Show that $x'=x\log(x)$, $x(0) = 0$ has an unique solution

Question

I have edited the question filling some details and reworking the errors. Is this right now? Thank you

I want to show that the only solution to the differential equation $$x'(t)= \begin{cases} x(t)\log|x(t)| \text{ if $x(t) \neq 0$} \\ 0 \text{ if $x(t) = 0$ } \end{cases}$$ is the null one. I was provided a lengthy proof by my professor but I'm asking myself if a shorter one is possible.

My idea is the following. Suppose that there exist a solution $x \neq 0$. This implies there exists a point $t_1$ such that $x(t_1) \neq 0$. Let's suppose $t_1 > 0$ and call $t_0 = \inf_{t_0 \in [0, t_0]} \{ x(\tau) \neq 0: \tau \in [t_0, t_1]\}$. It is easy to see that $x(t_0) = 0$. By continuity, we can also suppose there exists $t_2$ such that $|f(x)| \leq \frac{1}{2}$ in $(t_1, t_2]$

For $t \in (t_0, t_2]$ we can rearrange our original equation as $$ \frac{x'(t)}{x(t) \log(x(t))} = 1$$ since we have made sure that the denominator doesn't vanish.

Integrating at both sides: $$ \int \frac{x'(t)}{x(t) \log(x(t))} = \log |\log |x(t)|| = t + K$$ for some $K \in \mathbb{R}$.

Taking limits as $t \to t_0$: $$ \lim_{t \to t_0} \log |\log |x(t)|| = -\infty \neq \lim_{t \to t_0} t + K = t_0 + K$$ and we reach a contradiction. Repeating the argument for $t_1 < 0$ shows that we must have $x = 0$, ensuring uniqueness.

Is this proof right or I am missing something? I am especially dubious about the rearrangement of the equation. For example, $0$ satisfies the original equation, but not the rearranged one. This is logical since we can't divide by $0$, but I'm not entirely sure I have not removed other posible solutions.

Thanks in advance!

Answer 1

Let $f(x)=x\log(|x|)$ ($f(0)=0$). Then $f$ is strictly decreasing on $I:=[-1/e,1/e]$. If $x$ is a solution of the IVP then, as long as it does not leave $I$ $$ x'(t)x(t)=f(x(t))x(t) \le 0, $$ thus $x(t)^2 \le 0$, thus $x(t)=0$. I particular, it can't leave I, hence $x(t)=0$ for all $t$.

Answer 2

Assuming $x\ne1, x\ne 0$, we have $\frac{dx}{x \log x}=1$. Integrating this, we get $\int \frac{d(\log x)}{\log x}=t+c$ or $\log \log x=t+c$. Thus, $x=Ce^{e^t}$ where $C>0$. We can also see that $x=0$ is not a solution as $\log 0$ is not defined. $x=1$ is a solution but it does not satisfy the initial condition.

Answer 3

$$\frac{dx}{dt}=x \log x \implies \int \frac{dx}{x \log x}=\int dt$$ Integrating taking $x=e^u$ we have $$\log u=t+C \implies \log(\log x)=t+C$$ As this cannot satisfy $x(0)=0$, so no solution.