Mathematical Methods for Physics and Engineering - Matematica.NET

PARTIAL DIFFERENTIATIONTo establish just what constitutes sufficient conditions we first note that, sincef is a function of two variables and ∂f/∂x = ∂f/∂y = 0, a Taylor expansion ofthe type (5.18) about the stationary point yieldsf(x, y) − f(x 0 ,y 0 ) ≈ 1 2![(∆x) 2 f xx +2∆x∆yf xy +(∆y) 2 f yy],where ∆x = x − x 0 and ∆y = y − y 0 and where the partial derivatives have beenwritten in more compact notation. Rearranging the contents of the bracket asthe weighted sum of two squares, we findf(x, y) − f(x 0 ,y 0 ) ≈ 1 2( [f xx ∆x + f ) ( )]2xy∆y+(∆y) 2 f yy − f2 xy.f xx f xx(5.22)For a minimum, we require (5.22) to be positive for all ∆x and ∆y, and hencef xx > 0andf yy − (fxy/f 2 xx ) > 0. Given the first constraint, the second can bewritten f xx f yy >fxy. 2 Similarly for a maximum we require (5.22) to be negative,and hence f xx < 0andf xx f yy >fxy 2 . For minima and maxima, symmetry requiresthat f yy obeys the same criteria as f xx . When (5.22) is negative (or zero) for somevalues of ∆x and ∆y but positive (or zero) for others, we have a saddle point. Inthis case f xx f yy