PARTIAL DIFFERENTIATION
To establish just what constitutes sufficient conditions we first note that, sincefis a function of two variables and∂f/∂x=∂f/∂y= 0, a Taylor expansion of
the type (5.18) about the stationary point yields
f(x, y)−f(x 0 ,y 0 )≈1
2![
(∆x)^2 fxx+2∆x∆yfxy+(∆y)^2 fyy]
,where ∆x=x−x 0 and ∆y=y−y 0 and where the partial derivatives have been
written in more compact notation. Rearranging the contents of the bracket as
the weighted sum of two squares, we find
f(x, y)−f(x 0 ,y 0 )≈1
2[fxx(
∆x+fxy∆y
fxx) 2
+(∆y)^2(fyy−f^2 xy
fxx)].
(5.22)For a minimum, we require (5.22) to be positive for all ∆xand ∆y, and hence
fxx>0andfyy−(f^2 xy/fxx)>0. Given the first constraint, the second can be
writtenfxxfyy>fxy^2. Similarly for a maximum we require (5.22) to be negative,
and hencefxx<0andfxxfyy>fxy^2. For minima and maxima, symmetry requires
thatfyyobeys the same criteria asfxx. When (5.22) is negative (or zero) for some
values of ∆xand ∆ybut positive (or zero) for others, we have a saddle point. In
this casefxxfyy<fxy^2. In summary, all stationary points havefx=fy= 0 and
they may be classified further as
(i) minima if bothfxxandfyyare positiveandf^2 xy<fxxfyy,(ii) maxima if bothfxxandfyyare negativeandfxy^2 <fxxfyy,(iii) saddle points iffxxandfyyhave opposite signsorf^2 xy>fxxfyy.Note, however, that iffxy^2 =fxxfyythenf(x, y)−f(x 0 ,y 0 )canbewritteninone
of the four forms
±1
2(
∆x|fxx|^1 /^2 ±∆y|fyy|^1 /^2) 2
.For some choice of the ratio ∆y/∆xthis expression has zero value, showing
that, for a displacement from the stationary point in this particular direction,
f(x 0 +∆x, y 0 +∆y) does not differ fromf(x 0 ,y 0 )tosecondorderin∆xand
∆y; in such situations further investigation is required. In particular, iffxx,fyy
andfxyare all zero then the Taylor expansion has to be taken to a higher
order. As examples, such extended investigations would show that the function
f(x, y)=x^4 +y^4 has a minimum at the origin but thatg(x, y)=x^4 +y^3 has a
saddle point there.