Euler theorem for homogeneous functions (${\sum }_i{x}_i{\partial }_{i}f=kf$)

Let $f$ be a differentiable function on an open domain stable under positive rescaling, with $f(\lambda x)={\lambda }^{k}f(x)$ for $\lambda >0$. The chain rule differentiates this homogeneous function identity at $\lambda =1$ to giveIf $f$ is twice differentiable, differentiating once more gives ${\sum }_i{x}_i{\partial }_i{\partial }_{j}f=(k-1){\partial }_{j}f$. In degree one its Hessian matrix therefore annihilates the radial vector $x$. This is the identity used in the no-scale identity from degree-one homogeneity. For an extensive quantity, the degree-one formula also supplies the thermodynamic identity behind the Gibbs-Duhem equation.

For a twice continuously differentiable homogeneous function $h(x,y)$ of degree one, Euler theorem for homogeneous functions gives $x{h}_x+y{h}_y=h$. Differentiation first with respect to $x$ and then, separately, with respect to $y$ givesMultiplying by $x$ and $y$, respectively, shows that $x^2{h}_{xx}=y^2{h}_{yy}=-xy{h}_{xy}$. This identity can verify a second-order partial differential equation without expanding a complicated homogeneous expression twice.

Let $f$ be a twice continuously differentiable function on an open domain stable under positive scaling, with $f(\lambda x)={\lambda }^{n}f(x)$. Apply the chain rule twice to the scalar function $\lambda \mapsto f(\lambda x)$ and set $\lambda =1$. The result isIn two variables, this reads $x^2{f}_{xx}+2xy{f}_{xy}+y^2{f}_{yy}=n(n-1)f$. In matrix notation it is $x^T({\nabla }^{2}f)x=n(n-1)f$, where ${\nabla }^{2}f$ is the Hessian matrix. Equivalently, the radial differential operator $D={\sum }_i{x}_i{\partial }_i$ obeys $D^{2}f=Df+{\sum }_{i,j}{x}_i{x}_j{\partial }_i{\partial }_{j}f$; substituting $Df=nf$ explains the factor $n(n-1)$.

On an open domain invariant under positive scaling, $D{f}_x(x)=cf(x)$ implies $f(\lambda x)={\lambda }^{c}f(x)$. Indeed, $\frac{d}{d\lambda }[{\lambda }^{-c}f(\lambda x)]={\lambda }^{-c-1}[D{f}_{\lambda x}(\lambda x)-cf(\lambda x)]=0$. The converse follows by differentiating the scaling identity at $\lambda =1$. Connectedness of the whole domain is unnecessary: the argument works along each positive ray.

For a sufficiently regular homogeneous function $f$ of degree $n$ on a three-dimensional region, Euler theorem for homogeneous functions gives $x\cdot \nabla f=nf$. Hence $\nabla \cdot (fx)=(n+3)f$, and the divergence theorem gives the displayed reduction when $n\ne -3$. At degree $-3$ only the zero-flux identity follows, so division by $n+3$ is unavailable. Singular functions additionally require an excision or limiting argument; the surface contribution at a removed singularity must not be silently dropped. On the sloping side of a cone with its apex at zero, $x$ is tangent to the side and $x\cdot n=0$, so only the cap contributes to this formula.