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I'm trying to find the limit of the ratio of two functions $ \lim_{t \rightarrow \infty} \frac{f(t)}{g(t)} $ but only have the initial conditions and the differential equations they solve, but the equations don't have easy closed form solutions:

$f'(t) = -\beta f(t)^2 + (\beta - \gamma) f(t) + \gamma e^{-\delta t} \\ g'(t) = -\beta g(t)^2 + (\beta - \gamma) g(t) $

$f(0)=1, g(0)=1$ and $f(t) \rightarrow 0$ and $g(t) \rightarrow 0$.

L'hopital's is no use, and trying to numerically solve this for long time by solving the top and bottom first is no good, it breaks down after a point. I've consider the method of dominant balance, but I'm not sure it is applicable. I don't know what else to try, I'm a statistician out of his depth.

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    $\begingroup$ Where does $t$ tend in your limit conditions? Why do you think that solutions exist on the positive ray? This is unlikely. $\endgroup$
    – Alexandre Eremenko
    Commented Jan 28, 2015 at 21:09
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    $\begingroup$ I understand neither the votes to close nor Alexandre Eremenko's comment. If $\beta$ and $\gamma$ are assumed positive, it is clear that solutions with positive initial data exist globally for positive time. Moreover, if $\gamma>\beta$ and $\delta>0$, then both f and g tend to zero. Finding the limit of their ratio is certainly not easy, although it may be impossible except in special cases. $\endgroup$
    – Michael Renardy
    Commented Jan 29, 2015 at 16:13

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Your differential equations do have closed-form solutions. $f(t)$ is a rather complicated expression involving Bessel functions, while $$ g \left( t \right) ={\frac {{{\rm e}^{- \left( \gamma-\beta \right) t }} \left( \gamma-\beta \right) }{{\gamma-{\rm e}^{- \left( \gamma-\beta \right) t}}\beta}} $$

Since you want $g(t) \to 0$ (presumably as $t \to + \infty$), I suppose you want $\gamma > \beta$. I'll also assume $\beta, \gamma, \delta, c > 0$. As $t \to \infty$ we have $$g(t) \sim e^{-(\gamma-\beta)t} (1 - \beta/\gamma)$$

Now suppose $f(t) = u(t) g(t)$. I get

$$ u'(t) = -{\frac { \left( u \left( t \right) \right) ^{2} \left( \gamma-\beta \right) \beta}{\gamma\,{{\rm e}^{ \left( \gamma-\beta \right) t}}- \beta}}+{\frac {u \left( t \right) \left( \gamma-\beta \right) \beta }{\gamma\,{{\rm e}^{ \left( \gamma-\beta \right) t}}-\beta}}-{\frac {c \left( 2\,{{\rm e}^{ \left( \gamma-\beta-\delta \right) t }}\gamma\, \beta-{{\rm e}^{\left( 2\,\gamma-2\,\beta-\delta \right) t}}{\gamma}^ {2}-{{\rm e}^{-\delta\,t}}{\beta}^{2} \right) }{ \left( \gamma-\beta \right) \left( \gamma\,{{\rm e}^{ \left( \gamma-\beta \right) t}}- \beta \right) }} $$

For large $t$, the coefficients of the first two terms on the right go to $0$ exponentially, while the largest contribution to the third is $$ \dfrac{c \gamma}{\gamma-\beta} e^{(\gamma-\beta-\delta)t}$$

So if $\gamma - \beta - \delta > 0$, $u$ will grow exponentially, while if $\gamma - \beta - \delta < 0$, $u'$ will decay exponentially and $u$ will approach a constant.

One nice special case is $\beta = \gamma/2$, $\delta = \gamma$, where I get $$ \lim_{t \to \infty} \dfrac{f(t)}{g(t)} = \dfrac{4\,\sqrt {c\gamma}\cosh \left( {\sqrt{2c/\gamma}} \right) +4\,\sqrt {2}c\,\sinh \left( {\sqrt{2c/\gamma}} \right) }{\sqrt{2}\,\gamma\sinh \left( {\sqrt{2c/\gamma}} \right) +2\,\sqrt {c\gamma}\cosh \left( {\sqrt{2c/\gamma}} \right) } $$

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  • $\begingroup$ c in this answer is equal to $\gamma$. Fixed in OP. $\endgroup$
    – OctaveCello
    Commented Jan 29, 2015 at 14:37
  • $\begingroup$ Using the closed form solution in terms of Bessel functions and Mathematica, I was able to get an expression for the limit of the ratio. Unfortunately the expression is too long to fit into a comment. $\endgroup$
    – Michael Renardy
    Commented Jan 29, 2015 at 20:03
  • $\begingroup$ @MichaelRenardy, that is surely the most 21st-century update imagineable of Fermat's marginal remark. :-) $\endgroup$
    – LSpice
    Commented Mar 11, 2015 at 1:49
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This expands on the answer given by Robert Israel. Using the closed form solution in terms of Bessel functions and Mathematica, I found the following expression for $\lim_{t\to\infty} f(t)e^{(\gamma-\beta)t}$: $$\frac{\sqrt{\gamma } (\gamma -\beta ) \delta ^{\frac{2 (\beta -\gamma )}{\delta }} (\beta \gamma )^{-\frac{2 \beta +\delta -2 \gamma }{2 \delta }} \Gamma \left(\frac{\beta +\delta -\gamma }{\delta }\right) \left(\sqrt{\beta } I_{\frac{\beta -\gamma }{\delta }}(p)+\sqrt{e^{-\delta }} \sqrt{\gamma } I_{\frac{\beta +\delta -\gamma }{\delta }}(p)\right)}{\sqrt{\beta } \Gamma \left(\frac{-\beta +\delta +\gamma }{\delta }\right) \left(\sqrt{e^{-\delta }} \sqrt{\gamma } I_{-\frac{\beta +\delta -\gamma }{\delta }}(p)+\sqrt{\beta } I_{\frac{\gamma -\beta }{\delta }}(p)\right)},$$ where $$p=\frac{2\sqrt{\beta\gamma e^{-\delta}}}{\delta}.$$

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