2.1.10.1 Existence and uniqueness analysis

\[\begin {aligned} x^{\prime }&=\sqrt {x^{2}-1}\\ x \left (0\right ) &= 1\\ \end {aligned}\]

This is non linear first order ODE. In canonical form it is written as

\begin{align*} x^{\prime } &= f(t,x)\\ &= \sqrt {x^{2}-1} \end{align*}

The \(x\) domain of \(f(t,x)\) when \(t=0\) is

\begin{align*} \{1\le x \le \infty , -\infty \le x \le -1\} \end{align*}

And the point \(x_0 = 1\) is inside this domain. Now we will look at the continuity of

\begin{align*} \frac {\partial f}{\partial x} &= \frac {\partial }{\partial x}\left (\sqrt {x^{2}-1}\right ) \\ &= \frac {x}{\sqrt {x^{2}-1}} \end{align*}

The \(x\) domain of \(\frac {\partial f}{\partial x}\) when \(t=0\) is

\begin{align*} \{-\infty \le x <-1, -1<x <1, 1<x \le \infty \} \end{align*}

But the point \(x_0 = 1\) is not inside this domain. Hence existence and uniqueness theorem does not apply. Solution exists but no guarantee that unique solution exists.