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I have explicitly known expressions for:

$f(x,y), u(x,y), v(x,y), w(x,y)$

$\frac{\partial f}{\partial x}(x,y), \frac{\partial f}{\partial y}(x,y), \frac{\partial u}{\partial x}(x,y), \frac{\partial u}{\partial y}(x,y), \frac{\partial v}{\partial x}(x,y), \frac{\partial v}{\partial y}(x,y), \frac{\partial w}{\partial x}(x,y), \frac{\partial w}{\partial y}(x,y)$

, from which I want to obtain explicit expression for:

$\frac{\partial f}{\partial u}(x,y)$ and $\frac{\partial f}{\partial v}(x,y)$

At first, I thought I should go for chain-rule, but it seems I cannot get the answer (3 unknowns from 2 equation as below)

$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}+\frac{\partial f}{\partial w}\frac{\partial w}{\partial x}$

$\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial f}{\partial w}\frac{\partial w}{\partial y}$

I couldn't find anything similar to my problem (searched for multivariate derivative chain-rule, functional derivative), and couldn't come up with better keywords for the questions.

Can anyone let me know how to solve my problem, or more relevant keywords for my problem? Thank you.

P.S. For better understanding of my problem, explicit expressions are:

$u(x,y) = \sin(x)\cos(y)$

$v(x,y) = \sin(x)\sin(y)$

$w(x,y) = \cos(x)$

$f(x,y) = Y_l^m(x,y)$ is spherical harmonics

José Carlos Santos
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MJH
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1 Answers1

0

You have to pick two of $u,v,w$ to be free variables (noting that $u^2+v^2+w^2=1$, whence all three of them cannot be treated as free variables). I am guessing that $u$ and $v$ are free. However, the assignment $(x,y)\mapsto (u,v)$ is an almost everywhere $4$-to-$1$ map from $(\mathbb{R}/2\pi\mathbb{Z})\times(\mathbb{R}/2\pi\mathbb{Z})$ to $[-1,+1]\times[-1,+1]$. In other words, $$u(x,y)=u(-x,y+\pi)=u(x+\pi,y+\pi)=u(-x+\pi,y)$$ and $$v(x,y)=v(-x,y+\pi)=v(x+\pi,y+\pi)=v(-x+\pi,y)\,,$$ where the operations $z\mapsto -z$ and $z\mapsto z+\pi$ are considered modulo $2\pi$. Consequently, it is quite problematic to go from $(u,v)$ back to $(x,y)$. However, if $x$ and $y$ are both known, then you can see that $$x=s\,\text{arcsin}\big(\sqrt{u^2+v^2}\big)+t\text{ and }y=\text{arctan}\left(\frac{v}{u}\right)+\tau\,,$$ where $s\in\{-1,+1\}$ and $t,\tau\in\{0,\pi\}$ (indeed, $s$ is the sign of $\sin(x)$). The values of $s$, $t$, and $\tau$ are fixed in a small neighborhood of each noncritical point. That is, $$\frac{\partial f}{\partial u}(x,y)=\left(\frac{\partial f}{\partial x}(x,y)\right)\,\left(\frac{\partial x}{\partial u}\right)+\left(\frac{\partial f}{\partial y}(x,y)\right)\,\left(\frac{\partial y}{\partial u}\right)$$ and $$\frac{\partial f}{\partial v}(x,y)=\left(\frac{\partial f}{\partial x}(x,y)\right)\,\left(\frac{\partial x}{\partial v}\right)+\left(\frac{\partial f}{\partial y}(x,y)\right)\,\left(\frac{\partial y}{\partial v}\right)$$ can be determined from $$\frac{\partial x}{\partial u}=\frac{su}{\sqrt{(1-u^2-v^2)(u^2+v^2)}}=\frac{\cos(y)}{\big|\cos(x)\big|}\,,$$ $$\frac{\partial x}{\partial v}=\frac{sv}{\sqrt{(1-u^2-v^2)(u^2+v^2)}}=\frac{\sin(y)}{\big|\cos(x)\big|}\,,$$ $$\frac{\partial y}{\partial u}=-\frac{v}{u^2+v^2}=-\frac{\sin(y)}{\sin(x)}$$ and $$\frac{\partial y}{\partial v}=+\frac{u}{u^2+v^2}=+\frac{\cos(y)}{\sin(x)}\,.$$

Batominovski
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  • Thank you for tips - It worked for some test conditions! However, it did not work for other conditions (possibly due to numerical error because I tested using some extreme values where reference answers exist). – MJH Sep 02 '18 at 12:58
  • Actually, the question above was reformulated by myself. The original problem involves gradient in reciprocal k-space: $\vec{k} \times \frac{d}{d\vec{k}} Y_{lm}(\hat{k})$, $\frac{d}{d\vec{k}}=k_x \hat{x}+k_y \hat{y}+k_z \hat{z}$, which I have never met before. I expressed the k-vectors as $u,v,w$ in the original question. I wonder if my approach was acceptable. – MJH Sep 02 '18 at 13:05