TITLE: Why integral curves cannot be tangent to each other?
QUESTION [8 upvotes]: I am watching Lecture 1 of the differential equation course on MIT OpenCourseWare.
The teacher said, for $y'= f(x,y)$, if $f(x,y)$ is continuous around a point $(x0,y0)$, it would guarantee at least one solution, and if $\frac{\partial f(x,y)}{\partial {y}}$  is continuous, it would ensure uniqueness of the solution, i.e. there would be only one integral curve passing through  $(x0,y0)$, meaning two integral curves could not be tangent to each other at $(x0,y0)$.
Can anyone explain why there would be only one integral curve passing through $(x0,y0)$, i.e. two integral curves could not be tangent to each other at $(x0,y0)$, if $\frac{\partial f(x,y)}{\partial {y}}$ is continuous at $(x0,y0)$? 
If 2 integral curves are tangent to each other at $(x0,y0)$, then they should have the same derivative at  $(x0,y0)$, which is equal to $f(x0,y0)$, right? How would this make $\frac{\partial f(x,y)}{\partial {y}}$ discontinuous at $(x0,y0)$?
Thanks!

REPLY [3 votes]: If 
$f_y(x, y) = \dfrac{\partial f(x, y)}{\partial y} \tag 1$
is continuous in some neighborhood $U$of $(x_0, y_0)$, then $f(x, y)$ is Lipschitz continuous in $y$ on some, perhaps smaller, open ball $B((x_0, y_0), \epsilon) \subset U$ (we note $(x_0, y_0) \in B((x_0, y_0), \epsilon)$); this follows from the observation that, since the closure $\bar B((x_0, y_0), \epsilon)$ of $B((x_0, y_0), \epsilon)$ is compact and $f_y(x, y)$ is continuous, $\vert f_y(x, y) \vert$ is bounded by some real $L > 0$ on $\bar B((x_0, y_0), \epsilon)$; then for 
$(x, y_1), (x, y_2) \in B((x_0, y_0), \epsilon), \tag 2$
we have
$\vert f(x, y_2) - f(x, y_1) \vert = \left \vert \displaystyle \int_{y_1}^{y_2} f_s(x, s) \; ds \right \vert \le \left \vert \displaystyle \int_{y_1}^{y_2} \vert f_s(x, s) \vert \; ds \right \vert \le \left \vert \displaystyle \int_{y_1}^{y_2} L \; ds \right \vert = L\vert y_2 - y_1 \vert, \tag 3$
which shows that $f(x, y)$ is Lipschitz on $B((x_0, y_0), \epsilon)$; thus we may invoke the 
Picard-Lindeloef theorem to affirm that there exists precisely one solution passing through any point in $B((x_0, y_0), \epsilon)$; in particular, there is exactly one integral curve $y(x)$ of
$y' = f(x, y) \tag 4$
with  
$y(x_0) = y_0. \tag 5$
Thus, "two integral curves could not be tangent to each other at $(x0,y0)$" since there is in fact only one integral curve through this point! 
Our colleague Christian Blatter has, in his answer, provided some good examples of how uniqueness may fail in the absence of Lipschitz continuity.