Path planning problem of the form $Ae^{-2x}+Be^{-x}-Cx+D=0$

Question

For a path planning I startet out with using an $e$-Function which approximately suits the system behaviour in reality for the first part of the movement:

(1):

$F(t)=(1-e^{-\frac{t}{T}})*s$,

with $s, T$ being constant.

Though, due to different requirements, only the first part of the path should be described by an $e$-function, going into a linear path with a constant velocity $v$ reaching given $s$ after given time $t_{total}$.

So the new path is described with

(2):

For $t<=t_1$: $P(t)=(1-e^{-\frac{t}{T}})*s$

For $t_1<t<t_{total}$: $P(t)=(1-e^{-\frac{t_1}{T}})*s+v(t-t_1)$

Also we know at $t_{total}$ following shall be true

(3):

$P(t_{total})=s=(1-e^{-\frac{t_1}{T}})*s+vt_{total}-vt_1$

Looking for a solution for unknown $t_1$, this leads to

(4):

$t_1=(W(-\frac{s}{Tv}e^{-\frac{vt_{total}+s}{Tv}})+\frac{vt_{total}+s}{Tv})*T$,

with $W()$ being the Lambert function.

Unfortunately, this is a discountinous path, why the part between the $e$-function and the linear part shall be a path with a constant deceleration $a$. First, I get the derivative of $F(t)$

(5):

$F'(t) = \frac{s}{T}*e^{-\frac{t}{T}}$

Now, formulating the problem leads to:

(6):

$s=F(t_1)+at_2^2+F'(t_1)t_2+vt_3$, with

$v=2at_2+F'(t_1)$ and $t_{total}=t_1+t_2+t_3$

Substitute

(7):

$x=\frac{t_1}{T}$,

leads to a problem of the form

(8):

$Ae^{-2x}+Be^{-x}-Cx+D=0$,

with $x, A, C > 0$ and real, which I want to solve for $x$.

In this more specific case, the problem should look like this (if I did not miscalculate)

(9):

$\frac{s^2}{4aT^2}e^{-\frac{2t_1}{T}}+s(\frac{v}{2aT}-1)e^{-t_1/T}-vt_1+vt_{total}-\frac{v^2}{4a}=0$, where

$s,v,T, t_1, t_{total} > 0$ and $a < 0$. One example would be

$s=0.04[m]; v=0.04[m/s], T=0.16[s], t_{total} = 0.65[s], a = -3[m/s^2]$, though $a$ might be chosen differently. Also, I am looking for a solution not only for this example.

No matter how hard I try, it seems like I am hitting a brick wall. So, I will be very happy for your help.

Thanks!

Answer 1

If the zero of function $$f(x)=a\,e^{-2x}+b\,e^{-x}-cx+d$$ exists, it is bounded by $$x_1=\frac{d}{c}+W\left(\frac{a+b }{c}e^{-\frac{d}{c}}\right)\quad \quad \text{and} \quad \quad x_2=\frac{d}{c}+\frac 12W\left(2\frac{a+b }{c}e^{-\frac{2d}{c}}\right)$$ Since you will need some numerical method, this could be of interest.

Using prime numbers $(a=3,b=7,c=13;d=19)$, this gives $$1.49985 ~<~x~<~1.61459$$ while the solution is $x=1.58198$.

If you use Newton method, select as starting point the $x$ to which correspond $$f(x)\times f''(x)>0.$$ This is by the Darboux theorem.