De Giorgi Estimates and Elliptic Equations

De Giorgi Estimates and Elliptic Equations

Esteban Adam Navas

Department of MathematicsUniversity of California, Riverside

May 18, 2012 / Graduate Student Seminar

Esteban Adam Navas De Giorgi Estimates and Elliptic Equations

Introduction to Elliptic Equations

Elliptic Partial Differential Equations (PDEs) aregeneralizations of Laplace’s Equation, −∆u = 0 andPoisson’s equation, −∆u = f

where ∆ = ∂2

∂x21

+ ∂2

∂x22

+ · · ·+ ∂2

∂x2n

u : U ⊂ Rn → R is a scalar function of n variablesu(x) := u(x1, . . . , xn)

U is an open, bounded subset of Rn


Introduction to Elliptic EquationsGeneral Elliptic PDEs

General Elliptic PDEs look like:

Lu = −n∑

i,j=1

Dj(aij(x)Diu) +n∑

i=1

bi(x)Diu + c(x)u = f (x)

or

Lu = −div(A(x) · Du) + b · Du + c(x)u = f (x)

Notation:

Diu =∂u∂xi

Du = (D1u, · · · ,Dnu)

A(x) = (aij (x)) b(x) = (b1(x), · · · ,bn(x))


Introduction to Elliptic EquationsLu = −Dj(aij(x)Diu) + bi(x)Diu + c(x)u = f (x)

A(x) = (aij(x)) is a symmetric, positive definite coefficientmatrix.This imposes the uniform ellipticity condition onthe operator L, that there exists λ > 0 such that:

n∑i,j=1

aij(x)ξiξj ≥ λ|ξ|2

for all ξ ∈ Rn

That is, A(x) has smallest eigenvalue ≥ λ.



Our goal is to prove Holder continuity of weak solutions toa special case of the general elliptic equation:b ≡ 0,c ≡ 0,f ≡ 0 and aij ∈ L∞.

This special case is the equation −n∑

i,j=1

Dj(aij(x)Diu) = 0



Example:

Take A(x) = (δij(x)) = In, b ≡ 0, c ≡ 0.


Weak Solutions to Lu = f

From here on out we deal with the equation

−Dj(aij(x)Diu) + c(x)u = f (x)

b ≡ 0summation convention over repeated indices is used

(e.g. cix i =n∑

i=1

cix i )

Unless aij are differentiable (C1) functions, this equationmakes no sense in the classical sense.If we want to consider more general coefficients, we willneed a more general derivative.






(e.g. cix i =n∑

i=1

cix i )

Unless aij are differentiable (C1) functions, this equationmakes no sense in the classical sense.

If we want to consider more general coefficients, we willneed a more general derivative.






(e.g. cix i =n∑

i=1

cix i )

Unless aij are differentiable (C1) functions, this equationmakes no sense in the classical sense.If we want to consider more general coefficients, we willneed a more general derivative.


Weak Solutions to Lu = fThe Weak Derivative

Suppose u,v ∈ L1loc(U).

v is a weak partial derivative of u if for any testfunction φ ∈ C∞

c (U),∫U

uDiφdx = −∫

Uvφdx

We say v = Diu exists weakly if such v exists.

That is, Diu exists in the weak sense if we can integrate itby parts against any such test function and "move thederivative" to u.





c (U),∫U

uDiφdx = −∫

Uvφdx







c (U),∫U

uDiφdx = −∫

Uvφdx





Example:

Let n = 1 and U = (0,2). Then the weak derivative of

u(x) =

{x , 0 < x ≤ 11, 1 ≤ x < 2

is

v(x) =

{1, 0 < x ≤ 10, 1 < x < 2



Example:

Let n = 1 and U = (0,2). Then the function

u(x) =

{x , 0 < x ≤ 12, 1 ≤ x < 2

has no weak derivative.



Fact:

Weak partial derivatives are unique.


Weak Solutions to Lu = fThe Sobolev Space H1(U)

Suppose u : U ⊂ Rn → R and u ∈ L1loc(U).

The Sovolev Space H1(U) = W 1,2(U) is the space of allu such that Diu exists weakly and u, Diu ∈ L2(U) for each1 ≤ i ≤ n.

H1(U) = {u ∈ L1loc(U) ∩ L2(U) :

Diu exists weakly and ||Diu||L2(U) <∞}


Weak Solutions to Lu = fThe Sobolev Space H1(U)

Suppose u : U ⊂ Rn → R and u ∈ L1loc(U).

The Sovolev Space H1(U) = W 1,2(U) is the space of allu such that Diu exists weakly and u, Diu ∈ L2(U) for each1 ≤ i ≤ n.

H1(U) = {u ∈ L1loc(U) ∩ L2(U) :

Diu exists weakly and ||Diu||L2(U) <∞}


Weak Solutions to Lu = fDefinition

A weak solution to Lu = −Dj(aij(x)Diu) + c(x)u = f (x)is a function u ∈ H1(U) such that, for all test functionsφ ∈ H1

0 (U),

−∫

UDj(aij(x)Diu)φdx +

∫U

c(x)φdx =

∫U

f (x) dx

⇒∫

Uaij(x)DiuDjφ+ c(x)φdx =

∫U

f (x) dx

Notice that we no longer need differentiability of aij for thisPDE to make sense.


Weak Solutions to Lu = fDefinition

A weak solution to Lu = −Dj(aij(x)Diu) + c(x)u = f (x)is a function u ∈ H1(U) such that, for all test functionsφ ∈ H1

0 (U),

−∫

UDj(aij(x)Diu)φdx +

∫U

c(x)φdx =

∫U

f (x) dx

⇒∫

Uaij(x)DiuDjφ+ c(x)φdx =

∫U

f (x) dx

Notice that we no longer need differentiability of aij for thisPDE to make sense.


Results and RegularityLu = −Dj(aij(x)Diu) + c(x)u = f (x)

Unique weak solutions u ∈ H1(U) to Lu = f exist forgeneral elliptic PDEs.

If aij ∈ C∞(U) then the weak solution is C∞(U) smooth.If aij ∈ L∞(U) then the best regularity we can hope for isHolder continuous with exponent α ∈ (0,1).

that is, supx,y∈B1/2

|u(x)− u(y)||x − y |α

≤ C.



Unique weak solutions u ∈ H1(U) to Lu = f exist forgeneral elliptic PDEs.If aij ∈ C∞(U) then the weak solution is C∞(U) smooth.

If aij ∈ L∞(U) then the best regularity we can hope for isHolder continuous with exponent α ∈ (0,1).


|u(x)− u(y)||x − y |α

≤ C.



Unique weak solutions u ∈ H1(U) to Lu = f exist forgeneral elliptic PDEs.If aij ∈ C∞(U) then the weak solution is C∞(U) smooth.If aij ∈ L∞(U) then the best regularity we can hope for isHolder continuous with exponent α ∈ (0,1).


|u(x)− u(y)||x − y |α

≤ C.



Unique weak solutions u ∈ H1(U) to Lu = f exist forgeneral elliptic PDEs.If aij ∈ C∞(U) then the weak solution is C∞(U) smooth.If aij ∈ L∞(U) then the best regularity we can hope for isHolder continuous with exponent α ∈ (0,1).


|u(x)− u(y)||x − y |α

≤ C.


Theorem (De Giorgi)Special Case c ≡ 0, f ≡ 0, U = B1

Suppose −Dj(aij(x)Diu) = 0 weakly in B1, whereaij ∈ L∞(B1) and ||aij ||∞L (B1) ≤ Λ. Then

supB1/2

|u(x)|+ supx ,y∈B1/2

|u(x)− u(y)||x − y |α

≤ C||u||L2(B1)


Proof:

Step 1: Let u+ = max{u,0}. Then u+ is locally bounded:

supBr/2

u+ ≤ C||u+||L2(Br )

for any r .

Step 2: (Density Theorem) Suppose u is a positivesolution in B2 with |{x ∈ B1; u ≥ 1}| ≥ ε|B1|. Then there isa constant C such that infB1 u ≥ C.Step 3: (Oscillation Theorem) DefineoscBr u = supBr

u − infBr u. Suppose that u is a boundedsolution of Lu = 0 in B2. Then there exists a γ ∈ (0,1)independent of r such that

oscB1/2u ≤ γ oscB1u


Proof:


supBr/2

u+ ≤ C||u+||L2(Br )

for any r .Step 2: (Density Theorem) Suppose u is a positivesolution in B2 with |{x ∈ B1; u ≥ 1}| ≥ ε|B1|. Then there isa constant C such that infB1 u ≥ C.

Step 3: (Oscillation Theorem) DefineoscBr u = supBr




Proof:


supBr/2

u+ ≤ C||u+||L2(Br )

for any r .Step 2: (Density Theorem) Suppose u is a positivesolution in B2 with |{x ∈ B1; u ≥ 1}| ≥ ε|B1|. Then there isa constant C such that infB1 u ≥ C.Step 3: (Oscillation Theorem) DefineoscBr u = supBr




Proof:(Continued)

Step 4: Holder continuity proven from the OscillationTheorem.

supx ,y∈B1/2

|u(x)− u(y)||x − y |α

≤ C


Thank you for your attention!References:

Evans, Partial Differential EquationsLin, Han, Elliptic Partial Differential Equations


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