# Uniqueness of solution of initial value problem

Let $$F(x,y)=y\ln{\left(\frac{1}{y}\right)}$$ for $$0, and $$F(x,y)=0$$ for $$y=0$$. Show that $$y'=F(x,y)$$ has at most one solution satisfying $$f(0)=c$$, even though $$F$$ does not satisfy a Lipschitz condition

The above problem is taken from Garrett Birkhoff, Gian-Carlo Rota - Ordinary differential equations-Wiley(1989). It is on page 29, Exe F. Without solving the D.E, we can see that $$-y\ln{y}$$ is non-negative for $$0\le y<1$$, hence the solution will be an increasing function.

Suppose $$F$$ is Lipschitz in $$y$$ in the interval $$[0,1)$$ then there exist $$L$$ such that $$\left|-y_2\ln{y_2}+y_1\ln{y_1}\right|$$ $$\le L|y_2-y_1|$$ for all $$y_1,y_2 \in [0,1)$$. In particular it should also hold when $$y_2=0$$ and $$y_1\ne 0$$. This implies $$\left|\ln{y_1}\right|$$ $$\le L$$ for all $$y_1 \in (0,1)$$, but this is not possible as $$\left|\ln{y_1}\right| \rightarrow \infty$$ as $$y_1 \rightarrow 0$$. Hence $$F$$ is not Lipschitz. But if we restrict the domain for $$y$$ to be $$[ \epsilon,1)$$ where $$\epsilon>0$$ then we can see that F(x,y) is Lipschitz, as $$|F(x,y_2)-F(x,y_1)|$$ $$=\left|\frac{\partial F}{\partial y}\right||y_2-y_1|$$ and $$\left|\frac{\partial F}{\partial y}\right| \le 1+|\ln{\epsilon}|=L$$. But i am not sure whether this is good enough to show that if two function $$f$$ and $$g$$ satisfy $$f(0)=g(0)=c$$, then are equal for all $$x$$. Am i missing something ?

1st case: $$c>0$$.
In that case we rewrite the ODE $$f'(t) = -f(t)\log(f(t))$$ as $$-\log(f(t)) = \frac{f'(t)}{f(t)} = (\log(f(t)))'$$ with initial value $$\log(f(0))=\log(c)$$. This is possible, because $$f(t)\neq 0$$ in a neighbourhood of $$t=0$$. The ODE $$\begin{cases} u'(t)=-u(t) \\ u(0)=u_0\end{cases}$$ has the unique and global solution $$u(t)=u_0 e^{-t}$$ so that we get a unique solution $$\log(f(t))=\log(c)e^{-t}$$, i.e. $$f(t) = c^{e^{-t}}$$. In particular, $$f$$ is non-zero everywhere.
2nd case: $$c=0$$.
In that case $$f(t)=0$$ is a solution and it is in fact the only solution, because the ODE is autonomous and therefore the first case shows that $$f$$ is non-zero everywhere if it is non-zero anywhere.
• "Autonomous" means that the RHS, your F, does not depend on the time-parameter. Therefore the solution is always a shift of the same function: If $u$ is the solution with $u(0)=c$, then $v(t):=u(t-t_0)$ is the solution of the same ODE with $v(t_0)=c$. – Johannes Hahn Feb 18 at 16:36