π“œ

Math Notes

University Mathematics

ProofComplete

Proof of the Picard-Lindelof Theorem

We present the complete proof of the fundamental existence and uniqueness theorem for ODEs using the Banach fixed-point theorem (contraction mapping principle).

Setup

Theorem8.1Picard-Lindelof theorem

Let F:Ξ©β†’Rn\mathbf{F}: \Omega \to \mathbb{R}^n be continuous on an open set Ξ©βŠ‚RΓ—Rn\Omega \subset \mathbb{R} \times \mathbb{R}^n and Lipschitz in x\mathbf{x} with constant LL on Ξ©\Omega. Let (t0,x0)∈Ω(t_0, \mathbf{x}_0) \in \Omega and R=[t0βˆ’a,t0+a]Γ—Bb(x0)β€ΎβŠ‚Ξ©R = [t_0-a, t_0+a] \times \overline{B_b(\mathbf{x}_0)} \subset \Omega. Set M=max⁑Rβˆ₯Fβˆ₯M = \max_R \|\mathbf{F}\| and Ξ΄=min⁑(a,b/M)\delta = \min(a, b/M).

Then there exists a unique solution x∈C([t0βˆ’Ξ΄,t0+Ξ΄];Rn)\mathbf{x} \in C([t_0-\delta, t_0+\delta]; \mathbb{R}^n) of xβ€²=F(t,x)\mathbf{x}' = \mathbf{F}(t, \mathbf{x}), x(t0)=x0\mathbf{x}(t_0) = \mathbf{x}_0.

Complete Proof

Proof

Step 1: Reformulation as a fixed-point problem.

The IVP is equivalent to the integral equation x(t)=x0+∫t0tF(s,x(s)) ds\mathbf{x}(t) = \mathbf{x}_0 + \int_{t_0}^{t} \mathbf{F}(s, \mathbf{x}(s))\,ds. Define the operator:

T[x](t)=x0+∫t0tF(s,x(s)) ds.T[\mathbf{x}](t) = \mathbf{x}_0 + \int_{t_0}^{t} \mathbf{F}(s, \mathbf{x}(s))\,ds.

A solution of the IVP is precisely a fixed point of TT.

Step 2: Define the complete metric space.

Let I=[t0βˆ’Ξ΄,t0+Ξ΄]I = [t_0 - \delta, t_0 + \delta] and consider:

X={x∈C(I;Rn):βˆ₯x(t)βˆ’x0βˆ₯≀bΒ forΒ allΒ t∈I}.X = \{\mathbf{x} \in C(I; \mathbb{R}^n) : \|\mathbf{x}(t) - \mathbf{x}_0\| \leq b \text{ for all } t \in I\}.

Equip XX with the weighted supremum norm:

βˆ₯xβˆ₯Ξ»=sup⁑t∈Ieβˆ’Ξ»βˆ£tβˆ’t0∣βˆ₯x(t)βˆ’x0βˆ₯,\|\mathbf{x}\|_\lambda = \sup_{t \in I} e^{-\lambda|t-t_0|}\|\mathbf{x}(t) - \mathbf{x}_0\|,

where Ξ»>L\lambda > L is to be chosen. The space (X,βˆ₯β‹…βˆ₯Ξ»)(X, \|\cdot\|_\lambda) is a closed subset of a Banach space, hence complete.

Step 3: TT maps XX to XX.

For x∈X\mathbf{x} \in X and t∈It \in I:

βˆ₯T[x](t)βˆ’x0βˆ₯=βˆ₯∫t0tF(s,x(s)) dsβˆ₯≀M∣tβˆ’t0βˆ£β‰€Mδ≀b.\|T[\mathbf{x}](t) - \mathbf{x}_0\| = \left\|\int_{t_0}^{t} \mathbf{F}(s, \mathbf{x}(s))\,ds\right\| \leq M|t - t_0| \leq M\delta \leq b.

So T[x]∈XT[\mathbf{x}] \in X. Moreover, T[x]T[\mathbf{x}] is continuous (as the integral of a continuous function).

Step 4: TT is a contraction.

For x,y∈X\mathbf{x}, \mathbf{y} \in X:

βˆ₯T[x](t)βˆ’T[y](t)βˆ₯β‰€βˆ«t0tβˆ₯F(s,x(s))βˆ’F(s,y(s))βˆ₯ ds≀L∫t0tβˆ₯x(s)βˆ’y(s)βˆ₯ ds.\|T[\mathbf{x}](t) - T[\mathbf{y}](t)\| \leq \int_{t_0}^{t} \|\mathbf{F}(s, \mathbf{x}(s)) - \mathbf{F}(s, \mathbf{y}(s))\|\,ds \leq L\int_{t_0}^{t} \|\mathbf{x}(s) - \mathbf{y}(s)\|\,ds.

Using the weighted norm, βˆ₯x(s)βˆ’y(s)βˆ₯≀eλ∣sβˆ’t0∣βˆ₯xβˆ’yβˆ₯Ξ»\|\mathbf{x}(s) - \mathbf{y}(s)\| \leq e^{\lambda|s-t_0|}\|\mathbf{x} - \mathbf{y}\|_\lambda, so (for tβ‰₯t0t \geq t_0):

βˆ₯T[x](t)βˆ’T[y](t)βˆ₯≀Lβˆ₯xβˆ’yβˆ₯λ∫t0teΞ»(sβˆ’t0) ds=LΞ»βˆ₯xβˆ’yβˆ₯Ξ»(eΞ»(tβˆ’t0)βˆ’1).\|T[\mathbf{x}](t) - T[\mathbf{y}](t)\| \leq L\|\mathbf{x} - \mathbf{y}\|_\lambda \int_{t_0}^{t} e^{\lambda(s-t_0)}\,ds = \frac{L}{\lambda}\|\mathbf{x} - \mathbf{y}\|_\lambda (e^{\lambda(t-t_0)} - 1).

Therefore:

eβˆ’Ξ»βˆ£tβˆ’t0∣βˆ₯T[x](t)βˆ’T[y](t)βˆ₯≀LΞ»βˆ₯xβˆ’yβˆ₯Ξ».e^{-\lambda|t-t_0|}\|T[\mathbf{x}](t) - T[\mathbf{y}](t)\| \leq \frac{L}{\lambda}\|\mathbf{x} - \mathbf{y}\|_\lambda.

Choosing Ξ»>L\lambda > L gives q=L/Ξ»<1q = L/\lambda < 1, so βˆ₯T[x]βˆ’T[y]βˆ₯λ≀qβˆ₯xβˆ’yβˆ₯Ξ»\|T[\mathbf{x}] - T[\mathbf{y}]\|_\lambda \leq q\|\mathbf{x} - \mathbf{y}\|_\lambda.

Step 5: Apply the Banach fixed-point theorem.

Since (X,βˆ₯β‹…βˆ₯Ξ»)(X, \|\cdot\|_\lambda) is complete and T:Xβ†’XT: X \to X is a contraction, there exists a unique xβˆ—βˆˆX\mathbf{x}^* \in X with T[xβˆ—]=xβˆ—T[\mathbf{x}^*] = \mathbf{x}^*. This is the unique solution of the IVP on II. β– \blacksquare

β– 

Remarks on the Proof

RemarkAlternative proof using the standard supremum norm

One can also use the standard supremum norm βˆ₯xβˆ₯∞=sup⁑t∈Iβˆ₯x(t)βˆ’x0βˆ₯\|\mathbf{x}\|_\infty = \sup_{t \in I}\|\mathbf{x}(t) - \mathbf{x}_0\| by choosing Ξ΄\delta small enough that LΞ΄<1L\delta < 1. The weighted norm trick with Ξ»>L\lambda > L avoids this restriction and allows Ξ΄\delta to be determined purely by the geometry (Ξ΄=min⁑(a,b/M)\delta = \min(a, b/M)) without needing LΞ΄<1L\delta < 1.

RemarkConstructive nature

Unlike the Peano theorem (which uses compactness arguments), this proof is constructive: the Banach fixed-point theorem gives explicit iterates Tn[x0]T^n[\mathbf{x}_0] that converge to the solution. These are exactly the Picard iterates. The convergence rate is geometric: βˆ₯Tn[x0]βˆ’xβˆ—βˆ₯λ≀qn/(1βˆ’q)β‹…βˆ₯T[x0]βˆ’x0βˆ₯Ξ»\|T^n[\mathbf{x}_0] - \mathbf{x}^*\|_\lambda \leq q^n/(1-q) \cdot \|T[\mathbf{x}_0] - \mathbf{x}_0\|_\lambda.

ExampleApplying the theorem

For xβ€²=sin⁑(x)+tx' = \sin(x) + t, x(0)=0x(0) = 0: F(t,x)=sin⁑x+tF(t,x) = \sin x + t is C∞C^\infty and globally Lipschitz in xx with L=1L = 1 (since βˆ£βˆ‚F/βˆ‚x∣=∣cos⁑xβˆ£β‰€1|\partial F/\partial x| = |\cos x| \leq 1). By Picard-Lindelof, the IVP has a unique solution. Since ∣Fβˆ£β‰€1+∣t∣|F| \leq 1 + |t| (linear growth), the solution exists for all t∈Rt \in \mathbb{R}.