Finding multivariate extrema with degenerate hessian matrix

[talolard]

Homework Statement

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)

Homework 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

The Attempt at a Solution

 

[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!

 

[LCKurtz]

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!

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.

 

[micromass]

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.

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.

 

[Ray Vickson]

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.

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

RGV