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

  1. Oct 29, 2012 #1

    lo2

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

    Ok I have this general homogeneous function, which is a [itex]C^1[/itex] function:

    [itex]f(tx,ty)=t^k f(x,y)[/itex]

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

    [itex]x\frac{\partial f}{\partial x}(x,y)+y\frac{\partial f}{\partial y}(x,y)=k\cdot f(x,y)[/itex]

    2. Relevant equations



    3. The attempt at a solution

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

    [itex]x(1\cdot t+1\cdot 0) + y(1\cdot 0+1\cdot t)=xt+yt[/itex]

    But that does not really do the trick, so am I on the right way? And if so what more should I do?
     
  2. jcsd
  3. Oct 29, 2012 #2
    Differentiate f(tx, ty) with respect to t
     
  4. Oct 29, 2012 #3

    lo2

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    But I have this equation:

    [itex]x\frac{\partial f}{\partial x}(x,y)+y\frac{\partial f}{\partial y}(x,y)=k\cdot f(x,y)[/itex]

    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.
     
  5. Oct 29, 2012 #4
    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.
     
  6. Oct 29, 2012 #5

    lo2

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    Ok so I get this when I differentiate [itex]f(xt,yt)[/itex]:

    [itex]\frac{\partial f}{\partial t}(xt,yt) = 1\cdot x + 1\cdot y = x + y[/itex]

    But I am not sure how to differentiate:

    [itex]t^k f(x,y)[/itex]

    Shall I once again differentiate with regards to t?
     
  7. Oct 29, 2012 #6
    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))
     
  8. Oct 29, 2012 #7

    lo2

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

    [itex]\frac{\partial f}{\partial tx}(tx,ty)[/itex]

    The other one I think I got correct?

    [itex]\frac{\partial f}{\partial t}(tx) = x[/itex]
     
  9. Oct 29, 2012 #8

    lo2

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    I do not want to seem rude.

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

    Would be most appreciated! :)
     
  10. Oct 29, 2012 #9
    Yeah I guess that looks a little tricky. If you want, you can also write it as [itex]\frac{\partial f(a,b)}{\partial a}[/itex] evaluated at a=tx.
     
  11. Oct 29, 2012 #10

    lo2

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    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...
     
  12. Oct 29, 2012 #11

    lo2

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    Ok I think I have got something:

    If we first differentiate

    [itex]f(tx,ty)[/itex]

    We get:

    [itex]x\frac{\partial f}{\partial xt}(xt,yt)+y\frac{\partial f}{\partial yt}(xt,yt)[/itex]

    And since this has to be equal to

    [itex]k\cdot f(x,y)[/itex]

    We have that

    [itex]t^k[/itex]

    Can only be a constant when [itex]t=1[/itex], so if we do that we get:

    [itex]\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)[/itex]
     
  13. Oct 29, 2012 #12

    lo2

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    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!
     
  14. Oct 29, 2012 #13

    epenguin

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    Gold Member

    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
  15. Oct 29, 2012 #14

    lo2

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    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 [itex]C^1[/itex] function:

    [itex]f(tx,ty)=t^k f(x,y)[/itex]

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

    [itex]x\frac{\partial f}{\partial x}(x,y)+y\frac{\partial f}{\partial y}(x,y)=k\cdot f(x,y)[/itex]
     
  16. Oct 30, 2012 #15
    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.
     
  17. Oct 30, 2012 #16

    lo2

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