Solve $x = a(1-c \exp(-bx)) $

Question

I am trying to solve the following equation \begin{align} x = a(1-c \exp(-bx)) \end{align} for some $a>0$, $c \in (0,1)$ and $b\in (0,1)$.

One can find an exact solution for this equation in terms of Lambert_W function, which is given by \begin{align} x=\frac{W\left(-a b c e^{-a b}\right)}{b}+a \end{align} However, this not very useful for me.

I was wondering if instead, we can find an upper bound on the solution in terms of standard functions.

Answer 1

You can use a linear approximation for the exponential to get an upper bound for the solution. Note that $-e^{-bx} \leq -1+bx$. This gives $$x=a(1-ce^{-bx}) \\ x \leq a(1-c(1-bx)) \\ x \leq \frac{a(1-c)}{1-abc}$$

This works well when b is small. You could do an asymptotic expansion about $0$ for b if the answer when b is small is important, or a cubic approximation. A quadratic approximation is easily solvable but would give a lower bound instead.

Answer 2

You are looking for the zero(s) of function $$f(x)=x-a \left(1-c e^{-b x}\right)$$ for which $$f'(x)=1-a b c e^{-b x}\qquad \text{and} \qquad f''(x)=a b^2 c e^{-b x}\quad >0 \quad \forall x$$ The first derivative cancels at $$x_*=\frac{\log (a b c)}{b}$$ which, by the second derivative test, corresponds to a minimum. So, to have roots the requirement is $$f(x_*)\leq 0 \implies \frac{1+\log (a b c)}{b}-a \leq0$$

If this is the case, you can approximate the roots using a Taylor expansion around $x_*$. This would give $$f(x)=f(x_*)+\frac 12 f''(x_*) (x-x_*)^2+O\big((x-x_*)^3\big)$$ which would give $$x_\pm=x_*\pm \sqrt{-2\frac{f(x_*)}{f''(x_*)}}=\frac{\log (a b c)\pm \sqrt{2} \sqrt{a b-\log (a b c)-1}}{b}$$ from which you could start Newton method.

Now, the question is : are $x_\pm$ upper or lower bounds of the solution ? The answer is not so obvious : by Darboux theorem, since $f''(x) > 0 \,\, \forall x$, if $f(x_\pm) > 0$ then $x_\pm$ is a upper bound of the solution and if $f(x_\pm) < 0$ then $x_\pm$ is an lower bound of it (but this must be switched depending on the sign of $x_\pm$).

Let us try with $a=2$, $b=\frac 12$ and $c=\frac 13$. This will give $$x_*=-2 \log (3)$$ $$x_-=-4 \left(\sqrt{\frac{\log (3)}{2}}+\frac{\log (3)}{2}\right)\approx -5.16\qquad x_+=4 \left(\sqrt{\frac{\log (3)}{2}}-\frac{\log (3)}{2}\right)\approx 0.77$$ $$f(x_-)\approx 1.64432\qquad\qquad f(x_+)\approx -0.77839$$

Using Lambert function or Newton iterations, the solutions are $-4.57856$ and $1.71755$.

So, effectively $x_- <0$ and $x_+>0$ make them both lower bounds of the solution.