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

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

## References

- Blatter, Christian (1979). "20. Mehrdimensionale Differentialrechnung, Aufgaben, 1.".
*Analysis II (2nd ed.)* (in German). Springer Verlag. pp. p. 188. ISBN 3-540-09484-9.

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