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I was wondering whether there is an explicit solution to the following differential equation

$$f'(x) = g'(t)\left(f(t)\left(\frac{a}{g(t)} -1 \right)-\frac{a}{g(t) \lambda}\left( 1+ W_{-1}(-e^{-1-\lambda f(t)} \right) \right) $$

All the functions are well-behaved, $W_{-1}$ is the -1 branch of the Lambert W function. I am not sure how to solve this one.

Thanks a lot for any help!

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$$f'(t) = g'(t)\left(f(t)\left(\frac{a}{g(t)} -1 \right)-\frac{a}{g(t) \lambda}\left( 1+ W_{-1}(-e^{-1-\lambda f(t)} \right) \right) $$ Let : $W_{-1}( -e^{-1-\lambda f(t) })=-y(t) \quad\to\quad -ye^{-y}=-e^{-1-\lambda f(t)}\quad\to\quad f(t)=\frac{1}{\lambda}\left( y-\ln(y)-1\right)$

$f'(t)=\frac{1}{\lambda}\left( 1-\frac{1}{y}\right) y'(t)$

$$y'(t)=\lambda \frac{y}{y-1} g'(t)\left(\frac{1}{\lambda}\left( y-\ln(y)-1\right)\left(\frac{a}{g(t)} -1 \right)-\frac{a}{g(t) \lambda}\left( 1-y \right) \right)$$

$$y'(t)=\frac{y}{y-1} g'(t)\left(\left( y-\ln(y)-1\right)\left(\frac{a}{g(t)} -1 \right)-\frac{a}{g(t)}\left( 1-y \right) \right)$$

$$y'(t)=y\: g'(t)\left(\left( 1-\frac{\ln(y)}{y-1}\right)\left(\frac{a}{g(t)} -1 \right)+\frac{a}{g(t)} \right)$$

So, we got rid of the Lambert function.

Supposing that we could solve this non-linear ODE, the result $y(t)$ gives : $$f(t)=\frac{1}{\lambda}\left( y-\ln(y)-1\right)$$

Without knowing what exactly is the function $g(t)$ it seems not possible to go further.

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  • $\begingroup$ Thank you so far! $g(t)$ is given by $$ g(t) = \frac{\lambda \overline{g} e^{-\lambda \cdot t}}{\lambda \overline{g} e^{-\lambda \cdot t} + (1-\overline{g})\mu e^{-\mu \cdot t}} $$ $\endgroup$ – ctmntlds May 3 '16 at 13:13
  • $\begingroup$ An explicit expression of $g(t)$ seems complicated. So, solving the ODE appears even more arduous. I am afraid there is few hope to solve it analytically. It seems more realistic to rely on numerical methods. $\endgroup$ – JJacquelin May 3 '16 at 13:32
  • $\begingroup$ Thanks again, now, I simplified it such that $g'(t)$ does not appear anymore in the non-linear ODE and we get to $$ y'(t) = a y \left( -1 - \frac{g(t)}{b} (1-\frac{\ln(y)}{y-1}-b) \right)$$ Is your feeling that then there is some hope to obtain $y(t)$ analytically? $\endgroup$ – ctmntlds May 4 '16 at 9:51
  • $\begingroup$ The analytical solving is possible in some very particular cases, depending on the function $g(t)$. For example it is possible in the case $g(t)=1$. But it is probably impossible in the case $g(t)=t$. One can found some complicated functions $g(t)$ allowing analytical solution. But in the majority of functions $g(t)$, I am quite sure that it is impossible. $\endgroup$ – JJacquelin May 4 '16 at 18:07

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