Antiderivative of a matrix

Question

Assume an $M$ by $N$ matrix $Y$ whose entries are given by $$Y_{i,j} = \mbox{tr} \left( C_{i,j}^T X \right)$$ where $\{C_{i,j}\}_{i,j=1}^{M,N}$ and $X$ are $M$ by $N$ matrices. What is the antiderivative of $Y$ with respect to $X$?

Answer 1

Replace the matrix $Y$ by $F$ (since $y$ is used for something different in the update section), and write it as the double-dot product between the 4th order tensor ${\cal C}$ and the matrix $X$, i.e. $$F= {\cal C}:X \quad\implies F_{ij}= \sum_{pq} {\cal C}_{ijpq} X_{pq}$$ This can be flattened into a matrix-vector relationship $$x={\rm vec}(X),\quad C={\rm vecvec}({\cal C}),\quad f=C\cdot x={\rm vec}(F)$$ Your question is analogous to Hooke's Law in a linearly elastic medium.
Consider the (scalar) stress-strain energy and its differential. $$\eqalign{ \varepsilon &= \tfrac{1}{2}x\cdot C\cdot x \\ d\varepsilon &= (C\cdot x)\cdot dx = f\cdot dx \\ }$$ Integrating yields $$\eqalign{ \int f\cdot dx &= \int d\varepsilon = \varepsilon &= \tfrac{1}{2}x\cdot C\cdot x \\ &= \int F:dX &= \tfrac{1}{2}X:{\cal C}:X \\ }$$ For this analysis to hold, the stiffness tensor must be symmetric, i.e. $\;{\cal C}:X = X:{\cal C}$

Update

When $C$ is constant but not symmetric, you can pull it out of the integral and write $$\eqalign{ \int f\cdot dx = C^T:\int x\,dx^T \\ }$$ Consider the matrix integral on the RHS in $2\times 2$ case. $$\eqalign{ M=\int x\,dx^T &= \int\left[\matrix{x\\y}\right] \left[\matrix{dx&dy}\right] \\ &= \left[\matrix{\int x\,dx&\int x\,dy\\\int y\,dx&\int y\,dy}\right] =\left[\matrix{\frac{1}{2}x^2&\int x\,dy\\\int y\,dx&\frac{1}{2}y^2}\right]\\ }$$ The problem is that those off-diagonal terms are path-dependent, i.e. you cannot integrate them until you specify the path of integration $-$ the functions $y=g(x)$ and $x=h(y)$.

In physics such path-dependent line integrals indicate a non-conservative field. You can still perform the integration, but you must specify a path, and if the path is not simple then you must resort to numerical techniques.

However, if $C$ is symmetric then you have a conservative field. In this case, there exists a potential $(\varepsilon)$ and the integral is simply $\;\int\nabla\varepsilon\cdot dx=\int d\varepsilon = \varepsilon\;$ the value of which is the same for all paths, i.e. it's path-independent.

Note that $\,\Big(\int x\,dy + \int y\,dx\Big) = \int d(xy) = xy\;$ and therefore $$\left(M+M^T\right) = \left[\matrix{x^2&xy\\xy&y^2}\right] \;=\; xx^T$$ which is another route to solving the symmetric case.

Answer 2

Rephrasing, we have the following problem.

Given matrix $\mathrm A \in \mathbb R^{n \times n}$, find the scalar field $f : \mathbb R^n \to \mathbb R$ such that $\nabla_{\rm{x}} f ( \rm{x} ) = \rm{A} \rm{x}$.

If matrix $\rm{A}$ is not symmetric, I do not believe function $f$ exists.

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$2 \times 2$ case

$$\begin{aligned} \partial_1 f (x_1, x_2) &= a_{11} x_1 + a_{12} x_2\\ \partial_2 f (x_1, x_2) &= a_{21} x_1 + a_{22} x_2\end{aligned}$$

Integrating the first equation,

$$f (x_1, x_2) = \frac12 a_{11} x_1^2 + a_{12} x_1 x_2 + g (x_2)$$

Taking the partial derivative with respect to $x_2$ and using the 2nd equation,

$$g' (x_2) = a_{22} x_2 + (\underbrace{a_{21} - a_{12}}_{=0}) x_1 = a_{22} x_2$$

Note that if $a_{21} \neq a_{12}$ then $g$ is also a function of $x_1$, which is nonsense.