# How to check the convexity of this function?

I have a function $$g(x,y,\alpha)$$ that for $$g(x,y,\alpha)=0$$ defines a closed surface of $$x$$ and $$y$$. Depending on the varable $$\alpha$$ the shape of the surface changes, see the figure below$g$ on the parameter $$\alpha$$">.

As you can see in the figure, upon increasing the parameter $$\alpha$$, the shape of the closed surface changes and corners are created. I would like to find the critical value of $$\alpha$$ for which the closed surface $$g(x,y,\alpha)=0$$ is not convex anymore.

The function $$g$$ is a bit complex but I provide its expression just in case it is needed

$$g=-0.049(\frac{1}{1+\alpha^3})^\frac{1}{3}+((x^2 +x y + y^2)^\frac{3}{2}+\frac{3\sqrt{3} x y (x+y)(-1+\alpha^3)}{2(1+\alpha^3)})^\frac{1}{3}$$.

$$\textbf{What I did so far:}$$

I know that for the function $$g(x,y,\alpha)$$ to be convex its Hessian matrix should be positive semi-definite. So I calculated the Hessian as follows

$$\mathbf{H}=\begin{pmatrix} \frac{\displaystyle \partial^2 g}{\displaystyle \partial x^2} & \frac{\displaystyle\partial^2 g}{\displaystyle \partial x \partial y} \\ \frac{\displaystyle \partial^2 g}{\displaystyle \partial y \partial x} & \frac{\displaystyle \partial^2 g}{\displaystyle \partial y^2} \end{pmatrix}$$.

In order to ensure the positive semi-definiteness of the Hessian $$\mathbf{H}$$, its eigenvalues must be positive. Thus, it should satisfy the following conditions:

1. $$\det(\mathbf{H})>0$$.

2. $$\frac{\displaystyle \partial^2 g}{\displaystyle \partial x^2}>0$$.

However, after calculating I understood that $$\det(\mathbf{H})$$ is zero for every $$\alpha$$ and also $$\frac{\displaystyle \partial^2 g}{\displaystyle \partial x^2}$$ is too complex. Therefore I couldn't determine the positive semi-definiteness of $$\mathbf{H}$$ with this procedure.

I would like to know if I am on the right path for the determination of the critical value of $$\alpha$$ for which the surface $$g$$ is not convex. If yes, then how can I continue.

Switch to polar coordinates $$x = r \cos(\theta)$$, $$y=r \sin(\theta)$$, and you can explicitly parametrize your curve as $$r = R(\theta)$$. The criterion for a smooth polar curve to be convex is $$r^2 + 2 (r')^2 - r r'' \ge 0$$ for all $$\theta$$, which is rather a mess here. Numerically minimizing $$\alpha$$ subject to the constraint $$R^2 + 2 (R')^2 - R R'' = 0$$, I find that the critical $$\alpha$$ is approximately $$1.44224957028754$$.
• Thanks for your answer. In the last equation, there should be $R^2+2(R')^2-R R''=0$, right? And also the second question. How did you parametrize the curve as $r=T(\theta)$? Isn't is too complex? Commented Jul 9, 2019 at 12:54
• But I didn't understand exactly what you did that you got $\alpha=1.4422495...$. What did you minimize exactly? Could you explain more please? Commented Jul 9, 2019 at 14:44
I would introduce $$z=-x-y$$ so it is a shape on the plane $$x+y+z=0$$. Then there are $$(r,\theta)$$ with $$x=r\cos\theta\\y=r\cos(\theta+2\pi/3)\\ z=r\cos(\theta-2\pi/3)$$ Then the equation is $$(x^2+y^2+z^2)^{3/2}-cxyz=d\\ r^3(1-p\cos3\theta)=q$$ You can work out $$p$$ and $$q$$ in terms of $$\alpha$$.
Then $$x=r\cos\theta$$ so $$x^3=\frac{q\cos^3\theta}{1-p\cos3\theta}$$ The shape is convex if $$x$$ has a local maximum at $$\theta=0$$ Find the value of $$p$$ for which the second derivative $$d^2(1/x^3)/d\theta^2=0$$.