Diblík, Josef

Nowak, Christine

Siegmund, Stefan

Electronic Journal of Qualitative Theory of Differential Equations

Siegmund Stefan

Nowak Christine

Diblík Josef

University of Szeged

Electronic Journal of Qualitative Theory of Differential Equations2016, No. 28, 1–8; doi: 10.14232/ejqtde.2016.1.28 http://www.math.u-szeged.hu/ejqtde/A generalized Picard–Lindelöf theoremStefan SiegmundB 1, Christine Nowak2 and Josef Diblík3, 41Institute for Analysis & Center for Dynamics, Department of Mathematics,Technische Universität Dresden, 01062 Dresden, Germany2 Institute for Mathematics, University of Klagenfurt, 9020 Klagenfurt, Austria3Brno University of Technology, Department of Mathematics and Descriptive Geometry,Faculty of Civil Engineering, 602 00 Brno, Czech Republic4Brno University of Technology, Department of Mathematics,Faculty of Electrical Engineering and Communication, 616 00 Brno, Czech RepublicReceived 22 December 2015, appeared 20 May 2016Communicated by Mihály PitukAbstract. We generalize the Picard–Lindelöf theorem on the unique solvability of ini-tial value problems ẋ = f (t, x), x(t0) = x0, by replacing the sufficient classical Lipschitzcondition of f with respect to x with a more general Lipschitz condition along hy-perspaces of the (t, x)-space. A comparison with known results is provided and thegenerality of the new criterion is shown by an example.Keywords: Picard–Lindelöf theorem, initial value problem, generalized Lipschitz con-dition, unique solvability.2010 Mathematics Subject Classification: 34A12, 34A34.1 IntroductionWe consider the initial value problemẋ = f (t, x), x(t0) = x0, (1.1)where f : D → Rn is defined on an open set D ⊆ R × Rn and (t0, x0) ∈ D. We assumethroughout the paper that f is continuous. Problem (1.1) is called locally uniquely solvable ifthere exists an open interval I containing t0 such that (1.1) has exactly one solution on I.The unique solvability problem of (1.1) is not fully solved up to now as simple examplesshow (see [2] and the references therein, see also [1]). The classical Lipschitz condition mea-sures the vector field differences with respect to the x variable and is assumed in the classicalPicard–Lindelöf theorem to prove unique solvability for (1.1). By introducing a Lipschitz con-dition along a hyperspace of the extended state space R×Rn, we establish a new uniquenesstheorem which generalizes the classical Picard–Lindelöf theorem and Theorem 3.2 in the pa-per by Cid [2]. It is also an n-dimensional generalization of the scalar criterion in [6] and ofthe uniqueness theorem in [3] if the functions ϕ and ψ are constants. The advantage of ourresult is shown by an example.BCorresponding author. Email: stefan.siegmund@tu-dresden.de2 S. Siegmund, Chr. Nowak and J. DiblíkDefinition 1.1 (Lipschitz continuity along a hyperspace). Let D ⊆ R×Rn be open, f : D → Rnbe continuous and let V ⊂ R×Rn be a hyperspace, i.e. V is an n-dimensional linear subspaceof R1+n. We say that f is Lipschitz continuous along V on an open set U ⊆ D if there exists aconstant L ≥ 0 such that for all (t, x), (s, y) ∈ U‖ f (t, x)− f (s, y)‖ ≤ L‖(t, x)− (s, y)‖ if (t, x)− (s, y) ∈ V .2 Main resultIn the following let F(t, x) = (1, f (t, x))T be the vector of the direction field of (1.1) determinedby f at the point (t, x) ∈ D.Theorem 2.1 (Generalized Picard–Lindelöf theorem). Consider the initial value problem (1.1), letV ⊂ R×Rn be a hyperspace and assume that the following two conditions hold:(A1) Transversality condition: F(t0, x0) /∈ V ,(A2) Generalized Lipschitz condition: f is Lipschitz continuous along V on an open neighborhoodU ⊆ D of (t0, x0).Then (1.1) is locally uniquely solvable.The proof of Theorem 2.1 uses only Peano’s theorem and the implicit function theorem.Since the classical Picard–Lindelöf theorem is a special case of Theorem 2.1, the followingproof also offers an alternative proof of Picard–Lindelöf’s theorem.Proof. Let ‖ · ‖ denote the Euclidean norm and its induced matrix norm, respectively. SinceV is a hyperspace in R1+n, there exist linearly independent vectors v(1), . . . , v(n) ∈ R1+n withV = span{v(1), . . . , v(n)} ⊆ R1+n. Writev(i) = (v(i)t , v(i)1 , . . . , v(i)n )T for i = 1, . . . , n,and define vt := (v(1)t , . . . , v(n)t ) ∈ Rn, v(i)x := (v(i)1 , . . . , v(i)n )T ∈ Rn, Vx := (v(1)x | · · · |v(n)x ) ∈Rn×n. Then forV :=(v(1) | · · · | v(n))=v(1)t · · · v(n)tv(1)1 · · · v(n)1......v(1)n · · · v(n)n =v(1)t · · · v(n)tv(1)x | · · · | v(n)x = (vtVx)we have V ∈ R(1+n)×n and rank V = n. Peano’s theorem guarantees that (1.1) has at leastone solution x : [t0 − α, t0 + α] → Rn for some α > 0. By shrinking α > 0 if necessary, wecan assume that graph x ⊂ U and, by assumption (A1) and continuity of f , F(t, x(t)) /∈ Vfor all t ∈ I := (t0 − α, t0 + α). To prove that (1.1) is locally uniquely solvable with solutionx on I, assume to the contrary that there exists a solution y : I → Rn of (1.1) and x ≡/ y on[t0, t0 + α) (the case x ≡/ y on (t0 − α, t0] is treated similarly). For t1 := sup{t ∈ [t0, t0 + α) :x(s) = y(s) for s ∈ [t0, t]} we have t1 ∈ [t0, t0 + α), x(t1) = y(t1) =: x1 by continuity andF(t1, x1) /∈ V .A generalized Picard–Lindelöf theorem 3We show that the equationy(t + vtk(t)) = x(t) + Vxk(t) (2.1)is uniquely solvable with respect to k = k(t) = (k1(t), . . . , kn(t))T on a subinterval of I whichcontains t1. The problem suggests to apply the implicit function theorem. Choose ε > 0 suchthatH(t, k) := y(t + vtk)− x(t)−Vxkis well-defined on [t1 − ε, t1 + ε]× [−ε, ε]n. Then H(t1, 0) = 0,∂H∂k(t, k) =(fi(t + vtk, y(t + vtk))v(j)t − v(j)i)i,j=1,...,nand therefore ∂H(t1, 0)/∂k = WV withW := f (t1, x1)∣∣∣∣∣∣∣−1. . .−1 ∈ Rn×(1+n).By the rank-nullity theorem (see e.g. [4, p. 199]) dim im(V) + dim ker(V) = n and, usingthe fact that dim im(V) = rank V = n, we get ker V = {0}. Assume that WV is not invertible.Then there exists v ∈ Rn \ {0} such that WVv = 0. Hence w := Vv 6= 0 and w ∈ V , as wellas w ∈ ker W = span{F(t1, x1)}. Therefore F(t1, x1) ∈ V leads to a contradiction, proving thatWV is invertible.The implicit function theorem (cf. e.g. [5, Theorem 9.28]) yields a unique C1 function k : J →[−ε, ε]n on an open interval J ⊆ I containing t1 such that k(t1) = 0 and H(t, k(t)) = 0 for allt ∈ J. Using the fact that ∂H(t1, 0)/∂k is invertible, we get by shrinking J if necessary, that(∂H(t, k(t))/∂k)−1 exists and is bounded for t in J, i.e. there exists η ≥ 0 such that∥∥∥∥∂H∂k (t, k(t))−1∥∥∥∥ ≤ η for t ∈ J.Since ∂H(t, k)/∂t = f (t + vtk, y(t + vtk)) − f (t, x(t)), (A2) implies, together with (2.1) andVk(t) ∈ V , that ∥∥∥∥∂H∂t (t, k(t))∥∥∥∥ ≤ L‖Vk(t)‖.Now we consider u(t) := ‖k(t)‖2 = 〈k(t), k(t)〉. We getu̇(t) =ddt〈k(t), k(t)〉 = 2〈k(t), k̇(t)〉 .Using the fact thatk̇(t) = −∂H∂k(t, k(t))−1∂H∂t(t, k(t)),we conclude thatu̇(t) ≤∥∥∥∥2k(t)T ∂H∂k (t, k(t))−1 ∂H∂t (t, k(t))∥∥∥∥ ≤ 2‖k(t)‖ηL‖V‖‖k(t)‖and henceu̇(t) ≤ 2ηL‖V‖u(t)4 S. Siegmund, Chr. Nowak and J. Diblíkwhich is equivalent toddt[e−2ηL‖V‖(t−t1)u(t)]≤ 0.Since u(t1) = ‖k(t1)‖2 = 0, we get u(t) = ‖k(t)‖2 ≡ 0, and hence from (2.1) we concludex(t) ≡ y(t) on J, which contradicts the definition of t1.Remark 2.2. (a) The classical Picard–Lindelöf theorem which requires a Lipschitz conditionwith respect to x is a special case of Theorem 2.1 withV =(vtVx), vt = 0 ∈ Rn and Vx = In, (2.2)where In denotes the n × n identity matrix. Cid [2] introduces the notion of Lipschitz conti-nuity when fixing component i0 ∈ {0, 1, . . . , n} where the component i0 = 0 corresponds to thevariable t, i.e. Lipschitz continuity when fixing i0 = 0 is equivalent to Lipschitz continuitywith respect to x. Lipschitz continuity when fixing another component is defined similarly.Under the assumption that f is Lipschitz continuous when fixing a component i0, Cid canshow uniqueness provided that either i0 = 0 or fi0 6= 0. Thus Theorem 3.2 by Cid can beinterpreted as a special case of our Theorem 2.1 with matrices V of the form (2.2) where inthe case of i0 6= 0 the corresponding column of V is replaced by a vector v(i0) with v(i0)t = 1and all other components equal 0. Note that [3, Theorem 1] is a special case of Theorem 2.1for n = 1 if the functions ϕ and ψ are constants.(b) Let V = span{v(1), . . . , v(n)} ⊂ R1+n and U ⊆ D be a convex open neighborhood of(t0, x0) ∈ D ⊆ R×Rn. If the directional derivatives∂ f∂v(t, x) = limh→0f ((t, x) + hv)− f (t, x)h‖v‖ , v ∈ V ,exist and are continuous and bounded on U, then f is Lipschitz continuous along V on U.Proof. With (t, x) = (s, y) + v, v ∈ V , and g(τ) := f ((s, y) + τv) we getf (t, x)− f (s, y) = g(1)− g(0) =∫ 10g′(τ)dτ=∫ 10limh→0g(τ + h)− g(τ)hdτ=∫ 10limh→0f ((s, y) + (τ + h)v)− f ((s, y) + τv)hdτ=∫ 10(limh→0f ((s, y) + (τ + h)v)− f ((s, y) + τv)h‖v‖)‖v‖dτ=∫ 10∂ f∂v((s, y) + τv)‖v‖dτand therefore‖ f (t, x)− f (s, y)‖ ≤ L‖v‖, L := supτ∈[0,1]∂ f∂v((s, y) + τv).Example 2.3. Consider the 2-dimensional initial value problemẋ = f (t, x), x(0) = 0,A generalized Picard–Lindelöf theorem 5where f (t, x) = ( f1(t, x1, x2), f2(t, x1, x2))T withf1(t, x1, x2) ={x1 + g(x2), x1 < t,x1 + g(x2) + 3√x1 − t, x1 ≥ t,f2(t, x1, x2) = 1 + h(x1),g(x2) and h(x1) are Lipschitz continuous functions and g(0) 6= 1. The classical Lipschitzcondition is not fulfilled, and we cannot show uniqueness with the hyperspace V being the(t, x1)-plane or (t, x2)-plane. Therefore the result by Cid cannot be applied.With the basis vectors v(1) = (1, 1, 0)T, v(2) = (0, 0, 1)T and V = span{v(1), v(2)} we canshow uniqueness of the given problem.(A1) is satisfied, as (1, g(0), 1+ h(0))T /∈ V if g(0) 6= 1. The only numbers α, β, γ, satisfyingα(1, f (0, 0))T + βv(1) + γv(2) = 0 are α = β = γ = 0 if g(0) 6= 1.Now (A2) is shown. With vt = (1, 0) and Vx =(1 00 1)we have to show that‖ f (t + vtk, x + Vxk)− f (t, x)‖ = ‖ f (t + k1, x1 + k1, x2 + k2)− f (t, x1, x2)‖≤ L‖(vtk, Vxk)T‖with k = (k1, k2)T. For x1 < t we get∥∥∥∥( x1 + k1 + g(x2 + k2)− x1 − g(x2)1 + h(x1 + k1)− 1− h(x1))∥∥∥∥which can be estimated by L‖(k1, k1, k2)T‖ with L ≥ 0. For x1 ≥ t we get∥∥∥∥( x1 + k1 + g(x2 + k2) + 3√x1 + k1 − t− k1 − x1 − g(x2)− 3√x1 − t1 + h(x1 + k1)− 1− h(x1))∥∥∥∥which can also be estimated by L‖(k1, k1, k2)T‖ with L ≥ 0.3 Alternative proofWe provide an alternative proof for Theorem 2.1 by transforming (1.1) into a system to whichthe classical Picard–Lindelöf theorem can be applied.Alternative proof of Theorem 2.1. Choose a unit vector a0 ∈ R1+n such that V = a⊥0 and also〈a0, F(t0, x0)〉 > 0, which is possible due to assumption (A1). Since R1+n = 〈a0〉 ⊕ V is thedirect sum of 〈a0〉 = {sa0 ∈ R1+n : s ∈ R} and V , there exist unique s0 ∈ R and v0 ∈ V with(t0, x0) = s0a0 + v0. We divide the proof into three steps.Step 1: We show that the nonautonomous initial value problem on Vdvds= g(s, v) :=F(sa0 + v)− σ(s, v)a0σ(s, v), v(s0) = v0, (3.1)with σ(s, v) := 〈a0, F(sa0 + v)〉 is well-posed and locally uniquely solvable.The functionσ : R× V → R, (s, v) 7→ σ(s, v) = 〈a0, F(sa0 + v)〉6 S. Siegmund, Chr. Nowak and J. Diblíkis continuous and satisfies σ(s0, v0) = 〈a0, F(s0a0 + v0)〉 = 〈a0, F(t0, x0)〉 > 0. As a consequencethere exists an η > 0 and a bounded open neighborhood U ⊆ R× V of (s0, v0) such thatσ(s, v) ≥ η for all (s, v) ∈ U.Using assumption (A2) and by shrinking U if necessary, we can w.l.o.g. assume that f isLipschitz continuous along V on the open neighborhood {sa0 + v ∈ R1+n : (s, v) ∈ U} of(t0, x0). Using this fact, we get for (s, v), (s, v̄) ∈ U|σ(s, v)− σ(s, v̄)| = |〈a0, F(sa0 + v)〉 − 〈a0, F(sa0 + v̄)〉|= |〈a0, F(sa0 + v)− F(sa0 + v̄)〉| ≤ ‖a0‖ · ‖F(sa0 + v)− F(sa0 + v̄)‖= ‖F(sa0 + v)− F(sa0 + v̄)‖ = ‖ f (sa0 + v)− f (sa0 + v̄)‖≤ L‖v− v̄‖,proving that σ is Lipschitz continuous on U. With σ also the quotient 1/σ is Lipschitz contin-uous with respect to v. Thus we get‖g(s, v)− g(s, v̄)‖ =∥∥∥∥F(sa0 + v)σ(s, v) − F(sa0 + v̄)σ(s, v̄)∥∥∥∥≤∣∣∣∣ 1σ(s, v)∣∣∣∣ · ‖F(sa0 + v)− F(sa0 + v̄)‖+∣∣∣∣ 1σ(s, v) − 1σ(s, v̄)∣∣∣∣ · ‖F(sa0 + v̄)‖ .By shrinking U again if necessary, we can assume w.l.o.g. that Ū ⊆ D. Then boundedness ofF and of 1/σ on Ū imply Lipschitz continuity of g with respect to v on the neighborhood Uof (s0, v0). Since V is isomorphic to Rn, the classical Picard–Lindelöf theorem can be appliedto (3.1) to prove local unique solvability.Step 2: We show that the autonomous initial value problem on R× Vṡ = σ(s, v), s(t0) = s0,v̇ = F(sa0 + v)− σ(s, v)a0, v(t0) = v0,(3.2)is locally uniquely solvable.By Peano’s theorem (3.2) admits a solution. Assume that (ŝ1, v̂1), (ŝ2, v̂2) : J → R× V , aretwo solutions of (3.2) on an open interval J containing t0. Then the solution identities˙̂si(t) = σ(ŝi(t), v̂i(t)),˙̂vi(t) = F(ŝi(t)a0 + v̂i(t))− σ(ŝi(t), v̂i(t))a0(3.3)for t ∈ J and the initial conditionsŝi(t0) = s0, v̂i(t0) = v0 (3.4)are fulfilled for i=1, 2. By shrinking J if necessary, we can w.l.o.g. assume that (ŝi(t), v̂i(t))∈Uand therefore ˙̂si(t) = σ(ŝi(t), v̂i(t)) ≥ η for t ∈ J. As a consequence the functions ŝi : J → Rare strictly monotonically increasing, and hence the inverse functions ŝ−1i : ŝi(J)→ J exist andsatisfyŝ−1i (s0) = t0 (3.5)A generalized Picard–Lindelöf theorem 7for i = 1, 2. With the bijection t = ŝ−1i (s) both solution curves through (s0, v0) can be repara-metrized in the form{(ŝi(t), v̂i(t)): t ∈ J}={(ŝi(ŝ−1i (s)), v̂i(ŝ−1i (s)): s ∈ ŝi(J)}={(s, v̂i(ŝ−1i (s)): s ∈ ŝi(J)}for i = 1, 2. Thenvi : ŝi(J)→ V , vi(s) := v̂i(ŝ−1i (s)),solve (3.1) for i = 1, 2, sincedvids(s) =˙̂vi(ŝ−1i (s))˙̂si(ŝ−1i (s))(3.3)=F(sa0 + vi)− σ(s, vi)a0σ(s, vi)andvi(s0) = v̂i(ŝ−1i (s0))(3.5)= v̂i(t0)(3.4)= v0.By shrinking J if necessary, we can apply Step 1 to conclude that v1 = v2 on J and hencev̂1(ŝ−11 (s)) = v̂2(ŝ−12 (s)) for all s ∈ ŝ1(J) ∩ ŝ2(J), proving that ŝ1 = ŝ2 and v̂1 = v̂1 on J.Step 3: We show that (1.1) is locally uniquely solvable.By Peano’s theorem (1.1) admits a solution. Assume that x1, x2 : I → Rn are two solutionsof (1.1). For t ∈ I we have Xi(t) := (1, xi(t)) ∈ R1+n = 〈a0〉 ⊕ V and therefore there existunique functions si : I → R and vi : I → V such thatXi(t) = si(t)a0 + vi(t).Moreover, (si(t0), vi(t0)) = (s0, v0), and using the fact that ‖a0‖ = 1 and a⊥0 = V , si(t) =〈a0, Xi(t)〉 and vi(t) = Xi(t)− si(t)a0 for t ∈ I and i = 1, 2. Now (si, vi) : I → R×V solve (3.2),sinceṡi(t) = 〈a0, Ẋi(t)〉 = 〈a0, F(t, xi(t))〉 = 〈a0, F(si(t)a0 + vi(t))〉= σ(si(t), vi(t)),v̇i(t) = Ẋi(t)− 〈a0, Ẋi(t)〉a0 = F(t, xi(t))− 〈a0, F(t, xi(t))〉a0= F(si(t)a0 + vi(t))− 〈a0, F(si(t)a0 + vi(t))〉a0= F(si(t)a0 + vi(t))− σ(si(t), vi(t))a0for t ∈ I and i = 1, 2. By shrinking I if necessary, we can apply Step 2 to conclude that s1 = s2and v1 = v2 on I, proving that x1 = x2.AcknowledgmentsThe third author is supported by the Grant P201/11/0768 of the Czech Grant Agency (Prague).References[1] R. P. Agarwal, V. Lakshmikantham, Uniqueness and nonuniqueness criteria for ordinarydifferential equations, World Scientific Publishing, 1993. MR13368208 S. Siegmund, Chr. Nowak and J. Diblík[2] J. Á. Cid, On uniqueness criteria for systems of ordinary differential equations, J. Math.Anal. Appl. 281(2003), 264–275. MR1980090; url[3] J. Diblík, C. Nowak, S. Siegmund, A general Lipschitz uniqueness criterion for scalarordinary differential equations, Electron. J. Qual. Theory Differ. Equ. 2014, No. 32, 1–6.MR3250025[4] C. D. Meyer, Matrix analysis and applied linear algebra, SIAM, Philadelphia, 2000.MR1777382[5] W. Rudin, Principles of mathematical analysis, third edition, McGraw-Hill Book Co., 1976.MR385023[6] H. Stettner, C. Nowak, Eine verallgemeinerte Lipschitzbedingung als Eindeutig-keitskriterium bei gewöhnlichen Differentialgleichungen (in German) [A generalizedLipschitz condition as criterion of uniqueness in ordinary differential equations], Math.Nachr. 141(1989), 33–35. MR1014412

A generalized Picard–Lindelöf theorem

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A generalized Picard–Lindelöf theorem

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