Maximum Principles

Consider the coefficients of L,

the function f and

the set r regular enough so that we C2Cn) n C CI ).

Because of the regularity assumptions the Pole can

be equivalently written in divergence or non - divergence

form ( already remarked ) .We find it convenient to

choose the form

Lu = T.FI,

aiinxi,

+ II. biua.

+ cu

Since the proof we are going to do works under

weaker regularity assumptions onaid

,bi

,C

,

namely that aid,

bi,

c e C CR ),

we will only assume

that and further suppose that he Ctcnnccr ) .

We will work under the assumption aid'

= aji,

RCR"

bounded and open .

Theorem ( Week maximum principle )

Assume we Ctcr ) n CCI ) and czo in R.

( ^ ) Lee e 0,

CIO =p mox in = moxie

( u is said to be a of ZR

sub solution )

(2) 4,

C Z 0 =D

hexin E Max µ+

r anwhere utcx ) = mex { uCx3o] .

To prove the theorem we first need the following result

from linear algebra :

Lemma;

A,

B positive semi - definite symmetric uxu

n

matrices.

Then I aid'

bit zo .

Proof ,

ijae

F Re Ocn ) : A = RTDR for some D= okay 6h,

- ,dn)with di 30 . Being B positive semi - definite ,

the

metric C : = RBRT is positive semi - definite ,too

.

This implies that Cii 30,

it { 1,2,

- in } .

Now using the invariance of the trace under. .

transposition and under commutation namely the

fact that tr ( JK ) = tr ( KJ ) = tr KKJJT ) we have

A=AT

p

£

aid

bid'

=tr ( ABT ) = tr ( BAT )=tr(# B) = tr # B) =

iij 1 ↳v commutation

transposition .

= tr ( EDRB ) = tr ( DRBRT ) = to CDC ) =

Lo commutation

=

.§ dicii zo

Da

Proof of the theorem

ake E > 0 and t > o so big that

- 0×2+11 ball at + Uc Ha < °. Define rfk ) : = ucx ) + set "

and note that

Lrf = Lute e

"'

( T.gl aijoeisnjt'

+ bat c)

s o + set 't

( - 05+11 ballot + Hello ) < °

If F x. e R : 0 s E ( xD=

mzex nf =D

Lrrecxo ) =-

;§aiicx

. ) NI.

,( ×

. ) t ? bicx. ) which )tcQ° ) ntcx

. )

Z ccx . ) Nscxo )

Z ° because in a mex point TN{xo ) - 0

thanks to the lemme applied to A = @iicxo) )ijand B = - DIE Ho ) noting that A is pos ohfby the uniform elliphieity and the B is

positive definite since at ×.

DTEKO) is non - pos .

definite as this is a mex point .

This contradiction yields , mox we=/° %) < °

-e

R\ wax or

2R

Now mox u s wax NE f° 4×0 ) - ° E utans ngonxut

- -

+r R

mexv 's mex @E)an ar

The thesis follows as s -00 since Nz → u

uniformly in I.

=D met U £ me × µ+.

.

r an

Note that if CIO there is no need of introducing @4+

Ten fact mex u = gains nog×NE = max we → mexu

an an e→o are

and the chain of inequalities is actually a chain ofequalities .

Riney : Of course one has also an equivalentstatement for min instead of mex .

This reads as

( 1'

) Luz 0 in R,

c=o ⇒ min u =

minusan

( 2' ) "

, CZO =D min re z - maxein an

where ii. - win { Mio } .

Theorem ( Strong maximum Principle )

let CZO in RCIR"

openand connected

.

Assume

we C2Cn)n CCI ) be such that Lee so.

F Xo ER : ucxo ) = mgxw > 0 =D U = const in R

r

Jn the case CEO : F xo c- Rieecxo ) = moxie =D u= const in RI

Jn order to provethis result we make use of

the following Lemme by Hopf .

Hopf lemme

Let czo in R,

we C2Cn)n Ctr ) and Luso . let xo ← 2R

such that Ucxo ) > UCH Fx E R. Suppose moreover that

F B CR such that x. ← JB ( we also saythat R satisfy

an interior ball condition at xo ) .Then

u( xo ) 70 =D 21 ( x . ) > 0,

%o

where V×.

is the unitary outer normal to B at xo.

In the case CIO then 21 ( ×. ) > o holds without

2v

assuming be ( x. ) 30

.

×0

Proof of the twpt lemme

without loss of generality wemay assume that

�1� = B Car ) for some r > 0.

We set vcx ) :=£tk[ It r2

txe B ( qr ) with X to be chosen later.

By uniform elliptieihy : P.g.,

aii 5i5j 301512

Using that ¢" "1) × ,= - ztxi it * 12

and that

( it " 2)a g.

= 47k.

.×

;it ""

2

- ztoijetk"

we have that Lv = - PL,

aiirxixj to¥,

birxitc =

= e-" " "

III. aid ( -41in .× ; tztoi ;) -ITYIETbizxa . + of it "Ie→m)

g e-" *

( -451×120+21 ltrA 1+271511×1 + Icl )

where A= ( ais )..j

and b = ( be,

- ,b" ) .

Consider now the region Ri= BG ,r)\B%, %) : txcr rz < kkr

Lira e-" " "

( -4×2 ⇐ To tztlltrallatztllbllortlkll .)

Provided 1 is big enough we have Lrrso. O*

By the assumption ucx . ) > lecx ) txer we have that

F e > ° such that ulx. ) 7 MCA + ercx ) kxe 213,26) . @

r.

RCKD

0¥ •µ.

TJBGMD°←¥in•JR Xo I i >

• rg r 1×1

Moreover,

since v = 0 we have UK.) 3 Uk ) + e Ncx ) �2�

12136 ,r)

ttxe 2136, r ) .We now have that tx€R

L ( utsr - ucxo ) ) = Keyt ELCN ) - Lcucxoj ) = - cucxo ) EO

A /×" either CEO

O 0 CUCXO ) or czo ,MK

. ) > 0by assumptionby *O

By the weak maximum principle

mfx ( utev - ucx . ) ) E wax ( let er - ucxo ))+ = O

a ar" by @& �2�

2136 ,r)U2Bl9%)

As a result we now have that txe R went even ) Eucx. )

Since for ×=×.

we have ll(xo)+ ENCX. ) = llcx

. ) we

:can write

( ucxteyrcx) -uCxo))y,

?ucx ) + ercx ) - ucx

. ) t×eR

0

Fnperlienlor Ju ( n + Ev - ucx . ) )| 30 or in other

x=Xo

words 2g ( ×. ) + Efg ( × . ) 30 ⇐D ( here w×o=×f )

- In

2£ ( x . ) 3- E @s ;×f)= Zetre > 0

-

zµc⇒=-Heitktx Tvcx. ) = -

zietkdx.

)@vk.7.ge )= - zxe→r4×÷P= -

siren↳ ix. 12am

a

Proof ( of the Strong maximum principle )

Let m : =

mgxin and set M :-. { xer : uk ) = in } .

r

Zf u¥ m consider N : = { xeri ucx ) < m } .

Pick a

point in N such that dist ( y ,M ) < dint ( q ,

24.

Such a point y exists since N is supposed Ntf

andR

is connectedNote that N is open by the continuity of n .

Joke the

largest ball centered aty

and contained in N

( the set of such balls is von empty again by the feet

that N isopen ) .

Denote such a ball by B.

There exists x. e Mr JB.

At the point xo N satisfies

an in tenor ball condition and by Hopf 's lemme

N.N2£yC xD 20 .

But this contradicts the fact that

u attains a wax point at xo.

BeThe contradiction implies re = in

.