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Greens functions, Math 362, 2011
Lecturer: Jari Kaipio (G09, 70 Symonds Street), email: [email protected]
Recommended books:
Stakgold, Greens functions and boundary value problems;
Friedman, Principles and techniques of applied mathematics.
Introduction
In this part of the course we are going to be looking at linear boundary valueproblems; one typical example, that we shall spend a lot of time on, is the steady-
state distribution of heat in one spatial dimension.
d2u
dx2= f(x), 0 < x < 1,
with the boundary conditions
u(0) = ,
u(1) = .
Here, u represents the steady distribution of heat in a one-dimensional regionthat is held at temperature on the left, on the right, and with a heat inputof f(x) along the length of the bar. Its a bit of a toy problem, but not all thatunrealistic for all that.
Note that this is a boundary value problem, where the boundary conditions are
specified, not an initial value problem like the ones you saw in 260. If youve
done 361 you will have seen a lot of boundary value problems, but if you havent
done 361, you probably havent seen them before much.
Because we dont want always to have to work with this particular example,
we use a more general notation. The derivative d2u
dx2can be written as L[u], where
L stands for a linear operator. For our simple example, L = d2
dx2 , but L couldbe any linear differential operator. Another example is L = a2
d2
dx2+ a1
ddx
+ a0,for any constants ai.
1Lecture notes by James Sneyd.
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Derivation of the steady-state heat equation. The full heat, or diffusion
equation isu
t= D
2u
x2+ f(x).
Set the time derivative to be zero (because its at steady state) to get
D2u
x2= f(x),
or, in one spatial dimension with D = 1,
d2u
dx2 = f(x).
So, in general, well be looking at equations of the form
L[u] = f(x),
on some appropriate domain. Often well drop the square brackets and just write
Lu = f.
Now, lets take a short detour. The following is not rigorous, so dont think it
is. Its just a kludgy, intuitive argument to try and make it a bit easier to see the
big picture.
Consider a matrix equation
Ax = b,
where A is a matrix, x is the solution vector, and b is a given vector. When A1
exists, we can solve this explicitly as
x = A1b.
Well, the cool thing is that a similar thing works for linear differential opera-tors. The solution ofLu = f can often (not always) be written as
u = L1f.
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Check that L = a2d2
dx2 + a1d
dx + a0 is a linear operator. (Its clearlya differential operator as it contains derivatives. That is, check thatL[a1u1 + u2] = a1L[u1] + L[u2]).
But what is the inverse of a differential operator? Well, its no surprise that it turns
out to be an integral. In fact, it turns out that, when = = 0 (i.e., the boundaryconditions are zero),
L1f =10
g(x, s)f(s) ds,
for some function g, called the kernel, or the Greens function. In this course (andin much of 763 also), we are going to be looking at ways of calculating g. Its noteasy to do in general.
Note:
u = LL1u = L
g(x, s)u(s) ds
=10
L[g(x, s)]u(s) ds.
SoLg
is some weird function such that
u(x) =
L[g(x, s)]u(s) ds.
This property of Lg is so important that its given its own symbol and name: thedelta function, written as (x s), has the fundamental property that
u(x) =
(x s)u(s)ds.
Youve maybe all seen various versions of this function in other courses. By
putting u(x) = 1 we see that (s) ds = 1. However, we also see that (x) = 0for any x = 0 (prove this if you can). Clearly, no traditional mathematical functioncan have these properties, so the delta function is a weird beast. Stakgolds book,
Chapter 2, shows how this function can be treated rigorously by means of distri-
bution theory, and this is what you learn in 763. We wont go into so much detail
in this course; mostly well just focus on how to calculate the Greens function.
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For our purposes, it suffices to think of (x s) as being zero everywhere
except x = s, but being infinitely large at x = s so that it integrates to 1. Like aninfinitely concentrated impulse.Note that g satisfies its own differential equation:
Lg(x, s) = (x s),
where all the derivatives are taken with respect to x.
Summary
To solve
Lu = f, u(0) = u(1) = 0,
it suffices to solve
Lg(x, s) = (x s),
and then we can write
u(x) =10
g(x, s)f(s) ds.
The Greens function, g(x, s), is the solution to a point impulse applied atx = s.
Example
Lets see how this works in a simple example. We cant yet calculate the Greens
function, but lets see how one works anyway. So Ill have to cheat a little bit.
Consider the equation
d2u
dx2= 1, u(0) = u(1) = 0.
We can solve this directly, to get
u(x) = x2
2+
x
2.
However, we can also solve this equation using the Greens function. You just
have to trust me that, for this equation,
g(x, s) =
(1 s)x, 0 x < s,(1 x)s, s < x 1.
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Well be working this out a bit later. However, just for now lets check to see how
it works. 10
g(x, s)f(s) ds =10
g(x, s) ds
=x0
(1 x)s ds +1x
(1 s)x ds
= x2
2+
x
2,
just as before. Note a couple of things. Firstly, the Greens function is piecewise
defined, which is very often the case, and so, secondly, the integral has to be
broken up into two pieces. You also have to be careful whether or not s < x, and
use the correct bit of the Greens function in each bit of the integral.
Inuitive explanation
Why does this work? Well, one can give an approximate, intuitive explanation,
although dont make the mistake of thinking its rigorous.
We can think of f(x) has consisting of a number of small concentrated im-pulses f(s1)s, f(s2)s, and so on, as in the picture. Remember that f is a heatinput (at least according to the simple model were using) and so a continuous
input of heat along the bar can be approximated by the summation of a lot of con-
centrated inputs. Often, f(x) is called a forcing function, which also makes theintuitive picture reasonably clear. Instead of pushing something continuously, you
can push it by a lot of concentrated little pushes, to get the same overall result.
The response to the impulse at s1 is g(x, s1)f(s1)s, and so on. Thus, sum-
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ming over all the impulses, we get the total response to be
u(x) =n
k=1
g(x, sk)f(sk)s.
Letting n and s 0 we get
u(x) =10
g(x, s)f(s) ds.
Note how this relies absolutely on the fact that L is a linear operator, so thatLu = f is a linear differential equation. This means that the solutions to the sumof two inputs is just the sum of the solutions to each input. Thus, we can solve for
each impulse separately, and then just sum the solutions.This does not not not not not not not happen for nonlineardifferential equa-
tions. Greens functions work ONLY for linear differential equations, just like a
matrix inverse works ONLY for systems of linear equations. If you have a system
of nonlinear equations, a matrix inverse cant be used to calculate the solution, as
youve all seen before.
An initial example in detail
Lets go back to our simple example
d2u
dx2= f(x), 0 < x < 1,
with the boundary conditions
u(0) = 0,
u(1) = 0,
and well suppose that f(x) is applied only over a small region:
f(x) =
0, 0 x < x0 /2,
p, |x x0| /2,0, x0 + /2 < x 1.
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Then,
u(x)
p=
(1 x0)x, 0 x < x0 /2,x2
2 + xx0
x0 +12
12
x0
2
2, |x x0| /2,
(1 x)x0, x0 + /2 < x 1.
Details in class. (You have to assume that both u and u are continuous at x0 /2and x0 + /2.)
Sketch of solution:
What happens as the force gets larger, but is applied over a smaller area? That
is, what happens when p and 0, but in such a way that p = 1?This is the solution to an impulse applied at x = x0, and so this is the Greens
function, or impulse response;
g(x, x0) =
(1 x0)x, 0 x < x0,(1 x)x0, x0 < x 1.
Sketch of solution:
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More direct method
Usually, one doesnt want to have to calculate g(x, s) in this roundabout method.If we think ofg(x, s) as the solution of
d2g(x, s)
dx2= (x s), g(0, s) = g(1, s) = 0, (1)
then (proceeding in a non-rigorous manner) we know that g = 0 everywhereexcept at x = s. Furthermore, if we integrate from x = s to x = s + (i.e.,we just integrate across the place where the impulse occurs), then we get
s+
s
d2g(x, s)
dx2dx = 1,
since the function integrates to 1. Taking the limit as 0 then gives
dg
dx
x=s+
dg
dx
x=s
= 1. (2)
We also assume that g is continuous at x = s.We can now solve for g directly. On region I, 0 < x < s we have g = Ax + B,
for some A and B to be determined. Similarly, on region II, s < x < 1 we haveg = Cx + D. Now apply the boundary conditions and the jump condition, todetermine the unknown constants. Details in class.
We get the same solution as before:
g(x, s) =
(1 s)x, 0 x < s,(1 x)s, s < x 1.
Note that we can use different symbols for the second variable ( s or x0 or ,which is what Stakgold uses). Dont get confused by this variability; the Greens
function is always just doing the same thing.
This trick of integrating the differential equation (1) in order to obtain
the jump condition (2) is one that we are going to be seeing quite a lot,
so make sure youre comfortable with it.
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An important check
I said before that, given g(x, s) as derived above, we can use it to solve the inho-mogeneous problem
d2u
dx2= f(x), 0 < x < 1,
with zero boundary conditions. (If the boundary conditions are non-zero, things
get a little more complicated, and well be dealing with that in a few lectures.)
I claimed that
u(x) =10
g(x, s)f(s) ds.
However, I didnt give a proper proof, not by a long way. So here lets just check
that this formula works, at least in this simple example.First, calculate the derivative of u. Be careful, because g is not differentiableat x = s.
du
dx=
d
dx
x0
g(x, s)f(s) ds +1x
g(x, s)f(s) ds
.
Now use the general formula
d
dx
b(x)a(x)
h(x, s) ds =b(x)a(x)
h
xds + h(x, b(x))
db
dx h(x, a(x))
da
dx
to get
du
dx=
x0
g
xf(s) ds +
1x
g
xf(s) ds + g(x, x)f(x) g(x, x)f(x)
=x0
g
xf(s) ds +
1x
g
xf(s) ds.
Note how the last two terms cancel out since we are assuming that f and g arecontinuous.
However, when we differentiate again, the terms evaluated at x = s no longercancel out, as the derivative ofg is not continuous at x = s.
Differentiating again gives
d2u
dx2=
x0
2g
x2f(s) ds +
1x
2g
x2f(s) ds +
g(x, s)
x
s=x
f(x) g(x, s)
x
s=x+
f(x)
= 0 + 0 + f(x)
g(x, s)
x
s=x
g(x, s)
x
s=x+
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= f(x)g(x, s)
xs=x
g(x, s)
xs=x+
.
Since s = x is equivalent to x = s+ and s = x+ is equivalent to x = s, wehave the desired result.
So weve shown that we do indeed have the correct solution. Just as well.
Examples of Greens functions
Variable conductivity or diffusion
Lets suppose that the heat is diffusing in a rod where the conductivity, or heat
diffusion, is a function of x, k(x) say. Suppose further that one end is insulated.The Greens function then satisfies
d
dx
k(x)
dg
dx
= (x s),
with the boundary conditions
g(0, s) = 0,
g(1, s) = 0. This is the insulated boundary.
Note that Im using a shorthand notation, g
, to denote the derivative ofg withrespect to x, NOT with respect to s.In order to find g we need to convert this to a classical differential equation by
integrating over the domain.
10
d
dx(k(x)g) dx = (x s)
k(x)g|x=s+
x=s = 1
k(s)[g(s+, s) g(s, s)] = 1
g(s+, s) g(s, s) =1
k(s). (3)
This jump condition is almost the same as the previous jump condition (2), it
just has the function k(x) getting in the way.
On each subinterval 0 < x < s and s < x < 1 we have ddx
k(x) dg
dx
= 0.
However, each subdomain has a different boundary condition.
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Domain I: 0 x < s. For convenience, lets call the Greens function on Domain
I g1. On this domain we havedg1dx
=A
k(x), g1(0, s) = 0,
for some unknown constant A. (Well be determining A from the boundary andjump conditions).
Hence x0
dg1d
d =x0
A
k()d,
and so
g1(x, s) g1(0, s) =x0
A
k() d.
Applying the boundary condition, g1(0, s) = 0 we get
g1(x, s) =x0
A
k()d.
Domain II: s < x 1. For convenience, lets call the Greens function onDomain II g2. On this domain we have
dg2dx
= Ak(x)
, g2(1, s) = 0,
and thus g2 = B, for some constant B.We now work out the unknown constants A and B by using the fact that g has
to be continuous, and satisfy the jump condition (3). Firstly, continuity says that
g1(s, s) = g2(s, s),
which gives
B = s
0
A
k()d.
Next, writing (3) in terms of g1 and g2 gives
g2(s, s) g1(s, s) =
1
k(s),
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from which it follows thatA
k(s) =1
k(s) ,
and thus A = 1.Thus, finally, the solution for g is
g(x, s) =
x0
1k()
d, 0 x < s,s0
1k() d, s < x 1.
Infinite domain with diffusion and absorption
Substance u that is diffusing at the same time it is being degraded (at rate q2u; likeradioactive decay, for example), and produced at a rate f(x). For convenience wespecify that q > 0. Steady state is given by
d2u
dx2+ q2u = f(x), < x < .
The equation for the Greens function is
d2g
dx2 + q2
g = (x s), < x, s < .
We can put the impulse at s = 0. Since the domain is infinite the solution willbe the same no matter where the impulse goes, just moved along to left or right.
Domain I: x < 0. Call this part of the Greens function g1. Then
d2g1dx2
+ q2g1 = 0, g bounded at .
So
g1 = Aeqx + Beqx, some unknown constants A,B.
The fact that g1 must be bounded as x means that B = 0. So g1 = Aeqx.Domain II: x > 0. Call this part of the Greens function g2. Then, just as for g1,
g2 = Beqx.
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Now, match up the solutions at x = 0. The condition g1(0) = g2(0) gives
A = B. Finally, what is the jump condition? Integrate the original ODE for gfrom to , where is some small number.
d2g1dx2
dx +
q2g1 dx =
(x) dx = 1.
Letting 0 we see that the second integral tends to zero, since g1 is continuous.Hence, we get the usual jump condition
dg
dx
x=0+
dg
dx
x=0
= 1.
In terms ofg1 and g2 this becomes
g2(0) g1(0) = 1,
and thus
qA + qB = 1.
It follows that A = B = 12q , and thus the Greens function is
g(x, 0) =eq|x|
2q.
If the impulse is not at s = 0 but at a general s, convince yourself that theGreens function is
g(x, s) =eq|xs|
2q.
Question: As q 0 there is no solution for g. Why not? What is thephysical basis for this result?
Picture ofg:
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Method of images
When we have to calculate Greens functions on half-infinite domains, we can use
a nice trick called the method of images. To see how the method works, think of
putting an impulse at x = s and another impulse, of identical amplitude but ofopposite sign, at x = s. Exactly halfway between the impulses, at x = 0, thetwo Greens functions will cancel out, giving zero.
This sum of Greens functions (opposite signs at x = s and x = s couldthus work as a Greens function on the half-infinite domain 0 x < , with theboundary condition g(0, s) = 0.
Lets take a specific example. We want to calculate the Greens function for
the previous problem, but this time on only x > 0. That is, we want to solve
d2g
dx2+ q2g = (x s), 0 < s, x < , g(0, s) = 0.
We could do this by a specific calculation like weve done before. Solving
the equations, getting unknown constants, and using the boundary and jump con-
ditions to determine the constants, but we dont need to. All we need to do is
add a negative Greens function, centred at x = s to a positive Greens functioncentreed at x = s, to get the Greens function to be
g(x, s) =eq|xs|
2q eq|x+s|
2q . (4)
Im not going to prove that this works, but it certainly seems intuitively reason-
able. One thing to note is that the first term in (4) has a discontinuous derivative
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at x = s, but the second term is perfectly well behaved on x > 0. So, although
youre adding two Greens function, you only get one place where the derivativeis discontinuous, as the other discontinuity is outside the domain.
The method can be extended to other types of boundary conditions. Suppose
we wanted a Greens function for the above problem with g(0, s) = 0? Well, ifwe just put positive Greens functions at x = s and x = s the point between, atx = 0, will have zero slope.
We can thus immediately write down the Greens function for
d2g
dx2
+ q2g = (x s), 0 < s, x < , g(0, s) = 0
as
g(x, s) =eq|xs|
2q+
eq|x+s|
2q. (5)
If you dont believe that (5) is the correct Greens function, check for
yourself. Show that it has the correct kink at x = s and check that itdoes indeed have a zero derivative at x = 0. Its often handy to plotthese functions using Maple, just to have a look at them.
A very brief look at distributions
Our treatment of the function has been very sloppy so far. Now well tidy itup a little bit. Not too much, mind you. For a lot more details, see Chapter 2 of
Stakgold.
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The usual way of defining a function y = f(x) is to say that, for every value of
x, we can look up the corresponding value of y. In other words, f is defined in apointwise manner. However, this way of thinking isnt very useful for a functionlike the function, which doesnt behave in this way.
Instead of defining a function like this, we could define a function by how it
acts on other functions. For example, let (x) be some nice, infinitely differen-tiable function. If we know the value of
f(x)(x) dx
for every possible such , then this characterises f completely. The neat thing isthat a definition like this works even when f is notdefined pointwise.
For this to work, the s need to be rather special kinds of functions, calledtest functions.
Definition: A test function (x) is a function that is infinitely differentiable (C)in < x < and has compact support (i.e., it is equal to zero outside somefinite interval). The space of test functions is called D
Example:
(x) =
exp
1
x21
, |x| < 1,
0, |x| 1.
Just for giggles: (i) Show that 1 D, 2 D c11 + 2 D,
where c1, c2 R. (ii) Show that D D.
Definition: Let n be a sequence in D. Then
n 0 (in D)
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if and only if
1. All the n vanish outside a common finite interval, and
2. Every n, and every derivative ofn, tends to zero uniformly in x.
If youre not familiar with the concept of uniform convergence, its not so
important for this course. I just state the definition here for completeness, so that
those of you who know about uniform convergence can use it if they wish.
Definition: IfF is a linear mapping from D to the real or complex numbers, thenF is said to be continuous if
n 0 (in D) F(n) 0.
A continuous linear mapping on D is called a distribution. The set of all distribu-tions is denoted by D.
Notation: Iff is some distribution, well write the action off on as f, .
At this stage this definition might seem a little arbitrary and not very useful.
Actually, the opposite is true. Although we dont deal with the formal theory
of distributions in this course, they are a vital concept for many areas of appliedmathematics, and youll see them a lot more in 763, should you do that course.
Mainly the distributions well be using in this course is the function and itsderivatives. So lets use this as our first example.
Example: The distribution is defined by
, = (0).
Actually, this statements needs proof. In other words, if we define this way, doesit turn out to be a distribution? Well, yes it does, but we have to prove this. We
have to show that this is a continuous linear functional from D to R. Linearityis immediate. Continuity follows since if n 0 D it follows that (0) 0(since the n are converging uniformly to zero). Hence , 0 also, and thus D.
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Usually we wont bother with these sorts of proofs.
Example:
H, =0
(x) dx
defines H D. (If you want to, prove this for yourself.) This distribution H isgenerated by the classical Heaviside function
H(x) =
0, x < 0,1, 0 < x.
Differentiation of distributions
The neat thing about this definition of functions like or H, is that it lets usdefine rigorously what we mean by derivatives of these functions. For motivation,
suppose f(x) is a nice smooth integrable function with |f(x)| dx < . We
can use f to define a distribution as
f, =
f(x)(x) dx.
(The fact that this defines a distribution is something you can prove for yourself.)Then,
f, =
f(x)(x) dx
=
f(x)(x) dx + f(x)(x)|
=
f(x)(x) dx.
Note how the boundary terms disappear because is zero outside some boundeddomain, and thus certainly zero at . Thus, the nice properties of have let us
transfer the derivatives from f to , at only the cost of a change of sign. This is areally important concept that we shall see again and again in this course. Its done
by integration by parts, and so we all love integration by parts.
Now, we use this to motivate our definition of the derivative of a distribution.
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Definition: Ift D (i.e., t is a distribution), we define the derivative of t by
for every D, t, = t, = t, .
Example: Lets calculate the derivative of the distribution H we saw before. Forany D
H, = H, = 0
(x) dx = (0).
Since for every DH, = ,
we say that H = (in the sense of distributions). Indeed, this is the definitionof what we mean when we say that two distributions are equal. They cant be
compared at each point, as they are not necessarily defined pointwise, but they
have the same action on any test function.
Example: What about the derivative of? For any D
, = , = (0).
Note that we have done something very striking. We have defined the deriva-
tives of functions that are not differentiable in the classical sense.
Example: One final example.
|x|, =
|x|
= 0
x 0
(x)
=0
0
=
(sgn x)
= sgn, .
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Thus
|x|
= sgn x.Here, we can think of sgn (an abbreviation of sign) either as being the distri-
bution defined by
sgn, =0
0
,
or as the classical function
sgn (x) =
1, x < 0,1, 0 < x.
Weak solutions
We now use this idea of distributions to define special kinds of solutions to dif-
ferential equations, solutions that are not necessarily even differentiable (just like
the Greens function).
Suppose we want to solve
u = f(x), a < x < b. (6)
We can either find some differentiable function u, such that u(x) = f(x) forevery x in the domain (this is the usual classical solution). OR, we could find adistribution, u, such that, for any D,
u, = f, ,
i.e., such that
u, = f, .
The distribution u is called a weak solution of the differential equation, and doesnot have even to be differentiable in the classical sense.
Weve already seen an example of a weak solution; the Greens function. As
another example, lets calculate a weak solution to (6).
Example: Find a weak solution of
g = (x s), 0 < x, s < , g(0) = 0.
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Use the usual trick of integrating from s to s + and let 0 to get
g(s+) g(s) = 1.
Hence, g = 0 everywhere except at x = s, and there is a jump of 1 in the valueof g at x = s. Clearly, the solution is g = 0 when x < s and g = 1 when x > s,i.e., g(x, s) = H(x s).
Check: this is a weak solution if, for every D, g, = (x s), =(s). But
g, = g, = 0
H(t s)(s) ds = (s),
as required. So, it all works out.
Note that, since weve already shown that H = (page 19), we couldhave written this answer down immediately, without going through all
those details. This is a dangerous procedure, though, and one we shall
usually avoid. But be warned; it is not easy to find weak solutions in
general. It can be a lot easier to check them, once they are found, but
finding them in the first place is not easy at all. In this course we are
only looking at some of the simplest and best known.
More general operators
So far weve constructed weak solutions for first order and second order ODEs.
We now want to define what is a weak solution for a more general linear operator,
and show how to find them. Before we do this, we need to do a little preliminary
work to define the adjoint of a differential operator. Well only look at second
order operators, as higher orders introduce a lot more algebra but no more under-
standing.
The adjoint operatorLet L be the linear differential operator
L = a2(x)d2
dx2+ a1(x)
d
dx+ a0(x).
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Then
ba
(Lu)v =ba
(a2u + a1u + a0)v
=ba
u(a2v) +ba
u(a1v) +ba
a0uv
= ba
u(a2v)
ba
u(a1v) +
ba
a0uv + [u(a2v) + ua1v]|
b
a
=ba
u(a2v)
ba
u(a1v) +
ba
a0uv + [u(a2v) + ua1v u(a2v)
]|b
a .
This is usually written as
Lagranges Identity:
ba
(Lu)v ba
uLv = J(u, v)ba
,
where
Lv = (a2v) (a1v)
+ a0v,
and
J(u, v) = a2(vu uv) + uv(a1 a
2).
L is called the formal adjoint of L, and J is called the conjunct of uand v. If L = L, L is said to be formally self-adjoint. Note that thisis not the same as being self-adjoint, which requires consideration of
boundary conditions also, as we shall see shortly.
Since Lagranges identity is valid for any b, we can set b = x and then differ-entiate with respect to x to get
Greens formula:
(Lu)v uLv =d
dxJ(u, v).
Example: If L = ddx
then L = ddx
. This is because
Lu =
u =u() =
uL.
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Example: If L = d2
dx2then L = d
2
dx2= L, so L is formally self-adjoint. Again,
just use integration by parts, ignoring the boundary terms.
Lu =
u
=
u() =
u =
uL.
Weak solutions of more general operators
Given a second order linear differential operator, L, a weak solution of
Lu = v
is a distribution, u D, such that, for every D
u, L
= v, .
Question: can we even write down u, L? Is it defined? Well, yes,it is, since L D as long as is. Think about why this is so. Welldiscuss this in class also.
Actually, its not strictly accurate to call all such u weak solutions. For in-stance, ifu corresponds to a nice differentiable function, then its also a solutionin the classical sense. However, if u corresponds to a function for which Ludoesnt exist in the classical sense, then u is called a weak solution. (If u doesntcorrespond to any function at all, its called a distributional solution, but we wont
be looking at any such solutions in this course.) All three types of solutions are
called generalised solutions.
Fundamental solutions
Weve seen some examples of Greens functions, that satisfy
Lu = (x s), a < x < b,
together with some boundary conditions at x = a and x = b. If the domain is thewhole real line, then the Greens function is given a special name; its called the
fundamental solution. This might seem like a bit of a silly distinction to make, butits actually quite useful.
In a more formal definition, a fundamental solution for L with pole at s is asolution of
Lu = (x s),
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where this solution is to be interpreted in the sense of distributions, i.e., that for
every D, u, L = (s).
The fundamental solution is not unique, as the addition of any solution of the
equation Lu = 0 will give another fundamental solution.
The fundamental solution is usually constructed using the general ho-
mogeneous solution, with unknown constants. To get a Greens func-
tion the boundary conditions are then used to determine the unknown
constants. So
Greens function = fundamental solution + boundary conditions.
Lets see how this works in detail in a particular example.
Example:
d2E
dx2+ q2E = (x s), < x < .
We have already calculated the fundamental solution for this equation (see page
12) as
E =
eq|xs|
2q .
At the time we called this solution a Greens function, but we now know its
actually a fundamental solution as its on an infinite domain.
Now, this solution isnt unique, as we could add any solution of d2Ehdx2
= q2Ehto get another fundamental solution, i.e., we could add Eh = Ae
qx+Beqx for anyconstants A and B. Mind you, since every Eh is unbounded on < x < wed get an unphysical and nasty fundamental solution, but lets not worry about
that right now.
How would we go about constructing a Greens function, for, say,
d
2
gdx2 + q2g = (x s), g(0) = g(1) = 0?
Well, we could just follow the usual procedure of splitting the domain into two
regions, one where x < s the other where x > s, solving for g in each separate
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region and then joining up the solutions at x = s by using the continuity and jump
conditions.However, weve already done most of that work on page 12 so theres little
point in doing it again. Instead, we use the fact that we already know the funda-
mental solution, which automatically satisfies the continuity and jump conditions
at x = s. Thus, we use the fact that
g = E+ Eh =eq|xs|
2q+ Aeqx + Beqx,
for some A and B to be determined from the boundary conditions.At x = 0 we have
0 = g(0) = A + B +
eqs
2q ,
while at x = 1 we have
0 = g(1) =eq(1s)
2q+ Aeq + Beq.
Two equations, two unknowns, solve for A and B.
Boundary value problems for second order equations
So far, we havent paid much attention to the question of whether or not a solu-
tion to a boundary value problem actually exists. This is what we look at now.
Obviously, its an important question.
We shall consider the differential equation
Lu = f(x), a < x < b,
together with the boundary conditions
B1u = 11u(a) + 12u(a) + 11u(b) + 12u
(b) = 1,
B2u = 21u(a) + 22u(a) + 21u(b) + 22u(b) = 2.
This serves as the definition of B1 and B2. We use this notation purely becauseits easier to write B1u = 1 than to write out the whole boundary condition everytime. This is called laziness. As a side note, the vectors (11, 12, 11, 12) and
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(21, 22, 21, 22) have to be independent, otherwise the boundary conditions are
either identical or inconsistent, neither of which case is attractive.Boundary conditions of this type occur frequently in applications, and give the
most easily developed theory, which is why we restrict ourselves to this type.
The corresponding homogeneous equation is
Lu = 0, B1u = B2u = 0.
Now, lets look at an example that shows how things can get tricky.
Example:
u = f(x), 0 < x < 1, u(0) = 1, u(1) = 2.
If f(x) = 1 = 2 = 1 then no solution exists. The general solution is u =x2/2 + Bx + C; if you try to solve for the boundary conditions you then getB = 1 and B = 2, which is inconsistent.
However, if f(x) = sin(2x) then there are many solutions, u(x) = A x/(2)+(1/(42)) sin(2x), for any value ofA. Clearly, sometimes such bound-ary value problems have solutions, and other times they dont. So care must be
taken.
Question: In the previous example, show that
10 f(x) = 1 + 2, and
thus f(x) = 1 and 1 = 2 = 0 still wont work. Whats the physicalinterpretation of this?
Nice facts
1. Any two solutions ofLu = f, B1(u) = B2(u) = 0 differ by a solution ofthe homogeneous equation. (For, if Lu1 = f and Lu2 = f, then L(u1 u2) = 0.)
2. IfLu = 0, B1(u) = B2(u) = 0, has only the trivial solution, the solution toLu = f, B1(u) = 1, B2(u) = 2, exists and is unique. (Im not going toprove this, but its by construction. Full details in Stakgold.)
3. If Lu = 0, B1(u) = B2(u) = 0, has a nontrivial solution, then Lu = fhas either no solution or many solutions. This is a version of the Fredholm
Alternative that (I hope) we shall have time to look at later.
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Example: Note that, it follows from these nice facts that the Greens function
doesnt always exist. For example, suppose we try and solve
u = 1, 0 < x < 1, u(0) = 0, u(1) = 0,
by using a Greens function. Well, the Greens function satisfies
g = (x s), g(0, s) = g(1, s) = 0,
which says that g = 0 everywhere except x = s and that g has a jump in thederivative at x = s. However, g = 0, g(0) = 0 implies that g is a constant onx < s, and similarly for x > s. So g has to be piecewise constant, and can neversatisfy the jump condition on the derivative. So you cant solve for g.
Of course the difficulty is that the completely homogeneous problem
u = 0, 0 < x < 1, u(0) = 0, u(1) = 0,
has an infinite number of nontrivial solutions, i.e., any constant function, in which
case we see from our nice results that solutions to Lu = f do not necessarily exist.In this case, they dont.
Nonzero boundary conditions
Recall that the solution to
Lu = f, B1u = B2u = 0
can be written as u =
g(x, s)f(s)ds (as long as there are no nontrivial solutionswhen f = 0), where
Lg = (x s), B1g = B2g = 0.
A word of warning: This does NOT work if the boundary conditions are
nonzero. For example, if we want to solve
Lu = f, B1u = 1, B2u = 2
then we might indeed be able to solve Lg = , B1g = 1, B2g = 2, but thisdoesnt help us, as it is NOT TRUE that the solution can then be written as u =
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g(x, s)f(s)ds. For although Lu = L
gf =
(Lg)f =
(x s)f(s) = f(x),
and thus this bit works, the boundary conditions do not, since B1u = B1
gf =(Bg)f = 1f which does not equal 1 necessarily.So we have to be cleverer. Or more clever, whichever is the better grammar.
First, let u1 and u2 be nontrivial solutions ofLu = 0, with B1u1 = B2u2 = 0.Note that, since the fully homogeneous problem has no nontrivial solutions, we
must have that B1u2 = 0 and B2u1 = 0.Now let g be the Greens function, i.e.,
Lg = (x s), B1g = B2g = 0,
and write
u(x) =ba g(x, s)f(s) ds + c
1u1 + c2u2,
for some constants c1 and c2 to be determined.Apply the boundary conditions to get
1 = B1(u) = B1
ba
g(x, s)f(s) ds + c1u1 + c2u2
=
B1(g)f(s) ds + c1B1u1c2B1u2
= c2B1u2,
since B1(g) = B1(u1) = 0.Similarly,2 = c1B2u1.
It thus follows that
c1 =2
B2u1
c2 =1
B1u2,
and so
u(x) =ba g(x, s)f(s) ds +
2
B2u1u1(x) +
1
B1u2u2(x).
Example:
d2u
dx2= f(x), 0 < x < 1,
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with the boundary conditions
u(0) = ,
u(1) = .
We can solve this as follows:
Step 1: Find a u1 such that u1 = 0, u1(0) = 0 (i.e., Lu1 = 0, B1u1 = 0). This
gives u1 = Ax, for any A.
Step 2: Find a u2 such that u2 = 0, u2(1) = 0 (i.e., Lu2 = 0, B2u2 = 0). This
gives u2 = B(1 x), for any B.
Step 3: Notice that B1u2 = B, while B2u1 = A, and thus get the solution
u(x) =ba
g(x, s) ds +2A
u1 +2B
u2
=ba
g(x, s) ds + (1 x) + x.
Self-adjoint operators and boundary conditions
We have already defined the adjoint to an operator (see page 21). Just to remind
you, the adjoint is defined by using integration by parts to switch derivatives, and
Lagranges identity is
ba
(Lu)v ba
uLv = J(u, v)|ba ,
where J(u, v), the conjunctofu and v is just all the stuff that is left over after theintegration by parts.
J(u, v) = a2(vu uv) + uv(a1 a
2).
IfL = L then L is said to be formally self-adjoint.The general second-order linear operator is formally self-adjoint if and only if
L =d
dx
a2(x)
d
dx
+ a0(x).
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Now, lets have a look at how the boundary conditions come into it.
Definition: Let the set of functions that satisfy B1u = B2u = 0 be denote byD. Then, J(u, v)|ba will vanish if and only if v satisfies a (possibly different) setof boundary conditions, which we shall call B1v = B
2v = 0. Call this set of
functions, D. B1 and B2 are called the adjoint boundary conditions.
IfL = L and D = D then L is said to self-adjoint.
Note that it doesnt make sense to say that L is or isnt self-adjoint, as theboundary conditions must always be taken into account also. So you can only say
that L, with a certain set of boundary conditions, is or is not self-adjoint.
Example: Lu = a2u + a1u + a0u, B1u = u(a) = 0, B2u = u(b) = 0. Then Dis the set of all functions that are zero at a and have zero derivative at b. Not veryexciting, really. A short calculation then gives
J(u, v)|ba =
[a1(b) a2(b)]v(b) a2(b)v
(b)
u(b) a2(a)v(a)u(a).
So, in order for J(u, v)|ba to vanish, we have to have
v(a) = 0,
[a1(b) a2(b)]v(b) a2(b)v
(b) = 0.
Hence, D is the set of all functions that satisfy these two equations. Since D =D, L (with these boundary conditions) is not self-adjoint.
Ifa1 = a2 then D = D
and L = L in which case L is self-adjoint.
Example: L = d2/dx2, u(0) = u(1) = 0. First, calculate L.
10
(Lu)v =10
uv =10
vu + (vu uv)|10 .
Thus, L = L and so L is formally self-adjoint. Also,
J(u, v)|10 = u(1)v(1) u(0)v(0)
and thus J(u, v)|10 vanishes only when v(1) = v(0) = 0. Thus, D = D, and so
L, with these boundary conditions, is self-adjoint.
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Nice things
Nice property The adjoint Greens function satisfies the equation
Lh(x, ) = (x ), B1h = B2h = 0.
Since the Greens function satisfies the equation
Lg(x, s) = (x s), B1g = B2g = 0
it follows that(hLggLh) =
h(x, )(xs)
g(x, s)(x) = h(s, )g(, s).
IfL is self-adjoint then h(x, y) = g(x, y) and
(hLg gL
h) = 0, in whichcase it follows that h(s, ) g(, s) = 0 and thus
g(x, s) = g(s, x).
Thus, the variables in the Greens function of a self-adjoint operator are
interchangeable.
This is known as the reciprocity principle. The effect at x of an impulse ats is the same as the effect at s of an impulse at x. Kind of makes physicalsense for a lot of systems.
Nice result Suppose that
Lu = f, B1u = , B2u = ,
and let h be the adjoint Greens function. Then
J(u, h)|ba =ba
(h(x, )Lu u(x)Lh(x, )) dx
=ba
h(x, )f(x) dx ba
u(x)(x ) dx
=ba
h(x, )f(x) dx u().
If = = 0 then we know that J(u, h)|b
a = 0 by definition, since usatisfies B1u = B2u = 0 and h satisfies B
1h = B
2h = 0. Hence, if the
boundary conditions are homogeneous we just get
u() =ba
h(x, )f(x) dx =ba
g(, x)f(x) dx,
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which is just the identity weve seen before.
However, if and are nonzero we get
u() =ba
g(, x)f(x) dx J(u(x), g(, x))|ba ,
which gives another way of solving the problem with inhomogeneous bound-
ary conditions.
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