uniform convergence etc

1.Preliminaries: uniform convergence Recall the following denition. Denition 1.1. A sequence (xn)n2NRis said to converge to L2Rif for each  >0, there exists N2Nsuch thatjxnLj<whenevernN. This denition is really the starting point for all of Calculus. It invests phrases like `as
xtends to zero' with a precise meaning and therefore allows us to speak about things like continuity, dierentiation, and integration in a rigorously logical and not just intuitive fash- ion. A fundamental property of real numbers is that of completeness . It can be formulated in a number of ways (e.g. every set of real numbers that is bounded above has a least upper bound), but the best (arguably) uses the following additional notion. Denition 1.2. A sequence (xn)n2NRis said to be Cauchy if for each>0, there exists N2Nsuch thatjxnxmj<whenevern;mN. Observe that a convergent sequence is necessarily Cauchy: if ( xn) converges to Land>0 is given, then we can choose N2Nsuch thatjxnLj<=2 whenever nN. Therefore if two indices n;m2Nareboth larger than N, we get jxnxmj=j(xnL)(xmL)jjxnLj+jxmLj< 2+ 2=: Hence (xn) is Cauchy. The completeness property of Ris the converse assertion: Completeness Axiom. Every Cauchy sequence of real numbers converges. This property of Rgives us a way of showing sequences converge without actually knowing anything about the limiting. Here we will need a notion of convergence for sequences of functions rather than real numbers. To this end, let us x a subset SR. For anyf:S!R, we dene kfk=kfkS:= sup x2Sjf(x)j: Of course it can happen that kfk=1. Consider for instance S= (0;1),f(x) = 1=x. To keep things more manageable, we restrict attention to the set B(S) =ff:S!R;kfk<1g ofbounded real-valued functions on S. Note thatB(S) is a vector space over R. We leave the reader to verify Proposition 1.3. k
kSis anorm onB(S). That is, for every f;g2B(S)and everyc2R, we have (positivity)kfk0with equality if and only if f(x) = 0 for everyx2S. (homogeneity)kcfk=jcjkfk. (triangle inequality) kf+gkkfk+kgk: In other words,k
kShas essentially the same properties on B(S) that the absolute value function has on R. In particular, it gives a way to measure the distance between functions onS. That is, given f;g2B(S), we can declare dist( f;g) =kfgkto be the distance betweenfandg. And whenever you have a notion of distance, there is a corresponding notion of convergence. 1 Denition 1.4. A sequence (fn)n2N2B(S)is said to converge uniformly to a function g2B(S)onSif for every  >0, there exists N2Nsuch thatkfngk<  whenever nN. In other words, ( fn) converges uniformly to fif lim n!1kfnfk= 0. The qualier `uni- formly' is used here because there are other notions of convergence for sequences of functions, useful in other contexts, and they are fundamentally dierent than the one specied here. One can also rewrite the denition of Cauchy sequence of real numbers to come up with a denition of `uniformly Cauchy' sequence of bounded functions. We leave doing this to the reader. Let us remark before continuing that if ( fn) converges uniformly to fonS, then we have thepointwise convergence lim n!1fn(t) =f(t) for every t2S: The converse, however, is not true. Fairly straightforward counterexamples show that one can havefn(t)!f(t) for every individual t2Syet still not get that fn!funiformly on S. There are three fundamental facts about uniform convergence. The rst is that bounded functions are complete with respect to uniform convergence. Theorem 1.5. A uniformly Cauchy sequence (fn)B(S)of functions is uniformly conver- gent. Proof. We rst produce a candidate for the limit function g:S!R. Let>0 be given, and use the fact that ( fn) is Cauchy to obtain N2Nsuch thatn;mNimplieskfnfmk<. Then for any xed x2S, we have as a consequence that jfn(x)fm(x)jkfnfmk< whenevern;mN. That is, ( fn(x))n2Nis a Cauchy sequence of real numbers . It follows then from completeness of Rthat (fn(x)) converges. We let g(x) = lim n!Nfn(x). Since we can do this for each x2S, we obtain a function g:S!R. Next we show that gis bounded (i.e. g2B(S)). Taking= 1, we again use the fact that (fn) is uniformly Cauchy to obtain N2Nsuch thatn;mNimplieskfnfmk<1. In particular, nNimplieskfNfnk<1. Thus ifM=kfNk<1, we get kfnk=k(fnfN) +fNkkfnfNk+kfNk<1 +M for allnN. Thus, for any x2S, we have jg(x)j=jlimfn(x)j= limjfn(x)j1 +M; becausejfn(x)jkfnk<1 +Mfor allnN. Thuskgk1 +M <1, which proves that gis bounded. Finally, we argue that fnconverges uniformly to g. Let >0 be given. Applying the fact that (fn) is uniformly Cauchy one last time, we take N2Nsuch thatn;mN implieskfnfmk<=2. Now ifx2Sis any particular point, we have already shown that limn!1fn(x) =g(x). So we can also nd N02Nsuch thatjfn(x)g(x)j<=2 whenever nN0. Therefore, if nNandmmaxfN;N0g, we estimate jg(x)fn(x)j=j(g(x)fm(x))+(fn(x)fm(x))jjg(x)fm(x)j+jfn(x)fm(x)j< 2+ 2=: 2 In fact, since N(as opposed to N0) was chosen independent of x, we get kfngk= sup x2Sjg(x)fn(x)j<: whenevernN. This proves that ( fn) converges uniformly to g.  The second fundamental fact about uniform convergence is that it cooperates well with continuity. Theorem 1.6. Suppose that (fn)B(S)is a sequence of continuous functions converging uniformly to f:S!R. Thenfis continuous. To see that this theorem is special, note that it is false if we replace `continuous' with `dierentiable' in the hypothesis and conclusion. Can you think of a counterexample? Proof. Letx2Sand>0 be given. Since fn!funiformly, there exists N2Nsuch that kfnfk<=3 whenever nN. SincefNis continuous at x, we also have >0 such that jfN(y)fN(x)j<=3 whenever y2Sandjyxj<. Thus jg(y)g(x)j=j(g(y)fN(y)) + (fN(y)fN(x)) + (fN(x)g(x))j  jg(y)fN(y)j+jfN(y)fN(x)j+jfN(x)g(x)j < 3+ 3+ 3 for ally2Ssuch thatjyxj<. Hencegis continuous at x.  Finally, we show that uniform convergence cooperates well with integration. Theorem 1.7. Suppose that (fn)B(S)is a sequence of continuous functions converging uniformly to f:S!R. If[a;b]Sis a closed and bounded interval, then lim n!1Zb afn(t)dt=Zb af(t)dt: Proof. We have Zb afn(t)dtZb af(t)dtZb ajfn(t)f(t)jdtZb akfnfkdt= (ba)kfnfk: Hence, lim n!1Zb afn(t)dtZb af(t)dt(ba) lim n!1kfnfk= 0; sincefn!funiformly on S.  2.The existence and uniqueness theorem for first order ODEs The fundamental fact about ordinary dierential equations is that, under suitably nice circumstances and subject to appropriate initial conditions, one gets unique solutions. Here we will discuss this fact in the particular case of rst order ODEs. The case of rst order systems of ODEs is quite similar and essentially contains all other possible cases. 3 Let us set up the problem before stating any results. We begin with an open set UR2 and a function F:U!R. Given any point ( t0;y0)2Uwe seek solutions to the initial value problem (1) y0(t) =F(t;y(t)); y(t0) =y0: The domain of the function yis not so important here, so we allow ourselves to consider any dierentiable function y:I!Rdened on an open interval Icontaining t0. If such a y satises (1), then we refer to y:I!Ras a solution of (1). Note that dierent solutions can have dierent domains, but the domains of any two solutions must intersect in an open interval containing t0. Theorem 2.1 (Existence and Uniqueness Theorem) .Suppose that F=F(t;y)is continuous onUand furthermore continuously dierentiable with respect to the second variable y. Then for any (t0;y0)2Uthere is a solution y:I!Rof the initial value problem (1). This solution is unique in the following sense: if ~y:~I!Ris another solution, then ~y(t) =y(t) for allt2I\~I. Proof. SinceUis open and ( t0;y0)2U, there exists a closed rectangle R= [t0a;t 0+a][y0b;y 0+b]U: A continuous function on such a rectangle will be bounded, so we have constants A;B > 0 such that jF(t;y)jA;@F @y(t;y)Bfor all (t;y)2R: In particular, by the mean value theorem, we have for all ( t;y 1);(t;y 2)2R jF(t;y 1)F(t;y 2)j=@F @y(t;c)(y1y2)Bjy1y2j wherecis a number between y1andy2. Lemma 2.2. Supposebis a positive number no larger thanb A. LetI= (t0;t0+) andy:I![y0b;y 0+b]is a continuous function. Then the function ~y:I!Rgiven by ~y(t) =y0+Zt t0F(s;y(s))ds is well-dened and satises ~y(t)2[y0b;y 0+b]for allt2I. Proof. The hypothesis on yimplies that ( t;y(t))2Rfor allt2I. This, and continuity of yandF, imply that the right side of the equation dening ~ ymakes sense for all t2I. We have moreover for such tthat j~y(t)y0j=Zt t0F(s;y(s))dsAjtt0jAAb; where the last inequality comes from the hypothesis on .  Continuing to let I= (t0;t0+), with>0 as in the lemma, we invoke the conclusion of the lemma to dene a sequence of functions yn:I!R,n2Ninductively as follows: y0(t)y0; y n+1(t) =y0+Zt t0F(s;yn(s))ds: 4 We will show that ( yn) converges uniformly to a solution of (1). The key step in doing so is our next lemma. Note that k
kmeansk
kIhere. Lemma 2.3. Suppose that <1=2B. Then for all n1, kyn+1ynk1 2kynyn1k: In particular, kyn+1ynk1 2nky1y0k: Proof. For anyn1 andt2I, we have jyn+1(t)yn(t)j=Zt t0(F(s;yn(s))F(s;yn1(s)))dsBZt t0jyn(s)yn1(s)jds Bjtt0jkynyn1kj<Bkynyn1k1 2kynyn1k Sincet2Ion the left side was arbitrary, the rst conclusion of the lemma follows. The second conclusion is obtained by iterating the rst: kyn+1ynk1 2kynyn1k1 4kyn1yn2k


1 2nky1y0k:  From now on we assume that minfa;b=A; 1=2Bgsatises the hypotheses of both of the above lemmas. We demonstrate convergence of the sequence ( yn) be rewriting the nth term as a telescoping sum: yn(t) =y0+nX j=1yj(t)yj1(t): Hence, the limit of the sequence (if it exists) may be written as a telescoping series y(t) = lim n!1yn(t) =y0+1X j=1yj(t)yj1(t): Note here that by the previous lemma, the jth term in this series satises jyj(t)yj1(t)jkyjyj1kC 2j forC=ky1y0k. SinceP1 j=1C 2jconverges, it therefore follows from the Weierstrass M-test that (yn) converges uniformly to some function y:I!R. Uniform convergence implies thatyis continuous, since all of the ynare continuous. We have moreover that Lemma 2.4. For anyt2I, lim n!1Zt t0F(s;yn(s))ds=Zt t0F(s;y(s))ds 5 Proof. Givent2I, we estimate as in the preceding lemmaZt t0(F(s;yn(s))F(s;y(s)))dsBjtt0jkynyk: Uniform convergence of yntoymeans exactly that lim n!1kynyk= 0. Hence the right side tends to zero as n!1 , and the lemma is proved.  Now we claim that y:I!Ris a solution of (1). First of all, y(t0) =y0+Zt0 t0F(s;y(s))ds=y0; soysatises the right initial condition. We further have from the previous lemma that y(t) = limyn(t) =y0+ lim n!1Zt t0F(s;yn1(s))ds=y0+Zt t0F(s;y(s)ds: Applying the fundamental theorem of calculus to the integral on the right side, we therefore obtain y0(t) = 0 +F(t;y(t)) =F(t;y(t)) for allt2I. This proves our claim and concludes the existence portion of the proof. Finally, we turn to uniqueness. Suppose that ~ y:~I!Ris another solution of (1). First we show that yand ~yagree neart0. Lemma 2.5. There exists an open interval JI\~Icontainingt0such that ~y(t) =y(t)for allt2J. Proof. By continuity of ~ y, there is an open interval J~I\Icontaining t0such that j~y(t)y0j<bfor allt2J. That is, ~y(t)2[y0b;y 0+b]. Also, by integrating both sides of ~y0=F(t;~y), we obtain ~y(t) =t0+Zt t0F(s;~y(s))ds for allt2J. Thus we may argue as in the proof of Lemma 2.3 that j~y(t)y(t)j=Zt t0(F(s;~y(s))F(s;y(s)))ds1 2k~yyk: Sincet2Jis arbitrary, this implies that k~yyk1 2k~yyk;which can only happen if k~yyk= 0|i.e. if ~ yyonJ.  To conclude the proof of uniqueness in Theorem 2.1, we let JI\~Ibe the largest open interval containing t0on whichyand ~yagree. By the previous lemma, we know at least thatJis not empty. We assume, in order to reach a contradiction, that J6=I\~I. Under this assumption, we have that one of the two endpoints t1ofJlies inI\~I. For the sake of deniteness, we assume that t1is the righthand (i.e. upper) endpoint of J. By continuity of yand ~y, we have y1:=y(t1) = lim t!t 1y(t) = lim t!t 1~y(t) = ~y(t1) Hencey:I!Rand ~y:~I!Rare also both solutions of (1) subject to the initial condition y(t1) =y1. By Lemma 2.5, it follows that y~yon an open interval J1I\~Icontaining 6 t1. ThusJ[J1is an open interval strictly larger than Jon which ~yy, contradicting the assumption that Jis the largest interval of agreement about t0. We conclude that y~y everywhere on I\~I.  In closing, let us observe that the uniqueness part of Theorem 2.1 ensures us that there is a solutiony:Imax!Rfor which the interval Imaxis as large as possible. To see that this is so, let Imax=[ fJR:Jis an open interval about t0on which (1) admits a solution g be the union of all solution intervals. Then Imaxis an open interval about t0(why?), and we can dene our solution y:Imax!Rat any point t2Imaxby settingy(t) = ~y(t) where ~y:J!Ris a solution whose domain Jcontainst. Since solutions agree on the intersection of their domains, it will not matter which solution ~ ywe use to dene y(t). We will have moreover that y~yon all ofJ, so that in particular y0=F(t;y) holds att. That is,y satises (1) on all of Imax. We have just shown Theorem 2.6 (Existence of solutions with maximal domain) .Under the hypotheses of The- orem 2.1, there exists a solution y:Imax!Rof(1)whose domain Imax contains the domains of all other solutions ~y:I!Rof(1). An important feature of the solution y:Imax!Ris that it persists until its graph `exits' the domain of existence Ufor the righthand side F(t;y) of (1). This can be stated more precisely in terms of `compact sets'. A subset KRnis called compact if it is closed and bounded. So for instance, an interval IRis compact if and only if I= [a;b] where a;b2R. Theorem 2.7. Lety:Imax!Rbe the solution of (1)with maximal domain Imax. IfKU is any compact set, then there is a compact interval IKImaxsuch that (t;y(t))=2Kfor anyt =2IK. The conclusion of this theorem is often summarized by saying that the graph of y: Imax!Ris a curve that is `properly embedded' in U. The proof of Theorem 2.7 depends on a renement of the existence and uniqueness theorem. The gist of the renement is that solutions to (1) vary continuously with the initial condition. In particular, the intervals on which these solutions are dened may be taken to vary continuously with the initial condition. Theorem 2.8 (Existence and uniqueness with stability) .Under the hypotheses of Theorem 2.1, letKUbe a compact subset. Then there exists  >0and a continuous function :K(;)!Rsuch that for any (t0;y0)2Kthe function y: (t0;t0+)!Rgiven byy(t) :=(t0;y0;tt0)is the unique solution of (1). This theorem follows from a slightly more careful version of the proof of existence used for Theorem 2.1. We omit the details here. For purposes of proving Theorem 2.7, the important part of the stability theorem is that it gives a positive lower bound on the `lifespan' of any solution that begins in K. Proof of Theorem 2.7. Suppose that the theorem is false for some compact set KU. WriteImax= (a;b) (note that either or both of aandbcould be innite). Let  >0 be the constant associated to Kin the Stability Theorem. Then taking the closed interval J= [a+;b]Imax, we may choose a point t12ImaxJsuch that (t1;y1)2K, where 7 y1:=y(t1). Thusyis a solution of y0=f(t;y) subject to the initial condition y(t1) =y1. The stability theorem guarantees us that there is another solution ~ y: (t1;t1+)!Rof the same initial value problem. By uniqueness, we therefore have ~ yyonImax\(t0;t0+). By setting, ^y(t) = y(t) ift2Imax ~y(t) ifjtt1j<; we obtain that ^ yis a solves (1) on Imax[(t1;t1+). Since this last interval is not contained inImax, we have contradicted the fact that Imaxis the maximal domain of existence for the solution of (1). So the theorem holds.  3.Asymptotic Behavior of Solutions to Autonomous 1st Order Equations In this section we consider solutions of the initial value problem (2) y0=f(y); y(t0) =y0 wheref:R!Ris aC1function. Dierential equations like the one here, in which the right side does not depend explicitly on t, are called autonomous . Such ODEs are both common in applications and important in theory1 The ODE in (2) is separable and therefore in principle solvable by integration. In prac- tice, however, the integration can be unmanageable and will only give an implicit and fairly unenlightening formula for the solution. Here we take a more qualitative approach to ana- lyzing the problem, and in particular, understanding what happens to the solution as ttends toward the ends of the maximal domain of existence. This is, in pedestrian terms, somewhat akin to plugging information about today's weather into the equations of uid mechanics to try and infer whether one ought to plan picnics in the year 10,000. In this light, it is somewhat remarkable that we will be able to say anything sensible at all. Iff(y0) = 0 then we call y0anequilibrium point of the ODE. In this case, one checks easily that y:R!Rgiven byy(t)y0is the solution of (2). In particular, the maximal domain of existence is all of R, and we have lim t!1y(t) =y0. Iff(y0)6= 0, then things are certainly more complicated. For deniteness' sake, let us suppose from now on that f(y0)>0, and consider the solution y:I!Rof (2) with maximal domain of existence I= (a;b). Lemma 3.1. y0(t)>0for allt2I. Proof. If the assertion is false, then there exists t12Isuch thatf(y(t1)) =y0(t1)0. Since f(y(t0))>0, the intermediate value theorem tells us there exists t2betweent0andt1such thatf(y(t2)) = 0. But then y(t2) =z(t2) wherez(t)y(t2) is a constant solution. So by uniqueness of solutions to initial value problems, we see that y(t) =z(t) for allt2I. In particular 0 <f(y(t0)) =f(y(t2)) = 0, which is a contradiction. So the assertion is true.  Lemma 3.2. limt!by(t)exists (and is possibly 1). Proof. By the previous lemma yis an increasing function. Letting L= supt2Iy(t), we claim limt!by(t) =L. We prove the claim only in the case L=1, leaving the case L <1to you. Given M2R, we know there exists T2Isuch thaty(T)> M . So ifT < t < b , we 1In somewhat the same way we reduced solving higher order ODEs to solving rst order systems, one can always reduce a non-autonomous ODE to a rst order autonomous system. 8 see thaty(t)y(T)> M , becauseyis increasing. Since Mis arbitrary, we conclude that limt!by(t) =1.  Lemma 3.3. IfL= lim t!by(t)<1, thenb=1andf(L) = 0 . Proof. LetL= lim t!by(t), and for any M > t 0consider the compact set K:= [t0;M] [y0;L]. Then by Theorem 2.7, there exists T2Isuch that (t;y(t))=2Kfor allt>T . Since (t0;y0)2K, it follows that t0< T < b . So ifT < t < b , we havey(t)2[y0;M]. It follows therefore that t>M . Thusb>t>M for anyM2R, which means that b=1. Now sincefandyare continuous, we have lim t!1f(y(t)) =f(lim t!1y(t)) =f(L). So given >0, we have T2Rsuch thatt > T impliesjf(y(t))f(L)j< =2. In fact, since Lf(y(t)), we have 0f(L)f(y(t))<=2. So if we choose any t>T , the mean value theorem gives us s2(t;t+ 1) such that f(y(s)) =f(y(t+ 1))f(y(t)) (t+ 1)t<f(L)(L=2) 1==2: That is, there exists s>T such that 0 <f(y(s))<. Thus, jf(L)j=jf(L)f(s) +f(s)jjf(L)f(s)j+jf(s)j<=2 +=2 =:  Let us combine the assertions above into a single statement. Theorem 3.4. Suppose that f(y0)>0and thaty: (a;b)!Ris the solution of (2)with maximal domain of existence I. Thenyis strictly increasing on all of (a;b), and there are two possibilities for the asymptotic behavior of y(t)astincreases. Either limt!by(t) =1; or b=1andlimt!1y(t) =LwhereLis an equilibrium point of the ODE. We leave it to the reader to puzzle out the statement of this theorem in the case f(y0)<0 and to draw the appropriate conclusions about the asymptotic behavior of the global solution y:I!Rastdecreases toward the leftendpoint of the domain I. In essence, Theorem 3.4 is telling us that solutions to rst order autonomous ODEs will either drift o to innity or settle down and become asymptotically constant. If the reader nds this unsurprising, then he or she should try to imagine what the analogous assertion should be for solutions of autonomous systems of 2 or 3 ODEs (Hint: don't even try when there are 3 or more equations involved.) 9