7

Calculus of Variations

Ref: Evans, Sections 8.1, 8.2, 8.4

7.1

Motivation

The calculus of variations is a technique in which a partial dierential equation can be reformulated as a minimization problem. In the previous section, we saw an example of this technique. Letting vi denote the eigenfunctions of () and dening the class of functions Yn = {w C 2 (), w 0, w = 0 for x , < w, vi >= 0, i = 1, . . . , n 1}, we saw that if u Yn is a minimizer of the functional I(w) = || w||2 2 () L ||w||2 2 () L v = v v=0 x x ,

over Yn , then u is an eigenfunction of (*) with corresponding eigenvalue m || u||2 2 () L ||u||2 2 () L .

Further, we showed that m is in fact the nth eigenvalue of (*). In other words to nd a solution of an eigenvalue problem, we reformulated the problem in terms of minimizing a certain functional. Proving the existence of an eigenfunction is now equivalent to proving the existence of a minimizer of I over the class Yn . Proving the existence of a minimizer requires more sophisticated functional analysis. We will return to this idea later. The example above could be reformulated equivalently to say that we are trying to minimize the functional I(w) = || w||2 2 () L over all functions w Yn such that ||w||2 2 () = 1. In particular, if v is in Yn , then the L 2 normalized function v v/||v||L2 () (which has the same Rayleigh quotient as v) is in Yn , and, of course, ||v||2 2 () = 1. Therefore, minimizing I over functions w Yn is equivalent L to minimizing I over functions w Yn subject to the constraint ||w||2 ( ) = 1. This type of L minimization problem is called a constrained minimization problem. We begin by considering a simple example of how a partial dierential equation can be rewritten as a minimizer of a certain functional over a certain class of admissible functions.

7.2

Dirichlets Principle

In this section, we show that the solution of Laplaces equation can be rewritten as a minimization problem. Let A {w C 2 (), w = g for x }. 1 | w|2 dx. 2 Theorem 1. (Dirichlets Principle) Let be an open, bounded subset of Rn . Consider Laplaces equation on with Dirichlet boundary conditions, I(w) () u = 0 u=g x x . Let

The function u A is a solution of (*) if and only if I(u) = min I(w).

Proof. First, we suppose u is a solution of (*). We need to show that I(u) I(w) for all w A. Let w A. Then 0=

u(u w) dx | u|2 dx +

u 1 2

w dx

| u|2 dx +

[| u|2 + | w|2 ] dx

1 = 2 Therefore,

1 | u|2 dx + 2 | u|2 dx

| w|2 dx.

| w|2 dx.

But w is an arbitrary function in A. Therefore, I(u) = min I(w).

Next, suppose u minimizes I over all w A. We need to show that u is a solution of (*). Let v be a C 2 function such that v 0 for x . Therefore, for all , u + v A. Now let i( ) I(u + v). By assumption, u is a minimizer of I. Therefore, i must have a minimum at therefore, i (0) = 0. Now i( ) = I(u + v) =

= 0, and,

| (u + v)|2 dx [| u|2 + 2

u 2

v+

| v|2 ] dx,

implies i( )=

[ u

v] + 2 | v|2 dx.

Therefore, i (0) =

v dx u v dS(x)

(u)v dx +

(u)v dx.

Now i (0) = 0 implies (u)v dx = 0.

Since this is true for all v C 2 () such that v = 0 for x , we can conclude that u = 0, as claimed.

7.3

Euler-Lagrange Equations

Laplaces equation is an example of a class of partial dierential equations known as EulerLagrange equations. These equations are dened as follows. Let be an open, bounded subset of Rn . Let L be a smooth function such that L : Rn R R. We will write L = L(p, z, x) where p Rn , z R and x . Associated with a function L, we dene the Euler-Lagrange equation
n

i=1

(Lpi ( u, u, x))xi + Lz ( u, u, x) = 0.

The function L is known as the Lagrangian. Example 2. Let 1 L(p, z, x) = |p|2 . 2 The associated Euler-Lagrange equation is just Laplaces equation u = 0.

Example 3. Let

1 L(p, z, x) = aij (x)pi pj zf (x) 2 i,j=1 3

where aij = aji . The associated Euler-Lagrange equation is

a generalization of Poissons equation.

(aij uxj )xi = f,

Recall that Dirichlets principle stated that a solution of u = 0 u=g is a minimizer of I(w) = 1 2 x Rn x | w|2 dx

over C 2 functions which satisfy the boundary condition. In other words, a harmonic function on is a minimizer of I(w) = L( w, w, x) dx

where L is the associated Lagrangian given by L(p, z, x) = 1 |p|2 . 2 For a given Lagrangian L dene IL (w) =

L( w, w, x) dx.

We will now show that if u is a minimizer of IL (w) over an admissible class of functions A, then u is a solution of the associated Euler-Lagrange equation
n

i=1

(Lpi ( u, u, x))xi + Lz ( u, u, x) = 0.

As in the proof of Dirichlets principle, suppose u is a minimizer of IL (w) =

L( w, w, x) dx

over an admissible class of functions A. Let v C () such that v has compact support within . We denote this space of functions by Cc (). Dene i( ) = I(u + v). If u is a minimizer of I, then i (0) = 0. i( ) = I(u + v) =

L( u +

v, u + v, x) dx.

Therefore,
n

i( )=
i=1

Lpi ( u +

v, u + v, x)vxi + Lz ( u +

v, u + v, x)v dx.

Now i (0) = 0 implies

0 = i (0) =
i=1

Lpi ( u, u, x)vxi + Lz ( u, u, x)v dx.

Integrating by parts and using the fact that v = 0 for x , we conclude that
n

i=1

(Lpi ( u, u, x))xi + Lz ( u, u, x) v dx = 0.

Since this is true for all v Cc (), we conclude that u is a solution of the Euler-Lagrange equation associated with the Lagrangian L. Consequently, to solve Euler-Lagrange equations, we can reformulate these partial dierential equations as minimization problems of the functionals L( w, w, x) dx. IL (w) =

Above we showed how solving certain partial dierential equations could be rewritten as minimization problems. Sometimes, however, the minimization problem is the physical problem which we are interested in solving. Example 4. (Minimal Surfaces) Let w : R. The surface area of the graph of w is given by I(w) =

(1 + | w|2 )1/2 dx.

The problem is to look for the minimal surface, the surface with the least surface area, which satises the boundary condition w = g for x . Alternatively, this minimization problem can be written as a partial dierential equation. In particular, the Lagrangian associated with I is L(p, z, x) = (1 + |p|2 )1/2 . The associated Euler-Lagrange equation is
n

i=1

uxi (1 + | u|2 )1/2

= 0.
xi

This equation is known as the minimal surface equation.

7.4

Existence of Minimizers
I(w) =

We now discuss the existence of a minimizer of L( w, w, x) dx

over some admissible class of functions A. We will discuss existence under two assumptions on the Lagrangian L: convexity and coercivity. We discuss these issues now. 5

7.4.1

Convexity I(w) =

Assume u is a minimizer of L( w, w, x) dx.

We discussed earlier that if u is a minimizer of I, then for any v Cc (), the function

i( ) = I(u + v) has a local minimum at = 0, and, therefore, i (0) = 0. In addition, if i has a minimum at = 0, then i (0) 0. We now calculate i (0) explicitly to see what this implies about I and L. By a straightforward calculation, we see that
n

i ( )=
i,j=1 n

Lpi pj ( u + Lpi z ( u +
i=1

v, u + v, x)vxi vxj v, u + v, x)vxi v + Lzz ( u + v, u + v, x)v 2 dx.

+2

Therefore, we conclude that

0 i (0) =
i,j=1

Lpi pj ( u, u, x)vxi vxj + 2

Lpi z ( u, u, x)vxi v

(7.1)

+Lzz ( u, u, x)v 2 dx
for all v Cc (). By an approximation argument, we can show that (7.1) is also valid for the function x v(x) = (x) where > 0, Rn , Cc () and : R R is the periodic zig-zag function 1 x 0x 2 (x) = 1 1 x x1 2

and (x + 1) = (x) elsewhere. Notice that the function satises | | = 1 almost everywhere (except on a set of measure zero). Calculating the partial derivative of v with respect to xi , we see that x vxi (x) = i (x) + O() as 0. Combining this fact with (7.1), using the fact that | | = 1 almost everywhere, and taking the limit as 0, we conclude that
n

0
i,j=1

Lpi pj ( u, u, x)i j dx. 6

 Since this estimate holds for all Cc (), we conclude that n

0
i,j=1

Lpi pj ( u, u, x)i j .

This is a necessary condition for a minimizer. Therefore, to guarantee the possibility of a minimizer, we will assume the following convexity condition on the Lagrangian L,
n

Lpi pj (p, z, x)i j 0.

(7.2)

That is, we will assume L is convex in the variable p. 7.4.2 Coercivity

In order to prove that the functional I has a minimum, we would need to know that I is bounded below. Of course, this is not enough to guarantee the existence of a minimizer. For 2 example, the function f (x) = ex is bounded below by zero. In addition, inf xR f (x) = 0. However, the inmum is never achieved. One way of guaranteeing that a continuous function f : R R achieves its inmum is to assume that f (x) + as |x| +. Using this same idea for our functional I, we will assume that I(w) + as |w| +. In particular, we will assume the following coercivity condition. Let 1 < q < be xed. > 0, 0 s.t. L(p, z, x) |p|q (p, z, x) Rn R . (7.3)

If the Lagrangian L satises this coercivity condition, then the functional I satises I(w) =

L( w, w, x) dx | w|q dx ||.

Therefore, I(w) + as 7.4.3

| w|q dx +.

Theorem on Existence of Minimizers

In this section, we state an existence result for minimizers. We will be considering the functional I(w) = L( w, w, x) dx

where the Lagrangian L satises the convexity and coercivity conditions described in the previous sections. We will prove the existence of a minimizer u of I over a certain class of admissible functions A. We describe this class of functions now. First, for 1 q < , let Lq () {u :

|u|q dx < }.

We dene the norm of a function u Lq () as

||u|| Now we let

Lq ()

|u| dx

W 1,q () {u : u Lq (), u Lq ()}. Remark. The space W 1,q is an example of a Sobolev space. For a more thorough introduction to Sobolev spaces, see Evans, Chapter 5. Now we let A {w W 1,q ()| w = g on }. Theorem 5. Assume L satises the coercivity condition (7.3) and the convexity condition (7.2). Assume the set A is nonempty. Then there exists at least one function u A such that I(u) = min I(w).
wA

Below, we give an outline of the proof. The complete details rely on functional analysis facts which are beyond the scope of this course. Outline of Proof. 1. By the coercivity assumption (7.3), we know that I is bounded below. Therefore, I has an inmum. Let m I(w).
wA

If m = , then were done. Therefore, we assume that m is nite. Let uk be a minimizing sequence. That is, assume I(uk ) m as k +. We now want to show that there exists a u A such that I(u) = m. First, we will show that uk is a bounded sequence in W 1,q . This will imply there exists a u W 1,q such that uk u (converges weakly). Then we need to show that u A and I(u) = m. 2. Assume = 0 in the coercivity assumption (7.3). (Otherwise, we could consider L = L + .) Therefore, L(p, z, x) |p|q . which implies I(w)

| w|q dx.

Therefore, for the minimizing sequence uk , I(uk )

| uk |q dx.

Now I(uk ) m which is nite implies that sup

| uk |q dx < . 8

3. Fix any w A. Since uk = g = w on , uk w = 0 for x . Therefore, we can use Poincares inequality. Therefore, we have ||uk ||Lq () ||uk w||Lq () + ||w||Lq () C|| uk w||Lq () + C C. Therefore, sup ||uk ||Lq () < +.
k

That is, uk is a bounded sequence in W 1,q (). Fact: This implies there exists a function u W 1,q () and a subsequence {ukj } such that ukj converges to u weakly, meaning (ukj u)v dx 0 u.

as k + for all v in the dual space of W 1,q (). We write this as ukj

4. Now we need to show that u A. From the previous step, one can show that u W 1,q (). Therefore, it just remains to show that u satises the necessary boundary conditions. See Evans, Sec. 8.2. 5. Using the fact that ukj able to conclude that u, and the fact that L is convex and bounded below, we are I(u) lim inf I(ukj ) = m. See Evans, Sec. 8.2. But, since u A, I(u) m. Therefore, I(u) = m.

7.5

Constrained Minimization Problems

In this section, we discuss constrained minimization problems: minimizing a functional I(w) subject to the constraint J(w) = 0. Example 6. Earlier we mentioned that the Minimum Principle for the nth Eigenvalue could be written as a constrained minimization problem. Let J(w) ||w||2 2 () . L Let A {w Yn |J(w) = 0}. Let I(w) = || w||2 2 () . L We showed that if u A satises I(u) = min I(w),
wA

then u satises

u = n u u=0

x x ,

where n = || u||2 2 () . L We will now state and prove a generalization of the result in the above example. Let I(w) 1 2 | w|2 dx.

Let G : R R be a smooth function. Let g = G . Let J(w)

G(w(x)) dx.

Let A {w C 2 (), w = 0 for x | J(w) = 0}. Theorem 7. Suppose there exists a u A such that I(u) = min I(w).
wA

Then there exists a real number such that u = g(u) u=0 x x .

The number is called the Lagrange multiplier corresponding to the integral constraint J(u) = 0. Proof. We would like to use a similar argument as used in the unconstrained minimization problem. In particular, we would like to nd w such that u + w A, and then look at i( ) = I(u + w). Unlike in the previous case, however, we cannot just choose w Cc (). Because, this does not necessarily imply that u + w is in A. We construct such a w as follows. Let v Cc (). As mentioned above u + v is not necessarily in A, because it does not necessarily satisfy the constraint J(w) = 0. But, maybe we could nd some function w C 2 () which would correct this problem. Maybe we could nd some w and some function ( ) such that u + v + ( )w A. Choose w such that g(u)w dx = 0.

(7.4)

Dene j( , ) J(u + v + w) =

G(u + v + w) dx.

We see that j(0, 0) =

G(u) dx = 0, 10

by the assumption that u A. We see that j ( , ) = g(u + v + w)w dx.

Now by the assumption (7.4), we see that j (0, 0) = 0. Therefore, by the implicit function, there exists a C 1 function : R R such that (0) = 0 and j( , ( )) = 0. In particular, this implies that j( , ( )) = J(u + v + ( )w) = 0, which means that u + v + ( )w A. Now we can use a similar argument as to the one we used earlier. Let i( ) = I(u + v + ( )w) = 1 2 | u+

v + ( ) w|2 dx.

By the assumption that u is a minimizer of I over all functions in A, we know that i (0) = 0. Now i ( ) = ( u + v + ( ) w) ( v + ( ) w) dx

implies i (0) =

( u

v + (0) u

w) dx = 0.

(7.5)

Using the fact that j( , ( )) = 0, we can calculate (0) as follows. j j ( , ( )) + ( , ( )) ( ) = 0 implies (0) = j (0, 0) = j (0, 0) g(u)v dx . g(u)w dx

(7.6)

Combining (7.5) with (7.6), we conclude that u

v dx =

g(u)v dx g(u)w dx u w dx = g(u)w dx

w dx

g(u)v dx.

11

Dening we conclude that u

u w dx , g(u)w dx g(u)v dx.

v dx =

Using the integration by parts formula on the left-hand side, and the fact that v vanishes for x , we conclude that

u v dx =

g(u)v dx.

Since this is true for all v C 2 () which vanish for x , we conclude that u = g(u) u=0 x x .

12