1. Jan 4, 2008

### esorolla

Hello:

I have a doubt because I don't find the general definition of the nabla operator in order to solve my matter.

I am working with dyadic analysis and I have to prove that

$$\nabla^{2}F = \nabla \nabla \bullet F$$

where F is a symmetric dyadic function.

My problem is when I have to get $$\nabla^{2}F$$

because I don't know how to calculate the gradient of a dyadic. I know the definition of the gradient of a vector what results in a dyadic, but no idea about the general definition of the gradient in cartesian coordinates for whatever order of cartesian tensors.

I 'd be very thankful for a little of help.

Last edited: Jan 4, 2008
2. Jan 5, 2008

### dextercioby

As far as i know, the gradient of a (n,0) tensor is a (n+1,0) tensor. Let's say you have a (3,0) tensor T

$$T=T_{ijk}e_{i}\otimes e_{j}\otimes e_{k}$$ (cartesian tensor and coordinates)

Then

$$\nabla T= \partial_{l}T_{ijk} e_{l}\otimes e_{i}\otimes_{j}\otimes e_{k}$$

is the gradient of the tensor T.

3. Jan 5, 2008

### esorolla

Ok

I only need to relate the gradient of a (n+1,0) tensor with its divergence now. Because I have to get the divergence of the gradient of a (2,0) tensor, and comparate with the gradient of the divergence of the same one.

I had no too problems with the gradient of the divergence, but when I put in reverse order the operators I had some troubles because I didn't get the same result and I have to prove it's the same.

I'll put my results in a few minutes, I have to leave now.

Thank you