Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

5.8 STATIONARY VALUES OF MANY-VARIABLE FUNCTIONS


Show that the functionf(x, y)=x^3 exp(−x^2 −y^2 )has a maximum at the point(


3 / 2 ,0),


a minimum at(−


3 / 2 ,0)and a stationary point at the origin whose nature cannot be
determined by the above procedures.

Setting the first two partial derivatives to zero to locate the stationary points, we find


∂f
∂x

=(3x^2 − 2 x^4 )exp(−x^2 −y^2 )=0, (5.23)

∂f
∂y

=− 2 yx^3 exp(−x^2 −y^2 )=0. (5.24)

For (5.24) to be satisfied we requirex=0ory= 0 and for (5.23) to be satisfied we require
x=0orx=±



3 /2. Hence the stationary points are at (0,0), (


3 / 2 ,0) and (−


3 / 2 ,0).


We now find the second partial derivatives:


fxx=(4x^5 − 14 x^3 +6x)exp(−x^2 −y^2 ),
fyy=x^3 (4y^2 −2) exp(−x^2 −y^2 ),
fxy=2x^2 y(2x^2 −3) exp(−x^2 −y^2 ).

We then substitute the pairs of values ofxandyfor each stationary point and find that
at (0,0)


fxx=0,fyy=0,fxy=0

and at (±



3 / 2 ,0)


fxx=∓ 6


3 /2exp(− 3 /2),fyy=∓ 3


3 /2exp(− 3 /2),fxy=0.

Hence, applying criteria (i)–(iii) above, we find that (0,0) is an undetermined stationary
point, (



3 / 2 ,0) is a maximum and (−


3 / 2 ,0) is a minimum. The function is shown in
figure 5.3.


Determining the nature of stationary points for functions of a general number

of variables is considerably more difficult and requires a knowledge of the


eigenvectors and eigenvalues of matrices. Although these are not discussed until


chapter 8, we present the analysis here for completeness. The remainder of this


section can therefore be omitted on a first reading.


For a function ofnreal variables,f(x 1 ,x 2 ,...,xn), we require that, at all

stationary points,


∂f
∂xi

= 0 for allxi.

In order to determine the nature of a stationary point, we must expand the


function as a Taylor series about the point. Recalling the Taylor expansion (5.20)


for a function ofnvariables, we see that


∆f=f(x)−f(x 0 )≈

1
2


i


j

∂^2 f
∂xi∂xj

∆xi∆xj. (5.25)
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