Closed form solution for the (easy at first glance) IVP $wu' =(2-w) u$, $ww'=u-w$

Consider the nonlinear autonomous 1st order IVP:

\begin{align*} u'(t) & = \frac{(2-w) u}{w} \\ w'(t) & = \frac{u-w}{w} \end{align*} $t > 0$, with $u(0)=w(0)=0$. We know that $t = 0$ is a singular point of the system and the asymptotic behaviors $u = 3w = 6t$ are valid for $t \ll 1$.

Looking at the phase portrait, we see that the critical point $u = w = 2$ is a stable focus of the system. That allows me to say that for $t \to \infty$, $u$ and $w$ converges to the value $2$. I can support this numerically but trying to derive an analytical solution for all $t$ seems impossible.

Both Matlab and Mathematica tell me that there does not exist such a solution to the system. I have tried a variety of change of variables, which led me to nowhere, and divide formally the equations to get:

$$ \frac{\mathrm{d} u}{\mathrm{d} w} = \frac{(2-w) u}{u-w}, \quad u(0) = 0$$

I am (and so is Mr. Mathematica) completely unable to solve this equation although it looks easier (at least to the untrained eye).

My question is,

• Should I give up on solving this equation algebraically and embrace numerics?
• Is there any change of variables that would make my life easier?

Any thoughts are deeply appreciated.

It is possible to separate $u$ and $w$ quite easily for your system:

$\begin{cases} wu'=(2-w)u \\ ww'=u-w \end{cases}\implies \begin{cases} u'=(2-w)(w'+1) \\ u=w(w'+1) \end{cases}$

We can then isolate $w$ like this : $\quad w'(w'+1)+ww''=(2-w)(w'+1)$

But I'm afraid this ODE is not much simpler to solve even though it is in single variable.

There are two different attempts to attack the equation. But both of them are unsuccessful.

$${\bf Attempt\ 1.\ To\ full\ differencial}$$ Let $\ v=w-2,\ $ then $$ \begin{cases} u'=-\,\dfrac{v}{v+2}\,u\\[4pt] u = (v'+1)(v+2), \end{cases}\rightarrow \begin{cases} v''(v+2) + v'(v'+v+1) + v = 0\\ u = (v'+1)(v+2), \end{cases} $$ $$\left(\left(v'+\frac{v}2\right)(v+2)\right)' + v = 0,$$ and this seems as the maximum of possible.

$${\bf Attempt\ 2.\ Substitution.}$$ Let $\dot a = \frac{da}{dt},$ then as shown $$u = w(\dot w +1),\quad \dot u = (2-w)(\dot w+1),$$ $$w\ddot w+ (\dot w + w - 2)(\dot w+1) = 0.$$ That means that $$w\dot y + (y+w-2)(y+1) = 0,$$ where $$y=\dot w.$$ If to consider $y$ as function of $w,$ then $$\dot y = \dfrac{dy}{dt} = \dfrac{dy}{dw}\cdot\dfrac{dw}{dt} = y'y,$$ $$wyy'+(y-w-2)(y+1) = 0,$$ $$y(wy'+ y-w-1) = w+2,$$ $$y((wy)'-w-1) = w+2.\qquad(1)$$ Homogenius equation is: $$y((wy)'-w-1) = 0.$$ Non-trivial solution: $$wy = \frac12w^2 + w + C_1,$$ $$y = \frac12w+1+\dfrac {C_1}w.$$ Using the constant variation method: $$y=\frac12w+1+\dfrac zw.$$ Substitution to $(1)$ gives: $$\left(\frac12w+1+\dfrac zw\right)z' = w+2,$$ $$(2z+w^2+2w)z' = 2(w^2+2w),$$

but further simplification seems impossible.