I am looking for a solution in $s$ to $$ \lambda -\frac{1}{s} +K e^t \log(\delta) \delta^s = 0 $$ Mathematica is not best pleased with this equation. If the equation were $$ 0- \frac{1}{s} +K e^t \log(\delta) \delta^s =0, $$ Mathematica tells me the solution is $s= \frac{ProductLog(\frac{1}{K e^t \log(\delta)})}{\log (\delta)}$, where ProductLog is Mathematica's implementation of the Lambert W Function, but there doesn't seem to be a closed form solution for my equation. All my variables are real; in fact $s\geq 0$ , $0<\delta<1$, $0<\lambda<1$, $t\geq 0$. $K$ is a constant of integration, will have to be chosen to satisfy another equation. A numerical approach would be almost as good as analytical; this is a first order condition to a boundary value problem. Are there any tricks I can use for convenient numerical implementation here?

For some background, the solution to this, as function of $t$, will be fed to an ODE and drive a particular value from a steady state to a boundary. So efficient numerical approaches would be highly prized.

  • $\begingroup$ Have you tried linearizing it? $\endgroup$ – artificial_moonlet Sep 16 '14 at 0:50
  • $\begingroup$ So you mean something like linearizing around $s=\frac{1}{\lambda}$? $\endgroup$ – Dennis Sep 16 '14 at 1:06
  • $\begingroup$ If that's reasonable-- from your problem description, I don't have a sense of how $s$ and $\lambda$ should relate. But if $s$ runs through all nonnegative values, then you'll have to be careful with the error. $f(x) = f(x_0) + f'(x_0)(x-x_0) + E$, and $E$ (error) for the general $f$ gets bigger the farther you are from $x_0$. Maybe a second-order approximation would be better, especially if you plan to do anything numerical with this. $\endgroup$ – artificial_moonlet Sep 16 '14 at 1:21
  • $\begingroup$ Well, in my problem you can think of $s$ as realizations of random variable that has mean $\frac{1}{\lambda}$, but are actually going to be chosen to be either above or below this mean. IE in instance, I will be driving a value above its mean, so $s>\frac{1}{\lambda}$, and in another I will be driving the value below its mean, so $s<\frac{1}{\lambda}$, but the "process" I am solving for is at rest when $s = \frac{1}{\lambda}$. The problem than the $s$ I solve for could be far from $\frac{1}{\lambda}$ is a concern, however. $\endgroup$ – Dennis Sep 16 '14 at 19:03
  • $\begingroup$ Interesting. This comes from stochastics, then? (Unfortunately, I know nothing about stochastics.) But I do think that approximating the LHS around $ s= {1\over\lambda} $ is reasonable. Just be sure to keep track of the error for whatever subsequent steps you need to run the results through. If you have an idea of the mean of the upper values or of the lower values you plan to use for $s$, then perhaps write two different approximations for each of those means. $\endgroup$ – artificial_moonlet Sep 17 '14 at 0:46

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