What does the following matrix expression equal to after differentiating it wrt. $\dot{\mathrm{x}}$ and then wrt time

Question

Suppose I have the following expression:

$$\dot{\mathbf{x}}^\intercal\left(\mathbf{A}-\mathbf{B}\mathbf{D}^{-1}\mathbf{B}^\intercal\right)\dot{\mathbf{x}}$$

where $\mathbf{x}(t)\in \mathbb{R}^{m}$ is a vector and $\mathbf{A}\in \mathbb{R}^{m\times m}, \mathbf{B}\in \mathbb{R}^{m\times (m-n)}$ and $\mathbf{D}\in \mathbb{R}^{(m-n)\times(m-n)}$ are matrices that are dependent on $\mathbf{x}(t)$.

Derivation of the expression with respect to $\dot{\mathbf{x}}$ yields

$$2\dot{\mathbf{x}}^\intercal\left(\mathbf{A}-\mathbf{B}\mathbf{D}^{-1}\mathbf{B}^\intercal\right)$$

according to the rules discussed in this Wikipedia article about matrix calculus.

What happens if I would like to further derive this expression w.r.t. time?

So my question in short: What does this expression equal to?

$$\dfrac{\text{d}}{\text{d}t}\left(\dfrac{\partial}{\partial \dot{\mathbf{x}}}\left(\dot{\mathbf{x}}^\intercal\left(\mathbf{A}(\mathbf{x}(t))-\mathbf{B}(\mathbf{x}(t))\mathbf{D}^{-1}(\mathbf{x}(t))\mathbf{B}^\intercal(\mathbf{x}(t))\right)\dot{\mathbf{x}}\right)\right)$$

Thank you in advance!

Answer 1

Define the variables $$\eqalign{ v &= {\dot x},\,\,\,\,\,M&=A-BD^{-1}B^T \cr {\mathcal E} &= v^TMv &= M:vv^T \cr }$$ where colon is a convenient product notation for the trace, i.e. $\,\,A:B={\rm Tr}(A^TB)$

Now calculate the differential and gradient of the scalar function $$\eqalign{ d{\mathcal E} &= M:(dv\,v^T+v\,dv^T) \cr &= (M+M^T):(dv\,v^T) \cr &= (M+M^T)v:dv \cr p=\frac{\partial{\mathcal E}}{\partial v} &= (M+M^T)v \cr }$$ NB:   If $M=M^T$ we can simplify this to $\,\,p = 2Mv$.

Now take the time derivative of the $p$ vector $$\eqalign{ {\dot p} &= 2{\dot M}v + 2M{\dot v} \cr }$$ Since you were vague on how your matrices depend on the $x$-vector, that's about as far as I can take it.