I am looking for explicit function descriptions $F_1(s)$ and $F_2(s)$, following the line plotted. The line is just a description, but $F_1$ should never exceed $F_m$ and start at $s_0$ with a tangent of $dF_0$.
$F(s) = \left\lbrace \begin{matrix} dF_0\cdot s & 0 \leq s < s_0 \\ F_1(s) & s_0 \leq s < s_m \\ F_2(s) & s_m \leq s < s_s \\ F_s & s \geq s_s \end{matrix} \right.$
Both $F$ and $\dfrac{dF}{ds}$ need to be continious, such that for $F_1(s)$ at $s_0$:
$F_1(s_0) = dF_0 \cdot s_0$
$\dfrac{dF_1(s_0)}{ds} = dF_0$
at $s_m$:
$F_1(s_m) = F_m$
$\dfrac{dF_1(s_m)}{ds} = 0$
$F_2(s_m) = F_m$
$\dfrac{dF_2(s_m)}{ds} = 0$
and at $s_s$:
$F_2(s_s) = F_s$
$\dfrac{dF_2(s_s)}{ds} = 0$
For $F_1$ I've tried to use quadratic and cubic polynomials and a sine, but without any clear result.
Edit:
The actual values are:
$s_0 = 0.015;$
$dF_0 = 120\cdot 10^3;$
$s_m = 0.11;$
$F_m = 4.4\cdot 10^3;$
$s_s = 0.5;$
$F_s = 4.25\cdot 10^3;$
Since you know both values and derivatives at each node, it is a Hermite interpolation problem.
The simplest solution to the problem can be obtained using cubic Hermite spline. Note, we are using spline, i.e. a piecewise-polynomial function, not a single polynomial. $$ F(x) = \begin{cases} dF_0 \cdot s, & 0 \leq s < s_0\\ F_2(s), & s_0 \leq s < s_m\\ F_3(s), & s_m \leq s < s_s\\ F_s, & s \geq s_s \end{cases} $$ Let $$ F_2(s_0 + (s_m - s_0) t) = f_2(t)\\ F_3(s_m + (s_s - s_m) t) = f_3(t). $$ Such parametrization rescales each interval into $t \in [0, 1]$. We have $$ f_2(0) = s_0 dF_0, \quad f_2(1) = F_m, \quad f_2'(0) = (s_m - s_0) dF_0, \quad f_2'(1) = 0\\ f_3(0) = F_m, \quad f_3(1) = F_s, \quad f_3'(0) = 0, \quad f_3'(1) = 0\\ $$ From the Wikipedia article a cubic Hermite spline is given as $$ g(t) = (2t^3 -3t^2 + 1)g(0) + (3t^2 - 2t^3) g(1) + (t^3 - 2t^2 + t) g'(0) + (t^3 - t^2) g'(1). $$ Thus $$ f_2(t) = (2t^3 -3t^2 + 1)s_0 dF_0 + (3t^2 - 2t^3)F_m + (t^3 - 2t^2 + t) (s_m - s_0) dF_0\\ f_3(t) = (2t^3 -3t^2 + 1)F_m + (3t^2 - 2t^3) F_s $$
Doing so, we end with the following spline (and it's not what you want)
The function $f_2(t)$ overshoots and we want it to be monotone.
Let's interpolate $f_2(t)$ but using another function. Recall that $f_2'(1) = 0$. Thus I've picked the following function $$ f_2(t) = a + b - b \cosh k (t - 1). $$ It has three parameters $a, b, k$, should be enough to satisfy three conditions. Also this function is monotone. $$ f_2(1) = a = F_m\\ f_2(0) = a + b - b \cosh k = s_0 dF_0\\ f_2'(0) = b k \sinh k = (s_m - s_0)dF_0. $$ Eliminating $a, b$ $$ a = F_m, b = \frac{F_m - s_0 dF_0}{\cosh k - 1} $$ we obtain equation for $k$: $$ \frac{F_m - s_0 dF_0}{\cosh k - 1} k \sin k =(s_m - s_0) dF_0 $$ Solving it numerically gives $$ k \approx 4.2628397074 $$
Now the approximating function looks better