Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

PARTIAL DIFFERENTIATION


To establish just what constitutes sufficient conditions we first note that, since

fis 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.

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