Problems in dimension h > 1

We want to study the heat flow on RCR"

open , for te [ at ] .

The kind of problem we are going to consider

are of Cauchy + bdy type . Namely we

will always give information on ucxit )

as t=o,theR and as xe an

,the [o⇒ .

We define Rt : = Rx ( ° it ] the parabolic interior

of the parabolic cylinder I,

= I × [ OF ] and

define if i. It lrt to be its parabolic bdy .

Cauchy date bdy date

÷+=Lsxitie Rxlo } } u lcxrt ) c- 2R× Git ) }

T- - -

ITII.an

In What follows we will examine the

uniqueness of the solution. ( The existence will

be proved in the context of the week - solutions

in a lecture to come

.)This will follow from

one of the maximum principles below -

Ln what follows we assume R to be bounded.

Theorem ( maximum principle for the heat egw)

let we q2(r,

)nC( I , ) be a

solution of Me =D u in RT .

Then

(1) mex il = Max u

I rT T

(2) if R is connected and F ( xo.to ) e Rt :

u ( x.,

to

)==axu

,then his const

.in Re

.

.

R,

We limit ourselves to the proof of (1)

( week mex. principle ) For a proof of (2)

( Strong mex . principle ) see Evans 2-3.3.

Proof : For E > o, define rcxit ) = u(x,t ) + EK12

( 2- - D) v = ( 2- . D) m + ( 2- D)ekl2= .

eAki2=- 2h E < 0

÷° 2h *

Suppose by contradiction that F ( xo,

to )e Rt mox

point for r.

Then 2- r GO.to ) 30 ( > 0 only for tst,

that is at the bdy of R+ ) and Dvcx . .to ) so .

This implies ( at - D) v zo

,which contradicts *

.

The following chain of inequalities follows :

ng.exu@nrf.trOenffexre

mf,×uternex

1×12

↳

bsatnik.int?.IinIEa.

Taking the limit as E → 0 we have

mgxuemnfxusmf.fm is

Corollary ( Uniqueness on bold sets )

Letg

€ C ( q ), fe C ( Rt ) .

Then there exists at

most one solution of the problem :

Me - B u=f in RT| u=g on F

Proof : Suppose by contradiction miu to be solutions,

wtu. By the maximum principle applied to

w= a - it we find that w=o in RT . �1�

Remark ( About the mex principle for t unbounded)

It R,

is unbounded in time,

the mex principle does

not hold true .Take for instance n =L and

ukcxit) = c. e- tttsinkx.

Note that these functions

solve in Co , # ) × R the problem

Mz - M×× =0

|u(qt)=u(it ,t)=o,

but they are unbounded in

R,

= ( °, A) × C- a ,T ) .

Remark ( IU - posed problem for the heat equation )The Dirichlet problem in space and time is in posed .

Let A = ( o, e) x ( at ) . We look for a solution uiA→R

of the problem me - u××=o in A[ u=q on aa, ye CC2A )

Note that if y has a unique mex at ( ×. ,T )

,then by

the max principle there is no solution to the problem -

( otherwise the mex should be on E,and Cxo ,t)¢ I ! )

. What if R is unbounded ?

Theorem ( Max principle for the Cauchy problem )

suppose me 4 ( IR "x ( o ,T])n C ( R "x [ at ] ) solves

at - Bu = 0,

R"×(o,T ]

| a =

g ,R "× Lt=o}

, ge CCR "× lt=o} )

and satisfies the growth assumption :

akpFA

,a > 0 i txe R

"

,te [ at ] ucx

, E) e Ae.

Then sup u = sup if .

R"

× [qT] Rn

RemoikliAs a result we obtain the uniqueness

result in the unbounded case as before .

Proof . Steph Assume 4aT< 1 and choose E20

such that 4a( Tte )< 1.

Fix yer"

,µ > o and define

vcxit ) : = eecx ,t ) - b-1×-912

( Tie - £51- ¥tt ) ,

×eRn,

tso

We have ( 2- - D) v=o. ( ohek for exercise )

Fix r > o and set R : = § ( y ,r )

,

so that we have

Rt=BlY,r)x( of ]. By the wax principle In R

,we obtain

( * ) mex r= moxv

I,

F

Step 2- k-yI

For xepr,

vcxio ) = eecx.dz?=gu,ze4ct+9sukio)=gey

It k-yI=r ,te Git ] vcx ,t)=uCx,t) €+1,

,zec%¥ 't )

zr2

Assnmph@aealxl_ee_e4Ttte.tt( T.IE - t)% r27,+ea( ' Y' +121 eactte)

( Ttc ) "E

ti→ the#

decreasing and 1×12<-41+42 .

1Aea("ltr[µ=jEca24aCTt e) < e => 1-= at y for some y > °

4 CTTE )

§ map F ,since him - = - a

R r→+x

Thus , since I = 213 ( y ,r ) × [°it )

, TI = 2134 ,r )x [ OF ]

,

we have proven that rcxit ) ± sup g txe Bcy , r)×[o,T]=RIand by the arbitrariness of YER

"

, we here that

vcx , E) e sup g tx c- R "x[qT]

Letting r -00 we have ucgt ) E supg .

To get rid of the condition T< fya it is enough to

apply the result above in the intervals [ o,

T,

],

[ I ,2T

, ],

- - , [ kt , (ktDT ] ,. . with I=fz (a lga)r

Remark 2 : If we remove the growth Gwlihiow�1�

we have existence of solution blowing -

up very

fast as 1×1 → + a ( see Tikhonov example later on

in the notes )

Remark 3 :

If ve ch ( Rt ) satisfies re - Breoinrt

N is said to be a sub - solution.

In this

case ( with the same proof as in the case

of - Dsao ) we have mex 5 =mox r

.

-

rRT T

Remark 4 :

Since the proof above is based on the wax

principle ,we can prove that the following

generalization holds :

me C2E ( Rnx ( oit ])n C ( R "x[oI]) s .t .

( D ut - Du so in R "x ( o ,t)

(2) ucxio ) EO fxe Rna 1×12

(3) ucxit ) s A e

Then we 0 in Rnx [ o ,T] .

corollaryLet me th ( Rnxco ,T])nC(R"x[oF]) be

such that F A > o,

a > o : lucx ,t)1 e A eakt.

Then lmcx , the supnlucx ,o)l + T . mp1 ut - Dul

,

R"x[oI]xe R R "xGiT]

Before provingthe corollary observe that

the bound is optimal .

.take ucxit ) = ttt

,

then Mt - In =L and hcxp ) =L and the

corollary gives little ttt which is the best

possible estimate ( equality for t=T ! )

Proof of the corollaryWe may suppose that mp lucx ,o)|Ec and

×eRh

that sup 1M£ - Dale C otherwise there

G ,t)eR"x[ of ]

is nothing to prove .

let us consider the

functions

w± cx, ⇒

i-ygplucxidlttj.znsggp.yglut-bultucxitfwehovei.f@Dt.bw±=mplUt-but if - Du 30 in IR"xG,t]IR "×[oI] [±l×' ° ) =

sandlucxio ) I Iueyo ) zo in R

"

Wtcxrt ) >,

- |µc×,t ) 1 >,

- Aea""

2

in IR "×[°iT]

This is to say that W : = - W± satisfies the

assumption of Remark 4 which gives weo

By w± 30 we have

lucxit) Issfeplncaolltntansfpnjht' " I D

Remark : By the corollary above uniquenessand continuous dependence on initial date for

ucxit ) i

fix:i÷¥*Fa > 0 : lucxit ) 1=0 ( eakr )

Backward Cauchy Problem

We observe here that there is lock of continuous

dependence w . r.

t. initial date for the problem :

[ at -

uxx= ° 112×(-6/0)

eecxio ) = gcx ) R.

consider uecxit ) = E I %2sin(x/e) .

It solves

@e)+

- @e) ××=o in Rxtx ,° )

| mecxio ) = esinfy ) in R

E sin ( %) -00 uniformly in R,

while for Tso

HUGH = E etttt → + a

C ( Dvx C- To ] ) E→o

Again a consequence of the time Irreversibility .