I have the following problem (1.4 from Qingkai Kong’s Short Course in ODEs) in my differential equations homework, but I think that the initial condition given makes it unsolvable. Please inspect my work and tell me if I am mistaken somewhere.

"Show that for $\alpha \in (0,1]$ the IVP \begin{equation*} x \,' \, = \, \begin{cases} x\, \ln^{\,\alpha}|x|, & x \neq 0,\\ 0, & x = 0; \end{cases} \,\,\,\,\, x(0) = 0 \end{equation*} has a unique solution,’'

I am taking cases on whether $\alpha < 1$ or $\alpha = 1$. In either case I separate variables. To integrate $\frac{dx}{x \ln^{\alpha}(x)}$, I make the substitution $u = \ln |x|$.

First, take $\alpha \in (0,1)$. Then there is a constant $c$ so that $$t + c = \int \frac{du}{u^{\alpha}} = \frac{1}{1-\alpha}u^{1-\alpha} = \frac{1}{1-a}(\ln|x|)^{1-\alpha}.$$ I believe that it is impossible to satisfy the given initial condition, because substituting $0$ for each of $x$ and $t$, this would imply that $c$ was $- \infty$.

In the $\alpha = 1$ case, there is some constant $c$ so that $$t + c = \int \frac{du}{u} = \ln|\ln |x||.$$ Note that if I take $x(0) = 0$, then I cannot even take a natural log twice on the right side.

I suspect that there was an error in the statement of the problem (which is not implausible, as the textbook was published in 2014). Perhaps it ought to have given $x(0) = 1$ as an initial condition instead? This would make sense in the case that $\alpha < 1$ case, as it would give a constant of integration equal to $0$ (which I find is often the case in such problems). Then in the $\alpha = 1$ case, I could say that $x \equiv 0$ is a solution.

Any thoughts? Thanks in advance.

  • $\begingroup$ What is the differential equation at $x=0$ and what can be said about that? Also what would you get if you write $x$ as a function of $t$ while $c=-\infty$? $\endgroup$ – Kwin van der Veen Sep 7 '15 at 11:13
  • $\begingroup$ When $x=0$, $x’ = 0$ as well. I suppose I could take $x \equiv 0$ as a solution, but surely this can’t be what the author intended?! $\endgroup$ – Jordan Green Sep 7 '15 at 13:53
  • $\begingroup$ sometimes exercises can be solved very easily without using all the given information and if you see this, then you do not have spent as much time solving it, which can especially useful during an exam. It also a way of testing if you understand the theory sufficiently. $\endgroup$ – Kwin van der Veen Sep 7 '15 at 14:02

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