1 Calculus of Variations 2 History - Weebly

Monday, August 26, 2019

1 Calculus of Variations

2 History

Calculus of variations is a branch of mathematical analysis that studies extrema

and critical points of functionals (or energies). One of the ...rst questions that

may be framed within this theory is the Dido's isoperimetric problem: to ...nd

the shape of a curve of prescribed perimeter that maximizes the area enclosed.

Dido was a Phoenician princess who emigrated to North Africa and upon arrival

obtained from the native chief as much territory as she could enclose with an

ox hide. She cut the hide into a long strip, and used it to delineate the territory

later known as Carthage, bounded by a straight coastal line and a semi-circle.

It is commonly accepted that the systematic development of the theory of

the calculus of variations began with the brachistochrone curve problem pro-

posed by Johann Bernoulli in 1696: consider two points A and B on the same

vertical plane but on di?erent vertical lines. Assume that A is higher than B,

and that a particle M is moving from A to B along a curve and under the action

of gravity. The curve that minimizes the time travelled by M is called brachis-

tochrone. The solution to this problem required the use of in...nitesimal calculus

and was later found by Jacob Bernoulli, Newton, Leibnitz and de l'H?pital.

The arguments thus developed led to the development of the foundations of the

calculus of variations by Euler. Important contributions to the subject are at-

tributed to Dirichlet, Hilbert, Lebesgue, Riemann, Tonelli, Weierstrass, among

many others.

The common feature underlying Dido's and the brachistochrone problems is

that one seeks to maximize or minimize a functional over a class of competitors

satisfying given constraints. In both cases the functional is given by an integral

of a density depending on an underlying ...eld and some of its derivatives, and

this will be the prototype we will adopt in what follows. Precisely, we consider

a functional

Z

u 2 X 7! F(u) := f (x; u(x); ru(x)) dx;

(1)

where X is a function space (usually a Lp space or a Sobolev-type space),

u: ! Rd, with

RN an open set, N and d are positive integers, and the

density is a function f (x; s; ), with (x; s; ) 2

Rd Rd N . Here, and in

what follows, ru stands for the d N matrix-value distributional derivative of

u.

Important examples are:

Fermat's principle in geometrical optics: N = d = 1 and p

f (x; s; ) := g(x; s) 1 + 2:

1

The Newton's problem: N = d = 1 and

3

f (x; s; ) := 2 s 1 + 2 :

The brachistochrone problem: N = d = 1 and

p

f (x; s; ) :=

p1 +

2

:

2gs

Dirichlet's principle: N 1, d = 1 and

f (x; s;

)

:=

1 2

k

k2 + g(x; s):

The p-Laplacian problem: 1 p < 1, N 1, d = 1 and

f (x; s;

)

:=

1 p

k

kp + g(x; s):

Non-parametric minimal surfaces: N 1, d = 1 and p

f (x; s; ) := 1 + k k2:

3 Critical Points

De...nition 1 Let (X; k kX ) and (Y; k kY ) be two normed spaces, let E X, let f : E ! Y , let x0 2 E and let v 2 X with kvkX = 1 be such x0 is an accumulation point for E \ Lv, where Lv is the line through x0 in the direction

v. The directional derivative of f in the direction v at x0 is given by

@f @v

(x0)

=

lim

t!0

f

(x0

+

tv) t

f (x0) ;

provided the limit exist. The function f is G?teaux di?erentiable at x0 if there

exists

@f @v

(x0)

for

every

direction

v

and

the

mapping

@f v 7! @v (x0)

can be extended to a linear and continuous function dGf (x0) : X ! Y .

Example 2 If X is an in...nite dimensional normed space, let fengn be a sequence of linearly independent unit vectors. De...ne T (en) := nkenkX . Complete this sequence fengn to a basis for X and de...ne T to be zero for the elements of the basis not in the sequence. This de...nes a linear functional T : X ! R. Note that T is discontinuous since it is not bounded on the unit sphere. For every

x 2 X and every direction v we have that

@T (x) = T (v): @v

Since T is not bounded on the unit sphere, we have that T is not G?teaux di? erentiable at any x.

2

Example 3 This example shows that G?teaux di? erentiability does not imply

continuity. Let

f (x; y) :=

1 if y = x2; x 6= 0; 0 otherwise.

Given a direction v = (v1; v2), the line L through (0; 0) in the direction v intersects the parabola y = x2 only in (0; 0) and in at most one point. Hence, if

t is very small,

f (0 + tv1; 0 + tv2) = 0:

It follows that

@f (0; 0) = lim f (0 + tv1; 0 + tv2)

f (0; 0)

0

= lim

0 = 0:

@v

t!0

t

t!0 t

However, f is not continuous in (0; 0), since f x; x2 = 1 ! 1 as x ! 0, while f (x; 0) = 0 ! 0 as x ! 0.

Example 4 This example shows that the existence of all the directional deriv-

atives does not imply G?teaux di? erentiability. Let

(

x2 y

f (x; y) := x4+y2 0

if (x; y) 6= (0; 0) ; if (x; y) = (0; 0) :

Let's ...nd the directional derivatives of f at (0; 0). Given a direction v = (v1; v2), with v12 + v22 = 1, we have

f (0 + tv1; 0 + tv2) = 0:

It follows that

f (0 + tv1; 0 + tv2)

f (0; 0)

=

(tv1 )2 tv2 (tv1 )4 +(tv2 )2

0

t

t

=

t3 t5v14

v12v2 + t3v22

:

If v2 = 0 then

f (0 + tv1; 0 + tv2) t

f (0; 0)

0

= t5v14 + 0 = 0 ! 0

as

t ! 0,

so

@f @x

(0; 0) = 0.

If

v2

6= 0,

then,

f (0 + tv1; 0 + tv2) t

f

(0; 0)

=

v12v2 t2v14 + v22

!

v12v2 0 + v22

=

v12 ; v2

so

@f (0; 0) = v12 :

@v

v2

3

In

particular,

@f @y

(0; 0)

=

0 1

=

0.

Thus,

f

is

not

G?teaux

di? erentiable

since

we

do not have linearity in v.

Now let's prove that f is not continuous at (0; 0). We have

0 f (x; 0) = 0 + y2 = 0 ! 0

as x ! 0, while

f

x; x2

x2x2

11

= x4 + x4 = 2 ! 2

as x ! 0. Hence, the limit lim(x;y)!(0;0) f (x; y) does not exists and so f is not continuous at (0; 0).

De...nition 5 Let (X; k kX ) and (Y; k kY ) be two normed spaces, let E X, let f : E ! Y , and let x0 2 E be an accumulation point of E. The function f is

Fr?chet di?erentiable at x0 if there exists T 2 L(X; Y ) such that

lim f (x) f (x0) T (x x0) = 0:

x!x0

kx x0kX

The linear function T is the di?erential of f at x0 and is denoted df (x0). The point x0 is a critical point of f if df (x0) = 0.

Here L(X; Y ) is the space of all linear and continuous functions T : X ! Y , endowed with the norm

kT kL(X;Y ) := sup fkT (x)kY : kxkX 1g :

In the special case in which Y = R, the space L(X; R) is denoted X0 and is called the topological dual space of X.

Remark 6 If f is Fr?chet di? erentiable at x0 2 E , then f is G?teaux di? erentiable at x0 and

@f @v (x0) = dGf (x0)(v) = df (x0)(v):

De...nition 7 Let (X; k kX ) and (Y; k kY ) be two normed spaces, let U X be an open set and let f : U ! Y . We say that f is of class C1 if f is Fr?chet di? erentiable at every x 2 U and the function

df : U ! L(X; Y ) x 7! df (x)

is continuous.

The following theorem will be useful in applications.

4

Theorem 8 Let (X; k kX ) and (Y; k kY ) be two normed spaces, let E X, let f : E ! Y , and let x0 2 E . Assume that f is G?teaux di? erentiable in BX (x0; r) E, for some r > 0, and that the function

dGf : BX (x0; r) ! L(X; Y ) x 7! dGf (x)

is continuous at x0. Then f is Fr?chet di? erentiable at x0 and df (x0) = dGf (x0).

The following lemma is an in...nite?dimensional version of the mean value theorem.

Lemma 9 Let (X; k kX ) and (Y; k kY ) be two normed spaces, let E X, let f : E ! Y , and let x0 2 E . Assume that f is G?teaux di? erentiable in BX (x0; r) E. Then for all x1; x2 2 BX (x0; r),

kf (x2) f (x1)kY kx1 x2kX sup kdGf (tx1 + (1 t)x2)kL(X;Y ) :

t2[0;1]

Wednesday, August 28, 2019

Proof. Let x1; x2 2 BX (x0; r) and assume that f (x1) 6= f (x2). By a corollary of the Hahn?Banach theorem there exists T 2 Y 0 such that kT kY 0 = 1 and

T (f (x2) f (x1)) = kf (x2) f (x1)kY :

(2)

Consider the function g : [0; 1] ! R de...ned

Since T is linear,

g(t) := T (f (tx1 + (1 t)x2)):

g(t) g(t0) = T (f (tx1 + (1 t)x2)) T (f (t0x1 + (1 t0)x2))

t t0

t t0

= T f (tx1 + (1 t)x2) f (t0x1 + (1 t0)x2) t t0

= kx1

x2kX T

f (xt0 + sv) s

f (xt0 ) ;

where xt0 := t0x1 + (1

t0)x2,

v

:=

x1 kx1

, x2

x2 kX

s

:=

(t

t0) kx1

x2kX . Since

f is G?teaux di?erentiable at xt0 and T is continuous, there exists the limit

g0(t0)

=

lim

t!t0

g(t) t

g(t0) t0

=

kx1

x2kX T

@f @v (xt0 ) :

5

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download