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Finding multivariate extrema with degenerate hessian matrix

  1. Jul 5, 2011 #1
    1. The problem statement, all variables and given/known data
    For what real values of the parameters a,b,c,d does the functiob [tex] f(x,y)=ax^3+by^3+cx^4+dy^4-(x+y)^5 [/tex] have a local minimum at (0,0)


    2. Relevant equations
    I calculated the gradient at (0,0) and it is always zero regardless of parameters.
    The problem is that the Hessian matrix is also zero so I don't know what kind of criticial point it is.
    I also noticed that if (0,0) is a minimum then [tex] ax^3+by^3+cx^4+dy^4>(x+y)^5 [/tex] in the nieberhood but that still hasnt taken me very far.
    I don't see how I can use Lagrange multipliers, the inverse or implicit function theorems, since the gradient is 0 which precludes using them in any direct way. So my arsenal seems rather depleted.
    Any ideas?
    Thanks
    Tal


    3. The attempt at a solution
     
  2. jcsd
  3. Jul 5, 2011 #2

    micromass

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    Hi talolard! :smile:

    Your Hessian matrix is indeed zero, which means that we can't use it here. We'll have to use more direct methods.

    Consider the lines [itex]y=\alpha x[/itex] and x=0 through (0,0). If f has a local minimum in (0,0), then it must also be a local minimum on these lines. So it suffices to show that [itex]f(x,\alpha x)[/itex] and f(0,y) have local minima in 0. And this is easy since the function has only one variable!!
     
  4. Jul 5, 2011 #3

    LCKurtz

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    Perhaps I misunderstand your point, but those conditions are necessary but not sufficient to guarantee a local minimum.

    f(x,y) = (y - x2)(y-2x2) has a local min along all rays through the origin but has a saddle point there.
     
  5. Jul 5, 2011 #4

    micromass

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    I see your point, and I must admit I haven't really thought about it. But the point is that your f is not differentiable in 0 (I think). I guess that the method would work for differentiable maps (although I can't seem to find a proof for it now). Correct me if I'm wrong!

    Edit: thinking some more about it leads me to think that it is not even true for differentiable maps. Can't really find a counterexample, but I don't think this is the way to prove it.
     
    Last edited: Jul 5, 2011
  6. Jul 6, 2011 #5

    Ray Vickson

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    No: the above f is just a product of polynomials, so is as differentiable as you could ever want.

    RGV
     
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