# Why is existence not guaranteed for this initial value problem using Existence and Uniqueness Theorem?

Given $\frac{dy}{dx}=\sqrt{x-y}$; $\ \$ $y(2)=2$

Why is existence not guaranteed using the Existence and Uniqueness Theorem for Differential Equations?

I thought that if $f(x,y)$ was continuous "near" the initial value, it guarantees existence of a solution of the given initial value problem.

In this case, wouldn't it exist? $f(2, 2)=\sqrt{2-2}=\sqrt{0}=0$

I understand, however, that it is not unique: $\frac{\partial f}{\partial y}=-\frac{1}{2\sqrt{x-y}}$, where it would not be continuous at $(2,2)$: $-\frac{1}{2\sqrt{2-2}}=-\frac{1}{0}$

Where am I going wrong?

• Your $f$ is not continuous in a neighborhood of $(2,2)$ even though it is in fact continuous at $(2,2)$. – Ian Sep 7 '15 at 21:20
• $\sqrt{x-y}$ is not even defined in a neighborhood of $(2, 2)$. – user99914 Sep 7 '15 at 21:24
• I guess I don't fully understand what "in the neighborhood of" means. Could someone explain how I would mathematically show that $f$ is not defined in a neighborhood of $(2,2)$ or what this exactly means? – bobtran12 Sep 7 '15 at 21:31
• One way of saying it is that there should be a rectangle centered at $(2,2)$ such that $f$ is continuous on the entire rectangle. But there is not. – Ian Sep 7 '15 at 21:36
• Oh, I see. Is there a way of showing this mathematically or would I have to picture the graph and/or plug in nearby points to prove this? – bobtran12 Sep 7 '15 at 21:41