Showing this Euler's equation with a homogeneous function via the chain rule

lo2
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Homework Statement



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)

Homework Equations





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?
 
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Differentiate f(tx, ty) with respect to t
 
clamtrox said:
Differentiate f(tx, ty) with respect to t

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.
 
lo2 said:
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.

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.
 
clamtrox said:
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.

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?
 
lo2 said:
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

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

lo2 said:
But I am not sure how to differentiate:

t^k f(x,y)

Shall I once again differentiate with regards to t?

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

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
 
I do not want to seem rude.

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

Would be most appreciated! :)
 
lo2 said:
Well I guess not, I am not sure what this differentiates up to be:

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

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
clamtrox said:
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.

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
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
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
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:
  • #14
epenguin said:
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.


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
lo2 said:
We get:

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

lo2 said:
And since this has to be equal to

k\cdot f(x,y)

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