What are the extrema and saddle points of #f(x, y) = xye^(-x^2-y^2)#?

1 Answer
May 3, 2018

Answer:

#(0,0)# is a saddle point

#(1/sqrt 2,1/sqrt 2) # and #(-1/sqrt 2,-1/sqrt 2) # are local maxima

#(1/sqrt 2,-1/sqrt 2) # and #(-1/sqrt 2,1/sqrt 2) # are local minima

#(0,pm 1/sqrt 2) # and #(pm 1/sqrt 2,0) # are points of inflection.

Explanation:

For a general function #F(x,y)# with a stationary point at #(x_0,y_0)# we have the Taylor series expansion

#F(x_0+xi,y_0+eta) = F(x_0,y_0) + 1/(2!)(F_{x x} xi^2+F_{yy}eta^2+2F_{xy}xi eta ) +ldots #
For the function

#f(x)=x y\ e^{-x^2-y^2}#

we have

#(del f)/(del x) = ye^{-x^2-y^2}+x y (-2x)e^{-x^2-y^2}#
#qquad = y(1-2x^2)e^{-x^2-y^2}#

#(del f)/(del y) = xe^{-x^2-y^2}+x y (-2y)e^{-x^2-y^2}#
#qquad = x(1-2y^2)e^{-x^2-y^2}#

It is easy to see that the both the first derivatives vanish at the following ponrs

  • #(0,0)#
  • #(0,pm 1/sqrt2)#
  • #(pm 1/sqrt2, 0)#
  • #(pm 1/sqrt2, pm 1/sqrt2)#

To examine the nature of these stationary points, we need to look at the behavior of the second derivatives there.

Now

#(del^2 f)/(del x^2) = y(-4x)e^{-x^2-y^2}+y(1-2x^2) (-2x)e^{-x^2-y^2}#
#qquad = x y(4x^2-6)e^{-x^2-y^2}#

and similarly

#(del^2 f)/(del y^2) = xy(4y^2-6)e^{-x^2-y^2}#

and

#(del^2 f)/(del xdel y) = (1-2y^2)e^{-x^2-y^2}+x(1-2y^2) (-2x)e^{-x^2-y^2}#
#qquad = (1-2x^2-2y^2+4x^2y^2)e^{-x^2-y^2}#
#qquad = (1-2x^2)(1-2y^2)e^{-x^2-y^2}#

So for #(0,0)# we have #(del^2 f)/(del x^2) = (del^2 f)/(del y^2) = 0# and #(del^2 f)/(del x del y)=1# - hence

#f(0+xi,0+eta) = f(0,0) + xi eta = xi eta#

If you approach #(0,0)# along the line #x=y#, this this becomes

#f(0+xi,0+xi ) = xi^2#

and so #(0,0)# is obviously a minimum if you approach from this direction. On the other hand, if you approach along the line #x=-y# we have

#f(0+xi,0-xi ) = -xi^2#

and so #(0,0)# is a maximum along this direction,

Thus #(0,0)# is a saddle point.

For #(1/sqrt2,1/sqrt2)# it is easily seen that

#(del^2 f)/(del x^2) = (del^2 f)/(del y^2) = -2e^{-1/2}<0# and #(del^2 f)/(del x del y)=0#

which means that

#f(1/sqrt 2 + xi,1/sqrt 2+eta) = f(1/sqrt 2,1/sqrt 2) -e^{-1/2(xi^2+eta^2)}#

So, the function decreases whichever way you move away from #(1/sqrt 2,1/sqrt 2) # and this is a local maximum. It is easily seen that the same goes for #(-1/sqrt2,-1/sqrt2)# (this should have been obvious, since the function stays the same under #(x,y) to (-x,-y)#!

Again, for both #(1/sqrt2,-1/sqrt2)# and #(-1/sqrt2,1/sqrt2)# we have

#(del^2 f)/(del x^2) = (del^2 f)/(del y^2) = 2e^{-1/2}>0# and #(del^2 f)/(del x del y)=0#

So, both these points are local minima.

The four points #(0,pm 1/sqrt2)# and #(pm 1/sqrt2, 0)# are more problematic - since all second order derivatives vanish at these points. We have to now look at higher order derivatives. Fortunately, we do not really need to work very hard for this - the very next derivative yields

#(del^3 f)/(del x^3) = -2y(3-12x^2+4x^4)e^{-x^2-y^2}#

which is non-zero for both #(0,pm 1/sqrt2)# and #(pm 1/sqrt2, 0)#. Now, this means that, for example

#f(0+xi,1/sqrt 2) = f(0,1/sqrt 2) +1/3((del^3 f)/(del x^3) )_{(0,1/sqrt2)}xi^3+... #

which shows that this will increase from # f(0,1/sqrt 2) # in one direction, and decrease from it in the other. Thus #(0,1/sqrt2)# is a **point of inflection. The same argument works for the other three points.