# Existence and Uniqueness Theorem

1. Feb 2, 2014

### Bashyboy

Hello Everyone.

I have a question. Suppose I have a differential equation for which I want to find the values at which $\displaystyle f(x,y)$ and $\displaystyle \frac{\partial f}{\partial y}$ are discontinuous, that I might know the points at which more than one solution exists. Suppose that $\displaystyle y_1$ is such a value. Now suppose we want to find a unique solution at $\displaystyle (x_0, y_0)$, and that it exists. My question is, can the region that encloses $\displaystyle (x_0,y_0)$ also include $\displaystyle y_1$?

Last edited: Feb 2, 2014
2. Jul 26, 2014

### crossfire234

Hi bashyboy,

Statement of the Existence and Uniqueness Theorem:

If $f(x,y)$ and $f_y(x,y)$ are continuous in a rectangle $R: |t|<= a, |y| <= b$, then there is some interval $|t| <= h <= a$ in which there exists a unique solution $y = \phi(t)$ of the initial value problem.

I find the wording of your question a bit confusing, so I'm coming in with the assumption that you're trying to find out if $y_1$ is the value at which the function and/or its y partial are discontinuous if it can exist in the rectangle $R$. If this assumption is correct then maybe I am the confusing one!

I'm also assuming here that you have created a line $y_1$ where the function and/or its derivative are continuous. Line, point, whatever :P

So if the function and its y partial are discontinuous at that point $y_1$, then it does not satisfy the first condition of the theorem where it must be continuous in that rectangle. Therefore your points $(x_0,y_0)$ and your line $y_1$ cannot live in the same rectangle. Bad blood bro.

I hope this is what you were looking for!