tailieunhanh - Handbook of mathematics for engineers and scienteists part 180

Handbook of mathematics for engineers and scienteists part 180. Tài liệu toán học quốc tế để phục vụ cho các bạn tham khảo, tài liệu bằng tiếng anh rất hữu ích cho mọi người. | . Second-Order Linear Equations 1221 8. yXx a -2q cos 2x y - Mathieu equation. 1 . Given numbers a and q there exists a general solution y x and a characteristic index j such that y x n e2n y x . For small values of q an approximate value of j can be found from the equation 2 1 _ nq2 _ A 4 cosh n 1 2sin2 2 - sin n ä O q . If y1 x is the solution of the Mathieu equation satisfying the initial conditions y1 0 1 and y 0 0 the characteristic index can be determined from the relation cosh 2nj y1 n . The solution y1 x and hence j can be determined with any degree of accuracy by means of numerical or approximate methods. The general solution differs depending on the value of y1 n and can be expressed in terms of two auxiliary periodical functions 1 x and x see Table . table The general solution of the Mathieu equation expressed in terms of auxiliary periodical functions 1 x and p2 x Constraint General solution y y x Period of 1 and r 2 Index yi n 1 C1 A 1 x C2r - xpyx n y is a real number yi n -1 C1e2px Mx Ce x 2n y p 2 i i2 -1 p is the real part of y yi n 1 C1 cos vx C2 sin vx i x C1 cos vx-C2 sin vx 2 x n y iv is a pure imaginary number cos 2nv y1 n yi n 1 C1 1 x C2x. .2 xj n y 0 2 . In applications of major interest are periodical solutions of the Mathieu equation that exist for certain values of the parameters a and q those values of a are referred to as eigenvalues . The most important periodical solutions of the Mathieu equation have the form tt tt ce2n x q E A m cos 2mx ce2n 1 x q E AXd cos 2m 1 x m 0 m 0 se2n x q E B2m sin 2mx se2n 1 x q E 2X1 sin 2m 1 x m 0 m 0 where Aj and Bj are constants determined by recurrence relations. The Mathieu functions ce2n x q and scq . x q are discussed in Section in more detail. 1222 Ordinary Differential Equations 9. yXx a tan xy X by 0. 1 . The substitution sin x leads to a linear equation of the form 2 - 1 y 1 - a y i - by 0. 2 . Solution for a -2 _ C1 sin kx C2 cos kx y C1 sinh kx C2 .