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Existence Uniqueness Theorem
We will now see that rather mild conditions on the right hand side of an ordi- nary differential equation give us local existence and uniqueness of solutions. Definition. Let f : D → Rn^ be a coninuous function defined in the open set D ⊆ Rn+1. We say that f is locally Lipschitz in the Rn^ variable if for each (t 0 , x 0 ) ∈ D, there is an open set U ⊆ D containing (t 0 , x 0 ) such that there is a constant K > 0 such that if (t, x), (t, y) ∈ U , then
| f (t, x) − f (t, y) | ≤ K| x − y |
If we write f as f (t, x) with t ∈ R, x ∈ Rn, we also say that f is locally Lipschitz in x. Remark. If f (t, x) is C^1 in x, with derivative depending continuously on t, then it is locally Lipschitz in x. Theorem (Existence and Uniqueness Theorem for ODE). Suppose f (t, x) is continuous in the open set D ⊆ Rn+1^ and is locally Lipschitz in x in D. Let (t 0 , x 0 ) ∈ D. Then, the initial value problem
x ˙ = f (t, x), x(t 0 ) = x 0 (1) has a unique solution defined in a small interval I about t 0 in R. Proof. Let U be an open neighborhood about (t 0 , x 0 ) in D so that
- f is continuous in U and Lipschitz in x in U with Lipschitz constant no larger than K > 0.
- | f (t, x) | ≤ M for (t, x) ∈ U.
Let Iα = {t : | t − t 0 | ≤ α}, Bβ = {x : | x − x 0 | ≤ β}. Choose α, β
small enough so that Iα × Bβ ⊆ U. Let α 0 be small enough so that
α 0 M < β (2) and
α 0 K < 1 (3)
Now, consider the set A of continuous functions φ from Iα 0 to Rn^ such that for t ∈ Iα 0
| φ(t) − x 0 | ≤ β (4)
. With the sup norm, A is a closed bounded subset of the Banach space of continuous functions from Iα 0 into Rn. Thus, A is a complete metric space with the metric d(φ, ψ) = supt∈Iα 0 | φ(t) − ψ(t) |. Consider again the integral operator
T φ(t) = x 0 +
∫ (^) t
t 0
f (s, φ(s))ds
We claim:
- T maps A into itself.
- T is a contraction mapping on A.
Proof that T maps A into itself: Let φ ∈ A. Then, clearly T φ is a continuous map defined on all of Iα 0. Also, for t ∈ Iα 0 ,
| T φ(t) − x 0 | ≤ M | t − t 0 | ≤ M α 0 < β
so T φ ∈ A. Proof that T is a contraction on A: Let φ, ψ ∈ A. The continuous function | φ(s) − ψ(s) | assumes its maxi- mum at some point s 0 in Iα 0. Let t ≥ t 0. Then,
| T φ(t) − T ψ(t) | = |
∫ (^) t t 0 f^ (s, φ(s))^ −^ f^ (s, ψ(s))ds^ | ≤
∫ (^) t
t 0
K| φ(s) − ψ(s) |ds
≤ K| φ(s 0 ) − ψ(s 0 ) |(t − t 0 ) ≤ K| φ − ψ |α 0
| φ(t 1 ) − φ(t 2 ) | ≤ M | t 2 − t 1 |
which implies that as t 1 , t 2 approach b from the left the norm | φ(t 1 ) − φ(t 2 ) | approaches 0. This proves the existence of the desired left limit limt→b− φ(t). A similar argument works for the right limit limt→a+ φ(t). The last statement follows from the integral equation and the Fundamen- tal Theorem of Calculus. QED. Definition. A maximal solution φ to a differential equation ˙x = f (t, x) is a solution defined on an interval I such that there is no solution defined on an interval Iˆ which properly contains I. Theorem. Suppose that f (t, x) is defined, continuous, and locally Lips- chitz in x in an open set D ⊆ Rn+1, and φ is a solution defined on an interval I. Then, there is a maximal solution φˆ on an interval Iˆ which contains I. As t approaches the boundary of Iˆ, either f (t, ˆφ(t)) becomes unbounded or (t, φˆ(t)) approaches the boundary of D. Proof. Let Iˆ be the union of all intervals containining I on which a solution exists. By uniqueness, they all patch together to give a maximal soution. Suppose φˆ is this solution. If Iˆ has a right boundary point, say b, and f (t, ˆφ(t)) remains bounded as t → b, then by the previous lemma, limt→b φˆ(t) = x 0 exists. If x 0 is in the interior of D, then patching φˆ together with a solution to the IVP ˙x, x(b) = x 0 , enables one to get a solution on an interval strictly larger than Iˆ which contradicts the defintion of Iˆ. Thus, x 0 must be in the boundary of D. QED.
