Check the convexity of a function with respect to a matrix.

Question

Suppose that I have the function $g(\varepsilon_{dev})$ as

$g(\varepsilon_{dev})=[(-I_2)^{3/2}-b I_3]^{1/3}-a$,

where $I_2=-\frac{1}{2} \operatorname{tr} (\varepsilon_{dev})^2$, $I_3 = \det(\varepsilon_{dev})$, $a$ and $b$ are constants and $\varepsilon_{dev}$ is the deviatoric part of a symmetric matrix $\varepsilon$ with $\operatorname{tr}(\varepsilon)=\operatorname{tr}(\varepsilon_{dev})=0$.

In mechanics, the function $g(\varepsilon_{dev})$ is a yield surface in the strain $\varepsilon$ field.

Now the question is how to check the convexity of $g(\varepsilon_{dev})$ with respect to $\varepsilon_{dev}$.

P.S. The deviatoric part of the $\varepsilon$ matrix is represented by

$\varepsilon_{dev}=\varepsilon - \frac{1}{3} \operatorname{tr}(\varepsilon)$.

Thanks in advance.

Answer 1

Since the matrix $\varepsilon$ is symmetric, thus its deviatoric part is also symmetric. The transformation surface, the function $g$, depends on the invariants of $\varepsilon_{dev}$, therefore it should be convex in any plane. In order to make the calculation easy, the principal planes of strain are considered in which the shear strains are zero, and since $\operatorname{tr}(\varepsilon_{dev})=0$, thus,

$\varepsilon_{dev} = \begin{bmatrix} \varepsilon_{dev_1} & 0 & 0 \\ 0 & \varepsilon_{dev_2} & 0 \\ 0 & 0 & -\varepsilon_{dev_1}-\varepsilon_{dev_2} \end{bmatrix}$

Note that in the above matrix, $\varepsilon_{dev}[3,3]$ is taken in such a way to give $\operatorname{tr}(\varepsilon_{dev})=0$.

Now that the problem has been simplified to a large extent, since $a$ and $b$ are constants, by simply calculating the Hessian matrix and consequently the Eigen values of the Hessian matrix, the convexity of the transformation surface $g$ with respect to $\varepsilon_{dev_1}$ and $\varepsilon_{dev_2}$ will be determined.

Answer 2

It seems that $g$ can be non-convex. For example, let $b=2, a=0,$

$$ A = \begin{pmatrix} 1 & 0 & 0\\ 0 & -1 &0 \\0 & 0 & 0 \end{pmatrix}, B = \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 &0 \\0 & 0 & -1 \end{pmatrix}, C = \frac 12A+\frac 12B. $$ Then, we have $g(A)=g(B)= 1$, and $g(C)= (\frac {3\sqrt 3}8 + \frac 12)^{\frac 13} > 1 $ which implies that $g$ isn't convex.