# Homogeneous of Degree N

## Homework Statement

Let n be a positive integer. A function F is called honogeneous of degree n if it satisfies the equation F(tx,ty) = tnF(x,y) for all real t. Suppose f(x,y) has continuous second-order partial derivatives.

Show that if F is homogeneous of degree n, then

x*F_x + y*F_y = n*F(x,y), where F_x,F_y are the partial derivatives
with respect to x,y.

## The Attempt at a Solution

Suppose I let u=tx and v=ty. Then,

F_t = F_u * x + F_v * y,

which should then be equal to

n*t^(n-1) * F(x,y).

This, I think, almost looks like what I want to prove. Dividing by
t^(n-1) gives n*F(x,y) and (F_u * x + F_v * y)/(t^(n-1)), which I want
to rewrite as

x*F_x + y*F_y.

But I have no idea if/why this should be true. Am I thinking about this correctly, or have I done it the wrong way?

## Answers and Replies

LCKurtz
Science Advisor
Homework Helper
Gold Member
Try setting t = 1 in your answer. If t=1, then n*t^(n-1)*F(x,y) becomes n * F(x,y), which still equals

F_u * x + F_v * y,

which must clearly be equal to

F_x * x + F_y * y

if our premise is true. But I don't find this solution particularly convincing. Why should t have to equal 1? As in the definition, t can be equal to any real number.

LCKurtz
Science Advisor
Homework Helper
Gold Member
You are given F(tx,ty)=tnF(x,y). You differentiated both sides with respect to t and got:

xFx(tx,ty)+yFy(tx,ty)=ntn-1F(x,y)

This is an identity in t,x,and y. In particular it holds for t=1, which gives your result. What is left to explain?

Tell me what an "identity" is, then maybe I'll understand better.

LCKurtz
Science Advisor
Homework Helper
Gold Member
Tell me what an "identity" is, then maybe I'll understand better.

An identity is an equation that is true for all values of its variables. For example:

(x-1)2= x2-2x+1

is an identity because it is a true statement no matter what value x has. This is different than an equation like this:

x2+x = 6

which is true only for some values of x.