# Homework Help: Showing this Euler's equation with a homogeneous function via the chain rule

1. Oct 29, 2012

### lo2

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

Ok I have this general homogeneous function, which is a $C^1$ function:

$f(tx,ty)=t^k f(x,y)$

And then I have to show that this function satisfies this Euler equation:

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

2. Relevant equations

3. The attempt at a solution

Ok so I have tried to take the derivative, and I get:

$x(1\cdot t+1\cdot 0) + y(1\cdot 0+1\cdot t)=xt+yt$

But that does not really do the trick, so am I on the right way? And if so what more should I do?

2. Oct 29, 2012

### clamtrox

Differentiate f(tx, ty) with respect to t

3. Oct 29, 2012

### lo2

But I have this equation:

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

Where I have to differentiate first with regards to x and then y. So am I not sure I can see how I should just differentiate with regards to t.

4. Oct 29, 2012

### clamtrox

It's a straightforward application of the chain rule... ∂f(tx,ty)/∂t = ∂f(tx,ty)/∂(tx) ∂(tx)/∂t + ∂f(tx,ty)/∂(ty) ∂(ty)/∂t ...

After you got that, then differentiate the other side, tk f(x,y) and then consider what happens when t→1.

5. Oct 29, 2012

### lo2

Ok so I get this when I differentiate $f(xt,yt)$:

$\frac{\partial f}{\partial t}(xt,yt) = 1\cdot x + 1\cdot y = x + y$

But I am not sure how to differentiate:

$t^k f(x,y)$

Shall I once again differentiate with regards to t?

6. Oct 29, 2012

### clamtrox

That is not right. Can you understand the formula in my previous post? That tells you how it goes.

Yes... The point is to keep the equality
f(tx, ty) = tk f(x,y), therefore
∂f(tx, ty)/ ∂t = ∂/∂t (tk f(x,y))

7. Oct 29, 2012

### lo2

Well I guess not, I am not sure what this differentiates up to be:

$\frac{\partial f}{\partial tx}(tx,ty)$

The other one I think I got correct?

$\frac{\partial f}{\partial t}(tx) = x$

8. Oct 29, 2012

### lo2

I do not want to seem rude.

But might someone else perhaps chip in with a little bit of help?

Would be most appreciated! :)

9. Oct 29, 2012

### clamtrox

Yeah I guess that looks a little tricky. If you want, you can also write it as $\frac{\partial f(a,b)}{\partial a}$ evaluated at a=tx.

10. Oct 29, 2012

### lo2

Well I must admit that I am still not sure how to compute that.

As you do not know what the function is, and thereby I find it hard differentiate...

11. Oct 29, 2012

### lo2

Ok I think I have got something:

If we first differentiate

$f(tx,ty)$

We get:

$x\frac{\partial f}{\partial xt}(xt,yt)+y\frac{\partial f}{\partial yt}(xt,yt)$

And since this has to be equal to

$k\cdot f(x,y)$

We have that

$t^k$

Can only be a constant when $t=1$, so if we do that we get:

$\frac{df}{dt}=x\frac{\partial f}{\partial x}(x,y)+y\frac{\partial f}{\partial y}(x,y)=1^k f(x,y)=k\cdot f(x,y)$

12. Oct 29, 2012

### lo2

Ok well I kind of have to go soon, so if you would please have a short glance at my suggested solution, I would be more than happy!

13. Oct 29, 2012

### epenguin

I couldn't quite follow that last post. Are we sure you are not assuming there what you have to prove?

I may have a mental blockage; am not sure that you can or are meant to prove 2 from 1.

The only 'homogeneous functions' I know are homogeneous polynomials which are things of form

f(x, y) = Ʃ arxryk-r (r from 0 to k)

Get the two derivatives of that and you'll see it's easy.

Last edited: Oct 29, 2012
14. Oct 29, 2012

### lo2

Ok well I am not exactly sure what you mean here, but if you are asking what it is I need to show then it is:

I have this general homogeneous function, which is a $C^1$ function:

$f(tx,ty)=t^k f(x,y)$

And then I have to show that this function satisfies this Euler equation:

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

15. Oct 30, 2012

### clamtrox

Good!

No! :) It has to be equal to k tk-1 f(x,y). This has to hold for any value of t, and in particular for t=1, which is the case you're interested in.

16. Oct 30, 2012

### lo2

Ah yeah ok, thanks a lot for the help! :)