Nonconvex increasing function with single tunable inflection point

Question

In 1D nonlinear elasticity, the stress $\sigma$ is given as a function of the strain $\varepsilon > -1$. A typical expression is $$ \sigma = E\varepsilon \left( 1 - \beta \varepsilon - \delta \varepsilon^2\right) , $$ where $E >0$ is Young's modulus, and $\beta$, $\delta$ are positive constants. This function has the following nice features :

1. smooth;
2. amounts to linear elasticity in the limit $\varepsilon\to 0$, where $\sigma \simeq E \varepsilon$;
3. has a single inflection point located at the strain $\epsilon^*$;
4. is convex for $\varepsilon < \epsilon^*$ and concave for $\varepsilon > \epsilon^*$;
5. the location $\epsilon^*$ of the inflection point is tunable, in the sense that it depends on the parameters of the model.

However, it has a very annoying feature related to wave propagation. Since the speed of sound is $\propto\sqrt{\sigma'(\varepsilon)}$ where $\sigma'$ denotes the derivative of $\sigma$, we require that the stress $\sigma$

6. is an increasing function (at least) over $]-1,\infty [$.

This condition guarantees hyperbolicity of elastodynamics (i.e., real sound speeds). With the present expression of $\sigma$, (6.) is not true.

Question. Find a real function which satisfies the above requirements (1.)-(6.) The simpler, the better.

Trial. I have played with expressions such as $ \sigma = {E} \arctan(B\varepsilon)/{B} $, where $B>0$. The requirements (1.)-(4.) and (6.) are satisfied with an inflection point located at $\epsilon^* = 0$. Thus, (5.) is not satisfied. The fact is that such function does not have enough parameters to tune the slope and the curvature of its first derivative independently. Any ideas?

Answer 1

Just a thought for now, I shall try to provide a full answer to this interesting question in due course.

Maybe the task you set yourself could be simplified if you looked at the strain energy $W$, instead of looking at the stress $\sigma = \frac {\mathrm{d}W}{\mathrm{d}\epsilon}$

Then your requirements should translate to:

1) smooth

2) finite second derivative for $\epsilon = 0$

3) 4) very loosely speaking, $W'' $ is growing for $ \epsilon < \epsilon^*$, and $W''$ decreasing for $ \epsilon > \epsilon^*$

5) analogous

6) W is convex

I find these requirements somehow easier to fiddle with, might well be subjective of course. I wonder if a function of the type $ W = \epsilon^{\omega{(\epsilon})}$ could be a useful starting point, with $\omega$ a suitable function whose definition is hinted at by conditions 3) 4): $\omega(0)>2$, and so on. Another possibility is to simply solve an ODE such as $W''' = z$, and the conditions on $z(\epsilon)$ should be transparent.

Answer 2

Using an $\arctan$ expression of $\sigma$, we set $$ \sigma'(\xi) = \frac{A}{1+B^2(\xi - {\epsilon^*})^2} \, , $$ where $A$, $B$ and $\epsilon^*$ are two constants. This ensures that (1.) and (3.)-(6.) are satisfied. Condition (2.) requires $\sigma'(0) = E$ and $\sigma(0)=0$. Thus, \begin{aligned} \sigma(\varepsilon) &= \int_0^\varepsilon E\,\frac{1+B^2 {\epsilon^*}^2}{1+B^2(\xi - {\epsilon^*})^2}\, \mathrm{d}\xi \, ,\\ &= E\,\frac{1+B^2 {\epsilon^*}^2}{B}\left(\arctan(B(\varepsilon-\epsilon^*)) + \arctan(B\epsilon^*)\right) , \end{aligned} but there are many other possible choices.