Efficient representations of the angular dependence of quadratic and quartic functions in 2d

Question

Consider the set of arbitrary quadratic functions in two dimensions $F(x_1, x_2) = \sum_{ij} A_{ij} x_i x_j$, as well as the set of arbitrary quartic functions $G(x_1, x_2) = \sum_{ijkl} B_{ijkl} x_i x_j x_k x_l$ in two dimensions. In general since $x_1 = r\cos(\theta)$ and $x_2 = r\sin(\theta)$ the polar representations of these functions will be of the form $F(r,\theta) = r^2 f(\theta)$ and $G(r,\theta) = r^4 g(\theta)$. Are there efficient (i.e. notationally compact) representations of $f$ and $g$ that are known? I'm looking for something that looks similarly simple to $f(\theta) = \alpha\cos(\omega \theta + \phi) + \beta$ with expressions for $\alpha,\beta,\omega,\phi$ in terms of the components $A_{ij}$ and $B_{ijkl}$.

Answer 1

I cannot speak about $G$, but I can about $F$.

$F$ is a quadratic form $x \mapsto F(x)$ for $x \in \mathbb R^2$. Only the symmetric part of $A$ is relevant, so we will assume that $A$ is symmetric; if it isn't, then replace $A$ with $\tfrac12(A + A^T)$. Denote by $x\cdot y$ the bilinear form associated to $F$: $$ x\cdot y = \frac12(F(x + y) - F(x) - F(y)) = x^TAy. $$ Find the eigenvectors $w_1, w_2$ of $A$; denote their eigenvalues by $\lambda_1, \lambda_2$ and assume $w_1, w_2$ are Euclidean unit vectors, i.e. $w_1^Tw_1 = w_2^Tw_2 = 1$. Also assume WLOG that $w_2$ is exactly $w_1$ rotated $90^\circ$ counterclockwise; we can achieve this by negating $w_2$ when necessary. They are necessarily Euclidean orthogonal, meaning $w_1^Tw_2 = 0$, but they are also $F$-orthgonal with $w_1\cdot w_2 = 0$. Note that $$ F(w_1) = \lambda_1,\quad F(w_2) = \lambda_2. $$

Let $e_1$ be the positively-oriented unit vector along the $x_1$-axis, and $e_2$ the one along the $x_2$-axis. Also let $\phi$ be the angle $w_1$ makes with $e_1$. By the above assuptions on $w_1, w_2$ we have $$ e_1 = w_1\cos\phi - w_2\sin\phi,\quad e_2 = w_1\sin\phi + w_2\cos\phi. $$ We then apply $F$ to an arbitary Euclidean unit vector parameterized by $\theta$: $$\begin{aligned} F(e_1\cos\theta + e_2\sin\theta) &= F((\cos\phi\cos\theta + \sin\phi\sin\theta)w_1 + (\cos\phi\sin\theta - \sin\phi\cos\theta)) \\ &= F(w_1\cos(\phi-\theta) + w_2\sin(\phi-\theta)) \\ &= \lambda_1\cos^2(\phi-\theta) + \lambda_2w_2\sin(\phi-\theta) \\ &= (\lambda_1 - \lambda_2)\cos^2(\phi-\theta) + \lambda_2. \end{aligned}$$ This is your $f$ function. Since $w_1$ is a unit vector we have $\cos\phi = (w_1)^1$ and $\sin\phi = (w_1)^2$ where $(w_1)^i$ is the $e_i$-component of $w_1$. So we can write $$ \cos(\phi-\theta) = (w_1)^1\cos\theta + (w_1)^2\sin\theta $$ and finally $$ f(\theta) = \lambda_2 + (\lambda_1-\lambda_2)\bigl[(w_1)^1\cos\theta + (w_1)^2\sin\theta\bigr]^2. $$ Note that the last bracketed term is $w_1^T\hat r$ where $\hat r = e_1\cos\theta + e_2\sin\theta$ is the unit position vector. In other words, we can write $$ F(\mathbf r) = F(r\hat r) = r^2\Bigl[\lambda_2 + (\lambda_1-\lambda_2)(w_1^T\hat r)^2\Bigr]. $$

When $\lambda_1 = \lambda_2$, meaning $F$ is positive- or negative-definite, we see that $f$ is constant; this outcome should be obvious.

In the particular case $\lambda_1 = -\lambda_2$ we can write $$ f(\theta) = \lambda_2(1 - 2\cos^2(\phi-\theta)) = \lambda_2\cos[2(\phi-\theta)]. $$