PhyslibAlpha.ClassicalMechanics.CoupledSpringPotential

The function maps a vector $x = (x_0, x_1)$ in the 2-dimensional Euclidean space $\mathbb{R}^2$ to the real-valued potential energy of a pair of coupled springs, defined by the expression $x_0^2 + x_0 x_1 + x_1^2$.

Let $U: \mathbb{R}^2 \to \mathbb{R}$ be the coupled spring potential defined by $U(x_0, x_1) = x_0^2 + x_0 x_1 + x_1^2$. For any point $z \in \mathbb{R}^2$, the function $U$ is analytic at $z$.

Let $U: \mathbb{R}^2 \to \mathbb{R}$ be the coupled spring potential defined by $U(x_0, x_1) = x_0^2 + x_0 x_1 + x_1^2$. For any vector $x = (x_0, x_1) \in \mathbb{R}^2$, the Fréchet derivative of $U$ at $x$ is the linear map $Df(x): \mathbb{R}^2 \to \mathbb{R}$ given by $$Df(x)(h) = (2x_0 + x_1)h_0 + (x_0 + 2x_1)h_1$$ for any increment vector $h = (h_0, h_1)$.

Let $U: \mathbb{R}^2 \to \mathbb{R}$ be the coupled spring potential defined by $U(x_0, x_1) = x_0^2 + x_0 x_1 + x_1^2$. The gradient $\nabla U$ evaluated at the origin $(0, 0)$ is equal to the zero vector.

Let $U: \mathbb{R}^2 \to \mathbb{R}$ be the coupled spring potential defined by $U(x_0, x_1) = x_0^2 + x_0 x_1 + x_1^2$. For any point $z \in \mathbb{R}^2$ and any two vectors $a, b \in \mathbb{R}^2$, the second Fréchet derivative of $U$ at $z$ evaluated on $(a, b)$ is given by $$D^2 U(z)(a, b) = 2 a_0 b_0 + a_0 b_1 + a_1 b_0 + 2 a_1 b_1,$$ where $a_i$ and $b_i$ denote the components of the vectors $a$ and $b$ respectively.

Let $U: \mathbb{R}^2 \to \mathbb{R}$ be the coupled spring potential defined by $U(x_0, x_1) = x_0^2 + x_0 x_1 + x_1^2$. The quadratic map $Q: \mathbb{R}^2 \to \mathbb{R}$ associated with the second Fréchet derivative of $U$ at the origin $(0, 0)$, given by $Q(y) = D^2 U(0, 0)(y, y)$, is positive definite.

Let $U: \mathbb{R}^2 \to \mathbb{R}$ be the potential function defined by $U(x_0, x_1) = x_0^2 + x_0 x_1 + x_1^2$, where $x_0$ and $x_1$ are the components of a vector in the 2-dimensional Euclidean space. Then $U$ has a local minimum at the origin $(0, 0)$.