calculus of variations summer term 2015 · 15. juli 2015 c daria apushkinskaya (uds) calculus of...

Calculus of VariationsSummer Term 2015

Lecture 19

Universität des Saarlandes

15. Juli 2015

c© Daria Apushkinskaya (UdS) Calculus of variations lecture 19 15. Juli 2015 1 / 25

Purpose of Lesson

Purpose of Lesson:

To introduce Sobolev spaces which would be suitable functionspaces in direct methods

To introduce the notions of convexity and lower semicontinuity andfind the connection between these notions.

To formulate the existence theorem.

To discuss several examples emphasizing the optimality of thehypothesis of the existence theorem.


Suitable Functional Spaces Sobolev Spaces

Sobolev spaces

Before giving the definition of Sobolev spaces, we need to weakenthe notion of derivative.

In doing so we want to keep the right to intergate by parts; this isone of the reasons of the following definition.



Definition (weak derivative)

Let Ω ⊂ Rn be open and u ∈ L1,loc(Ω).

We say that v ∈ L1,loc(Ω) is the weak partial derivative of u withrespect to xi if∫

Ω

v(x)ϕ(x)dx = −∫Ω

u(x)∂ϕ(x)

∂xidx , ∀ϕ ∈ C∞

0 (Ω).

By abuse of notations we write v =∂u∂xi

or uxi .

We say that u is weakly differentiable of the weak partial derivativesux1 , . . . ,uxn exist.



RemarksAll the usual rules of differentiation are easily generalized to thepresent context of weak differentiability.

If a function is C1, then the usual notion of derivative and the weakone coincide.

Not all measurable functions can be differentiated weakly. Inparticular, a discontinuous function of R cannot be differentiated inthe weak sense.



Definition (Sobolev spaces)

Let Ω ⊂ Rn be an open set and 1 6 p 6∞

We let W 1p (Ω) be the set of functions u : Ω→ R, u ∈ Lp(Ω), whose

weak partial derivatives uxi ∈ Lp(Ω) for every i = 1, . . . ,n.

We endow this space with the following norm

‖u‖W 1p

=(‖u‖pp + ‖∇u‖pp

)1/p if 1 6 p <∞

‖u‖W 1∞

= max ‖u‖∞, ‖∇u‖∞ if p =∞.

Here

‖u‖p :=

∫Ω

|u|pdx

1/p

if 1 6 p <∞

‖u‖∞ := ess supx∈Ω|u(x)| = inf α : |u(x)| 6 α a.e.in Ω .



Remarks

By abuse of notations we write W 0p = Lp.

If 1 6 p <∞, the set W 1p,0(Ω) is defined as the closure of

C∞0 (Ω)-functions in W 1

p (Ω).

We often say, if Ω is bounded, that u ∈W 1p,0(Ω) is such that

u ∈W 1p (Ω) and u = 0 on ∂Ω.

We also write u ∈ u0 + W 1p,0(Ω) meaning that u,u0 ∈W 1

p (Ω) andu − u0 ∈W 1

p,0(Ω).

We let W 1∞,0(Ω) = W 1

∞(Ω) ∩W 11,0(Ω).

Note that if Ω is bounded, then

C1(Ω) $ W 1∞(Ω) $ W 1

p (Ω) $ Lp(Ω) for every 1 6 p <∞.



Analogously we define the Sobolev spaces with higher derivatives asfolows.

Definition (Sobolev spaces with higher derivatives)

Let Ω ⊂ Rn be an open set and 1 6 p 6∞

If k > 0 is an integer we let W kp (Ω) to be the set of functions

u : Ω→ R, whose weak partial derivatives Dαu ∈ Lp(Ω), for everymulti-index α ∈ Am with

Am :=

α = (α1, ...αn), αj > 0 an integer andn∑

j=1

αj = m

,

0 6 m 6 k .



Definition (Sobolev spaces with higher derivatives - cont.)

The norm in W kp (Ω) is given by

‖u‖W kp

=

∑06|α|6k

‖Dαu‖pp

1/p

if 1 6 p <∞

max06|α|6k

‖Dαu‖∞ if p =∞

RemarkIf we denote by I = (a,b), we have, for p > 1,

C∞0 (I) ⊂ · · · ⊂W 2

p (Ω) ⊂ C1(I) ⊂W 1p (I)

⊂ C(I) ⊂ L∞(I) ⊂ · · · ⊂ L2(I) ⊂ L1(I)



Theorem 19.1Let Ω ⊂ Rn be an open set, 1 6 p 6∞ and k > 1 an integer.

W kp (Ω) equipped with its norm ‖ · ‖W k

pis a Banach space which is

separable if 1 6 p <∞ and reflexive if 1 < p <∞.

Remarks

Note that the space W 11 is not reflexive.

This is the main source of difficulties in the minimal surfaceproblem.


Convexity and Lower Semicontinuity

§13. Convexity and Lower Semicontinuity



Let X be a Banach space, J : X→ R, and consider the minimizationproblem

infu∈X

J[u].

Let us first consider the problem of the existence of a solution.

Proving of existence is usually achieved by the following steps,which constitute the direct method of the calculus of variations:



1 One constructs a minimizing sequence un ∈ X, i.e., a sequencesatisfying

limn→∞

J[un] = infu∈X

J[u].

2 If J is coercive(

lim|u|→∞

J[u] =∞)

, one can obtain a uniform

bound |un|X 6 C. If X is reflexive, then (by Theorem 18.1) onededuce the existence of u0 ∈ X and of subsequence unj such that

unj Xu0.

3 To prove that u0 is a minimum point of J it suffices to have theinequality

limunju0

J[unj ] = sup infunju0

J[unj ] > J[u0],

which obviously implies that J[u0] = minu∈X

J[u].


Convexity and Lower Semicontinuity Lower semicontinuity

Lower Semicontinuity

This last property, which appears here naturally, is called weak lowersemicontinuity. More precisely, we have the following definition:

Definition (lower semicontinuity)J is called lower semicontinuous (l.s.c.) for the weak topology if forall sequence un u0 we have

limunu0

J[un] = sup infunu0

J[un] > J[u0]

The same definition can be given with a strong topology.

In the direct method, the notion of weak l.s.c. emerges verynaturally.


Convexity and Lower Semicontinuity Convexity

Convexity

Unfortunately, it is difficult in general to prove weak l.s.c.

A sufficient condition that implies weak l.s.c. is convexity:

Definition (convexity)J is convex on X if

J[λu + (1− λ)v ] 6 λJ[u] + (1− λ)J[v ]

for all u, v ∈ X and λ ∈ [0,1].


Convexity and Lower Semicontinuity Integral functionals

Convexity and Integral Functionals

If J is an integral functional, we can even say more about the linkbetween convexity and l.s.c.

Let Ω ⊂ Rn be a bounded open set, and let f : Ω× R× Rn → R bea continuous function satisfying

0 6 f (x ,u, ξ) 6 a(x , |u|, |ξ|),

where a is increasing with respect to |u| and |ξ|, and integrable inx .

Let W 1p (Ω) be the Sobolev space.

For u ∈W 1p (Ω) we consider the functional

J[u] =

∫Ω

f (x ,u,∇u)dx



Convexity and Integral Functionals

Theorem 19.2 (l.s.c. and convexity)1

J[u] is (sequentially) weakly l.s.c. on W 1p (Ω), 1 6 p <∞

mf is convex in ξ.

2

J[u] is (sequentially) weakly* l.s.c. on W 1∞(Ω),

mf is convex in ξ.

We emphasize that convexity is a sufficient condition for existence.There exist nonconvex problems admitting a solution.



Theorem 19.3Let Ω ⊂ Rn be bounded and f : Ω× R× Rn → R continuous satisfying

1 f (x ,u, ξ) > a(x) + b|u|q + c|ξ|p for every (x ,u, ξ) and for somea ∈ L1(Ω), b > 0, c > 0, and p > q > 1.

2 ξ → f (x ,u, ξ) is convex every (x ,u).

3 There exists u0 ∈W 1p (Ω) such that J[u0] <∞.

Then the problem

inf

J[u] =

∫Ω

f (x ,u(x),∇u(x))dx , u ∈W 1p (Ω)

admits a solution. Moreover, if (u, ξ)→ f (x ,u, ξ) is strictly convex forevery x , then the solution is unique.



RemarksIn Theorem 19.3 the coercivity condition (1) implies theboundedness of the minimizing sequences.

Condition (2) permits us to pass to the limit on these sequences.

Condition (3) ensures that the problem has a meaning.


Convexity and Lower Semicontinuity Examples

We propose below some classical examples where either coercivity,reflexivity, or convexity is no longer true.

Example 19.1 (Weierstrass)

Let Ω = (0,1], let f be defined by f (x ,u, ξ) = xξ2 and let us set

J[u] =

1∫0

xu′2dx with u(0) = 1, u(1) = 0.

Then, we can show that this problem does not have any solution.



Example 19.1 (continued)

The function f is convex, but the W 12 (Ω)-coercivity with respect to

u is not satisfied because the integrand f (x , ξ) = xξ2 vanishes atx = 0.

Let us first prove that

m := inf J[u] = 0.

The idea is to propose the following minimizing seqience

un(x) =

1, if x ∈ (0,1/n),

− log (x)

log (n), if x ∈ (1/n,1).



Example 19.1 (continued)

It is then easy to verify that un ∈W 1∞(0,1), and that

J[un] =

1∫0

xu′2n dx =1

log (n)→ 0.

So we have m = 0.

If there exists a minimum u0, then we should have J[u0] = 0, thatis u′0 = 0 almost everywhere (a.e.) in (0,1).

But u′0 = 0 a.e. in (0,1) is clearly incompatible with the boundaryconditions.



Example 19.2 (Bolza)

Let Ω = (0,1], and let f be defined by f (x ,u, ξ) = u2 + (ξ2 − 1)2.The Bolza problem is

min← J[u] =

1∫0

[(1− u′2)2 + u2

]dx , u ∈W 1

4 (0,1)

with u(0) = u(1) = 0.

The functional J is clearly nonconvex.



Example 19.2 (continued)It is easy to see that

m := inf J[u] = 0.

Indeed, for n an integer and 0 6 k 6 n − 1, if we choose

un(x) =

x − k

n, if x ∈

(2k2n,2k + 1

2n

)−x +

k + 1n

, if x ∈(

2k + 12n

,2k + 2

2n

),

then un ∈W 1∞(0,1) and

0 6 un(x) 61

2nfor every x ∈ (0,1),

|u′n(x)| = 1 a.e. in (0,1),

un(0) = un(1) = 0.c© Daria Apushkinskaya (UdS) Calculus of variations lecture 19 15. Juli 2015 24 / 25


Example 19.2 (continued)Therefore,

0 6 infu

J[u] 6 J[un] 61

4n2 .

Letting n→∞, we obtain m = 0.

However, there exists no function u ∈W 14 (0,1) for which

u(0) = u(1) = 0 and J[u] = 0.

So the problem does not have a solution in W 14,0(0,1).


calculus of variations summer term 2015 · 15. juli 2015 c daria apushkinskaya (uds) calculus of...

Documents