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Second derivative composite function

  1. Sep 28, 2012 #1
    Hi guys, I have this function


    f(g(t)) and I have to find the second time derivative of f, is it correct the following solution?:

    f''=∂f/∂g*g'=∇f*g'
    f ''=∇^2f*|g'|^2+∇f*g''

    where ∇^2 is the laplacian function
     
  2. jcsd
  3. Sep 28, 2012 #2

    haruspex

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    Looks ok to me. Try it with some simple examples.
     
  4. Sep 28, 2012 #3

    pasmith

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    Your (standard) notational abuse has confused you.

    The chain rule for a vector-valued function (f) of a vector-valued function (g) of a scalar (t) is:
    [tex]
    \frac{d\mathbf{f}(\mathbf{g})}{dt} = (\nabla \mathbf{f}) \cdot
    \frac{d\mathbf{g}}{dt}
    [/tex]
    ie
    [tex]
    \frac{df_i(\mathbf{g})}{dt}(t) = \sum_j \frac{\partial f_i}{\partial x_j}(\mathbf{g}(t)) \frac{dg_j}{dt}(t).
    [/tex]

    Differentiating with respect to t and using the product rule gives:
    [tex]
    \frac{d^2\mathbf{f}(\mathbf{g})}{dt^2} = \left(\frac{d}{dt}(\nabla \mathbf{f})\right) \cdot \frac{d\mathbf{g}}{dt} + (\nabla \mathbf{f}) \cdot
    \frac{d^2\mathbf{g}}{dt^2}
    [/tex]
    Using the chain rule on the first term gives
    [tex]
    \frac{d}{dt}(\nabla \mathbf{f}) = (\nabla \nabla \mathbf{f})\cdot \frac{d\mathbf{g}}{dt}
    [/tex]
    and so
    [tex]
    \frac{d^2\mathbf{f}(\mathbf{g})}{dt^2} =
    \left(\left((\nabla \nabla \mathbf{f}) \cdot \frac{d\mathbf{g}}{dt}\right) \cdot
    \frac{d\mathbf{g}}{dt}\right) + (\nabla \mathbf{f}) \cdot
    \frac{d^2\mathbf{g}}{dt^2}
    [/tex]
    ie
    [tex]
    \frac{d^2f_i(\mathbf{g})}{dt^2} =
    \sum_j \sum_k \frac{dg_j}{dt} \frac {dg_k}{dt} \frac{\partial^2 f_i}{\partial x_j \partial x_k} + \sum_j \frac{d^2g_j}{dt^2} \frac{\partial f_i}{\partial x_j}.
    [/tex]

    You appear to have the first term as
    [tex]
    \sum_j \sum_k \left(\frac {dg_j}{dt}\right)^2 \frac{\partial^2 f_i}{\partial x_k^2}
    [/tex]
    which is incorrect.
     
  5. Sep 29, 2012 #4
    Thank you pasmith, actually I didn't get comething with the dimensions:
    if I have g as a vector 2x1
    [tex]
    \nabla \nabla \mathbf{f}
    [/tex]
    will be a 2x2 matrix. In the calculus of :
    [tex]
    \left(\left((\nabla \nabla \mathbf{f}) \cdot \frac{d\mathbf{g}}{dt}\right) \cdot
    \frac{d\mathbf{g}}{dt}\right)
    [/tex]
    I have a matrix 2x2 times a vector 2X1 which gives a vector 2x1 and this vector is multiplied by a vector 2x1 but the dimension are wrong.
     
  6. Oct 17, 2012 #5

    pasmith

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    Actually it's a 2x2x2 (ie, rank 3) tensor:
    [tex]
    (\nabla\nabla \mathbf{f})_{ijk} = \frac{\partial^2 f_i}{\partial x_j \partial x_k}
    [/tex]

    This is a rank 3 tensor dotted with two vectors (rank 1 tensors), which is equivalent to a
    rank 5 tensor contracted over two pairs of indices, yielding a rank 1 tensor (ie a vector).
    In cartesian components, it is

    [tex]
    \sum_{j}\sum_{k} \frac{\partial^2 f_i}{\partial x_j \partial x_k} \frac{dg_j}{dt} \frac{dg_k}{dt}
    [/tex]
     
  7. Oct 17, 2012 #6
    $$(a * \nabla_t) f = [(\{a * \nabla_t \} g) * \nabla_g] f(g)$$
    This is the chain rule. The stars just mean a generalized product. This form of the chain rule is valid for ##t## vector or scalar. Here, ##a## must be a scalar, so we'll just cancel it out.

    $$\nabla_t f = [(\nabla_t g) \cdot \nabla_g] f(g)$$

    Clearly, then,
    $$\nabla_t^2 f = [(\nabla_t g) \cdot \nabla_g]^2 f(g)$$
     
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