Question

This is about differential equations:

Answer 1

About the existence of a solution to \(\frac{dy}{dx}=f(x,y)\)
Theorem:
f:A\(\subseteq\mathbb{R}^2\to \mathbb{R} \) \((x_0,y_0)\in A\)
1. If f \(\in \cal{C}\) then \(\exists \) a solution to the ODE on a neighborhood of \((x_0,y_0)\)

\(\sf My\ question \ is, \ what \ is\ this\ neighborhood? \ (in\ this~ context)\)

Answer 2

Assuming you're asking what a neighborhood is (feel free to correct me):

A neighborhood of a point \((x_0,y_0)\) in \(\mathbb{R}^2\) is the set of points \((x,y)\) such that \(\|(x_0,y_0)-(x,y)\|=\sqrt{(x_0-x)^2+(y_0-y)^2}<\epsilon\) for any given \(\epsilon\). Here \(\|\cdot\|\) denotes the Euclidean norm in \(\mathbb{R}^2\), i.e. distance between two points in the plane.

Visually:

where every point inside the circle centered at \((x_0,y_0)\) with radius \(\epsilon\) but not on the boundary of the circle belongs to the neighborhood.

Answer 3

Oh I figured it out. I knew what a neighborhood is. But here for the differential equation we get some function, not a point. At first I thought "solution in the neighborhood of \((x_0,y_0)\)" meant some point in that neighborhood. But it really means the solution is \(f(x,y)=0\) where \(f:D\subseteq\mathbb{R}^2\to\mathbb{R}\) where \(D\) is that neighborhood of \((x_0,y_0)\). Am I right @SithsAndGiggles ?

Answer 4

I don't know about the \(f(x,y)=0\) part of your claim, but the theorem states that a (possibly zero) solution exists.

Answer 5

I believe this is the same theorem:
http://mathworld.wolfram.com/PicardsExistenceTheorem.html
Note there is only mention of the existence of a (unique) solution.

Answer 6

ok.