I do have a problem in continuum mechanics for which I kindly ask for your help. How can I obtain the isotropic scalar-valued functional $\psi(\boldsymbol{U})$ that satisfies the derivative


$$\frac{\partial \psi}{\partial \boldsymbol{U}}=\ln (\det(\boldsymbol{U})) ~\boldsymbol{1}$$


in which $\boldsymbol{1}$ represents the unit tensor. I already tried to solve the question in a systematic way: Construct the functional $\psi(I\_1,I\_2,I\_3)$ with the invariants $I\_1= \boldsymbol{U}:\boldsymbol{1}$, $I\_2= \frac{1}{2}\boldsymbol{U}:\boldsymbol{U}$ and $I\_3=\det(\boldsymbol{U})$ and calculate the derivative


$$\frac{\partial \psi}{\partial \boldsymbol{U}}=\psi\_1~\boldsymbol{1}+\psi\_2~\boldsymbol{U}+\psi\_3~I\_3~\boldsymbol{U}^{-1}$$


with $\psi\_j=\frac{\partial \psi}{\partial I\_j}$.


My "best" solution was to set $\psi=\ln(I\_3)I\_1$ but this results in an additional (unwanted) term:


$\psi\_1=\ln(I\_3)$ and $\psi\_3=I\_1 I\_3^{-1}$ and finally


$$\frac{\partial \psi}{\partial \boldsymbol{U}}=\ln(I\_3)~\boldsymbol{1}+I\_1~\boldsymbol{U}^{-1}$$.


Any help is kindly appreciated!



How to obtain an isotropic functional of a symmetric tensor which satisfies a given first derivative?
If I understand the question, we can try the case where $U$ is a two by two
matrix $$U = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$
and we want a function $\psi(U)$ that satisfies
$$ \psi\_a = \ln(ad-bc), \qquad \psi\_b = 0, \qquad \psi\_c = 0, \qquad \psi\_d = \ln(ad-bc).$$ There does not seem to be any solution unless you restrict $U$ to multiples of the identity matrix:


Suppose you have a solution. Equality of mixed partials gives $\psi\_{ab} = 0 = \frac{c}{ad-bc}$ so that $c = 0$. Similarly $b = 0$ and
we still need $$\psi\_a = \psi\_d = \ln(ad).$$ Now equality of mixed partials gives
$$ \psi\_{ad} = \frac{1}{a} = \frac{1}{d}$$. Therefore $a = d$.


