2
$\begingroup$

I am attempting to calculate the functional derivative of a functional $$E[\rho] = \int G(\rho(\mathbf{r}),\nabla\rho(\mathbf{r}),\mathbf{r})d\mathbf{r},$$ where $$G(\rho(\mathbf{r}),\nabla\rho(\mathbf{r}),\mathbf{r})=\rho(\mathbf{r})^{4/3}\left(\alpha-\frac{(\nabla\rho(\mathbf{r})\cdot\nabla\rho(\mathbf{r}))^{3/4}}{137 \rho(\mathbf{r})^{2}}\right),$$ and $\alpha$ is a constant. This is for use in a computational chemistry code.

To find the functional derivative I think I should use the Euler-Lagrange equation, $$\frac{\delta G}{\delta \rho}=\frac{\partial G}{\partial \rho} - \nabla\cdot\frac{\partial G}{\partial \nabla \rho}, $$ as given on the Wikipedia article on functional derivatives.

What I am struggling with is the second term in the E-L equation. Firstly, I am not sure how to approach the partial derivative with respect to $\nabla\rho$. So far, I have use the chain rule to obtain $$ \frac{\partial G}{\partial \nabla\rho}=-\frac{3}{4\times 137 \rho^{2/3}}\frac{1}{(\nabla\rho(\mathbf{r})\cdot\nabla\rho(\mathbf{r}))^{1/4}}\left(\frac{\partial}{\partial \nabla\rho}(\nabla\rho(\mathbf{r})\cdot\nabla\rho(\mathbf{r}))\right), $$ but I am not sure how to proceed with the differentiation of the dot product. Furthermore, it appears from the E-L equation that I must then find the divergence of this partial derivative. I think that the result of $\frac{\partial G}{\partial \nabla\rho}$ will be a scalar function, so am not sure how the divergence can be applied here.

I would appreciate some advice on how to tackle the partial derivative and subsequent divergence. Perhaps I am missing something, or there is a flaw in my reasoning.

$\endgroup$
3
  • $\begingroup$ Here $\frac{\partial}{\partial\nabla\rho}$ means, in some formal sense, $\nabla_{\nabla\rho}$ where $\rho$ and $\mathbf{r}$ are fixed constant while any instance of $\nabla\rho$ in $G$ is allowed to vary. So basically you're looking for $\nabla_{\mathbf{x}}\|\mathbf{x}\|^2$ and then substituting $\nabla\rho$ for $\mathbf{x}$. This is a vector field for which the divergence can be taken. $\endgroup$
    – anon
    Sep 8, 2011 at 10:01
  • 1
    $\begingroup$ You may find it easier to write $\sigma=\nabla \rho$ in the definition of $G$, giving $G(\rho,\sigma,r)$ and then differentiate with respect to $\sigma$, treating it as independent from $\rho$. $\endgroup$ Sep 8, 2011 at 10:02
  • $\begingroup$ @anon Yes, I agree, the substitution makes it clearer which parameters are fixed constant in the partial derivative. $\endgroup$ Oct 8, 2011 at 11:12

1 Answer 1

0
$\begingroup$

Denote ∇ρ(r) = a (whereas a is a vector).

Now, how to find: ∂( aa ) / ∂a .

You just have to remember that differentiating with respect to a vector is a symbolic notion of a vector whose components are the differentials with respect to the appropriate vector component.

That is,

∂f / ∂a = i ⋅ ∂f / ∂a[i], sum over all i.

wherease i is the i'th component vector component and i is the appropriate unit vector.

Now, we have

∂( aa ) / ∂a = i ⋅ ∂( aa ) / ∂a[i] =

i ⋅ ∂( a[j] ⋅ a[j] ) / ∂a[i] = i ⋅ { (∂a[j]/∂a[i])⋅a[j] + a[j]⋅(∂a[j]/∂a[i]) } =

i ⋅ { δ(i,j)⋅a[j] + a[j]⋅δ(i,j) } = i ⋅ { a[i] + a[i] } = 2 i ⋅ a[i] = 2 a

So, the answer is: ∂( aa ) / ∂a = 2 a

$\endgroup$
1
  • $\begingroup$ Thank you, very clear answer. The result of the differentiation is then a vector field, so taking the divergence is not a problem. $\endgroup$ Oct 8, 2011 at 11:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .