Variational Principles for
Equilibrium Physical Systems
1. Variational Principles
One way of deriving the governing equations for a physical system is the express the
relevant conservation statements and constitutive laws in terms of a set of state variables to
obtain a system of differential equations. A second approach is to postulate a variational
principle for the physical system. In this approach, one characterizes a scalar variable, one
that is often related to the energy of the system, in terms of the system state variables. The
variational principle then is the statement that the system will assume the state in which the
”energy” is minimized. Aside from any other reasons for considering the variational
approach to the derivation of governing equations, this approach provides a means for
weakening the formulation of the equations that govern the behavior of the physical system.
We will illustrate the approach with examples. First, we recall some notation and define an
alternative norm for the space H 10 ÝUÞ.
Equivalent Norms on H 10 ÝU Þ
Throughout this chapter we are going to use the notation that for a bounded set U,
Ýu, vÞ 0 = XU u v dx = L 2 ? inner product
and
Ýu, vÞ 1 = XU ßu v + 4u 6 4và = H 1 ? inner product
|| u|| 20 = Ýu, uÞ 0
and || u|| 21 = Ýu, uÞ 1 = || u|| 20 + ||4u|| 20
We recall also that the Poincare inequality asserts the existance of a constant C
depending only on U such that,
|| u|| 20 ² C ||4u|| 20
for all u 5 H 10 ÝUÞ, .
It follows from this inequality that for all u 5 H 10 ÝUÞ,
1 || u|| 2 ² ||4u|| 2 ² || u|| 2 .
1
0
1
1+C
This means that | u| 1 = ||4u|| 0 defines a new norm on H 10 ÝUÞ and this new norm is
equivalent to the old norm || u|| 1 in the sense that any sequence that is convergent in one of
the norms is also convergent in the other. It follows that H 10 ÝUÞ is a Hilbert space for the new
inner product and norm given by
Öu, v× 1 = XU 4u 6 4v dx
and
| u| 21 = Öu, u× 1
for u, v 5 H 10 ÝUÞ.
This is sometimes referred to as the Poincare inner product and norm on H 10 ÝUÞ. Note
that since | C| 1 = 0 for any constant, this is not a norm on H 1 ÝUÞ.
Transverse Deflection of an Elastic Membrane
Consider an elastic membrane, stretched over a rigid frame lying in the plane. Suppose the
frame forms a simple closed curve containing a region U in its interior. Then the curved
frame forms the boundary of the domain U containing the membrane.
In its relaxed state the membrane lies in the plane. If we denote the out of plane
deflection of the membrane by u = uÝx, yÞ then u = 0 corresponds to the relaxed state. Now
suppose the membrane is subjected to a transverse loading having force density described
by the function f = fÝx, yÞ. Then the potential energy stored in the stretched membrane
1
whose deformation state is uÝx, yÞ corresponding to the load fÝx, yÞ can be shown to be
given by the expression
EßuÝx, yÞà =
XU
1 TÝx, yÞ4u 6 4u ? uÝx, yÞfÝx, yÞ
2
dx dy
(1.1)
Here T = TÝx, yÞ denotes a bounded and strictly positive function representing the tension in
the membrane. If we suppose TÝx, yÞ is piecewise continuous, and f = fÝx, yÞ is in L 2 ÝUÞ,
then Eßuà is a functional whose domain could be taken to be the linear subspace H 10 ÝUÞ in
the Sobolev space of order one H 1 ÝUÞ. Clearly Eßuà is well defined on this subspace and
the condition that the boundary of the membrane is fixed to the rigid frame lying in the plane
(i.e., uÝx, yÞ = 0 for Ýx, yÞ 5 /U ) is incorporated into the definition of the domain of the
functional. The space H 1 ÝUÞ, is often reffered to as the ”energy space” since it contains
functions uÝx, yÞ for which the energy (1.1) is finite. In general, functions in L 2 ÝUÞ do not
have finite energy and would therefore not be feasible candidates for functions that describe
the deformation state of this membrane.
The problem of finding the deformation state uÝx, yÞ corresponding to a given loading
fÝx, yÞ 5 L 2 ÝUÞ can be stated as a variational principle
Find u 5 H 10 ÝUÞ, such that Eßuà ² Eßvà
for all v 5 H 10 ÝUÞ
(1.2)
Physically, this is equivalent to the assertion that in a state of elastic equilibrium, the
membrane will assume the deflection state that minimizes the energy Eßuà over all
admissible deformation states, u. Without speculating on why this should be true, we accept
this statement as a postulate. Our purpose is to consider the mathematical consequences
of the problem that ensues.
Note that
where
Eßuà =
1
2
aÝu, uÞ ? Fßuà
aÝu, vÞ = X TÝx, yÞ4u 6 4v dxdy
U
and
Fßuà = X uÝx, yÞfÝx, yÞ dxdy = Ýu, fÞ 0 .
U
Clearly aÝu, vÞ is bilinear and Fßuà is linear on H 10 ÝUÞ. Moreover, if the coefficient TÝx, yÞ
satisfies,
T 1 ³ TÝx, yÞ ³ T 0 > 0 for Ýx, yÞ 5 U,
then
| aÝu, vÞ| ² XU | TÝx, yÞ4u 6 4v| dxdy ² T 1 ||4u|| 0 ||4v|| 0 ² T 1 || u|| 1 || v|| 1
and
|Fßuà| ² ||f|| 0 || u|| 0 ² ||f|| 0 || u|| 1
from which it is evident that aÝu, vÞ is a bounded bilinear functional and Fßuà is a bounded
linear functional on H 10 ÝUÞ. In addition,
aÝd, dÞ ³ T 0 ||4d|| 20 = T 0 | d| 21 ³ T 0 || d|| 21
for all d 5 DÝUÞ
2C
and since the test functions are dense in H 10 ÝUÞ, the estimate extends to all u 5 H 10 ÝUÞ.
Then the bounded, symmetric bilinear form aÝu, vÞ is also positive on H 10 ÝUÞ and it follows
from lemma 3.1 that the problem of minimizing Eßuà over H 10 ÝUÞ is equivalent to the
2
following problem
find u 5 H 10 ÝUÞ such that aÝu, vÞ = Fßvà for all v 5 H 10 ÝUÞ
(1.3)
This equation asserts that u = uÝx, yÞ satisfies
aÝu, dÞ = X TÝx, yÞ4u 6 4d dxdy = Fßvà = X dÝx, yÞfÝx, yÞ dxdy
U
U
-d 5 DÝUÞ Ð H 10 ÝUÞ.
Formal integration by parts leads to
XU TÝx, yÞ4u 6 4d dxdy = X/U dÝT4uÞ 6 n dS ? XU d 4ÝTÝx, yÞ4uÞ dxdy,
-d 5 DÝUÞ Ð H 10 ÝUÞ
Since d = 0 on /U for all test functions d, the boundary integral vanishes and we see that
uÝx, yÞ satisfies
XU d ß4ÝTÝx, yÞ4uÞ + fÝx, yÞà dxdy = 0
-d 5 DÝUÞ Ð H 10 ÝUÞ
(1.4)
This is just the assertion that
4ÝTÝx, yÞ4uÞ + fÝx, yÞ = 0
in the sense of distributions on U.
Since the test functions are dense in H 10 ÝUÞ, it is evident that (1.4) holds not just for all test
functions but for all functions in H 10 ÝUÞ as well and then (1.4) asserts that
4ÝTÝx, yÞ4uÞ + fÝx, yÞ = 0 in a somewhat stronger sense than the distributional sense. In
addition, the fact that u 5 H 10 ÝUÞ means that u is the limit in the H 1 ÝUÞ ? norm of a
sequence of test functions (which all vanish on the boundary of U) so, in some sense u
can be said to vanish on the boundary of U. If it were known in addition that
u 5 H 10 ÝUÞ V C 0 ÝŪÞ then it would follow that uÝxÞ = 0 at each point x 5 /U. Without such
additional information we have to be content to say that u satisfies the boundary condition in
some generalized sense.Thus we would say that u 5 H 10 ÝUÞ satisfying (1.2) or (1.3) is a
distributional solution of the problem
? 4ÝTÝx, yÞ4uÞ = fÝx, yÞ
u=0
in U TCItag
on /U
In more advanced courses on PDE’s it is shown how to infer additional smoothness for the
solution u from smoothness assertions about the data f. If the solution has additional
smoothness then the solution of (1.2)-(1.3) may satisfy the abstract boundary value
problem (1.5) in some sense stronger than the distributional sense. We refer to (1.2), (1.3)
and (1.5) respectively as the variational formulation, the weak formulation and strong
formulation of the Dirichlet boundary value problem for the elliptic operator
Lßuà = ?4ÝTÝx, yÞ4uÞ
Next consider the variational problem
Find u 5 H 1 ÝUÞ, such that Eßuà ² Eßvà
for all v 5 H 1 ÝUÞ
(1.6)
where we minimize the energy over the space H 1 ÝUÞ instead of the closed subspace,
H 10 ÝUÞ. Physically, this corresponds to releasing the condition that the membrane is
attached to the rigid frame on the boundary of U. Here the solution space, H 1 ÝUÞ, contains
no mention of the boundary conditions and it remains to be seen what conditions, if any,
apply on the boundary.
We continue to have Eßuà = 12 aÝu, uÞ ? Fßuà with aÝu, vÞ and Fßuà both bounded on the
new domain, H 1 ÝUÞ. Note, however, that
3
aÝu, uÞ ³ T 0 ||4u|| 20
for u 5 H 1 ÝUÞ
does not imply that aÝu, vÞ is positive on H 1 ÝUÞ, since constants belong to H 1 ÝUÞ and
aÝC 1 , C 1 Þ = 0 for any constant, C 1 . We remove this difficulty by letting Ĥ 1 ÝUÞ denote the
quotient space of equivalence classes of functions in H 1 ÝUÞ where two functions belong to
the same equivalence class if they differ by a constant. Then the Poincare norm is a norm
on this quotient space and we have
aÝu, uÞ ³ T 0 ||4u|| 20 = T 0 | u| 21 > 0
for u 5 Ĥ 1 ÝUÞ, u ® 0.
Now it follows that problem (1.6) (with H 1 ÝUÞ replaced by Ĥ 1 ÝUÞ) is equivalent to the
problem
find u 5 Ĥ 1 ÝUÞ such that aÝu, vÞ = Fßvà for all v 5 Ĥ 1 ÝUÞ
(1.7)
To see what the strong version of this problem is, write
XU TÝx, yÞ4u 6 4v dxdy = XU vÝx, yÞfÝx, yÞ dxdy
-v 5 Ĥ 1 ÝUÞ.
Formal integration by parts (i.e., Green’s second identity) leads to
XU TÝx, yÞ4u 6 4v dxdy = X/U vÝT4uÞ 6 n dS
? X v 4ÝTÝx, yÞ4uÞ dxdy, -v 5 Ĥ 1 ÝUÞ
U
Ý1.8Þ
First choose v 5 Ĥ 1 ÝUÞ to be an arbitrary test function. The test functions are not dense in
Ĥ 1 ÝUÞ but they are contained in the space so this is perfectly legal. For v = vÝx, yÞ a test
function, the boundary integral vanishes and we are left with
XU TÝx, yÞ4u 6 4v dxdy = ? XU v 4ÝTÝx, yÞ4uÞ dxdy, -v 5 C Kc ÝUÞ,
and for u solving (1.7) this implies
XU v ß4ÝTÝx, yÞ4uÞ + fÝx, yÞà dxdy = 0, -v 5 C Kc ÝUÞ.
This is just the statement that
? 4ÝTÝx, yÞ4uÞ = fÝx, yÞ
in the sense of distributions on U.
Now returning to Ý1.7Þ, we have
aÝu, vÞ ? Fßvà = X
/U
vÝT4uÞ 6 n dS + X v ß4ÝTÝx, yÞ4uÞ + fà dxdy = 0
U
-v 5 Ĥ 1 ÝUÞ.
Now the integral over U is zero if v 5 Ĥ 1 ÝUÞ is chosen to be a test function. Then this
integral must also vanish if v 5 Ĥ 1 ÝUÞ is chosen to be an arbitrary function in C K ÝUÞ, since
the boundary is a set of measure zero and cannot alter the value of the integral. Then this
last equation becomes
aÝu, vÞ ? Fßvà = X
/U
vÝT4uÞ 6 n dS = 0
-v 5 C K ÝUÞ Ð Ĥ 1 ÝUÞ.
Now the subspace C K ÝUÞ is dense in Ĥ 1 ÝUÞ so this last equation implies that
ÝT4uÞ 6 n = 0 on /U, at least in some generalized sense we will not be able to precisely
define. At any rate, it appears that the equivalent problems, (1.6)-(1.7), are the variational
and weak formulations of the following strong problem
? 4ÝTÝx, yÞ4uÞ = fÝx, yÞ
ÝT4uÞ 6 n = 0
for Ýx, yÞ 5 U TCItag
on /U
4
which we recognize as the Neumann problem for the elliptic operator, Lßuà = ?4ÝTÝx, yÞ4uÞ.
Note that the Neumann boundary conditions were not built into the definition of the solution
space but appeared only later as a necessary consequence of being a weak or variational
solution of the problem. Boundary conditions that must be incorporated into the definition of
the solution space, like the Dirichlet boundary condition of the first example, are called
stable boundary conditions. Conditions that are not built into the definition of the solution
space but appear naturally, like the Neumann boundary conditions, are called natural
boundary conditions. In the context of the elastic membrane, the Neumann boundary
condition means that the edges of the membrane are free to move in the out of plane
direction since there is no constraining force exerted at the boundary.
We have considered the problem of minimizing the quadratic functional Eßuà over H 10 ÝUÞ
and over the larger space, H 1 ÝUÞ. In the first case the variational problem is equivalent to
the weak form of the Dirichlet boundary value problem, while in the second case it is the
weak form of the Neumann boundary value problem that is equivalent to the variational
problem. Now consider a space, V, that is ”between” H 1 ÝUÞ and H 10 ÝUÞ, in the sense that
H 10 ÝUÞ Ð V Ð H 1 ÝUÞ. We have in mind the space obtained by completing, in the norm of
H 1 ÝUÞ, the subspace of infinitely differentiable functions which vanish on a part of /U. More
precisely, let /U be comprised of two disjoint pieces /U 1 and /U 2 ; i.e., /U = /U 1 W /U 2 .
Then let V denote the H 1 ÝUÞ ? closure of the infinitely differentiable functions which are
zero on /U 1 . When /U 1 = /U, /U 2 = empty, we have V = H 10 ÝUÞ and, when
/U 1 = empty, /U 2 = /U, we have V = H 1 ÝUÞ. We will suppose that neither /U 1 nor /U 2 is
empty so that V is strictly between H 10 ÝUÞ and H 1 ÝUÞ. Then we consider the problem
Find u 5 V, such that Eßuà ² Eßvà
for all v 5 V.
(1.10)
Now it is not hard to show that the Poincare norm defines a norm on V since V does not
contain any nonzero constants. Then aÝu, vÞ is symmetric bounded and positive on V and
(1.10) is equivalent to the weak problem,
Find u 5 V such that
aÝu, vÞ = Fßvà for all v 5 V
(1.11)
Since V contains the test functions C Kc ÝUÞ it follows that a solution of (1.11) satisfies the
equation ?4ÝTÝx, yÞ4uÞ = f, in the sense of distributions on U. Then it also follows that
aÝu, vÞ ? Fßvà = X
/U
vÝT4uÞ 6 n dS = 0
But for v 5 C K ÝUÞ V v = 0 on /U 1
-v 5 C K ÝUÞ V v = 0 on /U 1
Ð V.
the integral over /U 1 vanishes,
X/U vÝT4uÞ 6 n dS = X/U vÝT4uÞ 6 n dS + X/U vÝT4uÞ 6 n dS = X/U vÝT4uÞ 6 n dS = 0,
1
2
2
and since the behavior of the smooth function v on /U 2 is completely free, we can conclude
that ÝT4uÞ 6 n must equal zero on /U 2 in some generalized sense. Then the problems
(1.10),(1.11) are the variational and weak formulations of the following strong boundary
value problem,
? 4ÝTÝx, yÞ4uÞ = fÝx, yÞ
u=0
ÝT4uÞ 6 n = 0
in
U
TCItag
on /U 1
on /U 2
Note that the Dirichlet condition is incorporated into the definition of the solution space V,
while the Neumann boundary condition is a natural boundary condition and is automatically
satisfied by solutions of the weak and variational problems.
5
2. Weak Formulations
We have just seen that the following problems are alternative formulations of the same
problem:
for all v 5 H 10 ÝUÞ
1Þ Find u 5 H 10 ÝUÞ, such that Eßuà ² Eßvà
2Þ Find u 5 H 10 ÝUÞ such that aÝu, vÞ = Fßvà for all v 5 H 10 ÝUÞ
3Þ ? 4ÝTÝx, yÞ4uÞ = fÝx, yÞ
in U and u = 0 on /U
Similarly, the following are alternative formulations of one problem:
for all v 5 Ĥ 1 ÝUÞ
4Þ Find u 5 Ĥ 1 ÝUÞ, such that Eßuà ² Eßvà
1
5Þ find u 5 Ĥ ÝUÞ such that aÝu, vÞ = Fßvà for all v 5 Ĥ 1 ÝUÞ
6Þ ? 4ÝTÝx, yÞ4uÞ = fÝx, yÞ in U
and ÝT4uÞ 6 n = 0 on /U,
as are:
7Þ Find u 5 V, such that Eßuà ² Eßvà
for all v 5 V
8Þ find u 5 V such that aÝu, vÞ = Fßvà for all v 5 V
9Þ ? 4ÝTÝx, yÞ4uÞ = fÝx, yÞ in U, u = 0 on /U 1 and ÝT4uÞ 6 n = 0 on /U 2 ,
Recall that in problems 3),6) and 9), we are interpreting the partial differential equation in
the distributional sense and the boundary condition in some unspecified sense that is
weaker than the pointwise sense. Of course, it may be the case that both the equation and
the boundary condition are valid in some stronger sense but we have not done the analysis
needed to establish this.
Problems 1,2,3 are the variational, weak and strong formulations for the Dirichlet
problem, while 4,5,6 are the variational, weak and strong formulations for the Neumann
problem. Problems 7,8,9 are the variational, weak and strong formulations for a mixed BVP
with Dirichlet conditions on part of the boundary and Neumann conditions on the remainder
of the boundary. The weak and variational problems are equivalent in all three examples
and the strong form of the problem is related to the other two in that if u = uÝx, yÞ is a strong
solution then it is also a weak and variational solution but the converse does not necessarily
hold. Lemma 3.1 implies the existence of a unique solution to the variational/weak problems
in these examples. We have no information in any of the examples about whether a strong
solution exists or, what is the same thing, if the weak solution is also a strong solution. That
a weak solution is, in fact, also a strong solution can be proved under appropriate
hypotheses on the data but it requires techniques that are beyond the scope of this class.
In order to see whether it is always the case that a problem has all three formulations,
consider the following strong formulation:
Find u = uÝx, yÞ such that LßuÝx, yÞà = fÝx, yÞ
and
u=0
in U
on /U.
where
and
LßuÝx, yÞà = ?4ÝTÝx, yÞ4uÞ + 3
cÝx, yÞ 6 4u + bÝx, yÞ u
3
cÝx, yÞ 6 4u = c 1 Ýx, yÞ/ x u + c 2 Ýx, yÞ/ y u.
This strong problem has a corresponding weak formulation but it does not have a
variational formulation. To obtain the weak formulation, multiply both sides of the PDE by
an arbitrary function v = vÝx, yÞ 5 H 10 ÝUÞ, and integrate over U. Then
XU vÝx, yÞß?4ÝTÝx, yÞ4uÞ + 3cÝx, yÞ 6 4u + b uà dx = XU f v dx,
6
and
?X
/U
vÝTÝx, yÞ4uÞ 6 3
n dS + X ßT 4u 6 4v + v 3
c 6 4u + b u và dx = X f v dx.
U
U
Now the boundary integral vanishes for v 5
aÝu, vÞ = Fßvà
H 10 ÝUÞ,
and then we have
for v 5 H 10 ÝUÞ,
where
c 6 4u + b u và dx, for u, v 5 H 10 ÝUÞ.
aÝu, vÞ = X ßT 4u 6 4v + v 3
U
Note that, because of the term, v 3
c 6 4u, the bilinear form aÝu, vÞ is not symmetric; i.e.,
aÝu, vÞ ® aÝv, uÞ. For a nonsymmetric bilinear form there can be no quadratic functional
whose gradient equals aÝu, vÞ ? Fßvà. To see why this happens, recall that the quadratic
functional
®ÝuÞ =
1
2
aÝu, uÞ ? Fßuà+C
satisfies
®Ýu + tvÞ = ®ÝuÞ + tßaÝu, vÞ ? Fßvàà + 12 t 2 aÝu, uÞ
= ®ÝuÞ + t d®Ýu, vÞ + 12 t 2 d 2 ®Ýu, uÞ.
Evidently, when the quadratic functional is defined from a bilinear form, then d®Ýu, vÞ, the
”gradient of the functional at u in the direction v” is equal to aÝu, vÞ ? Fßvà. On the other
hand, any bilinear form can be written as the sum of a symmetric and an antisymmetric part
as follows,
aÝu, vÞ =
1
2
ÝaÝu, vÞ + aÝv, uÞÞ +
1
2
ÝaÝu, vÞ ? aÝv, uÞÞ = a S Ýu, vÞ + a A Ýu, vÞ.
But then
aÝu, uÞ = a S Ýu, uÞ + 0 so the antisymmetric part of the bilinear form cannot
contribute to the quadratic functional. Now u minimizes the functional ®ÝuÞ over H 10 ÝUÞ if
and only if u satisfies
a S Ýu, vÞ = Fßvà for all v 5 H 10 ÝUÞ,
and this is not the same as
aÝu, vÞ = Fßvà for all v 5 H 10 ÝUÞ
unless aÝu, vÞ = a S Ýu, vÞ; i.e., unless aÝu, vÞ is symmetric.
Even when the bilinear form is not symmetric so there is no variational formulation for
the problem, the weak problem
Find u 5 H 10 ÝUÞ such that aÝu, vÞ = Fßvà
for v 5 H 10 ÝUÞ
Ý2.2Þ
is still uniquely solvable. We assume that the coefficients satisfy,
0 < T 0 ² TÝx, yÞ ² T 1 ,
0 < b 0 ² bÝx, yÞ ² b 1 ,
cÝx, yÞ| ² c 0
|3
Ýx, yÞ 5 Ū,
Ý2.3Þ
We will show first that the bilinear form aÝu, vÞ is bounded. Write
c 6 4u + b u v| dx
| aÝu, vÞ| ² X | T 4u 6 4v + v 3
U
² T 1 ||4u|| 0 ||4v|| 0 + | 3
c | || v|| 0 ||4u|| 0 + |b| || v|| 0 || u|| 0
² ÝT 1 + c 0 + b 1 Þ || u|| 1 || v|| 1
for u, v 5 H 10 ÝUÞ.
This establishes that the bilinear form is bounded on H 10 ÝUÞ × H 10 ÝUÞ. Next, write
7
| aÝu, uÞ| ³ | X ßT 4u 6 4u + u 3
c 6 4u + b u 2 à dx|.
U
| aÝu, uÞ| ³ T 0 ||4u|| 20 ? c 0 X | u| |4u| dx + b 0 || u || 20 .
U
Now for all real numbers A and B and P > 0,
PA? B
2 P
2
2
= PA 2 ? AB + B ³ 0.
4P
It follows that
XU | u| |4u| dx ² P||4u|| 20 + 1 ||u|| 20 ,
4P
and then we have
| aÝu, uÞ| ³ ÝT 0 ? c 0 PÞ ||4u|| 20 + b 0 ? c 0 ||u|| 20 .
4P
If we choose P < T 0 /c 0 , then under certain additional assumptions on the coefficients in the
problem, there exists a positive constant a 0 such that
| aÝu, uÞ| ³ a 0 ||u|| 21
-u 5 H 10 ÝUÞ.
This shows that the nonsymmetric bilinear form is positive or coercive.
Problem 1 Show that if the coefficients T, b, and c are such that
4T 0 b 0 > c 20
Ý2.4Þ
then P can be chosen such that for some constant a 0 ,
T0 ? c0P ³ a0 > 0
and
b 0 ? c 0 ³ a 0 > 0.
Ý2.5Þ
4P
It now follows from (2.5) that for coefficients satisfying (2.4) the nonsymmetric bilinear form
a = aÝu, vÞ satisfies the hypotheses of the Lax-Milgram lemma. In addition,
Fßvà = Ýf, vÞ 0 for v 5 H 10 ÝUÞ
is a bounded linear functional on H 10 ÝUÞ. Then the Lax-Milgram lemma implies that for each
f 5 L 2 ÝUÞ, there exists a unique weak solution for (2.1). That is, there exists a unique
u 5 H 10 ÝUÞ satisfying (2.2).
3. Additional Variational Problems
It becomes evident at some point that the fundamental ingredient in the variational
formulation of boundary value problems is the Green’s identity. This identity makes it
possible to reduce certain partial differential equations and associated boundary conditions
to a corresponding variational statement. Of course not every boundary value problem can
be treated in this way but the class of problems to which it applies includes a large number
of problems of practical importance.
Interface Problems
Consider a domain U Ð R n composed of complementary subdomains, U 1 W U 2 = U, and
let @ denote the interface between U 1 and U 2 . Let F 1 = /U 1 \ @ and F 2 = /U 2 \ @ denote the
”non-interfacial” portions of the boundary of U 1 and U 2 respectively. Now consider the
problems
8
?4ÝA k ÝxÞ4u k Þ = f k ÝxÞ x 5 U k
u k = 0 on F k
For k = 1, 2
Ý3.1Þ
and
on @.
u1 = u2
ßA 1 ÝxÞ4u 1 ? A 2 ÝxÞ4u 2 à6n = 0 on @
Ý3.2Þ
Here A k ÝxÞ denotes an n by n matrix whose entries are in H 10 ÝU k Þ and f k 5 H 0 ÝU k Þ. This
type of problem corresponds to an elliptic problem in which the domain is composed of two
parts having distinctly different physical properties.
H = H 0 ÝU 1 Þ × H 0 ÝU 2 Þ,
V = H 1 ÝU 1 Þ × H 1 ÝU 2 Þ,
V 0 = H 10 ÝU 1 Þ × H 10 ÝU 2 Þ,
Define spaces
v = ßv 1 , v 2 à in V, define
and, for 3
u = ßu 1 , u 2 à, 3
3, 3
aßu
v à = a 1 Ýu 1 , v 1 Þ + a 2 Ýu 2 , v 2 Þ = X
U1
4v 1 6 A 1 4u 1 dx + X
U2
4v 2 6 A 2 4u 2 dx.
v = ßv 1 , v 2 à in C K ÝUÞ × C K ÝUÞ, the Green’s identity leads to
Then, for 3
u = ßu 1 , u 2 à, 3
3, 3
aßu
và = ? X
U1
?X
U2
v 1 4ÝA 4u 1 Þ dx + X
/U 1 \ @
v 2 4ÝA 4u 2 Þ dx + X
v1 3
n 1 6 A 1 4u 1 dS + X v 1 3
n 1 6 A 1 4u 1 dS
/U 2 \ @
@
v2 3
n 2 6 A 2 4u 2 dS + X v 2 3
n 2 6 A 2 4u 2 dS.
@
Now let
W=
3
u = ßu 1 , u 2 à 5 V : u k = 0 on /U k \ @, k = 1, 2 and u 1 = u 2 on @ .
Then
V 0 Ð W Ð V Ð H,
and, for any 3
u, 3
v 5 W V C K ÝUÞ,
3, 3
aßu
và = ? X
because
U1
v 1 4ÝA 4u 1 Þ dx ? X
v 1 = 0 on /U 1 \ @,
Now it follows that
(3.1),(3.2) if
3, 3
aßu
và = ? X
U1
U2
v 2 4ÝA 4u 2 Þ dx + X v 2 3
n 1 6 ßA 1 4u 1 ? A 2 4u 2 à dS,
@
v 2 = 0 on /U 2 \ @,
3
3 2 on @.
n 1 = ?n
3
u = ßu 1 , u 2 à 5 W is a weak solution of the transmission problem,
v 1 f 1 dx ? X
U2
v 2 f 2 dx = 3f, 3
v
0
-v3 5 W.
Then, since ßu 1 , u 2 à 5 W, it follows that in some generalized sense which we have not
clearly defined,
9
u k = 0 on /U k \ @, k = 1, 2
and
u 1 = u 2 on @.
In addition, based on the previous calculations, it is clear that we have,
?4ÝA k ÝxÞ4u k Þ = f k in U k ,
k = 1, 2 ,
where the equality here is in the sense of distributions. Finally, we can show in the usual
way that the weak solutions satisfy the natural boundary condition,
3
n 1 6 ßA 1 4u 1 ? A 2 4u 2 à = 0 on @.
The precise sense in which this equality holds has not been defined.
Problem 2 Show that W is a Hilbert space for the H 1 ÝUÞ ? norm.
Use the Poincare inequality to show that the Poincare norm is equivalent to the V norm on
W.
Is W a Hilbert space for the Poincare norm?
3, 3
Problem 3 Show that aßu
v à is a bounded bilinear form on W (which norm must you
use?)
3, 3
u| 2 for all u 5 W (which norm must you use?)
Problem 4 Show that aßu
uà ³ C | 3
Problem 5 Use the results of problems 3 and 4 to show that the weak form of the
transmission problem is uniquely solvable for all 3f = ßf 1 , f 2 à 5 H
A Higher Order Equation
Consider the following boundary value problem for the so called biharmonic equation
? 4 2 4 2 uÝxÞ = fÝxÞ
u = 0 and
in
3
n 6 4u = 0
U Ð Rn,
on /U.
Ý3.3Þ
4 2 4 2 uÝx, yÞ = / xxxx uÝx, yÞ + 2/ xxyy uÝx, yÞ + / yyyy uÝx, yÞ
In R 2 ,
Problem (3.3) is the Dirichlet problem for the biharmonic equation. We will now
consider the weak formulation of this problem.
The weak/variational solution of this fourth order problem will reside in the Hilbert space
H 2 ÝUÞ =
u 5 H 0 ÝUÞ : / J uÝxÞ 5 H 0 ÝUÞ, |J| ² 2 ;
It is the space of all functions in H 0 ÝUÞ whose distributional derivatives of order less than or
equal to two are also in H 0 ÝUÞ. This linear space is a Hilbert space for the inner product
10
Ýu, vÞ 2 = > |J|²2 Ý/ J u, / J vÞ 0
||u|| 22 = > |J|²2 ||/ J u|| 20 .
and
In the case n = 1 this becomes
Ýu, vÞ 2 = Ýu, vÞ 0 + Ýu v , v v Þ 0 + Ýu”, v”Þ 0
and
|| u|| 22 = || u|| 20 + || u v || 20 + || u”|| 20 .
The proof that H 2 ÝUÞ is complete is a slight generalization of the proof that showed H 1 ÝUÞ is
complete.
Let H 20 ÝUÞ denote the completion of the test functions in the norm ||6|| 2 . A Poincare
inequality holds for H 20 ÝUÞ
> |J|=2 || / J u|| 20 ³ C || u|| 2 -u 5 C K0 ÝUÞ.
Ý3.4Þ
This inequality can be proved by a slight extension of the proof used for the Sobolev space
of order one. Since we didn’t give this proof, we can, without loss of generality, skip this
proof as well. The inequality implies that the Poincare norm,
| u| 22 = > |J|=2 || / J u|| 20 , u 5 H 20 ÝUÞ,
defines a norm on H 20 ÝUÞ. In 2 dimensions this becomes
| u| 22 = ||/ xx u|| 20 + ||/ xy u|| 20 + ||/ yy u|| 20
u 5 H 20 ÝUÞ.
The functions in H 20 ÝUÞ are those functions in H 2 ÝUÞ that satisfy (in some sense, not
specified here), u = / N u = 0 on /U.
In addition to the Poincare inequality, (3.4), we also have a Green’s identity for the
biharmonic operator,
XU v 4 2 4 2 u dx = X/U áv / N Ý4 2 uÞ + / N v 4 2 uâdS + XU 4 2 v 4 2 u dx
-u, v 5 C K ÝUÞ.
Ý3.5Þ
Now let
and
Then
H = H 0 ÝUÞ, V = H 2 ÝUÞ
V 0 = H 20 ÝUÞ = completion of test functions in the H 2 ÝUÞ ? norm.
C K0 ÝUÞ Ð V 0 Ð V Ð H,
and we can define a weak solution of (3.3) to be a function u 5 V 0 such that
aÝu, vÞ = X 4 2 v 4 2 u dx = Ýf, vÞ 0
U
-v 5 V 0
Ý3.5Þ
The associated variational problem is to minimize
11
Eßuà =
1
2
aÝu, uÞ ? FÝuÞ = X ß 12 | 4 2 u | 2 ? f uà dx
U
over V 0 . In order to apply the Hilbert space lemma to prove existence of a unique solution,
we have to show that the bilinear form is bounded and positive.
Problem 6 Show that for u 5 H 20 ÝUÞ,
a)
||/ xy u|| 20 ² ||/ xx u|| 0 ||/ yy u|| 0 ²
b)
|Ý/ xx u, / yy uÞ 0 | = |Ý/ xy u, / xy uÞ 0 |
Problem 7 Show that for u, v 5 H 20 ÝUÞ,
1
2
||/ xx u|| 20 + ||/ yy u|| 20
| aÝu, vÞ| ² C | u| 2 | v| 2
Problem 8 Use the result of problem 6 b) to show that for u 5 H 20 ÝUÞ,
aÝu, uÞ ³ ||/ xx u|| 20 + 2||/ xy u|| 20 + ||/ yy u|| 20 ³ C | u| 22
Problem 9 What strong boundary value problem is solved by the solution of the variational
problem obtained by minimizing Eßuà over the space H 2 ÝUÞ V H 10 ÝUÞ ?
Problem 10 What strong boundary value problem is solved by the solution of the
variational problem obtained by minimizing Eßuà over the space H 2 ÝUÞ ?
4. Approximation Methods
The conditions under which it is possible to construct an analytical (exact) solution for a
BVP are rather special and it is often the case that no analytical solution is possible. In such
cases the only option is then to try to construct an approximate solution to the BVP (after
first ascertaining by abstract methods that the problem has a unique solution in a specific
solution class).
Finite difference methods are one way of approximating the solution to a BVP for which
analytic methods fail. These methods replace derivatives by difference quotients, thereby
changing a differential equation plus boundary conditions into a system of algebraic
equations. However, this is not the only way in which a BVP can be transformed into a
system of algebraic equations. We will consider three apparently different but ultimately
equivalent ways for approximating the solution of a BVP.
The Rayleigh-Ritz Method
The Rayleigh-Ritz method applies only to problems with a variational formulation.
Therefore, consider the variational problem,
Find u 5 V, such that Eßuà ² Eßvà
for all v 5 V
Ý4.1Þ
where H 10 ÝUÞ Ð V Ð H 1 ÝUÞ. Let ád 1 , u, d N â denote N linearly independent functions in V,
and let M = spanßd 1 , u, d N à. Then consider the following approximation to the problem (4.1)
12
Find u M 5 M, such that Eßu M à ² Eßvà
for all v 5 M Ð V
Ý4.2Þ
We refer to M as the approximating subspace and the functions d k are called ”trial
functions” or ”shape functions”. The approximate solution, u M , belongs to M, and since the
functions d k ÝxÞ span M, we can write
N
u M ÝxÞ = > C k d k ÝxÞ.
k=1
Now we can define a function of N real variables, C 1 , ..., C N , by FÝC 1 , ..., C N Þ = Eßu M à and
seek to minimize the function F over all of R N . This will have the effect of minimizing Eßu M à
over M as required by (4.2). Since we are minimizing F over all of R N , all minima for F must
be interior minima hence we must have
/F Ĉ 1 , ..., Ĉ N = 0
for
k = 1, ..., N
Ý4.3Þ
/C k
This is a system of N linear equations for the N unknowns Ĉ 1 , ..., Ĉ N . To see what these
equations might look like, recall that quadratic functional, Eßuà, in the variational problem
has the form,
1
2
Eßuà =
aÝu, uÞ ? Ýf, uÞ 2 .
Then
Eßu M à =
1
2
aÝu M , u M Þ ? Ýf, u M Þ 2
=
1
2
a > C j d j ÝxÞ, > C k d k ÝxÞ
N
j=1
N
k=1
N
f, > C k d k ÝxÞ
?
N
k=1
2
V
> > C j C k aÝ d j ÝxÞ, d k ÝxÞÞ ? > C k Ýf, d k ÝxÞÞ 2
1
2
=
i.e.,
N
j=1 k=1
FÝC 1 , ..., C M Þ =
1
2
N
k=1
N
N
> > C j C k A jk ? > C k F k ,
j=1 k=1
k=1
where
A jk = aÝ d j ÝxÞ, d k ÝxÞÞ = A kj and F k = Ýf, d k ÝxÞÞ 2 .
Then (1.3) is equivalent to the system
/F =
/C k
1
2
N
> A jk C k +
k=1
N
1
2
N
> A jk C j ? F k
j=1
= > A jk C k ? F k = 0, k = 1, ..., N.
k=1
Ý4.3 v Þ
If the bilinear form aÝu, vÞ is positive then the matrix A is positive definite and the system
(4.3’) has a unique solution. The corresponding u M is then the Rayleigh-Ritz approximation
to the solution of the variational problem, (4.1).
Problem 11 Show that if the bilinear form aÝu, vÞ is positive and symmetric, then the matrix
ßA jk à = ßaÝd j , d k Þà is symmetric and positive definite.
The Galerkine Procedure
Since not every BVP has a variational formulation, the Rayleigh-Ritz procedure can not
13
always be applied. Consider then the following weak formulation of a BVP
Find u 5 V such that
aÝu, vÞ = Ýf, vÞ 2 = Fßvà for all v 5 V
Ý4.4Þ
Let ád 1 , u, d N â and M be as they were defined in the previous example and let u M denote
the solution of the problem,
Find u M 5 M such that
aÝu M , vÞ = Fßvà for all v 5 M.
Ý4.5Þ
Obviously, (4.5) is equivalent to
Find u M 5 M such that
Now
Ý4.5 v Þ
aÝu M , d k Þ = Fßd k à for k = 1, ..., N.
N
u M ÝxÞ = > B k d k ÝxÞ
k=1
satisfies
N
a > B j d j ÝxÞ, d k
= Fßd k à,
j=1
i.e.,
N
Mn
j=1
j=1
> B j aÝd j , d k Þ = > A jk B j = F k , for k = 1, ..., N.
Clearly these are the same equations as the system (4.3’). The Rayleigh-Ritz procedure
only applies to problems having a variational formulation, while the Galerkine procedure
applies to the weak formulation whether or not there is a variational formulation. In the case
that there is no variational formulation, the matrix ßA jk à is not symmetric. However, if the
original bilinear form aÝu, vÞ is positive, then the matrix will be positive definite and the
approximate problem is uniquely solvable.
The Method of Weighted Residuals
Consider the following strong formulation of a BVP
? 4ÝTÝx, yÞ4uÞ + 3
cÝx, yÞ 6 4u + bÝx, yÞ u = fÝx, yÞ
in
U
u=0
on /U 1
ÝT4uÞ 6 n = 0
on /U 2
We can write this as
Find u 5 V such that
LßuÝx, yÞà = fÝx, yÞ
Ý4.6Þ
Where /U is comprised of two disjoint pieces /U 1 and /U 2 ; i.e., /U = /U 1 W /U 2 , and V
denotes the H 1 ÝUÞ ? closure of the infinitely differentiable functions which are zero on
/U 1 . We will suppose that neither /U 1 nor /U 2 is empty so that V is strictly between
H 10 ÝUÞ and H 1 ÝUÞ. To construct an approximate solution by the method of weighted
residuals, the trial functions ád 1 , u, d N â are chosen to be N linearly independent functions in
V with the additional constraint that Lßd k à 5 L 2 ÝUÞ for each k. Then let S denote the
N-dimensional subspace spanned by the trial functions. In the previous two methods the
trial functions were not required to have this additional smoothness.
With these trial functions, let
N
u S ÝxÞ = > c k d k ÝxÞ
k=1
be define by
ÝLßu S à, d j Þ 2 = Ýf, d j Þ 2
for j = 1, ..., N
Ý4.7Þ
14
Note that
N
N
L > c k d k ÝxÞ , d j
ÝLßu S à, d k Þ 2 =
k=1
= > c k ÝLßd k à, d j Þ 2 ,
2
k=1
and
ÝLßd k à, d j Þ 2 = aÝd k , d j Þ
for d k , d j 5 S Ð V.
Then (4.7) is equivalent to
N
> A kj c k = F j ,
k=1
for j = 1, ..., N
Ý4.7 v Þ
Evidently, the method of weighted residuals leads to the same equations as the other two
methods although the trial functions carry an extra smoothness constraint that is not
required of the Galerkine and Rayleigh-Ritz trial functions.
The success of these approximation methods depends on making a good choice of trial
functions. If the trial functions are chosen from families of piecewise polynomials called
finite element spaces then several advantages arise:
¾
it is possible to deal systematically with irregular regions, even ones having curved
boundaries
¾
the accuracy of the approximation can be estimated in a systematic way in terms of
the adjustable parameters characterizing the finite element family
¾
the ingredients of the approximation problem, including the coefficient matrix ßA jk à
and the data vector ßF k à in the system of algebraic equations and even the mesh
decomposition of the region can be efficiently generated by packaged software.
Employing one of these three approximation techniques in concert with a finite element
family of trial functions is referred to as the ”finite element method”.
We should note that the Hilbert space H 1 ÝUÞ is the ”best” solution space in which to look
for solutions to second order elliptic boundary value problems for several reasons:
¾ an elliptic BVP of order 2 may fail to have any solution in a smaller space (e.g.
H 2 ÝUÞ ). For example, this could be due to irregularities in /U
¾
solutions in a space larger than H 1 ÝUÞ may not make sense physically. For example,
the functions may fail to have finite energy.
¾
it is easy to build finite dimensional subspaces of H 1 ÝUÞ which are convenient for
constructing approximate solutions to the BVP. This is less easy for spaces like
H 2 ÝUÞ or CÝŪÞ.
We should also be aware that there are various possible formulations we could imagine for
a weak solution to the problem,
?4 2 u = F
in U,
u = 0 on /U.
(a) Ultra-regular Weak solution Find u 5 H 2 ÝUÞ V H 10 ÝUÞ such that
XU Ý4 2 u + FÞ v dx = 0 for all v 5 L 2 ÝUÞ
(b) Weak solution Find u 5 H 10 ÝUÞ such that
XU 4u 6 4v dx = XU F v dx for all v 5 H 10 ÝUÞ
15
(c) Ultra-Weak solution Find u 5 L 2 ÝUÞ such that
XU Ý4 2 v + FÞu dx = 0 for all v 5 H 2 ÝUÞ V H 10 ÝUÞ.
In formulation (a) we are obliged to construct a family áV n â of approximate solution spaces
which approach H 2 ÝUÞ V H 10 ÝUÞ with increasing n. Then the family áW n â of test function
spaces is just L 2 ÝUÞ for every n. In formulation (c) the situation is reversed with the space
V n of approximate solutions equal to L 2 ÝUÞ for all n and the family áW n â of test function
spaces approaching H 2 ÝUÞ V H 10 ÝUÞ with increasing n. In case (b) we make the choice
áV v â = áW n â and these spaces must approach H 10 ÝUÞ with increasing n. This compromise
leads to an approximate problem with a symmetric and positive definite matrix as an
approximation to the operator in the BVP. This does not happen in the other two cases.
This fact, coupled with the difficulty in building a sequence of finite dimensional spaces
tending to H 2 ÝUÞ V H 10 ÝUÞ, suggests that (b) is the optimal weak formulation of the BVP.
16