MATH 401: GREEN’S FUNCTIONS AND VARIATIONAL METHODS
The main goal of this course is to learn how to “solve” various
differential equation) problems arising in science. Since only very
problems have solutions we can write explicitly, “solve” here often
find an explicit approximation to the true solution, or (b) learn
qualitative properties of the solution.
PDE (= partial
few special such
means either (a)
some important
A number of techniques are available for this. The method of separation of variables
(together with Fourier series), perhaps familiar from earlier courses, unfortunately
only applies to very special problems. The related (but more general) method of
eigenfunction expansion is very useful – and will be discussed in places in this course.
Integral transform techniques (such as the Fourier transform and Laplace transform)
will also be touched on here. If time permits, we will also study some perturbation
methods, which are extremely useful if there is a small parameter somewhere in the
problem.
The bulk of this course, however, will be devoted to the method/notion of Green’s
function (a function which, roughly speaking, expresses the effect of the “data” at one
point on the solution at another point), and to variational methods (by which PDE
problems are solved by minimizing (or maximizing) some quantities). Variational
methods are extremely powerful – and even apply directly to some nonlinear problems
(as well as linear ones). Green’s functions are for linear problems, but in fact play a key
role in nonlinear problems when they are treated as – in some sense – perturbations
of linear ones (which is frequently the only feasible treatment!).
We begin with Green’s functions.
A. GREEN’S FUNCTIONS
I. INTRODUCTION
As an introductory example, consider the following initial-boundary value problem
for the inhomogeneous heat equation (HE) in one (space) dimension

2
 ∂u
= ∂∂xu2 + f (x, t) 0 < x < L, t > 0 (HE)
∂t
u(0, t) = u(L, t) = 0
(BC)

u(x, 0) = u0 (x)
(IC)
which, physically, describes the temperature u(x, t) at time t and at point x along a
“rod” 0 ≤ x ≤ L of length L subject to (time- and space-varying) heat source f (x, t),
and with the ends held fixed at temperature 0 (boundary condition (BC)) and with
initial (time t = 0) temperature distribution u0 (x) (initial condition (IC)).
1
This problem is easily solved by the (hopefully) familiar methods of separation of
variables and eigenfunction expansion. Without going into details, the eigenfunctions
of the spatial part of the (homogeneous) PDE (namely d2 /dx2 ) satisfying (BC) are
sin(nπx/L), n = 1, 2, 3, . . . (with corresponding eigenvalues −n2 π 2 /L2 ), and so we
seek a solution of the form
u(x, t) =
∞
X
an (t) sin(nπx/L).
n=1
We similarly expand the data of the problem – the source term f (x, t) and the initial
condition u0 (x) – in terms of these eigenfunctions; that is, as Fourier series:
Z
∞
X
2 L
f (x, t) =
fn (t) sin(nπx/L),
fn (t) =
f (x, t) sin(nπx/L)dx
L 0
n=1
Z
∞
X
2 L
u0 (x) =
gn sin(nπx/L),
gn =
u0 (x) sin(nπx/L)dx.
L 0
n=1
Plugging the expression for u(x, t) into the PDE (HE) and comparing coefficients
then yields the family of ODE problems
a0n (t) + (n2 π 2 /L2 )an (t) = fn (t),
an (0) = gn
which are easily solved (after all, they are first-order linear ODEs) by using an integrating factor, to get
Z t
−(n2 π 2 /L2 )t
(n2 π 2 /L2 )s
an (t) = e
gn +
e
fn (s)ds ,
0
and hence the solution we sought:
Z
∞
2 X −(n2 π2 /L2 )t L
e
g(y) sin(nπy/L)dy
u(x, t) =
L n=1
0
Z t
Z L
(n2 π 2 /L2 )s
+
e
f (y, s) sin(nπy/L)dyds sin(nπx/L).
0
0
No problem. But it is instructive to re-write this expression by exchanging the order
of integration and summation (note we are not worrying here about the convergence
of the sum or the exchange of sum and integral – suffice it to say that for reasonable
(say continuous) functions g and f all our manipulations are justified and the sum
converges beautifully due to the decaying exponential) to obtain
Z L
Z tZ L
u(x, t) =
G(x, t; y, 0)u0 (y)dy +
G(x, t; y, s)f (y, s)dyds
(1)
0
0
2
0
where
∞
G(x, t; y, s) =
2 X −(n2 π2 /L2 )(t−s)
e
sin(nπy/L) sin(nπx/L).
L n=1
Expression (1) gives the solution as an integral (OK, 2 integrals) of the “data” (the
initial condition u0 (x) and the source term f (x, t)) against the function G which is
called, of course, the Green’s function for our problem (precisely, for the heat equation
on [0, L] with 0 boundary conditions).
Our computation above suggests a few observations about Green’s functions:
• if we can find the Green’s function for a problem, we have effectively solved the
problem for any data – we just need to plug the data into an integral like (1)
• a Green’s function is a function of 2 sets of variables – one set are the variables
of the solution (x and t above), the other set (y and s above) gets integrated
• one can think of a Green’s function as giving the effect of the data at one point
((y, s) above) on the solution at another point ((x, t))
• the domain of a Green’s function is determined by the original problem: in the
above example, the spatial variables x and y run over the interval [0, L] (the
“rod”), and the time variables satisfy 0 ≤ s ≤ t – here the condition s ≤ t
reflects the fact that the solution at time t is only determined by the data at
previous times (not future times)
The first part of this course will be devoted to a systematic study of Green’s functions,
first for ODEs (where computations are generally easier), and then for PDEs, where
Green’s functions really come into their own.
3
II. GREEN’S FUNCTIONS FOR ODEs
1. An ODE boundary value problem
Consider the ODE (= ordinary differential equation) boundary value problem
Lu := a0 u00 + a1 u0 + a2 u = f (x) x0 < x < x1
u(x0 ) = u(x1 ) = 0.
(2)
Here
d2
d
+ a2 (x)
+
a
(x)
1
dx2
dx
is a (first-order, linear) differential operator. As motivation for problem (2), one can
think, for example, of u(x) as giving the steady-state temperature along a rod [x0 , x1 ]
with (non-uniform) thermal conductivity ρ(x), subject to a heat source f (x) and with
ends held fixed at temperature 0, which leads to the problem
L := a0 (x)
(ρ(x)u0 )0 = f,
u(x0 ) = u(x1 ) = 0
of the form (2). Zero boundary conditions are the simplest, but later we will consider
other boundary conditions (for example, if the ends of the rod are insulated, we should
take u0 (x0 ) = u0 (x1 ) = 0).
We would like to solve (2) by finding a function G(x; z), the Green’s function, so
that
Z x1
u(x) =
G(x; z)f (z)dz =: (Gx , f )
x0
where we have introduced the notations
Gx (z) := G(x; z)
(when we want to emphasize the dependence of G specifically on the variable z,
thinking of x as fixed), and
Z x1
(g, f ) :=
g(z)f (z)dz ( inner product ).
x0
Then since u solves Lu = f , we want
u(x) = (Gx , f ) = (Gx , Lu).
Next we want to “move the operator L over” from u to Gx on the other side of the
inner-product – for which we need the notion of adjoint.
Definition: The adjoint of the operator L is the operator L∗ such that
(v, Lu) = (L∗ v, u) + “ boundary terms “
4
for all smooth functions u and v.
The following example illustrates the adjoint, and explains what is meant by “boundary terms”.
2
d
d
Example: Let, as above, L = a0 (x) dx
2 + a1 (x) dx + a2 (x) acting on functions defined
for x0 ≤ x ≤ x1 . Then for two such (smooth) functions u(x) and v(x), integration by
parts gives (check it!)
Z x1
v[a0 u00 + a1 u0 + a2 u]dx
(v, Lu) =
Zx0x1
x
u[a0 v 00 + (2a00 − a1 )v 0 + (a2 + a000 − a01 )v]dx + [a0 vu0 − a00 v 0 u + a1 uv]x10
=
x0
x
= a0 v 00 + (2a00 − a1 )v 0 + (a2 + a000 − a01 )v, u + [a0 (vu0 − v 0 u) + (a1 − a00 )uv]x10 .
The terms after the integral (the ones evaluated at the endpoints x0 and x1 ) are what
we mean by “boundary terms”. Hence the adjoint is
L ∗ = a0
d2
d
+ (2a00 − a1 ) + (a2 + a000 − a01 ).
2
dx
dx
The differential operator L∗ is of the same form as L, but with (in general) different
coefficients.
An important class of operators are those which are equal to their adjoints.
Definition: An operator L is called (formally) self-adjoint if L = L∗ .
2
d
d
Example: Comparison of L and L∗ for our example L = a0 dx
2 + a1 dx + a2 shows that
0
L is formally self-adjoint if and only if a0 = a1 . Note that in this case
Lu = a0 u00 + a00 u0 + a2 u = (a0 u0 )0 + a2 u
which is an ordinary differential operator of Sturm-Liouville type.
Now we can return to our search for a Green’s function for problem (2):
u(x) = (Gx , f ) = (Gx , Lu) = (L∗ Gx , u) + BT
where we know from our computations above that the “boundary terms” are
x
x
BT = a0 Gx u0 − a00 G0x u + a1 Gx u x10 = a0 Gx u0 x10
where we used the zero boundary conditions u(x0 ) = u(x1 ) = 0 in problem (2). We
can make the remaining boundary term disappear if we also impose the boundary
conditions Gx (x0 ) = Gx (x1 ) = 0 on our Green’s function G. Thus we are led to the
problem
u(x) = (L∗ Gx , u),
Gx (x0 ) = Gx (x1 ) = 0
5
for G.
So L∗ Gx should be a function g(z) which satisfies
Z x1
g(z)u(z)dz
u(x) = (g, u) =
x0
for all (nice) functions u. What kind of function is this? In fact, it is a (Dirac) delta
function, which s not really a function at all! Rather, it is a generalized function, a
notion we need to explore further before proceeding with Green’s functions.
6
2. Generalized functions (distributions).
The precise mathematical definition of a generalized function is:
Definition: A generalized function or distribution is a (continuous) linear functional acting on the space of test functions
C0∞ (R) = { infinitely differentiable functions on R which vanish outside some interval }
(an example of such a test function is
−1/(a2 −x2 )
e
−a < x < a
φ(x) =
0
|x| ≥ a
for any a > 0). That is, a distribution f maps a test function φ to a real number
Z
f (φ) = (f, φ) = “ f (x)φ(x)dx“
which it is useful to think of as an integral (as in the parentheses above) – hence the
common notation (f, φ) for f (φ) – but is not in general an integral (it cannot be –
since f is not in general an actual function!). Further, this map should be linear: for
test functions φ and ψ and numbers α and β,
f (αφ + βψ) = αf (φ) + βf (ψ)
or (f, αφ + βψ) = α(f, φ) + β(f, ψ) .
Some examples should help clarify.
Example:
1. If f (x) is a usual function (say, a piecewise continuous one), then it is also
a distribution (after all, we wouldn’t call them “generalized functions” if they
didn’t include regular functions), which acts by integration,
Z
f (φ) = (f, φ) = f (x)φ(x)dx
which is indeed a linear operation. This is why we use inner-product (and
sometimes even integral) notation for the action of a distribution – when the
distribution is a real function, its action on test functions is integration.
2. The (Dirac) delta function denoted δ(z) (or more generally δx (z) = δ(z − x)
for the delta function centred at a point x) is not a function, but a distribution,
whose action on test functions is defined to be
Z
(δ, φ) := φ(0)
= “ δ(z)φ(z)dz“
Z
(δx , φ) := φ(x)
= “ δ(z − x)φ(z)dz“
7
(where, again, the integral here is just notation). That is, δ acts on test functions
by picking out their value at 0 (and δx acts on test functions by picking out their
value at x).
Generalized functions are so useful because we can perform on them many of the
operations we can perform on usual functions. We can
1. Differentiate them: if f is a usual differentiable function, then for a test function
φ, by integration by parts,
Z
Z
0
0
(f , φ) = f (x)φ(x)dx = − f (x)φ0 (x)dx = (f, −φ0 )
(there are no boundary terms because φ vanishes outside of some interval). Now
if f is any distribution, these integrals make no sense, but we can use the above
calculation as the definition of how the distribution f 0 acts on test functions:
(f 0 , φ) := (f, −φ0 )
and by iterating, we can differentiate f n times:
(f (n) , φ) = (f, (−1)n φ(n) ).
Example:
(a) the derivative of a delta function:
(δ 0 , φ) = (δ, −φ0 ) = −φ0 (0)
(b) the derivative of the Heavyside function
0 x≤0
H(x) :=
1 x>0
(which is a usual function, but is not differentiable in the classical sense at
x = 0):
Z
Z ∞
0
0
0
φ0 (x)dx
(H , φ) = (H, −φ ) = H(x)(−φ (x))dx) = −
0
=
−φ(x)|x=∞
x=0
= φ(0)
(since φ vanishes outside an interval). Hence
d
H(x) = δ(x).
dx
8
The fact that we can always differentiate a distribution is what makes them so
useful for differential equations.
2. Multiply them by smooth functions: if f is a distribution and a(x) is a smooth
(infinitely differentiable) function we define
(a(x)f, φ) := (f, a(x)φ)
(which makes sense, since aφ is again a test function). Note this definition
conincides with the usual one when f is a usual function.
3. Consider convergence of distributions: we say that a sequence {fj }∞
j=1 of distributions converges to another distribution f if
lim (fj , φ) = (f, φ)
for all test functions φ.
j→∞
This kind of convergence is called weak convergence.
R
Example: Let ψ(x) be a smooth, non-negative function with ψ(x)dx = 1, and
set ψj (x) := jψ(jx) for j = 1, 2, 3, . . .. Note that as j increases, the graph
of ψj (x) is becoming
both taller, and more concentrated near x = 0, while
R
maintaining ψj (x)dx = 1. In fact, we have
lim ψj (x) = lim jψ(jx) = δ(x)
j→∞
j→∞
in the weak sense – it is a nice exercise to show this!
4. Compose them with invertible functions: let g : R → R be a one-to-one and
onto differentiable function, with g 0 (x) > 0. If f is a usual function, then by
changing variables y = g(x) (so dy = g 0 (x)dx), we have for the composition
f ◦ g(x) = f (g(x)),
Z
Z
1
(f ◦g, φ) = f (g(x))φ(x)dx = f (y)φ(g −1 (y))dy/g 0 (g −1 (y)) = (f, 0 −1 φ◦g −1 )
g ◦g
and so for f a distribution, we define
1
(f ◦ g, φ) := (f, 0 −1 φ ◦ g −1 ).
g ◦g
Example: Composing the delta function with g(x) gives
(δ(g(x)), φ) = (δ,
1
φ(g −1 (0))
−1
φ
◦
g
)
=
,
g 0 ◦ g −1
g 0 (g −1 (0))
and in particular if g(x) = cx (constant c > 0)
1
1
(δ(cx), φ) = φ(0) = ( δ, φ)
c
c
1
and hence δ(cx) = c δ(x).
9
3. Green’s functions for ODEs.
Returning now to the ODE problem (2), we had concluded that we want our Green’s
function G(x; z) = Gz (z) to satisfy
u(x) = (L∗ Gx , u) ∀ u,
Gx (x0 ) = Gx (x1 ) = 0.
After our discussion of generalized functions, then, we see that what we want is really
L∗ Gx (z) = δ(z − x),
Gx (x0 ) = Gx (x1 ) = 0.
Notice that for z 6= x, we are simply solving L∗ Gx = 0. The strategy is to solve this
equation for x < z and for z > x, and then “glue the two pieces together”. Some
examples should help clarify.
Example: use the Green’s function method to solve the problem
u00 = f (x), 0 < x < L,
u(0) = u(L) = 0.
(3)
(which could, of course, be solved simply by integrating twice).
2
d
∗
(the operator is self-adjoint), so the problem for our
First note that L = dx
2 = L
Green’s function G(x; z) = Gx (z) is
G00x (z) = δ(z − x),
(here 0 denotes
d
).
dz
Gx (0) = Gx (L) = 0
For z < x and z > x, we have simply G00x = 0, and so
Az + B 0 ≤ z < x
Gx (z) =
Cz + D x < z ≤ L
The BC Gx (0) = 0 implies B = 0, and the BC Gx (L) = 0 implies D = −LC, so we
have
Az
0≤z<x
Gx (z) =
C(z − L) x < z ≤ L
Now our task is to determine the remaining two unknown constants by using matching
conditions to “glue” the two pieces together:
1. continuity: we demand that Gx be continuous at z = x: Gx (x−) = Gx (x+) (the
notation here is g(x±) := limε↓0 g(x ± ε)). This yields Ax = C(x − L).
2. jump condition: for any ε > 0, integrating the equation G00x = δ(z − x) between
x − ε and x + ε yields
Z x+ε
Z x+ε
0
0
0 x+ε
00
Gx (x + ε) − Gx (x − ε) = Gx x−ε =
Gx (z)dz =
δ(z − x)dz = 1
x−ε
and letting ε ↓ 0, we arrive at
G0x (x+) − G0x (x−) = 1.
This jump condition requires C − A = 1.
10
x−ε
Solving the two equations for A and C yields C = x/L, A = (x − L)/L, and so
z(x − L)/L 0 ≤ z < x
G(x; z) = Gx (z) =
x(z − L)/L x < z ≤ L
which gives our solution of problem (3)
Z L
Z
Z
x−L x
x L
G(x; z)f (z)dz =
u(x) =
zf (z)dz +
(z − L)f (z)dz.
L
L x
0
0
Remark:
1. whatever you may think of think of the derivation of this expression for the
solution, it is easy to check (by direct differentiation and fundamental theorem
of calculus) that it is correct (assuming f is “reasonable” – say, continuous).
2. notice the form of the Green’s function Gx (z) (graph it!) – it has a “singularity”
(in the sense of being continuous, but not differentiable) at the point z = x. This
“singularity” must be there, since differentiating G twice has to yield a delta
function.
Example: use the Green’s function method to solve the problem
x2 u00 + 2xu0 − 2u = f (x), 0 < x < 1,
u(0) = u(1) = 0.
(4)
Remark: Notice that the coefficients x2 and 2x vanish at the left endpoint x = 0 (this
point is a “regular singular point” in ODE parlance). This suggests that unless u is
very wild as x approaches 0, we will require f (0) = 0 to fully solve the problem. Let’s
come back to this point after we find a solution formula.
2
d
d
Notice the operator L = x2 dx
2 + 2x dx − 2 =
so the problem for the Green’s function is
d 2 d
x dx
dx
LGx = z 2 G00x + 2zG0x − 2Gx = δ(z − x),
− 2 is self-adjoint (L = L∗ ), and
Gx (0) = Gx (1) = 0.
For z 6= x, the equation z 2 G00 +2zG0 −2G = 0 is an ODE of Euler type, and so looking
for solutions of the form G = z r yields
0 = z 2 (rz r−1 )0 + 2z(rz r−1 ) − 2z r = (r(r − 1) + 2r − 2)z r = (r + 2)(r − 1)z r
and so we want r = −2 or r = 1. Thus
Az + B/z 2 0 ≤ z < x
Gx (z) =
.
Cz + D/z 2 x < z ≤ 1
The BCs Gx (0) = Gx (1) = 0 imply B = 0 and C + D = 0, so
Az
0≤z<x
Gx (z) =
.
2
C(z − 1/z ) x < z ≤ 1
The matching conditions are
11
1. continuity Gx (x−) = Gx (x+) implies Ax = C(x − 1/x2 )
2. jump condition
Z
Z x+
δ(z−x)dz =
1=
x+
x−
x−
x+
2 0 0
(z Gx ) −2Gx ]dz = z 2 G0x (z)x− = x2 (G0x (x+)−G0x (x−))
implies x2 C(1 + 2/x3 ) − x2 A = 1.
Solving the simultaneous linear equations for A and C yields
1 = C (x2 + 2/x) − x2 (1 − 1/x3 ) = 3C/x
=⇒ C = x/3, A = (x − 1/x2 )/3
and so the Green’s function is
1
G(x; z) =
3
z(x − 1/x2 ) 0 ≤ z < x
x(z − 1/z 2 ) x < z ≤ 1
and the corresponding solution formula is
Z 1
Z x
Z 1
1
1
2
u(x) =
G(x; z)f (z)dz = (x − 1/x )
zf (z)dz + x
(z − 1/z 2 )f (z)dz.
3
3 x
0
0
Does this really solve (4) (supposing f is, say, continuous)? For 0 < x < 1, differentiation (and fundamental theorem of calculus) gives
Z 1
Z x
2
0
3
(z − 1/z 2 )f (z)dz
zf (z)dz + (x − 1/x )xf (x) +
3u (x) = (1 + 2/x )
x
0
Z 1
Z x
2
3
(z − 1/z 2 )f (z)dz
zf (z)dz +
− x(x − 1/x )f (x) = (1 + 2/x )
0
x
so
Z x
4
3
2
3(x u + 2xu − 2u) = x −6/x
zf (z)dz + (1 + 2/x )xf (x) − (x − 1/x )f (x)
0
Z x
Z 1
3
2
(z − 1/z )f (z)dz
zf (z)dz +
+ 2x (1 + 2/x )
0
x
Z x
Z 1
2
2
− 2 (x − 1/x )
zf (z)dz + x
(z − 1/z )f (z)dz
0
x
Z x
2
2
2
= (−6/x + 2x + 4/x − 2x + 2/x )
zf (z)dz
0
Z 2
+ (2x − 2x)
(z − 1/z 2 )f (z)dz + (x3 + 2 − x3 + 1)f (x)
2 00
0
2
x
= 3f (x)
12
and we see that that the ODE is indeed solved. What about the BCs? Well, u(1) = 0
obviously holds. As alluded to above, the BC at x = 0 is subtler. We see that
Z 1
Z
1 x
f (z)/z 2 dz.
zf (z)dz − lim x
3 lim u(x) = − lim 2
x→0+
x→0+
x→0+ x
x
0
If f is smooth, we have f (z) = f (0) + O(z) for small z, so
3 lim u(x) = −f (0) − f (0) = −2f (0),
x→0+
and so we require f (0) = 0 to genuinely satisfy the boundary condition at x = 0.
13
4. Boundary conditions, and self-adjoint problems.
The only BCs we have seen so far have been homogeneous (i.e. 0) Dirichlet
(specifying the value of the function) ones; namely u(x0 ) = u(x1 ) = 0. Let’s make this
more general, first by considering an ODE problem with inhomogeneous Dirichlet
BCs:
Lu := a0 u00 + a1 u0 + a2 u = f
x0 < x < x 1
.
(5)
u(x0 ) = u0 , u(x1 ) = u1
Recall that by integration by parts, for functions u and v,
x
(v, Lu) = (L∗ v, u) + [a0 (vu0 − v 0 u) + (a1 − a00 )vu]x10 .
Suppose we find a Green’s function G(x; z) = Gx (z) solving the problem
L∗ Gx = δ(z − x)
Gx (x0 ) = Gx (x1 ) = 0
with the corresponding homogeneous (i.e. 0) BCs to the BCs in problem (5). Then
x
x
u(x) = (L∗ Gx , u) = (Gx , Lu) − [a0 G0x u]x10 = (Gx , f ) − [a0 G0x u]x10
Z x1
=
G(x; z)f (z)dz + a0 (x0 )G0x (x0 )u0 − a0 (x1 )G0x (x1 )u1
x0
a formula which gives the solution of problem (5) in terms of the Green’s function
G(x; z), and the “data” (the source term f (x), and the boundary data u0 and u1 ).
The other way to generalize boundary conditions is to include the value of the derivative of u (as well as u itself) at the boundary (i.e. the endpoints of the interval).
For example, if u is the temperature along a rod [x0 , x1 ] whose ends are insulated,
we should impose the Neumann BCs u0 (x0 ) = u0 (x1 ) = 0 (no heat flux through the
ends).
In general, for the following discussion, think of imposing 2 boundary conditions,
each of which is a linear combination of u(x0 ), u0 (x0 ), u(x1 ), and u0 (x1 ) equal to 0
(homogeneous case) or some non-zero number (inhomogeneous case).
Definition:
1. A problem
Lu = f
BCs on u
is called (essentially) self-adjoint if
(a) L = L∗ (so the operator is self-adjoint), and
(b) (v, Lu) = (Lv, u) (i.e. with no boundary terms) whenever both u and v
satisfy the homogeneous BCs corresponding to the BCs on u in the problem.
14
Remark: As in the above example, the Green’s function for a self-adjoint problem
should satisfy the homogeneous BCs corresponding to the BCs in the original
problem. More generally, the Green’s function for a problem should satisfy:
Lu = f
2. The homogeneous adjoint boundary conditions for a problem
BCs on u
are the BCs on v which guarantee that (v, Lu) = (L∗ v, u) (i.e. no boundary
terms) when u satisfies the homogeneous BCs corresponding to the BCs in the
original problem.
Remark:
1. these definitions are quite abstract – it is better to see what is going on by doing
some specific examples
2. a problem can be non-self-adjoint even if L = L∗ , for example (see homework)
u00 + q(x)u = f (x)
u0 (0) − u(1) = 0, u0 (1) = 0
3. if L 6= L∗ , we can make Lu = f self-adjoint by multiplying by a function (again,
see homework).
Example: (Sturm-Liouville problem)

 Lu := (p(x)u0 )0 + q(x)u = f (x)
α0 u(0) + β0 u0 (0) = 0

α1 u(1) + β1 u0 (1) = 0
0<x<1
(6)
where p(x) > 0, and α0 , α1 , β0 , β1 are numbers with α0 , β0 not both 0, and α1 , β1
not both 0.
First notice that L = L∗ (the operator is self-adjoint), and integration by parts (as
usual) gives
1
(v, Lu) = (Lv, u) + [p(vu0 − v 0 u)]0 .
If u and v both satisfy the BCs in problem (6) (which are homogeneous) then
α0 [v(0)u0 (0) − v 0 (0)u(0)] = u0 (0)(−β0 v 0 (0)) − v 0 (0)(−β0 u0 (0)) = 0
β0 [v(0)u0 (0) − v 0 (0)u(0)] = v(0)(−α0 u(0)) − u(0)(−α0 v(0)) = 0
and so (since α0 β0 6= 0), v(0)u0 (0) − v 0 (0)u(0) = 0. A similar computation shows that
v(1)u0 (1) − v 0 (1)u(1) = 0. Hence (v, Lu) = (Lv, u) (boundary terms disappear), and
the problem is, indeed, self-adjoint.
15
Thus a Green’s function G(x; z) = Gx (z) for problem (6) should satisfy

 LGx = (p(z)G0x (z))0 + q(z)Gx (z) = δ(z − x)
α0 Gx (0) + β0 G0x (0) = 0
.

α1 Gx (1) + β1 G0x (1) = 0
For z 6= x, we have LGx = 0, so
Gx (z) =
c0 w0 (z) 0 ≤ z < x
c1 w1 (z) x < z ≤ 1
where for j = 0, 1, wj denotes any fixed, non-zero solution of
Lwj = 0,
αj wj (j) + βj wj0 (j) = 0
(which is an initial value problem for a second-order, linear, ODE – hence there is a
one-dimensional family of solutions), and c0 , c1 are non-zero constants. Continuity of
G at x implies
c0 w0 (x) = c1 w1 (x).
The jump condition at x is (check it!) p(x)[G0x (x+) − G0x (x−)] = 1, so
c1 w10 (x) − c0 w00 (x) =
1
.
p(x)
Hence
c1 [w0 (x)w10 (x) − w1 (x)w00 (x)] =
w0 (x)
,
p(x)
which involves the Wronskian
W = W [w0 , w1 ](x) = w0 (x)w10 (x) − w1 (x)w00 (x).
Recall that since w0 and w1 satisfy Lw = 0,
p(x)W [w0 , w1 ](x) ≡ constant.
There are two possibilities:
1. If W ≡ 0, then we cannot satisfy the equations for the coefficients above – there
is no Green’s function! In this case, w0 and w1 are linearly dependent, which
means that w0 and w1 are both actually solutions of the homogeneous equation
Lu = 0 satisfying both BCs in (6). We’ll discuss this case more later.
16
2. Otherwise, W is non-zero everywhere in (0, 1) and so we have
c1 =
w0 (x)
,
p(x)W
c0 =
w1 (x)
p(x)W
and hence a Green’s function
1
Gx (z) =
W [w0 , w1 ]p(x)
w1 (x)w0 (z) 0 ≤ z < x
w0 (x)w1 (z) x < z ≤ 1
and a solution to our Sturm-Liouville problem (6)
Z 1
Z x
1
w1 (z)f (z)dz .
u(x) =
w0 (z)f (z)dz + w0 (x)
w1 (x)
W [w0 , w1 ]p(x)
x
0
Remark:
1. Using the variation of parameters procedure from ODE theory would result in
precisely the same formula.
2. Notice that the Green’s function here (and for all our examples so far) is symmetric in its variables: Gx (z) = G(x; z) = G(z; x) = Gz (x). This is no accident
– rather, it is a general property of Green’s functions for self-adjoint problems.
17
5. Modified Green’s functions.
We have seen that we run into trouble in constructing a Green’s function for the
Sturm-Liouville problem

0<x<1
 Lu := (p(x)u0 )0 + q(x)u = f (x)
0
α0 u(0) + β0 u (0) = 0
(7)

α1 u(1) + β1 u0 (1) = 0
(where p > 0, q, and f smooth functions, and α0 , α1 , β0 , β1 numbers with α0 , β0 not
both 0, and α1 , β1 not both 0) if the corresponding homogeneous problem ((7) with
f ≡ 0) has a non-trivial solution u∗ :
Lu∗ = 0,
α0 u∗ (0) + β0 u∗ 0 (0) = 0,
α1 u∗ (1) + β1 u∗ 0 (1) = 0.
(8)
In fact in this case, a simple integration by parts
0 = (u, Lu∗ ) = (Lu, u∗ ) = (f, u∗ )
leads to the solvability condition
∗
Z
(f, u ) =
1
f (x)u∗ (x)dx = 0
(9)
0
which the source term f must satisfy, to have any hope of a solution u. In fact:
Theorem: [Fredholm alternative (for the Sturm-Liouville problem)] Either
1. Problem (7) has exactly one solution; or,
2. There is a non-zero solution u∗ of the corresponding homogeneous problem (8).
In this case, problem (7) has a solution if and only if the solvability condition (9)
holds (and the solution is not unique – you can add any multiple of u∗ to get
another).
Proof:
We’ve already seen that if there is no such u∗ , we have a solution (constructed above
via Green’s function). Also, it is unique, since the difference of any two solutions
would be a solution of the homogeneous problem.
If there is a non-zero u∗ , we’ve seen already that the solvability condition (9) is
required. If it is satisfied, we will show below how to construct a solution u using a
modified Green’s function. So suppose now that u∗ is a non-zero solution to (8), and that the solvability condition (9) on f holds. A function G̃(x; z) = G̃x (z) satisfying
LG̃x (z) = δ(z − x) + c(x)u∗ (z)
same BCs on G̃x as in (7)
18
is called a modified Green’s function. Notice that we can choose the “constant”
c(x) here so that the solvability condition
0 = (G̃x , Lu∗ ) = (LG̃x , u∗ ) = (δx + c(x)u∗ , u∗ ) = u∗ (x) + c(x)(u∗ , u∗ )
holds – namely c(x) = −u∗ (x)/(u∗ , u∗ ). This allows the probem for G̃x to be solved
(though we won’t do it here in general – you can use the variation of parameters
method to do it). Given such a G̃, if we have a solution u of (7), then
u(x) = (δx , u) = (LG̃x − cu∗ , u) = (G̃x , f ) +
(u∗ , u) ∗
u
(u∗ , u∗ )
is a solution formula for u in terms of f . Note that the constant in front of u∗ doesn’t
matter – remember we can add any multiple of u∗ and still have a solution. In fact
using reciprocity (G̃ is symmetric in x and z), we can check this formula indeed solves
problem (7):
Lu = (Lx G̃z (x), f ) = (δz (x), f ) + (c(z)u∗ (x), f ) = (δx (z), f ) −
u∗ (x) ∗
(u (z), f ) = f
(u∗ , u∗ )
where we had to use (of course!) the solvability condition (9) on f . Similarly, the
boundary conditions hold.
Hopefully an example will clarify the use of modified Green’s functions.
Example: Solve
u00 (x) = f (x),
0<x<1
.
0
0
u (0) = u (1) = 0
Obviously u∗ (x) ≡ 1 solves the corresponding homogeneous problem, leading to the
solvability condition
Z 1
∗
0 = (f, u ) =
f (x)dx
0
(physically: there is no steady-state temperature distribution in a uniform rod with
insulated ends unless the (spatial) average heat source is zero).
R1
Noticing that (1, 1) = 0 1dx = 1, the modified Green’s function G̃x (z) should solve
G̃00x = δx (z) − 1,
G̃0x (0) = G̃0x (1) = 0,
leading (since G̃00 = −1 =⇒ G̃ = −z 2 /2 + c1 z + c2 , and using the BCs) to
−z 2 /2 + A
0≤z<x
G̃x (z) =
.
2
−z /2 + z + B x < z ≤ 1
19
x+
Continuity at x requires A = x + B. The jump condition G̃0x x− = 1 already holds.
(This leaves one free parameter, which is characteristic of modified Green’s function
problems, since a multiple of u∗ (inRthis case a constant) can always be added.) So,
1
assuming the solvability condition 0 f (x)dx = 0 holds, we arrive at a formula for
the general solution
u(x) = (G̃x , f ) + C
Z
Z x
Z 1
Z 1
1 1 2
=−
z f (z)dz + (B + x)
f (x)dz + B
f (z)dz +
zf (z)dz + C
2 0
0
x
x
Z 1
Z x
f (z)dz +
zf (z)dz + C
=x
0
x
R1
where we simplified using the solvability condition on f , and we absorbed 0 z 2 f (z)dz
into the general constant C. It is very easy to check that this expression solves our
original problem (of course, we could have solved this particular problem just by
integrating).
Remark: So far in this section we discussed the issues of solvability conditions, modified Greens functions, and so on, only for self-adjoint problems. For a general problem,
the obstacle to solvability comes from considering the homogeneous adjoint problem:
L∗ u∗ = 0
.
hom. adj. BCs on u
If this problem has a non-trivial solution u∗ , then the solvability condition for Lu = f
with homogeneous BCs is
0 = (u, L∗ u∗ ) = (Lu, u∗ ) = (f, u∗ ).
One can also construct a modified Green’s function in this case, but we won’t go into
it here.
20
6. Green’s functions and eigenfunction expansion.
Consider again the Sturm-Liouville problem (with homogeneous BCs)

)0 + q(x)u = f (x)
x0 < x < x 1
 Lu := (p(x)u0
0
α0 u(x0 ) + β0 u (x0 ) = 0
(BC)

α1 u(x1 ) + β1 u0 (x1 ) = 0
(10)
(p > 0, q, and f smooth functions). Recall that the Sturm-Liouville problem has
eigenvalues
λ0 < λ1 < λ2 < λ3 · · ·
and corresponding eigenfunctions φj (x) such that
Lφj = λj φj
φj satisfies (BC)
which can be taken orthonormal:
(φj , φk ) = δjk =
1 j=k
.
0 j=
6 k
Furthermore, the eigenfunctions are complete, meaning that “any” function g(x) on
[x0 , x1 ] can be expanded
g(x) =
∞
X
cj φj (x)
j=0
cj = (φj , g) = “Fourier coefficient”.
To be more precise, this series does not (in general) converge at every point x in
[x0 , x1 ] (for example, it cannot converge at the endpoints
R x1 2 if g does not satisfy (BC)
2
!), but rather converges in the “L -sense”: provided x0 g (x)dx < ∞,
Z
x1
g(x) −
lim
N →∞
"
x0
N
X
#2
cj φj (x)
dx = 0.
j=0
So let’s express the Green’s function G(x; z) = Gx (z) for problem (10) as such an
eigenfunction expansion
G(x; z) = Gx (z) =
∞
X
j=0
21
cj (x)φj (z)
and try to find the coefficients cj (x). Using LGx = δx , and applying the linear
operator L to the expression for Gx , we find
∞
X
δx = LGx =
cj (x)(Lφj )(z) =
j=0
∞
X
cj (x)λj φj (z)
j=0
and so the coefficients cj (x)λj of this expansion should satisfy
cj (x)λj = (φj , δx ) = φj (x),
and hence we arrive at an expression for the Green’s function as an eigenfunction
expansion:
∞
X
1
φj (x)φj (z).
Gx (z) =
λ
j
j=0
Remark:
1. The reciprocity (symmetry in x and z of G) is very clearly displayed by this
formula.
2. If, for some j, λj = 0, this expression is ill-defined. Indeed, in this case, the
eigenfunction φj is a non-trivial solution to the homogeneous problem (i.e. it is
a “u∗ ”), and we have already seen there is no (usual) Green’s function in this
case.
3. The corresponding solution formula for problem (10) is
Z
u(x) = (Gx , f ) =
∞
x1 X
x0
j=0
∞
X 1
1
φj (x)φj (z)f (z)dz =
φj (x)(φj , f ).
λj
λ
j
j=0
Notice that if λj = 0 (for some j), the only way for this expression to make
sense is if also (φj , f ) = 0 – which is exactly the solvability condition on f we
have seen earlier.
22
III. GREEN’S FUNCTIONS FOR ELLIPTIC (STEADY-STATE)
PROBLEMS
We turn now from ODEs to PDEs (partial differential equations).
1. Green’s functions for the Poisson equation.
As a motivating example, consider the electrostatic potential u(x) in a uniformly
conducting two-dimensional (or three-dimensional) region D ⊂ R2 (or D ⊂ R3 ),
which satisfies
∆u = f (x)
x ∈ D (Poisson equation)
(11)
u = g(x)
x ∈ S = ∂D
where f (x) is the charge density on the region D, g(x) is the potential applied on the
boundary S = ∂D of D, and
∂ 2u ∂ 2u
∂ 2u
+
∆u :=
+ 2
∂x21 ∂x22
∂x3
is the Laplacian of u. (Notice also our notation for vectors in R2 and R3 : x = (x1 , x2 )
(or x = (x1 , x2 , x3 )).) The Poisson problem (11) arises in other physical contexts as
well, for example:
• u(x) is the steady-state temperature distribution inside the (uniform) region
D subject to heat source f (x) and with the temperature fixed at g(x) on the
boundary
• u is the steady-state (small) deformation of an elastic membrane/solid from its
reference position D, when it is subject to applied force f (x) and held at fixed
deformation g(x) on the boundary.
If we are looking to solve problem (11) by finding a Green’s function G(x; y) = Gx (y)
(x, y ∈ D), then by analogy with the self-adjoint ODE problems studied earlier, we
might seek G which solves
∆Gx (y) = δx (y) y ∈ D
(12)
Gx (y) = 0
y∈S
where here δx (y) = δ(y − x) is a 2- or 3-dimensional delta function centred at x,
defined in the same way as in 1-dimension: for a test function φ(x),
Z
(δx , φ) = “
δx (y)φ(y)dy“ = φ(x)
R2 or 3
23
(one can think δ(y − x) = δ(y1 − x1 )δ(y2 − x2 ) in 2 dimensions, and δ(y − x) =
δ(y1 − x1 )δ(y2 − x2 )δ(y3 − x3 ) in 3 dimensions).
Supposing we can find a Green’s function solving (12), we would like to derive a
formula for the solution u of (11) in terms of G. To do this, we will use a simple
“integration by parts” formula, for which we need to define the normal derivative of
a function u at the boundary S = ∂D:
∂u
:= n̂ · ∇u
∂n
where n̂ denotes the outward unit normal vector to the curve/surface S.
Lemma 1 (Green’s second identity) Let D ⊂ R2 or 3 be bounded by a smooth
curve/surface S. Then if u1 and u2 are twice continuously differential functions on
D,
Z
Z ∂u2
∂u1
(u1 ∆u2 − u2 ∆u1 ) dV =
u1
− u2
dS.
(13)
∂n
∂n
D
S
Proof: subtracting the relations
∇ · (u1 ∇u2 ) = ∇u1 · ∇u2 + u1 ∆u2
∇ · (u2 ∇u1 ) = ∇u1 · ∇u2 + u2 ∆u1
and using the divergence theorem, we find
Z Z
Z
∂u2
∂u1
n̂ · (u1 ∇u2 − u2 ∇u2 )dS =
u1
(u1 ∆u2 − u2 ∆u1 ) dV =
− u2
dS.
∂n
∂n
S
S
D
Taking u1 = u (solution of (11)) and u2 = Gx (solution of (12)) in (13), we find
Z
Z
∂Gx
(u(y)δx (y) − Gx (y)f (y))dy =
g(y)
dS(y)
∂n
D
S
or
Z
u(x) =
Z
G(x; y)f (y)dy +
D
S
∂G(x; y)
g(y)dS(y)
∂n(y)
(14)
which is the solution formula we sought.
Free-space Green’s functions. For a general region D, we have no chance of finding
an explicit Green’s function solving (12). One case we can easily solve explicitly, is
for the whole space D = R2 or D = R3 , where there is no boundary at all.
24
• R2 : for a given x ∈ R2 , we want to find Gx (y) solving ∆Gx (y) = δ(y − x). It is
reasonable to try a G which depends only on the distance from the singularity:
G = h(r), r := |y − x|. So for y 6= x (r > 0), we need
2
∂
1
1 ∂
0 = ∆G =
h(r) = (rh0 (r))0
+
2
∂r
r ∂r
r
and so
h0 = c1 /r
=⇒ h = c1 log r + c2
for some constants c1 and c2 . Including c2 is simply adding an overall constant
to the Green’s function, so we will drop it: c2 = 0. The constant c1 is determined
by the condition ∆Gx = δx : for any ε > 0, we have, by the divergence theorem,
Z
Z
Z
δx (y)dy =
∇ · ∇Gx (y)dy =
∇Gx (y) · n̂ dS(y)
1=
|y−x|≤ε
|y−x|≤ε
|y−x|=ε
Z
y−x
y−x
c1
c1
=
·
dS(y)
=
2πε = 2πc1
|y − x|2 |y − x|
ε
|y−x|=ε
and so c1 = (2π)−1 , and our expression for the two-dimensional free-space
Green’s function is
1
GR2 (x; y) =
log |y − x|.
2π
• R3 : we play the same game in three dimensions. Try Gx (y) = h(r), r := |y − x|,
so for r > 0
2
∂
2 ∂
1
0 = ∆Gx =
+
h(r) = 2 (r2 h0 (r))0
2
∂r
r ∂r
r
and so
h0 = c1 /r2
=⇒ h = −c1 /r + c2
and again we set c2 = 0. Constant c1 is determined by
Z
Z
Z
δx (y)dy =
∇ · ∇Gx (y)dy =
∇Gx (y) · n̂dS(y)
1=
|y−x|=ε
|y−x|≤ε
|y−x|≤ε
Z
y−x
y−x
c1
=
c1
·
dS(y) = 2 4πε2 = 4πc1
3
|y − x| |y − x|
ε
|y−x|=ε
and so c1 = (4π)−1 , and our expression for the three-dimensional free-space
Green’s function is
−1
GR3 (x; y) =
.
4π|y − x|
25
Knowing the free-space Green’s functions, we are lead immediately to solution formulas for the Poisson equation ∆u(x) = f (x) in R2 and R2 :
1 R
log |x − y|f (y)dy R2
2π R2
R f (y)
u(x) =
.
(15)
1
R3
− 4π
R3 |x−y|
The following theorem makes precise the claim that these formulas indeed solve Poisson’s equation:
Theorem: Let f (x) be a twice continuously differentiable function on R2 (respectively
R3 ) with compact support (i.e. it vanishes outside of some ball). Then u(x) defined
by (15) is a twice continuously differentiable function, and ∆u = f (x) for all x.
Proof: we’ll do the R3 case (the R2 case is analogous). Notice first that by the change
of variable y 7→ x − y in the integral,
Z
Z
1
f (y)
1
f (x − y)
u(x) = −
dy = −
dy.
4π R3 |x − y|
4π R3
|y|
If êj denotes the j-th standard basis vector, then
Z u(x + hêj ) − u(x)
1
f (x + hêj − y) − f (x − y) dy
=−
,
h
4π R3
h
|y|
f (x+hê −y)−f (x−y)
∂f
j
and since
(x − y) as h → 0 uniformly in y (since f is continu→ ∂x
h
j
ously differentiable and compactly supported),
Z
u(x + hêj ) − u(x)
dy
1
∂f
∂u
(x) = lim
(x − y) .
=−
h→0
∂xj
h
4π R3 ∂xj
|y|
Similarly,
Z
∂ 2u
1
∂ 2f
dy
(x) = −
(x − y)
∂xj ∂xk
4π R3 ∂xj ∂xk
|y|
which is a continuous function of x (again, because f is twice continuously differentiable and compactly supported) - hence u is twice continuously differentiable. Now,
to find ∆u we want to “integrate by parts” twice. But the Green’s function 1/|y| is
not twice continuously differentiable (indeed not even continuous) at y = 0, so to do
this legally, we remove a small neighbourhood of the origin: for ε > 0,
Z
Z
1
dy
1
dy
∆u(x) = −
∆f (x − y) −
∆f (x − y)
=: A + B .
4π |y|>ε
|y| 4π |y|≤ε
|y|
Now,
1
|B| ≤
max |∆f (x)|
4π x
=
Z
|y|≤ε
2
dy
1
=
max |∆f (x)|
|y|
4π x
1
ε
max |∆f (x)| → 0 as ε → 0
4π x
2
26
Z
ε
rdr
0
(∆f is bounded because it is continuous and compactly supported). For the other
term, we apply Green’s second identity on the domain {|y| > ε}, and use the fact
that ∆(1/|y|) = 0 for |y| > 0 to find
1
A=
4π
Z
|y|=ε
∂ 1
1 ∂
f (x − y) −
f (x − y) dS(y) =: C + D.
∂n |y|
|y| ∂n
For the second term,
1
|D| ≤
max |∇f |
4π
Finally, using
1
C=
4π
Z
|y|=ε
1
1
1
dS(y) =
max |∇f | · 4πε2 → 0 as ε → 0.
|y|
4π
ε
∂ 1
∂n |y|
=
Z
1
1
f
(x
−
y)dS(y)
=
|y|2
4πε2
|y|=ε
1
,
|y|2
Z
f (x − y)dS(y) = f (x − y ∗ )
|y|=ε
for some y ∗ with |y ∗ | = ε, by the mean-value theorem for integrals. So as ε → 0,
y ∗ → 0, and so sending ε → 0 in all terms yields ∆u(x) = f (x), as required. Remark:
1. for the above theorem to hold true, we don’t really need f to be compactly
supported (just with enough spatial decay so the the integral defining u makes
sense), or twice continuously differentiable (actually, f just needs to be a “little
better than continuous” – although just continuous is not quite enough)
2. the free space Green’s functions are not just useful for solving Poisson’s equation
on R2 or R3 , but in fact play a key role in boundary value problems – as we will
soon see
3. a physical interpretation: the three-dimensional free-space Green’s function
−(4π|x − y|)−1 is the electrostatic potential generated by a point charge located
at x (i.e. the Coulomb potential).
27
2. The method of images.
Example: Solve the boundary value problem for Laplace’s equation on the 1/2-plane
{ (x1 , x2 ) | x2 > 0 } by the Green’s function method:
∆u = 0
x2 > 0
.
(16)
u(x1 , 0) = g(x1 ) x2 = 0
The corresponding Green’s function problem is: for x with x2 > 0,
∆Gx (y) = δx (y) y2 > 0
.
Gx (y) ≡ 0
y2 = 0
We know that the two-dimensional free-space Green’s function Gf (x; y) = log |y −
x|/2π will generate the right delta function at x, but most certainly does not satisfy
the boundary conditions. The idea of the method of images is to balance Gf (x, y)
with other copies of Gf with singularities at different points in hopes of satisfying
the boundary conditions. For this problem, the geometry suggests placing an “image
charge” (the name comes from the interpretation of free-space Green’s functions as
potentials generated by point electric charges) at x̄ := (x1 , −x2 ), the reflection of x
about the x1 axis. So we set
Gx (y) := Gf (x; y) − Gf (x̄; y) =
1
(log |y − x| − log |y − x̄|) .
2π
Notice that
y2 = 0 =⇒ |y − x| = |y − x̄| =⇒ Gx (y) = 0
so we have satisfied the boundary conditions. But have we satisfied the right PDE?
Yes, since for y with y2 > 0,
∆Gx (y) = δ(y − x) − δ(y − x̄) = δ(y − x)
since the singularity introduced at x̄ lies outside the domain, and so does not contribute. So we have our Green’s function. To write the corresponding solution formula, we need to compute its normal derivative on the boundary y2 = 0: for y with
y2 = 0 (using |y − x| = |y − x̄|),
∇Gx (y) =
1
1
(y − x − (y − x̄)) =
(x̄ − x)
2
2π|y − x|
2π|y − x|2
and since the outward unit normal on the boundary is n̂ = (0, −1),
∂
1
x2
Gx (y) =
(x2 − x̄2 ) =
.
2
∂n
2π|y − x|
π((y1 − x1 )2 + x22 )
28
Hence our formula for the solution of problem (16) is
Z
g(y1 )
x2 ∞
dy1 .
u(x) =
π −∞ (y1 − x1 )2 + x22
Now, does this expression really yield a solution to problem (16)? For x2 > 0, the
function x2 /(x21 + x22 ) is harmonic:
∆
x2
∂ −2x1 x2
∂ x21 − x22
=
+
x21 + x22
∂x1 (x21 + x22 )2 ∂x2 (x21 + x22 )2
−2x2 (x21 + x22 ) + 8x21 x2 − 2x2 (x21 + x22 ) − 4x2 (x21 − x22 )
=
= 0,
(x21 + x22 )3
and hence so is x2 /((y1 −x1 )2 +x22 ) for any y1 . So supposing the function g is bounded
and continuous, we may differentiate under the integral signRto conclude that ∆u(x) =
∞
0 for x2 > 0. What about the boundary conditions? Since −∞ (s2 + 1)−1 ds = π, and
changing variables y1 = x2 s, we have for fixed x1 , and for x2 > 0,
Z ∞
1
1
|u(x1 , x2 ) − g(x1 )| = [g(x
+
x
s)
−
g(x
)]
ds
1
2
1
π −∞ s2 + 1
Z ∞
Z M
4 max |g|
ds
1
1
≤
+
|g(x1 + x2 s) − g(x1 )|ds.
2
π
π −M 1 + s2
M s +1
Let ε > 0 be given, and then choose M large enough so that the first term above is
< ε/2. Then since a continuous function on a compact set is uniformly continuous,
we may choose x2 small enough so that the second term above is < ε/2. Thus we
have shown that
lim u(x1 , x2 ) = g(x1 )
x2 ↓0
– i.e., that the boundary conditions in problem (16) are also satisfied.
In practice, the method of images can only be used to compute the Green’s function
for very special, highly symmetric geometries. Examples include:
• half-plane, quarter-plane, half-space, octant, etc.
• disks, balls
• some combinations of the above, such as a 1/2-disk
Example: Solve the boundary value problem for Laplace’s equation in the 3D ball of
radius a { x ∈ R3 | |x| < a} using the method of images:
∆u = 0 |x| < a
.
(17)
u = g |x| = a
29
The Green’s function should solve
∆Gx (y) = δ(y − x) |y| < a
.
Gx (y) ≡ 0
|y| = a
Given x in the ball (|x| < a), a natural place to put an “image charge” is on the ray
from the origin through x. In fact, we choose the point
a2
x := 2 x
|x|
∗
and notice that |x∗ | > a, so x∗ is not in our domain, and hence ∆Gf (x∗ ; y) will
contribute nothing for |y| < a. Now suppose that y is on the boundary (|y| = a), and
notice that
2
2 a2
a
x
y
a2
|y − x∗ |2 = 2 |x| − a = 2 (|x|2 − 2y · x + a2 ) = 2 |x − y|2
|x|
a
|x|
|x|
|x|
so |y − x∗ | = (a/|x|)|x − y|. Thus if we define
−1
a
1
a f ∗
G (x ; y) =
−
,
Gx (y) = G (x; y) −
|x|
4π |y − x| |x||y − x∗ |
f
we have satisfied the boundary condition: |y| = a =⇒ Gx (y) = 0. The normal
derivative of G on the boundary is
y 1
a(y − x∗ )
∂Gx
y−x
a2 − |x|2
= ·
−
.
=
∂n
a 4π |y − x|3 |x||y − x∗ |3
4πa|y − x|3
Our solution formula is then
a2 − |x|2
u(x) =
4πa
Z
|y|=a
g(y)
dS(y),
|y − x|3
(18)
which is Poisson’s formula for harmonic functions in a 3D ball. Again, it is possible to
prove rigorously that for g a continuous function on the sphere |y| = a, this formula
produces a function u which is harmonic in the ball, and takes on the values g on
the sphere. Furthermore, any function harmonic on the ball and with continuous
boundary values g is given by formula (18) (we will address this issue later on).
A nice consequence of Poisson’s formula (though there are also direct proofs) is the:
Mean value formula for harmonic functions: let u be harmonic (∆u ≡ 0) in
the ball |y| < a in Rn and continuous for |y| ≤ a. Then the value of u at the centre
of the ball is equal to its average over the boundary of the ball.
30
Proof (for the R3 case): take x = 0 in (18) to find
Z
Z
a
g(y)
1
u(x) =
dS(y) =
g(y)dS(y)
4π |y|=a |y|3
4πa2 |y|=a
as required. Solvability conditions and modified Green’s functions.
Example: Consider the problem
∆u = f
∂u
=g
∂n
in D ⊂ R2
on ∂D
(19)
with so-called Neumann boundary conditions. Notice that the corresponding
homogeneous problem
∆u∗ = 0 in D ⊂ R2
∂u∗
=0
on ∂D
∂n
has the non-trivial solution u∗ (x) ≡ 1. As a result, there is a solvability condition on
the data f and g in problem (19), which can be derived using Green’s second identity:
Z
Z
Z
∂u
∂u∗
∗
∗
u − u∗ )dS
0=
∆u udx =
u ∆udx +
(
∂n
D
D
∂D ∂n
and since ∂u∗ /∂n ≡ 0, we arrive at the solvability condition
Z
Z
g(x)dS(x).
f (x)dx =
D
∂D
Problem (19) will not have a Green’s function, but will have a modified Green’s
function, satisfying
∆G̃x (y) = δ(y − x) + C y ∈ D
∂
G̃ (y) = 0
y ∈ ∂D
∂n x
for some constant C, which can be determined using the divergence theorem:
Z
Z
Z
∂
1=
δ(y − x)dy = (∆G̃x − C)dy =
G̃x dS − C|D| = −C|D|
D
D
∂D ∂n
(where |D| denotes the area/volume of D), and hence C = −1/|D|. If we can find the
modified Green’s function, we can construct the family of solutions of problem (19)
(assuming the solvability condition holds), as
Z
Z
u(x) =
G̃x (y)f (y)dy −
G̃x (y)g(y)dS(y) + A
D
∂D
for any constant A.
31
3. Green’s functions for Laplacian: some general theory.
So far we have mainly discussed methods for explicit computation of Green’s functions and solutions for certain problems involving the Laplacian operator. For general
regions, of course, explicit computations are not possible. Nevertheless, some general
theory tells us that Green’s functions exist, and that the corresponding solution formulas do yield solutions to the problems of interest. We describe some of this theory
here.
Let D be a region in R2 or R3 bounded by a smooth curve/surface S = ∂D. Let
Gf (x; y) denote the free-space Green’s function (i.e. log |x − y|/(2π) for R2 and
(4π|y − x|)−1 for R3 ).
Definition: a (Dirichlet) Green’s function G(x; y) for the operator ∆ and the region
D is a function continuous for x, y ∈ D̄ := D ∪ S, x 6= y, satisfying
1. the function Hx (y) := Gx (y) − Gfx (y) is harmonic (i.e. ∆y Hx (y) = 0) for y ∈ D
2. Gx (y) ≡ 0 for y ∈ S
Remark:
• the first property implies that Gx solves ∆Gx = δx , and also implies that Gx (y)
is a smooth function on D\{x} (since Gfx is, and since harmonic functions are
smooth).
• the terminology “Dirichlet Green’s function” refers to the “Dirichlet boundary
conditions” – specifying the value of the function (rather than, say, its normal
derivative) on the boundary.
Properties of Green’s functions:
1. Existence: a Green’s function as above exists. (The proof of this fact calls for
some mathematics a little beyond the level of this course – eg., some functional
analysis – so we will have to simply take it as given.)
2. Uniqueness: the Green’s function is unique – in particular, we may refer to
“the” Green’s function (rather than “a” Green’s function).
3. Symmetry (“reciprocity”): G(x; y) = G(y; x).
4. Relation to solution of BVP for Poisson: let f be a smooth function on
D, and let g be a smooth function on S. Then u solves the boundary value
problem for Poisson’s equation
∆u = f D
(20)
u = g ∂D
32
if and only if
Z
u(x) =
Z
G(x; y)f (y)dy +
D
∂D
∂Gx (y)
g(y)dS(y).
∂n
(21)
Some (sketches of ) proofs:
• Symmetry: a formal (i.e. “cheating”) argument giving the symmetry is:
Z
G(x; y) − G(y; x) =
(G(x; z)δy (z) − G(y; z)δx (z)) dz
D
Z
(Gx (z) ∆Gy (z) − Gy (z)∆Gx (z)) dz
=
D
Z ∂Gy (z)
∂Gx (z)
Gx (z)
=
− Gy (z)
dS(z) = 0.
∂n(z)
∂n(z)
S
This is not a rigorous argument, because we applied Green’s second identity
to functions (i.e. the Green’s functions) which are NOT twice continuously
differentiable (indeed, not even continuous). To make the argument rigorous,
try applying this identity on the domain D with balls of (small) radius ε around
the points x and y removed – the Green’s functions are smooth on this new
domain. Then take ε ↓ 0. This is left as an exercise.
• Representation of solution of Poisson: suppose u is twice continuously differentiable on D, continuous on D̄ := D ∪ S, and satisfies (20). We will show that u
is given by formula (21). Assume first that f = 0, so that u is harmonic in D.
We claim then that
Z ∂ f
f ∂
u(x) =
u Gx − Gx u dS.
(22)
∂n
∂n
S
Assuming this for a moment, applying Green’s second identity to the harmonic
functions u and Hx := Gx − Gfx in D, we arrive at
Z ∂Hx
∂u
0=
u
− Hx
dS
∂n
∂n
S
which we add to (22) to arrive at
Z Z
∂Gx
∂u
∂Gx
u(x) =
u
− Gx
dS =
gdS
∂n
∂n
S
S ∂n
as required. Now let’s prove (22) (we will just do the three-dimensional case
here, for simplicity). Let Dε be D with a ball of (small) radius ε about x
33
removed, and apply Green’s second identity to u and Gfx on Dε (both functions
are smooth and harmonic here) to find
Z
∂Gfx
f ∂u
0=
u
− Gx
dS
∂n
∂n
∂Dε
Z Z
∂Gfx
y−x
y−x
1
f ∂u
=
u
− Gx
dS −
· u
∇u dS.
−
∂n
∂n
4π|y − x|3 4π|y − x|
∂D
|y−x|=ε |y − x|
Now the second term in the second integral is bounded by
1
max |∇u| · 4πε2 = (max |∇u|)ε → 0 as ε ↓ 0,
4πε
while the first term is, in the limit as ε ↓ 0,
Z
1
lim
u(y)dS(y) = u(x)
ε↓0 4πε2 |y−x|=ε
(by the mean value theorem for integrals), and we recover (22), as desired.
Finally, suppose f 6= 0 in (20). Let
Z
G(x; y)f (y)dy.
w(x) :=
D
Notice that for x ∈ S, G(x; y) = 0 (reciprocity), and so w(x) = 0. Just as we
did for the Poisson equation in R3 , we can show that ∆w = f , and so
∆(u − w) = ∆u − f = f − f = 0,
and we may write
Z
u(x) − w(x) =
∂D
∂Gx
[g(y) − w(y)]dS(y) =
∂n
Z
∂D
∂Gx
g(y)dS(y)
∂n
and we recover (21) as needed.
• Formula (21) solves the Poisson BVP: it remains to show (21) really does solve
the BVP (20). We already argued that the first term in (21) (what we called
w above) solves ∆w = f and has zero boundary conditions. It remains to show
that the boundary integral is harmonic, and has boundary values g. We won’t
do this here – the proof can be found in rigorous PDE texts.
• Uniqueness of the Green’s function: we will prove this using the “maximum
principle” in the next section.
34
4. The maximum principle.
Let D be a (open, connected) region in Rn , bounded by a smooth surface ∂D.
Theorem: [maximum principle for harmonic functions]. Let u be a function
which is harmonic in D (∆u = 0) and continuous on D = D ∪ ∂D. Then
1. u attains its maximum and minimum values on the boundary ∂D
2. if u also attains its maximum or minimum value at an interior point of D, it
must be a constant function
Remark:
1. u must attain its max. and min. values somewhere on D, since it is a continuous
function on a closed, bounded set
2. the “maximum” part of the theorem extends to subharmonic functions (∆u ≥ 0),
and the “minimum” part to superharmonic functions (∆u ≤ 0).
We’ll give two proofs of the maximum principle.
Proof #1: suppose u does, in fact attain its maximum (say) at an interior point
x0 ∈ D. Then for any r > 0 such that the ball Br of radius r about x0 lies in D, the
mean-value property of harmonic functions implies that
Z
1
u(x0 ) =
u dS ≤ u(x0 )
|∂Br | ∂Br
since u(x0 ) is the maximum value of u, and so we must have equality above, and u
must be equal to u(x0 ) everywhere on ∂Br . Hence u is constant on Br . Repeating
this argument, we can “fill out” all of D, and conclude u is constant on D. (A slick
argument for this: the set where u(x) = u(x0 ) is both closed and open in D, hence is
all of D, since D is connected.) Our second proof does not rely so heavily on the mean-value property (at least for the
first statement of the maximum principle), and hence generalizes to other “Laplacianlike” (a.k.a elliptic) operators, for which the mean-value property does not hold.
Proof # 2: again, suppose u were to have a maximum (say) at an interior point
x0 ∈ D. Then we know from vector calculus that ∆u(x0 ) ≤ 0, which is almost a
contradiction to ∆u = 0, but not quite (because it is ≤ rather than <). We can “fix”
this by introducing a new function, for any ε > 0,
v(x) := u(x) + ε|x|2 .
Notice that
∇v = ∇u + 2εx,
∆u = ∆u + 2εn = 2εn > 0
35
(n is the space dimension), and so v really cannot have a maximum at an interior
point of D. That means that for any x ∈ D,
u(x) ≤ v(x) < max v(y) = max[u(y) + ε|y|2 ] ≤ max u(y) + ε max |y|2
y∈∂D
y∈∂D
y∈∂D
y∈∂D
and now letting ε ↓ 0, we see
u(x) ≤ max u(y),
y∈∂D
which is the first statement of the maximum principle (sometimes called the weak
maximum principle. For the second statement (sometimes called the strong maximum
principle), we can use the mean value property as before. Consequences of the maximum principle:
• Uniqueness of solutions of the Dirichlet problem for the Poisson equation:
Theorem: Let f and g be continuous functions on D and ∂D respectively. There
is at most one function u which is twice continuously differentiable in D, continuous on D̄, and solves
∆u = f D
.
u = g ∂D
Proof: If u1 (x) and u2 (x) are both solutions, then their difference w(x) :=
u1 (x) − u2 (x) solves
∆w = 0 D
.
w = 0 ∂D
The maximum principle says that w attains its maximum and minimum on the
boundary ∂D. Since w ≡ 0 on ∂D, we must have w ≡ 0 in D, and hence u1 ≡ u2
in D. • Uniqueness of the (Dirichlet) Green’s function:
Theorem: There is at most one Dirichlet Green’s function for a given region D.
Proof: if G(x; y) is a (Dirichlet) Green’s function for D, then the function
Hx (y) := G(x; y) − Gf (x; y) (where Gf denotes the free-space Green’s function)
solves the problem
∆Hx (y) = 0
y∈D
H x (y) = −Gf (x; y) y ∈ ∂D
and so is unique, by the previous theorem. 36
5. Green’s functions by eigenfunction expansion.
We will just do one example.
Example: Find the Dirichlet Green’s function in the infinite wedge of angle α, described in polar coordinates (r, θ) by 0 ≤ θ ≤ α, r ≥ 0.
So given a point x in the wedge, we are looking for a function Gx solving ∆Gx = δx
inside the wedge, and Gx = 0 on the rays θ = 0 and θ = α. It is natural to work in
polar coordinates. Let (r0 , θ0 ) be the polar coordinates of x. Then the delta function
centred at x is written in polar coordinates as r10 δ(r − r0 )δ(θ − θ0 ) (the factor 1/r0 is
there since integrating in polar coordinates, dy becomes rdrdθ – check it). Also, the
Laplacian in polar coordinates is
1 ∂2
∂2
1 ∂
+
∆= 2 +
∂r
r ∂r r2 ∂θ2
and so we are looking for G(r, θ) which solves
Grr + 1r Gr + r12 Gθθ = r10 δ(r − r0 )δ(θ − θ0 )
G(r, 0) = G(r, α) = 0
2
∂
The eigenfunctions of ∂θ
2 (the angular part of the Laplacian) which satisfy zero boundary conditions at θ = 0 and θ = α are sin(nπθ/α), n = 1, 2, 3, . . ., so we will try to
find G in the form of an eigenfunction expansion
G(r, θ) =
∞
X
cn (r) sin(nπθ/α)
n=1
(in fact, a Fourier sine series since the eigenfunctions are sines), and our job is to
find the coefficients cn (r).
Applying ∆ term-by-term to the expansion, we find
∞ X
1 0
n2 π 2
1
00
cn + cn − 2 cn sin(nπθ/α) = 0 δ(r − r0 )δ(θ − θ0 ),
r
α
r
n=1
and integrating (in θ) both sides against sin(kπθ/α) yields
Z α
1 0 k2π2
21
2
00
0
δ(r −r )
sin(kπθ/α)δ(θ −θ0 )dθ = 0 sin(kπθ0 /α)δ(r −r0 ).
ck + ck − 2 ck =
0
r
α
αr
αr
0
For r 6= r0 , this is
1
k2π2
c00k + c0k − 2 ck = 0
r
α
37
an equation of “Euler type” whose solutions are of the form rb , where plugging rb in
yields
k2π2
0 = b(b − 1) + b − 2 = (b − kπ/α)(b + kπ/α),
α
so b = ±kπ/α. Thus we find
Arkπ/α + Br−kπ/α 0 < r < r0
ck (r) =
.
Crkπ/α + Dr−kπ/α r0 < r < ∞
For finiteness at r = 0, we take B = 0, and to avoid growth as r → ∞, we take
C = 0, hence
Arkπ/α 0 < r < r0
ck (r) =
.
Dr−kπ/α r0 < r < ∞
We will find A and D through matching conditions at r = r0 . Continuity at r = r0
implies A(r0 )kπ/α = D(r0 )−kπ/α , and so defining a new constant à = A(r0 )kπ/α , we
have
)
(
r kπ/α
0
min(r, r0 )
0
<
r
<
r
0
kπ/α
r
=
Ãρ
,
ρ
:=
.
ck (r) = Ã
kπ/α
r0
max(r, r0 )
r0 < r < ∞
r
The jump condition is
Z r0 + kπθ0
2
1 0 0 k2π2
2kπ Ã
0 r 0 +
sin
(rc
c
=
)
−
dr
=
c
k
k
k r 0 − = −
0
2
αr
α
r
α
αr0
r0 −
so à = − sin(kπθ0 /α)/(kπ), and ck (r) = − sin(kπθ0 /α)ρkπ/α /(kπ). So our expression
for the Green’s function is
∞
1X1
G=−
sin(nπθ0 /α) sin(nπθ/α)ρnπ/α ,
π n=1 n
and using the cosine summation law to write
nπ
i
1 h nπ
sin(nπθ0 /α) sin(nπθ/α) =
cos
(θ − θ0 ) − cos
(θ + θ0 ) ,
2
α
α
we find
∞
nπ
i
1 X 1 h nπ
0
nπ/α
0
nπ/α
G=−
cos
(θ − θ ) ρ
− cos
(θ + θ ) ρ
2π n=1 n
α
α
∞
i
X
1
1h
0
0
= − Re
(ρei(θ−θ ) )nπ/α − (ρei(θ+θ ) )nπ/α .
2π
n
n=1
38
0
π/α
0
π/α
Now setting z1 := ρei(θ−θ ) , z2 := ρei(θ+θ ) , noting that |z1 | = |z2 | = ρπ/α < 1 (for
r 6= r0 ), and recalling the Taylor series
∞
X
1 n
z
− log(1 − z) =
n
n=1
is convergent for |z| < 1, we arrive at
!
π/α
1 − z π/α 1 − z π/α 2
1
1
1
−
z
1
1
1
1
G(r, θ; r0 , θ0 ) =
=
Re log
log log =
π/α
π/α
π/α
1 − z2 4π
1 − z2 2π
2π
1 − z2
1
1 + ρ2π/α − 2ρπ/α cos(π(θ − θ0 )/α)
=
log
4π
1 + ρ2π/α − 2ρπ/α cos(π(θ + θ0 )/α)
π/α
(ρ
+ ρ−π/α )/2 − cos(π(θ − θ0 )/α)
1
log
=
4π
(ρπ/α − ρ−π/α )/2 − cos(π(θ + θ0 )/α)
and so finally
1
G(r, θ; r , θ ) =
log
4π
0
0
cosh( απ log ρ) − cos( απ (θ − θ0 ))
cosh( απ log ρ) − cos( απ (θ + θ0 ))
,
ρ :=
min(r, r0 )
,
max(r, r0 )
a rather attractive formula for the (Dirichlet) Green’s function in the wedge of angle
α. (And we should pause and appreciate it when we are lucky enough to find an
explicit formula – it is pretty rare!)
Amusing exercise: check that we recover the formulas that we previously derived
(using the method of images) for the half-plane (α = π) and the quarter-plane (α =
π/2) .
39
IV. GREEN’S FUNCTIONS FOR TIME-DEPENDENT PROBLEMS
1. Green’s functions for the heat equation.
Suppose we measure the temperature u(x, t) at each point x in a bounded region V
in Rn (for us, usually n = 1, 2, or 3), and at each time t. The region is subjected to
heat sources f (x, t), and at its boundary, it is held at temperature g(x, t) (x ∈ ∂V ).
Initially (at time t = 0, say), the temperature distribution in V is u0 (x). The following
initial-boundary value problem for the heat equation describes the temperature
distribution u(x, t) at later times t > 0:

∂u
− D∆u = f (x, t),
x ∈ V, t > 0
 heat equation:
∂t
,
(23)
boundary value: u(x, t) = g(x, t), x ∈ ∂V, t > 0

initial condition: u(x, 0) = u0 (x), x ∈ V
where D > 0 is the diffusion rate constant.
We would like to represent the solution to this problem using a Green’s function.
The first observation is that the differential operator appearing in the equation is not
self-adjoint:
∂
∂
L :=
− D∆u
=⇒
L∗ = − − D∆u.
∂t
∂t
Next, some notation. We fix a time interval [0, T ] (some T > 0), and let CT denote
the space-time cylinder CT = V × [0, T ].
As before, a Green’s function for our problem should be a function of two sets of
variables: G(x, t; y, τ ), x, y ∈ V , t, τ ≥ 0. To determine the problem that G should
solve, we suppose u(y, τ ) solves problem (23), and integrate G against Lu over the
space-time cylinder, and integrate by parts:
Z
Z
G(uτ − D∆u)dydτ
G Lu dydτ =
CT
CT
Z
Z TZ ∂G
∂u
∗
=
L G u dydτ +
u
−G
dS(y)dτ
∂n
∂n
CT
0
∂V
Z
+
(Gu|τ =T − Gu|τ =0 ) dy.
V
So supposing we demand that our Green’s function G(x, t; y, τ ) solves
 ∂G
 − ∂τ − D∆G = δ(y − x)δ(τ − t)
G ≡ 0 for y ∈ ∂V

G ≡ 0 for τ > t (“causality”) ,
40
(24)
we arrive at a representation formula for u(x, t), x ∈ V , 0 ≤ t < T ,
Z TZ
Z
u(x, t) =
G(x, t; y, τ )f (y, τ )dydτ +
G(x, t; y, 0)u0 (y)dy
0
V
V
Z TZ
∂G
−
g(y, τ )dS(y)dτ.
0
∂V ∂n
(25)
Note that the above “causality” condition implies that the solution at time t should
not depend on any of its values at later times – i.e. we are solving forward in time.
Problem (24) for the Green’s function looks a little odd. It is a backwards heat
equation, which would be nasty, except that is also solved backwards in time (starting
from time τ = t and going down to τ = 0). So, to straighten it out, it is useful to
change the time variable from τ to σ := t − τ . Problem (24) then becomes
 ∂G
 ∂σ − D∆G = δ(y − x)δ(σ)
.
(26)
G = 0 for y ∈ ∂V

G = 0 for σ < 0
This is the problem we will try to solve in various situations. The simplest case is
when there is no boundary – the free-space case.
41
2. Free-space Green’s function for the heat equation.
We will find the free-space Green’s function for the heat equation by using the Fourier
transform, so let us first recall the definition and some key properties of it.
R
Definition: Let f be an integrable function on Rn (that means Rn |f (x)|dx < ∞).
The Fourier transform of f is another function, fˆ, defined by
Z
−n/2
ˆ
e−iξ·x f (x)dx.
f (ξ) := (2π)
Rn
Here are some useful properties of the Fourier transform. Let f , g be smooth functions
with rapid decay at ∞.
\
1. F.T. is linear: (αf
+ βg)(ξ) = αfˆ(ξ) + βĝ(ξ)
ˇ
2. F.T. is (almost) its own inverse: fˆ = f , where
Z
−n/2
ǧ(x) := (2π)
eiξ·x g(ξ)dξ
Rn
is the inverse Fourier transform.
3. F.T. is unitary (its inverse is its adjoint): (fˆ, g) = (f, ǧ), and in particular
(taking g = fˆ), preserves the “L2 -norm”:
Z
Z
2
ˆ
|f (ξ)| dξ =
|f (x)|2 dx.
Rn
Rn
4. F.T. “interchanges differentiation and coordinate multiplication”:
∂ ˆ
(x\
f (ξ).
j f (x))(ξ) = i
∂ξ
d
∂f
(ξ) = (iξj ) fˆ(ξ),
∂xj
5. F.T “interchanges convolution and multiplication”:
Z
f[
∗ g(ξ) = (2π)n fˆ(ξ)ĝ(ξ),
(f ∗ g)(x) :=
f (x − y)g(y)dy.
Rn
6. F.T interchanges coordinate translation and multiplication by an exponential:
for a ∈ Rn ,
\
f (x
− a)(ξ) = e−ia·ξ fˆ(ξ).
7. F.T maps Gaussians to Gaussians: for a > 0,
|ξ|2
\
a|x|2
e− 2 (ξ) = a−n/2 e− 2a .
42
It is property 4 which makes the Fourier transform so useful for differential equations
– it converts differential equations into algebraic ones.
Property 2 is deep, and difficult to prove. See an analysis textbook for this. The
other properties are easy to show. We will do 4 and 7 – two that we will use shortly.
Proof of property 4: integrating by parts,
Z
Z
d
∂f
−ix·ξ ∂f
−n/2
−n/2
e
(−iξj )e−ix·ξ f (x)dx = iξj fˆ(ξ).
(ξ) = (2π)
(x)dx = −(2π)
∂xj
∂x
n
n
j
R
R
And for the other one,
∂ ˆ
∂
i
f (ξ) = i
(2π)−n/2
∂ξj
∂ξj
Z
e
−ix·ξ
−n/2
Z
f (x)dx = i(2π)
Rn
(−ixj )e−ix·ξ f (x)dx = (x\
j f (x))(ξ).
Rn
Proof of property 7: the higher-dimensional cases follow easily from the case n = 1,
so we’ll do that one. Completing the square,
Z ∞
Z ∞
2
a
2
\
−1/2
−ix·ξ − a2 x2
−1/2
− ξ2a
− a2 |x|2
e
(ξ) = (2π)
e
e
e− 2 (x+iξ/a) dx.
dx = e (2π)
−∞
−∞
2
The last integral is the integral of the entire complex function f (z) = e−az /2 along
the contour z = x + iξ/a, −∞ < x < ∞ in the complex plane. We can “shift” the
contour to the real axis using Cauchy’s theorem.
Z
Z
Z ∞
Z
− a2 (x+iξ/a)2
f (z)dz
e
dx = lim
f (z)dz = lim
f (z)dz +
R→∞
−∞
R→∞
[−R,R]+iξ/a
AR ∪BR
[−R,R]
where AR denotes the contour z = −R +iy, y : eiξ/a → 0, and BR denotes the contour
z = R + iy, y : 0 → eiξ/a (draw a picture!). Along AR and BR , we have
a
a
2
|f (z)| = e− 2 Re(z ) ≤ e 2 (R
and so
Z
AR ∪BR
Hence
Z
∞
ξ2
a 2
f (z)dz ≤ 2e 2a e− 2 R → 0 as R → ∞.
− a2 (x+iξ/a)2
e
2 −ξ 2 /a2 )
Z
∞
dx =
− a2 x2
e
r
dx =
−∞
−∞
2π
a
(the last equality is a standard fact which can be proved, for example, by squaring
the integral, interpreting it as a two-dimensional integral, and changing to polar
coordinates). Finally, then, we arrive at
2
ξ
− a2 |x|2
e\
(ξ) = a−1/2 e− 2a
43
as needed.
Now let’s return to the problem of finding the free-space Green’s function for the heat
equation. That is, solving (26) when V = Rn . Let Ĝ(x, t; ξ, σ) denote the Fourier
transform of G(x, t; y, σ) in the variable y. Property 4 of the Fourier transform shows
that ∆ in the variable y corresponds to multiplication by −|ξ|2 . Also, note that
Z
−n/2
e−iy·ξ δ(y − x)dy = (2π)−n/2 e−ix·ξ ,
δ[
x (y)(ξ) = (2π)
Rn
and so we have to solve
∂
Ĝ + D|ξ|2 Ĝ = (2π)−n/2 e−ix·ξ δ(σ).
∂σ
For σ > 0, this is Ĝσ + D|ξ|2 Ĝ = 0, an ODE which is easily solved to find
2
Ĝ = Ce−D|ξ| σ ,
where C = C(x, t, ξ) can be found by a jump condition:
Z
0+
n/2 ix·ξ
δ(σ)dσ = (2π)
1=
0−
Z
0+
e
0−
h
i
σ=0+
Ĝσ + D|ξ| Ĝ = (2π)n/2 eix·ξ Ĝσ=0− = (2π)n/2 eix·ξ C
2
where we used the fact that Ĝ is bounded, so that the second term in the integral
contributes nothing, and the “causality” condition that Ĝ = 0 for σ < 0. Thus we
arrive at
2
Ĝ = (2π)−n/2 e−ix·ξ e−D|ξ| σ .
Finally, then, inverting the Fourier transform, and using properties 7 and 6, we find
for τ > 0,
G(x, t; y, σ) = (4πDσ)−n/2 e−
|y−x|2
4Dσ
and so, changing back to τ = t − σ, our free-space Green’s function (also called the
fundamental solution) of the heat equation in Rn is
(
|y−x|2
−n/2 − 4D(t−τ )
(4πD(t
−
τ
))
e
τ <t .
G(x, t; y, τ ) =
0
τ >t
We obtain a solution formula for the heat equation in Rn ,
∂u
− D∆u = f (x, t),
x ∈ Rn , t > 0
∂t
,
u(x, 0) = u0 (x),
x ∈ Rn
44
(27)
by substituting our expression for G into (25):
Z tZ
u(x, t) =
|y−x|2
−n/2 − 4πD(t−τ )
(4πD(t−τ ))
0
e
−n/2
Z
f (y, τ )dy dτ +(4πDt)
V
e−
|y−x|2
4Dt
u0 (y)dy.
V
(28)
It is not hard to prove the following (though we will not do it here):
Theorem: Suppose f (x, t) is continuously differentiable and bounded on Rn × [0, T ],
and u0 (x) is continuous on Rn . Then for 0 < t < T , u(x, t) given by formula (28) is
continuously differentiable in t and twice continuously differentiable in x, and solves
the heat equation in (27). Furthermore, for any x ∈ Rn , limt↓0 u(x, t) = u0 (x).
Based on our free-space solution, we can infer a couple of important properties of
diffusion:
• Instantaneous smoothing: in the absence of sources, solutions of the heat
equation become instantaneously smooth (even if the initial data is not). In
the free-space case, this property is reflected in the fact that the fundamental
2
solution (4πDt)−n/2 e−|x| /4Dt is infinitely differentiable for t > 0 (though it is a
delta function at t = 0!), and in the second integral of (28) (the one containing
the intial data u0 (x)), derivatives in x and t will fall on the fundamental solution.
• Infinite propagation speed: suppose f ≡ 0 (no sources), and u0 (x) ≥ 0 is
positive in a ball of radius 1 about the origin, and vanishes outside a ball of
radius 2. Then for any t > 0, and any x ∈ Rn ,
Z
2
−n/2
e−|y−x| /(4Dt) u0 (x)dx > 0.
u(x, t) = (4πDt)
Rn
That is, the solution becomes positive at all points in space instantaneously for
t > 0 – hence the initial, localized disturbance is propagated with infinite speed.
45
3. Maximum principle for the heat equation.
We have already seen that the (elliptic) maximum principle is a powerful tool for
analysing solutions of Laplace and Poisson equations. An analogous (parabolic) maximum principle plays the same role for heat equations.
Theorem: [Maximum principle for the heat equation]. Let V be a bounded
(and open, and connected) region in Rn , and let T > 0. Suppose u(x, t) is continuous
on the closed cylinder V̄ × [0, T ], and continuously differentiable (once in t, twice in
x) and solving the heat equation ∂u/∂t = D∆u in V × (0, T ]. Then the maximum
(and the minimum) of u over V̄ × [0, T ] is attained either initially (at t = 0) or on
the spatial boundary (x ∈ ∂V ).
Remark: Just as in the “elliptic” case, there is also a “strong maximum principle”
which says that if the max. (or min.) of u is also attained at an interior point (x0 , t0 )
(x0 ∈ V , t0 > 0), then u ≡ const. for t ≤ t0 . (Notice it doesn’t say anything for
t > t0 .) We will not prove this here.
Proof of the maximum principle: the same argument as for harmonic functions – i.e.
considering the function
v(x, t) := u(x, t) + |x|2
for > 0 – shows that the max. (and min.) is attained somewhere on the boundary
of the cylinder (since ∂u/∂t = 0 at an interior max. or min.). We only have to show
that the max. (or min.) is attained somewhere on the boundary other than at the
“final time” t = T . So suppose v has a max. (say) at (x0 , T ), for some x0 ∈ V . Then
∆v(x0 , T ) ≤ 0,
and
∂v
v(x0 , T ) − v(x0 , T − h)
(x0 , T ) = lim
≥ 0.
h→0+
∂t
h
Hence
0≤
∂v
(x0 , T ) − D∆v(x0 , T ) = −2D < 0,
∂t
a contradiction. So for any (x, t) ∈ V̄ × [0, T ], we have
u(x, t) = v(x, t) − |x|2 ≤ v(x, t) ≤
max
v≤
max
u
∂CT \{t=T }
and letting ↓ 0, we find
u(x, t) ≤
∂CT \{t=T }
as desired. 46
max
∂CT \{t=T }
u + (const.),
Just as for the Laplace/Poisson equation, an immediate consequence of the maximum
principle is the uniqueness of solutions of the initial-boundary-value problem for the
heat equation.
As usual let V be a bounded (open, connected) domain in Rn . Fix any T > 0. Let
f (x, t), g(x, t), and u0 (x) be continuous functions (on V × (0, T ), ∂V × [0, T ], and V ,
respectively), and consider the problem


∂u
∂t
− D∆u = f (x, t),
x ∈ V, 0 < t ≤ T
.
u(x, t) = g(x, t), x ∈ ∂V, 0 ≤ t ≤ T

u(x, 0) = u0 (x), x ∈ V
(29)
Theorem: There is at most one function u(x, t) which is continuous on V̄ × [0, T ],
continuously differentiable (once in t, twice in x) in V ×(0, T ], and solves problem (29).
Proof: if there are 2 solutions, their difference w(x, t) satisfies


∂w
∂t
− D∆w = 0,
x ∈ V, t > 0
.
w(x, t) = 0, x ∈ ∂V, t > 0

w(x, 0) = 0, x ∈ V
Applying the (parabolic) maximum principle, we conclude that the max. and min.
values of w on the cylinder are 0. Hence w ≡ 0. 47
4. Methods of images and eigenfunction expansion.
Notation: to make the writing easier, we will often denote derivatives using subscripts,
2
, uxx = ∂∂xu2 , etc.
eg. ut = ∂u
∂t
Example: (diffusion on the 1/2-line). Solve

 ut − uxx = f (x, t) x > 0, t > 0
u(0, t) = 0 t > 0

u(x, 0) = u0 (x) x > 0
(30)
by finding the Green’s function.
Recalling our notation σ = t − τ , the problem for the Green’s function G(x, t; y, σ)
is: for x > 0, t > 0,

 Gσ − Gyy = δ(y − x)δ(σ) y > 0, σ ≥ 0
G|y=0 = 0

G=0 σ<0
Since the geometry is so simple, let’s try the method of images. We know that the
free-space Green’s function (in one space dimension) with singularity at x is
Gf (x, t; y, σ) = √
(y−x)2
1
e− 4σ .
4πσ
If we put our “image charge” at −x (outside our domain!), we can also satisfy the
boundary conditions: set, for σ > 0,
(y+x)2
(y−x)2
1
f
f
− 4σ
− 4σ
G(x, t; y, σ) := G (x, t; y, σ) − G (−x, t; y, σ) = √
−e
.
e
4πσ
Then
Gσ − Gyy = δ(y − x)δ(σ) − δ(y + x)δ(σ) = δ(y − x)δ(σ)
(since x > 0 and y > 0), and furthermore
2
2
y
1
− 4σ
− y4σ
e
G|y=0 = √
−e
=0
4πσ
so we are in business! Replacing σ by t − τ , our Green’s function is, for τ < t (recall
it is zero for τ > t),
(y−x)2
(y+x)2
1
− 4(t−τ )
− 4(t−τ )
G(x, t; y, τ ) = p
e
−e
4π(t − τ )
48
and the resulting solution formula for our half-line problem (30) is
Z tZ
∞
0
+√
(y−x)2
− 4(t−τ )
(y+x)2
− 4(t−τ )
p
e
−e
f (y, τ )dy dτ
4π(t − τ )
Z ∞
(y−x)2
(y+x)2
− 4t
− 4t
e
−e
u0 (y)dy.
u(x, t) =
0
1
1
4πt
0
It can be checked rigorously that for reasonable functions f and u0 (precisely: bounded
and continuous, with f also continuously differentiable), this formula gives a continuously differentiable (once in t twice in x) function u which solves (30) – but we won’t
do it here.
Example: (diffusion on a rod). Solve

0 < x < L, t > 0
 ut − uxx = 0
u(0, t) = u(L, t) = 0
t>0
 u(x, 0) = u (x)
0
≤
x
≤L
0
(31)
by finding the Green’s function.
The method of images will not work so well here, so we try an eigenfunction expansion
instead. The eigenfunctions of the spatial part of the differential operator (d2 /dx2 )
with zero BCs at x = 0 and x = L are sin(nπx/L), n = 1, 2, 3, . . ., and so we seek
our Green’s function in the form
G(x, t; y, σ) =
∞
X
gn (x, t; σ) sin(nπy/L)
n=1
for σ > 0. We require Gσ − Gyy = δ(y − x)δ(σ), thus
∞ X
n=1
n2 π 2
(gn )σ + 2 gn sin(nπy/L) = δ(y − x)δ(σ)
L
and so
n2 π 2
2
(gn )σ + 2 gn =
L
L
Z
L
sin(nπy/L)δ(y − x)δ(σ)dy =
0
For σ > 0, we have (gn )σ + (n2 π 2 /L2 )gn = 0, and so
2 π 2 σ/L2 )
gn = Ce−(n
49
2
sin(nπx/L)δ(σ).
L
and we determine the constant C from a jump condition:
Z 0+
Z 0+
L
(gn )σ + (n2 π 2 /L2 )gn dσ
1=
δ(σ)dσ =
2 sin(nπx/L) 0−
0−
L
L
=
gn |σ=0+
C
σ=0− =
2 sin(nπx/L)
2 sin(nπx/L)
(where we used G = 0 for σ < 0). Hence C = 2 sin(nπx/L)/L, and we have an
expression for our Green’s function
∞
2 X −(n2 π2 σ/L2 )
e
sin(nπx/L) sin(nπy/L)
G=
L n=1
(which indeed satisfies the zero boundary conditions at y = 0 and y = L). The
corresponding formula for the solution of problem (31) is
Z L
∞
2 X −(n2 π2 t/L2 )
u(x, t) =
e
sin(nπx/L)
sin(nπy/L)u0 (y)dy.
L n=1
0
50
5. Green’s function for the 1D wave equation.
Suppose we measure the displacement u(x, t) from equilibrium, at each point x in a
bounded region V in Rn (representing an elastic string (n = 1), membrane (n = 2), or
solid (n = 3)) , and at each time t. The string/membrane/solid is subjected to forces
f (x, t), and at its boundary, it is held at fixed displacement g(x, t) (x ∈ ∂V ). At time
t = 0, at each x ∈ V , the initial displacement is u0 (x), and the initial velocity is v0 (x).
Assuming the displacements are small, the following initial-boundary value problem
for the wave equation is a reasonable description of the displacement u(x, t) at later
times t > 0:

∂2u
 wave equation:
− c2 ∆u = f (x, t),
x ∈ V, t > 0
∂t2
,
(32)
boundary value: u(x, t) = g(x, t), x ∈ ∂V, t > 0

∂u
initial conditions: u(x, 0) = u0 (x), ∂t (x, 0) = v0 (x), x ∈ V
where c > 0 is the wave speed. The wave equation also describes the propagation of
other waves, such as sound and light.
Just as for the heat equation, we would like to solve problems like this one via Green’s
functions. We will jump right in, by looking immediately for the free-space Green’s
functions (or fundamental solutions) in dimensions one, two, and three (unlike
for the heat equation, the form of the Green’s function depends significantly on the
dimension).
1D free-space Green’s function for the wave equation:
As with the heat equation, the Greens function should be a function of two sets
of space-time variables: x, t; y, τ . And as for the heat equation, it is sometimes
convenient to work with σ := t − τ , rather than τ . Hence, for x, t, y, σ ∈ R, we seek
G(x, t; y, σ) solving
Gσσ − c2 Gyy = δ(y − x)δ(σ),
−∞ < y < ∞, σ ≥ 0
.
(33)
G ≡ 0,
σ < 0 (causality)
You can think of G as the signal (eg. sound or light) emitted by a unit point source
at spatial point x and at time 0.
For the sake of variety, we will solve this problem using the Laplace transform.
Recall that the Laplace transform of a (say, bounded, continuous) function f defined
on [0, ∞) is another function defined on [0, ∞):
Z ∞
f˜(s) = L(f )(s) :=
e−st f (t)dt.,
0
and recall the key property (“Laplace transform turns differentiation into coordinate
multiplication”)
L(f 0 )(s) = sL(f )(s) − f (0),
51
which is easily verified by integration by parts. Thus if we let G̃(x, t; y, s) be the
Laplace transform of G in the variable σ, and use L(δ) = 1, and the causality condition, we find
s2 G̃ − c2 G̃yy = δ(y − x).
Solving this ODE (in y) to the left and right of x, and imposing the conditions that
G̃ decay as y → ±∞, and that is be continuous at y = x, yields
s
G̃ = Ae− c |y−x| .
As usual, the remaining constant A is determined by the jump condition:
Z x+
Z x+ h
i
1=
δ(y − x)dy =
s2 G̃ − c2 Gyy dx = −c2 Gy |y=x+
y=x− = 2scA
x−
x−
hence A = (2sc)−1 , and
1 e−(|y−x|/c)s
G̃(x; y, s) =
.
2c
s
Now for any r ≥ 0, the Laplace transform of the Heavyside function
Z ∞
e−rs
0 t<0
e−st dt =
=⇒ L(H(t − r)) =
H(t) :=
,
1 t≥0
s
r
so by comparison, we must have
1
1
G = H σ − |y − x| .
2c
c
Restoring τ = t − σ, we arrive at
1
1
G(x, t; y, τ ) = H t − τ − |y − x| ,
2c
c
our 1D free-space Green’s function. It is instructive to sketch a space-time graph (i.e.
in the y-τ plane) of G!
Knowing the Green’s function, we can find the solution to the initial value problem
for the wave equation on the line:
utt = c2 uxx
− ∞ < x < ∞, t > 0
,
(34)
u(x, 0) = u0 (x), ut (x, 0) = v0 (x)
which is sometimes called the Cauchy problem for the wave equation.
52
Integrating by parts, we have, for any T > t,
Z TZ ∞
Z TZ ∞
u(y, τ )δ(y − x)δ(τ − t)dydτ =
u(y, τ )(Gτ τ − c2 Gyy )dydτ
u(x, t) =
0
−∞
0
−∞
Z TZ ∞
Z ∞
=
(uτ τ − c2 uyy )Gdydτ +
(uGτ − uτ G)|ττ =T
=0 dy
0
−∞
−∞
Z ∞
(v0 (y)G(x, t; y, 0) − u0 (y)Gτ (x, t; y, 0))dy
=
−∞
where we used the causality condition G ≡ Gτ ≡ 0 for τ > t. Now, since H 0 = δ,
Gτ = −Gσ = −
1
1
1
δ(σ − |y − x|) = − [δ(y − (x + ct)) + δ(y − (x − ct))] ,
2c
c
2
and so our solution to (34) is
1
1
[u0 (x + ct) + u0 (x − ct)] +
u(x, t) =
2c
2c
Z
x+ct
v0 (y)dy,
x−ct
which is known as D’Alembert’s formula.
Remark:
• Notice that D’Alembert’s formula represents the sum of two waves, one moving
to the left with speed c, one moving to the right with speed c:
Z
1
1 z
+
−
±
u(x, t) = f (x + ct) + f (x − ct),
f (z) := u0 (z) ±
v0 (z)dz.
2
2c 0
• It is simple to check D’Alembert’s formula does, in fact, solve problem (34).
Indeed, if f is any twice differentiable function, then f (x ± ct) solves the wave
equation: [ ∂t∂2 − ∂x∂ 2 ]f (x ± ct) = c2 f 00 − c2 f 00 = 0. It is also easy to check the
initial data are satisfied. Hence, assuming u0 is twice differentiable, and v0 is
once differentiable, we have solved problem (34).
• Finite speed of propagation: The solution at space-time point (x, t) depends
only on the initial data (u0 and v0 ) in the interval [x − ct, x + ct] (draw a graph!)
– that is, signals propagate with speed at most c.
• Sidenote: D’Alembert’s formula makes sense even if u0 and v0 are not differentiable (just continuous) – though then the PDE doesn’t hold in the classical
sense, only in the sense of distributions.
53
6. The wave equation in higher dimensions.
2D free-space Green’s function for the wave equation:
We look for a Green’s function depending on the spatial variable y only through
r := |y − x|, leading to the problem
r > 0, σ > 0
Gσσ − c2 Grr + 1r Grr = δ(σ)δ(y − x),
.
G ≡ 0,
σ<0
Taking again the Laplace transform in the variable σ we are lead to
1
2
2
s G̃ − c G̃rr + G̃r = δ(y − x).
r
For r > 0, the ODE
1
s2
G̃rr + G̃r − 2 G̃ = 0
r
c
is solved by G̃ = AK0 ((s/c)r), where K0 (z) is the modified Bessel function of order
0 which solves
 2 00
2
0
 z K0 (z) + zK0 (z) − z K0 (z) = 0
K0 (z) ∼ log z as z → 0
.
 K (z) ∼ (const) e√−z as z → ∞
0
z
The constant A can be determined using the divergence theorem: denoting the disk
of radius about x by B ,
Z
Z
Z
∂ G̃
2
2
1 = lim
δ(y − x)dy = −c lim
∆G̃dy = −c lim
dS
→0+ B
→0+ B
→0+ r= ∂n
Z
c
2 s
= −c A lim
dS = −2πc2 A
c →0+ r= sr
so A = −(2πc2 )−1 , and
s 1
K
r .
0
2πc2
c
It turns out we can invert the Laplace transform on the Bessel function explicitly. In
fact, for b > 0, we have
1
H(t − b) = −K0 (bs),
L √
t2 − b2
G̃ = −
and so
G=
1
1
p
H(σ − r/c)
2
2
2πc
σ − r2 /c2
54
and replacing τ = t − σ, r = |y − x|, our two-dimensional free-space Green’s function
for the wave equation is
1
1
q
G(x, t; y, τ ) =
H t − τ − |y − x| .
c
2
2πc (t − τ )2 − c12 |y − x|2
Remark: As with the 1D wave equation, the fact that the Green’s function is supported inside the cone { |y − x| ≤ c(t − τ ) } shows that signals propagate with speed
no greater than c. Notice also, as in the 1D case, signals do not propagate “sharply”
in 2D, since for fixed x, y, τ , the Green’s function is non-zero for times t beyond
τ + 1c |y − x| (though it does decay like 1/t).
It is left as an exercise to use this expression for the free-space Green’s function to
show that the solution of the Cauchy problem
utt = c2 ∆u
x ∈ R2 , t > 0
,
u(x, 0) = u0 (x), ut (x, 0) = v0 (x)
for the 2D wave equation, is


Z
Z
1 
v0 (y)
∂
u0 (y)
q
q
u(x, t) =
dy +
dy  ,
2
2πc
∂t
1
1
2
2
2
2
|y−x|≤ct
|y−x|≤ct
t − c2 |y − x|
t − c2 |y − x|
sometimes known as Poisson’s formula.
3D free-space Green’s function for the wave equation:
We again look for a Green’s function depending on the spatial variable y only through
r := |y − x|, leading to the problem
Gσσ − c2 Grr + 2r Grr = δ(σ)δ(y − x),
r > 0, σ > 0
.
G ≡ 0,
σ<0
Taking again the Laplace transform in the variable σ we are lead to
2
2
2
s G̃ − c G̃rr + G̃r = δ(y − x).
r
For r > 0, the ODE
2
s2
G̃rr + G̃r − 2 G̃ = 0
r
c
is solved by (we’ve seen this before in a homework problem)
s
e− c r
G̃ = A
.
r
55
Again, we find the constant A using the divergence theorem. Let B be the ball of
radius about x. Then
Z
Z
Z
∂ G̃
2
2
1 = lim
δ(y − x)dy = −c lim
∆G̃dy = −c lim
dS
→0+ B
→0+ B
→0+ r= ∂n
Z
s
s s e− c r e− c r
2
− 2 dS = 4πc2 A
= −c lim
A −
→0+ r=
c r
r
so A = (4πc2 )−1 , and
s
1 e− c r
.
G̃ =
4πc2 r
Now notice that for b > 0,
Z
L(δ(t − b)) =
∞
e−st δ(t − b)dt = e−bs
0
and so by comparison,
1
δ(σ − r/c).
4πc2 r
Reinstating τ = t − σ, we have the 3D free-space Green’s function for the wave
equation,
1
1
G(x, t; y, τ ) =
δ t − τ − |y − x| .
4πc2 |y − x|
c
G=
Remark: Notice that the Green’s function is supported exactly on the cone { |y − x| =
c(t − τ ) } (sketch a graph). So not only do signals propagate with speed c, they are
also crisp in 3D – standing at y, you receive a signal emitted at time τ = 0 and point
x, exactly at time t = |y − x|/c and then it is gone! It’s nice to live in 3D.
Again, it is left as an exercise to use this expression for the free-space Green’s function
to show that the solution of the Cauchy problem
utt = c2 ∆u
x ∈ R3 , t > 0
,
u(x, 0) = u0 (x), ut (x, 0) = v0 (x)
for the 3D wave equation, is
Z
Z
1
1
∂ 1
u(x, t) =
v0 (y)dS(y) +
u0 (y)dS(y) ,
4πc2 t |y−x|=ct
∂t t |y−x|=ct
sometimes known as Kirchoff ’s formula. Note again the crisp signal propagation,
reflected in the fact that the solution at (x, t) is given in terms of integrals of the data
over the sphere of radius ct about x.
56
B. VARIATIONAL METHODS
I. EIGENVALUE PROBLEMS
Let D ⊂ Rn be a bounded domain, and consider the following variable-coefficient
generalizations of the heat equation
r(x)ut = ∇ · [p(x)∇u] − q(x)u
x ∈ D, t > 0
(35)
∂u
+ bu = 0
x ∈ ∂D
∂n
and the wave equation
r(x)utt = ∇ · [p(x)∇u] − q(x)u
∂u
+ bu = 0
x ∈ ∂D
∂n
x ∈ D, t > 0
(36)
for (smooth) functions p(x) > 0 (the spatially variable diffusion rate in the first case,
wave speed in the second), r(x) > 0, and q(x) (a source of heat loss in the first case,
friction in the second), and constant b ≥ 0 (notice the boundary conditions are a mix
of Dirichlet and Neumann for the moment).
Suppose for some λ ≥ 0, we seek solutions of (35) of the separated-variables form
u(x,
t) = e−λt φ(x), or solutions of (36) of the (oscillatory in time) form u(x, t) =
√
ei λt φ(x). Then in both cases, as an easy substitution will show, we arrive at the
following problem for φ(x), involving only the variable x:
Lφ(x) := −∇ · [p(x)∇φ] + q(x)φ = λr(x)φ
in D
.
(37)
∂φ
+ bφ = 0
on ∂D
∂n
Problem (37) is an eigenvalue problem for the (differential) operator L. If there
is a non-zero solution φ(x) for some λ ∈ R, we call φ an eigenfunction, and λ the
corresponding eigenvalue.
We will be studying the eigenvalue problem (37) for the next few lectures.
1. Basic properties of eigenvalues and eigenfunctions
We begin by listing the fundamental properties of the eigenvalues and eigenfunctions
of problem (37). Our basic assumptions: D is a smooth, bounded, connected region
in Rn ; p, q, and r are smooth functions on D̄, with p(x) > 0 and r(x) > 0.
1. The eigenvalues form an infinite sequence, tending to plus infinity:
λ1 ≤ λ2 ≤ λ3 ≤ . . . ,
λj → ∞ as j → ∞.
The multiplicity of an eigenvalue λ is the dimension of the subspace of eigenfunctions corresponding to λ – that is, the maximum number of linearly independent solutions φ of Lφ = λrφ. It is always finite. Convention: we will
incorporate multiplicity in our ordered list of eigenvalues above by repeating
each eigenvalue according to its multiplicity.
57
2. Eigenfunctions corresponding to different eigenvalues are orthogonal in the
sense:
Z
λj 6= λk =⇒ (φj , φk )r :=
φj (x)φk (x)r(x)dx = 0.
D
This means that by normalizing, we may (and will!) assume that the eigenfunctions φj corresponding to the eigenvalues λj form an orthonormal set:
Z
(φj , φk )r =
φj (x)φk (x)r(x)dx = δjk =
D
1 j=k
0 j=
6 k
(for eigenvalues of multiplicity greater than 1, we can choose an orthonormal
set of eigenfunctions using the Gram-Schmidt orthogonalization procedure).
3. The eigenfunctions form a complete set. This means that any function u which
is continuous on D̄ can be expanded in terms of the eigenfunctions:
u(x) =
∞
X
Z
cj φj (x),
cj = (φj , u)r =
φj (x)u(x)r(x)dx
D
j=1
in the sense
N
X
lim u(x) −
cj φj (x) = 0
N →∞ j=1
where k · k denotes the L2 norm, defined by
2
Z
kf k :=
(f (x))2 dx.
D
4. If q(x) ≥ 0, then λ1 ≥ 0 (and hence λj ≥ 0 for all j) and λ1 = 0 only if q ≡ 0,
b = 0, and φ1 = constant.
Proofs of properties 1 (existence of eigenfunctions) and 3 (completeness of eigenfunctions) are beyond the scope of this course, although we will sketch some ideas a bit
later on. Properties 2 and 4 are easy to prove, and we will do that here.
First, notice that L is self-adjoint. That is, if u and v are smooth functions on D
∂u
∂v
both satisfying the boundary conditions: ∂n
+ bu = ∂n
+ bv = 0 on ∂D, then using
58
the divergence theorem,
Z
(v, Lu) − (Lv, u) =
[v(x)(Lu)(x) − u(x)(Lv)(x)] dx
ZD
[−v∇ · (p∇u) + qvu + u∇ · (p∇v) − quv] dx
Z Z
∂v
∂u
∇ · (p(u∇v − v∇u))dx =
p u
=
−v
dS
∂n
pn
D
∂D
Z
=b
[p(−uv + uv)] dS = 0.
=
D
∂D
Proof of property 2, orthogonality of eigenfunctions with different eigenvalues: using
Lφj = λj rφj , Lφk = λk rφk , the fact that φj and φk satisfy the BCs, and the selfadjointness of L, we have
(λj − λk )(φj , φk )r = (λj rφj , φk ) − (φj , λk rφk )r = (Lφj , φk ) − (φj , Lφk ) = 0.
Hence if λj 6= λk , then (φj , φk )r = 0. To prove property 4, we first derive a simple, but important identity: if u is a smooth
function satisfying the BCs, then
Z
u(x) [−∇ · (p(x)∇u(x)) + q(x)u(x)] dx
(u, Lu) =
D
Z
=
−∇ · (u(x)p(x)∇u(x)) + p(x)|∇u(x)|2 + q(x)u2 (x) dx
ZD
Z
(38)
∂u
2
2
=
p(x)|∇u(x)| + q(x)u (x) dx −
p(x)u(x) (x)dS(x)
∂n
∂D
ZD
Z
p(x)u2 (x)dS(x).
=
p(x)|∇u(x)|2 + q(x)u2 (x) dx + b
D
∂D
Proof of property 4, positivity of first eigenvalue for positive q: apply the above identity to the eigenfunction φ1 corresponding to the first eigenvalue λ1 :
λ1 = λ1 (φ1 , φ1 )r = (φ1 , λ1 rφ1 ) = (φ1 , Lφ1 )
Z
Z
2
2
=
p(x)|∇φ1 (x)| + q(x)φ1 (x) dx + b
D
φ21 (x)dS(x) ≥ 0
∂D
since p > 0, b ≥ 0, and assuming q ≥ 0. And the only way to get λ1 = 0 is if
φ1 ≡ constant , q ≡ 0, and b = 0. 59
2. The “energy”, and variational principles for eigenvalues.
The identity (38) turns out to be a very important and useful one. We will specialize
for now to the two most common boundary conditions:
∂u
∂n
• Neumann BCs: b = 0 – i.e.
= 0 on ∂D
• Dirichlet BCs: “b = ∞” – i.e. u = 0 on ∂D.
In both of these cases, the boundary integral in (38) vanishes. Let’s give this quantity
a name.
Definition: Given the operator L (i.e. the functions p and q), the (Dirichlet) energy
of a function u on D is
Z
E(u) :=
p(x)|∇u(x)|2 + q(x)u2 (x) dx.
D
Now suppose u is any (smooth) function satisfying the boundary conditions (Dirichlet
or Neumann). By completeness of the eigenfunctions, we may write
∞
X
u(x) =
cj φj (x),
j=1
and notice that
(u, u)r =
∞
X
cj φj ,
j=1
∞
X
!
=
ck φk
k=1
r
∞
X
cj ck (φj , φk )r =
∞
X
cj ck δjk =
c2j .
j=1
j,k=1
j,k=1
∞
X
Now, as computed in (38),
E(u) = (u, Lu) =
∞
X
cj ck (φj , Lφk ) =
j,k=1
=
∞
X
j,k=1
cj ck λk δjk =
∞
X
cj ck (φj , λk rφk ) =
j,k=1
∞
X
λk c2k .
j=1
And since λk ≥ λ1 for all k,
E(u) ≥ λ1
∞
X
c2k = λ1 (u, u)r ,
j=1
or
λ1 ≤
E(u)
,
(u, u)r
60
∞
X
j,k=1
cj ck λk (φj , φk )r
and we see that the only way to get equality here, is if the only non-zero coefficients
cj are those corresponding to the lowest eigenvalue (i.e. λj > λ1 =⇒ cj = 0).
That is, equality only holds if and only if u is an eigenfunction corresponding to the
lowest eigenvalue λ1 . This observation gives us our first example of a variational
principle:
Theorem: [variational principle for the first eigenvalue]: the lowest eigenvalue
of the operator L (with Dirichlet or Neumann BCs) is given by
R
[p(x)|∇u(x)|2 + q(x)u2 (x)] dx
E(u)
D
R
=
min
.
λ1 =
min
u2 (x)r(x)dx
u satisfying BCs
u satisfying BCs (u, u)r
D
Remark: The quantity
E(u)
(u, u)r
is sometimes called the Rayleigh quotient.
Here is a simple example of how a variational principle like this might be used.
2
d
Example: Find an upper bound for the lowest eigenvalue (with r ≡ 1) of L = − dx
2 +x
on the interval [0, 1] with Dirichlet (zero) BCs at the endpoints.
We have D = [0, 1], r(x) ≡ 1, p(x) ≡ 1, q(x) = x. According to the variational
principle, any function u(x) on [0, 1] which is zero at the endpoints, gives us an upper
bound on λ1 : λ1 ≤ E(u)/(u, u).
1. One simple such choice is u(x) = x(1 − x) = x − x2 . Then u0 (x) = 1 − 2x, and
we compute
Z
1
Z
2 2
(x − x ) dx =
(u, u) =
0
1
(x2 − 2x3 + x4 )dx = (x3 /3 − x4 /2 + x5 /5)|10 =
0
1
30
and
Z
1
E(u) =
Z0 1
=
0
(1 − 2x)2 + x(x − x2 )2 dx
1
(x5 − 2x4 + x3 ) + 4x2 − 4x + 1 dx = +
3 60
and hence
λ1 ≤
E(u)
1
= 10 + .
(u, u)
2
2. A little more thought will give us a better (i.e. smaller) upper bound, supposing d2
is small. For = 0, we know the (Dirichlet) eigenfunctions of − dx
2 are sin(kπx),
2 2
k = 1, 2, . . ., with eigenvalues π k . Let’s use the lowest = 0 eigenfunction
61
√
u(x) = 2 sin(πx) (notice we have normalized it so (u, u) = 1) as our “test
function” in the variational principle:
Z 1h √
i
√
E(u)
( 2π cos(πx))2 + x( 2 sin(πx))2 dx
λ1 ≤
=
(u, u)
0
Z 1
1
x sin2 (πx)dx = π 2 + ,
= π 2 + 2
2
0
which is indeed better (since π 2 < 10), and in particular becomes exact in the
limit → 0.
We can easily generalize the above variational principle to the higher eigenvalues.
Suppose u is a function on D satisfying the boundary conditions, and which is orthogonal to the first n − 1 eigenfunctions:
(u, φj )r = 0
j = 1, 2, . . . , n − 1.
Then c1 = c2 = · · · = cn−1 = 0, so
u(x) =
∞
X
cn φj (x),
(u, u)r =
∞
X
j=n
and
E(u) =
∞
X
c2j
j=n
λj c2j ≥ λn
j=n
∞
X
c2j ≥ λn (u, u)r
j=n
with equality if and only if cj = 0 for all j with λj > λn – i.e., if and only if u is an
eigenfunction with eigenvalue λn . Hence:
Theorem: [variational principle for higher eigenvalues]: the n-th eigenvalue of
the operator L (with Dirichlet or Neumann BCs) is given by
λn =
E(u)
.
satisfying BCs,(u,φ1 )r =···=(u,φn−1 )r =0 (u, u)r
min
u
Remark: Our variational principles apply also for the more general boundary condi∂u
tions ∂n
+ bu = 0 on ∂D, provided we use the full form of the energy from (38):
Z
Z
2
2
E(u) = (p|∇u| + qu ) + b
pu2 .
D
∂D
In practice, the variational principle above for higher eigenvalues (n > 1) is not so
useful, since we typically do not know what the eigenfunctions φ1 , . . . , φn−1 are! A
62
more useful version is the following, in which we minimize over functions orthogonal
to any set of n − 1 functions (not necessarily the eigenfunctions), and then maximize
over that set.
Theorem: [Courant max-min principle]
λn =
max
functions m1 ,...,mn−1 u
E(u)
min
satisfying BC,(u,mj )r =0 (u, u)r
Proof: let m1 , m2 , . . . , mn−1 by any functions on D. We will build a function u,
satisfying the BCs, and which is orthogonal to all the mk , of the form
u(x) =
n
X
aj φj (x).
j=1
That is, we want
0 = (u, mk )r =
n
X
aj (φj , mk )r ,
k = 1, 2, . . . , n − 1.
j=1
This is n − 1 linear equations in the n variables a1 , a2 , . . . , an . Since there are more
variables than equations, linear algebra tells us there is a non-zero solution (in fact,
at least a one-parameter family of solutions). Now as above,
E(u) = (u, Lu) =
n
X
λj a2j
j=1
hence
≤ λn
n
X
a2j = λn (u, u)r ,
j=1
E(u)
≤ λn .
satisfying BC,(u,mj )r =0 (u, u)r
min
u
But since this is true for any choice of m1 , . . . , mn−1 , it is true for the maximum over
all such choices:
E(u)
max
min
≤ λn .
functions m1 ,...,mn−1 u satisfying BC,(u,mj )r =0 (u, u)r
It remains to show that the maximum actually equals λn – that is, for some choice of
m1 , . . . , mn−1 , and for any u which is orthogonal to all the mj , E(u)/(u, u)r ≥ λn . For
this, we take mj = φj for j = 1, . . . , n − 1. Then any u orthogonal to these functions
can be written
∞
X
u(x) =
cj φj (x),
j=n
63
and, as above
E(u) =
∞
X
λj c2j
≥ λn
j=n
∞
X
c2j = λn (u, u)r
j=n
as required. .
Important remark: It turns out that in the case of Neumann BCs, the variational
principles for eigenvalues discussed above (including the max-min principle) remain
true if the minimization is done over all test functions u – i.e. we do not have to
impose the Neumann BCs on the functions u inserted into the Rayleigh
quotient. In the case of Dirichlet BCs, however, the test functions u do indeed need
to vanish on the boundary. We will return to this point when we discuss the calculus
of variations.
64
3. Bounds on eigenvalues.
We fix here Dirichlet (zero) BCs, and, as always, denote by λn the n-th (Dirichlet)
eigenvalue of L = −∇ · p(x)∇ + q(x) (with respect to r(x)).
Bounds on Coefficients
Suppose

 0 < pmin ≤ p(x) ≤ pmax
qmin ≤ q(x) ≤ qmax
 0<r
min ≤ r(x) ≤ rmax
in our domain D. Denote by λn,min the n-th eigenvalue of
−∇ · (pmin ∇φ) + qmin φ = λn,min rmax φ
φ=0
on ∂D
in D
and by λn,max the n-th eigenvalue of
−∇ · (pmax ∇φ) + qmax φ = λn,max rmin φ
φ=0
on ∂D
in D
,
.
Theorem:
λn,min ≤ λn ≤ λn,max .
Proof: notice that for any function u on D,
Z
Z
2
2
E(u) =
p|∇u| + qu dx ≤
pmax |∇u|2 + qmax u2 dx =: Emax (u)
D
D
and
Z
Z
2
u r dx ≥
(u, u)r =
D
u2 rmin dx = (u, u)rmin ,
D
so that
Emax (u)
E(u)
≤
.
(u, u)r
(u, u)rmin
Also notice that for any functions m1 , . . . , mn−1 on D,
(u, mj )r = 0 if and only if (u, m̃j )rmin = 0 where m̃j (x) :=
r(x)
mj (x).
rmin
So by the min-max principle,
E(u)
m1 ,...,mn−1 {u=0 ∂D ,(u,mj )r =0} (u, u)r
Emax (u)
≤ max
min
= λn,max .
m̃1 ,...,m̃n−1 {u=0 ∂D ,(u,m̃j )rmin =0} (u, u)rmin
λn =
max
min
65
The other inequality is really the same thing. 2
d
2
Example: Find upper and lower bounds for the n-th eigenvalue of L = − dx
2 + x on
[0, 1] with Dirichlet (zero) BCs. Here ≥ 0.
So here D = [0, 1], p ≡ 1, r ≡ 1, q(x) = x2 . Let’s use the easy bounds
qmin := 0 ≤ q(x) ≤ =: qmax
2
d
on [0, 1]. So λn is sandwiched between the n-th (Dirichlet) eigenvalue of − dx
2 , and
2
d
that of − dx2 + . That is
π 2 n2 ≤ λn ≤ π 2 n2 + .
Notice that we get a better (i.e. smaller) upper bound for the first eigenvalue by
d2
taking sin(πx) (the first Dirichlet eigenfunction of − dx
2 ) as a test function in the
variational principle:
R1 2
2
2
2
π
(πx)
dx
cos
(πx)
+
x
sin
E(sin(πx))
1
π 2 /2 + /6
λ1 ≤
= 0
= π 2 + .
=
R1 2
(sin(πx), sin(πx))
1/2
3
sin (πx)dx
0
Bounds on geometry
Now let’s fix the coefficients p, q, and r, and make the dependence on the domain D
explicit; that is, denote by λn (D) the n-th eigenvalue of our operator L = −∇ · p∇ + q
on D with Dirichlet BCs.
Theorem:
D̃ ⊂ D
=⇒
λn (D) ≤ λn (D̃).
That is, “the smaller the domain, the larger the eigenvalue”.
Proof: Again, we’ll use the max-min principle. Let m1 , . . . , mn−1 be any given functions on D. Notice that if u is a function on D̃ which vanishes on ∂ D̃ and satisfies
(with integrals restricted to D̃), (m1 , u)r,D̃ = · · · = (mn−1 , u)r,D̃ = 0 (i.e. u is an
“admissible” test function for the Rayleigh quotient in the min-max principle for D̃),
then its extension
u(x)
x ∈ D̃
û(x) :=
0
x ∈ D\D̃
is a function on D, vanishing on the boundary ∂D, and satisfying (with integrals now
over D) (m1 , u)r,D = · · · = (mn−1 , u)r,D = 0 (i.e. û is an “admissible” test function
for the Rayleigh quotient in the min-max principle for D). Further (denoting region
of integration with a subscript),
ED (û) = ED̃ (u),
(û, û)r,D = (u, u)r,D̃ .
66
So
E(û)
E(û)
≤
min
{û=0 ∂D, (mj ,û)r =0} (û, û)r
{û≡0 in D\D̃, (mj ,û)r =0} (û, û)r
E(u)
=
min
{u=0 ∂ D̃, (mj ,u)r =0} (u, u)r
min
≤
max
{fns. m1 ,...,mn−1
on
(minimizing over smaller set)
E(u)
= λn (D̃).
(mj ,u)r =0} (u, u)r
min
D̃} {u=0 ∂ D̃,
Since this inequality holds for all choices of m1 , . . . , mn−1 (functions on D), maximizing over all such choices yields λn (D) ≤ λn (D̃), as required. Example: Find upper and lower bounds for the Dirichlet eigenvalues of −∆ on the
following 2D domain, which is the union of 2 rectangles: D := ( [0, a] × [0, b] ) ∪
( [a, a + h] × [c, c + g] ) with a, b, c, h, g positive, and 0 < c < c + g < b (draw it!).
Well, we have
[0, a] × [0, b] =: Dmin ⊂ D ⊂ Dmax := [0, a + h] × [0, b].
By separation of variables, we know the Dirichlet eigenvalues of −∆ in a rectangle [0, a] × [0, b] are π 2 (n2 /a2 + m2 /b2 ), m, n = 1, 2, 3, . . . (with eigenfunctions
sin(nπx1 /a) sin(mπx2 /b)). It is a bit annoying, though, to notate them. One way is
to denote
µk (a, b) := k − th element of the set {n2 /a2 + m2 /b2 | n, m = 1, 2, 3, . . .}
ordered by size, and counting multiplicity (so, eg., µ1 (1, 1) = 1, µ2 (1, 1) = 5, µ3 (1, 1) =
5, µ4 (1, 1) = 8, etc.). In this notation, λn ([0, a] × [0, b]) = π 2 µn (a, b), and so we can
conclude for the original problem
π 2 µn (a + h, b) ≤ λn ≤ π 2 µn (a, b).
In particular, for the first eigenvalue,
1
1
1
1
2
2
π
+
≤ λ1 ≤ π
+
.
(a + h)2 b2
a2 b 2
One more observation, using the geometric bounds established in this section. As
usual, Let D ⊂ Rn be a bounded domain, and let λk (D) denote the k-th Dirichlet
eigenvalue of the operator L = −∇ · p(x)∇ + q(x) on D. Let pmin , qmin , and rmax
be constants such that 0 < pmin ≤ p(x), qmin < q(x), and r(x) ≤ rmax . Since D
is bounded, it fits inside an n-dimensional (hyper-) cube C of (sufficiently large)
67
side-length a. Separation of values shows easily that the Dirichlet eigenvalues (with
respect to rmax ) of Lmin := −∇ · pmin ∇ + qmin on C are
qmin
pmin π 2 2
,
(k1 + · · · + kn2 ) +
2
rmax a
rmax
k1 , . . . , kn = 1, 2, 3, . . . .
In particular, the eigenvalues go off to infinity: λmin,k (C) → ∞ as k → ∞. Then our
eigenvalue bounds above imply that the Dirichlet eigenvalues λk (D) of L on D satisfy
λk (D) ≥ λmin,k (C) → ∞,
k → ∞,
and so,
lim λk (D) = ∞.
k→∞
This was a general property of the eigenvalues we listed at the beginning – which our
eigenvalue bounds (based, in turn, on variational principles) give a nice slick proof of.
68
II. CALCULUS OF VARIATIONS
1. Euler-Lagrange equations.
We begin with an example.
Example: Let D ⊂ Rn be a bounded domain. Let q(x) and g(x) be given smooth
functions on D and ∂D respectively. Among all functions u(x) on D (which are
sufficiently smooth – say, twice continuously differentiable) and satisfy u(x) = g(x)
for x ∈ ∂D, find the one which minimizes the “functional”
Z 1
2
I(u) :=
|∇u(x)| + q(x)u(x) dx.
2
D
Well, suppose I(u) is minimized by a function u∗ (with u∗ = g on ∂D). Consider now
a one-parameter family of functions nearby u∗ :
u (x) := u∗ (x) + w(x)
where w is some fixed function. In order for u to be in our allowed class of functions
(twice continuously differentiable and equal to g on the boundary), we require that
w also be twice continuously differentiable, and w = 0 on ∂D. Key observation:
u∗ minimizes I(u)
=⇒
I(u ) is minimized at = 0.
Since I(u ) is a function of just the single variable , we are in the realm of simple
calculus, and we know that at a minimum, the derivative is zero:
Z d
1
d
2
0 = I(u )=0 =
|∇u (x)| + q(x)u (x) dx=0
d
d D 2
Z ∂
∂
=
∇u (x) · ∇ u (x) + q(x) u (x) dx=0
∂
∂
ZD
=
(∇u∗ (x) · ∇w(x) + q(x)w(x)) dx
ZD
=
(∇ · [w(x)∇u∗ (x)] + w(x)[−∆u∗ (x) + q(x)]) dx
ZD
Z
∂
∗
=
w(x) [−∆u (x) + q(x)] dx +
w(x) u∗ (x)dS(x)
∂n
∂D
ZD
=
w(x)[−∆u∗ (x) + q(x)]dx
D
where we used the divergence theorem toward the end, and the fact that w = 0 on
∂D (so the boundary term disappears).
69
To summarize: in order for u∗ to minimize I(u) (among sufficiently smooth functions
with fixed boundary values g(x)), the integral above must be zero for any (sufficiently
smooth) function w(x) vanishing on the boundary ∂D. It turns out that the only
way this can happen, is if the function multiplying w in the integral is zero – that is,
−∆u∗ (x) + q(x) = 0 – as the following lemma shows.
R
Lemma: if g(x) is a continuous function on D, and if D g(x)w(x)dx = 0 for all
smooth functions w on D vanishing on the boundary, then g(x) ≡ 0 on D.
Proof: suppose, for some x0 ∈ D, g(x0 ) 6= 0. Since g is continuous, in some ball
B around x0 , we have either g(x) > 0 or g(x) < 0. So let w(x) be a little “bump
function” at x0 : that is, a smooth,
non-negative function, with w(x0 ) > 0, and
R
vanishing outside B. Then 0 = D g(x)w(x)dx is contradicted. Hence we must have
g ≡ 0 in D. So, to conclude:
u∗ minimizes I(u) with fixed boundary data =⇒
∆u∗ (x) = q(x) in D.
That is, the PDE ∆u∗ = q is a necessary condition – known as the Euler-Lagrange
equation for I – for u∗ to be a minimizer of I.
Remarks:
1. Notice that we have not touched the question of whether or not a minimizer
u∗ exists, but merely written down a necessary condition that such a minimizer
would have to satisfy. (This example, however, is simple enough that we can be
sure of the existence of a minimizer – it is the solution of the Poisson equation
∆u∗ = q in D, u∗ = g on ∂D.) The situation is completely analogous to finding
the minimum of a function of several variables. There, the necessary condition
is that the gradient vanish, which gives an algebraic equation to solve to find the
“critical points”. The question of which (if any) of these minimize the function
is then addressed separately. The difference is that in the calculus of variations,
we are trying to minimize a function defined on a (infinite-dimensional) space
of functions, rather than (finite-dimensional) Rn – and as our example showed,
the necessary condition may be not just an algebraic equation, but in fact a
differential equation.
2. Indeed, the key point here is the relation between “variational problems” (minimizing functionals) and the (partial) differential equations which arise as their
Euler-Lagrange equations. This idea is ubiquitous in physics (and many other
applications) – think, for example, of the “principle of minimal action” in mechanics, and its relation to the equations of motion (Newton’s equations). In
practice, the relationship works both ways: we may try to solve a variational
problem by solving the corresponding Euler-Lagrange equation; or, we may try
70
to solve a PDE by recognizing it as the E-L equation of some variational problem, which we then try to solve directly (we got a hint of this latter approach
when we discussed variational problems for eigenvalues — more on this to come).
3. A natural question follows from the example above: suppose we indeed have a
solution u∗ of the problem
∆u∗ = q
in D
.
∗
u =g
on ∂D
Does it really minimize I(u) among functions u with boundary values g? For
this example, in fact, it does, since for any such u, using the divergence theorem,
Z 1
1
∗
2
∗ 2
∗
I(u) − I(u ) =
|∇u| − |∇u | + q(u − u ) dx
2
2
D
Z 1
∗ 2
∗
∗
∗
=
|∇(u − u )| + ∇u · ∇(u − u ) + q(u − u ) dx
2
D
Z 1
∗ 2
∗
∗
=
|∇(u − u )| + (u − u )(−∆u + q) dx
2
D
Z
∂u∗
dS(x)
+
(u − u∗ )
∂n
∂D
Z
1
|∇(u − u∗ )|2 dx ≥ 0.
=
2 D
4. Natural BCs: consider the same minimization problem as above, except now
we would like to minimize I(u) among all (smooth) functions u (without the
condition that u = g on ∂D). What problem should a minimizer u∗ solve in this
d
case? Proceeding as above, we conclude that for uε = u∗ + εw, dε
I(uε )|ε=0 = 0
where w now is any smooth function (it does not have to vanish on ∂D). This
leads, as above, to
Z
Z
∂
∗
w(x) u∗ (x)dS(x)
0=
w(x) [−∆u (x) + q(x)] dx +
∂n
D
∂D
(the boundary term remains). In particular, this must hold for all w vanishing
on the boundary, and so we conclude, as above,
equation ∆u∗ = q
R that the∂ E-L
must hold. Then in addition we require that ∂D w(x) ∂n u∗ (x)dS(x) = 0, again
∗
for any function w. This can only hold if ∂u
= 0 on ∂D. Thus
∂n
∆u∗ = q
in D
.
∂u∗
=0
on ∂D
∂n
These Neumann BCs are sometimes called the natural boundary conditions
for the minimization problem, since they arise naturally from the minimization
when no BCs are imposed in the problem.
71
Our next example generalizes our first one.
Example: Find the Euler-Lagrange equation for the problem
Z
min I(u), I(u) :=
F (x, u, ∇u) dx, H = {u ∈ C 2 (D) u = g on ∂D}
u∈H
D
(notation: C 2 (D) denotes the twice continuously differentiable functions on D).
As above, a necessary condition for u to be a minimizer is, for any smooth function
w vanishing on ∂D,
Z
d
d
0 = I(u + εw) ε=0 =
F (x, u + εw, ∇u + ε∇w)dxε=0
dε
dε D
Z
=
(Fu (x, u, ∇u)w + F∇u (x, u, ∇u) · ∇w)
ZD
=
(−∇ · F∇u (x, u, ∇u) + Fu (x, u, ∇u)) w dx
D
(where again we used the divergence theorem , and the fact that w vanishes on the
boundary). So the Euler-Lagrange equation for this problem is
∇ · F∇u (x, u, ∇u) = Fu (x, u, ∇u).
(39)
(Note that F∇u is a gradient – that is, it is a vector field.) As a special case, consider
the previous example above, which corresponds to F (x, u, ∇u) = 21 |∇u|2 + q(x)u, so
that Fu = q, F∇u = ∇u, ∇ · F∇u = ∆u, and the Euler-Lagrange equation is ∆u = q.
Remark: Notice that in general, the Euler-Lagrange equation (39) is a nonlinear
PDE – whereas so far this course has been concerned almost exclusively with linear
problems.
72
2. Some further examples.
Example: (“Brachistochrone problem”) A bead slides (frictionlessly) down a curve
from a point P to a (lower) point Q. What curve gives the shortest travel time?
Put point P at the origin of an xy-plane (with the positive y direction pointing down,
for convenience), and describe possible curves C from P to Q by functions y = f (x),
with Q at (x1 , f (x1 )) (draw a picture!). The speed of the bead along the curve is
v = ds/dt where s denotes the arc length along the curve (and t denotes time, of
course), so the travel time is
Z
Z
ds
T =
dt =
.
C
C v
The arc length differential is given by
p
p
p
ds = dx2 + dy 2 = dx2 + (f 0 (x)dx)2 = 1 + (f 0 (x))2 dx.
We can determine the velocity by conservation of energy:
1
energy = kinetic + potential = mv 2 − mgy = constant = 0,
2
where m is the mass of the bead, and g is the gravitational acceleration, so
p
p
v = 2gy = 2gf (x).
So now we have an expression for the travel time
Z x1 s
Z
1
ds
1 + (f 0 (x))2
=√
dx.
T =
f (x)
2g 0
C v
So now our variational problem is clear: among smooth functions f (x) defined on
[0, x1 ] with boundary conditions f (0) = 0, f (x1 ) = y1 , minimize
s
Z x1
1 + (f 0 )2
F (f (x), f 0 (x))dx,
F (f, f 0 ) :=
I(f ) :=
.
f
0
We can read the Euler-Lagrange equation for this problem directly off (39):
p
d
d −1/2
f
(1 + (f 0 )2 )−1/2 f 0
Ff 0 = 1 + (f 0 )2 (−1/2)f −3/2 −
dx
dx
1
0 2 2
0 2
0 2
0 2 00
0 2
00
=
−(1
+
(f
)
)
+
(f
)
(1
+
(f
)
)
+
2f
(f
)
f
−
2f
(1
+
(f
)
)f
2(1 + (f 0 )2 )3/2 f 3/2
1
(1 + (f 0 )2 ) + 2f f 00 .
=−
0
2
3/2
3/2
2(1 + (f ) ) f
0 = Ff −
73
So if there is a minimizing function f (x), it should satisfy the nonlinear ODE
2f f 00 + (f 0 )2 = −1.
It turns out that the cycloid (the curve traced out by a point on a circle rolling along
a line), which is given parametrically by
x(θ) = R(θ − sin θ),
y(θ) = R(1 − cos θ),
solves this ODE, as we now verify:
f0 =
R sin θ
sin θ
dy/dθ
=
=
dx/dθ
R(1 − cos θ)
1 − cos θ
(1 − cos θ) cos θ − sin2 θ
d 0 [sin θ/(1 − cos θ)]θ
1
f =
=
f =
=−
3
dx
R(1 − cos θ)
R(1 − cos θ)
R(1 − cos θ)2
00
−2(1 − cos θ) + (1 − cos θ)2 + sin2 θ
= 0.
(1 − cos θ)2
So the cycloid does indeed solve the E-L equation. Choosing R and θ1 so that (R(θ1 −
sin θ1 ), R(1 − cos θ1 ) = (x1 , y1 ), we can also satisfy the boundary conditions. The
question of whether or not the cycloid really minimizes the travel time is a more
difficult one to answer.
2f f 00 + (f 0 )2 + 1 =
Example: (Higher derivatives). Let p(x) be a given function on a bounded domain
D. Find the E-L equation for the minimization problem
Z 1
2
minu∈C 4 , u= ∂u =0 on ∂D
(∆u(x)) − p(x)u(x) dx,
∂n
2
D
which arises in elasticity (where u(x) represents the deflection of a plate D ⊂ R2 as
a result of a load p(x)).
If u∗ is a minimizer, and w(x) is any smooth function with w = ∂w
= 0 on ∂D, then
∂n
Z Z
d
1
∗
2
∗
0=
(∆(u + εw)) − p(u + εw) dx ε=0 =
(∆u∗ ∆w − pw) dx
dε
2
D Z
Z Z D
∂w
∂
= (∆∆u∗ − p)wdx +
∆u∗
−
∆u∗ w dS = (∆∆u∗ − p)wdx
∂n
∂n
∂D
D
D
using the divergence theorem twice (or, if you prefer, Green’s second identity) and
the boundary conditions on w. Employing the same argument we used above, we
conclude that the E-L equation for u∗ is ∆2 u∗ = ∆∆u∗ = p, sometimes known as the
biharmonic equation. That is, the problem for a minimizer u∗ is
∆∆u∗ (x) = p(x)
D
.
∂u∗
∗
u = ∂n = 0
∂D
74
3. Variational problems with constraints.
Here we consider the constrained variational problem
min
I(u)
(40)
u∈H, M (u)=C
where
H = {u ∈ C 2 (D) | u ≡ 0 on ∂D}
is the class of functions we work in,
Z
F (x, u, ∇u)dx
I(u) =
D
is the functional we are minimizing, and
Z
M (u) =
G(x, u, ∇u)dx
D
is another functional, which gives the constraint.
Just as for functions of several variables, the necessary condition (Euler-Lagrange
equation) for a constrained problem involves Lagrange multipliers:
Theorem: If u∗ solves the variational problem (40), then u∗ is a critical point (that
is, satisfies the Euler-Lagrange equation for) of the functional
I(u) + λ M (u)
for some λ ∈ R (called a Lagrange multiplier).
Sketch of proof: Suppose u∗ (x) is a minimizer of problem (40). That means M (u∗ ) =
c, and u∗ minimizes I(u) among functions u ∈ H with M (u) = c. In particular, let
uε be a one-parameter family of functions in H with M (uε ) = c, and u0 = u∗ . Then
I(uε ) (which is a function of ε) is minimized at ε = 0, and so
Z
d
d
F (x, uε , ∇uε )dx |ε=0
0 = I(uε )|ε=0 =
dε
dε D
Z ∂uε
∗
∗ ∂uε
∗
∗
=
Fu (x, u , ∇u )
|ε=0 + F∇u (x, u , ∇u ) · ∇
|ε=0 dx.
∂ε
∂ε
D
Let us denote
∂uε
|ε=0 ,
∂ε
notice that ξ ≡ 0 on ∂D, and use (as always!) the divergence theorem, to get
Z
0=
[Fu (x, u∗ , ∇u∗ ) − ∇ · F∇u (x, u∗ , ∇u∗ )] ξ(x)dx.
ξ :=
D
75
Now if ξ(x) could be any (smooth) function on D (with zero BCs), we would conclude,
as before, that
f (x) := Fu (x, u∗ , ∇u∗ ) − ∇ · F∇u (x, u∗ , ∇u∗ )
is zero, and that would be our E-L equation. However, because of the constraint
M (uε ) = 0, ξ(x) cannot be just any function. Indeed, by the same sort of calculation
we just did,
Z
d
0 = M (uε )|ε=0 =
g(x)ξ(x)dx,
g(x) := Gu (x, u∗ , ∇u∗ ) − ∇ · G∇u (x, u∗ , ∇u∗ ).
dε
D
That is, ξ must be orthogonal to the function g. In fact, it turns out that a one∂uε
parameter family of functions uε as
R above, with ∂ε |ε=0 = ξ, can be constructed for
any (smooth enough) ξ satisfying gξ = 0. So, we have
Z
Z
f (x)ξ(x)dx = 0 for any ξ(x) with
g(x)ξ(x)dx = 0.
D
D
What can we conclude about f (x) from this? The following:
Claim: f (x) + λg(x) = 0 for some λ ∈ R.
To see this, consider any (smooth) function η(x) on D, and write it as
Z
(g, η)
η(x) =
g(x) + η̃(x),
g(x)η̃(x)dx = (g, η̃) = 0
(g, g)
D
(i.e., in linear algebra language, η̃ is the orthogonal projection onto the subspace
perpendicular to g). Since η̃ is perpendicular to g, we have
Z
Z
Z
1
g(y)η(y)dy
0=
f (x)η̃(x)dx =
f (x) η(x) − g(x)
(g, g) D
D
D
Z (g, f )
=
f (x) −
g(x) η(x)dx
(g, g)
D
Finally, since η can be any function, we conclude, as in the earlier lemma, that
f (x) −
(g, f )
g(x) = 0,
(g, g)
or, in other words, u∗ is a critical point of the functional
λ=−
I(u) + λM (u),
76
(g, f )
.
(g, g)
Example: (Eigenvalue problem). Recall that the first (Dirichlet) eigenvalue of −∆ on
domain D is given by
R
Z
|∇u(x)|2 dx
D
R
=
min R
|∇u(x)|2 dx
λ1 = min
2 dx
u=0 on ∂D
u(x)
u=0 on ∂D, D u2 =1 D
D
(the second expression equals the first,
R since we may normalize any function u(x),
by multiplying by a number, to get u2 = 1, and this doesn’t change the Rayleigh
quotient). The second expression is a constrained minimization problem – let’s find
its Euler-Lagrange equation. The method of Lagrange multipliers tells us to consider
the functional
Z
Z
Z
2
2
|∇u| dx + µ
u dx =
|∇u|2 + µ u2 dx,
D
D
D
with Lagrange multiplier µ, whose Euler-Lagrange equation we can find (as usual) by
considering, for any (smooth) ξ(x) vanishing on the boundary, u + εξ:
Z
Z
d
2
2
0=
|∇u + ε∇ξ| + µ(u + εξ) dxε=0 =
2 [∇u · ∇ξ + 2µuξ] dx
dε D
D
Z
=2
ξ(x) [−∆u(x) + µu(x)] dx,
D
and therefore
−∆u(x) + µu(x) = 0,
that is, u is an eigenfunction of −∆ with eigenvalueR −µ. And notice, finally, that if
we integrate the equation above against u, and use u2 dx = 1, we find
Z
Z
Z
2
−µ = −µ u (x)dx = −
u(x)∆u(x)dx =
|∇u(x)|2 dx = λ1 ,
D
D
so indeed, the minimizer is an eigenfunction with eigenvalue λ1 (which, in any case,
we already knew).
Extensions:
1. (first Neumann eigenvalue). Notice that if we do not impose any boundary conditions in the minimization problem, the minimizer will still satisfy the natural
∂u
boundary conditions – namely Neumann conditions ∂n
≡ 0 on ∂D. Hence the
first Neumann eigenvalue of −∆ on D is given by
Z
λ1 = R min
|∇u|2 dx,
D
u2 dx=1
D
and the minimizer is a corresponding eigenfunction.
77
2. (higher eigenvalues). Recall that the n-th (back to Dirichlet again) eigenvalue
is given by
Z
λn =
min
|∇u|2 dx
R
u=0 ∂D,
D
u2 =1, (u,φ1 )=···=(u,φn−1 )=0
D
where φ1 , . . . , φn−1 are the first n − 1 eigenfunctions. This is a constrained
variational problem with several constraints – in fact n of them. In this case,
we need n Lagrange multipliers. That is, we are looking for critical points of
Z
Z
2
2
|∇u| dx + µ0
u dx +
D
D
n−1
X
Z
µj
j=1
uφj dx
D
for some numbers µ0 , µ1 , . . . µn−1 . The Euler-Lagrange equation (check it!) is
−∆u + µ0 u +
n−1
X
2µj φj = 0.
j=1
Integrating this equation against φk for some k ∈ {1, 2, . . . , n − 1}, and using
(φk , u) = 0,
(φk , ∆u) = (∆φk , u) = (−λk φk , u) = −λk (φk , u) = 0,
and the orthonormality of the eigenfunctions, we find µk = 0 for k = 1, 2, . . . , n−
1. And then integrating against u, we find
Z
|∇u|2 dx = −λn .
µ0 = −
D
Hence
−∆u = λn u
D
.
u=0
∂D
That is, the minimizer is indeed an eigenfunction corresponding to the eigenvalue
λn (which, again, we already knew).
Example: (Minimizing the surface area for fixed volume). Let D be a region
in R2 , and consider surfaces described by graphs of functions x3 = u(x) ≥ 0,
x ∈ D, with u = 0 on ∂D. The problem is to minimize the surface area
Z p
A(u) =
1 + |∇u(x)|2 dx
D
subject to the constraint of a fixed volume:
Z
V (u) =
u(x)dx = V0 .
D
78
Thus we consider the functional
Z p
A + λV =
1 + |∇u(x)|2 + λu(x) dx
D
for a Lagrange multiplier λ, and find its Euler-Lagrange equation:
Z 1
d
2 −1/2
(1 + |∇u(x)| )
2∇u(x) · ∇ξ(x) + λξ(x) dx
0 = (A + λV )(u + εξ) ε=0 =
dε
2
D
!
∇u(x)
= ξ(x) −∇ · p
+ λ dx.
1 + |∇u(x)|2
Thus a minimizing function should satisfy
∇u(x)
∇· p
= constant .
1 + |∇u(x)|2
As a special case, let D be the disk x21 + x22 ≤ r02 , and try
q
q
2
2
2
u(x1 , x2 ) = c0 − x1 − x2 − c20 − r02 ,
p
for r0 ≥ c0 , which is (part of) a sphere, of radius c0 , centred at (0, 0, − c20 − r02 ).
It is a good exercise to check that this satisfies our minimal surface equation
above. Notice that the radius c0 can be chosen to satisfy the volume constraint
V0 , provided V0 is not too large (exercise: find how large, in terms of r0 ).
79
4. Approximating minimizers: Rayleigh-Ritz method.
Since it is very rare to be able to solve variational problems (indeed, PDE problems in
general) explicitly, one usually needs to try and find approximate solutions – whether
“by hand”, or (more commonly) by computer. The Rayleigh-Ritz method is an
elementary, classical technique for approximating solutions to variational problems.
The idea could not be simpler. Suppose we want to find an approximation to the
solution of
min I(u)
u∈H
where H is some class of functions (eg. with some given BCs), and I is a functional.
Let’s assume that H is a vector space (i.e. closed under linear combinations), which
is often the case. Let
v1 (x), v2 (x), . . . , vm (x)
be m “trial functions”, and consider linear combinations of these m functions:
u(x) =
n
X
cj vj (x),
cj ∈ R for j = 1, 2, . . . , m.
j=1
Such functions lie in H (i.e. satisfy the given BCs), and if we try to minimize the
functional I over all such functions (rather than over all functions in H), we get a
finite-dimensional minimization problem,
!
m
X
min I
cj vj ,
c1 ,...,cm
j=1
which we can solve by standard multi-variable calculus (i.e. set the partial derivatives
with respect to each cj equal to 0).
Example: Find an approximation to the solution of
Z 1
2
min
|∇u| + f (x)u dx
u=0 on ∂D D
2
where
D is the rectangle [0, 1]2 . (Recall, the minimizer satisfies Poisson’s equation
∆u = f
in D
.)
u=0
on ∂D
Convenient trial functions vj for computation are eigenfunctions of the Laplacian
(with zero BCs) – namely, products of sines. That is, let’s minimize the given functional among functions of the form
N X
N
X
cjk vjk (x),
vjk (x) = sin(jπx1 ) sin(kπx2 )
j=1 k=1
80
for some positive integer N . Then a nice way to compute the functional is, by the
divergence theorem, and the orthogonality of the vjk ,
1
2
Z
1
|∇u| dx = −
2
D
2
Z
N
1 X
vj 0 k0 (x)∆vjk (x)dx
cjk cj 0 k0
u(∆u)dx = −
2 j,k,j 0 ,k0 =1
D
D
Z
Z
N
N
π2 X 2
1 X
vj 0 k0 (x)π 2 (j 2 + k 2 )vjk (x) =
=
cjk cj 0 k0
(j + k 2 )c2jk .
2 j,k,j 0 ,k0 =1
8 j,k=1
D
And
Z
f (x)udx =
D
N
X
Z
vjk (x)f (x)dx =:
cjk
j,k=1
D
N
X
cjk fˆn,k .
j,k=1
So our job is to find the choice of the coefficients cjk which minimizes
N 2
X
π 2
2 2
ˆ
(j + k )cj,k + fj,k cj,k .
I(u) =
8
j,k=1
That is, we want to minimize a function of the N 2 variables cjk . So, as we know from
calculus, we should take the partial derivatives and set them equal to zero:
N ∂ X π2 2
π2
2 2
0=
(j + k )cj,k + fˆj,k cj,k = (j 2 + k 2 )cj,k + fˆj,k .
∂cjk j,k=1 8
4
So we take
cj,k = −
4fˆj,k
,
π 2 (j 2 + k 2 )
and our approximate minimizer is
N
4 X fˆj,k
ũ(x) = − 2
sin(jπx1 ) sin(kπx2 ).
π j,k=1 j 2 + k 2
Remark: In this example, the approximate minimizer turns out to be nothing but the
eigenfunction expansion solution of the corresponding Poisson equation, truncated at
j, k ≤ N .
Remark: In general it is somewhat difficult to assess how good an approximation the
Rayleigh-Ritz method generates. In this particular example, from what we know
about the completeness of eigenfunctions, we can at least conclude that our approximation should get better as N gets larger, and, in particular, should approach the
true solution as N → ∞.
81
Example: Let D be the 2D region bounded by the ellipse x21 /a2 + x22 /b2 = 1. Approximate the solution u of
(
2 2 # )
Z "
∂u
∂u
min
I(u) =
− x2 +
+ x1
dx .
u∈C 2 (D)
∂x1
∂x2
D
Notice that no boundary conditions are imposed in the problem.
Remark: Before we start, notice something: clearly I(u) ≥ 0. Can we simply get a
minimizer with I(u) = 0 by setting ux1 = x2 and ux2 = −x1 ? Well, let’s try: the first
equation requires u = x1 x2 + f (x2 ), and then the second requires x1 + f 0 (x2 ) = −x1
so f 0 (x2 ) = −2x1 which we cannot satisfy. So, no. In vector calculus language, we
are trying to find a function u(x) so that ∇u = (x2 , −x1 ). A necessary condition for
a vector-field to be a gradient is that its curl vanishes. But curl(x2 , −x1 ) 6= 0, so we
cannot solve the equation.
OK, back to the question of approximating the minimizer. To keep things simple, let’s
use just one trial function (m = 1). One computationally nice choice is v1 (x) = x1 x2
since a multiple of this (v1 itself) makes the first term in the integral vanish, while a
different multiple (−v1 ) makes the second term vanish. So, we are lead to
min I(αx1 x2 ),
α∈R
i.e. just minimizing a function of the single variable α. Let’s compute
Z
Z
Z
2 2
2 2
2
2
2
I(αx1 x2 ) =
(α − 1) x2 + (α + 1) x1 dx = (α − 1)
x2 dx + (α + 1)
x21 dx.
D
D
D
Before proceeding, let’s pause and ask what we expect in the case of a disk: a = b.
The symmetry suggests that neither positive nor negative α should be favoured, and
so α = 0 should be the minimizer. We’ll keep this idea in mind as a “check” on our
computation at the end.
Now
Z
x21 dx
D
Z
a
=4
x21 dx1
0
= 4a3 b
Z b√1−x21 /a2
Z
0
0
Z
a
dx2 = 4b
π/2
3
sin2 (θ) cos2 (θ)dθ =
0
and similarly,
Z
x22 dx =
D
82
πab3
.
4
πa b
,
4
x21
q
1 − x21 /a2 dx1
So our problem is
min
α
πab 2
b (α − 1)2 + a2 (α + 1)2 ,
4
which we solve as usual:
0=
d 2
b (α − 1)2 + a2 (α + 1)2 = 2b2 (α − 1) + 2a2 (α + 1) = 2(a2 + b2 )α + 2(a2 − b2 )
dα
2
2
hence α = ab 2−a
gives the minimum value (it must be a minimum, since the graph of
+b2
this function of α is an upward parabola), and so our (admittedly crude) approximate
minimizer is
b 2 − a2
ũ(x) = 2
x1 x 2
a + b2
which gives a value of
" 2
2 #
2
πa3 b3
πab 2 −2a2
2b
2
b
.
+
a
=
I(ũ) =
4
a2 + b 2
a2 + b 2
a2 + b 2
(And note that our guess that α = 0 when a = b is indeed confirmed.)
83
5. Approximating eigenvalues and eigenfunctions
Recall that the first eigenvalue of the problem
Lφ := −∇ · [p(x)∇φ] + q(x)φ = λr(x)φ
φ = 0 on ∂D
in D
is given by minimizing the Rayleigh quotient:
R
(p|∇u|2 + qu2 ) dx
D
R
λ1 = min
,
u=0 on ∂D
ru2 dx
D
and the minimizing function is a corresponding eigenfunction. So we can use a
Rayleigh-Ritz approach to approximate the eigenfunction.
Start with a set of m “trial functions”:
v1 (x), v2 (x), . . . , vm (x)
with the correct boundary conditions (in this case Dirichlet: vj ≡ 0 on ∂D), and we
will minimize the Rayleigh quotient among linear combinations
u(x) = c1 v1 (x) + c2 v2 (x) + . . . + cm vm (x).
Let’s compute the relevant quantities. Using the convention of summation over repeated indices:
Z
Z
−c · B →
−c
2
ru dx = cj ck
rvj vk dx = →
D
D
−c denote the m-vector
where →
→
−c = (c , c , . . . , c ),
1 2
m
and B is the m × m matrix
B :=
(Bjk )m
j,k=1 ,
Z
Bjk =
rvj vk dx = (vj , vk )r .
D
Similarly,
Z
2
2
p|∇u| + qu
Z
dx = cj ck
D
−c · A→
−c
(p∇vj · ∇vk + qvj vk ) dx = →
D
where
A :=
(Ajk )m
j,k=1 ,
Z
(p∇vj · ∇vk + qvj vk ) dx = (vj , Lvk ) = (vk , Lvj ) = Akj
Ajk =
D
84
where we have noted that the matrix A is symmetric, because the operator L is
self-adjoint. Notice B is symmetric, too. Hence our problem is:
→
−c · A→
−c
min →
−
→
−,
−
→
c ∈Rm c · B c
a simple minimization of a function of m variables. Of course, by calculus, the
minimizer will satisfy (going back to summation over repeated indices notation)
0=
→
−c · B →
−c (A + A )c − →
−c · A→
−c (B + B )c
∂ Akl ck cl
jl
lj l
nj
jn n
=
,
−c · B →
−c )2
∂cj Bnr cn cr
(→
j = 1, 2, . . . , m
which, using the symmetry of A and B, yields
−c · B →
−c )A→
−c = (→
−c · A→
−c )B →
−c
(→
or
→
−c · A→
−c
λ := →
−c · B →
−c ,
which is a (generalized) matrix eigenvalue problem.
−c = λB →
−c ,
A→
2
d
Example: Approximate the first Dirichlet eigenfunction of − dx
2 + αx on [0, 1] using
trial functions v1 (x) = sin(πx), v2 (x) = sin(2πx) (notice that v1 and v2 are the first
d2
two eigenfunctions of − dx
2 on [0, 1] with 0 BCs – that is, of the problem with α = 0).
Easy computations give
Z 1
Z 1
1
1
2
(v1 , v1 ) =
sin (πx)dx = , (v2 , v2 ) =
sin2 (2πx)dx = ,
2
2
0
Z0 1
(v1 , v2 ) =
sin(πx) sin(2πx)dx = 0,
0
so our matrix B is
B=
1
2
0
0
1
2
.
Next,
v10 = π cos(πx),
so
Z
0
1
(v10 )2 dx
π2
= ,
2
Z
v20 = 2π cos(2πx),
1
(v20 )2 dx
0
2
Z
= 2π ,
1
v10 v20 dx = 0.
0
Slightly more involved computations (using integrations by parts) yield
Z 1
Z 1
Z 1
1
1
1
2
2
xv1 dx = ,
xv2 dx = ,
xv1 v2 dx =
.
4
4
2π
0
0
0
85
Hence our matrix A is
A=
π2
2
α
4
+
α
2π
α
2π
2
2π +
α
4
.
−c = 0, we set
To solve the eigenvalue problem (A − λB)→
0 = det(A−λB) = π2
2
α
4
α
2π
+
−
λ
2
α
2π
2
2π + α4
−
λ
2
2
α
λ
α
λ
α2
π
2
=
+
−
2π
+
−
−
,
2
4
2
4
2
4π 2
and the quadratic formula leads to
r
2λ = 5π 2 + α ±
(5π 2 + α)2 − 4[(π 2 + α/2)(4π 2 + α/2) −
α2
].
π2
Since we are trying to approximate the lowest eigenvalue/eigenfunction, we take the
negative sign here, to obtain the approximate eigenvalue λ. Then one can find an
−c which will produce an approximate eigenfunction c v + c v . Of
eigenvector →
1 1
2 2
course, the expressions are a little messy. One thing we can do, supposing α is small,
is expand the approximate eigenvalue in a Taylor series in α to find, for example,
λ = π2 +
α
+ O(α2 ).
2
Remark: Notice that this approximate eigenvalue λ is actually an upper-bound for
the true first eigenvalue:
λ1 ≤ λ,
since we obtain it by minimizing the Rayleigh quotient over a sub-set of functions
(the true eigenvalue is the minimizer over all admissible functions, hence cannot be
larger).
86