I have the following system of ODEs $$ -4 (u a(u)+1) f'(u)=2 f(u) \left(u a'(u)+a(u)\right)+\left(u \left(3 a(u)^2+\frac{1}{4}\right) a'(u)+\left(a(u)^3+\frac{a(u)}{4}\right)\right),$$ $$a'(u)=\frac{1}{2} u f'(u),$$ $$a(0)=-1/2, f(0)=0.$$
I need to find the behaviour of solutions at $u\rightarrow\infty$. Numerical experiments indicate that $f(u)\approx 3/8$ and $a(u)\approx -1/u$. But how can it be rigorously shown? I would be even happy to be able to show that $\lim_{u\rightarrow\infty}a(u)=0.$ How to achieve that?
These equations have some relevance for Hubbard model of strongly correlated system (2-sites case). $-1/2\le a(u)\le 1/2$ implies that some approximate method fulfils the Pauli exclusion principle.
The first equation can be completely integrated. I a first step only using the first equation one can combine $$ -2(ua(u)+1)f'(u)=\frac{d}{du}\left[2(ua(u)+1)f(u) + u\left(a(u)^3+\frac{a(u)}4\right)\right] $$ Using also the second equation this results in $$ C = 2a(u)^2+2f(u)+2(ua(u)+1)f(u) + u\left(a(u)^3+\frac{a(u)}4\right) $$ With the initial conditions at $u=0$ we get $C=\frac12$.
Multiply the first equation with $f(u)$ and integrate, or reversely consider $$ \frac{d}{du}\left[2(ua(u)+1)f(u)^2+uf(u)\left(a(u)^3+\frac{a(u)}4\right)\right] = uf'(u)\left(a(u)^3+\frac{a(u)}4\right) \\= 2a'(u)\left(a(u)^3+\frac{a(u)}4\right)=\frac{d}{du}\left(\frac{a(u)^4}2+\frac{a(u)^2}4\right) $$ giving another first integral. Together these should be sufficient to compute values for the limits $f_0$ and $a_{-1}$ of $f(u)$ and $ua(u)$ for $u\to\infty$, under the empirical assumption that they exist and that $a_0=a(\infty)=0$.
The remaining system is $$ \frac12=\frac94a_{-1}+4f_0\iff a_{-1}=\frac{2-16f_0}9\\ -\frac{3}{32}=2(a_{-1}+1)f_0^2+f_0\frac{a_{-1}}4 $$ which can be reduced to a cubic equation in $f_0$ $$ -\frac{27}{8}=8(11-16f_0)f_0^2+f_0(2-16f_0)=-128f_0^3+72f_0^2+2f_0 $$ This has its only real root at $f_0=0.64914270034929$, which is not the measured value.