# The explicit expression of $\frac{δF}{δP}$

I'm writing simulation code of ferroelectric domain, and there is a math problem that I can't solve.

The expression of $$F$$ is $$F = \frac{|\vec{k} \cdot \vec{P}(\vec{k})|^2}{k^2}.$$ $$\vec{k}$$ is a wave-vector in Fourier space, i.e. $$\vec{k}=(k_x,k_y)$$, and $$\vec{P}(\vec{k}) = (P_x,P_y)$$ is the Fourier transform of Polarization in real space.

What's the explicit expression of $$\frac{δF}{δP_x}?$$

More details are available here.

## 1 Answer

Disclaimer: I only know variational calculus from Physics, so this won't be rigorous.

For fixed $$\vec{k}$$ you have $$F = \frac{(k_x P_x + k_y P_y)(\bar{k}_x \bar{P}_x + \bar{k}_y \bar{P}_y)}{k_x^2 + k_y^2}$$ and therefore $$d F = \frac{\bar{k}_x \bar{P}_x + \bar{k}_y \bar{P}_y}{k_x^2 + k_y^2} (k_x \,d P_x + k_y \,dP_y) + \frac{k_x P_x + k_y P_y}{k_x^2 + k_y^2} (\bar{k}_x \,d \bar{P}_x + \bar{k}_y \,d \bar{P}_y)$$

According to this, $$\frac{\delta F}{\delta P_x} = \frac{\vec{\bar{k}} \cdot \vec{\bar{P}}}{k^2} k_x$$

• But, $|\vec{k} \cdot \vec{P}(\vec{k})|^2$, this one is not just the square of(kxPx+kyPy), it's the square of its module, Px and Py is complex. Dose this kind of situation still satisfy the derivative principle in real function? – Kurt Friedman Feb 18 at 1:34
• @KurtFriedman Yes, this changes things. I've edited my answer accordingly – 0x539 Feb 18 at 13:20