How Does the Lemma Support the Uniqueness Theorem in ODEs?

[yifli]

I'm reading the differential equations chapter of Advanced Calculus by Loomis, and have some questions.

First it proved the following theorem:
Let A be and open subset of a Banach space W, let I be an open interval in R, and let F be a continuous mapping from I X A to W which has a continuous second partial differential. Then for any point <t_0, \alpha_0> in I X A, from some neighborhood U of \alpha_0 and for any sufficiently small interval J containing t_0, there is a unique function f from J to U which is a solution of the differential equation passing through the point <t_0, \alpha_0>

Then it states the following Lemma:
Let g_1 and g_2 be any two solutions of d\alpha/dt=F(t,\alpha) through <t_0, \alpha_0>. Then g_1(t)=g_2(t) for all t in the intersection J=J_1\cap J_2 of their domains.

What is the above lemma useful for? The theorem says there is only one solution through <t_0, \alpha_0>, why does the lemma say "g1 and g2 be any two solutions"?

 

[rmehta]

I think the difference is that the theorem says the solution exists and is unique in some "sufficiently small interval". The lemma asserts that the uniqueness (but not the existence) is more robust.

The lemma could be rephrased as: "if a solution exists on any interval, then it is the unique solution on that interval." And this is true even for large intervals, possibly even for all of R (if a solution exists for all t).

 

[yifli]

I think the difference is that the theorem says the solution exists and is unique in some "sufficiently small interval". The lemma asserts that the uniqueness (but not the existence) is more robust.

The lemma could be rephrased as: "if a solution exists on any interval, then it is the unique solution on that interval." And this is true even for large intervals, possibly even for all of R (if a solution exists for all t).

Thanks.

I just noticed the book says that "the lemma allows us to remove restriction on the range of f", so the theorem can be stated as follows:
Let A, I, and F be as in the above theorem. Then for any point <t_0,\alpha_0> in I X A and any sufficiently small interval neighborhood J of t_0, there is a unique solution from J to A passing through <t_0,\alpha_0>

Why does the lemma allow us to remove the restriction on the range of f?

 

[csco]

I don't understand your question. The lemma allow us to remove the restriction on the range of f because that what the lemma is about. Uniqueness on arbitrary intervals follows from local uniqueness that's what the lemma says.

There's a short proof of this fact. Let there be two solutions defined on interescting intervals. Let J be their intersection. Let A be the subset of J such that the solutions are equal there and B the subset of J such that the solutions are not equal. Observe that A and B are disjoint sets and that their union is J. We have to prove B is the empty set. Take a in A. Both solutions take the same value on a so by local existence and uniqueness the solutions are equal on a small interval around a. Then A is open. But B is open too because the difference of the solutions is a continuous function and takes a value different from zero for any b in B so there is a small interval around b where the solutions are different. Now A is not empty because the initial time is in A, if B where non empty too then this would imply J is not connected and this absurd so B is empty and this completes the proof.

A more important result is the existence of a maximal solution and that also follows from local uniqueness and existence easily.