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Can anyone help solve this non-linear ODE? I've been attempting to solve it using asymptotics, however I have found this approach falls apart under many parameter regimes of interest, I was wondering if there is a way to solve the full system analytically.

The problem can be defined as the following:

\begin{align} %%% vitreous %%% \frac{\text{d}y}{\text{d}\tau} &= - \mu y % +\Delta \left[ (1-z)-Ezy \right], \\ %%% vitreal network %%% \frac{\text{d}z}{\text{d}\tau} &= (1-z) -Ezy. \end{align}

Where $z(0)=y_0\in(0,1)$ and $z(0)=z_0\in(0,1)$, and

$$0<\mu\ll E,\Delta$$ $$E,\Delta \sim O(1)$$

This is roughly what the system looks like when numerically solved, the 2 rightmost panels are dimensional $y,z$ from middle to right. (note, $E=$1/eps)enter image description here

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    I couldn't even solve the case where everything (all parameters and $\Delta$) is set to $1$. Have you looked at numerical solutions to see what the solution curves look like? – 2017-02-06
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    @Moo: Thanks for your input, it looks like a lambert's function of a decreasing exponential. – 2017-02-06
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    @Moo: Posted a screenshot of numerics. – 2017-02-06
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    Why haven't you eliminated $y$ by using the first equation? Do you want to keep it in first order form? – 2017-02-06
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    @MrYouMath: Would you be able to elaborate? This is concerned the two equation system? – 2017-02-06

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This looks quite similar to the Lorenz system. I wonder if it can be transformed to that system. In any case, a closed-form solution is not to be expected. You may get chaotic behaviour in some parameter regimes.

EDIT: Depending on parameter values, there may be stable fixed points: your screenshot may be evidence of approach to some fixed point. In particular, $x=0, y=0, z=1$ is always a fixed point. Depending on the parameters, this may be stable or unstable. In particular, this is likely to depend on $AD-BC$ which is the determinant of the Jacobian matrix of the linearlization at that fixed point.

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    Hmm, ok! That's interesting. And yes, I entirely agree, the value of $\Delta$ is that which mainly controls the behavior, $z$ it is monotonically increasing past small time when $\Delta\gg1$. – 2017-02-06
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    The determinant of the matrix [A,B;C,D] is always positive, which can be proven with more information I have omitted here. And I agree, this is something I have been considering. – 2017-02-06