Tutorial 9 : Selected

problems

of Assignment 9

Leon Li

21/11/2018

Recall the Initial Value Problem :

Def An Initial Value Problem consists os the following

{¥= th

'× '

( zvp )

XHD = Xo

where f : R 's [ to - a. total × [ x.- b. Xotb ] → R is a function

.

An IVP is Uniquely ) solvable for a' ECO , a) if there exists ( unique )

XH )

:[to -

a'

, total ] → [ xo - b ,Xotb ] st . × ( t ) is CL and solves IVP :

←alto ) Ibko )

{XHKJH

. x ),

f

te I alto )

Xto ) = Xo

Thm ( Picard . Lindelof ) Given an IVP as above,

�1� If

FEAR) satisfies a Lipschitz condition ( Uniform in t ) ,

i.e.

7- L > O st .

V. ( t.xd.lt , xD ER,

HH, xD - FH .×< HE Llxixd

then IVP is uniquely solvable for O< a 's min { a. Fi,

't },

M :=s£PHHxH( assuming M > O )

�2� If in addition TECKIR ),

-71<>-1,

then

XCHECK"

( Ia .to ) )

Q1 ) l HW 9,

Q 5)

Using

the

perturbation

of identity ,

prove

①

for Osa 's min { a,

b-

,

-13Moe sup I -5ft

. xosl

Mot Lb L ,

Where

t.CIa.toI

PT ) Recall that by Prop .3.11

, it suffices to solve the integral equation

Xlt ) = Xo tf ! Tls,XIs ) ) Is

,

where X : Ia .Ho ) → Ib l Xo ) is continuous

.

Equivalently : Xlt ) t ( - Xo - SIKHS. xcss ) - fax odds ) = Is

.

#ds

Applying the perturbation of identity with ( X. Hill ) = ( Gto - a'i total

, Hello)

Io :X → X by Back D= Xlt ) t C- Xo - f Is . xcss ) - fax odds )

= ( It E) CXHD,

where ZCXHD = - Xo - Sit ( Tls . xcss ) - As

. xo ) )ds

Let Xoth, Yott ) EX be defined as Xolttxo

,htt E Iad to )

{yous -0

then 8 Hoth ) = You ).

Checking I :X → X is a contraction : V-x.CA,

KH ) E X,

V

TE Ia .

to)

II Kita ) - Exits ) I =/ 4THG. x. on - flex . asDds I S Sit.

Lil Hss - xdsslds

E L . llxi-xdts.lt - Tol Ska' ) . Hx .

-

kilos = 811k -

Hhs, where 8 La 's I

.

'

.HIM - Elwha E 811k - Hls

.

'

. By the perturbation of identity .-V r > o

,R It - Hr

.

✓ Y

theWot,

It !

XHIEBr

Hoths - t .

BATHE

Yet)

In particular ,choose r=b

,

12=11 - La 't b,

checking 4th 's fttoflsxdds E

Bplyoth) i ft E Ia .to)

,

ly th - yo Hit = I Sfo -56 , xD dsl S

Mo. It - tot E Mia

'sCl - La

'

) b -

- R

I

( since a 'sM¥4⇐

Moa't Lba

'

Cb ⇐

Moa's H - La

'

)b)

Therefore,

I ! XTHEBbcxoth) s . t .

faith ) = Stotts ,Xo ) ds

i. e . I !XHI: Ia ,

to ) → Ibl Xo ) satisfying the integral equation .

- FA

Q2 ) C Hwa,

Qb ) Prove ②.

Sol ) Induction on kzoi FE d '

CR ) ⇒ XH EC" "

I Ia ,Ho ))

k=O : Follows fwm the integral equation Xltkxot Is , Xis Ddt

by the FundamentalTheorem of Calculus I FTC ) .

Suppose the statement holds for k = K,

then for k - Kt I,

assume FECK "

( R ),

then TE CYR),

hence by Inductive hypothesis

Xlt ) E C' " '

( Ia . Ho ) ).

Therefore.

TH.

xth ) : Taito ) → IR is Ck "

.

then Xlt ) = Xo -1 Stuffs . xcsslds E C' "

( Ia . Hot ) by FTC

."

. By Induction.

Hk > O,

FE d '

CR ) ⇒ XHIEC" "

I Ia ,Ho )) - FA

Q 3) I HW 9. Q 9) Show that the following integral equation

has = It IS !,+µ handy ,where he CE - 2. D

has a unique solution.

Moreover,

show that h is nonnegative .

Sol : Define ( X ,d) = (E he CI - 2,13 I hlxszo

,Axe I -2.133

,

Hello)

Define Ti X → CE -123 by th ) Cx ) =It ! ftp.hlyldy

then Fx E I - III, th ) Ix ) 2 I 20

. .

'

.

T : X → X

Also,

T is a contraction : tf hi,

h.

E X,

V x EE - I,

I ],

I Th )K ) - Thd KY =

I ÷ I,

¥⇒ . ( him - him ) dy I

E IT . till h.

- hello . ( I - I.

- It ,

= ¥ 11h,

- halls

:.

H Th.

- Thall E Y 11h

. - hello

,where f- ¥51

.

'

. By Contraction Mapping Principle . T has a unique fixed point .

I . e - the above integral equation has a unique non-

negative solution.

- HI