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! :)