LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS DONE RIGHT KYLE ORMSBY

Transcription

1 LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS DONE RIGHT KYLE ORMSBY INTRODUCTION TO THE PROBLEM Considr a continous function F : R n R n W will think of F as a vctor fild, and can think of F x as a vlocity vctor positiond at x R n Our goal is to find a path γ : R R n which starts at a point p R n so γ0 p, our initial condition and, for ach tim t R, satisfis th diffrntial quation γ t F γt If γ x, x 2,, x n, thn w may rwrit as th systm of scalar quations x F x,, x n x 2 F 2 x,, x n x n F n x,, x n W may think of this systm as th tim-drivativs of n dpndnt variabls simultanously satisfying n position-dpndnt quations A solution γ is thn a flow lin, th path of a particl pushd around by th vctor fild F so that its vlocity matchs F at any givn point Exampl As an xampl, considr th vctor fild F : R 2 R 2 taking x, y to F x, y x, y If our initial position is p, p 2, thn w ar sking γ : R R 2 taking t to γt xt, yt satisfying x0 p, y0 p 2, x t xt, and y t yt Figur dpicts th vctor fild F by drawing scald down vrsions of th vctors F x, y at a sampling of points in th squar [0, 2] [0, 2] Th curv drawn within th figur is a solution satisfying th initial condition γ0, This particular systm of diffrntial quations may b solvd on variabl at a tim sinc x only dpnds on x and y only on y Indd, sparation of variabls quickly givs th solution x p t, y p 2 t W may thn not that xy p p 2 is a constant, so th curv tracd by γ is a hyprbola Th abov xampl fits into a class of xampls of th form x a x + a 2 y + b y a 2 x + a 22 y + b in which th a ij and b k ar all scalar constants This may thn b rphrasd in matrix form as x a a y 2 x b + a 22 y b 2 a 2 This admits th furthr gnralization to n variabls x Ax + b

2 FIGURE Th vctor fild F x, y x, y and a flow lin with initial position, whr x a a 2 a n b x 2 x, A a 2 a 22 a 2n, b b 2, x n a n a n2 a nn b n and x is th column vctor consisting of th tim drivativs of x,, x n This is a vry spcial sort of diffrntial quation in which th vctor fild F is an affin transformation If w spcializ yt furthr and tak b 0 so that F is linar, thn w arriv at our currnt objct of study Dfinition 2 A homognous linar systm of diffrntial quations is on of th form whr x x,, x n and A is an n n matrix x Ax Rmark 3 Whn b 0, th litratur calls th systm x Ax + b an inhomognous linar systm of diffrntial quations From our point of viw, it would b nicr to call this an affin systm of diffrntial quations, but w shall quail in th fac of cnturis worth of tradition In ths nots but prhaps not lswhr! w shall assum that all linar systms ar homognous unlss xplicitly notd othrwis Exampl 4 Hr is on of thos patntly ridiculous practical xampls that should nonthlss illustrat th typs of problms which fit into th framwork of linar systms of diffrntial quations: A picklmakr has two brin tanks, on containing 00 gallons of brin, th othr containing 200 gallons of brin A systm of pips and pumps conncts th tanks so that th following procsss happn: a frsh watr ntrs th first tank at a rat of 20 gallons pr minut, b solution movs from th first tank to th scond at a rat of 30 gallons pr minut, on who maks pickls 2

3 c solution movs via a sparat pip from th scond tank to th first at a rat of 0 gallons pr minut, and d solution is dumpd from th scond tank at a rat of 20 gallons pr minut Not that total solution volum is consrvd in this systm If th first tank contains x pounds of salt and th scond tank contains y pounds of salt, thn th dvlopmnt of th systm may b dscribd by th following systm of quations: x 30 x y 200 y 30 x y 200 Simplifying fractions and translating into matrics, w hav x 3/0 /20 x y 3/0 3/20 y Th picklmakr rsts this systm at midnight on January, 206 so that th first tank contains 0 pounds of salt and th scond contains 5 pounds of salt If h lts th systm run for a yar, how much salt will b in th tanks? 2 THE OPERATOR NORM AND MATRIX EXPONENTIATION In ordr to hlp th picklmakr, w ar going to dvlop a thory of matrix xponntiation In ordr to nsur that ths mthods ar lgitimat th picklmakr apprciats our hlp but dmands authnticity, w will tak a dtour through th world of oprator norms Rcall that th complx xponntial function taks any complx numbr t to t k! tk If L : R n R n is a linar transformation, dfin L 0 id, th idntity linar transformation, and inductivly dfin L k L L k for k > 0 Thus L L, L 2 L L, L 3 L L L, tc Dfinition 5 Th xponntial of a linar transformation L : R n R n is L : k! Lk Som commnts on intrprtation ar ncssary, som trivial, som dp First, not that ach factor /k! is a scalar, and if M is a linar transformation and λ is a scalar, thn λm taks v to λmv; this givs maning to th trms k! Lk Also rcall that w ar prfctly comfortabl adding togthr finitly many linar transformations: L + M : v Lv + Mv Thus for any intgr N, th sum N k! Lk maks sns as a linar transformation What, though, is mant whn w lt N? In ordr to mak sns of such a limit, w will nd a notion of distanc btwn linar transformations This notion is providd by th oprator norm In ordr to motivat it, rcall th following lmma which w provd arlir in th trm 3

4 Lmma 6 For any linar transformation L : R n R m, thr xists a constant c R such that for all v R n, Lv c v By taking th smallst possibl such c, w masur th maximal factor by which L dilats vctors; this is th oprator norm, dfind formally blow Dfinition 7 Th oprator norm of a linar transformation L : R n R m is It follows immdiatly that for any v R n, L : inf{c 0 Lv c v for all v R n } Lv L v In your homwork, you will prov that L may b computd in th following ways as wll Proposition 8 For any linar transformation L : R n R m, L sup{ Lv v } sup{ Lv v } sup{ Lv / v v R n {0}} Th oprator norm is, in fact, a norm, maning that it satisfis th following proprtis as wll Proposition 9 For any linar transformations L, M : R n R m and any scalar λ R, i L 0 with quality if and only if L 0, ii λl λ L, iii L + M L + M Proof Th first proprty is obvious, and you will prov th scond on in your homwork For part iii, obsrv that for any v R n, L + Mv Lv + Mv Lv + Mv L v + M v L + M v Sinc L+M is th gratst lowr bound on all c 0 such that L+Mv c v and L + M is such a c, w s that w must hav L + M L + M Whnvr w hav a norm, w gt a notion of distanc This is xactly how w dvlopd th notion of distanc in R n Indd, with th following dfinition and th proprtis listd blow in Proposition, w hav all th ncssary notions ndd to talk intlligntly about convrgnc of th sris dfining L Dfinition 0 Th oprator distanc btwn two linar transformations L, M : R n R m is dl, M : L M Proposition For any linar transformations L, M, N : R n R m w hav i dl, M 0 with quality if and only if L M, ii dl, M dm, L, and iii dl, N dl, M + dm, N Proof Ths proprtis follow immdiatly from thir analogus in Proposition 9 4

5 Additionally, th oprator norm has an xtra fatur that will prov spcially usful in our analysis of th xponntial on linar oprators; namly, th oprator norm is submultiplicativ Thorm 2 For any linar transformations L : R n R m and M : R k R n, L M L M Thinking in trms of dilation factors, this thorm should sm fairly obvious Th proof is a on-linr using Proposition 8 Proof W hav L M sup LMv sup L Mv L sup Mv L M v v v Hr th first inquality follows from our old obsrvation that Lw L w with w Mv Th othr manipulations ar standard W now com upon a surprising proprty of convrgnc with rspct to th oprator distanc W will only sktch th proof sinc a full tratmnt dpnds on Cauchy compltnss of th oprator mtric spac, and w can only do so much in th short tim w v bn givn Proposition 3 Lt L k : R n R n k 0 b a squnc of linar transformations If k 0 L k convrgs as a squnc of ral numbrs, thn k 0 L k convrgs with rspct to th oprator distanc Sktch of proof W would lik to show that thr is a linar transformation L : R n R n such that for all ε > 0 thr xists N N such that n L k L < ε whnvr n > N Lacking a candidat for L, w will instad show that n n L k L k < ε whnvr n, n ar sufficintly larg In othr words, th squnc of partial sums gts arbitrarily clos to itslf for larg indics This is in fact an quivalnt condition by Cauchy compltnss of th oprator mtric spac, but w will not prov it Tak ε > 0 By hypothsis, k 0 L k convrgs, so by Cauchy compltnss of th ral numbrs, w may choos N N such that whn n n > N, For th sam n, n, n L k ε > L k n n L k n L k n kn+ L k so L k convrgs with rspct to th oprator distanc n kn+ n kn+ L k L k < ε, W ar now wll-poisd to prov convrgnc of th oprator xponntial 5

6 Thorm 4 For any linar transformation L : R n R n, L k! Lk convrgs to a linar transformation Proof By Proposition 3, it suffics to prov that k 0 k! Lk convrgs By submultiplicativity Thorm 2 and Proposition 9ii, k! L k But k! Lk k! L k L convrgs so th comparison tst for sris implis that k 0 k! Lk convrgs and w ar don Implicit in our sktch proof of Proposition 3 is th fact that linar transformation R n R m form a complt mtric spac undr th distanc function inducd by th oprator norm In othr words, Cauchy squncs rlativ to th oprator norm of linar transformations convrg to linar transformations Givn Thorm 4, this mans that L is a linar transformation R n R n for any linar transformation L : R n R n As such, L has a matrix, which w provid notation for blow Dfinition 5 If L : R n R n is a linar transformation with matrix A, thn w lt A dnot th matrix associatd with L and call it th xponntial matrix of A Of cours, A has a sris dscription as A k! Ak whr A k dnots itratd matrix multiplication But w must b carful in how w intrprt this sris as it stands, it only maks sns whn w translat back to linar transformations and considr convrgnc rlativ to th oprator norm It is natural to ask for mor Qustion 6 Suppos L k is a squnc of linar transformations R n R m with associatd squnc of m n matrics A k Furthr suppos that L k L as k rlativ to th oprator norm Lt a ij k dnot th ij-ntry of A k, and lt a ij dnot th ij-ntry of th matrix A associatd with L Undr ths circumstancs, is lim k aij k aij? Answr 7 Ys It turns out that all norms on finit dimnsional vctor spacs induc th sam topology, and thus hav th sam convrgnt squncs W will rturn to ths idas at th nd of ths nots, tim prmitting W now comput a coupl of asy xampls of xponntial matrics a 0 a Exampl 8 Suppos A is a 2 2 ral matrix Thn A 0 b k k 0 0 b k chck this and A a k! Ak k /k! 0 a 0 0 b k /k! 0 b 6

7 Similarly, if A is an n n diagonal matrix with diagonal ntris a,, a n, thn A is a diagonal matrix with diagonal ntris a,, an Exampl 9 An n n matrix A is calld nilpotnt if thr is a positiv intgr l such that A l 0 Hr 0 mans th n n matrix with all 0 ntris If A is nilpotnt, thn th infinit sum giving A has only finitly many nonzro trms and may b computd xplicitly by just adding up thos trms, i, A l k! Ak thn For instanc, if Thus A , A and A A I + A + 2 A SOLVING LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS WITH EXPONENTIAL MATRICES 3 Th main thorm Lt M n n R dnot th st of n n ral matrics For a fixd A M n n R, w may dfin a matrix-valud function γ A : R M n n R t At whr At is th scalar multipl of A by t Thus At k! Atk k! Ak t k Hr w ar using th fact that scalar multiplication commuts with matrix multiplication Idntifying M n n R with R n2 why can w do this?, w may viw γ A as a path As such, it maks 7

8 sns to diffrntiat: γ At d dt At d dt k A k! Ak t k k! Ak d dt tk k k! Ak t k k! Ak t k k k! Ak t k A At Svral stps hr rquir justification First and formost, w hav not justifid th third quality, which claims that diffrntiation commuts with infinit sris This is a dpr point that w will rturn to latr Th final stp also dsrvs attntion Obsrv that k rangs from to, but all of th apparancs of k in th sum ar in fact k s This rcovrs th xponntial function Givn this computation, w may prov th following thorm Thorm 20 Lt A M n n R and lt p 0 R n Thn th function x : R R n t At p 0 is a solution to th diffrntial quation x Ax with initial condition x0 p 0 Proof First not that by Exampl 8, 0 I whr 0 is th n n matrix of all 0 s Now obsrv that x0 A 0 p 0 Ip 0 p 0, so x satisfis th initial condition Finally, w may comput x d dt At p 0 A At p 0 Ax, as dsird Exampl 2 Considr th matrix A of Exampl 9 It is asy to comput At I + At + 3t 4t + 9t2 2 A2 t 2 0 6t 0 0 As such, w can us Thorm 20 to solv th diffrntial quation x 3y + 4z y 6z z 0 8

9 FIGURE 2 A vctor fild and flow lin givn by xponntiation of a nilpotnt matrix In particular, if our initial condition is γ0,,, thn γt At + 7t + 9t 2, + 6t, is a flow lin for this systm Figur 2 sktchs th vctor fild A x y and th flow lin γ z 32 Proprtis of th matrix xponntial W will us th following thorm to comput mor complicatd xponntial matrics and thus solv mor complicatd linar systms of diffrntial quations Thorm 22 If A, B M n n R and AB BA, thn A+B A B Proof Th ssntial obsrvation is that whn A and B commut, th binomial thorm applis so that k k k A + B k A l B l k k! l l!k l! Al B l k Hnc l0 A+B A + Bk k! k k! l0 l0 l! Al r+sk 9 l0 k! l!k l! Al B l k k l! Bl k r! Ar s! Bs

10 Obsrv that in this final doubl sum, th pairs r, s rang through all of N N xactly onc Thus laving som algbraic vrifications to th radr w may writ A+B r! Ar s! Bs r! Ar s! Bs A B, as dsird r0 s0 Rmark 23 Th abov thorm is gnrally fals if A and B do not commut! r0 Corollary 24 For all A M n n R, A is invrtibl with A A I Proof W hav A A AA, so Thorm 22 implis that I 0 A A A A Exampl 25 Lt s us this thorm to comput th xponntial of a 0 0 b + and 0 0 Thus a 0 0 b 0 0 a b b 0 0 a 0 0 b b s0 a b First not that a 0 a b a 0 Sinc is diagonal, w hav a a 0 0 a Sinc 2 0 b 0, w hav 0 0 Thus w may conclud that a b Mor gnrally, Thus th diffrntial quation 0 b 0 0 I + 0 b 0 0 b 0 a 0 b a 0 a a b 0 0 a a b t at at bt 0 at x a b x y y 0

11 FIGURE 3 Vctor fild and flow lin whn a b, p /4, and p 2 /4 with initial condition p, p 2 at tim 0 has solution a b t p p 2 at at bt p p 0 at at + p 2 at bt p 2 p 2 at x In Figur 3, w st a b Th plot dpicts th vctor fild along with a flow lin 0 y passing through /4, /4 at t 0 Explicitly, this solution is givn by t t /4 + t t/4, t /4 33 Priodic matrics Exampl 26 W will now considr a systm of diffrntial quations w hav sn bfor: x y y x W hav alrady sn that th solutions to this systm ar circls Dos th matrix xponntial rcovr ths solutions? W may r-xprss th abov systm as x y Obsrv that , 0 0 x 0 y , Sinc th fourth powr of this matrix is th idntity matrix, th pattrn now rpats, i, if A 0, thn A 0 4k+j A j whr k N and j is 0,, 2, or 3 Thus w may comput th xponntial At k k! Ak t k 2k! t2k k 2k+! t2k+ cos t sin t sin t cos t k 2k+! t2k+ k 2k! t2k

12 FIGURE 4 An llips as a solution to a homognous linar systm of diffrntial quations? Th sums hr look complicatd, but thy ar just rcording th various ± s, 0 s, factorials, and powrs of t which appar in th summation If at tim 0 w ar at th point r, 0, thn Thorm 20 tlls us that our systm has solution t At r 0 r cos t r sin t This prcisly dscribs a circl of radius r cntrd at th origin, which matchs our old solution of this problm Qustion 27 Do you think thr ar homognous systms of linar quations whos solutions ar llipss? If so, how might you form thm? If this sms mystrious right now, com back to this qustion aftr you hav larnd what a chang of basis is in Linar Algbra /2 0 Lt M which has invrs M /2 x Figur 4 plots th vctor fild MAM x y chck this! and considr th systm and a solution to this systm x y MAM y passing through, t t 0 givn by t cost + 2 sint, 5 sint This should sm quit tantalizing 34 A linar algbraic approach to th gnral 2 2 cas Lt A b an n n matrix and suppos A QBQ for Q, B n n matrics with Q invrtibl Thn A 2 QBQ QBQ QBQQ BQ QB 2 Q sinc QQ I Similarly, for any k 0, A k QB k Q 2

13 Thus At k! Ak t k k! QBk Q t k Q k! Bk t k Q Q Bt Q As such, if w can find B and Q such that Bt is computabl and A QBQ, thn w will b abl to comput At In th 2 2 cas, th following thorm prmits us to do just this a b Thorm 28 Lt A b a 2 2 matrix Thn A QBQ whr B is of th form c d λ 0 0 λ 2 or λ 0 λ Hr th ral or complx numbrs λ i ar th roots of th charactristic polynomial of A: x 2 a + dx + ad bc Proof This is th 2 2 cas of a powrful thorm from linar algbra about so-calld Jordan canonical form of matrics W will not commnt on th proof hr Rmark 29 As statd, th thorm dos not tll us anything about how to find Q In th first cas, λ 0 whr B this is rlativly simpl In this cas, it is possibl to find a basis of R 0 λ 2 of 2 ignvctors of B An ignvctor of B is a vctor v such that Bv λv for som scalar λ; th scalar λ 0 λ in such an quation is calld an ignvalu Th ignvalus of ar prcisly λ 0 λ and λ 2 2 chck this! If v and v 2 ar linarly indpndnt vctors such that Bv λ v and Bv 2 λ 2 v 2, thn w may tak Q to b th matrix with columns v and v 2 λ In th scond cas, in which B, th matrix Q is constructd in a diffrnt fashion 0 λ Th abov mthod dos not work bcaus B has only on ignvalu, λ, and is not alrady diagonal, and thus cannot hav a basis of ignvctors Whil it is not particularly complicatd, w will not study this mthod in ths nots 0 Exrcis Us ths mthods to comput th xponntial matrix of You will find λ 0 i, λ 2 i, so that Bt it it Aftr finding an invrtibl matrix Q such that 0 QBQ and computing At Q Bt Q you will b confrontd with a matrix of th form 2 it + it i 2 it it i 2 it + it 2 it + it Rcalling th idntity it cost + i sint you should s how this matrix simplifis to prvious computation of At Exampl 30 Considr th systm of diffrntial quations x 4x + 2y y 3x y 4 2 which is inducd by th matrix A This matrix has charactristic polynomial 3 x 2 3x 0 λ + 2λ 5 3

14 with roots λ 2, λ 2 5 Thus 2 0 B 0 5 W can comput Q and Q to b 2 Q 3 so At Q Bt Q 7 and Bt and Q 2t 0 0 5t /7 2/7 3/7 /7 2t + 6 5t 2 2t + 2 5t 3 2t + 3 5t 6 2t + 5t Th solution to th systm of diffrntial quations with initial conditions x0 7 and y0 0 is thus xt 2t + 6 5t 2 2t + 2 5t 7 yt 7 3 2t + 3 5t 6 2t + 5t 2t + 6 5t 0 3 2t + 3 5t It would b nic to additionally undrstand th curv dscribd by this solution To driv this, first not that th systm givn by B with initial condition p, p 2 has solution xt, yt p 2t, p 2 5t In this cas, x 5 y 2 p 5 p 2 2 which givs us a handl on th curv dscribd Th nxt thing to obsrv is that γt is a solution to th systm givn by B if and only if Qγt is a solution for th systm givn by A Why is this th cas? Hnc th solution curvs for A diffr from th ons for B via th linar chang of coordinats Q Now would b an opportun momnt to rturn to Qustion 27 and to considr what curvs ar dscribd whn a linar chang of coordinats is applid to a circl 4 THEORETICAL ODDS AND ENDS Givn th matrix xponntial s facility in solving diffrntial quations, w can now s why it was important to prov its convrgnc Rcall, though, that w only provd convrgnc with rspct to th oprator norm? Is this th sam thing as convrgnc with rspct to a mor traditional norm, lik th on inducd by considring n n matrics as points in R n2? W will sktch th proof of a thorm which, lik Voltair s Pangloss, assurs us that w liv in th bst of all possibl worlds Loosly spaking, it tlls us that all topologis on finit dimnsional ral vctor spacs which ar inducd by norms ar quivalnt Thus convrgnc with rspct to on norm is quivalnt to convrgnc with rspct to all othr norms In ordr to put mat on this statmnt, lt us rcall th dfinition of a norm Dfinition 3 A norm on R n is a function : R n R satisfying th following proprtis: for all x R n, x 0 with quality if and only if x 0, 2 for all x R n and λ R, λx λ x, and 3 for all x, y R n, x + y x + y Th Euclidan norm on R n is, of cours, a norm A priori, th oprator norm is dfind on th st of linar transformations R n R n Using our dictionary btwn linar transformations and matrics, though, w s that ach linar map R n R n corrsponds to a uniqu point in R n2 Thus it maks sns to think of th oprator norm as a norm on R n2 4

15 Whnvr w hav a norm, it inducs a distanc function d x, y x y, and w can dfin opn balls and a topology with rspct to this distanc function W can dfin limits of squncs with rspct to this norm via th usual ε-n formalism Dfinition 32 Two norms and ar quivalnt if thr ar positiv constants c and C such that c x x C x Rmark 33 Th apparnt asymmtry in this dfinition is an illusion Givn th chain of inqualitis c x x C x, w gt C x x c x for fr As such, quivalnc of norms is in fact an quivalnc rlation on th st of norms chck this! Th significanc of this dfinition is mad apparnt by th following thorm Thorm 34 If two norms and on R n ar quivalnt, thn a squnc a k is convrgnt with rspct to if and only if it is convrgnt with rspct to, and 2 a subst of R n is opn with rspct to if and only if it is opn with rspct to Th proof follows dirctly from th dfinitions chck this! and w will not prsnt it hr W can now proprly apprciat our Panglossian thorm: Thorm 35 For n 0, any two norms on R n ar quivalnt Corollary 36 Th matrix xponntial convrgs with rspct to all norms on R n2 Euclidan on including th usual Sinc it will b ssntial in our proof of th thorm, lt us rcall th Bolzano-Wirstrass thorm which you larnd in som for in Math 2 bfor procding Thorm 37 Bolzano-Wirstrass Evry boundd squnc in R has a convrgnt subsqunc Hr w man convrgnc with rspct to th standard absolut valu on R For x R n, w dfin x x + + x n ; it is calld th l norm or th taxi-cab norm think of th distanc your taxi must tak to travl btwn points in Manhattan Chck that is in fact a norm W will nd th following nhancd vrsion of th Bolzano-Wirstrass thorm Thorm 38 Bolzano-Wirstrass on R n with rspct to th norm In R n, any -boundd squnc has a -convrgnt subsqunc Proof Lt {x k } b a -boundd squnc in R n Lt x j k b th j-th coordinat of x k, j n Not that x k x k, so {x k } is a boundd squnc in R and thus has a convrgnt subsqunc {x lk } by Thorm 37 Now considr th squnc {x 2 l k } W again hav that x 2 l k x lk is boundd, so thr is a convrgnt subsqunc x hlk If n 2, w ar don with convrgnt subsqunc x h lk, x 2 h lk chck this! Othrwis, kp rpating th procss until it has bn don to all of th coordinats Rmark 39 It is a consqunc of Thorm 35 that th Bolzano-Wirstrass thorm on R n holds with rspct to any norm W ar now prpard to prov our main thorm 5

16 Proof Sinc quivalnc of norms is transitiv, it suffics to show that an arbitrary norm is quivalnt to W first chck that thr is a constant c such that c x x for all x R n Obsrv that x x i i, so th triangl inquality implis that x x i i If m max{ i }, thn w furthr hav x x i m m x Hnc w may tak c /m and driv that c x x It rmains to show that thr is a positiv constant C such that x C x for all x R n Suppos for contradiction that no such C xists In that cas, for ach k N w can find x k R n such that x k > k x k Lt y k x k / x k Thn y k so th squnc {y k } is boundd in th norm and thus has a subsqunc y lk convrging in to an lmnt y Sinc y lk, w also hav y chck this! and, in particular, y 0 Not, though, that y lk < /l k, so y 0, whnc y 0, a contradiction Thus thr must b som C > 0 such that x C x for all x, complting our proof 6