ) dt, G(x, t )is a Green’s function given by
G(x, t )= ∞
n=1
ϕn(x)ϕn(t ) λ2n−λ2 . Show that the solution is
ϕ(x)= ∞
n=1
λ2n−λ2 λ2 ϕn(x)
b a
f (t )ϕn(t ) dt.
10.5.14 Show that the Green’s function integral transform operator b
a
G(x, t )[ ]dt is equal to−L−1, in the sense that
(a) Lx b
a
G(x, t )y(t ) dt= −y(x), (b)
b a
G(x, t )Lty(t ) dt= −y(x).
Note. TakeLy(x)+f (x)=0, Eq. (10.92).
10.5.15 Substitute Eq. (10.87), the eigenfunction expansion of Green’s function, into Eq. (10.88) and then show that Eq. (10.88) is indeed a solution of the inhomogeneous Helmholtz equation (10.82).
10.5.16 (a) Starting with a one-dimensional inhomogeneous differential equation (Eq. (10.89)), assume that ψ (x) and ρ(x) may be represented by eigenfunction expansions.
Without any use of the Dirac delta function or its representations, show that ψ (x)=
∞
n=0
b
a ρ(t )ϕn(t ) dt λn−λ ϕn(x).
Note that (1) if ρ=0, no solution exists unless λ=λn and (2) if λ=λn, no solution exists unlessρis orthogonal toϕn. This same behavior will reappear with integral equations in Section 16.4.
(b) Interchanging summation and integration, show that you have constructed the Green’s function corresponding to Eq. (10.90).
10.5.17 The eigenfunctions of the Schrửdinger equation are often complex. In this case the orthogonality integral, Eq. (10.40), is replaced by
b a
ϕi∗(x)ϕj(x)w(x) dx=δij. Instead of Eq. (1.189), we have
δ(r1−r2)= ∞
n=0
ϕn(r1)ϕ∗n(r2).
Show that the Green’s function, Eq. (10.87), becomes G(r1,r2)=
∞
n=0
ϕn(r1)ϕn∗(r2)
k2n−k2 =G∗(r2,r1).
Additional Readings
Byron, F. W., Jr., and R. W. Fuller, Mathematics of Classical and Quantum Physics. Reading, MA: Addison- Wesley (1969).
Dennery, P., and A. Krzywicki, Mathematics for Physicists. Reprinted. New York: Dover (1996).
Hirsch, M., Differential Equations, Dynamical Systems, and Linear Algebra. San Diego: Academic Press (1974).
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