Homogeneous Functions and Euler's Theorem

Definition 8.12 (Homogeneous Function).

1. $F:A\to\mathbb R$, $A\subset\mathbb R^2$, is homogeneous of degree $p$ on $A$ if there is a constant $p$ with $F(\lambda x,\lambda y)=\lambda^p F(x,y)$ for all $\lambda\in\mathbb R$ and suitably restricted $\lambda,x,y$ with $(\lambda x,\lambda y)\in A$.
2. Likewise for three variables: $G:B\to\mathbb R$, $B\subset\mathbb R^3$, is homogeneous of degree $p$ if $G(\lambda x,\lambda y,\lambda z)=\lambda^pG(x,y,z)$ for all suitably restricted $\lambda,x,y,z$. ("Suitably restricted" only excludes values that would divide by zero.) Homogeneous functions matter especially in solving ordinary differential equations (a later chapter). Worked examples of the definition. $F(x,y)=x^3-2y^3+5xy^2$: $F(\lambda x,\lambda y)=(\lambda x)^3-2(\lambda y)^3+5(\lambda x)(\lambda y)^2=\lambda^3(x^3-2y^3+5xy^2)=\lambda^3F(x,y)$ — homogeneous of degree $3$. By contrast, $G(x,y)=e^{x^2}+3y^2$ is not homogeneous: $G(\lambda x,\lambda y)=e^{(\lambda x)^2}+3(\lambda y)^2\ne\lambda^pG(x,y)$ for any single $p$, for any $\lambda\ne1$, because $e^{(\lambda x)^2}$ does not scale as a power of $\lambda$ at all. A second example: $F(x,y)=\dfrac{x^2+5xy-10y^2}{3x+7y}$; here $F(\lambda x,\lambda y)=\dfrac{\lambda^2(x^2+5xy-10y^2)}{\lambda(3x+7y)}=\lambda F(x,y)$ — homogeneous of degree $1$ (numerator degree $2$ minus denominator degree $1$).
   Tip

   Fastest practical test: if every term of a numerator (and every term of a denominator, if present) has the same total power-count in $x,y,(z)$ — including arguments of $\sin,\cos,\log,e^{(\cdot)}$ being themselves a degree-$0$ ratio — the function is homogeneous, with degree = (numerator degree) $-$ (denominator degree), or the shared degree of a homogeneous sum. Mixing terms of different degree — including an added bare constant, "degree $0$" — breaks homogeneity, since no single power $\lambda^p$ can correctly rescale two differently-scaling pieces for every $\lambda$.

   Theorem (Euler, on Homogeneous Functions — stated as Definition 8.13 in the text). If $F:A\to\mathbb R$ ($A\subset\mathbb R^2$) has continuous partial derivatives and is homogeneous of degree $p$, then

   $x\frac{\partial F}{\partial x}(x,y) + y\frac{\partial F}{\partial y}(x,y) = pF(x,y) \qquad \forall\,(x,y)\in A.$

   Three variables: if $F:B\to\mathbb R$ ($B\subset\mathbb R^3$) is homogeneous of degree $p$ with continuous partials, $\ x F_x+yF_y+zF_z=pF(x,y,z)$ for all $(x,y,z)\in B$. (Proof omitted; the theorem extends to any number of variables, and is very useful for first-order-partial-derivative identities without differentiating term-by-term.) Worked example (direct application). $F(x,y)=x^3-2x^2y+3xy^2+y^3$: every term has degree $3$, so $F$ is homogeneous of degree $3$; Euler's Theorem then gives $xF_x+yF_y=3F$ immediately (and this can be double-checked by computing $F_x=3x^2-4xy+3y^2,\,F_y=-2x^2+6xy+3y^2$ directly and confirming $xF_x+yF_y$ does simplify to $3F$). …