A function in $\mathbb{R}^n$ is equal to its linearization in each point

Question

I have a function $P: \mathbb{R}^n \to \mathbb{R}^n$. This function satisfies: $$ P(\vec{x}) = J_P(\vec{x}) \cdot \vec{x}$$ where $\vec{x}\in \mathbb{R}^n$, $J_P$ is the Jacobian of $P$ and "$\cdot$" is the matrix-vector product. I would rougly describe it as in the title.

I guess that it is a well known property. Do you know what is its name? Can you suggest some literature about it, or any known property?

Also, I would like to to know about the symplectic equivalent: $$ P(\vec{x}) = -\Omega \cdot J_P(\vec{x}) \cdot \Omega \cdot \vec{x}$$ where $\Omega$ defines the symplectic form.

Answer 1

As for the symplectic case. Let $J$ be the symplectic matrix $J:=\left[ \matrix{ 0 & I_n \\ -I_n & 0 }\right]$. The characteristic lines for the (first order, linear, partial differential, vector) equation $$P(x)=-J \,{\rm d}P(x)J x, \qquad x\in\mathbb{R}^{2n}\setminus\{0\},$$ are the solutions of the ODE $\dot \xi=J\xi$, that is circles $\xi(t)=e^{tJ}x_0$. Along these characteristic lines, the equation is $\partial_t P(\xi(t))= {\rm d}P(\xi(t))J \xi(t)=JP(\xi(t))$, meaning that $P\circ \xi$ satisfies the same ODE, so that we have that $P(e^{tJ}x)=e^{tJ}P(x)$, for all $t\in\mathbb{R}$ and $x\in\mathbb{R}^{2n}\setminus\{0\}$, is a necessary and sufficient condition, for a differentiable map $P:\mathbb{R}^{2n}\setminus\{0\}\to\mathbb{R}^{2n}$, to satisfy your equation. We may use complex notation and identify $\mathbb{R}^{2n}$ with $\mathbb{C}^n$, $J$ with the multiplication by $-i$, and the operators $e^{tJ}$ with multiplication by the complex scalars of modulus $1$ (Warning: in doing so, we still consider real differentiability, not complex differentiability, i.e. the Fréchet differential ${\rm d}P(x)$ is an $\mathbb{R}$-linear map, not necessarily $\mathbb{C}$-linear, i.e. not assumed to commute with $J$). The condition then reads: $P$ is an equivariant map w.r.to the group action of $\mathbb{S}^1$:

$$P(\theta x)=\theta P(x)\quad\text{for all } x\in\mathbb{C}^n\setminus\{0\} \text{ and } \theta\in\mathbb{C} \text{ with } |\theta|=1.$$

for instance, in dimension $2$ (i.e. $n=1$) these differentiable maps $P:\mathbb{R}^2\setminus\{0\}\to\mathbb{R}^2$ are exactly those that in polar coordinates write as $P(r e^{it})=e^{it}\phi(r)$, where $\phi:\mathbb{R}_+\to\mathbb{R}$ is a differentiable map.

Rmk If one takes $J$ to be the identity map, and $P$ differentiable for any $x\ne0$, the same computation gives $P(tx)=tP(x)$ for all $t>0$ and $x\ne0$, that is, $P$ is (positively) homogeneous of degree $1$, (equivariant w.r.to the action of homoteties) If $P$ is also assumed to be defined and differentiable at $0$, this implies $P$ linear, as observed in comments.

Answer 2

Such functions are linear. In dimension $1$, your equation means $$P(x)=P'(x)x$$ Solving this differential equation we obtain $P(x)=cx$. Now in arbitrary dimension, your condition can be written as $$x_j\sum_j \partial P/\partial x_j=P$$ for each coordinate $P$, which means that $P$ is a homogeneous function of degree $1$, by Euler's theorem,

http://www.its.caltech.edu/~kcborder/Notes/EulerHomogeneity.pdf

Answer 3

I think this is the equation for a homogeneous degree one function; see "Euler's homogeneous function theorem".

If such a function is differentiable at the origin, then it has to be linear.