initial value problem with differential Does this initial value problem have a solution which is valid on the domain 
$\mathbb{R}$?  
$$y'=\sqrt{x^2-y^2}\\ y(1) = 1$$
If not, does it have a solution which is valid on the domain $(1-\epsilon, 1+\epsilon)$? I can't use the Picard Lindelöf theorem here since $\sqrt{x^2-y^2}$ is not Lipschitz continuous with respect to $y$ when $x=y=1$.
 A: First, I will allow myself to refine the statement of the question slightly (I can only hope that you are not trying to determine an answer to a more general question).
Question. Does there exist $\epsilon \in \mathbb{R}$, $\epsilon > 0$ and a function $y:(1-\epsilon, 1+\epsilon)\longrightarrow\mathbb{R}$ such that it is differentiable on $(1-\epsilon, 1+\epsilon)$, $y(1)=1$ and $(D_{x} y)(x) = \sqrt{x^2-y(x)^2}$ for all $x \in (1-\epsilon, 1+\epsilon)$?
Answer
Suppose that $y$ and $\epsilon$ satisfy the conditions stated above. Then, given $x_0 = 1$, $y(x_0) = x_0 = 1$. Therefore, $(D_x y)(x_0)=\sqrt{1-1}=0$.
Given that $y(x_0)>0$, by differentiability of $y$, we can find $\epsilon_0$ such that $0 < \epsilon_0 < \epsilon$, $\epsilon_0 < 1$ and $0\leq y(x)$ for all $x \in (1-\epsilon_0, 1+\epsilon_0)$. 
The square root is only defined for nonnegative real numbers. Therefore, $x^2 - y(x)^2 \geq 0$ for all $x \in (1-\epsilon_0, 1+\epsilon_0)$. Equivalently, $y(x)^2 \leq x^2$ for all $x \in (1-\epsilon_0, 1+\epsilon_0)$. Taking into account that $0 \leq x$ and $0 \leq y(x)$ for all $x \in (1-\epsilon_0, 1+\epsilon_0)$, this can be simplified to $y(x)\leq x$.
The derivative of $y$ at $x_0$ is equal to $(D_{x} y)(x_0) = \lim_{h\uparrow 0}\frac{y(x_0+h)-y(x_0)}{h}=\lim_{h\uparrow 0}\frac{y(x_0+h)-x_0}{h}$ (for convenience, I use one sided derivative). However, from above, also $y(x_0+h) \leq x_0 + h$ for all $-\epsilon_0 < h < 0$. Therefore, $\frac{y(x_0+h)-x_0}{h} \geq 1$ for all $-\epsilon_0 < h < 0$. Thus, $(D_{x} y)(x_0)=\lim_{h\uparrow 0}\frac{y(x_0+h)-x_0}{h} \geq 1$. However, this is in contradiction with $(D_{x} y)(x_0) = 0$, as shown above.
Thus, $\epsilon$ and $y$ that satisfy the conditions do not exist.
A: It is interesting to note that while the domain of y can’t be extended to the left at all ( as xanonec showed), it can be extended infinitely to the right, with only the right-hand derivative defined for y(1).the resulting function is asymptotic to the hyperbola x^2-y^2=1.
