# 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?

LCKurtz
Homework Helper
Gold Member

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
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