# Finding multivariate extrema with degenerate hessian matrix

1. Jul 5, 2011

### talolard

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

### micromass

Hi talolard!

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 $y=\alpha x$ 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 $f(x,\alpha x)$ and f(0,y) have local minima in 0. And this is easy since the function has only one variable!!

3. Jul 5, 2011

### LCKurtz

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.

4. Jul 5, 2011

### micromass

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
5. Jul 6, 2011

### Ray Vickson

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

RGV