Partial of the divergence of a gradient?

In summary, the expression ##\frac{\partial}{\partial \phi_i}\left(\nabla \phi_i \cdot \nabla \phi_j \right)## can be expanded in two ways, but both are equivalent. It can also be rewritten as ##\nabla^2(\phi_i \phi_j)##, which is a possible answer for the expression in parentheses. This can be obtained through the calculus of variations, by taking the variation of a functional involving the expression.
  • #1
Hypatio
151
1
I am dealing with an expression in a large amount of literature usually presented as:

[itex] \frac{\partial}{\partial \phi_i}\left(\nabla \phi_i \cdot \nabla \phi_j \right)[/itex]

I'm looking at tables of vector calculus identities and cannot seem to find one for the exact expression given, even if I remove the outside partial. Is it correct to expand this as:

[itex] \frac{\partial}{\partial \phi_i}\left[\frac{\partial \phi_i}{\partial x}\left(\frac{\partial \phi_j}{\partial x}\right)\right]+\frac{\partial}{\partial \phi_i}\left[\frac{\partial \phi_i}{\partial y}\left(\frac{\partial \phi_j}{\partial y}\right)\right][/itex]

or this:

[itex] \frac{\partial}{\partial \phi_i}\left[\frac{\partial}{\partial x}\left(\phi_i \frac{\partial \phi_j}{\partial x}\right)\right]+\frac{\partial}{\partial \phi_i}\left[\frac{\partial}{\partial y}\left(\phi_i\frac{\partial \phi_j}{\partial y}\right)\right][/itex]

Or are these the same?

I'm trying to construct the correct forward explicit, space centered, finite-difference of this expression but I can't find the correct form. Any help is appreciated.

EDIT: Looking at the wiki on vector calculus identities, it looks like this is a possible answer for the expression in parentheses:

[itex]\nabla^2(\phi_i \phi_j) = \phi_i\nabla^2\phi_j+2\nabla\phi_j\cdot\nabla\phi_j+\phi_j\nabla^2\phi_i[/itex]
rearranging:
[itex]\nabla\phi_j\cdot\nabla\phi_j = \frac{1}{2}\left(\nabla^2(\phi_i\phi_j)-\phi_i\nabla^2\phi_j-\phi_j\nabla^2\phi_i\right)[/itex]

Also, there is:
[itex]\nabla\cdot\left(\phi_i\nabla\phi_j\right) = \phi_i\nabla^2\phi_j + \nabla\phi_i\cdot \nabla\phi_j[/itex]
rearranging:
[itex]\nabla\phi_i\cdot \nabla\phi_j = \nabla\cdot\left(\phi_i\nabla\phi_j\right)- \phi_i\nabla^2\phi_j[/itex]
 
Last edited:
Physics news on Phys.org
  • #3
T
jedishrfu said:
I think you're dealing with the gradient of a vector field here ie ##\phi_{i}## is a vector function so you're taking the gradient of a vector function and dotting itself before taking the partials,

https://math.stackexchange.com/questions/156880/gradient-of-a-vector-field

and that then implies tensor derivatives:

https://en.wikipedia.org/wiki/Tensor_derivative_(continuum_mechanics)#Curvilinear_coordinates

I am confident that they are both scalar fields.
 
  • #4
Can you cite some of the large amount of literature?
 
  • #5
martinbn said:
Can you cite some of the large amount of literature?
It's from phase field literature. For example, Eq 2 in Miyoshi and Takaki (2017).

Not all literature presents the term with the dot. For example, Eq. 9 in Steinbach and Pezzolla, 1999.

Apparently the term (without the outer partial) is equal to or generalized by the following expression (Eq. 6, Steinbach et al., 1996; Eq. 51, Moelans et al., 2008), but I can't see exactly how they differ:

[itex]|\phi_i\nabla\phi_j - \phi_j\nabla\phi_i |^2[/itex]
References:
https://www.sciencedirect.com/science/article/pii/S0022024816308144?via=ihub
Miyoshi and Takaki (2017), Multi-phase-field study of the effects of anisotropic grain-boundary propertis on polycrystalline grain growth, Journal of Crystal Growth.

https://www.sciencedirect.com/science/article/pii/S0364591607000880
Moelans et al. (2008), An introduction to phase-field modeling of microstructure evolution, Computer coupling of phase diagrams and thermochemistry.

https://www.sciencedirect.com/science/article/pii/S0167278999001293?via=ihub
Steinbach and Pezzolla (1999), A generalized field method for multiphase transformations using interface fields, Physica D: Nonlinear Phenomena.

https://www.sciencedirect.com/science/article/pii/0167278995002987
Steinbach et al. (1996), A phase field concept for multiphase systems, Physica D: Nonlinear Phenomena.
 
  • #6
I cannot find the expression you wrote in any of these papers. Can you cite the equation number?

The expression that appears there is ##\nabla\phi_i\cdot\nabla\phi_j## (or without the dot). This is just the dot product of the two vectors.
 
  • #7
martinbn said:
I cannot find the expression you wrote in any of these papers. Can you cite the equation number?

The expression that appears there is ##\nabla\phi_i\cdot\nabla\phi_j## (or without the dot). This is just the dot product of the two vectors.
Consider Miyoshi and Takaki (2017). ##\nabla\phi_i\cdot\nabla\phi_j## appears in Eq. 2, then variational derivatives of a function including the term are shown in Eq. 4. The apparent result is ##\nabla^2\phi_j## in Eq. 5. It's not clear to me how it is obtained.
 
  • #8
I see. When you wrote ##\frac{\partial F}{\partial \phi}##, you meant ##\frac{\delta F}{\delta \phi}##. You need to look up calculus of variations.

Roughly it is the following. You have functional ##F## and you want to vary with respect to ##\phi_i##. In your case

##
F[\phi_i]=\int \left(-\nabla\phi_i\cdot\nabla\phi_j\right)dV.
##

Then the variation is

##
\frac{\delta F}{\delta \phi}=\frac{d}{d\varepsilon}F[\phi_i+\varepsilon\varphi]|_{\varepsilon=0}
##

That leads to

##
\int \left(-\nabla\varphi\cdot\nabla\phi_j\right)dV.
##

Here you use the identity ##\nabla(\varphi\nabla\phi_j)=\nabla\varphi\cdot\nabla\phi_j+\varphi\nabla^2\phi_j##, the divergence theorem, some boundary or decay conditions that make the boundary integral zero and you are left with.

##
\int \varphi\nabla^2\phi_jdV
##

and since ##\varphi## is any, and you are looking for stationary point, for the equations you have just ## \nabla^2\phi_j=0##.
 
  • Like
Likes Hypatio

1. What is the partial of the divergence of a gradient?

The partial of the divergence of a gradient is a mathematical operation that involves taking the partial derivatives of a vector field, followed by taking the divergence of that resulting vector field. It is often used in vector calculus to describe the behavior of a vector field in three-dimensional space.

2. Why is the partial of the divergence of a gradient important?

The partial of the divergence of a gradient is important because it can help us understand the behavior of a vector field and how it changes in different directions. It is also used in many physical and engineering applications, such as in fluid dynamics and electromagnetics.

3. How is the partial of the divergence of a gradient calculated?

The partial of the divergence of a gradient is calculated by taking the partial derivatives of each component of the vector field with respect to its corresponding variable, and then adding these derivatives together. This resulting vector is then operated on by the divergence operator, resulting in a scalar value.

4. What are some real-world examples of the partial of the divergence of a gradient?

The partial of the divergence of a gradient can be seen in many physical phenomena, such as the flow of water in a river, the movement of air in a weather system, or the distribution of electric charge in an electric field. It is also used in image processing and computer graphics, such as in edge detection algorithms.

5. Are there any limitations to the partial of the divergence of a gradient?

While the partial of the divergence of a gradient is a useful tool in vector calculus, there are some limitations to its use. It can only be applied to vector fields that are defined in three-dimensional space and are continuously differentiable. Additionally, the resulting scalar value may not always have a physical interpretation, so it is important to consider the context in which it is being used.

Similar threads

  • Linear and Abstract Algebra
2
Replies
41
Views
3K
  • Linear and Abstract Algebra
Replies
1
Views
787
  • Differential Equations
Replies
2
Views
1K
Replies
9
Views
1K
Replies
5
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
385
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
18
Views
2K
Replies
8
Views
743
Back
Top