Does this initial value problem have a solution which is valid on the domain


$$y'=\sqrt{x^2-y^2}\\ y(1) = 1$$

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 $x=y=1$.

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    $\begingroup$ what have you tried? $\endgroup$ Nov 1, 2019 at 10:16
  • $\begingroup$ Didn't you read the post ? $\endgroup$ Nov 1, 2019 at 14:11
  • $\begingroup$ 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. $\endgroup$ Nov 1, 2019 at 21:33
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    $\begingroup$ @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 $y>x>0$. $\endgroup$ Nov 1, 2019 at 23:04

2 Answers 2


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.


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.


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