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.

  • $\begingroup$ 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 '11 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 $\rho$. $\endgroup$ – Chris Taylor Sep 8 '11 at 10:02
  • $\begingroup$ @anon Yes, I agree, the substitution makes it clearer which parameters are fixed constant in the partial derivative. $\endgroup$ – James Womack Oct 8 '11 at 11:12

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.