Let $\alpha \in \mathbb{R}$ and $f$ be a real function defined by $f(u)=-ue^{\alpha u}\ln(\lvert u \rvert)$ if $u \neq 0$ and $f(0)=0$. Let $u_0 \in \mathbb{R}^+$ for the following problem:
\begin{equation}
\begin{cases}
u'(t)=f(u(t)), t\in \mathbb{R}\\
u(0)=0
\end{cases}
\end{equation}
Has the Cauchy problem a unique maximal solution?
I tried:
Let $g$ be a function such that:
\begin{equation}
\begin{cases}
g:\mathbb{R}\times\mathbb{R} \rightarrow \mathbb{R}\\
g:(t,u) \rightarrow f(u(t))
\end{cases}
\end{equation}
I try to verify the Cauchy-Lipschitz theorem. $f$ is continuous so $g$ is also continuous.
Now I need to prove that $g$ is locally Lipschitz with respect to its second variable. I can't manage to prove it. I tried to use the mean value theorem, but without success…
Maybe it does not verify Cauchy-Lipschitz conditions.
Best Answer
The assumptions of the Cauchy-Lipschitz theorem are in fact not fulfilled. Note that $f'(u)\to\infty$ as $u\to 0$ and this implies that $f$ is not locally Lipschitz in the neighborhood of $0$. However, we still can prove that there exists a unique solution of \begin{equation} \begin{aligned} u'(t) &= f(u), \\ u(0) &= 0. \end{aligned} \tag{*}\end{equation}
Obviously $u(t)\equiv 0$ is a solution, defined on the whole real line. Therefore we only need to prove that there is no solution starting from $0$ which is not identically $0$. Assume $u$ is a solution and there exists $t_0$ such that $u(t_0)=u_0 \in (0,1)$. Whenever $u(t)\neq 0$ and $1$, we can divide equation (*) by $f(u)$ and obtain $$ \frac{u'}{f(u)} = 1 $$ and then integrate both sides from $t_0$ to $t$. By the change of variables, we have $$ \int_{t_0}^t \frac{u'(s)}{f(u(s))}\mathrm{d}s = \int_{u_0}^{u(t)} \frac{\mathrm{d}x}{f(x)}, $$ therefore $$ \int_{u_0}^{u(t)} \frac{\mathrm{d}x}{f(x)} = t-t_0. $$ Now if $u$ is a solution with $u(0)=0$, in particular $u(t)\to 0$ as $t\to 0$. Thus taking the limit in the integral we need to have $$ \int_0^{u_0} \frac{\mathrm{d}x}{f(x)} = t_0 <\infty, $$ but $\int_0^{u_0} \frac{\mathrm{d}x}{f(x)} = \infty$, which gives a contradiction.
To show that indeed the integral $\int_0^{u_0} \frac{\mathrm{d}x}{f(x)}$ is divergent, you can use comparison criterion: if $\alpha>0$, then on the interval $(0,u_0)$ $$ \frac{-1}{u\ln u}e^{-\alpha u} \geq \frac{-1}{u\ln u}e^{-\alpha u_0}$$ and if $\alpha\leq 0$ then $$ \frac{-1}{u\ln u}e^{-\alpha u} \geq \frac{-1}{u\ln u} $$ and the integral of $\frac{-1}{u\ln u}$ is easy to calculate by hand.
The analogous reasoning will work if we assume $u_0\in (-1,0)$.