Calculus of Variations: Suggested Exercises

Instructor: Robert Kohn. Solutions by the Fall 09 class on Calculus of Variations.

December 9, 2009

Contents

1 Lecture 1: The Direct Method 1

2 Lecture 2: Convex Duality 7

3 Lecture 3: Geodesics 11

4 Lecture 4: Geodesics 19

5 Lecture 5: Optimal Control 20

6 Lecture 7: 34

7 Lecture 8 40

1 Lecture 1: The Direct Method

1. Two examples were given where the Direct Method fails because theboundary condition is lost in the limit (examples A and C). Which stepsin our proof of the direct method fail in each case, and why?

Solution 1 (Sinziana) (A) was an example of the direct method failing tosolve the Laplace equation ��u + f = 0 with Neumann boundary conditions@u@n = g at @. In the 1D case, take = [0; 1]; f = 0; @u@n (0) = �1; @u@n (1) = 1and we solve

minux(0)=�1;ux(1)=1

Z 1

0

1

2u2xdx

A minimizing sequence is

u� =

8<: �x+ � 0 � x � �0 � � x � 1� �x� (1� �) 1� � � x � 1

and E[u�] = �. Taking � ! 0, the minimum value is 0, so the boundary condi-tions are lost.

1

Following the steps of the direct method with W (ru) = 12u

2x, we see that

W is well de�ned, continuous on H1[0; 1], convex and bounded below by 0. LetS be the space of functions which satisfy the boundary condition. Since W isbounded below, there exists a uniformly bounded minimizing sequence u�k 2 Ssuch that ku�kkW 1;p � M ; by weak compactness of the unit ball for p > 1 thissequence has a weakly convergent subsequence, ju�kj �u

�jH1 ! 0. However, theabove shows that u� = 0 =2 S, (since u�k ! 0) i.e. the space S is not closed,since u� escapes from it and loses the boundary conditions. This is where theDirect Method fails.Remark: The map f ! @f

@n

���@= g is not a continuous functional in H1 so

this has no hope of working. This is because if u 2 H1 ) ru 2 L2 ) tr(ru) 2H� 1

2 and we need at least H12 for continuity.

(C) was a minimal surface problem with Dirichlet boundary conditions, thesurface given by the graph of u. This example shows that even Dirichlet bound-ary conditions may be lost.

minuj@='

Z

(1 + jruj2) 12

To use the Direct Method let W (ru) =p1 + jruj2, so W is convex and there-

fore lower semicontinuous. Let be the annulus fa < r < bg in R2, noting that is not convex. Assuming u is radial, we impose the boundary condition

' =

�A r = aB r = b

with A;B su¢ ciently large.We observe that if jruj is very large

p1 + jruj2 � jruj, soW (ru) behaves

like the L1 norm of ru. Following the steps of the Direct Method, this impliesthat the lower bound of the growth condition holds only for p = 1, i.e. if � islarge

C1(k�k1 � 1) �W (�) � C2(k�kp + 1)We can show the upper bound still holds for p > 1 by applying Jensen�s inequal-ity. W is well de�ned, condinuous and bounded below. Call u� the minimumvalue and take uk 2 W 1;p a minimizing sequence. Since for p = 1 the unit ballis no longer weakly compact, we cannot extract a convergent subsequence.What happens here is that when A�B gets large, the minimizer exists but

it looses the boundary conditions. For A � B large enough the surface has avertical piece so it is no longer a graph. We can show this in 2D. Changing topolar coordinates

W (ru) = 2�Z b

a

rp1 + u2rdr

so the minimization problem becomes

minu(a)=A;u(b)=B

Z b

a

rp1 + u2rdr

2

Solving this is equivalent to solving the Euler-Lagrange equations

d

dr

�d

dprp1 + p2

�= 0

where p = ur. This gives

d

dr

r

urp1 + u2r

!= 0) urp

1 + u2r+

rurr�p1 + u2r

�3 = 0Calling y = ur and using separation of variables this becomes

(1 + y2)y + ry0 = 0) dy

y(1 + y2)= �dr

r

so that

ln j y

1 + y2j = � log r + C ) j y

1 + y2j = C

r, y = �

rC

r2 � C

and integrating we obtain

u(r) = �pC log

�r +

pr2 � C

�+D

Assuming A > B, only the negative solution makes sense. Replacing C by C2

and plugging in the boundary conditions

�C log�a+

pa2 � C2

�+D = A

�C log�b+

pb2 � C2

�+D = B

thus

A�B = � log a+

pa2 � C2

b+pb2 � C2

!If A � B is very large the above equation cannot hold for a; b �xed and C isdetermined by the boundary conditions. If A�B is very large the solution weobtain is discontinuous. Namely, we �nd C = a;D = B + a log

�b+

pb2 � a2

�,

so that

u(r) =

(B � a log

�r+pr2�a2

b+pb2�a2

�r > a

A r = a

In this case the surface given by the graph of u has a vertical region.

2 It was mentioned by the end of class that we can solve a variational prob-lem numerically by minimizing the functional over a �nite dimensional

3

subspace (for example, piecewise linear functions on a particular triangu-lation). Justify this for the special case:�

��u = f in u = 0 in @

�by showing that for any subspace V � H1

0 (); the solution w� of

minw2V

Z

1

2krwk2 + wf

satis�es Z

1

2kru�rw�k2 = min

w2V

Z

1

2kru�rwk2

(The right hand side is easy to estimate, using the smoothness of u). Afood for thought: how can this type of result be generalized to

RW (ru)+

fu when W is convex but not quadratic?

Solution 2 (Eduardo) This result can be generalized replacing krwk2 witha(w;w); where a is a symmetric, continuous and elliptic bilinear form, and itis known as Cea�s lemma (the �nite dimensional minimization is known as theGalerkin-Ritz method). First, we note that the minimizer over the entire space,that is,

u� = arg minuj@='

Z

1

2kruk2 + uf

Satis�es the Euler-Lagrange equations:Z

ru�>rv + fv = 0 8v 2 H10 ()

For the exact same reason, the solution w� satis�es:Z

rw�>rw + fw = 0 8w 2 V

Now, since V is a subspace of H10 (); we can subtract these two to obtain:Z

r(u� � w�)>rw = 0 8w 2 V

We note that in this case, this is an orthogonality statement: u� � w� 2 V ?;and hence w� is the orthogonal projection of u� onto V . In any case, we cannow study the quantity:

1

2

Z

kru� �rwk2 =1

2

Z

hr(u� � w);r(u� � w)i

=1

2

Z

hr(u� � w� + w� � w);r(u� � w� + w� � w)i

=1

2

Z

hr(u� � w�);r(u� � w�)i+ hr(w� � w);r(w� � w)i

=1

2

Z

kru� �rw�k2 + krw� �rwk2

4

Which is essentially Pithagoras theorem. We can easily see that the minimumis attained when krw� �rwkL2() = 0 () w = w�.Food for thought: Although we can easily generalize last result for quadraticfunctionals, it is not straightforward to do so for a general convex W . Theidea here is to use convex duality or something else to show that w� such thatRW (rw�) is nearly optimal =) w� is close to u�. If the Hessian of W is

strictly positive de�nite, then we may use a quadratic approximation for W .Assuming W 2 C2; the convex minimization problem on V is:

minw2V

�ZW (rw)� fw

�= min�2Rn

(ZW (

nXi=1

�ir�i)�nXi=1

�if�i

)3 Does the variational problem

min (u)='

�Z

1

2kruk2 + fu dx

�achieve its minimum when f = �x0 for some x0 2 ? Hint: dim = n � 2is di¤erent from dim = 1.

4 What PDE and boundary condition holds at the critical points of:Z

1

2kruk2 + fu dx + �

Z@

juj2

Solution 3 (Russell) Letting w = u+ �v:

@

@�

�����=0

Z

1

2r(u+ �v) � r(u+ �v) + (u+ �v) � f dx+ �

Z@

(u+ �v) � (u+ �v) =Z

rv � ru+ v � f dx+ 2�Z@

v � u =Z@

vru+ 2�v � u+Z

�v�u+ v � f : dx = Z@

v(ru+ 2�u) +Z

v(��u+ f) : dx

Hence, we have u is a weak solution to the PDE�u = f in , with Robinboundary condition ru+ 2�u = 0 on @.

5 Suppose W is strictly convex and C2; and u is a classical solution of:

div(@W

@(ru) (ru)) = 0 on

With Dirichlet boundary conditions (u = ' at @). Show that u achieves:

min (u)='

Z

W (ru)

and that it is the unique (classical) minimizer.

5

Solution 4 (Evan) First we show that u is a critical point of I(u) =RW (ru)

by showing that I 0(u) = 0 (�rst variation). Taking any v 2 C2 such that (v) = 0, we compute

I 0(u) =d

d"

����"=0

I(u+ "v)

=

Z

@W

@ru (ru) � rv

= �Z

div

�@W

@ru (ru)�v

= 0

since u satis�es div�@W@ru (ru)

�= 0 on . Thus u is a critical point of I. Since

I is a strictly convex functional, it follows that it must be the unique minimizerover all functions in C2 satisfying the boundary conditions.

Proof. We can also verify this by writing for any w 2 C2 satisfying the Dirichletboundary conditions,

I(w) =

Z

W (ru+r(w � u))

=

Z

W (ru) + @W

@ru (ru) � (r(w � u)) + (strictly positive term)

= I(u)�Z

div

�@W

@ru (ru)�(w � u) + (strictly positive term)

= I(u) + (strictly positive term)

> I(u)

noting that w�u is zero on the boundary, so the divergence term vanishes sincethe �rst variation vanishes. Thus u is a global minimizer over functions in C2

satisfying the Dirichlet boundary conditions.

6 Consider W (ux) = (u2x � 1)2; and suppose that W 00(b � a) > 0. Showthat the linear function u�(x) = a + (b � a)x is a C1-local minimizer ofR 10W (ux)dx subject to u(0) = a and u(1) = b, in the sense that for any

v 2 C1(0; 1) with the same boundary conditions and kv � ukC1 su¢ cientlysmall we have: Z

W (vx)dx >

Z

W (u�x)dx

(Hint: start by showing that the function t!R 10W (u�x + t(vx � u�x))dx

is convex).

Solution 5 (Tarek) Notice that u� solves the Euler Lagrange equation associ-ated with the functional in question:

(W 0(ux))x = 0:

6

For t 2 [0; 1] and v 2 C1[0; 1] de�ne

E(t) =

Z 1

0

W (u�x + t(vx � u�x)):

Claim: There exists � > 0 so that if 0 < jjv�u�jjC1 := jjv�u�jjL1 + jjvx�u�xjjL1 < �; and if v(0) = a; v(1) = b E is a strictly convex function of t:

Proof of Claim: Since W is continuous and W 00(b� a) = � > 0; there exists� > 0 so that if jx� (b� a)j < �; W 00(x) > �=2:

Thus since u�x = b � a and if jjv � ujjC1 < �; W 00�x + t(vx � u�x)) > �=2

uniformly in [0; 1]: Now compute E00(t) :

E0(t) =

Z 1

0

W 0�x + t(vx � u�x))(vx � u�x)

E00(t) =

Z 1

0

W 00�x + t(vx � u�x))(vx � u�x)2 � �=2

Z 1

0

(vx � u�x)2

Thus,E00(t) > 0

unless vx � u�x ( which implies v � u using the boundary conditions) whichwe assumed not to be the case. Thus the claim is proven. Notice now thatE0(0) = 0 since u� satis�es the Euler Lagrange Equation (multiply both sidesof the equation by vx � u�x and integrate by parts). Thus E(0) < E(t) forall t 2 [0; 1] by the strict convexity of E: Taking t = 1 establishes that if0 < jjv � u�jjC1 < �; Z 1

0

W (u�x) <

Z 1

0

W (vx)

2 Lecture 2: Convex Duality

1. Show that the convex dual of

minuj@=u0

�Z1

2kruk2

�is

maxdiv(�)=0

�Z@

(� � n)u0 �1

2

Z

k�k2�

Solution 6 (Jim) We will use that the convex conjugate of the function fde�ned by

f(x) = 12 jxj

2;

isf�(y) = sup

x

�hx; yi � 1

2 jxj2�= 1

2 jyj2:

7

We follow the procedure of writing down the minimization problem, substitut-ing the integrand by the maximization problem involving its complex conjugate,interchanging the min and the max, applying partial integration and inspectingthe max-min problem to �nd the constraints involved in the dual problem:

minu=u0@@

Z12 jruj

2 = minu=u0@@

max�

Zh�;rui � 1

2 j�j2

(1)

� max�

minu=u0@@

Z

h�;rui � 12 j�j

2

= max�

minu=u0@@

Z

(�div �)u+Z@

(� � n)u0 �Z

12 j�j

2

= maxdiv �=0

Z@

(� � n)u0 �Z

12 j�j

2:

In case u� is the minimizer of

minu=u0@@

Z

jruj2;

we may de�ne �� = ru� and see

max�h�;ru�i � 1

2 j�j2 = h��;ru�i � 1

2 j��j2 = 1

2 j��j2:

It follows that whenever div � = 0,

Z@

(� � n)u0 �1

2

Z

j�j2

=

Z

h�;ru�i �1

2j�j2

�Z

h��;ru�i �1

2j��j2

=1

2

Zjru�j2:

Hence, inequality (1) was actually an equality. Summarizing,

minu=u0@@

Z12 jruj

2 = maxdiv �=0

Z@

(� � n)u0 �Z

12 j�j

2:

2 Show that the variational problems

(P ) min

�Z

k�k : div(�) = F in , � � nj@ = f

�and

(D) max

�Z@

ufdS �Z

uFdx : kruk1 � 1�

8

are a dual pair, if Z

Fdx =

Z@

fdS

How should � and ru be related if equality is to hold? Hint:

max�2Rn

fh�; �i � k�kg =�0 ; k�k � 11 ; k�k > 1

Remark: if � R2 and f = 0 then (P ) can be solved explicitly in simplecases using the coarea formula. Why?

Solution 7 We �rst note that for any � 2 Rn that satis�es j�j < 1,

h�; �i � j�j:

This accounts for the �rst of the following inequalities. The second is the usualinterchange of the max and the min. Then we apply partial integration andobserve that the resulting expression does not depend on � anymore.

mindiv(�)=F in ��n=f@@

Z

j�j � mindiv(�)=F in ��n=f@@

maxjruj1�1

Z

h�;rui

� maxjruj1�1

mindiv(�)=F in ��n=f@@

Z

h�;rui

= maxjruj1�1

mindiv(�)=F in ��n=f@@

Z@

ufds�Z

uFdx

= maxjruj1�1

Z@

ufds�Z

uFdx:

Would the condition (??) not hold, the maximum in the last line would notexist, as we could always add a constant to u and make the result arbitrarylarge.The condition for equality is

h�;rui = j�j:

At any point where � does not vanish, this means,

ru = �

j�j :

Indeed, let �� be the optimal � and let u� be related via

h��;ru�i = j��j:

9

Then,

maxjruj1�1

mindiv(�)=F in ��n=f@@

Z

h�;rui

= maxjruj1�1

mindiv(�)=F in ��n=f@@

Z

h��;rui

= maxjruj1�1

mindiv(�)=F in ��n=f@@

Z

h��;ru�i

= mindiv(�)=F in ��n=f@@

maxjruj1�1

Z

h��;ru�i

= mindiv(�)=F in ��n=f@@

Z

j��j

= mindiv(�)=F in ��n=f@@

Z

j�j;

and thus equality holds.

Remark 8 We will now turn to the remark that was made after the exercise,that the minimization problem (P ) could actually be solved explicitly with � R2and F = 0 in some simple cases, using the coarea formula. We will assume that is connected and convex. We note that div(�) = F = 0 implies that thereexists a function : R2 ! R such that

�1 = @2

�2 = �@1 :

Consequently,j�j = jr j;

and the minimization problem can be written as

minr �t=f@@

Z

jr j

The value of the integral is invariant under addition of a constant to . More-over, if is admissible, plus an arbitrary constant is, as well. Therefore,we might just restrict ourselves to those that have a particular value for someparticular point on the boundary. The condition r �t = f can then be integratedacross the boundary, and can be replaced by

min =g

Z

jr j (1)

for some function g that is uniquely determined by f .

10

In case is the indicator function of some area, we can interpret the latterintegral as the perimeter of its boundary. In general, it is advantageous tominimize the length of the contour lines of the function , as the coarea formulaapplied to the case at hand reads

Z

jr j =Z 1

�1

Z �1(t)

dH1dt

=

Z 1

�1flength of curve (x) = tgdt:

As a simple example, we might consider the case in which is the openunit disk in R2. We assume that g is smooth, that it attains its maximum andminimum once and the values in between only twice. Then, the contour lines ofthe that minimizes (1) are straight lines connecting the corresponding valueson the boundary.

3 When we study homogenization we will consider a periodic "conductivity"A(x), and we will learn that the associated e¤ective conductivity Aeffsatis�es:

hAeff�; �i = min' periodic

�1

�()

Z

hA(x)(� +r'); � +r'i�

(Aeff is in general a matrix). Show using duality that:

hAeff�; �i = max�

1

�()

Z

2 h�; �i �A�1(x)�; �

�: div(�) = 0, � periodic

�This can alternatively be written as:

2 he�; �i�hAeff�; �i = min� 1

�()

Z

A�1(x)�; �

�: div(�) = 0, � periodic,

1

�()

Z� = e��

for any �; e� 2 Rn. Optimizing over �; we get:A�1(x)�; �

�= min

�1

�()

Z

A�1(x)�; �

�: div(�) = 0, � periodic,

1

�()

Z� = �

�

3 Lecture 3: Geodesics

1. Show directly (using the Euler-Lagrange equations) that any extremal forR baj _xj2 dt has constant speed. Here j _xj2 = (

Pgij(x(t)) _xi _xj)

1=2.

Solution 9 (Logan) We are dealing with the variational problem

minu(a)=�u(b)=�

Z b

a

F (t; u(t); _u(t))dt;

11

where F : R � Rd � Rd ! R is de�ned F (t; u; p) =Pdi;j=1 gi;j(u)pipj, and g

denotes a Riemannian metric, i.e., a symmetric, positive de�nite bilinear formon Rd, depending smoothly on u. If u(t) is a minimizer for this variationalproblem, then it must satisfy the Euler-Lagrange equations:

Fu(t; u(t); _u(t)) =d

dtFp(t; u(t); _u(t)):

Now in order to show that the speed of u is constant, we examine the derivativeof j _u(t)j2:

d

dtj _u(t)j2 = d

dtF (t; u(t); _u(t))

= Fu _u+ Fp�u

= (d

dtFp) _u+ Fp�u (due to the E-L equations)

=d

dt(Fp _u):

Here we observe that

Fp =

�d

dp1F; � � � ; d

dpdF

�=

"dXi=1

2g1;i(u)pi; � � � ;dXi=1

2gd;i(u)pi

#;

so that

Fp(t; u(t); _u(t)) _u =dX

i;j=1

2gj;i(u(t)) _ui(t) _uj(t)

= 2j _u(t)j2:

Substituting this into our chain of equalities tells us that

d

dtj _u(t)j2 = 2 d

dtj _u(t)j2;

and hence ddt j _u(t)j

2 = 0, meaning the speed of u (with respect to g) is constant.

2 Show that if b is a conjugate to a; then:

min

8<:bZa

�' _'

��Fu;u Fu; _uFu; _u F _u; _u

��'_'

�: '(a) = '(b) = 0

9=; = 0

Solution 10 (Dorian)

12

Theorem 11 Let F : R �Rn �Rn be smooth, and assume further that p 7!F (t; z; p) strictly convex and that F has uniformly bounded second derivativesin p and z. Let u(t) be a smooth solution to the minimization problem,

u 7! I[u] :=

Z b

a

F (t; u(t); _u(t))dt;

with �xed boundary data u(a) and u(b). Then the minimum of the second vari-ation of I[u] over all ' with supp ' � (a; b) vanishes:

min'(a)='(b)=0

Z b

a

Fuu' '+ Fu _u _' '+ F _u _u _' _'dt = 0; (2)

whenever a and b are conjugate points.

Proof: Since a and b are conjugate points, by de�nition there exists a curve' : [a; b] ! Rn which achieves the minimum on the left side of (2). We de�nethe integrand to be the function G(t; '; _') as in class:

G(t; '; _') := Fuu' '+ Fu _u _' '+ F _u _u _' _':

We note that G(t; �'; � _') = �2G(t; '; _') so that

2

Z b

a

Gdt =

Z b

a

'jG'j + _'jG _'jdt

The right side of the above is just the weak form of the Euler Lagrange equationfor G in the direction '. Since G is quadratic and smooth in ', _', it satis�esthe growth conditions:

jG(p; z; x)j � C(jpj2 + jqj2 + 1);

jDpG(p; z; x)j � C(jpj+ jzj+ 1);jDzG(p; z; x)j � C(jpj+ jzj+ 1);

the constant C depending on the second derivatives of F which are boundedby assumption. Hence G indeed weakly solves it�s Euler Lagrange equation,and moreover ' is an admissible test function, belonging to W 1;2

0 ((a; b)). Thisfollows from strict convexity of p 7! F (t; z; p) and the fact that ' is a minimizer.Indeed,

1 >

Z b

a

F _u _u _� _� � c

Z b

a

j _'j2dt;

the �rst inequality following from the fact that ' is a minimizer of (1), andthe constant c above can be taken to be the smallest eigenvalue of F _u _u which ispositive by strict convexity. Since �(a) = �(b) = 0, a Poincare inequality insuresthat indeed ' 2 W 1;2

0 ((a; b)). Consequently the right side of (??) vanishes,establishing (1).

13

3 When studying waves it is useful to consider paths that minimize traveltime, where the wave speed (x) is a speci�ed function of the location x.Show that this amounts to considering geodesics in the metric gij(x) :=1

(x)2 �ij .

Solution 12 (Danny) This problem is related to Fermat�s principle of leasttime, which postulates that rays of light follow the path which takes the leasttraveling time. The aim of this exercise is to prove that the wave path, which isgoverned by Fermat�s principle, in an isotropic medium (i.e., a medium in whichthe wave speed (x) at each point x is direction-independent) are geodesics withrespect to the metric gij(x) := 1

(x)2 �ij. In other words, we will show

Theorem 13 In an isotropic medium the paths that minimize the travelingtime, and the geodesics in the metric

gij(x) :=1

(x)2�ij (3)

are exactly the same locally, provided that (x) � 0 > 0 for some constant 0,where (x) is the wave speed at location x given by the medium.

Remark 14 (Generalization of Theorem 1)(i) In order to simplify our argument, we assume

(x) � 0 > 0;

which is natural in the physical world, but not necessary to ensure Theorem 13.(ii) The full version of Fermat�s principle states that the optical paths are ex-tremals corresponding to the traversal-time (as action). One can prove that suchextremals and the geodesics described in Theorem 13 are exactly the same; seesection 33.3 in [1] for instance.

Proof of Theorem 13.

Element of A.

14

For any two given �xed points P and Q 2 Rn, we denote the space of curvesconnecting P and Q as

A :=f� � Rn; � = Image (X); for some continuous functionX : [a; b]! Rn such that X(a) = P and X(b) = Qg:

See �gure 3 for an example of an element of A.On this space A, we can de�ne an action

I[�] :=

Z�

jdxjg =Z�

1

(x)

qdx21 + � � �+ dx2n:

Indeed, I[�] is the arc-length of � under the metric gij(x) := 1 (x)2 �ij . In the

following, we will show that the traveling time of the wave from P to Q alongthe path � is I[�], and hence, Theorem 13 is proved.Let X : [0; T ]! Rn be a wave path (parameterized by the time t) traveling

from P to Q in the given isotropic medium, and so, X satis�es:8><>:�� ddtX(t)

�� = (X(t))

X(0) = P

X(T ) = Q:

(4)

Under these notations, T is the traveling time of the path X. Using theabove parametrization and (4), we compute

I[Image(X)] =ZImage(X)

1

(x)

qdx21 + � � �+ dx2n

=

Z T

0

1

(X(t))

s�dX1(t)

dt

�2+ � � �+

�dXn(t)

dt

�2dt

=

Z T

0

1

(X(t)) (X(t)) dt

= T:

On the other hand, for any � 2 A, since (x) � 0 > 0, by applying anappropriate reparametrization, there exist a constant T 2 R+ and a functionX : [0; T ] ! Rn satisfying (4) such that � = Image(X). Therefore, by theprevious computations,

I[�] = T;

which is the traveling time of the wave from P to Q along �.

In conclusion,traveling time (�) = I[�];

15

and hence, when P and Q are closed enough,

the path that minimizes the traveling time from P to Q

= minimizer of fmin�2A

traveling time (�)g

= minimizer of fmin�2A

I[�]g

= minimizer of fmin�2A

arc-length (�) under the metric gij(x) =1

(x)2�ijg

= geodesic from P to Q under the metric gij(x) =1

(x)2�ij :

Remark 15 (Application of Theorem 1) In a uniform medium, in which (x) � c � constant, the metric in (3) becomes gij(x) = 1

c2 �ij, so the wavepaths are straight lines.

References

[1] Dubrovin, B. A.; Fomenko, A. T.; Novikov, S. P. Modern geometry �methods and applications. Part I. The geometry of surfaces, transforma-tion groups, and �elds. Translated from the Russian by Robert G. Burns.Graduate Texts in Mathematics, 93. Springer-Verlag, New York, 1984.

4 In this lecture we focused on Dirichlet boundary conditions. Supposeinstead we impose u(a) = �; and u(b) 2 M where M is a submanifold.What end condition does the Euler-Lagrange equation get at t = b? Whatis the proper notion of a Jacobi �eld in this case?

Solution 16 (Russell) The ordinary derivation of the E-L equation comesfrom the evaluation of functions of the type u(t)+ s�(t) at critical points for thevariable s, where �(a) = �(b) = 0. For simplicity, a = 0 and b = 1. In the caseof �(b) 2M , so that the �nal endpoint is nonconstant, we consider all solutionsof the ordinary Euler-Lagrange equation as the variation, and for a minimizerof the above problem, we look at variations of the form u(t; s) : [0; 1]�U ! Rn,where 0 2 U � Rd, d the dimension of the submanifold, and n the overall di-mension of the space and u0(t) = u(t; 0) minimizes I[u]. (We assume that thespace is geodesically complete, so that such a U exists.) Let �(t) = du

ds js=0: Then,since u0(t) is a minimizer:

16

0 =d

ds

����s=0

I[u(t; s)] =d

ds

����s=0

Z 1

0

F (t; u(t; s); _u(t; s))dt

=

Z 1

0

"Fu(t; u0(t); _u0(t))

d

ds

����s=0

u(s; t) + Fp(t; u0(t); _u0(t))_d

ds

����s=0

_u(t; s)

#dt

=

Z 1

0

[Fu(t; u0(t); _u0(t))�(t) + Fp(t; u0(t); _u0(t)) _�(t))] dt

=

Z 1

0

��Fu �

�d

dtFp

��� �(t) + d

dt(Fp(t; u0(t); _u0(t)) � �(t))

�dt (5)

= Fp(1; u0(1); _u0(1)) � �(1) (6)

The term Fu ��ddtFp

�� 0 in for all s; t in (1) since u(t; s) is assumed to

be a minimizer for all s and therefore satis�es the Euler-Lagrange equationsof the �xed-endpoint problem. As well, since the endpoint � is also �xed,ddsu(0; s) � 0 giving the result in (2). Now, since u(1; s) is a curve along M in aneighborhood of �, d

dsu(1; s) corresponds to an element of T�M . In particular,it can be easily shown that � : U !M , �(s) = u(1; s) is a di¤eomorphism, andhence d�( @

@si )(0) spans T�M . Thus, for the free boundary condition on u(1);we have the constraint that 0 = Fp(1; u0(1); _u0(1)) � �; 8� 2 TM . In a simpleexample, if F (t; u(t); _u(t)) is the energy of a curve on a Riemannian manifold,then Fp(t; u(t); _u(t)) = gij _u

i(t), and the minimizer is a geodesic parameterizedby arc length which satis�es the endpoint condition h _u(1); �i = 0 8� 2 T�M .,namely that the geodesics from a point to a submanifold are orthogonal to thesubmanifold.To understand the notion of a Jacobi Field, we must rede�ne the acces-

sory function �(t; �; ) corresponding to the second variation d2

ds2F (t; u(t); _u(t))

where �; correspond to duds js=0;

d2uds2 js=0, respectively. Following above, with

also d2uds2

����s=0

= (t):

d2

ds2

����s=0

u(t; s) =d2

ds2

����s=0

F (t; u(t; s); _u(t; s))

=d

ds

����s=0

�Fu(t; u(t; s); _u(t; s))

d

dsu(t; s) + Fp(t; u(t; s); _u(t; s))

d

ds_u(t; s)

�= Fuu� � + Fu + 2Fup� _� + Fpp _� _� + Fp _

If u0does in fact minimize the original variational problem, then the secondvariation as well as its integral along u0will be non-negative. Since �(t) =0��u(t)integrates to zero, the lower bound on the integral of the second variation

17

along u0must be identically zero:

0 =

Z 1

0

�Fuu� � + 2Fup� _� + Fpp _� _� + Fu + Fp _

�dt

=

Z 1

0

d

dt

�(Fpipj _�

i + Fuipj�i) � �j

�� d

dt

�Fpipj _�

i + Fuipj�i�� �j + (Fuiuj�i + Fpiuj _�i) � �j

+d

dt(Fp ) +

�Fu �

d

dtFp

� (7)

= Fpp _� � + Fup� � + Fp ����10

(8)

= Fpp _� � + Fup� �����10

(9)

The, the second, third, and �fth terms we be removed in (3) because �satis�ed the E-L equations for the second variation and u0 satis�ed the originalE-L equations. Finally, the third term on the (4) could be removed as well froman interpretation of the �rst result. Given the coordinates sion U , we have thefollowing �trivial�Jacobi �elds �k(t) = d

dsku(t; s)

��s=0

:

0 =d

dsk

�Fu(t; u(t; s); _u(t; s))�

d

dtFp(t; u(t; s); _u(t; s))

�=

d

dskFu(t; u(t; s); _u(t; s))�

d

dt

d

dskFp(t; u(t; s); _u(t; s))

����s=0

= Fuu(t; u0(t); _u0(t)) � �k + Fup(t; u0(t); _u0(t)) � _�k � d

dt

hFpu(t; u0(t); _u0(t)) � _�

k+ Fpp(t; u0(t); _u0(t)) � �k

iHence, the endpoint conclusion from the �rst result implies that for a Jacobi

�eld � on u0, the variation � + t�k is also a Jacobi �eld, and hence the �k areexactly the set of candidates for above. But the �kspan T�M and so the�rst endpoint condition eliminates the term in (4). The expression (5) above,

Fpp _� � + Fup� �

����10

, can still be expressed in more reasonable terms If � is

expressed as a function of both s and t as in the examples of the functions �k;then

d

ds

����s=0

(Fp(t; u(t; s); _u(t; s)) � �(t; s))����1t=0

= Fpp(t; u0(t); _u0(t)) _�(t)�+Fpu��+Fp� (t)����10

. As before, the Fp � term is zero, so we get the expression above. Thisversion has a better physical interpretation if one thinks of the �interesting�(i.e. vanishing endpoint-type) Jacobi �elds as in�nitessimal �pushes� of thecurve u0(t) in directions that don�t change I[u(t; s)] (though this is not alwaysthe case). Then just as solutions of the E-L equation depend only on theirvalues at the endpoints and Jacobi �elds which vanish at the end change thecurve without changing the endpoints, in the free boundary case the Jacobi

18

�elds are somewhat relaxed by the fact that given a minimizing curve (whichmust have its �nal endpoint orthogonal to the submanifold), they only need�push�the minimizing curve onto another solution of the E-L equation whichstill satis�es the orthogonality at the endpoint condition.

4 Lecture 4: Geodesics

1. Let u be a critical point ofRW (ru)dx: Assume u is scalar valued and

C2; and suppose W is nonconvex at ru(x) for some x 2 . Show thatthere exists ' with compact support on such that:

d2

d"2j"=0

�Z

W (r(u+ "'))dx�< 0

Solution 17 (Dorian)

Theorem 18 Let u be a critical point ofRW (Du)dx. Assume u is scalar

valued, C2 and that W is non-convex at some point ru(x0) for x0 2 . Thenthere exists a compactly supported � 2W 1;1() such that

d2

ds2 js=0

Z

W (r(u+ s�))dx < 0:

Proof: Since W is not convex at the point ru(x0) there necessarily exists a� 2 Rn with �T � D2W (ru(x0)) � � < ��2 < 0 for some � 2 R. Moreover,since x 7! ru(x) is continuous and W is smooth, there exists a � > 0 such that�T � D2W (ru(x)) � � < ��2 for all x 2 R�(x0) (rectangle centered at x0 withsides � > 0).By possibly translating and changing coordinates, we may assume w.l.o.g

that � == e1 and x0 = 0. De�ne the function ~� : ! R such that ~� iscontinuous and

r~�(x1; � � � ; xn) =(+� on [��; 0]� Rn�1 \ �� on [0; �]� Rn�1 \

(10)

Now let � 2 C1(;R) be a smooth cuto¤ function such that � � 1 on R� and� � 0 on R�+� for � > 0. We �nally de�ne

� := �~�;

and note that r� = r~� = �� Ld a.e on R�. Now ru is a critical point and wecompute the second variation:

d2

ds2 js=0

Z

W (r(u+ s�))dx =Z

r�T �D2W (ru(x0)) � r�:

19

Then we have thatZ

r�T �D2W (ru(x0)) � r� =ZR�

�T �D2W (ru(x0)) � � +ZRc�

r�T �D2W (ru(x0)) � r�

� ��n�2 + C(�; n)�n: (11)

Our choice of � > 0 however was arbitrary and consequently we may choose itsu¢ ciently small so that (11) is indeed negative. �

5 Lecture 5: Optimal Control

1. Consider the �nite-horizon utility maximization problem with discountrate �. The dynamical law is thus:

_y = f(y(s); �(s)) y(t) = x

and the optimal utility discounted to time 0 is

u(x; t) = max�2A

(Z T

t

e��sh(y(s); �(s))ds+ e��T g(y(T ))

)It is often more convenient to consider, instead of u, the optimal utilitydiscounted to time t; this is,

v(x; t) = e�tu(x; t) = max�2A

(Z T

t

e��(s�t)h(y(s); �(s))ds+ e��(T�t)g(y(T ))

)

Solution 19 (Sean) (a) Due to the discount rate, we no longer have that util-ity is additive. Instead, we have to discount the contribution from t+�t onwardby e���t when calculating u(x; t). This give us

v(x; t) � h(x; a)�t+ e���tv(x+ f(x; a)�t; t+�t) + error

Taylor expanding e���t and v(x+ f(x; a)�t; t+�t), we get

v(x; t) � h(x; a)�t+ (1� ��t)[v(x; t) + (rv � f(x; a) + vt)�t] + error� h(x; a)�t+ v(x; t)� �v(x; t)�t+ (rv � f(x; a) + vt)�t+ error;

where we have put the superlinear �t terms in the error. This becomes anequality with a optimal, and in the limit as �t! 0 we get

0 = vt � �v(x; t) + maxa2A

frv � f(x; a) + h(x; a)g = vt � �v +H(x;rv)

whereH(x; p) = max

a2Aff(x; a) � p+ h(x; a)g

In the case when t = T , the dynamics are irrelevant and the discounting cancelsout the discount rate making start time irrelevant as well. Thus, we have thatv(x; T ) = g(x).

20

(b)Suppose v(x; t0) < v(x; t) where t0 < t. Then there exists some strategy �

de�ned on [t;1) and an � > 0 such that the in�nite-horizon problem, startedat t, using the strategy �, gives a payout that is at least v(x; t0) + �, that isZ 1

t

e��(s�t)h(y(s); �(s)) ds � max�2A

Z 1

t0e��(s�t

0)h(y(s); �(s)) ds+ �:

De�ne the strategy �0(s) = �(s+ t� t0) on [t0;1). Let y denote the trajectorywith strategy � and initial condition y(t) = x and y0 denote the trajectorywith strategy �0 and initial condition y0(t0) = x. As the state equation _y =f(y(t); �(t)) is autonomous, we have that y0(s) = y(s+ t� t0). Upon a changeof variables, we get

v(x; t0) �Z 1

t0e��(s�t

0)h(y0(s); �0(s)) ds =

Z 1

t

e��(s�t)h(y(s); �(s)) ds � v(x; t0) + �;

a contradiction. Thus, v(x; t0) � v(x; t) for t0 < t. Similarly, one can provev(x; t0) � v(x; t) and so v(x; �) is constant for each x.

We can run the same analysis of part a to show that v satis�es the equation

vt � �v +H(x;rv) = 0

with no �nal-time data. However, as vt � 0, we have that v satis�es the PDE

��v +H(x;rv) = 0:

2 Recall example 1 of the notes: the state equation is _y = ry � � withy(t) = x, and the value function is:

u(x; t) = max��0

�Z �

t

e��sh(�(s))ds

�with h(a) = a for some 2 (0; 1), and � the �rst time when y = 0 if thisoccurs before time T; or T otherwise.

Solution 20 (Logan) (a) In the notes, we found u(x; t) to be of the formu(x; t) = g(t)x for some g. Then we derived the following ODE for G(t) =e�tg(t):

G0 =( �1) = 0;

with end condition G(T ) = 0. Assuming that ��r = 0, then the above equationsimpli�es to

G0 =( �1) = 0:

As in the notes, multiply by (1 � )�1G =(1� ), and let H(t) = G(t)1=(1� ).Then we �nd

H 0 + 1 = 0;

21

with end condition H(T ) = 0. This is a very easy ODE to solve: the solution isH(t) = T � t. Unraveling the changes of variable yields

g(t) = e��t [T � t]1� :

Thus we�ve solved the HJB equation for �� r = 0.

(b) We see that (for �� r 6= 0)

limT!1

v(x; t) = limT!1

e�tg(t)x

= limT!1

�1� �� r

�1� e�

(��r )(T�t)1�

��1� x :

When ��r > 0, we see that the above limit goes to G1x , and when ��r < 0,we see that it goes to in�nity. For the case that �� r = 0, we use part (a) andsee that

limT!1

v(x; t) = limT!1

e�tg(t)x = limT!1

[T � t]1� x =1:

Thus we�ve shown that v(x; t) converges to the desired function as T goes toin�nity.

(c) The Hamiltonian given in problem 1(a), speci�ed to this example (and usingv in place of �v, because it�s easier to TeX), is

H(x; v0) = maxa�0

ff(x; a)v0 + h(x; a)g = rxv0 +maxa�0

f�av0 g:

Now as in the �nite-horizon problem, v0 must be positive if v is �nite (intuitively,more money means more fun), and so the above expression can be calculated(using basic calculus) to be

H(x; v0) = rxv0 ��v0

� 1 �1

v0 +

�v0

� �1

= rxv0 � (v0 �1

1

�1+

�v0

� �1

= rxv0 +h(1= )

�1 � (1= )

1 �1

i(v0

�1 :

Now the full HJB equation comes out to be

0 = ��v0 + rxv0 +h(1= )

�1 � (1= )

1 �1

i(v0

�1 :

Now if � � r > 0, then we see that v(x) = G1x satis�es the above HJB

equation (by plugging it in to the above expression and simplifying). If � � r were zero, then G1 wouldn�t make sense, and if �� r were negative, then G1would not be real, so G1x is not a possible solution for such values of �� r .In the case of �� r � 0, we can, however, �nd controls for which the utility isarbitrarily large. Take, for instance, c 2 (0; 1), and let

�(s) = rcxer(1�c)s:

22

For this control, we �nd that

y(s) = xer(1�c)s;

and so we �nd that the utility for this control isZ 1

0

e��s�(s) ds =

Z 1

0

e��s+ log(rcxer(1�c)s)ds

= (rcx) Z 1

0

e(��+ r(1�c))sds

= (rcx) �

1

��+ r(1� c)e(��+ r(1�c))s

�10

:

We see that we can choose c so that the above utility is as large as we wish.Thus for �� r � 0, we see that v =1, and we already showed v = G1x

for�� r > 0.

(d) In calculating the Hamiltonian speci�ed to this problem (which we did in part(c)), we �nd that the value for the Hamiltonian is obtained for a = (v01=( �1)

(by simply taking the derivative and setting it equal to zero, i.e., by �nding thecritical points). This is a feedback law for a, and so now that we know v0, wesimply plug it in to �nd a:

a(s) = (G1 s

�1

)1=( �1) =

�� r 1� s:

3 Consider the analogue of Example 1 with the power-law utility replacedby the logarithm: h(a) = ln a. To avoid confusion let us write u for thevalue function obtained in the notes using h(a) = a , and ulog for thevalue function obtained using h(a) = ln a. Recall that u (x; t) = g (t)x

with

g (t) = e��t�1� �� r

�1� e�

(��r )(T�t)1�

��1� :

(a) Show, by a direct comparison argument, that

ulog(�x; t) = ulog(x; t) +1

�e��t(1� e��(T�t)) ln�

for any � > 0. Use this to conclude that

ulog(x; t) = g0(t) lnx+ g1(t)

where g0(t) = 1�e��t(1�e��(T�t)) and g1 is an as-yet unspeci�ed function

of t alone.

(b) Pursue the following scheme for �nding g1: Consider the utility h = 1 (a

�1). Express its value function uh in terms of u . Now take the limit ! 0.Show this gives a result of the expected form, with

g0(t) = g (t)j =0

23

and

g1(t) =dg d (t)

���� =0

:

(This leads to an explicit formula for g1 but it�s messy; I�m not asking youto write it down.)

(c) Indicate how g0 and g1 could alternatively have been found by solvingappropriate PDE�s. (Hint: �nd the HJB equation associated with h(a) =ln a, and show that the ansatz ulog = g0(t) lnx+ g1(t) leads to di¤erentialequations that determine g0 and g1.)

Solution 21 (Jim)

(a) Let � > 0. Let �x denote the optimal control associated with ulog(x; t),that is, the optimal control for starting point x. We will denote the as-sociated trajectory by yx;t;�x . For starting point �x, we can then trythe control ��, which is also admissible, but maybe not optimal. Theassociated trajectory is

y�x;t;��x = �yx;t;�x ;

and is therefore positive. For ulog(�x; t) we �nd

ulog(�x; t) = max��0

Z T

t

e��sh(y�x;t;�(s); �(s))ds

�Z T

t

e��s ln(��x(s))ds

=

Z T

t

e��s ln�+

Z T

t

e��s ln�x(s)

=1

�e��t[1� e��(T�t)] ln�+ ulog(x; t):

On the other hand, replacing � by 1=� and �x by z, we observe that also

ulog(z; t) � ulog(�z; t)�1

�e��t[1� e��(T�t)] ln�

for all � > 0. Hence, we �nd

ulog(�x; t) = ulog(x; t) +1

�e��t(1� e��(T�t)) ln�

for any � > 0. By choosing � = 1=x,

ulog(x; t) = ulog(1; t) +1

�e��t(1� e��(T�t)) lnx

= g0(t) lnx+ g1(t);

24

where g0(t) = 1�e��t(1 � e��(T�t)) and g1(t) = ulog(1;t) is an as-yet un-

speci�ed function of t alone.

(b) This part of the exercise is based on the observation that for any a > 0,

lim !0

1 (a

� 1) = ln a:

Let us calculate the value function uh associated to the cost functionh = 1

(a � 1). We make use of the notes, in which it was calculated that

u (x; t) = g (t)x , with

g (t) = e��t�1� �� r

�1� e�

(��r )(T�t)1�

��1� :

We �nd

uh(x; t) = max��0

Z T

t

e��s1

(a � 1)ds

=1

u (x; t)�

1

� e��t

h1� e��(T�t)

i=1

(x � 1)g (t) +

1

g (t)�

1

� e��t

h1� e��(T�t)

i;

We note thatg0(t) =

1

�e��t

h1� e��(T�t)

i:

Therefore, when we take the limit ! 0, we �nd

lim !0

uh(x; t) = g0(t) ln(x) + g1(t);

where

g1(t) =dg dt

����t=0

:

(c) We can �nd ordinary di¤erential equations for g0 and g1 by writing downthe Hamilton-Jacobi-Bellman equation for ulog:

ut +maxa�0

�ux(rx� a) + e��t ln a

�= 0:

The maximizing value of a is found by di¤erentiation and equals

amax =e��t

ux:

Hence the HJB equation becomes

ut + ux

�rx� e��t

ux

�� �te��t � e��t lnux = 0

ut + rxux � e��t (1 + �t+ lnux) = 0:

25

We now substitute the Ansatz

u(x; t) = g0(t) lnx+ g1(t)

in the equation to �nd

g01(t) lnx+ g01(t) + rg0(t)� e��t

�1 + �t+ ln

�g0(t)

x

��= 0�

g0��t0

�lnx+ g01(t) + rg0(t)� e��t [1 + �t+ ln g0(t)] = 0:

So, �rst we may solveg0��t0 = 0;

which yields with the correct boundary condition u(x; T ) = g0(T ) lnx +g1(T ) = 0 the usual form of g0. Given g0, we can solve for g1 via

g01(t) + rg0(t)� e��t [1 + �t+ ln g0(t)] = 0:

4 Our example 1 considers an investor who receives interest (at constantrate r) but no wages. Lets consider what happens if the invesotr alsoreceives wages at constant rate w. The equation of state becomes:

_y = ry + w � � with y(t) = x

And the value function is

u(x; t) = max��0

(Z T

t

e��sh(�(s))ds

)

With h(a) = a for some 2 (0; 1) (power law utility). Since the investorearns wages, we now permit y(s) < 0; however we insist that the �nal-timewealth be nonnegative (y(T ) � 0).

(a) Which pairs (x; t) are acceptable? The strategy that maximizes y(T )is clearly to consume nothing (�(s) = 0 for all t < s < T ). Show thisresults in y(T ) � 0 exactly if:

x+ �(t)w � 0

where�(t) =

1

r(1� e�r(T�t))

Notice for future reference that � solves �0�r�+1 = 0 with �(T ) = 0.(b) Find the HJB equation that u(x; t) should satisfy in its natural do-

main f(x; t) : x+ �(t)w � 0g: Specify the boundary conditions whent = T and where x+ �(t)w = 0.

26

(c) Substitute into this HJB equation the ansatz

v(x; t) = e��tG(t)(x+ �(t)w)

Show v is a slution when G solves the familiar equation

Gt + (r � �)G+ (1� )G =( �1) = 0

Deduce a formula for v.

(d) In view of (a), a more careful de�nition of the value function for thiscontro problem is:

u(x; t) = max��0

�Z �

t

e��sh(�(s))ds

�where � is the �rst time when y(s) + �(s)w = 0 if this occurs beforeT; and is T otherwise. Use a veri�cation argument to prove that thefunction v obtained in (c) is indeed the value function u de�ned thisway.

Solution 22 (Eduardo) (a) Solving for the equation of state, we obtain thefollowing:

_y � ry = w � �(s)d

ds

�e�rsy

�= e�rs(w � �(s))

Integrating from t to T yields:

e�rT y(T )� e�rtx = w

r[e�rt � e�rT ]�

Z T

t

e�rs�(s)ds

Multiplying by ert on both sides:

e�r(T�t)y(T ) = x+w

r[1� e�r(T�t)]�

Z T

t

e�r(s�t)�(s)ds

Since �(s) � 0, the integral in the right hand side is always positive. Hence,

e�r(T�t)y(T ) � x+w

r[1� e�r(T�t)] = x+ �(t)w

And equality is achieved if and only if �(s) = 0 8s 2 (t; T ). Hence, y(T ) � 0 ifand only if x+�(t)w � 0.(b) The HJB equation takes the form:

ut +H(ux; x) = 0

WhereH(p; x) = max

a2Af[rx+ w � a]p+ e��sa g

27

We compute this maximum directly by computing the critical points of this ex-pression with respect to a. If �(a) = [rx+ w � a]p+ e��sa ,

d�

da= �p+ e��sa �1 = 0

a� = [ �1e�sp]1=( �1)

We have that �00(a�) = ( � 1)e��s[ �1e�sp]( �2)=( �1) < 0. Hence,

H(p; x) = (rx+ w)p+ e�s=( �1)p =( �1)h � =( �1) � �1=( �1)

iNow, we already know that u(x; T ) should be the �nal time utility g(y(T )) (whichin this case is 0). Also, if x + �(t)w = 0; we know that using our optimalstrategy y(T ) = 0. Hence, for such pairs we must also impose u(x; t) = 0.(c) Now, we substitute the ansatz in our HJB equation. We have that:

vx(x; t) = e��tG(t) (x+ �(t)w) �1

vt(x; t) = e��t(x+ �(t)w) [��G(t) +Gt(t)� G(t)(x+ �(t)w)�1e�r(T�t)w]

Then, since =( �1)� � =( �1) � �1=( �1)

�= (1� ),

H(vx; x) = e��t(x+ �(t)w) [(rx+ w) G(t)(x+ �(t)w)�1 + (1� )G =( �1)]

And so,

vt+H(vt; x) = e��t(x+�(t)w) [Gt(t)+(1� )G =( �1)+G(t)f��+( rx+ w� we�r(T�t))(x+�(t)w)�1g

Since � we�r(T�t) = � wr�(t)� w,

rx+ w � we�r(T�t) = r (x+ �(t)w)

And so, we have:

vt +H(vt; x) = e��t(x+ �(t)w) [Gt(t) + (1� )G =( �1) +G(t)f��+ r g] = 0

We already have that v is 0 whenever x + �(t)w = 0. Hence, this implies thatG must satisfy:

Gt(t) + (1� )G =( �1) +G(t)f��+ r g = 0 ; G(T ) = 0

The solution to this equation may be found in the notes, as part of the analysisof example 1. The solution for G is then:

G(t) = [��1(1� e��(T�t))]1� where � = (r � �)=(1� )

Hence, a formula for v is:

v(x; t) = e��t[��1(1� e��(T�t))]1� (x+ �(t)w)

28

(d) Veri�cation argument: The proof of the veri�cation theorem, as provided inthe notes for the case without wages and without discounting can be essentiallycopied for this case. When � = T; both variational problems are the same (havethe same value function), and hence u(x; t) = v(x; t) is trivial. The case inwhich � < T (such that x+�(�)w = 0) is also straightforward, since u(x; �) = 0and v(x; �) = 0: (?)

5 (An example of nonexistence of an optimal control) Consider the followingcontrol problem: the state is y(s) 2 R with y(t) = x; the control is�(s) 2 R; the dynamics is _y = �; and the goal is:

min

(Z T

t

y2(s) + (�2(s)� 1)2)

The value function u(x; t) is the value of this minimum.

Solution 23 (Katarina)

u(x; t) = inf�

Z T

t

y2(s) + (�2(s)� 1)2 dsdynamics: dy=dt = �; y(t) = x

(a) Show that when x = 0 and t < T , the value is u(0; t) = 0. For positiveinteger n, consider the sequence of controls

�n = (�1)j on�t+ j

T � tn

; t+ (j + 1)T � tn

�; j = 0; 1; 2; : : : ; n� 1:

Then the solution of dyn=dt = �n is piecewise linear with alternating slopes of�1, each for the interval of length 1=n. Hence, starting at y(t) = x = 0, thesolution stays within 0 � y(s) � 1=n, and

u(0; t) �Z T

t

y2n(s) + (�2n(s)� 1)2 �

Z T

t

1

nds =

T � tn

:

Letting n ! 1, we get u(0; t) � 0. On the other hand, clearly u(0; t) � 0(integrand is always nonnegative), so u(0; t) = 0. If we want to consider onlysmooth � and y, we can easily adjust the solution above and still get 0 � y � 1,R(�2n(s)� 1)2 ! 0.

(b) Show that when x = 0 and t < T there is no optimal control �(s). Assumethe contrary, i. e. there is an optimal control � so thatZ T

t

y2(s) + (�2(s)� 1)2 ds = u(0; t) = 0:

Since the integrand is nonnegative, we must have y = 0 a. e. and dy=ds = � =�1 a. e. That is clearly impossible.

29

Food for thought: What is the HJB equation? Is there a modi�ed goal leading tothe same Hamiltonian and value funcion, but for which optimal controls exist?Our minimization problem is equivalent to maximizing

u(x; t) = max�

Z T

t

�y2(s)� (�2(s)� 1)2 dsdy=dt = �; y(t) = x:

The HJB equation for this problem is

ut +H(ru; x) = 0 for t < T; u(x; T ) = 0 at t = T;

withH(p; x) = max

afap� x2 � (a2 � 1)2g:

Since H(p) is the Legendre transform of h(x; a) = (a2 � 1)2 + x2 (x is just aconstant here, the transform is with respect to a), it will stay the same if weconsider the convexi�cation hc of h instead of h:

hc(x; a) = h(x; a) for jaj � 1;= x2 for jaj � 1:

For this modi�ed goal and x = 0, the optimal control will be �(s) = 0: we willhave y(s) = 0 and

0 = u(0; t) =

Z T

t

�hc(y(s); �(s)) ds:

6 The Hopf-Lax solution formula solves the �nite-horizon problem with stateequation _y = � and value function:

u(x; t) = max�

(Z T

t

h(�(s))ds+ g(y(T ))

)

with h concave. The key step was to show that an optimal trajectoryhas constant velocity. Give an alternative justi�cation of this fact usingPontryagin�s maximum principle.

Solution 24 (Logan) In this problem, the Hamiltonian H(p; x) does not de-pend on x, and hence we know from the Pontryagin maximum principle that

d�

ds= �dH

dx(�(s); y(s)) = 0:

That is, �(s) is constant over all s (just call the constant � from here on). Wealso have from the Pontryagin maximum principle that

�(s) = maxa2A

�(s) � a+ h(a) = maxa2A

� � a+ h(a):

We now see that the above expression does not depend on s, and hence it isconstant.

30

8 This problem is a special case of the "linear-quadratic regulator" widelyused in engineering applications. The state is y(s) 2 Rn and the controlis �(s) 2 Rn. There is no pointwise restriction on the values of �(s). Theevolution law is

_y = Ay + �; y(t) = x

for some constant matrix A; and the goal is to minimizeZ T

t

jy(s)j2 + j�(s)j2 ds+ jy(T )j2

(In words: we prefer y = 0 along the trajectory and at the �nal time, butwe also prefer not to use too much control).

Solution 25 In order to have a better presentation, let us restate our linear-quadratic problem again with the same notations as in the notes:

We consider the control problem whose system dynamics are governed by aset of linear di¤erential equations8<:

dy

ds(s) = f(y(s); �(s)) := Ay(s) + �(s) in t < s � T

y(t) = x;(12)

for some constant matrix A 2Mn�n(R).Our goal is to minimize the quadratic cost

Cx;t[�] :=

Z T

t

h(y(s); �(s)) ds+ g(y(T )) :=

Z T

t

jy(s)j2 + j�(s)j2 ds+ jy(T )j2;

where y(�) depends on x; t and �(�), and solves (12), over the sets of admissiblecontrols

A := f� : [0; T ]! Rn; �(�) is measurableg:

Lastly, we de�ne the value function

v(x; t) := min�2A

Cx;t[�]

= min�2A

(Z T

t

jy(s)j2 + j�(s)j2 ds+ jy(T )j2):

8(a)Following the dynamic programming principle, one may �nd that the asso-

ciated Hamilton-Jacobi-Bellman equation is

vt +H(rv; x) = 0; (13)

31

where

H(p; x) := mina2Rn

ff(x; a) � p+ h(x; a)g

= mina2Rn

f(Ax+ a) � p+ jxj2 + jaj2g

= Ax � p+ jxj2 � 14jpj2 + min

a2Rnja+ 1

2pj2

= Ax � p+ jxj2 � 14jpj2:

Therefore, the associated Hamilton-Jacobi-Bellman equation is

vt +Ax � rv + jxj2 �1

4jrvj2 = 0: (14)

Next, we are going to study the feedback control by the principle of optimal-ity: for any �xed (x; t), we denote y� the optimal trajectory and �� the optimalcontrol corresponding to this (x; t), then

v(x; t) =

Z T

t

h(y�(s); ��(s)) ds+ g(y�(T )):

By the principle of optimality 1 ,

v(y�(s); s) =

Z T

s

h(y�(~s); ��(~s)) d~s+ g(y�(T )) (15)

for all s 2 (t; T ].Di¤erentiating (15) with respect to s, and using (12)1 and (13), we obtain

�H(rv(y�(s); s); y�(s)) +rv(y�(s); s) � f(y�(s); ��(s)) = vt(y�(s); s) +rv(y�(s); s) � dy

�

ds= �h(y�(s); ��(s))

and hence,

��Ay� � rv(y�; s) + jy�j2 � 1

4jrv(y�; s)j2

�+rv(y�; s) � fAy� + ��g = �

�jy�j2 + j��j2

j��(s) + 1

2rv�(y�(s); s)j2 = 0:

Therefore, we should expect

��(s) = �12rv�(y�(s); s)

1Bellman�s Principle of Optimality: An optimal policy has the property that, whateverthe initial state and initial decision are, the remaining decisions must constitute an optimalpolicy with regard to the state resulting from the �rst decision (see [1] for instance).

32

provided that those functions are smooth enough.More discussions on the feedback control �� can be found in section 10.3.3

of [2].8(b)De�ne the function u(x; t) :=< K(t)x; x > for some symmetric n�n matrix-

valued function K(t). Then

ut(x; t) =<dK

dtx; x > and ru(x; t) = 2Kx:

Substituting u into (14), we have

<dK

dtx; x > + < Ax; 2Kx > + < x; x > � < Kx;Kx > = 0

<

�dK

dt+�KTA+ATK

�+ I �K2

�x; x > = 0:

Comparing the coe¢ cients of the quadratic polynomials of x, we obtain

dK

dt= K2 � I � (KTA+ATK): (16)

Lastly, putting u into the boundary condition v(T ) = 0 implies

< K(T )x; x > = 0 8x 2 Rn;

which is equivalent toK(T ) = I: (17)

8(c)In this part, we assume that K is symmetric and satis�es the ODE (16) and

the boundary conditions (17). Our aim is to prove u(x; t) = v(x; t). For anyarbitrary � 2 A, we can obtain a unique y� by solving (12). Now that we cancompute, by using (12)1 and (16),

d

dsfu(y�(s); s)g

= < _Ky�; y� > + < K _y�; y� > + < Ky�; _y� >

= < fK2 � I � (KTA+ATK)gy�; y� > + < K(Ay� + �); y� > + < Ky�; Ay� + � >

= < Ky�;Ky� > � < y�; y� > +2 < Ky�; � >

=� jy�j2 � j�j2 + jKy� + �j2

�� jy�j2 � j�j2:

Integrating s from t to T , using (12)2 and (17), we have

jy�(T )j2 = u(y�(T ); T ) � u(x; t)�Z T

t

jy�(s)j2 + j�(s)j2 ds

33

so,

u(x; t) �Z T

t

jy�(s)j2 + j�(s)j2 ds+ jy�(T )j2 =: Cx;t[�]:

Taking minimum over � 2 A, we obtain

u(x; t) � min�2A

Cx;t[�] =: v(x; t):

On the other hand, by the ODE existence theorem, there exists a unique~y 2 C1[t; T ] which solves 8<:

d~y

ds= (A�K)~y

~y(t) = x:(18)

Setting ~� := �K~y 2 A, we know that ~y solves (12) with the control ~�, namely~y(�) = ~y~�(�). By the same direct computations,

d

dsfu(~y~�(s); s)g = �j~y~�(s)j2 � j~�(s)j2:

Therefore, integrating s from t to T , and using (17) and (18)2, we obtain

u(x; t) =

Z T

t

j~y~�(s)j2 + j~�(s)j2 ds+ j~y~�(T )j2

=Cx;t[~�]

�min�2A

Cx;t[�]

=v(x; t):

Q.E.D.

References

[1] Bellman, Richard, Dynamic programming. Princeton University Press,Princeton, N. J., 1957

[2] Evans, Lawrence C, Partial Di¤erential Equations, GSM 19, AmericanMathematical Society, Providence, RI, 1998

6 Lecture 7:

1. We showed that if xn ! x1 weakly in W 1;p, p > 2, (for maps R2 !R2), then detDxn ! detDx1 weakly in Lp=2. When the xn are moresingular strange things can happen, due to "cavitation". Explore this by

34

considering the map xn : B1 ! B1 (where B1 is the unit ball in R2), suchthat

xn(X) =

( 1��n�n

X for jXj � �n;�1�2�n1��n

+ �n1��n

jXj�

XjXj for �n � jXj < 1:

So xn "blows up" B�n to B1��n and "squashes" B1nB�n to B1nB1��n .(Here �n can be any sequence ! 0).

(a) Show that xn(X)! XjXj a.e.

(b) Show that detDxn converges as a distribution to a point measure locatedat 0.

(c) For which p doesRB1jDxnjpdX stay bounded as n!1?

Solution 26 (Jim)

(a) Because �n ! 0, for every X 2 B1nf0g, it holds that jXj > �n for largeenough n. Then,

xn(X) =

�1� 2�n1� �n

+�n

1� �njXj�X

jXj

! (1 + 0jXj) XjXj =X

jXj :

As the point-set f0g has Lebesgue measure 0, xn(X)! X=jXj a.e.

(b) Let us �rst calculate Dxn. In case jXj < �n,

Dxn(X) =

1��n�n

0

0 1��n�n

!:

For �n � jXj < 1

Dxn(X) =

�n1��n

+ 1�2�n1��n

X22

jXj3 � 1�2�n1��n

X1X2

jXj3

� 1�2�n1��n

X1X2

jXj3 : �n1��n

+ 1�2�n1��n

X21

jXj3

!:

Thus,

detDxn(X) =

8><>:�1��n�n

�2if jXj < �n;�

�n1��n

�2+ 2�n(1�2�n)(1��n)2

1jXj if �n � jXj < 1:

Now take � : B1 ! R continuous and bounded and considerZB1

�(X) detDxn(X)dX � 2��(0) =ZB�n

�(X)

�1� �n�n

�2dX

+

ZB1nB�n

"��n

1� �n

�2+2�n(1� 2�n)(1� �n)2

1

jXj

#�(X)dX:

35

Let us consider the latter term �rst. Since the integration is over B1nB�n ,for every X in the integration area it holds that �n=jXj < 1. Hence,the integrand is bounded and as it approaches zero pointwise, the wholeintegral approaches zero as n ! 1. Considering the �rst term, we canwrite

�����ZB�n

�(X)

�1� �n�n

�2dX � ��(0)

����� ==

�����ZB�n

�(0)

�1� �n�n

�2dX � ��(0) +

ZB�n

(�(X)� �(0))�1� �n�n

�2dX

������ ��(0)[(1� �n)2 � 1] + sup

X2B�n

j�(X)� �(0)j(1� �n)2

! 0; as n!1;

where we have used the continuity of � to take the limit in the last line.Summarizing, we �nd detDxn ! ��0 in distribution.

(c) We will interpret the norm on Dxn as the Frobenius norm, but due tothe equivalence of norms on �nite dimensional vectors spaces, this is nota restriction. We claim that

RB1jDxnjpdX remains bounded if and only

if p < 2. Consequently, we cannot use the convergence result for thedeterminant as stated in the formulation of the exercise. Parts (a) and(b) of the exercise already showed that the conclusion of the convergenceresult did not hold in the case at hand.

Let p < 2. ThenZB1

jDxnjp =ZB�n

jDxnjp +ZB1nB�n

jDxnjp

�ZB�n

2

�1� �n�n

�2!p=2+

ZB1

~C +

~D

jXj2

!p=2���2n��pn + C + C

ZB1

jXj�p <1:

For all n large enough, and ~C; ~D;C some positive constants. Now let

36

p � 2. It follows that for some c > 0,ZB1

jDxnjp � c

ZB1nB�n

�X41

jXj6

�p=2= c

ZB1nB�n

�X21

jXj3

�p= c

Z 1

�n

Z 2�

0

�r2 cos2 �

r3

�prd�dr

= c

Z 2�

0

(cos2p �)d�

Z 1

�n

r1�pdr

!1;

for n!1.

2 Show that ifRg(detF ) is lower semicontinuous then g must be convex.

Hint: from the discussion at the end of lecture 4, if W (F ) = g(detF )is lower semicontinuous, then it must be rank-one convex, i.e. thefunction h :: t 7! g(det(F + t�!a �!v )) is convex in t for every �!a ;�!v 2 Rn.Show that this forces g00 > 0.

Solution 27 (Russell) Following the hint, if F is of full-rank, then given any�!v 2 Rn, �!v 2 Im(F ), so without loss of generality, we may consider g(det(F +t�!a �!v )) = g(det((1 + at)F ))) = g((1 + at) detF );where a is an arbitraryconstant. Here, nonconvexity in t implies that there exists a t1; t2so that g((1+�at1+(1��)at2) detF ) > �g((1+at1) detF )+ (1��)g((1+at2) detF ):Undersuch an assumption, it is clear that g((1+t) detF ) is not lower semicontinuous.For example, on the interval [0;K), let us consider a sequence of functions �N (t)such that

D�N (t) =

(t1

n�1NK � t < n�1+�

NK

t2n�1+�NK � t < n

NK

;

for n 2 Z+; 0 < n � NK. ThenZ K

0

h(D�N (t)) = K � (�g((1 + at1) detF ) + (1� �)g((1 + at2) detF ))

<

Z K

0

g((1 + �at1 + (1� �)at2) detF )

=

Z K

0

h(D�(t))

where �(t) = limN!1 �N (t) (in the supremum norm), violating the lower semi-continuity condition. As well, if F is not of full rank, either g(det(F+t�!a �!v ) =g(tdet(F 0)) where F 0 is of full rank, or it is identically zero. In the former case,the argument above applies; the latter case is trivial.

37

Once it is known that h(t) is convex, showing the case for g(t) is fairlystraightforward. Di¤erentiating, h0(t) = g0((1 + at) detF ) � adetF , and oncemore we get h00(t) = g002, so because h00(t) � 0; g00((1 + at) detF ) � 0 as well,and since(1 + at) detF is just a linear transformation in t, g00(t) � 0 as well,and hence g(t) is convex. This can be shown using the standard de�nition ofconvexity, di¤erentiability is not necessary (to come).

3

Solution 28 (Sean)

It can be nontrivial to determine whether a given function W (F ) is poly-convex or not. As an example, show that in the 2� 2 setting

W (F ) =

8><>:1 + jF j2; �(F ) � 12�(F )� 2jdetF j; �(F ) � 1

is polyconvex, when

�(F ) =pjF j2 + 2jdetF j:

Hint: show that W (F ) = g(F;detF ) where

g(F; t) = max�=�1

nf�p

jF j2 + 2� detF�� 2�t

oin which

f(t) =

(1 + t2; t � 12t; t � 1:

Then check that g(F; t) is a convex function of F and t.

As 2x � 1 + x2 always, we have that

f�p

jF j2 + 2� detF�� 2� detF � 1 + jF j2;

and so g(F;detF ) � 1 + jF j2 always. Suppose �(F ) � 1. Then there is asign � such that jF j2 + 2� detF � 1 and so we see that the maximum isachieved, g(F;detF ) = 1 + jF j2. Suppose �(F ) � 1. Then no matter the sign,jF j2+2� detF � 1 and so we must choose the maximum of 2

pjF j2 + 2jdetF j�

2jdetF j and 2pjF j2 � 2jdetF j+ 2jdetF j. We have that

d

dx

�2pjF j2 + 2x� 2x

�=

2pjF j2 + 2x

� 2

38

which is positive when jF j2 + 2x � 1. Thus, we see that the maximal value iswhen x is positive, that is

2pjF j2 + 2jdetF j � 2jdetF j = 2�(F )� 2jdetF j:

Thus, we have that W (F ) = g(F;detF ).

Consider the map(s)

�� : R4 ! R20BB@abcd

1CCA 7!�a� db� c

�:

As this is a linear map, we have that ��(�x+(1��)y) = ���(x)+(1��)��(y)for � 2 (0; 1) and x; y 2 R4. We know that the Euclidean norm N is convex andso the function N � �� =

pjF j2 � 2 detF is convex. It is easy to verify that

f(x) =

(1 + t2; t � 1;2t; t � 1:

is convex (by say second derivative test). It is straightforward to verify that thesupremum, sum, and composition of convex functions are convex. This givesthat g is convex in F and t as it is a max of sums of compositions of convexfunctions which is convex.

4

Solution 29 (Bob Kohn) In the simpli�ed 2D, �two-well�version of marten-sitic transformation considered in the Section 7 notes, � is a possible twin normalexactly if there exist R1; R2 in SO(2) such that

R1

�1 �0 1

��R2

�1 ��0 1

�= a �

for some a 2 R2. Show that (for �xed �) there are just two possible twinning di-rections, and make them explicit. Actually, we�ll show that the possible twinningdirections are � k (1; 0) and � k (0; 1), regardless of the value of �. Let

U+ =

�1 �0 1

�and U� =

�1 ��0 1

�:

To show that if � is a twinning direction then � must be parallel to either(1; 0) or (0; 1), observe that if R1U+�R2U� = a � then R1U+�? = R2U��

?,so jU+�?j = jU��?j. But if �? = (x; y), this gives

(x+ �y)2 + y2 = (x� �y)2 + y2

39

which implies xy = 0.To show that � = (1; 0) is a possible twinning direction, it su¢ ces to �nd a

solution to �cos � sin �� sin � cos �

� �1 �0 1

���1 ��0 1

�=

�a1 0a2 0

�:

Writing c = cos � and s = sin �, the left hand side is�c� 1 c� + s+ ��s �s� + c� 1

�:

Therefore we need c� + s + � = c � 1 � s� = 0, i.e. � = �sc+1 =

c�1s , with

c2 + s2 = 1. It�s easy to see that a solution exists for any �.To show that � = (0; 1) is a possible twinning direction, we observe that�

1 �0 1

���1 ��0 1

�=

�0 a10 a2

�with a1 = 2� and a2 = 0.

7 Lecture 8

Solution 30 (Dorian) The ordinary derivation of the E-L equation comesfrom the evaluation of functions of the type u(t)+ s�(t) at critical points for thevariable s, where �(a) = �(b) = 0. For simplicity, a = 0 and b = 1. In the caseof �(b) 2M , so that the �nal endpoint is nonconstant, we consider all solutionsof the ordinary Euler-Lagrange equation as the variation, and for a minimizerof the above problem, we look at variations of the form u(t; s) : [0; 1]�U ! Rn,where 0 2 U � Rd, d the dimension of the submanifold, and n the overall di-mension of the space and u0(t) = u(t; 0) minimizes I[u]. (We assume that thespace is geodesically complete, so that such a U exists.) Let �(t) = du

ds js=0: Then,since u0(t) is a minimizer:

0 =d

ds

����s=0

I[u(t; s)] =d

ds

����s=0

Z 1

0

F (t; u(t; s); _u(t; s))dt

=

Z 1

0

"Fu(t; u0(t); _u0(t))

d

ds

����s=0

u(s; t) + Fp(t; u0(t); _u0(t))_d

ds

����s=0

_u(t; s)

#dt

=

Z 1

0

[Fu(t; u0(t); _u0(t))�(t) + Fp(t; u0(t); _u0(t)) _�(t))] dt

=

Z 1

0

��Fu �

�d

dtFp

��� �(t) + d

dt(Fp(t; u0(t); _u0(t)) � �(t))

�dt (19)

= Fp(1; u0(1); _u0(1)) � �(1) (20)

40

The term Fu ��ddtFp

�� 0 in for all s; t in (1) since u(t; s) is assumed to

be a minimizer for all s and therefore satis�es the Euler-Lagrange equationsof the �xed-endpoint problem. As well, since the endpoint � is also �xed,ddsu(0; s) � 0 giving the result in (2). Now, since u(1; s) is a curve along M in aneighborhood of �, d

dsu(1; s) corresponds to an element of T�M . In particular,it can be easily shown that � : U !M , �(s) = u(1; s) is a di¤eomorphism, andhence d�( @

@si )(0) spans T�M . Thus, for the free boundary condition on u(1);we have the constraint that 0 = Fp(1; u0(1); _u0(1)) � �; 8� 2 TM . In a simpleexample, if F (t; u(t); _u(t)) is the energy of a curve on a Riemannian manifold,then Fp(t; u(t); _u(t)) = gij _u

i(t), and the minimizer is a geodesic parameterizedby arc length which satis�es the endpoint condition h _u(1); �i = 0 8� 2 T�M .,namely that the geodesics from a point to a submanifold are orthogonal to thesubmanifold.To understand the notion of a Jacobi Field, we must rede�ne the acces-

sory function �(t; �; ) corresponding to the second variation d2

ds2F (t; u(t); _u(t))

where �; correspond to duds js=0;

d2uds2 js=0, respectively. Following above, with

also d2uds2

����s=0

= (t):

d2

ds2

����s=0

u(t; s) =d2

ds2

����s=0

F (t; u(t; s); _u(t; s))

=d

ds

����s=0

�Fu(t; u(t; s); _u(t; s))

d

dsu(t; s) + Fp(t; u(t; s); _u(t; s))

d

ds_u(t; s)

�= Fuu� � + Fu + 2Fup� _� + Fpp _� _� + Fp _

If u0does in fact minimize the original variational problem, then the secondvariation as well as its integral along u0will be non-negative. Since �(t) =0��u(t)integrates to zero, the lower bound on the integral of the second variation

along u0must be identically zero:

0 =

Z 1

0

�Fuu� � + 2Fup� _� + Fpp _� _� + Fu + Fp _

�dt

=

Z 1

0

d

dt

�(Fpipj _�

i + Fuipj�i) � �j

�� d

dt

�Fpipj _�

i + Fuipj�i�� �j + (Fuiuj�i + Fpiuj _�i) � �j

+d

dt(Fp ) +

�Fu �

d

dtFp

� (21)

= Fpp _� � + Fup� � + Fp ����10

(22)

= Fpp _� � + Fup� �����10

(23)

The, the second, third, and �fth terms we be removed in (3) because �satis�ed the E-L equations for the second variation and u0 satis�ed the original

41

E-L equations. Finally, the third term on the (4) could be removed as well froman interpretation of the �rst result. Given the coordinates sion U , we have thefollowing �trivial�Jacobi �elds �k(t) = d

dsku(t; s)

��s=0

:

0 =d

dsk

�Fu(t; u(t; s); _u(t; s))�

d

dtFp(t; u(t; s); _u(t; s))

�=

d

dskFu(t; u(t; s); _u(t; s))�

d

dt

d

dskFp(t; u(t; s); _u(t; s))

����s=0

= Fuu(t; u0(t); _u0(t)) � �k + Fup(t; u0(t); _u0(t)) � _�k � d

dt

hFpu(t; u0(t); _u0(t)) � _�

k+ Fpp(t; u0(t); _u0(t)) � �k

iHence, the endpoint conclusion from the �rst result implies that for a Jacobi

�eld � on u0, the variation � + t�k is also a Jacobi �eld, and hence the �k areexactly the set of candidates for above. But the �kspan T�M and so the�rst endpoint condition eliminates the term in (4). The expression (5) above,

Fpp _� � + Fup� �

����10

, can still be expressed in more reasonable terms If � is

expressed as a function of both s and t as in the examples of the functions �k;then

d

ds

����s=0

(Fp(t; u(t; s); _u(t; s)) � �(t; s))����1t=0

= Fpp(t; u0(t); _u0(t)) _�(t)�+Fpu��+Fp� (t)����10

. As before, the Fp � term is zero, so we get the expression above. Thisversion has a better physical interpretation if one thinks of the �interesting�(i.e. vanishing endpoint-type) Jacobi �elds as in�nitessimal �pushes� of thecurve u0(t) in directions that don�t change I[u(t; s)] (though this is not alwaysthe case). Then just as solutions of the E-L equation depend only on theirvalues at the endpoints and Jacobi �elds which vanish at the end change thecurve without changing the endpoints, in the free boundary case the Jacobi�elds are somewhat relaxed by the fact that given a minimizing curve (whichmust have its �nal endpoint orthogonal to the submanifold), they only need�push�the minimizing curve onto another solution of the E-L equation whichstill satis�es the orthogonality at the endpoint condition.

42