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Let $F(x,y)=y\ln{\left(\frac{1}{y}\right)}$ for $0<y<1$, 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 ?

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We should consider two cases:

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.

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  • $\begingroup$ What do you mean by "autonomous" ? $\endgroup$ – Sabhrant Feb 18 at 14:45
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    $\begingroup$ "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 $v(t_0)=c$. $\endgroup$ – Johannes Hahn Feb 18 at 16:36

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