# 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 $$x=y=1$$.

• what have you tried? Nov 1, 2019 at 10:16
• Didn't you read the post ? Nov 1, 2019 at 14:11
• The Peano existence theorem implies that there exists an epsilon such that your initial value problem has a solution on the domain $(1-\epsilon,1+\epsilon)$. I still don't know whether there is a solution valid on the whole of the reals. Nov 1, 2019 at 21:33
• @AngelaRichardson I am not certain how would one apply the Peano existence theorem here. I believe that there is no open set $D$ such that it contains $(1,1)$, $f:D\longrightarrow \mathbb{R}$ and $f(x,y)=\sqrt{x^2-y^2}$ for all $(x,y) \in D$ because $f$ is undefined for $y>x>0$. Nov 1, 2019 at 23:04

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 $$x \in (1-\epsilon, 1+\epsilon)$$?

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.