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Let $f(x,y)$ has continuous second partial derivative. Define $$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-f_{xy}(x,y)^2.$$

If $(x_0,y_0)$ is a stationary point of a function $f(x,y$, then the second partial derivative test asserts the following:

(1) If $D(x_0,y_0)>0$ and $f_{xx}(x_0,y_0)>0$, then $(x_0,y_0)$ is a minimum point.

(2) If $D(x_0,y_0)>0$ and $f_{xx}(x_0,y_0)<0$, then $(x_0,y_0)$ is a maximum point.

(3) If $D(x_0,y_0)<0$, then $(x_0,y_0)$ is a saddle point.

(4) If $D(x_0,y_0)=0$, then this test is inconclusive, and $(x_0,y_0)$ could be any of a minimum, maximum or saddle point.

My question is: Could one give a function $f(x,y)$ with exactly four different stationary points that satisfy $(1),(2),(3)$, and, $(4)$?

I try some function but I haven't found such function. For example, $$f(x,y)=x^3-\frac{1}{2}x^2+y^3-\frac{1}{2}y^2$$ has four different stationary points but it doesn't satisfy $(4)$. Meanwhile, $$g(x,y)=x^3-x^2+\frac{1}{4}y^4-\frac{1}{3}y^3$$ has four different stationary points but it doesn't satisfy $(3)$.

Mark Fantini
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beginner
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1 Answers1

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Not the simplest example in appearance, but pretty simple in analysis: $$f(x,y) =xy^2+ \int (x+1) x^3 (x-1)(x-2)\,dx $$ Indeed, the partial derivative $f_y$ vanishes only when $y=0$, which means all critical points are on the $x$-axis. Their $x$-coordinates are $x=-1,0,1,2$.

Since $f_{xy} = 2y=0$ on the $x$-axis, and $f_{yy} (x,y)= 2x$ everywhere, we have $$D(x,y) = 2x \frac{d}{dx}((x+1) x^3 (x-1)(x-2)) \tag{1}$$ The derivative in (1) is easy to evaluate at critical points $x_0=-1,1,2$: just divide by $(x-x_0)$. At $x_0=0$ the derivative is $0$ because multiple root. Thus,

  • at $(-1,0)$ there is a maximum: $D = 2x\cdot x^3 (x-1)(x-2) >0$ and $f_{yy}>0$
  • at $(2,0)$ there is a minimum: $D = 2x\cdot (x+1)x^3 (x-1) >0$ and $f_{yy}<0$
  • at $(1,0)$ there is a saddle point: $D = 2x\cdot (x+1)x^3 (x-2) < 0$
  • at $(0,0)$ the test is inconclusive: $D=0$.

Explicitly, $$f(x,y) = xy^2+ \frac17 x^7-\frac13 x^6- \frac15 x^5+\frac12x^4$$

user127096
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  • effective diet pill, Thank you for your example. I hope that I can find simpler example. I think that we can ask, what happen the stationary point if $D=0$ just like $(0,0)$ is the saddle point of $f(x,y)=x^3-y^3$? – beginner Apr 17 '14 at 01:05
  • @beginner In the particular example, $(0,0)$ is not an extremum: the function is positive at $(\epsilon,\epsilon)$ and is negative at $(-\epsilon,\epsilon)$. More generally, the behavior of $f$ matches the behavior of the lowest nonzero homogeneous polynomial in its Taylor expansion $f(a+x,b+y) \sim \sum_{d=0}^\infty P_d(x,y)$. Playing with homogeneous polynomials of high degree (3 or more) you'll find they have different kinds of zero sets and $\pm $ regions: the variety is much greater than for quadratic polynomials. This becomes more like algebraic geometry than calculus. – user127096 Apr 17 '14 at 01:11