Just testing out MathJax 3 with a lovely example. Consider

\[\frac{dy}{dt} = \sqrt{y}+1 \;\; y(0)=0 \;\; t \in [0,1] \;\; (y(t) \in \mathbb{R})\]

We will show that solutions are unique eventhough it is not Lipschitz. There are multiple ways to solve this. First of all we need to make sure that a solution exists. \(\sqrt{y}+1\) is continuous in a neighbourhood of zero Hence, a solution exists. Let’s write the differential equation in its integral form

\[\begin{aligned} \int^{y}_0 \frac{ds}{\sqrt{1+s}} &= \int^{t}_0 1 d\tau \\ \int^{y}_0 \frac{ds}{\sqrt{1+s}} &= t \end{aligned}\]

For convenience let’s define \(G(z) = \int^{z}_0 \frac{ds}{\sqrt{1+s}}\) (\(G: \mathbb{R} \rightarrow \mathbb{R}\)). Observe that \(G\) is an increasing function. Therefore, for all \(t\) there exists at most one \(z\) such that \(G(z) = t\).

Note that MathJax 3 is a major re-write of MathJax that brought a significant improvement to the loading and rendering speed, which is now on par with KaTeX.