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I'm using the following definition of a homogeneous function.

A function $f(x,y)$ is homogeneous of degree n if it satisfies the following equation $$f(tx, ty) = t^n f(x,y) \quad (1) $$ for all $t$ where $n>0$

Problem

Show that if $f$ is homogeneous of degree n, then $$x \frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} = n f(x,y) $$

Attempted Solution

I differentiated $(1)$ w.r.t $t$ giving me $$ \begin{align*}\frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t} &= nt^{n-1} f(x,y) \\ x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y}&=nt^{n-1} f(x,y) \end{align*} $$

The LHS looks okay, but I'm not sure how to handle the $t^{n-1}$ term on the RHS.

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3 Answers 3

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Note that, in your calculation, you have hidden an important information. What you have shown is $$xf_x(tx, ty)+yf_y(tx,ty) = nt^{n-1}f(x,y)$$.

Replace $v=tx, w=ty$ to arrive at $$\frac{1}{t}(vf_x(v,w)+wf_y(v,w))= nt^{n-1}f(\frac{v}{t},\frac{w}{t}) $$ and use homogeneity on the RHS again.

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This is also known as Euler’s theorem.

Euler’s theorem Let $f : \mathbb{R}^n_+ \to \mathbb{R}$ be continuous, and also differentiable on $\mathbb{R}^n_{++}$. Then $f$ is homogeneous of degree $k$ if and only if for all $x \in > \mathbb{R}^n_{++}$, $$kf(x) = \sum^n_{i=1} D_if(x)x_i \, \, \ldots \, \,\,(∗)$$

Proof: ($\Rightarrow$) Suppose $f$ is homogeneous of degree $k$. Fix $x \in \mathbb{R}^n_{++}$, and define the function $g : [0, \infty) \to \mathbb{R} $ (depending on x) by $$g(\lambda) = f(\lambda x) − \lambda ^kf(x)$$ and note that for all $\lambda ⩾ 0$, $$g(\lambda ) = 0$$ Therefore, $$g′(\lambda ) = 0$$ for all $\lambda > 0$. But by the chain rule, $$g′(\lambda ) = \sum^n_{i=1} D_if(x)x_i − k\lambda ^{k−1}f(x)$$ Evaluate this at $\lambda = 1$ to obtain $(∗)$.

($\Leftarrow$) Suppose $$kf(x) = \sum^n_{i=1} D_if(x)x_i$$ for all $x \in \mathbb{R}^n_{++}$. Fix any $x ≫ 0$ and again define $g : [0, \infty) \to \mathbb{R} $ (depending on $x$) by $$g(\lambda ) = f(\lambda x) − \lambda ^kf(x)$$ and note that $g(1) = 0$. Then for $\lambda > 0$, $$g′(\lambda ) = \sum^n_{i=1} D_if(\lambda x)x_i − k\lambda ^{k−1}f(x)$$ $$= \lambda^{-1}\sum^n_{i=1} D_if(\lambda x)\lambda x_i − k\lambda ^{k−1}f(x)$$ $$= \lambda^{-1} kf(\lambda x) − k\lambda ^{k−1}f(x)$$ So $$\lambda g′(\lambda ) = kf(\lambda x) − \lambda ^kf(x))= kg(\lambda )$$ Since $\lambda $ is arbitrary, $g$ satisfies the following differential equation: $$g′(\lambda ) −\frac{k}{\lambda }g(\lambda ) = 0$$ and the initial condition $g(1) = 0$. By theorem below, $$g(λ) = 0 \cdot e^{A(\lambda )} + e^{−A(\lambda )}\int_1^{\lambda} 0 \cdot e^{A(t)}dt = 0$$ where, irrelevantly, $$A(\lambda ) = −\int_1^{\lambda} \frac{k}{t} dt = −k \ln \lambda $$ This implies $g$ is identically zero, so $f$ is homogeneous on $\mathbb{R}^n_{++}$. Continuity guarantees that $f$ is homogeneous on $\mathbb{R}^n_{+}$.

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  • $\begingroup$ I appreciate you sharing the origin of this result. The book I'm using tends to just throw definitions/equations at the reader without any background so it's nice to know where this comes from. $\endgroup$
    – rgarci0959
    Dec 27, 2015 at 9:02
  • $\begingroup$ You are welcome. $\endgroup$ Dec 27, 2015 at 9:06
  • $\begingroup$ @SchrodingersCat Could you explain how did you solve the ODE, please? $\endgroup$ Aug 4, 2019 at 0:26
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For $t \in \mathbb{R}$, what you have shown is $$x\frac{\partial f}{\partial x}(tx, ty)+y\frac{\partial f}{\partial y}(tx,ty) = nt^{n-1}f(x,y)$$ then just put $t:=1$ to get $$x\frac{\partial f}{\partial x}(x, y)+y\frac{\partial f}{\partial y}(x,y) = nf(x,y).$$

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