Mathematical Methods for Physics and Engineering - Matematica.NET

5.9 STATIONARY VALUES UNDER CONSTRAINTSwhere the a r are coefficients dependent upon ∆x. Substituting this into (5.26), wefind∑∆f = 1 2 ∆xT M∆x = 1 2λ r a 2 r .Now, for the stationary point to be a minimum, we require ∆f = 1 ∑2 r λ ra 2 r > 0for all sets of values of the a r , and therefore all the eigenvalues of M to begreater than zero. Conversely, for a maximum we require ∆f = 1 ∑2 r λ ra 2 r < 0,and therefore all the eigenvalues of M to be less than zero. If the eigenvalues havemixed signs, then we have a saddle point. Note that the test may fail if some orall of the eigenvalues are equal to zero and all the non-zero ones have the samesign.◮Derive the conditions for maxima, minima and saddle points for a function of two realvariables, using the above analysis.For a two-variable function the matrix M is given by( )fxx fM =xy.f yx f yyTherefore its eigenvalues satisfy the equation∣ f xx − λ f xyf xy f yy − λ ∣ =0.Hence(f xx − λ)(f yy − λ) − fxy 2 =0⇒ f xx f yy − (f xx + f yy )λ + λ 2 − fxy 2 =0√⇒ 2λ =(f xx + f yy ) ± (f xx + f yy ) 2 − 4(f xx f yy − fxy),2which by rearrangement of the terms under the square root gives√2λ =(f xx + f yy ) ± (f xx − f yy ) 2 +4fxy.2Now, that M is real and symmetric implies that its eigenvalues are real, and so for botheigenvalues to be positive (corresponding to a minimum), we require f xx and f yy positiveand also√f xx + f yy > (f xx + f yy ) 2 − 4(f xx f yy − fxy),2⇒ f xx f yy − fxy 2 > 0.A similar procedure will find the criteria for maxima and saddle points. ◭r5.9 Stationary values under constraintsIn the previous section we looked at the problem of finding stationary values ofa function of two or more variables when all the variables may be independently167