# Euler's Theorem, Homogenous Functions

1. Jul 31, 2010

### psholtz

1. The problem statement, all variables and given/known data

I'm working on Problem 8, Chapter 1 in the classic Ince text on ODEs.

The question has me somewhat confused.

It involves Euler's Theorem on Homogeneous functions. For reference, this theorem states that if you have a function f in two variables (x,y) and homogeneous in degree n, then you have:

$$x\frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = nf(x,y)$$

The proof of this is straightforward, and I'm not going to review it here.

The problem asks the reader to prove the following theorem (I'm going to state the problem more or less verbatim, so I don't "read" anything into it that I'm not supposed). If f is a function homogeneous and of degree m in x1, x2, and homogeneous and of degree n in y1, y2, then:

$$\left(y_1 \frac{\partial}{\partial x_1} + y_2 \frac{\partial}{\partial x_2}\right)\left(x_1 \frac{\partial f}{\partial y_1} + x_2 \frac{\partial f}{\partial y_2}\right) - \left(x_1 \frac{\partial}{\partial y_1} + x_2 \frac{\partial}{\partial y_2} \right)\left(y_1 \frac{\partial f}{\partial x_1} + y_2 \frac{\partial f}{\partial x_2}\right) = (n-m)f$$

In other words, the reader is asked to prove an extension to Euler's theorem.

I have some "attempts" at a solution, which I'll post in the next post, but to keep this post from growing too long and complicated, I'm going to end this one here.

2. Jul 31, 2010

### psholtz

OK, now this is one of those problems where I have to stare at it for a while, just to try and figure out what it's asking.

At first, I wasn't sure if we were dealing w/ a function in 2 variables, or 4.

I'm pretty sure that we're dealing w/ a function in two variables, and that we're doing a change in variables from x1,x2 to y1, y2.

So we start w/ something like this:

$$f(tx_1,tx_2) = t^mf(x_1,x_2)$$

$$f\left(1,\frac{x_2}{x_1}\right) = F\left(\frac{x_2}{x_1}\right) = x_1^m f(x_1,x_2)$$

and likewise, after a change in variables to y1,y2:

$$f(ty_1,ty_2) = t^nf(y_1,y_2)$$

$$f\left(1,\frac{y_2}{y_1}}\right) = F\left(\frac{y_2}{y_1}\right) = y_1^nf(y_1,y_2)$$

By Euler's theorem, we would have:

$$x_1\frac{\partial f}{\partial x_1} + x_2 \frac{\partial f}{\partial x_2} = mf(x_1,x_2)$$

and

$$y_1\frac{\partial f}{\partial y_1} + y_2\frac{\partial f}{\partial y_2} = nf(y_1,y_2)$$

Subtracting these expressions suggests a way of attacking the problem:

$$\left(y_1 \frac{\partial f}{\partial y_1} + y_2 \frac{\partial f}{\partial y_2}\right) - \left(x_1 \frac{\partial f}{\partial x_1} + x_2\frac{\partial f}{\partial x_2}\right) = (n-m)f$$

at least the RHS are equal.

However, there's one thing that's still confusing..

I'll address that in the next post (don't want any one post to get too long).

3. Jul 31, 2010

### psholtz

Usually, in order to grapple and wrap my head around problems like this, I try to work out a concrete example first, and that's where I'm getting stuck here.

Suppose, for concreteness, we have:

$$f(x_1,x_2) = x_1^2x_2^2$$

so that in x1, x2, f is homogeneous in degree 4.

Now suppose we make the change of variables:

$$y_1 = x_1^2, y_2 = x_2^2$$

so that:

$$f(y_1,y_2) = y_1y_2$$

Now, in y1,y2, f is homogeneous in degree 2.

If we plug these expressions into the expression above, and solve, we should expect to arrive an answer of:

$$(n-m)f = (2-4)f = -2f$$

However, I don't get that.

I'm going to step through the calculations very slowly, partly to triple verify that I wasn't making an error. The expression above is complex, soo... startly slowly..

First, expanding the first "term" on the LHS above, we would get an expression w/ four terms:

$$y_1\frac{\partial}{\partial x_1}\left(x_1\frac{\partial f}{\partial y_1}\right) + y_1\frac{\partial}{\partial x_1}\left(x_2\frac{\partial f}{\partial y_2}\right) + y_2\frac{\partial}{\partial x_2}\left(x_1\frac{\partial f}{\partial y_1}\right) + y_2\frac{\partial }{\partial x_2}\left(x_2\frac{\partial f}{\partial y_2}\right)$$

Substituting the "concrete" expressions I suggested above, this reduces to:

$$y_1\frac{\partial}{\partial x_1}\left(x_1y_2\right) + y_1 \frac{\partial}{\partial x_1}\left(x_2y_1\right) +y_2\frac{\partial}{\partial x_2}\left(x_1y_2\right) + y_2\frac{\partial }{\partial x_2}\left(x_2y_1\right)$$

$$y_1y_2 + y_1\frac{\partial}{\partial x_1}\left(x_1^2x_2\right) + y_2\frac{\partial}{\partial x_2}\left(x_1x_2^2\right) + y_1y_2$$

$$2f + 2x_1^3x_2 + 2x_1x_2^3$$

Similarly, if we take the second term on the LHS of the original equation, we have:

$$x_1\frac{\partial}{\partial y_1}\left(y_1\frac{\partial f}{\partial x_1}\right) + x_1\frac{\partial }{\partial y_1} \left(y_2\frac{\partial f}{\partial x_2}\right) + x_2\frac{\partial }{\partial y_2}\left(y_1\frac{\partial f}{\partial x_1}\right) + x_2\frac{\partial }{\partial y_2}\left(y_2\frac{\partial f}{\partial x_2}\right)$$

which reduces to:

$$x_1\frac{\partial}{\partial y_1}\left(2x_1x_2^2y_1\right) + x_1\frac{\partial}{\partial y_1}\left(2x_1^2x_2y_2\right) + x_2\frac{\partial}{\partial y_2}\left(2x_1x_2^2y_1\right) + x_2\frac{\partial}{\partial y_2}\left(2x_1^2x_2y_2\right)$$

$$x_1\frac{\partial}{\partial y_1} \left(2x_1^3x_2^2\right) + x_1 \frac{\partial}{\partial y_1}\left(2x_1^2x_2^3\right) + x_2 \frac{\partial }{\partial y_2}\left(2x_1^3x_2^2\right) + x_2 \frac{\partial}{\partial y_2}\left(2x_1^2x_2^3\right)$$

$$x_1\frac{\partial}{\partial y_1}\left(2y_1^{3/2}y_2\right) + x_1 \frac{\partial}{\partial y_1} \left(2y_1y_2^{3/2}\right) +x_2 \frac{\partial}{\partial y_2}\left(2y_1^{3/2}y_2\right) + x_2 \frac{\partial}{\partial y_2}\left(2y_1y_2^{3/2}\right)$$

$$3x_1y_1^{1/2}y_2 + 2x_1y_2^{3/2} + 2y_1^{3/2}x_2 + 3y_1y_2^{1/2}x_2$$

$$6x_1^2x_2^2 + 2x_1x_2^3 + 2x_1^3x_2$$

$$6f + 2x_1^3x_2 + 2x_1x_2^3$$

Subtracting these two expressions, we get:

$$2f-6f = -4f$$

In other words, it's not -2f, which is what we would expect.

Am I making a very basic mistake somewhere?

4. Aug 1, 2010

### psholtz

OK, I figured it out.. although I still didn't answer my own earlier question, that is, what to do when you have:

$$f(x_1,x_2) = x_1^2x_2^2$$

$$f(y_1,y_2) = y_1y_2$$

I'll start by posting the solution, which is actually quite straightforward if you just go through the mechanics. Starting w/ the first term on the LHS:

$$\left(y_1 \frac{\partial}{\partial x_1} + y_2 \frac{\partial }{\partial x_2}\right) \left(x_1 \frac{\partial f}{\partial y_1} + x_2 \frac{\partial f}{\partial y_2}\right)$$

$$y_1 \frac{\partial}{\partial x_1} \left(x_1 \frac{\partial f}{\partial y_1}\right) + y_1 \frac{\partial}{\partial x_1} \left(x_2 \frac{\partial f}{\partial y_2}\right) + y_2 \frac{\partial }{\partial x_2} \left( x_1 \frac{\partial f}{\partial y_1} \right) + y_2 \frac{\partial}{\partial x_2} \left(x_2 \frac{\partial f}{\partial y_2}\right)$$

$$y_1 \left(\frac{\partial f}{\partial y_1} + x_1 \frac{\partial ^2f}{\partial x_1\partial y_1}\right) + x_2 y_1 \frac{\partial ^2f}{\partial x_1\partial y_2} + x_1y_2 \frac{\partial^2 f}{\partial x_2\partial y_1} + y_2\left(\frac{\partial f}{\partial y_2} + x_2\frac{\partial^2f}{\partial x_2\partial y_2}\right)$$

Applying Euler's theorem then gives:

$$nf + x_1y_1 \frac{\partial^2f}{\partial x_1\partial y_1} + x_1y_2 \frac{\partial^2f}{\partial x_2\partial y_1} + x_2y_1 \frac{\partial ^2f}{\partial x_2\partial y_1} + x_2y_2 \frac{\partial ^2f}{\partial x_2\partial y_2}$$

Applying the same logic to the second term on the LHS, gives mf along (w/ continuity arguments) the same "cross" second derivative products. These "cross" products cancel, and you're just left with:

$$(n-m)f$$

However, still not sure why the "concrete" example I cited above doesn't seem to work..