Apparent violation of fundamental theorem of ODEs, how to resolve?

Question

Consider, in the $(x, y)$-plane, the family of curves given by $y = (x - c)^3$, for the various possible values of the number $c$. Denote by $v$ the unit vector field everywhere tangent to this family of curves. Then it would seem that there are two integral curves of this $v$ beginning at the origin, namely, the one that goes along the $x$-axis, and the one that goes along the curve $y = x^4$. Thus, we apparently have a violation of the fundamental theorem of ordinary differential equations. How do we resolve this?

Answer 1

The curve through the point $(x_0, y_0)$ has$$c = x_0 - y_0^{1/3}.$$Its tangent vector at this point is$$\left(1, 3(x_0 - c)^2\right) = \left(1, 3y_0^{2/3}\right).$$But the unit vector field given by$$v^x = 1, \quad v^y = 3y^{2/3}$$is not smooth; and unitizing it is not going to make it smooth. Nonuniqueness of integral curves of this nonsmooth vector field does not violate the fundamental theorem of ordinary differential equations.

Remark. This is a trick question I am fond of. The first thing to do is draw these curves in the plane, just to see that there is a real issue here. It all looks so innocent. Everything in sight is smooth, and there is no reason to believe that the final vector field will not be smooth. I think the key to getting this problem is to really believe in your heart that the fundamental theorem of ordinary differential equations is true. Faced with this belief, and your diagram, there should be a real puzzle. Finally, out of pure frustration, you say the following.

OK, I am just going to write out explicitly what this vector field is.

Once you get to that point, you are in good shape. But you must actually write down the vector field, i.e. say exactly what its components are at each point. No $c$'s are allowed here! When you do so you discover, lo and behold, that is not smooth.

And this example illustrates, existence and uniqueness of integral curves fails in general for a vector field that is merely continuous, as this one is. So how nice, you might ask, must a vector field be to guarantee existence and uniqueness of integral curves? It turns out that $C^1$ is plenty good enough. In fact, there is a condition that is weaker than $C^1$ but stronger than $C^0$ that still suffices. It is that, in any local neighborhood, there be a number $a$ such that$$\left|\vec{v}(\vec{x}) - \vec{v}(\vec{x}')\right| \le a\left|\vec{x} - \vec{x}'\right|.$$This is called a local Lipschitz condition.