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 $$ \frac{δF}{δP_x}? $$

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

  • $\begingroup$ 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? $\endgroup$ – Kurt Friedman Feb 18 at 1:34
  • $\begingroup$ @KurtFriedman Yes, this changes things. I've edited my answer accordingly $\endgroup$ – 0x539 Feb 18 at 13:20

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