- #1

aliekor

- 7

- 0

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

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- Thread starter aliekor
- Start date

- #1

aliekor

- 7

- 0

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

- 39,188

- 8,496

Looks ok to me. Try it with some simple examples.

- #3

pasmith

Homework Helper

2022 Award

- 2,516

- 1,117

The chain rule for a vector-valued function (

[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.

- #4

aliekor

- 7

- 0

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.

- #5

pasmith

Homework Helper

2022 Award

- 2,516

- 1,117

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.

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]

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.

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]

- #6

Muphrid

- 834

- 2

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

$$(a * \nabla_t) f = [(\{a * \nabla_t \} g) * \nabla_g] f(g)$$

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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