
#1
Dec512, 10:50 AM

P: 1,339

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)  x_{0}  δ < x < x_{0} + δ, y_{0}  ε < y < y_{0} + ε } containing the point (x_{0}, y_{0}). Then there exists δ_{1} ≤ δ so that the solution y = f(t) is defined for x_{0}  δ_{1} < x < x_{0} + δ_{1}. Uniqueness : Suppose f(t,y) and f_{y} are continuous in a region R as above. Then there exists δ_{2} ≤ δ_{1} such that the solution y = f(t) whose existence is guaranteed from the theorem above is also a unique solution for x_{0}  δ_{2} < x < x_{0} + δ_{2}. 3. The attempt at a solution Okay, I'll start by discussing the first dot y' = 1 + y + y^{2}cos(t), y(t_{0}) = y_{0} on I = ℝ. Suppose f(t,y) = 1 + y + y^{2}cos(t), then f_{y} = 1 + 2ycos(t). Notice both f and f_{y} 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 t_{0} and the solution will be unique in some possibly smaller interval also centered at t_{0}. 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
Dec512, 01:42 PM

P: 1,339

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 δ_{1} and δ_{2} 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 + y^{2} with the condition y(0) = 0 and I = ℝ. Then f(t,y) = 1 + y^{2} and f_{y} = 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 δ_{1} = δ_{2} = π/2 as to make our solution unique. I'm having a bit of trouble relating this to the much more difficult equations given. 


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