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Chapter 16
Partial Differential
Equations
16.1 Examples of Partial Differential Equations and Boundary Conditions
Let us continue our discussion of partial differential equations (PDEs) by men- tioning examples of PDEs in physics. Among the most frequently encountered PDEs are the following:
- Laplace’s equation, ∇^2 ψ( r ) = 0. This very common and very important equation occurs in studies of a. electromagnetic phenomena, including electrostatics in regions con- taining no electric charges, dielectrics, steady currents, and magneto- statics; b. hydrodynamics (irrotational flow of perfect fluids and surface waves); c. heat flow and diffusion; and d. gravitation in regions containing no masses.
- Poisson’s equation, ∇^2 ψ( r ) = −ρ/ε 0. In contrast to the homogeneous Laplace equation, Poisson’s equation is nonhomogeneous, with a source term −ρ( r )/ε 0 in electrostatics and (with 0 → 1) in Newtonian gravity.
- The wave (Helmholtz) and time-independent diffusion equations, ∇^2 ψ( r ) ± k^2 ψ = 0. These equations also appear in such diverse phenomena as a. elastic waves in solids, including vibrating strings, bars, and mem- branes; b. sound or acoustics; c. electromagnetic waves; and d. nuclear reactors.
756
16.1 Examples of Partial Differential Equations and Boundary Conditions 757
- The diffusion or heat flow equation
∇^2 ψ( t , r ) =
a^2
∂ψ ∂ t
- The corresponding four-dimensional form in the time-dependent wave equation of electrodynamics involving the d’Alembertian, a four-dimen- sional analog of the Laplacian in Minkowski space (Chapters 2 and 4),
∂ μ^ ∂ (^) μ ϕ( x ) = ∂^2 ϕ =
[
c^2
∂^2
∂ t^2
− ∇^2
]
ϕ = 0.
- The time-dependent scalar potential equation, ∂^2 ψ( x ) = ρ/ε 0. Like Poisson’s equation, this equation occurs in electrodynamics and is nonhomogeneous with a source term ρ/ε 0.
- The Klein–Gordon equation, ∂^2 ψ( x ) = −μ^2 ψ( x ) for mass μ, and the cor- responding vector equations in which the scalar function ψ is replaced by a vector field. Other more complicated forms are common.
- The Schr ¨odinger wave equation,
h ¯^2 2 m
∇^2 ψ + V ψ = ih ¯
∂ψ ∂ t and, for the time-independent case,
h ¯^2 2 m
∇^2 ψ + V ψ = E ψ.
- The equations for elastic waves and viscous fluids and the telegraphy equa- tion containing the operators ∇^2 , ∂^2 /∂ t^2 , and ∂/∂ t.
- Maxwell’s coupled partial differential equations for electric and magnetic fields and Dirac’s equation for relativistic electron wave functions. For Maxwell’s equations, see Section 1.9.
Some general techniques for solving second-order PDEs were discussed in Chapter 8 and are further discussed in this chapter:
- Separation of variables (Section 8.8), where the PDE is split into ordinary differential equations (ODEs) that are related by common constants that appear as eigenvalues of linear operators, Lψ = l ψ, usually in one vari- able. This method is closely related to symmetries of the PDE and a group of transformations (see Section 4.2). The Helmholtz equation, given as example 3 above, has this form, where the constant k^2 may arise by separation of the time t from the spatial variables. Likewise, in exam- ple 8, the energy E is an eigenvalue that arises in the separation of t from r in the Schr ¨odinger equation. The resulting separated ODEs are discussed in Chapter 9 in greater detail.
- Conversion of a PDE into an integral equation using Green’s functions ap- plies to inhomogeneous PDEs, such as examples 2 and 7 given previously. An introduction to the Green’s function technique is given in Section 16.3.
16.1 Examples of Partial Differential Equations and Boundary Conditions 759
Table 16.
TYPE OF PDE Boundary condition Elliptic Hyperbolic Parabolic
Laplace, Poisson in ( x , y ) Wave equation in ( x , t ) Diffusion equation in ( x , t ) Cauchy Open surface Unphysical results (instability) Unique, stable solution Too restrictive Closed surface Too restrictive Too restrictive Too restrictive Dirichlet Open surface Insufficient Insufficient Unique, stable solution Closed surface Unique, stable solution Solution not unique Too restrictive Neumann Open surface Insufficient Insufficient Unique, stable solution Closed surface Unique, stable solution Solution not unique Too restrictive
Table 16.
Solutions in Spherical Polar Coordinates a
ψ =
∑ l , m
alm ψ lm
∇^2 ψ = 0 ψ lm =
{ r l r − l −^1
} { Pl m (cos θ) Q ml (cos θ)
} { cos m ϕ sin m ϕ
}b
∇^2 ψ + k^2 ψ = 0 ψ lm =
{ j (^) l ( kr ) n (^) l ( kr )
} { Pl m (cos θ) Q ml (cos θ)
} { cos m ϕ sin m ϕ
}b
∇^2 ψ − k^2 ψ = 0 ψ lm =
{ i (^) l ( kr ) k (^) l ( kr )
} { Pl m (cos θ) Q ml (cos θ)
} { cos m ϕ sin m ϕ
}b
a References for some of the functions are Pl m (cos θ), m = 0, Sec- tion 11.1; m = 0, Section 11.5; Q ml (cos θ), irregular Legendre solution; j (^) l ( kr ), n (^) l ( kr ), i (^) l ( kr ) ≡ i − l^ j (^) l ( ix ), and k (^) l ( kr ) ≡ − il^ h (1) l ( ix ). b (^) cos m ϕ and sin m ϕ may be replaced by e ± i m ϕ (^).
Dirichlet conditions lead to a unique stable solution. Neumann conditions likewise lead to a unique, stable solution independent of the Dirichlet solution. Therefore, Cauchy boundary conditions (meaning Dirichlet plus Neumann) could lead to an inconsistency. The term boundary conditions includes the concept of initial conditions. For instance, specifying the initial position x 0 and the initial velocity v 0 in some dynamical problem would correspond to the Cauchy boundary conditions. The only difference in the present usage of boundary conditions in these one- dimensional problems is that we are going to apply the conditions on both ends of the allowed range of the variable. For convenient reference, the forms of the solutions of Laplace’s equation, Helmholtz’s equation, and the diffusion equation for spherical polar coordi- nates are shown in Table 16.2. The solutions of Laplace’s equation in circular cylindrical coordinates are presented in Table 16.3.
760 Chapter 16 Partial Differential Equations
Table 16.
Solutions in Circular Cylindrical Coordinates a
ψ =
∑ m , α
am α ψ m α , ∇^2 ψ = 0
ψ m α =
{ Jm (αρ) Ym (αρ)
} { cos m ϕ sin m ϕ
} { e −α z e α z
}
ψ m α =
{ Im (αρ) K (^) m (αρ)
} { cos m ϕ sin m ϕ
} { cos α z sin α z
}
If α = 0 (no z -dependence) ψ m =
{ ρ m ρ− m
} { cos m ϕ sin m ϕ
}
a References for the radial functions are Jm (αρ), Section 12.1; Ym (αρ), Section 12.3; Im (αρ) ≡ i − m^ Jm ( ix ) and K (^) m (αρ) ≡ π/ 2 im +^1 H m (1) ( ix ).
For the Helmholtz and the diffusion equation (see Section 16.2) the constant ± k^2 is added to the separation constant ±α^2 to define a new parameter γ 2 or −γ 2. For the choice +γ 2 (with γ 2 > 0), we get Bessel functions Jm (γρ) and Ym (γρ). For the choice −γ 2 (with γ 2 > 0), we get modified Bessel functions Im (γρ) and K (^) m (γρ). These ODEs and two generalizations of them will be examined and system- atized in the following sections. General properties following from the form of the differential equations are discussed in Chapter 9. The individual solutions are developed and applied in Chapters 11–13. The practicing physicist will probably meet other second-order ODEs, some of which may possibly be transformed into the examples studied here. Some of these ODEs may be solved by the techniques of Sections 8.5 and 8.6. Others may require a computer for a numerical solution.
16.2 Heat Flow or Diffusion PDE
Here, we address the full time-dependent diffusion PDE for an isotropic medium. Assuming isotropy is actually not much of a restriction because, in case we have different (constant) rates of diffusion in different directions (e.g., in wood), our heat flow PDE takes the form ∂ψ ∂ t
= a^2
∂^2 ψ ∂ x^2
∂^2 ψ ∂ y^2
∂^2 ψ ∂ z^2
if we put the coordinate axes along the principal directions of anisotropy. Now we simply rescale the coordinates using the substitutions x = a ξ, y = b η, z = c ζ to get back the original isotropic form of Eq. (16.1),
∂ ∂ t
∂^2
∂ξ 2
∂^2
∂η^2
∂^2
∂ζ 2
for the temperature distribution function (ξ, η, ζ, t ) = ψ( x , y , z , t ).
762 Chapter 16 Partial Differential Equations
with coefficients (see Section 14.1)
a (^) l =
− 1
1 · cos
π lx 2
l π
sin
π lx 2
1
x =− 1
=
π l
sin
l π 2
4(−1) m (2 m + 1)π
, l = 2 m + 1;
a (^) l = 0, l = 2 m.
Including its time dependence, the full solution is given by the series
ψ( x , t ) =
π
∑^ ∞
m = 0
(−1) m 2 m + 1
cos
[
(2 m + 1)
π x 2
]
e − t ((2 m +1)π a /2)
2 , (16.5)
which converges absolutely for t > 0, but only conditionally at t = 0, as a result of the discontinuity at x = ± 1.
Without the restriction to zero temperature at the end points of the previous finite interval, the Fourier series is replaced by a Fourier integral. The general solution is then given by Eq. (16.4). At t = 0, the given temperature distribution ψ 0 = 1 gives the coefficients as (see Section 15.2)
A (ω) =
π
− 1
cos ω x dx =
π
sin ω x ω
1
x =− 1
2 sin ω πω
, B (ω) = 0.
Therefore,
ψ( x , t ) =
π
0
sin ω ω
cos(ω x ) e − a
(^2) ω (^2) t d ω. (16.6)
■
In three dimensions the corresponding exponential ansatz ψ = e i k · r / a +β t leads to a solution with the relation β = − k^2 = − k^2 for its parameter, and the three-dimensional form of Eq. (16.3) becomes
∂^2 ψ ∂ x^2
∂^2 ψ ∂ y^2
∂^2 ψ ∂ z^2
which is called the Helmholtz equation and may be solved by the separa- tion method just like the previously discussed Laplace equation in Cartesian, cylindrical, or spherical coordinates. In Cartesian coordinates, with the product ansatz of Eq. (8.114) the sepa- rated x and y ODEs from Eq. (16.3) are the same as Eqs. (8.117) and (8.120), whereas the z ODE [Eq. (8.121)] generalizes to
1 Z
d^2 Z dz^2
= − k^2 + l^2 + m^2 = n^2 > 0, (16.8)
where we introduce another separation constant n^2 , constrained by
k^2 = l^2 + m^2 − n^2 (16.9)
16.2 Heat Flow or Diffusion PDE 763
to produce a symmetric set of equations. Now our solution of Helmholtz’s Eq. (16.7) is labeled according to the choice of all three separation constants l , m , n subject to the constraint Eq. (16.9). As before, the z ODE [Eq. (16.8)] yields exponentially decaying solutions ∼ e − nz. The boundary condition at z = 0 fixes the expansion coefficients alm by the same Eq. (8.123). In cylindrical coordinates, we now use the separation constant l^2 for the z ODE with an exponentially decaying solution in mind, d^2 Z dz^2
= l^2 Z > 0, (16.10)
so that Z ∼ e − lz^ because the temperature goes to zero at large z. Setting k^2 + l^2 = n^2 , Eqs. (8.132)–(8.133) stay the same so that we end up with the same Fourier–Bessel expansion [Eq.(8.137)], as before. In spherical coordinates, the separation method leads to the same angular ODEs in Eqs. (8.143) and (8.146), whereas the radial ODE now becomes 1 r^2
d dr
r^2
dR dr
Q R
r^2
= 0, Q = l ( l + 1) (16.11)
instead of Eq. (8.147), whose solutions are the spherical Bessel functions of Section 12.7. They are listed in Table 16.2. The restriction that k^2 be a constant is unnecessarily severe. The separation process will still work with Helmholtz’s PDE for k^2 as general as
k^2 = f ( r ) +
r^2
g (θ) +
r^2 sin 2 θ
h (ϕ) + k ′^2. (16.12)
In the hydrogen atom we have k^2 = f ( r ) in the Schr ¨odinger wave equation, and this leads to a closed-form solution involving Laguerre polynomials.
Biographical Data Helmholtz Hermann Ludwig Ferdinand von. Helmholtz, a German physiologist and physicist, was born 1821 in Potsdam near Berlin and died in 1894 in Charlottenburg, now Berlin. He studied medicine in Berlin, gradu- ating in 1842. In 1849, with Humboldt’s support, he was appointed professor of physiology at the University of K ¨onigsberg in Prussia, now Russia. He studied the function of the human eye and ear, defining the quality of a tone in terms of overtones (more rapid vibrations than the basic one). However, he is best known for his contributions to physics and energy conservation in particular. He showed that the earth’s age would be less than 25 million years if the sun’s energy came from gravitational contraction. He suggested to his student Heinrich Hertz that he should prove that the electromagnetic spectrum extends well beyond the visible light.
Alternate Solutions
In a new approach to the heat flow PDE suggested by experiments, we now return to the one-dimensional PDE [Eq. (16.3)], seeking solutions of a new functional form ψ( x , t ) = u ( x /
t ), which is suggested by Example 15.2.2.
16.2 Heat Flow or Diffusion PDE 765
Again, these solutions have to be generalized to adapt them to boundary con- ditions. Also, there is another method of generating new solutions of a PDE with constant coefficients: We can translate a given solution, for example, ψ 1 ( x , t ) → ψ 1 ( x − α, t ), and then integrate over the translation parame- ter α. Therefore,
ψ( x , t ) =
2 a
t π
−∞
C (α) e −^
( x −α)^2 4 a^2 t (^) d α (16.19)
is again a solution, which we rewrite using the substitution
ξ =
x − α 2 a
t
, α = x − 2 a ξ
t , d α = − 2 ad ξ
t. (16.20)
Thus, we find that
ψ( x , t ) =
π
−∞
C ( x − 2 a ξ
t ) e −ξ^
2 d ξ (16.21)
is a solution of our PDE. In this form we recognize the significance of the weight function C ( x ) from the translation method because, at t = 0, ψ( x , 0) = C ( x ) = ψ 0 ( x ) is determined by the boundary condition, and
−∞ e
−ξ (^2) d ξ = √π. Therefore, we can write the solution as
ψ( x , t ) =
π
−∞
ψ 0 ( x − 2 a ξ
t ) e −ξ^ 2 d ξ, (16.22)
displaying the role of the boundary condition explicitly. From Eq. (16.22) we see that the initial temperature distribution ψ 0 ( x ) spreads out over time and is damped by the Gaussian weight function.
EXAMPLE 16.2.2 Special Boundary Condition Again^ Let us express the solution of Exam- ple 16.2.1 in terms of the error function solution of Eq. (16.16). The bound- ary condition at t = 0 is ψ 0 ( x ) = 1 for − 1 < x < 1 and zero for | x | > 1. From Eq. (16.22) we find the limits on the integration variable ξ by setting x − 2 a ξ
t = ±1. This yields the integration end points ξ = (± 1 + x )/ 2 a
t. Therefore, our solution becomes
ψ( x , t ) =
π
∫ x + 1 2 a √ t x − 1 2 a √ t
e −ξ^
2 d ξ.
Using the error function defined in Eq. (16.16) we can also write this solution as follows:
ψ( x , t ) =
[
x + 1 2 a
t
x − 1 2 a
t
)]
Comparing this form of our solution with that from Example 16.2.1, we see that we can express Eq. (16.23) as the Fourier integral of Example 16.2.1, an identity that gives the Fourier integral [Eq. (16.6)], in closed form of the tabulated error function. ■
766 Chapter 16 Partial Differential Equations
Finally, we consider the heat flow case for an extended spherically symmetric medium centered at the origin, which prescribes polar coordinates r , θ, ϕ. We expect a solution of the form ψ( r , t ) = u ( r , t ). Using Eq. (2.77) we find the PDE
∂ u ∂ t
= a^2
∂^2 u ∂ r^2
r
∂ u ∂ r
which we transform to the one-dimensional heat flow PDE by the substitution
u =
v( r , t ) r
∂ u ∂ r
r
∂v ∂ r
v r^2
∂ u ∂ t
r
∂v ∂ t
∂^2 u ∂ r^2
r
∂^2 v ∂ r^2
r^2
∂v ∂ r
2 v r^3
This yields the PDE
∂v ∂ t
= a^2
∂^2 v ∂ r^2
EXAMPLE 16.2.3 Spherically Symmetric Heat Flow^ Let us apply the one-dimensional heat flow PDE with the solution Eq. (16.16) to a spherically symmetric heat flow under fairly common boundary conditions, where x is relaced by the radial variable. Initially, we have zero temperature everywhere. Then, at time t = 0, a finite amount of heat energy Q is released at the origin, spreading evenly in all directions. What is the resulting spatial and temporal temperature distribution?
Inspecting our special solution in Eq. (16.18), we see that, for t → 0, the temperature
v( r , t ) r
C
t^3
e −^
r^2 4 a^2 t (^) (16.27)
goes to zero for all r = 0 so that zero initial temperature is guaranteed. As t → ∞, the temperature v/ r → 0 for all r , including the origin, which is implicit in our boundary conditions. The constant C can be determined from energy conservation, which gives the constraint
Q = σρ
v r
d^3 r =
4 πσρ C √ t^3
0
r^2 e −^
r^2 4 a^2 t (^) dr = 8
π^3 σρ a^3 C , (16.28)
where ρ is the constant density of the medium and σ its specific heat. Here, we have rescaled the integration variable and integrated by parts to get ∫ (^) ∞
0
e −^
r^2 4 a^2 t (^) r^2 dr = (2 a
t )^3
0
e −ξ^
2 ξ 2 d ξ,
∫ (^) ∞
0
e −ξ^
2 ξ 2 d ξ = −
ξ 2
e −ξ^
2
∞
0
0
e −ξ^
2 d ξ =
π 4
768 Chapter 16 Partial Differential Equations
When the initial temperature is prescribed for the one-dimensional or three-dimensional heat equation (with spherical or cylindrical symmetry), it becomes a weight function of the solution in terms of an integral over the generic Gaussian solution. The three-dimensional heat equation, with spheri- cal or cylindrical boundary conditions, is solved by separation of the variables leading to eigenfunctions in each separated variable and eigenvalues as sepa- ration constants. For finite boundary intervals in each spatial coordinate, the sum over separation constants leads to a Fourier series solution, whereas in- finite boundary conditions lead to a Fourier integral solution. The separation of variables method attempts to solve a PDE by writing the solution as a prod- uct of functions of one variable each. General conditions for the separation method to work are provided by the symmetry properties of the PDE to which continuous group theory applies.
EXERCISES
16.2.1 By letting the operator ∇^2 + k^2 act on the general form a 1 ψ 1 ( x , y , z ) + a 2 ψ 2 ( x , y , z ), show that it is linear; that is, (∇^2 + k^2 )( a 1 ψ 1 + a 2 ψ 2 ) = a 1 (∇^2 + k^2 )ψ 1 + a 2 (∇^2 + k^2 )ψ 2. 16.2.2 Show that the Helmholtz equation
∇^2 ψ + k^2 ψ = 0
is separable in circular cylindrical coordinates if k^2 is generalized to k^2 + f (ρ) + (1/ρ^2 ) g (ϕ) + h ( z ). 16.2.3 Separate variables in the Helmholtz equation in spherical polar coor- dinates, splitting off the radial dependence first. Show that your sepa- rated equations have the same form as Eqs. (8.143) and (8.146), whereas Eq. (8.147) is modified. Describe in your words what this exercise tells you. 16.2.4 Verify that
∇^2 ψ( r , θ, ϕ) +
[
k^2 + f ( r ) +
r^2
g (θ) +
r^2 sin^2 θ
h (ϕ)
]
ψ( r , θ, ϕ) = 0
is separable (in spherical polar coordinates). The functions f , g , and h are functions only of the variables indicated; k^2 is a constant. 16.2.5 For a homogeneous spherical solid with constant thermal diffusivity, K , and no heat sources, the equation of heat conduction becomes ∂ T ( r , t ) ∂ t
= K ∇^2 T ( r , t ).
Assume a solution of the form
T = R ( r ) T ( t )
16.3 Inhomogeneous PDE---Green’s Function 769
and separate variables. Show that the radial equation may take on the standard form
r^2
d^2 R dr^2
dR dr
- [α^2 r^2 − n ( n + 1)] R = 0; n = integer.
The solutions of this equation are called spherical Bessel functions. 16.2.6 Separate variables in the thermal diffusion equation of Eq. (8.112) in cir- cular cylindrical coordinates. Assume that you can neglect end effects and take ψ = T (ρ, t ).
16.3 Inhomogeneous PDE---Green’s Function
The series substitution of Section 8.5 and the Wronskian double integral of Sec- tion 8.6 provide the most general solution of the homogeneous , linear, second- order ODE. The specific solution, yp , is linearly dependent on the source term [ F ( x ) of Eq. (8.44)] and may be cranked out by the variation of parameters method. In this section, we discuss a different method of solution for PDEs— Green’s functions. For a brief introduction to the Green’s function method as applied to the solution of a nonhomogeneous PDE, it is helpful to use the electrostatic analog. In the presence of charges, the electrostatic potential ψ satisfies Poisson’s nonhomogeneous equation (compare Section 1.13)
∇^2 ψ = −
ρ ε 0
(mks units) (16.35)
and Laplace’s homogeneous equation,
∇^2 ψ = 0, (16.36) in the absence of electric charge (ρ = 0). If the charges are point charges q (^) i located at r i , we know that the solution is
ψ( r ) =
4 πε 0
i
q (^) i | r − r i |
a superposition of single-point charge solutions obtained from Coulomb’s law for the force between two point charges q 1 and q 2 a distance r apart,
F =
q 1 q 2 r ˆ 4 πε 0 r^2
By replacement of the discrete point charges with a smeared out distributed charge, charge density ρ [Eq. (16.37)] becomes
ψ( r = 0) =
4 πε 0
ρ( r ) r
d τ (16.39)
or, for the potential at r away from the origin and the charge at r 2 ,
ψ( r ) =
4 πε 0
ρ( r 2 ) | r − r 2 |
d τ 2. (16.40)
16.3 Inhomogeneous PDE---Green’s Function 771
in complete agreement with Eq. (16.40). Actually, ψ( r ) [Eq. (16.49)] is the particular solution of Poisson’s equation. We may add solutions of Laplace’s equation [compare Eq. (8.113)]. Such solutions could describe an external field. These results can be generalized to the second-order linear but inhomoge- neous, differential equation L y ( r 1 ) = − f ( r 1 ), (16.50) where L is a linear differential operator. The Green’s function is taken to be a solution of
L G ( r 1 , r 2 ) = −δ( r 1 − r 2 ) (16.51) [analogous to Eq. (16.41)]. Then the particular solution y ( r 1 ) becomes
y ( r 1 ) =
G ( r 1 , r 2 ) f ( r 2 ) d τ 2 , (16.52)
which can be verified by applying L to y ( r ). (There may also be an integral over a bounding surface depending on the conditions specified.) In summary, the Green’s function, often written G ( r 1 , r 2 ) as a reminder of the name, is a solution of Eq. (16.41) or Eq. (16.51) more generally. It enters in an integral solution of our differential equation, as in Eq. (16.45). For the sim- ple, but important, electrostatic case, we obtain Green’s function G ( r 1 , r 2 ) by Gauss’s law, comparing Eqs. (16.41) and (16.47). Finally, from the final solution [Eq. (16.49)] it is possible to develop a physical interpretation of Green’s func- tion. It occurs as a weighting function or propagator function that enhances or reduces the effect of the charge element ρ( r 2 ) d τ 2 according to its distance from the field point r 1. The Green’s function, G ( r 1 , r 2 ), gives the effect of a unit point source at r 2 in producing a potential at r 1. This is how it was introduced in Eq. (16.41); this is how it appears in Eq. (16.49).
EXAMPLE 16.3.1 Quantum Mechanical Scattering—Neumann Series Solution^ The quan- tum theory of scattering provides a good illustration of integral equation tech- niques and an application of a Green’s function. Our physical picture of scatter- ing is as follows. A beam of particles moves along the negative z -axis toward the origin. A small fraction of the particles are scattered by the potential V ( r ) and go off as an outgoing spherical wave, as shown schematically in Fig. 16.1. Our wave function ψ( r ) must satisfy the time-independent Schr ¨odinger equation
h ¯^2 2 m
∇^2 ψ( r ) + V ( r )ψ( r ) = E ψ( r ) (16.53a) or
∇^2 ψ( r ) + k^2 ψ( r ) = −
[
2 m h ¯^2
V ( r )ψ( r )
]
, k^2 =
2 mE h ¯^2
. (16.53b)
From the physical picture just presented, we search for a solution having an asymptotic form
ψ( r ) ∼ e i k^0 · r^ + fk (θ, ϕ)
e ikr r
772 Chapter 16 Partial Differential Equations
V
q k z
Detector
Figure 16.
Incident Plane Wave Scattered by a Potential V into an Outgoing Spherical Wave
which we shall derive from the integral equation following from the Schr ¨odinger equation. Here, e i k^0 · r^ is the incident plane wave, 2 with k 0 the prop- agation vector carrying the subscript 0 to indicate that it is in the θ = 0 ( z -axis) direction. The magnitudes k 0 and k are equal (for eleastic scattering ignoring recoil of the target), and e i kr^ / r is the outgoing spherical wave with an angu- lar (and energy)-dependent amplitude factor fk (θ, ϕ). 3 In quantum mechanics texts, it is shown that the differential probability of scattering, d σ/ d , the scattering cross section per unit solid angle, is given by | fk (θ, ϕ)|^2.
Identifying [−(2 m / h ¯^2 ) V ( r )ψ( r )] with f ( r ) of Eq. (16.50), we have
ψ( r ) = −
2 m h ¯^2
V ( r 2 )ψ( r 2 ) G ( r , r 2 ) d^3 r 2 (16.55)
by Eq. (16.52). This does not have the desired asymptotic form of Eq. (16.54), but we may add e i k^0 · r^ to Eq. (16.55), a solution of the homogeneous equation, and put ψ( r ) into the desired form:
ψ( r ) = e i k^0 · r^ −
2 m h ¯^2
V ( r 2 )ψ( r 2 ) G ( r , r 2 ) d^3 r 2. (16.56)
Our Green’s function is the inverse of the operator L = ∇^2 + k^2 [Eq. (16.53)] satisfying the boundary condition that it describe an outgoing wave. Then, from Example 16.3.2,
G ( r 1 , r 2 ) =
exp( ik | r 1 − r 2 |) 4 π| r 1 − r 2 |
and
ψ( r ) = e i k^0 · r^ −
2 m 4 π h ¯^2
V ( r 2 )ψ( r 2 )
e ik | r − r^2 | | r − r 2 |
d^3 r 2. (16.57)
(^2) For simplicity, we assume a continuous incident beam. In a more sophisticated and more realistic treatment, Eq. (16.54) would be one component of a Fourier wave packet. (^3) If V ( r ) represents a central force, fk will be a function of θ only, independent of azimuth.
774 Chapter 16 Partial Differential Equations
where k ≡ k rˆ is the scattered wave vector, and q = k 0 − k is the momen- tum transfer. In other words, the elastic scattering amplitude in first-order Born approximation is proportional to the Fourier transform of the potential at q. ■
EXAMPLE 16.3.2 Quantum Mechanical Scattering—Green’s Function^ Again, we consider the Schr ¨odinger wave equation [Eq. (16.53b)] for the scattering problem. This time we use Fourier transform techniques and we derive the desired form of the Green’s function by contour integration.
We solve the PDE for the Green’s function corresponding to Eq. (16.53b)
(∇^2 + k^2 ) G ( r , r 2 ) = −δ( r − r 2 ) = −
d^3 p (2π)^3
e i p ·( r − r^2 )^ (16.62)
in terms of a Fourier integral
G ( r , r 2 ) =
d^3 p (2π)^3
g 0 ( p ) e i p ·( r − r^2 )^ (16.63)
of the same type as the delta function driving term. Substituting the Fourier integral into the PDE, we see that
∇ e i p ·( r − r^2 )^ = i p e i p ·( r − r^2 )
so that the PDE becomes the trivial algebraic equation ( k^2 − p^2 ) g 0 ( p ) = − 1 for the Fourier transform g 0 of G. To find G , we need to evaluate the Fourier integral ∫ d^3 p
e i p ·ρ p^2 − k^2
where p^2 /( p^2 − k^2 ) is part of the radial integral, and d^3 p = p^2 dp sin θ d θ d ϕ. Here, p ρ cos θ has replaced p · ( r − r 2 ) for simplicity. With the substitution x = cos θ, dx = − sin θ d θ, the angular part in polar coordinates in momentum space is elementary. Integrating over ϕ by inspection, we pick up a 2π. The θ integration then leads to ∫ (^2) π
0
d ϕ
∫ (^) π
θ= 0
e ip ρ^ cos^ θ^ sin θ d θ = 2 π
− 1
e ip ρ x^ dx =
2 π ip ρ
( e ip ρ^ − e − ip ρ^ ). (16.65)
The remaining radial integral is given by
G ( r , r 2 ) =
4 π^2 ρ i
0
e ip ρ^ − e − ip ρ p^2 − k^2
pdp , (16.66)
and since the integrand is an even function of p , we may set
G ( r , r 2 ) =
8 π^2 ρ i
−∞
( e i κ^ − e − i κ^ ) κ^2 − σ 2
κ d κ. (16.67)
The latter step is taken in anticipation of the evaluation of G (^) k ( r , r 2 ) as a contour integral. The symbols κ and σ (σ > 0) represent p ρ and k ρ, respectively.
16.3 Inhomogeneous PDE---Green’s Function 775
If the integral in Eq. (16.67) is interpreted as a Riemann integral, the integral does not exist. This implies that L−^1 does not exist, and in a literal sense it does not. L = ∇^2 + k^2 is singular since there exist nontrivial solutions ψ for which the homogeneous equation Lψ = 0. We avoid this problem by intro- ducing a parameter γ , defining a different operator L− γ 1 , and taking the limit as γ → 0.
Splitting the integral into two parts so each part may be written as a suitable contour integral gives us
G ( r , r 2 ) =
8 π^2 ρ i
C 1
κ e i κ^ d κ κ^2 − σ 2
8 π^2 ρ i
C 2
κ e − i κ^ d κ κ^2 − σ 2
Contour C 1 is closed by a semicircle in the upper half-plane, and C 2 is closed by a semicircle in the lower half-plane. These integrals were evaluated in Chap- ter 7 by using appropriately chosen infinitesimal semicircles to go around the singular points κ = ±σ. As an alternative procedure, let us first displace the singular points from the real axis by replacing σ by σ + i γ and then, after evaluation, taking the limit as γ → 0 (Fig. 16.2).
For γ positive, contour C 1 encloses the singular point κ = σ + i γ and the first integral contributes
2 π i ·
e i (σ^ + i γ^ ).
g > 0
s + ig
s + ig
J ( k ) C 1
C 2
R ( k )
R ( k )
J ( k )
Figure 16.
Possible Green’s Function Contours of Integration
