Picard’s Existence and Uniqueness Theorem

1 1 Picard s Existence and Uniqueness TheoremDenise GutermuthThese notes on the proof of Picard s Theorem follow the textFundamentals of Di erentialEquations and Boundary Value Problems, 3rd edition, by Nagle, Sa , and Snider, Chapter13, Sections 1 and 2. The intent is to make it easier to understand the proof by supplementingthe presentation in the text with details that are not made explicit there. By no means isanything here claimed to be original of the most important theorems in Ordinary Di erential Equations is Picard sExistence and Uniqueness Theorem for first-order ordinary di erential equations. Why isPicard s Theorem so important? One reason is it can be generalized to establish existenceand Uniqueness results for higher-order ordinary di erential equations and for systems ofdi erential equations.

2 Another is that it is a good introduction to the broad class of existenceand Uniqueness theorems that are based on fixed s Existence and Uniqueness TheoremConsider the Initial Value Problem (IVP)y0=f(x, y),y(x0)= (x, y)and@f@y(x, y)are continuous functions in some open rectangleR={(x, y):a<x<b,c<y<d}that contains the point(x0,y0). Then the IVP has a unique solutionin some closed intervalI=[x0 h, x0+h], whereh>0. Moreover, thePicarditerationdefined byyn+1(x)=y0+xZx0f(t, yn(t))dtproduces a sequence of functions{yn(x)}that converges to this solution uniformly 1: Consider the IVPy0=3y2/3,y(2) = 0 Thenf(x, y)=3y2/3and@f@y=2y 1/3,sof(x, y) is continuous wheny= 0 but@f@yis the hypothesis of Picard s Theorem does not hold. Neither does the conclusion; theIVP has two solutions,y1/3=x 2 andy are many ways to prove the Existence of a solution to an ordinary di erentialequation.

3 The simplest way is to find one explicitly. This is a good approach for separableor exact equations, or linear equations with constant coe cients. But unfortunately there2are many equations that cannot be solved by elementary methods, so attempting to provethe Existence of a solution with this approach is not at all practical. An alternative approachis to approximate a solution to an IVP by constructing a sequence of functions that convergesto a solution. This is precisely the approach we will use for the proof of Picard s we discuss the idea behind successive approximations, let s first express a first-order IVP as an integral equation. For the IVPy0=f(x, y),y(x0)=y0, suppose thatfiscontinuous on some appropriate rectangle and that there is a solutiony(x) that is continuouson some intervalI.

4 Then we may integrate both sides of the DE to obtain integral equation:y(x)=y0+xZx0f(t, y(t))dtThus, under the assumptions of Existence and continuity, the IVP is equivalent to the integralequation. This fact seems convenient and useful at first glance, but upon closer inspectionwe notice two problems: The integral equation is not well-defined unless we know that a solution The integral equation is very hard to solve, except for very elementary we define an operatorTthat maps a functiony(x) to a functionT[y](x), givenbyT[y](x) :=y0+xZx0f(t, y(t)dtThen the integral equation is simplyy=T[y], and any solution to the IVP must be find fixed points, approximation methods are often useful. See Figure 1, below, foran illustration of the use of an approximation method to find a fixed point of a find a fixed point of the transformationTusing Picard iteration, we will start with thefunctiony0(x) y0and then iterate as follows:yn+1(x)=yn(x)+xZx0f(t, yn(t))dtto produce the sequence of functionsy0(x),y1(x),y2(x).)

5 If this sequence converges, thelimit function will be a fixed point term well-defined means essentially that (t,y(t)) must be in the domain off(x,y). For example,the functionf(x)=11 xis well-defined for allx6= 1. But ifx= 1, thenf(x) is fixed point of an operator or a transformation is an element in the domain that the operator ortransformation maps to itself. In terms of functions, a pointx=a2domgis a fixed point ofgi g(a)= , a fixed point is a point where the graph ofy=g(x) intersects the straight liney= 2: Consider the IVPy0=2y, y(0) = 1 This IVP is equivalent toy=1+xR02y dt, so the Picard iterates arey0(x) 1,y1(x)=1+xZ02y0(t)dt=1+2xy2(x)=1+xZ02(1 + 2t)dt=1+2x+(2x)22!and so on. It can be shown by induction that thenth iterate isyn(x)=1+2x+(2x)22!

6 +..+(2x)nn!=nXi=1(2x)ii!which is thenth partial sum of the Maclaurin series fore2x. Thus, asn !1,yn(x) ! carry out a rigorous test for convergence (which is especially necessary when we don trecognize the sequence of Picard iterates), we need some idea of distance between distance measure used in the proof of the Picard Theorem is based on the norm of afunction, as given in the following [a, b]denote the set of all functions that are continuous on[a, b]. Ify2C[a, b], then thenormofyiskyk:= maxx2[a,b]|y(x)|.The norm of a functiony(x) may be regarded as the distance betweeny(x) andy 0, thefunction that is identically this in mind we may define the distance betweentwo functions,y, z2C[a, b] to be the norm ofy z, orky zk= maxx2[a,b]|y(x) z(x)|.

7 Using this measure of distance, we can define convergence of a sequence of functions toa limiting sequence{yn(x)}of functions inC[a, b]converges uniformly to a functiony(x)2C[a, b]i limn!1kyn yk= illustrate uniform convergence, recall Example 2, where we found thatyn(x)=nPi=0(2x)ii!andy(x)=e2xon the interval [0,1]. Thenlimn!1kyn yk= limn!1"maxx2[0,1] nXi=0(2x)ii! e2x #= limn!124maxx2[0,1] 1Xi=n+1(2x)ii! 35= limn!11Xi=n+12ii!=03 See Nagel, Sa , and Snider, page 837, for properties of this ,yn !y. One important detail to note in this example is that the uniformconvergence of the sequence{yn(x)}toy(x)=e2xon [a, b] occurs only when the interval isbounded on the right. In other words,bmust be finite. If we were to takebto be infinite for example, if the interval under consideration were the whole real line then the sequencewould not converge uniformly.

8 The reason is that, if the value ofxis unbounded, then forany finiten, the norm of1Pi=n+1(2x)ii!is convergence is particularly useful in that if a sequence of di erentiable (andtherefore continuous) functions is uniformly convergent, then the function to which it con-verges is also Fixed Point Theorem for OperatorsLetSdenote the set of continuous functions on[a, b]that lie within a fixed distance >0ofa given functionyt(x)2C[a, b], {y2C[a, b]:ky ytk }. LetGbe an operatormappingSintoSand suppose thatGis acontractiononS, that is9k2R,0 k<1s. [w] G[z]k kkw zk8w, the operatorGhas a unique fixed point solution inS. Moreover, the sequence ofsuccessive approximations defined byyn+1:=G[yn],n=0,1, uniformly tothis fixed point, for any choice of starting prove the Theorem we first show that the functions in the sequenceyn+1=G[yn] arewell-defined; that is, every function in the sequence{yn(x)}is in the setS.

9 Next we showthat this sequence converges uniformly to a functiony12S. Finally, we show that thislimit is a fixed point ofG, that is,y1=G[y1].Proof:Take any starting functiony02S. Sincey02 DomG,theny1=G[y0] is defined. SinceGmapsSto itself, induction,yn2 SandG[yn] is well-defined, for alln rewriteyn=y0+(y1 y0)+(y2 y1)+..+(yn yn 1), so thatyn(x)=y0(x)+n 1Xj=0(yj+1(x) yj(x))(1)We now show that the sequence{yn}converges uniformly to an element inS. We do thisby applying theWeierstrass M-Test, an extension of the Comparison the textIntroduction to Analysisby James R. Kirkwood, pages 206-212, for the definitions andproofs of some properties of uniform M-TestLet{fn}be a sequence of functions defined on a setE. Suppose thatfor alln2N, there existsMn2 Rsuch that|fn(x)| Mn8x2 EThen ifPMnconverges,Pfnmust converge uniformly we need to do, then, is to find a boundMon the terms (actually functions, ofcourse) of the series (1).

10 Claim:kyj+1 yjk kjky1 claim is evidently true forj= 0. Suppose that it is true forj=q, whereq2N, q [yq+2] G[yq+1]k=kG[G[yq+1]] G[G[yq]]k kkG[yq+1] G[yq]k kq+1ky1 y0k,proving the to the series (1), it is clear from the claim that maxx2[a,b]|yj+1(x) yj(x)|=kyj+1 yjk kjky1 y0k. LetMj:=kjky1 y0k. Because1Pj=1Mj=ky0 y1k1Pj=1kjconverges5,the Weierstrass M-Test shows that{yn}converges uniformly to a continuous function, ,y12 Sbecause the assumption thatky1 ytk> implies thatkyn ytk> for somen, contradicting the fact a contraction, we have thatkG[y1] G[yn]k kky1 ynkfor anyn. Butky1 ynk !0 asn !1,soky1 yn+1k !0 asn !1. Of course,G[yn]=yn+ , limn!1kG[y1] G[yn]k= limn !1kG[y1] yn+1k limn !1kkG[y1] yn+1k= 0. Finally,G[y1] y1=(G[y1] yn+1)+(yn+1 y1), so that6kG[y1] y1k kG[y1] yn+1k+kyn+1 both terms on the right side of the above equation approach zero asnapproaches1,it follows thatkG[y1] y1k= 0, orG[y1]=y1.