# Existence and Uniqueness of solutions

1. Dec 5, 2012

### Zondrina

1. The problem statement, all variables and given/known data

These questions were on my midterm a while ago. I want to understand this concept fully as I'm certain these will appear on my final tomorrow and I didn't do as well as I would've liked on these questions.

http://gyazo.com/205b0f7d720abbcc555a5abe64805b62

2. Relevant equations

Existence : Suppose f(t,y) is a continuous function defined in some region R, say :

R = { (x,y) | x0 - δ < x < x0 + δ, y0 - ε < y < y0 + ε }

containing the point (x0, y0). Then there exists δ1 ≤ δ so that the solution y = f(t) is defined for x0 - δ1 < x < x0 + δ1.

Uniqueness : Suppose f(t,y) and fy are continuous in a region R as above. Then there exists δ2 ≤ δ1 such that the solution y = f(t) whose existence is guaranteed from the theorem above is also a unique solution for x0 - δ2 < x < x0 + δ2.

3. The attempt at a solution

Okay, I'll start by discussing the first dot y' = 1 + y + y2cos(t), y(t0) = y0 on I = ℝ.

Suppose f(t,y) = 1 + y + y2cos(t), then fy = 1 + 2ycos(t). Notice both f and fy are continuous for all (t,y) in I. Thus by our theorems, we can conclude that a solution exists in some open interval centered around t0 and the solution will be unique in some possibly smaller interval also centered at t0.

This looks like a Riccati equation to me. I'm not sure if I should solve it, or continue my argument here.

Any pointers would be great.

2. Dec 5, 2012

### Zondrina

Sorry for the double post, but I think I have something to add. What I was confused about is if I had to pick values for δ1 and δ2 as to make the solution continuous throughout the interval. Solving the equations he gave is actually quite difficult, so I'll give a simpler example to get my point across.

So for example, suppose I had y' = 1 + y2 with the condition y(0) = 0 and I = ℝ.

Then f(t,y) = 1 + y2 and fy = 2y are defined and continuous everywhere for all (t,y) in I. Thus by our theorems, we can conclude that a solution exists in some open interval centered around 0 and the solution will be unique in some possibly smaller interval also centered at 0.

Solving the equation by separation yields y = tan(x). Notice that tan(x) has discontinuities at ±π/2, ±3π/2, ±5π/2, etc. So the solution is not continuous through the whole interval I.

So for y=tan(x) to be a solution, we have to restrict the domain of t to be in the open interval (-π/2, π/2).

So while we can pick δ as large as we like since f(t,y) is continuous, we would have to pick δ1 = δ2 = π/2 as to make our solution unique.

I'm having a bit of trouble relating this to the much more difficult equations given.