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Second derivative and tensors

  1. Nov 26, 2013 #1
    Given a function f(x(t, s) y(t, s)), if is possible to compact
    [tex]\frac{∂f}{∂t}=\frac{∂f}{∂x} \frac{∂x}{∂t}+\frac{∂f}{∂y} \frac{∂y}{∂t}[/tex]
    by
    [tex]\frac{df}{dt}=\bigtriangledown f\cdot \frac{d\vec{r}}{dt}[/tex]

    So, analogously, isn't possible to compact the sencond derivate
    [tex]\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}[/tex]
    using matrix with the matrix Hessian(f) ?
     
  2. jcsd
  3. Nov 27, 2013 #2
    [tex]\frac{\partial}{\partial t} = \sum_{i} \frac{\partial x_i}{\partial t}\frac{\partial}{\partial x_i} [/tex]
    [tex]\frac{\partial}{\partial s} = \sum_{i} \frac{\partial x_i}{\partial s}\frac{\partial}{\partial x_i} [/tex]
    [tex]\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} [/tex]
    Next I applied the product rule:
    [tex]\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) [/tex]
    The second term above involves the Hessian matrix, which is sandwiched between [itex]\partial \mathbf{x}^T/\partial s[/itex] and [itex]\partial \mathbf{x}/\partial t[/itex]. I've moved this to become the first term.
    [tex]\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} [/tex]
    Finally I reverse chain ruled the second term.
    [tex]\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} [/tex]

    You may want to check my math.
     
  4. Nov 27, 2013 #3
    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.
     
  5. Nov 27, 2013 #4
    How's this?

    [tex]\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[/tex]
     
  6. Nov 27, 2013 #5
    Oh yeah! This is what I was looking for! I never would be capable to deduce this!
    By the way, it's complicated!
    thx!
     
  7. Dec 5, 2013 #6
    Moreover... in matricial form, could be wrote so:
    [tex]\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}[/tex]

    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...
     
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