Picard-Lipschitz Theorem: Existence and Uniqueness of Solutions

Posted on Jan 23, 2025 in French

Given a differential equation (1), if A is an open set, f is continuous in A, and f∈Lloc(A,x), then, for a fixed (t0,x0)∈A, there exists a unique solution of (1) passing through (t0,x0).

Since (t0, x0)∈A and A is open, it is possible to take U1, an open neighborhood of (t0,x0), in which f is bounded. Let H be a bound of f in U1, that is, ||f(t, x)||≤ H, ∀(t, x) ∈ U1. We choose U2, an open neighborhood of (t0, x0), in which f∈L(U2,x), where K is a Lipschitz constant. The set U=U1∩U2 is an open set that contains (t0,x0), and in this set, f is continuous, bounded (with H a bound), and f∈LK(U,x).

Let d:=d∞((t0,x0),Fr(U))>0 (since U is open). We choose 0<δ<min{d,d/H,1/K} and consider the rectangle (compact) R = {(t,x):|t−t0|≤δ, ||x−x0||≤Hδ}⊂U, and the interval I=[t0−δ,t0+δ]. Consider the set M:={φ:I→Rn:φ is continuous and such that ||φ(t)−x0||≤ Hδ,∀t∈I}. We define a distance in M as follows: d(φ,ψ):=sup(t∈I)||φ(t)−ψ(t)||, ∀φ,ψ∈M.

(M,d) is a complete metric space (every Cauchy sequence in this space converges to an element of the space). M=BC(I)[x0,Hδ] since ||φ − x0||C(I) =sup(t∈I)||φ(t) − x0||, thus M is a closed set contained in C(I,Rn) (complete), then (M,d) is a complete metric space. We define T:φ∈M→Tφ∈M , by [Tφ](t)=x0+∫t0t(f(s,φ(s))ds).

Verify that T maps M into itself, that is, Tφ∈M ∀φ∈M. Tφ is continuous. Furthermore, ||[Tφ](t)−x0||=||∫t0t(f(s,φ(s)))ds||≤H|t−t0|≤Hδ, ∀t∈I.

T is a λ-contraction, that is, there exists a real number λ with 0≤λ<1 such that, ∀φ, ψ∈M, d(Tφ,Tψ)≤λd(φ,ψ).

Indeed, d(Tφ,Tψ)=sup||Tφ(t)−Tψ(t)|| = sup||x0+∫t0t(f(s,φ(s)))ds-(x0+∫t0t(f(s,ψ(s)))ds)|| = sup||∫t0t(f(s,φ(s)))-(f(s,ψ(s)))ds|| ≤ sup |∫t0t||(f(s,φ(s)))-(f(s,ψ(s)))||ds| ≤ sup|∫t0tK||φ(s)-ψ(s)||ds| ≤ sup|∫t0tKsup||φ(s)-ψ(s)||ds| = sup|∫t0tKd(φ,ψ)ds| = sup Kd(φ,ψ)|t–t0| = kd(φ,ψ)δ = λd(φ,ψ), where 0≤λ=Kδ<K1/K=1.

Since T is a λ-contraction, by the Contraction Mapping Theorem, there exists a unique function φ∈M such that Tφ=φ, which means that φ is continuous, (t,φ(t))⊂R⊂U⊂A, ∀t∈I, and [Tφ](t)=φ(t), thus φ(t)=x0+∫t0t(f(s,φ(s))ds). Therefore, φ is a solution of x‘=f(t,x) and φ(t0)=x0.

Could there exist a function that is a solution passing through the point (t0,x0) such that its graph is inside U but not inside R, that is, it does not belong to M? No. We will prove that if ψ : J→Rn is a solution of x’=f(t,x) passing through (t0,x0) with its graph inside U, then ∀t∈I∩J,(t,ψ(t))∈R (that is, ||ψ(t) − x0||≤Hδ). We will prove, by contradiction, that ||ψ(t)−x0||≤Hδ, ∀t∈I∩J. Suppose there exists t1∈I∩J such that ||ψ(t1)−x0||>Hδ and consider the case t1>t0 (since the case t1<t0 is analogous). The function t→||ψ(t)-x0|| is continuous, takes the value 0 at t0, and at t1 takes, by hypothesis, a value greater than Hδ. By the Bolzano Theorem, there exists t2∈(t0 , t1) such that ||ψ(t2)−x0||=Hδ. Let t∗=inf{t:||ψ(t)−x0||=Hδ,t>t0}. By continuity and the definition of the infimum, we deduce that ||ψ(t∗)−x0||=Hδ. Using that ||ψ’(t)||=||f(t,ψ(t))||≤H, we would have ||ψ(t∗)−x0||=||ψ(t∗)−ψ(t0)||≤H|t∗−t0|<Hδ, reaching a contradiction.

Suppose there exist φ1:I1→Rn and φ2:I2→Rn, solutions of x’=f(t,x) passing through (t0,x0), and suppose there exists t1∈I1∩I2 such that φ1(t1)≠φ2(t1). This is not possible. Suppose t1>t0. Let t̃:=inf{t∈I1∩I2:t≥t0 and φ1(t)≠φ2(t)}. By the definition of the infimum, φ1(t)=φ2(t),∀t∈[t0,t̃) and, by continuity, φ1(t̃)=lim(t→t̃–)φ1(t)=lim(t→t̃–) φ2(t)=φ2(t̃), which implies that φ1(t)=φ2(t), ∀t∈[t0,t̃]. Then, we consider x̃:=φ1(t̃)=φ2(t̃) and repeat for (t̃, x̃) the same process carried out for (t0,x0), which will lead, by the Contraction Mapping Theorem, to φ1(t)=φ2(t) ∀t∈[t̃−δ̃ ,t̃+δ̃] (with δ̃ > 0), which contradicts the choice of t̃.