Math 742: Geometric Analysis - UMD

Math 742: Geometric Analysis

Lecture 5 and 6 Notes

Jacky Chong jwchong@math.umd.edu

The following notes are based upon Professor Yanir Rubenstein's lectures with reference to Variational Methods 4th edition by Struwe and A User's Guide to Optimal Transport by Ambrosio and Gigli.

1 Direct Method in the Calculus of Variations

THEOREM 1. Let X be a topological Hausdorff space, and suppose E : X R {+} satisfies the bounded compactness condition: For any R the sublevel set, K = {u X : E(u) } is compact. Then there exists a uniform real constant CX on X such that E CX on X and E attains its infimum.

Proof. If E + on X, then there is nothing to prove. So, assume E +. Denote = infuX E(u)

- and choose a strictly decreasing sequence (m) such that m as m . Let Km = {u X : E(u) m}, then by assumption Km is compact. Moreover, it's clear that Km+1 Km for all m N, i.e. we have a decreasing nesting of compact sets. By Cantor's Intersection Theorem, we know

that

K = Km = .

m=1

Let u K X, then we observe (i) CX := E(u) = - and (ii)

E(u) m for all m.

Hence it follows E(u) when passing through the limit which means E(u) = . Thus E attains its infimum in X.

DEFINITION 1. Let (V, ? V ) be a reflexive Banach space and X V a subspace. Then E : X R {?} is coercive provided E(u) if u V .

DEFINITION 2. E : X R + is said to be sequentially weakly lower semicontinuous over X provided for every sequence (un) n=1 X and (un) n=1 converges weakly V , un u then

E(u) lim inf E(un).

n

Math 742 - Lecture 5 and 6

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EXAMPLE 1. Consider the function E : R R given by

1 E(x) = 1 + x2

2

if x = 0 if x = 0.

It's clear that E is not coercive since E(x) 0 when |x| . Moreover, it is also clear that E is not lower semi-continuous (in particular, not weakly l.s.c.) since

2

=

E(0)

lim inf

n

E(xn)

=

1

where xn 0 as n . Also, note that E does not attain its minimium.

THEOREM 2 (Special Case of Theorem 1). Let (V, ? V ) be a reflexive Banach Space and let X V be a weakly closed subset of V . Assuming

(i) E : X R {+} is coercive,

(ii) E is sequentially weakly lower semi-continuous.

Then E is bounded from below on X and attains its infimum in X.

Proof. Let := infuX E(u) -. Similar to the proof of Theorem 1, we begin by choose a minimizing sequence (m) X such that E(um) as m . Now, let us first make two claims:

(i) the sequence (um) is uniformly bounded in V ,

(ii) there exists a subsequence such that umk u for some u V .

Since umk u by (ii) and X is weakly closed, then we have that u X. Applying the fact that E is weakly l.s.c., we have that

E(u)

lim inf

k

E(umk

)

=

Hence E attains its minimium in X. Thus, to complete the proof we need to justify (i) and (ii). Proof of Claim (i): This is immediate from the definition of coerciveness of E: E(um) implies

um V . Proof of Claim (ii): Since V is reflexive, then by Theorem 8 in the Appendix we know that any closed ball, B(0, R), of V is weakly compact. Since (um) is bounded in V by (i), then (um) B(0, R) for some R. Now, apply Theorem 6 in the Appendix, we see that (um) has a subsequence which converges weakly to some u B(0, R) V .

2 Some Examples and Applications

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2.1 p-Laplacian in Rn

Let u W01,p(Rn), we informly defined the p-Laplacian as following

pu = ? (|u|p-2u).

(1)

Observe, we have that p = the usual Laplacian on Rn when p = 2. Note: We must check pu is indeed well-defined on W01,p(Rn) since the p-Laplacian "requires" second order (weak) derivatives of u. Observe for any C0(Rn) we have that

pu(x)(x) dx :=

Rn

|u(x)|p-2 u(x), (x) Rn dx

Rn

|u(x)|p-1|(x)| dx

Rn

L

|u(x)|p-1 supp()(x) dx

Rn

Csupp()

L

u

p-1 Lp

<

.

Thus, to define the pu in the distributional sense, we only need that u Lp(Rn). Hence it is clear that pu is well-defined on W01,p(Rn).

Assuming Rn is a bounded domain and consider the Dirichlet problem:

-pu = f in (2)

u = 0 on .

Claim: We claim -pu = f is the Euler-Lagrange equation of the function

E(u) = 1 |u(x)|p - u(x)f (x) dx p

where the Lagrangian, L : Rn ? Rn ? R - R, is given by

L(x, y, z) = 1 |y|p - zf (x). p

Since the Euler-Lagrange equation to a function E[u] = L(x, Du(x), u(x)) dx is given by

n

Lz(x, Du, u) - (Lyi (x, Du, u))xi = 0

(3)

i=1

then indeed we have that

n

-f (x) - ? (|u|p-2u) = -f (x) -

|u|p-1 uxi

= 0.

i=1

|u| xi

Thus, informly, solving the Dirichlet problem (2) is similar to finding a minimizer for E in W01,p().

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THEOREM

3.

Let

p [2, )

with

conjugate

q

satisfying

1 p

+

1 q

= 1.

Assume

also

f

W -1,q()(:=

W 1,p()). Then there exists a weak solution u W01,p() solving the Dirichlet problem (2).

Proof. Base upon the above discussion, let us show that E attains its minimium in W01,p(). We shall invoke Theorem 2, by checking that E satisfies coercivity and sequentially weakly l.s.c. over W01,p(). Coercivity of E: Observe by corollary 1 and the definition of the dual norm

1 E(u) =

p

u

- p

W01,p ()

f (x)u(x) dx

1

p

u

- p

W01,p ()

f

u W -1,q ()

W01,p ()

1 "Young's Inequality"

p

u

p W01,p ()

-

1 2pp

u

p W01,p ()

-

2q q

f

q W -1,q ()

=C

u

p W01,p ()

-

2q q

f

q W -1,q ()

where

C

>

0

and

2q q

f

q W -1,q ()

is

independent

of

u.

This

shows

that

E

is

coercive

over

W01,p().

Sequentially Weakly L.S.C. of E: Consider a sequence (um) W01,p() such that um u W01,p().

Applying proposition 2(d) in the appendix with (x) = (x) and the fact that f W -1,q(), then it

follows

E(u) lim inf E(um).

n

Thus, it is now clear that E is both coercive and weakly l.s.c. which means E attains its infimum in W01,p(). Lastly, we need to check that a minimium of E, say u, is a weak solution to problem (2). Let C0() and u a minimizer of E, since

d

1 |u(x) +

(x)|p - (u(x) +

(x))f (x)

dp

=|u(x) + (x)|p-2 u(x) + (x), (x) Rn - (x)f (x)

is bounded by some integrable function then by some variant of the Lebesgue Dominated Convergence Theorem we have that

d

d E(u + ) =

1 |u(x) +

(x)|p - (u(x) +

(x))f (x) dx

d =0

d p

=0

= |u(x) + (x)|p-2 u(x) + (x), (x) Rn - (x)f (x) dx

=0

= |u(x)|p-2 u(x), (x) Rn - (x)f (x) dx = 0

which indeed solves the weak formulation of problem (2).

2.2 Harmonic Maps into Rn -Generalizing Geodesics

Another important class of problem is finding minimal hypersurfaces in a Riemannian manifold (M, g) (geodesics, etc) which has a variational formulation. Let be a bounded domain in Rd, and M be a compact subset of Rn with a symmetric positive-definite Riemannian metric g = (gij)1i,jN . Consider

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the u H1(, Rn) and the functional

1 E(u) =

2

gij(u(x))ui(x)uj(x) dx

we shall compute its Euler-Lagrange equation.

PROPOSITION 1. The Euler-Lagrange equation of E(u) is equivalent to

uk + jikuiuj = 0

(4)

in .

Proof. First, observe

ui

uj ui

2ui

Lpkl = gik xl (Lpkl )xl = j gik xl xl + gik (xl)2

and

Lzk

=

1 2

k

gij

ui

uj

.

Using the Euler-Lagrange equations for systems

n

(Lpkl (x, u, Du))xl - Lzk (x, u, Du) = 0

l=1

we get that or for k = 1, . . . , n.

-

1 2

k

gij

ui

uj

+

j gikuiuj

+

gik ui

=

0

uk + gik

1 j gik - 2 kgij

uiuj = uk + jikuiuj = 0

EXAMPLE 2. If u is a solution to (4), then we say that u is a harmonic mapping from Rd to (M, g). In particular, if d = 1, then we have that (4) is just the geodesic equations

u?k + jiku iu j = 0

where u : (0, 1) M Rn.

2.3 Optimal Transport in Rn

The subject of Optimal Transport will be "explain" in more details in subsequential lecture notes, we shall only introduce the different formulations of the subject in this section.

DEFINITION 3. Given a Polish space (X, d) (i.e. a complete and separable metric space), we shall denote P(X) the set of Borel probability measures on X. The support supp(?) of a measure ? P(X) is the smallest closed set, C on which C d? = 1.

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