Download Green's functions 1 The delta function and distributions and more Exams Differential Equations in PDF only on Docsity!
Green’s functions
Suppose that we want to solve a linear, inhomogeneous equation of the form Lu(x) = f (x) (1)
where u, f are functions whose domain is Ω. It happens that differential operators often have inverses that are integral operators. So for equation (1), we might expect a solution of the form
u(x) =
Ω
G(x, x 0 )f (x 0 )dx 0. (2)
If such a representation exists, the kernel of this integral operator G(x, x 0 ) is called the Green’s function. It is useful to give a physical interpretation of (2). We think of u(x) as the response at x to the influence given by a source function f (x). For example, if the problem involved elasticity, u might be the displacement caused by an external force f. If this were an equation describing heat flow, u might be the temperature arising from a heat source described by f. The integral can be though of as the sum over influences created by sources at each value of x 0. For this reason, G is sometimes called the influence function.
1 The delta function and distributions
There is a great need in differential equations to define objects that arise as limits of functions and behave like functions under integration but are not, properly speaking, functions themselves. These objects are sometimes called generalized functions or distributions. The most basic one of these is the so-called δ-function. For each > 0 , define the family of ordinary functions
δ(x) =
π
e−x (^2) / 2
. (3)
When is small, the graph of δ (figure 1) is essentially just a spike at x = 0, but the integral of δ is exactly one for any . For any continuous function φ(x), the integral of φ(x)δ(x − x 0 ) is concentrated near the point x 0 , and therefore
lim → 0
−∞
φ(x)δ(x − x 0 )dx = lim → 0 φ(x 0 )
−∞
δ(x − x 0 )dx = φ(x 0 ).
On the other hand, taking the limit → 0 inside the integral makes no sense: the limit of δ is not a function at all! To get around this, we define a new object, δ(x − x 0 ), to behave as follows: ∫ (^) ∞
−∞
φ(x)δ(x − x 0 )dx = φ(x 0 ). (4)
Informally speaking, the δ-function “picks out” the value of a continuous function φ(x) at one point. There are δ-functions for higher dimensions also. We define the n-dimensional δ-function to behave as (^) ∫
Rn
φ(x)δ(x − x 0 )dx = φ(x 0 ),
for any continuous φ(x) : Rn^ → R. Sometimes the multidimensional δ-function is written as a product of one dimensional ones: δ(x) = δ(x) · δ(y) ·.. ..
Figure 1: Approximations of the δ-function.
1.1 A more precise definition
To be concrete about distributions, one needs to talk about how they “act” on smooth functions. Note that (4) can be thought of as a linear mapping from smooth functions φ(x) to real numbers by the process of evaluating them at x 0. Linear mappings from a vector space (in this case, a space of smooth functions like φ) to the real numbers are often called linear functionals. Now we come to the precise definition. A distribution is a continuous linear functional on the set of infinitely differentiable functions with bounded support; this space of functions is denoted by D. We can write d[φ] : D → R to represent such a map: for any input function φ, d[φ] gives us a number. Linearity means that
d[c 1 φ 1 (x) + c 2 φ 2 (x)] = c 1 d[φ 1 (x)] + c 2 d[φ 2 (x)].
Since integrals have properties similar to this, distributions are often defined in terms of them. The class of distributions includes just plain old functions in the following sense. Suppose that g(x) is integrable, but not necessarily continuous. We define the corresponding linear functional to be
g[φ] =
−∞
g(x)φ(x)dx.
(notice the subtlety of notation: g(x) means the function evaluated at x, whereas the bracket g[φ] means multiply by function φ and integrate). Conversely, while all distributions cannot be associ- ated with functions, they can be approximated by smooth, ordinary functions. This means that, for any distribution d, there exists a sequence of smooth functions dn(x) ∈ D so that
d[φ] = lim n→∞
dn(x)φ(x)dx, for all φ ∈ D.
For example, the sequence δ that we first investigated comprises a set of smooth approximations to the δ-distribution. We can now define what it means to integrate a distribution (notated as if it were a regular function d(x)) simply by setting ∫ (^) ∞
−∞
d(x)φ(x)dx ≡ d[φ], for any φ ∈ D.
1.3 The distributional Laplacian
In higher dimensions, one can make similar definitions of distributional derivatives by using Green’s identities. For a twice differentiable function u(x) and φ(x) ∈ D, one has ∫
Rn
(∆u)φ dx =
Rn
u ∆φ dx,
since φ(x) vanishes at infinity. This motivates a definition of the distributional Laplacian for func- tions u(x) which are not entirely twice differentiable, which is a distribution with linear functional
(∆u)[φ] = u[∆φ] =
Rn
u ∆φ dx. (5)
Example. Let’s compute the distributional Laplacian of f (x) = 1/|x|, where x ∈ R^3. In spherical coordinates,
∆(1/|x|) = (∂rr +
r ∂r)
r
except when |x| = 0. In order to capture the behavior at the origin, we need distributional deriva- tives instead. Using definition (5),
∆f [φ] = f [∆φ] = lim → 0
R^3 /B(0)
∆φ |x| dx,
where B(0) is a three dimensional ball of radius centered at the origin. The limit is needed because the integrand in unbounded, but for any > 0 , the Green’s identity can be applied. Let ∂/∂n denote the normal derivative in the inward radial direction ˆn = −x/|x|, so that
∆f [φ] = lim → 0
∂B(0)
|x|
∂φ ∂n
dx − φ
∂n
|x|
dx
= lim → 0
∂B(0)
∂φ ∂n
^2
∂B(0)
φ dx
where we used the fact that 1 /|x| = 1/ on the boundary ∂B(0). Since ∂B(0) is the surface of a sphere, we have (^) ∫
∂B(0)
φ dx ∼ 4 π^2 φ(0),
∂B(0)
∂φ ∂n dx = O(^2 ),
for small . The limit → 0 therefore yields
∆f [φ] = − 4 πφ(0).
We have discovered that ∆f = − 4 πδ(x). The emergence of the delta function could not have been predicted without applying the definition!
1.4 Relationship to Green’s functions
Part of the problem with the definition (2) is that it doesn’t tell us how to construct G. It is useful to imagine what happens when f (x) is a point source, in other words f (x) = δ(x − xi). Plugging into (2) we learn that the solution to
Lu(x) = δ(x − xi) + homogeneous boundary conditions (6)
should be
u(x) =
Ω
G(x, x 0 )δ(x 0 − xi)dx 0 = G(x, xi). (7)
In other words, we find that the Green’s function G(x, x 0 ) formally satisfies
LxG(x, x 0 ) = δ(x − x 0 ) (8)
(the subscript on L is needed since the linear operator acts on x, not x 0 ). This equation says that G(x, x 0 ) is the influence felt at x due to a point source at x 0. Equation (8) is a more useful way of defining G since we can in many cases solve this “almost” homogeneous equation, either by direct integration or using Fourier techniques. In particular,
LxG(x, x 0 ) = 0, when x 6 = x 0 , (9)
which is a homogeneous equation with a “hole” in the domain at x 0. To account for the δ-function, we can formally integrate both sides of LxG(x, x 0 ) = δ(x − x 0 ) on any region containing x 0. It is usually sufficient to allow these regions to be some ball Br(x 0 ) = {x||x − x 0 | < r}, so that ∫
Br (x 0 )
LxG(x, x 0 )dx = 1. (10)
Equation (10) is called the normalization condition, and it is used to get the “size” of the singularity of G at x 0 correct. In one dimension, this condition takes on a slightly different form (see below). In addition to (9-10), G must also satisfy the same type of homogeneous boundary conditions that the solution u does in the original problem. The reason for this is straightforward. Take, for example, the case of a homogeneous Dirichlet boundary condition u = 0 for x ∈ ∂Ω. For any point x on the boundary, it must be the case that ∫
Ω
G(x, x 0 )f (x 0 )dx 0 = 0. (11)
Since this must be true for any choice of f , it follows that G(x, x 0 ) = 0 for boundary points x (note that x 0 is treated as a constant in this respect, and can be any point in the domain).
2 Green’s functions in one dimensional problems
It is instructive to first work with ordinary differential equations of the form
Lu ≡ u(n)(x) + F (u(n−1)(x), u(n−2)(x),.. .) = f (x),
subject to some kind of boundary conditions, which we will initially suppose are homogeneous. Following the previous discussion, the Green’s function G(x, x 0 ) satisfies (8), which is
G(n)^ + F (G(n−1), G(n−2),.. .) = δ(x − x 0 ), (12)
where G(n)^ = ∂n/∂xn. This means that away from the point x 0
G(n)(x) + F (G(n−1)(x), G(n−2)(x),.. .) = 0, x > x 0 (13) G(n)(x) + F (G(n−1)(x), G(n−2)(x),.. .) = 0, x < x 0 , (14)
It follows that the solution to (16) can be written using G as
u(x) =
L
(∫ (^) x
0
x 0 (x − L)f (x 0 )dx 0 +
∫ L
x
x(x 0 − L)f (x 0 )dx 0
Notice the careful choice of integrands: the first integral involves the piece of the Green’s function appropriate for x 0 < x, not the other way around.
Example. The one dimensional Helmholtz problem is
uxx − k^2 u = f (x), lim x→±∞ u(x) = 0. (22)
The corresponding Green’s function therefore solves
Gxx(x, x 0 ) − k^2 G = 0 for x 6 = x 0 , (23)
together with limx→±∞ G(x, x 0 ) = 0 and connection conditions (18). The general solution of (23) is G = c 1 exp(−kx) + c 2 exp(kx). For the piecewise Green’s function to decay, the part for x < x 0 has c 1 = 0 and the part for x > x 0 has c 2 = 0, therefore
G(x, x 0 ) =
c 2 ekx^ x < x 0 , c 1 e−kx^ x > x 0.
Conditions (18) imply
c 2 exp(kx 0 ) = c 1 exp(−kx 0 ), −kc 1 exp(−kx 0 ) − kc 2 exp(kx 0 ) = 1,
which yield c 1 = − exp(kx 0 )/ 2 k and c 2 = − exp(−kx 0 )/ 2 k. The entire Green’s function may then be written compactly as G(x, x 0 ) = − exp(−k|x − x 0 |)/ 2 k,
and the solution to (22) is
u(x) = −
2 k
−∞
f (x 0 )e−k|x−x^0 |dx 0.
Notice that the Green’s function in this problem simply depends on the distance between x and x 0 ; this is a very common feature in problems on infinite domains which have translation symmetry.
2.1 Using symmetry to construct new Green’s functions from old ones
Green’s functions depend both on a linear operator and boundary conditions. As a result, if the problem domain changes, a different Green’s function must be found. A useful trick here is to use symmetry to construct a Green’s function on a semi-infinite (half line) domain from a Green’s function on the entire domain. This idea is often called the method of images. Consider a modification of the previous example
Lu = uxx − k^2 u = f (x), u(0) = 0, lim x→∞ u(x) = 0. (25)
We can’t use the “free space” Green’s function
G∞(x, x 0 ) = − exp(−k|x − x 0 |)/ 2 k,
because it doesn’t satisfy G(0, x 0 ) = 0 as required in this problem. Here’s the needed insight: by subtracting G∞ and its reflection about x = 0
G(x, x 0 ) = G∞(x, x 0 ) − G∞(−x, x 0 )
does in fact satisfy G(0, x 0 ) = 0. On the other hand, does this proposed Green’s function satisfy the right equation? Computing formally,
LG(x, x 0 ) = LG∞(x, x 0 ) − LG∞(−x, x 0 ) = δ(x − x 0 ) − δ(−x − x 0 ).
The second delta function δ(−x − x 0 ) looks like trouble, but it is just zero on the interval (0, ∞). Therefore Gxx − k^2 G = δ(x − x 0 ) when restricted to this domain, which is exactly what we want. The solution of (25) is therefore
u(x) = −
2 k
0
f (x 0 )
[
e−k|x−x^0 |^ − e−k|x+x^0 |
]
dx 0.
2.2 Dealing with inhomogeneous boundary conditions
Remarkably, a Green’s function can be used for problems with inhomogeneous boundary con- ditions even though the Green’s function itself satisfies homogeneous boundary conditions. This seems improbable at first since any combination or superposition of Green’s functions would always still satisfy a homogeneous boundary condition. The way in which inhomogeneous boundary condi- tions enter relies on the so-called ”Green’s formula”, which depends both on the linear operator in question as well as the type of boundary condition (i.e. Dirichlet, Neumann, or a combination). Suppose we wanted to solve
uxx = f, u(0) = A, u(L) = B, (26)
using the Green’s function G(x, x 0 ) we found previously (21) when A = 0 = B. For this prob- lem, the Green’s formula is nothing more than integration by parts twice (essentially just the one dimensional Green’s identity)
∫ (^) L
0
uv′′^ − vu′′^ dx = [uv′^ − vu′]L 0. (27)
To solve (26), we set v(x) = G(x, x 0 ) in (27) and obtain ∫ (^) L
0
u(x)Gxx(x, x 0 ) − G(x, x 0 )u′′(x)dx = [u(x)Gx(x, x 0 ) − G(x, x 0 )u′(x)]xx==0L. (28)
Using uxx(x) = f (x), Gxx(x, x 0 ) = δ(x − x 0 ) and noting that G = 0 if x = 0 or x = L, (28) collapses to
u(x 0 ) =
∫ L
0
G(x, x 0 )f (x)dx + [u(x)Gx(x, x 0 )]xx==0L
∫ L
0
G(x, x 0 )f (x)dx + BGx(L, x 0 ) − AGx(0, x 0 ). (29)
The first term on the right looks like the formula we had for homogeneous boundary conditions, with an important exception: x and x 0 are in the wrong places. We will resolve this apparent difference in section (3.3). The two terms at the end of this formula account for the inhomogeneous boundary conditions.
The strange looking far-field condition is needed because in general, solutions to ∆u = f behave like u ∼ A ln r + B as r → ∞. The condition just ensures that B = 0. Again we look for a Green’s function of the form G = g(|x − x 0 |) = g(r), so that in polar coordinates
1 r (rg′(r))′^ = 0 if r 6 = 0, (^) rlim→∞
g(r) − r ln rg′(r)
The general solution is G = c 1 ln r + c 2 , (37)
where c 2 = 0 by using the far-field condition in (36). The normalization condition (32) gives
1 =
∂B
∂xG ∂n (x, x 0 ) dx =
∂B
c 1 dx = 2πc 1 , (38)
where B is the unit disk, so that c 1 = 1/ 2 π. Thus the Green’s function is G(x, x 0 ) = ln |x − x 0 |/ 2 π, and the solution to (35) is
u(x) =
R^2
ln |x − x 0 |f (x 0 ) 2 π dx^20.
It is sometimes useful to write G in polar coordinates. Using the law of cosines for the distance |x − x 0 |, one gets
G(r, θ; r 0 , θ 0 ) =
4 π ln(r^2 + r^20 − 2 rr 0 cos(θ − θ 0 )) (39)
(the semicolon is used here to visually separate sets of coordinates).
Example: The Helmholtz equation A two dimensional version of problem (22) is
∆u − u = f (x), (^) rlim→∞ u = 0, u : R^2 → R. (40)
We can look for the Green’s function for the Helmholtz operator L = ∆ − 1 just as we did for the Laplacian, by supposing G(x, x 0 ) = g(r) where r = |x − x 0 |. Then since ∆G − G = 0 when x 6 = x 0 , it follows that g′′^ +
r
g′^ − g = 0, lim r→∞ g(r) = 0,
which is known as the modified Bessel equation of order zero. There is a single linearly indepen- dent solution which decays at infinity referred to as K 0 , which happens to be represented by a nice formula
K 0 (r) =
0
cos(rt) √ t^2 + 1
dt.
The Green’s function is therefore G(x, x 0 ) = cK 0 (|x − x 0 |) where c is found from a normalization condition. Applying the divergence theorem to (10) for L = ∆ − 1 gives something slightly different than (32):
1 =
∂Br (x 0 )
∂xG ∂n (x, x 0 ) dx −
Br (x 0 )
G(x, x 0 )dx ∼
∂Br (x 0 )
∂xG ∂n (x, x 0 ) dx, (41)
as r → 0. It can be shown (c.f. the venerable text of Bender & Orszag) that K 0 ∼ − ln(r) when r is small, and therefore ∫
∂Br (x 0 )
∂xG ∂n
(x, x 0 ) dx ∼ −c
∂Br (x 0 )
r
dx = − 2 πc. (42)
Using (41), it follows that c = − 1 / 2 π. The solution to (35) is therefore
u(x) = −
R^2
K 0 (|x − x 0 |)f (x 0 ) 2 π dx 0.
3.1 Dealing with boundaries and the method of images
The examples in the previous section are free space Green’s functions, since there are no domain boundaries. Recall that the Green’s function must satisfy all the same homogeneous boundary conditions as underlying linear problem. Free space Green’s functions typically satisfy conditions at infinity, but not along true boundaries. With a little cleverness, however, we can still employ them as a starting point to find Green’s functions for certain bounded or semi-infinite domains.
3.1.1 Arbitrary bounded domains
Suppose that we wish to solve the Poisson equation for u : Ω → R
∆u = f (x, y), u = 0 on ∂Ω (43)
where Ω is a bounded, open set in R^2. We need a Green’s function G(x, x 0 ) which satisfies
∆xG = δ(x − x 0 ), G(x, x 0 ) = 0 when x ∈ ∂Ω. (44)
It’s tempting to use the free space Green’s function G 2 (x, x 0 ) = ln |x−x 0 |/ 2 π, which does indeed satisfy the equation in (44), but G 2 is not zero on the boundary ∂Ω. We should be thinking of (44) as an inhomogeneous equation, and use the method of particular solutions. In fact, G 2 is a particular solution, so if we write G(x, x 0 ) = G 2 (x, x 0 ) + GR(x, x 0 ), then for each x 0 ∈ Ω GR solves a homogeneous equation with nonzero boundary data
∆xGR = 0, GR(x, x 0 ) = −G 2 (x, x 0 ) = −
2 π ln |x − x 0 | for x ∈ ∂Ω. (45)
The function GR is called the regular part of the Green’s function, and it just a nice, usual solution of Laplace’s equation, without any singular behavior. In other words, the desired Green’s func- tion has exactly the same logarithmic singularity as the free space version, but is quantitatively different far away from x 0. In general, this method is used in conjunction with other techniques as a way to characterize Green’s functions without actually solving for them.
3.1.2 Method of images
In section (2.1), we found that symmetry of domains can be exploited in constructing Green’s functions using their free space counterparts. The basic inspiration is a simple observation about even and odd functions: for continuously differentiable f (x) : R → R,
g(x) = f (x) − f (−x) is an odd function and g(0) = 0, (46) h(x) = f (x) + f (−x) is an even function and h′(0) = 0. (47)
Therefore subtracting mirror images of a function will satisfy a Dirichlet-type boundary condition at x = 0, whereas their sum satisfies a Neumann-type boundary condition.
Example. Consider finding the Green’s function for the upper-half space problem for u : R^3 ∩{z > 0 } → R: ∆u = f, lim |x|→∞
u(x) = 0, u(x, y, 0) = 0. (48)
What if u, v don’t necessarily satisfy homogeneous boundary conditions? Then something like (52) would still be true, but terms involving boundary values of u, v would appear: ∫
Ω
(Lv)u dx −
Ω
(Lu)v dx = boundary terms involving u and v. (53)
What this formula actually looks like depends on the linear operator in question and is known as the Green’s formula for the linear operator L. For the Laplacian, the associated Green’s formula is nothing more than Green’s identity, which reads ∫
Ω
u∆v − v∆u dx =
∂Ω
u
∂v ∂n − v
∂u ∂n dx. (54)
Suppose we wish to solve the problem with the inhomogeneous boundary condition
∆u = f in Ω, u(x) = h(x) x ∈ ∂Ω.
Let G be the Green’s function that solves ∆G = δ(x − x 0 ) with homogeneous, Dirichlet boundary conditions. Substituting v(x) = G(x, x 0 ) in (54), we have (being careful to keep x as the variable of integration) ∫
Ω
u(x)δ(x − x 0 ) − G(x, x 0 )f (x) dx =
∂Ω
u(x) ∂xG ∂n
(x, x 0 ) − G(x, x 0 ) ∂u ∂n
dx. (55)
Using the definition of the δ-function and the fact that G is zero on the domain boundary, this simplifies to
u(x 0 ) =
Ω
G(x, x 0 )f (x)dx +
∂Ω
h(x) ∂xG ∂n
(x, x 0 ) dx. (56)
The first term is just the solution we expect for homogeneous boundary conditions. The second term is more surprising: it is a derivative of G that goes into the formula to account for the inho- mogeneous Dirichlet boundary condition.
Example: the Poisson integral formula revisited. In the case that Ω is a disk of radius a, we have found the Green’s function which is zero on the boundary in equation (50). The boundary value problem ∆u = 0, u(a, θ) = h(θ)
has a solution given by (56), where f (x) = 0. The normal derivative to the boundary of G is just the radial derivative
∂xG ∂n (x, x 0 ) = Gr(r, θ; r 0 , θ 0 )
=
4 π
2 r − 2 r 0 cos(θ − θ 0 ) r^2 + r^20 − 2 rr 0 cos(θ − θ 0 )
2 rr^20 − 2 r 0 a^2 cos(θ − θ 0 ) r^2 r^20 + a^4 − 2 rr 0 a^2 cos(θ − θ 0 )
which at r = a is a 2 π
1 − (r 0 /a)^2 r^20 + a^2 − 2 ar 0 cos(θ − θ 0 )
The integral on the boundary can be parameterized using θ, which produces an extra factor of a from the arclength differential |dx| = adθ 0. Formula (56) becomes
u(r 0 , θ 0 ) =
2 π
∫ (^2) π
0
(a^2 − r^20 )h(θ) a^2 + r^20 − 2 ar 0 cos(θ − θ 0 ) dθ.
We obtained this result already using separation of variables; it is the Poisson integral formula.
Example: Neumann type boundary conditions. Suppose we want to solve Laplace’s equation in the upper half space {(x, y, z)|z > 0 }, with both a far-field boundary condition and a Neumann condition on the xy-plane:
∆u = 0, (^) zlim→∞ u(x, y, z) = 0, uz (x, y, 0) = h(x, y). (57)
Notice that the Green’s formula (54) has boundary terms than involve both Dirichlet and Neu- mann data. Of course, we only know the derivative of u on the boundary, so we need to make sure that boundary terms involving Dirichlet data will vanish. To make this happen, the Green’s function substituted in for v must have ∂xG/∂n = 0 on the boundary. In other words, we must respect the “boundary condition principle”:
The Green’s function must have the same type of boundary conditions as the problem to be solved, and they must be homogeneous.
For (57), we need a Green’s function which has Gz (x, y, 0; x 0 , y 0 , z 0 ) = 0, and vanishes at infin- ity. The method of images tells us this can be done using the even reflection across the xy-plane. In other words, we want
G(x, y, z; x 0 , y 0 , z 0 ) = G 3 (x, y, z; x 0 , y 0 , z 0 ) + G 3 (x, y, z; x 0 , y 0 , −z 0 )
=
4 π
(x − x 0 )^2 + (y − y 0 )^2 + (z − z 0 )^2
(x − x 0 )^2 + (y − y 0 )^2 + (z + z 0 )^2
It is easy to check Gz = 0 when z = 0. Now substituting v(x) = G(x, x 0 ) into (54), we obtain after collapsing the δ-function integral
u(x 0 ) = −
∂Ω
G(x, x 0 ) ∂u ∂n (x) dx, (58)
since ∆u = 0 and ∂G/∂z vanishes both at infinity and at z = 0. Notice that since ˆn is directed outward relative to Ω, ∂u/∂n = −uz (x, y, 0), so that in coordinates (58) becomes
u(x 0 , y 0 , z 0 ) =
2 π
−∞
−∞
h(x, y) √ (x − x 0 )^2 + (y − y 0 )^2 + z 02
dxdy. (59)
This is another kind of Poisson’s formula suitable for the upper half space.
3.3 Symmetry (reciprocity) of the Green’s function
Occasionally we need to rearrange the arguments of a Green’s function for self-adjoint opera- tors. It turns out that the associated integral operator is self-adjoint as well, which means that G(x, x 0 ) = G(x 0 , x). To demonstrate this fact, use the adjoint identity (52) with v(x) = G(x, x 1 ) and u(x) = G(x, x 2 ). Because of (6), we have Lv = δ(x − x 1 ) and Lu = δ(x − x 2 ). Plugging these into (52) gives (^) ∫
Ω
δ(x − x 1 )G(x, x 2 )dx −
Ω
G(x, x 1 )δ(x − x 2 )dx = 0. (60)
Using the basic property of the δ-function, this simplifies to
G(x 1 , x 2 ) − G(x 2 , x 1 ) = 0. (61)
