This is the theorem often given in given a differential equations book:
Given that $\frac{dx}{dt}=f(t,x)$, a solution-curve $x(t)$ passing through some initial point $(t_0,x_0)$ exists and is unique if $f(t,x)$ & $\frac{d}{dx}(f(t,x))$ are both continuous around that point.
However (at least as far as I understand, although I may be incorrect) for uniqueness, we don't actually need $f(t,x)$ to be differentiable with respect to $x$ for the solution curve to be unique. We need it to be Lipschitz continuous in the $x$ direction.
I understand that an author may not want to write "Lipschitz Continuous" as a condition in his textbook since a lot of students (myself included) may have not taken an analysis course when first studying differential equations.
However, after researching a little more on Lipschitz continuity, it seems to me that its guaranteed by $\frac{d}{dx}f(t,x)$ not tending towards $\displaystyle \pm \infty$ at the point.
If we look at the picture above, the function all the way on the left is not differentiable at $a$, but it IS Lipschitz continuous, while the one all the way to the right isn't Lipschitz continuous since the derivative tends towards infinity.
Is there a historical reason, or some other reason I may be missing (I could just be incorrect about what Lipschitz continuity implies), why differential equations don't say the following:
Given that $\frac{dx}{dt}=f(t,x)$, a solution-curve $x(t)$ passing through some initial point $(t_0,x_0)$ exists and is unique if and only if $f(t,x)$ is continuous around that point and $\frac{d}{dx}(f(t,x))$ does not tend towards $\displaystyle \pm \infty$ around that point.
Thanks!
This is wrong. There are lots of other ways a continuous function can fail to be locally Lipschitz near a point without having $df/dx$ tending to $\pm \infty$ there. For example, $df/dx$ might not exist at all.
BTW, locally Lipschitz is not an "if and only if" condition either. It is possible to have uniqueness even though the function is not locally Lipschitz. See for example this post