This is a two part question. On one hand, I am trying to find positive solutions of the following equation: $$1\frac{d}{dx}f=\frac{f\log(f)}{x}$$ for $x>1$. If that is not possible, I would at least like to know their behavior as $x\to\infty$. I don't think the above has a simple closed form solution, but I think it would be very useful to have a set of arbitrarily close approximations that can be calculated in some systematic fashion. The only thing that I have managed to do so far is bring the equation in the following form: $$e^ye^{u}=u+\frac{du}{dy}$$ where $y=\log(x)$ and $u=\log(f)$, and then define recursively $$e^ye^{u_n}=u_{n+1}+\frac{du_{n+1}}{dy}$$ which is solved by $u_{n+1}=Ce^{y} + e^{y}\int_0^y e^{2y'}e^{u_n}\,dy'$. The problem is that it is impossible to calculate the sequence $\{u_n\}$ past the first few elements. At least, this shows what I am trying to dofind a series (of the appropriate kind) approximation to the actual solution. As to the asymptotic behavior, I feel it should be given by the solution of $$1=\frac{f_0\log(f_0)}{x}$$ (related to the inverse Lambert W function). My intuitive thinking is that $$\frac{df_0}{dx}=\frac{f_0\log(f_0)}{x}\frac{1}{\log(x)+1}\sim\frac{f_0}{x}=\frac{1}{\log(f_0)}\sim0$$ for large $x$, thus $f_0$ roughly satisfies the differential equation. I have close to none experience with nonlinear ODEs, so any help or pointers to the correct literature would be appreciated.

$\begingroup$ Maple can't solve it so I doubt there is a closed form solution but if you just want to know the behavior why don't you solve it numerically? $\endgroup$– Sergio ParreirasCommented Nov 14, 2013 at 3:28

$\begingroup$ I would like to use the analytical approximation to the actual solution (if there is one), or at least justify the use of $f_0$. In general, since I know very little about the subject, I would like to know what could be done with equations of this form. $\endgroup$– IvanCommented Nov 14, 2013 at 3:45
1 Answer
Setting $x:=z^{1}$ and $f:=e^u$ your differential equations is transformed into
$$z^2\mathrm{d}u=(zue^{u})\mathrm{d}z$$
The curves integrating the above differential relation define a regular foliation along the line $\{z=0\}$. This line is such a solution curve. Therefore $u(z)$ cannot be asymptotic to a finite value (if any) as $z\to0$. On the contrary the inverse solutions $u\mapsto z(u)$ define analytic solutions. Pluging $z(u):=\sum_{n=0}^{\infty}a_nu^n$ you can solve for $a_n$ and derive a convergent power series, which is well approximated by the truncated Taylor polynomials. For instance depending on the initial data $a_0=z(0)$ we find
$$a_1=a_0^2 ~\mathrm{and}~ a_2=a_0^2(a_01)$$
PS: the curve $\{f_0(x)\log f_0(x)=x\}$ you mention is the locus of "horizontal" tangency of solution curves, so can't relate to the asymptotic behavior of solutions $x\mapsto f(x)$ as $x\to \infty$ since the more $x$ goes to $\infty$ the more "vertical" the solution curve becomes.