# Homogeneous function

In mathematics, a homogeneous function is a function with multiplicative scaling behaviour: if the argument is multiplied by a factor, then the result is multiplied by some power of this factor.

## Formal definition

Suppose that $f:V\rightarrow W\qquad \qquad$ is a function between two vector spaces over a field $F\qquad \qquad$ .

We say that $f\qquad \qquad$ is homogeneous of degree $k\qquad \qquad$ if

$f(\alpha \mathbf {v} )=\alpha ^{k}f(\mathbf {v} )$ for all nonzero $\alpha \in F\qquad \qquad$ and $\mathbf {v} \in V\qquad \qquad$ .

## Examples

• A linear function $f:V\rightarrow W\qquad \qquad$ is homogeneous of degree 1, since by the definition of linearity
$f(\alpha \mathbf {v} )=\alpha f(\mathbf {v} )$ for all $\alpha \in F\qquad \qquad$ and $\mathbf {v} \in V\qquad \qquad$ .

• A multilinear function $f:V_{1}\times \ldots \times V_{n}\rightarrow W\qquad \qquad$ is homogeneous of degree n, since by the definition of multilinearity
$f(\alpha \mathbf {v} _{1},\ldots ,\alpha \mathbf {v} _{n})=\alpha ^{n}f(\mathbf {v} _{1},\ldots ,\mathbf {v} _{n})$ for all $\alpha \in F\qquad \qquad$ and $\mathbf {v} _{1}\in V_{1},\ldots ,\mathbf {v} _{n}\in V_{n}\qquad \qquad$ .

• It follows from the previous example that the $n$ th Fréchet derivative of a function $f:X\rightarrow Y$ between two Banach spaces $X$ and $Y$ is homogeneous of degree $n$ .
• Monomials in $n$ real variables define homogeneous functions $f:\mathbb {R} ^{n}\rightarrow \mathbb {R}$ . For example,
$f(x,y,z)=x^{5}y^{2}z^{3}$ is homogeneous of degree 10 since

$(\alpha x)^{5}(\alpha y)^{2}(\alpha z)^{3}=\alpha ^{10}x^{5}y^{2}z^{3}$ .
$x^{5}+2x^{3}y^{2}+9xy^{4}$ is a homogeneous polynomial of degree 5. Homogeneous polynomials also define homogeneous functions.

## Elementary theorems

• Euler's theorem: Suppose that the function $f:\mathbb {R} ^{n}\rightarrow \mathbb {R}$ is differentiable and homogeneous of degree $k$ . Then
$\mathbf {x} \cdot \nabla f(\mathbf {x} )=kf(\mathbf {x} )\qquad \qquad$ .

This result is proved as follows. Writing $f=f(x_{1},\ldots ,x_{n})$ and differentiating the equation

$f(\alpha \mathbf {y} )=\alpha ^{k}f(\mathbf {y} )$ with respect to $\alpha$ , we find by the chain rule that

${\frac {\partial }{\partial x_{1}}}f(\alpha \mathbf {y} ){\frac {\mathrm {d} }{\mathrm {d} \alpha }}(\alpha y_{1})+\cdots {\frac {\partial }{\partial x_{n}}}f(\alpha \mathbf {y} ){\frac {\mathrm {d} }{\mathrm {d} \alpha }}(\alpha y_{n})=k\alpha ^{k-1}f(\mathbf {y} )$ ,

so that

$y_{1}{\frac {\partial }{\partial x_{1}}}f(\alpha \mathbf {y} )+\cdots y_{n}{\frac {\partial }{\partial x_{n}}}f(\alpha \mathbf {y} )=k\alpha ^{k-1}f(\mathbf {y} )$ .

The above equation can be written in the del notation as

$\mathbf {y} \cdot \nabla f(\alpha \mathbf {y} )=k\alpha ^{k-1}f(\mathbf {y} ),\qquad \qquad \nabla =({\frac {\partial }{\partial x_{1}}},\ldots ,{\frac {\partial }{\partial x_{n}}})$ ,

from which the stated result is obtained by setting $\alpha =1$ .

• Suppose that $f:\mathbb {R} ^{n}\rightarrow \mathbb {R}$ is differentiable and homogeneous of degree $k$ . Then its first-order partial derivatives $\partial f/\partial x_{i}$ are homogeneous of degree $k-1\qquad \qquad$ .

This result is proved in the same way as Euler's theorem. Writing $f=f(x_{1},\ldots ,x_{n})$ and differentiating the equation

$f(\alpha \mathbf {y} )=\alpha ^{k}f(\mathbf {y} )$ with respect to $y_{i}$ , we find by the chain rule that

${\frac {\partial }{\partial x_{i}}}f(\alpha \mathbf {y} ){\frac {\mathrm {d} }{\mathrm {d} y_{i}}}(\alpha y_{i})=\alpha ^{k}{\frac {\partial }{\partial x_{i}}}f(\mathbf {y} ){\frac {\mathrm {d} }{\mathrm {d} y_{i}}}(y_{i})$ ,

so that

$\alpha {\frac {\partial }{\partial x_{i}}}f(\alpha \mathbf {y} )=\alpha ^{k}f(\mathbf {y} )$ and hence

${\frac {\partial }{\partial x_{i}}}f(\alpha \mathbf {y} )=\alpha ^{k-1}f(\mathbf {y} )$ .

## Application to ODEs

The substitution $v=y/x$ converts the ordinary differential equation

$I(x,y){\frac {\mathrm {d} y}{\mathrm {d} x}}+J(x,y)=0,$ where $I$ and $J$ are homogeneous functions of the same degree, into the separable differential equation

$x{\frac {\mathrm {d} v}{\mathrm {d} x}}=-{\frac {J(1,v)}{I(1,v)}}-v$ . 