Mathematical Methods for Physics and Engineering : A Comprehensive Guide

5.9 STATIONARY VALUES UNDER CONSTRAINTS

where thearare coefficients dependent upon ∆x. Substituting this into (5.26), we

find

∆f=^12 ∆xTM∆x=^12

∑

r

λra^2 r.

Now, for the stationary point to be a minimum, we require ∆f=^12

∑

rλra

2
r>^0
for all sets of values of thear, and therefore all the eigenvalues ofMto be

greater than zero. Conversely, for a maximum we require ∆f=^12

∑

rλra

2
r<0,
and therefore all the eigenvalues ofMto be less than zero. If the eigenvalues have

mixed signs, then we have a saddle point. Note that the test may fail if some or

all of the eigenvalues are equal to zero and all the non-zero ones have the same

sign.

Derive the conditions for maxima, minima and saddle points for a function of two real

variables, using the above analysis.

For a two-variable function the matrixMis given by

M=

(

fxx fxy

fyx fyy

)

.

Therefore its eigenvalues satisfy the equation
∣∣
∣
∣

fxx−λfxy

fxy fyy−λ

∣∣

∣

∣=0.

Hence

(fxx−λ)(fyy−λ)−f^2 xy=0

⇒ fxxfyy−(fxx+fyy)λ+λ^2 −f^2 xy=0

⇒ 2 λ=(fxx+fyy)±

√

(fxx+fyy)^2 −4(fxxfyy−fxy^2 ),

which by rearrangement of the terms under the square root gives

2 λ=(fxx+fyy)±

√

(fxx−fyy)^2 +4f^2 xy.

Now, thatMis real and symmetric implies that its eigenvalues are real, and so for both
eigenvalues to be positive (corresponding to a minimum), we requirefxxandfyypositive
and also

fxx+fyy>

√

(fxx+fyy)^2 −4(fxxfyy−fxy^2 ),

⇒ fxxfyy−fxy^2 > 0.

A similar procedure will find the criteria for maxima and saddle points.

5.9 Stationary values under constraints

In the previous section we looked at the problem of finding stationary values of

a function of two or more variables when all the variables may be independently