# Do continuously differentiable functions which are not Lipschitz have uniqueness of solutions of ODE

In the undergrad textbook on ODEs by Blanchard, Devaney, and Hall it states that the solution to the ordinary differential equation $dy/dt=f(t,y)$ with initial value problem $y(t_0)=y_0$ has a unique solution on some interval containing $t_0$ if $f$ and $\partial f/\partial y$ are continuous on an open rectangle containing $(t_0,y_0)$.

On the other hand, the Picard–Lindelöf theorem guarantees uniqueness under the hypothesis that $f$ be uniformly Lipschitz in $y$, rather than continuously differentiable.

At first I assumed that the textbook had substituted continuous differentiability of $f$ as a stronger condition but which would be more familiar to beginning students without a background in real analysis. But there exist functions which are differentiable but not Lipschitz, such $e^y$ (on an unbounded domain) or $y^{3/2}\sin(1/y)$. If I've understood correctly, these functions would meet the criteria of the undergrad textbook, but not the Picard-Lindelöf theorem. The Wikipedia article and some M.SE answers mention stronger versions of the theorem, but none that seem to apply. Have I misunderstood the theorem? Should we expect $dy/dt = y^{3/2}\sin(1/y)$ to to have unique solutions at 0?

• I think you just need f(t,y) to be locally Lipschitz to have existence and uniqueness in some open interval of y0 – LJSilver Feb 4 '17 at 17:07
• @LJSilver ok, but I believe the function I listed is not even locally Lipschitz? – ziggurism Feb 4 '17 at 17:15
• @LJSilver I guess the example function $y^{3/2}\sin(1/y)$ is differentiable but not continuously differentiable. I guess any function which is continuously differentiable on an open interval is locally Lipschitz within that interval by Glitch's argument, so the two versions of the theorem really are the same. Thanks. – ziggurism Feb 5 '17 at 14:24

## 1 Answer

The key, as @LJSilver pointed out in the comments, is that local existence and uniqueness are guaranteed by $f$ satisfying the local Lipschitz condition, which says that $f$ is Lipschitz when restricted to compact sets. For the sake of simplicity let's further restrict our attention to compact intervals, i.e. $[a,b]$. If we assume that $f$ is continuous differentiable, then its restriction to $[a,b]$ is differentiable and $f'$ is continuous on $[a,b]$. Continuous functions on compact sets are bounded, so we can pick $M>0$ such that $|f'(x)| \le M$ for all $x \in [a,b]$. Now we use the mean value theorem. For $a \le x < y \le b$ we pick $z \in (x,y)$ such that $f'(z)(y-x) = f(y) -f(x)$. Then $$|f(y) -f(x)| = |f'(z)| |x-y| \le M |x-y|.$$ A similar estimate holds for $y < x$ and holds trivially when $x=y$. Thus $f$ is Lipshitz when restricted to $[a,b]$.

• ok yes if it's continuously differentiable on a closed bounded interval, then it is Lipschitz and the two statements agree. However this textbook assumes that the function is continuously differentiable on an open interval. There are functions which are continuously differentiable but not Lipschitz on open intervals. Are they still locally Lipschitz? – ziggurism Feb 4 '17 at 19:50
• If $f$ is continuously differentiable on $(A,B)$ then the argument I gave above will show that it is Lipschitz when restricted to any compact subinterval $[a,b] \subset (A,B)$. This is enough to use the basic local existence and uniqueness theory when the initial condition belongs to the open interval. Note, though, that the solution may not exist globally if $f$ is not globally Lipschitz. – Glitch Feb 4 '17 at 21:18
• Ok I see. If it's continuously differentiable on some bounded open interval, then it's also continuously differentiable on some smaller closed interval, where differentiability implies Lipschitz, so the hypotheses are the same on this smaller interval, And we never had any control over the size of the interval where solution defined, so we're not bothered we had to shrink the domain. Thanks. – ziggurism Feb 5 '17 at 14:20