Can the Second Derivative of a Function be Compact?

[Jhenrique]

Given a function f(x(t, s) y(t, s)), if is possible to compact
\frac{∂f}{∂t}=\frac{∂f}{∂x} \frac{∂x}{∂t}+\frac{∂f}{∂y} \frac{∂y}{∂t}
by
\frac{df}{dt}=\bigtriangledown f\cdot \frac{d\vec{r}}{dt}

So, analogously, isn't possible to compact the sencond derivate
\frac{\partial^2 f}{\partial s \partial t} = \frac{\partial^2 f}{\partial x^2} \frac{\partial x}{\partial s }\frac{\partial x}{\partial t } + \frac{\partial^2 f}{\partial x \partial y}\left( \frac{\partial y}{\partial s }\frac{\partial x}{\partial t } + \frac{\partial x}{\partial s }\frac{\partial y}{\partial t }\right) + \frac{\partial^2 f}{\partial y^2} \frac{\partial y}{\partial s }\frac{\partial y}{\partial t }+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial s \partial t}+\frac{\partial f}{\partial y}\frac{\partial^2 y}{\partial s \partial t}
using matrix with the matrix Hessian(f) ?

 

[MisterX]

\frac{\partial}{\partial t} = \sum_{i} \frac{\partial x_i}{\partial t}\frac{\partial}{\partial x_i}
\frac{\partial}{\partial s} = \sum_{i} \frac{\partial x_i}{\partial s}\frac{\partial}{\partial x_i}
\frac{\partial^2}{\partial s\partial t}f = \sum_{i,j} \frac{\partial x_i}{\partial s}\frac{\partial}{\partial x_i} \frac{\partial x_j}{\partial t}\frac{\partial f}{\partial x_j}
Next I applied the product rule:
\frac{\partial^2}{\partial s\partial t}f = \sum_{i,j} \frac{\partial x_i}{\partial s}\left( \frac{\partial^2 x_j}{\partial x_i \partial t}\frac{\partial f}{\partial x_j} +\frac{\partial x_j}{\partial t}\frac{\partial f}{\partial x_i \partial x_j} \right)
The second term above involves the Hessian matrix, which is sandwiched between \partial \mathbf{x}^T/\partial s and \partial \mathbf{x}/\partial t. I've moved this to become the first term.
\frac{\partial^2}{\partial s\partial t}f = \sum_{i,j} \frac{\partial x_i}{\partial s} H_{ij}(f)\frac{\partial x_j}{\partial t} + \sum_{i,j}\frac{\partial x_i}{\partial s}\frac{\partial^2 x_j}{\partial x_i \partial t}\frac{\partial f}{\partial x_j}
Finally I reverse chain ruled the second term.
\frac{\partial^2}{\partial s\partial t}f = \sum_{i,j} \frac{\partial x_i}{\partial s} H_{ij}(f)\frac{\partial x_j}{\partial t} + \sum_{j}\frac{\partial^2 x_j}{\partial s \partial t}\frac{\partial f}{\partial x_j}

You may want to check my math.

 

[Jhenrique]

Your math appears to be correct. But, I don't know how you and outhers are capable to comprehend this summation notation!

I'm looking for matrix, without summation.

 

[MisterX]

How's this?

\frac{\partial^2 f}{\partial s \partial t} = \frac{\partial \mathbf{x}^T}{\partial s} \mathbf{H}(f)\frac{\partial \mathbf{x}}{\partial t} + \frac{\partial^2 \mathbf{x}}{\partial s \partial t} \cdot \nabla f

 

[Jhenrique]

Oh yeah! This is what I was looking for! I never would be capable to deduce this!
By the way, it's complicated!
thx!

 

[Jhenrique]

Moreover... in matricial form, could be wrote so:
\begin{bmatrix} \frac{\partial^2 f}{\partial x\partial x} & \frac{\partial^2 f}{\partial x\partial y}\\ \frac{\partial^2 f}{\partial y\partial x} & \frac{\partial^2 f}{\partial y\partial y} \end{bmatrix} \cdot \begin{bmatrix} \frac{\partial x}{\partial s} \frac{\partial x}{\partial t} & \frac{\partial x}{\partial s} \frac{\partial y}{\partial t}\\ \frac{\partial y}{\partial s} \frac{\partial x}{\partial t} & \frac{\partial y}{\partial s} \frac{\partial y}{\partial t} \end{bmatrix} + \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{bmatrix} \cdot \begin{bmatrix} \frac{\partial^2 x}{\partial s\partial t}\\ \frac{\partial^2 y}{\partial s\partial t} \end{bmatrix}

But, I ask... exist dot product between matrices (tensors of rank 2)?

And is possible call the second matrix of something more simplificated? The first is H(f), the third is ∇f and the fourth is ∂²x/∂s∂t...