Let ${\bf{x}} \in \mathbb{R}^{n}$ be a column vector: $${\bf{x}} = \begin{pmatrix} x_{1} \\ \vdots \\ x_{n} \end{pmatrix} $$ and ${\bf{y}} \in \mathbb{R}^{n}$ analogously. Also, consider an $n \times n$ matrix $S$ and the weighted Laplacian: $$\Delta_{S} := \sum_{i,j=1}^{n}\frac{\partial}{\partial y_{i}}S_{ij}\frac{\partial}{\partial x_{i}}$$ where $S_{ij}$ are the entries of $S$.
Question: If $A$ is another $n\times n$ matrix, what is the result of the application: $$\Delta_{S}({\bf{y}}^{T}A{\bf{x}})^{2}?$$
This is not homework assignment. This question arose during a conversation with a friend. I calculated: $$\Delta_{S}({\bf{y}}^{T}A{\bf{x}}) = \operatorname{Tr}(SA)$$ but I was not able to find a nice closed expression for the former.
We notice that
$$\frac{\partial^2 (f^2)}{\partial x_i\partial y_j}=2f.\frac{\partial^2 f}{\partial x_i\partial y_j} +2\frac{\partial f}{\partial x_i}\frac{\partial f}{\partial y_j}$$
Then
$$\Delta_S(f^2)=2f\Delta_S(f)+2(\nabla_y(f))^TS(\nabla_x(f))$$
With $f = y^TAx$, we have $$\nabla_x(f)=\nabla_x((y^TA)x)=A^Ty$$ $$\nabla_y(f)=\nabla_x((x^TA^T)y)=Ax$$
Hence
$$ \begin{align} \Delta_{S}({\bf{y}}^{T}A{\bf{x}})^{2}=2{\bf{y}}^TA{\bf{x}}.\text{Tr}(SA)+2(A^Ty)^TS(Ax)\\ =2.{\bf{y}}^TA{\bf{x}}.\text{Tr}(SA)+2y^TASAx\\ \end{align} $$
Q.E.D