NOTES ON EXISTENCE AND UNIQUENESS THEOREMS FOR ODES

NOTES ON EXISTENCE AND UNIQUENESS THEOREMS FOR ODES JONATHAN LUK These notes discuss theorems on the existence, uniqueness and extension of solutions for ODEs. None of these results are original. The proofs of Theorems.2, 3.3 are contained in the textboo although in a slightly different setting and with a slightly different language. The statement and proof of Theorem. are not in the textboo, although the textboo does contain a discussion of Euler s broen line method. The purpose of these notes is to collect the various results scattered throughout different parts of the boo and also to provide slightly more details. This is a preliminary version: If you have any comments or corrections, no matter how minor, please send them to me.. Main theorems Throughout, we consider the following initial value problem: { u t = F t, ut, u = u, where F : I U R n is a continuous function with I R an open interval and u U R n an open set. We will loo for a solution u : I R for some open sub-interval I which is a C function satisfying.. A solution to this initial value problem always exists: Theorem. Peano. There exists ɛ > depending on u such that there exists a solution ut for t ɛ, ɛ. In general, uniqueness is false Exercise, or recall from wee. However, uniqueness holds if an extra condition is imposed on F : Theorem.2 Picard Lindelöf, Cauchy Lipschitz. Suppose that, in addition, F satisfies the condition that for any open interval I with Ī I and any open ball Bx, r with Bx, r U, there exists a constant L > such that F t, x F t, y L x y for every t, x, t, y I Bx, r. Then, there exists ɛ > depending on u such that there exists a unique solution ut for t ɛ, ɛ. Remar.3. The additional condition for F required by Theorem.2 is satisfied if F is C Exercise. We will prove Theorems. and.2 in Sections 4 and 2 respectively. We will also discuss another important theorem, the extension theorem Theorem 3.3 in Section 3. 2. Existence and uniqueness by Picard iteration In this section, we prove Theorem.2. Throughout, assume the hypotheses of Theorem.2. By openness of U, given u U, there exists r such that Bu, 2r U. Pic an arbitrary open interval I I such that I I. The assumption of Theorem.2 guarantees that there exists L > such that. F t, x F t, y L x y 2. for every t, x, t, y I Bu, 2r. On the other hand, since F is continuous, there exists M > such that F t, x M for t, x I Bu, 2r. 2.2

2 JONATHAN LUK Now, we pic the ɛ in the theorem to so that ɛ, ɛ I for I as above and ɛ < min{ r M, }. 2.3 2L Define a sequence of functions u : ɛ, ɛ Bu, 2r by and for, u t = u + u t = u F s, u s ds. 2.4 These u s are nown as Picard s iterates. Our goal is to show that they converge to a solution to.. This is accomplished by the following sequence of lemmata. Lemma 2.. For, u t Bu, r Bu, 2r for t ɛ, ɛ. Proof. We prove this by induction. The = case holds by definition. Assume the lemma holds for replaced by for some. Then, we estimate, using 2.2 and 2.4, u t u ɛ sup s ɛ,ɛ F s, u s ɛm r. Lemma 2.2. For every, u t u t 2 r for t ɛ, ɛ. Proof. We prove this by induction. The = case follows from 2. and the triangle inequality. Assume the inequality holds for replaced by for some 2. Then, by 2. and 2.4 u t u t F s, u s F s, u 2 s ds for every t ɛ, ɛ. ɛl sup s ɛ,ɛ u s u 2 s ɛl2 2 r 2 r, Lemma 2.3. There exists a continuous function u : ɛ, ɛ Bx, 2r such that u t converges to ut uniformly on ɛ, ɛ. Proof. By Lemma 2.2, for every t ɛ, ɛ, u t is a Cauchy sequence. We can then define ut = lim u t. Using Lemma 2.2 again, we see that u u uniformly. Therefore, u is continuous on ɛ, ɛ. Lemma 2.4. The function u in Lemma 2.3 solves for all t ɛ, ɛ. ut = u + Proof. Tae the limit lim of both sides of 2.4. F s, us ds 2.5 Lemma 2.5. Let u be a continuous function that solves 2.5. Then u is continuously differentiable and solves.. Proof. Exercise. Combining the above lemmata yields the existence part of Theorem.2. Note that we have not proven Theorem. since the above proof uses the assumptions in Theorem.2. It remains to prove uniqueness. Proposition 2.6. Suppose there is another solution ũ : ɛ, ɛ + R n for some ɛ ±, ɛ]. Then ũt = ut for all t ɛ, ɛ +. At this point, we do not yet now that the image of u is in Bu, 2r. This will be justified in Lemma 2..

NOTES ON EXISTENCE AND UNIQUENESS THEOREMS FOR ODES 3 Proof. Let us just prove that ũ and u agree on [, ɛ +. It is easy to chec Exercise that the same argument can be applied to show that they also agree on ɛ, ]. Step. First, we note that on any sub-interval [, ɛ [, ɛ +, if we assume that ũt Bu, 2r, then ũt = ut on [, ɛ. This is because for any t [, ɛ, d dt ũt ut 2 = ũt ut, F t, ũt F t, ut 2L ũt ut 2. Therefore, e 2Lt ũt ut 2 is a non-increasing function on [, ɛ. However, it is non-negative and is when t =. Hence, ũt ut 2 = for t [, ɛ. Step 2. Now, let τ := sup{t, ɛ + : ũs = us, s [, t}. If τ = ɛ +, we are done. So assume τ < ɛ +. Since uτ Bu, r, by continuity of ũ, there exists δ, ɛ + τ such that ũt Bu, 2r for t [τ, τ + δ. But then by Step, ũt = ut on [, τ + δ, contradicting the choice of τ. 2.. Aside: the contraction mapping theorem. Theorem.2 can also be realized as a special case of a more general theorem: the contraction mapping theorem. In fact, this is the point of view of the textboo. Let us briefly outline this. First, we recall the contraction mapping theorem from 6CM: Theorem 2.7 Contraction mapping theorem. Let X, d be a complete metric space. Suppose Φ : X X is a map such that for some λ <, dφx, Φy λdx, y for every x, y X, then Φ has a unique fixed point. Recall that existence can be proved by an iteration argument. Pic x X arbitrarily. Define x = Φx for. One then shows that there exists x X such that x x. Exercise This can be applied to the initial value problem as follows: Let X be the set of bounded and continuous functions u : ɛ, ɛ Bu, r. Let d be a distance defined on X by du, v = One checs that this is a complete space Exercise. Define now Φ : X X by Φut = u + sup ut vt. t ɛ,ɛ F s, us ds. It can be shown Exercise using methods similar to the proof of Theorem.2 that Φ is well-defined i.e., the image is indeed in X and is a contraction if t ɛ, ɛ for ɛ sufficiently small. 3. Extension theorem In this section, we continue to wor under the assumptions of Theorem.2. Recall that even in the case I = R, it is not always possible to have a solution on the whole of R. Our next goal is to understand what goes wrong. It is useful to eep in mind the example u t = ut 2, u =, where the solution ut = t can only be defined on,. Our goal is to characterize the maximal interval of existence in general. The main result is given in Theorem 3.3. In order to tal about the maximal interval of existence, we first need a slightly stronger uniqueness statement than that in Theorem.2. Roughly, it says that as long as a solution exists, it is unique. Proposition 3.. Let J and J 2 both be open sub-intervals of I containing. Suppose u : J R n and u 2 : J 2 R n are both continuously differentiable and both solve., then u u 2 on J J 2. Proof. As in the proof of Proposition 2.6, it suffices to consider J J 2,. The case J J 2, is similar. Assume uniqueness does not hold. Let τ := inf{t J J 2, : u t u 2 t}. By assumption, τ J J 2. Apply Theorem.2 but now to the initial value problem { u t = F t, ut, uτ = u τ,

4 JONATHAN LUK there exists δ > such that there is a unique solution for t τ δ, τ + δ. This contradicts the choice of τ. We can now define the maximal interval of existence J by J := {t I : there is a solution to. in an open interval including t}. By Proposition 3., there is a unique solution defined on J. Moreover, it is indeed maximal in the sense that there exists no open interval J with J J I such that a solution exists on J. Before we get to our main result Theorem 3.3, we first need a lemma. Note that the ey point of the lemma is that the same ɛ can be chosen for all t, u I K. Lemma 3.2. Given an open interval I such that I I and a compact subset K U, there exists an ɛ > such that for every t, u I K, there is a unique solution to { u t = F t, ut, 3. ut = u, on t t ɛ, t ɛ. Proof. Recall that we chose ɛ depending only on the interval I, and the constants L and M in 2. and 2.2. Therefore, for x U and r > such that Bx, 2r U, ɛ can be chosen uniformly for every t, u I Bx, r. Now for every x K, there is an r x > depending on x such that Bx, 2r x U. Obviously, x K Bx, r x K. By compactness of K, there exists a finite set {x,..., x N } such that N i=bx i, r xi K. For each i {,..., N}, we can choose an ɛ i such that for any u Bx i, r xi, the unique solution exists for t ɛ i, ɛ i. Taing ɛ = min{ɛ,..., ɛ N } yields the result. Theorem 3.3 Extension theorem. Let u : J R n be the maximal interval of existence. Suppose J = T, T +, where T + I i.e., T + is not the right end-point of I. Then lim inf t T + distut, U + ut = +. Here U is the boundary of U, and distut, U is defined by distut, U = sup{r : But, r U}. Proof. Suppose not. Then there is a sequence t n T + such that for every n N, distut n, U + ut n <. In particular, we now that {ut n } n= is a compact subset of U. By Lemma 3.2, there exists an ɛ > independent of n! such that a unique solution exists on the interval t n ɛ, t n + ɛ for every n N. This contradicts the fact that t n T +. Remar 3.4. Of course, in a completely analogous manner, if T I, then lim inf t T + distut, U + ut = +. This is called an extension theorem because it implies that one can extend the solution beyond a time T I if lim inf t T distut, U + ut < +. As we will see, this is quite useful, since we can often show that a unique solution exists for all time by proving certain estimates. 4. Existence by Euler s broen line method and compactness In this section, we prove Theorem.. Nevertheless, before we get to that, we need an important theorem, the Arzela Ascoli theorem, which gives a criterion such that one can extract a subsequential limit of continuous functions.

NOTES ON EXISTENCE AND UNIQUENESS THEOREMS FOR ODES 5 4.. Arzela Ascoli theorem. Theorem 4.. Let K R be a compact interval and {f n } n N be a sequence of continuous functions f n : K R n such that Uniform boundedness there exists an M > so that f n t M for all t K, n N, Equicontinuity for every ɛ >, there exists δ > such that f n t f n s < ɛ for all n N whenever t s < δ. Then there exists a subsequence {f n } = i.e., there is a subset {n, n 2,... } N with n < n 2 <... and a continuous function f : K R such that f n f uniformly as i.e., for every ɛ >, there exists N such that sup t K f n t ft < ɛ for every. Proof. Step. We first find the subsequence and the limiting function f. Let {t i } i= be a countable dense subset of K. Step a. By the assumptions of the theorem, {f n t } n= is a bounded subset of R n. By the Bolzano Weierstrass theorem, there exists a subsequence {f n,l } l= such that lim l f n,l t exists. Step b. Given {f ni,l } l=, we now construct a sequence {f n i+,l } l=, which is a subsequenc of {f ni,l } l=. By uniform boundedness, {f n i,l t i+ } l= is bounded. Using the Bolzano Weierstrass theorem, there exists a subsequence {f ni+,l } l= of {f n i,l } l= such that lim l f ni+,l t i+ exists. Step c. Define now {f n } = to be the diagonal sequence f n = f n,. By definition, lim f n t i exists for all i N. Step 2. We now show that indeed f n f uniformly as. Let ɛ >. By equicontinuity, there exists δ > such that if t s < δ, then f n t f n s < ɛ 3 for all n. By the density of {t i } i=, there exists a finite subset {t i} N i= such that for every t K, there is an i {,..., N} such that t t i < δ. Now since lim f n t i exists for all i {,..., N}, {f n t i } = is a Cauchy sequence in Rn. Hence, there exists such that whenever, 2, f n t i f n2 t i < ɛ 3. We now claim that for every t K, whenever, 2, f n t f n2 t < ɛ. Indeed, since for every t K, we can find i {,..., N} such that t t i < δ, we have f n t f n2 t f n t f n t i + f n t i f n2 t i + f n2 t i f n2 t < ɛ 3 + ɛ 3 + ɛ = ɛ. 4. 3 Step 3. Step 2 shows that {f n t} = is Cauchy for every t K. Hence, ft = lim f n t exists for every t K. Moreover, 4. implies that ft is a uniform limit of f n t, and it is therefore also continuous. 4.2. Euler s broen line method. The goal of this subsection is to prove Theorem.. From now on, suppose the hypotheses of Theorem. hold. In particular, we do not assume the hypotheses of Theorem.2. Lie the proof of Theorem.2, we prove existence by constructing a sequence of functions 2 u : [ ɛ, ɛ] U. The construction this time is different: For each, we divide the interval [ ɛ, ɛ] into 2 sub-intervals of equal length, and u t is a linear function on each of these sub-intervals hence the name broen line. More precisely, define u t on the intervals [ ɛ, ] and [, ɛ ] by For j =, 2,...,, for t [ jɛ, j+ɛ u t = u + tf, u. ], define u t = u jɛ + Similarly, for t [ j+ɛ 2 +, jɛ 2 + ], define u t = u jɛ + Let us note that this can be rephrased as u t = u + t jɛ jɛ F, u jɛ. t + jɛ F jɛ, u jɛ. F w s, v s ds, 4.2 2 Note that we define u on the closed interval so as to apply Arzela Ascoli Theorem later.

6 JONATHAN LUK where w : [ ɛ, ɛ] R is defined by for t = jɛ w t = jɛ for t jɛ, j+ɛ for t [ j+ɛ ], j =,,..., 4.3, jɛ, j =,,..., and v : [ ɛ, ɛ] U is defined by u for t = v t = u jɛ jɛ for t, j+ɛ ], j =,,..., 4.4 u jɛ j+ɛ for t [, jɛ, j =,,...,. We now choose our small ɛ as follows: Since u U, there is an r > such that Bx, r U. Fix a closed interval I I. By continuity, there exists M > such that F t, x M for t, x I Bx, r. We choose ɛ such that [ ɛ, ɛ] I and ɛ < r M 4.5 Lemma 4.2. u t is uniformly bounded and equicontinuous on [ ɛ, ɛ]. Hence, by Theorem 4., there exists a subsequence u l t and a continuous function u : [ ɛ, ɛ] U such that u l u uniformly as l. Proof. Step. We first prove uniform boundedness. In fact, we prove a stronger statement that u t Bu, r for all t [ ɛ, ɛ]. This follows since by 4.5, Step 2. By the definition of u and 4.5, Equicontinuity follows from this. u t u M ds Mɛ r. u t u s M t s. Lemma 4.3. Let w t : [ ɛ, ɛ] R be as in 4.3. Then w t converges uniformly to the function t for t ɛ, ɛ as. Proof. It is easy to chec that w t t ɛ, which as. Lemma 4.4. Let v t be as in 4.4 and v l be the subsequence with l as in Lemma 4.2. Then v l t converges uniformly to ut for t [ ɛ, ɛ]. Proof. Let δ >. We want to show that for l sufficiently large, v l t ut < δ for every l l for every t [ ɛ, ɛ]. It is easy to chec that v t u t ɛm. On the other hand, by Lemma 4.2, there exists l such that u l t ut < δ 2 for every l l and for every t [ ɛ, ɛ]. Now, taing l l such that ɛm l < δ 2, we have v l t ut v l t u l t + u l t ut < δ 2 + δ 2 = δ for every l l for every t [ ɛ, ɛ]. Lemma 4.5. ut as in Lemma 4.2 satisfies 2.5. Proof. Consider 4.2 for the subsequence l. Tae l on both sides. Lemma 4.6. ut as in Lemma 4.2 satisfies.. Proof. Apply Lemma 2.5.

NOTES ON EXISTENCE AND UNIQUENESS THEOREMS FOR ODES 7 5. Example: the exponential function Both the Picard s iteration method and the Euler s broen line method give us more than mere examples. They provide a way of computing the solution. Let us loo at a simple example. We now by explicit computation that e t is the unique solution to u t = ut, u =. By the above discussion, we have two ways to compute e t. First, by Picard s iteration, we have Explicit computation gives u =, u t = + u s ds. u t =, u t = + t, u 2 t = + t + t2 2,..., u t = Theorem.2 implies that for small t at least, e t = n= t n n!. Of course, we now from 6CM that in fact this is true for all t R! On the other hand, we can use Euler s broen line method to compute u jɛ = + ɛ j, which implies Exercise e t = lim + ɛ t ɛ. n= t n n!.