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EXAMPLE 1.33 Find the ®rst three successive approximations y 1 , y 2 , y 3 for the solution of the equation dyadx 1 xy 2 , y00, in a rectangle D: À 1 2 x 1 2 , À 1 2 y 1 2 . m Solution We can write that f xY y1 xy 2 , max D j f xY yj 1 1 8 9 8 , that is, M 9 8 ; h minaY baM min 1 2 Y 4 9 4 9 , df ady 2xy A max D jdf adyj 1 2 , that is, L 1 2 . If we choose y 0 x0, then y 1 x 0 f xY y 0 dx x 0 dx xY y 2 x 0 f xY y 1 dx x 0 1 x 3 dx x x 4 4 Y y 3 x 0 f xY y 2 dx 1 0 1 xx x 4 4 2 45 dx x x 3 3 x 7 14 x 10 160 X The absolute error in the third approximation does not exceed the value jy À y 3 j ML 3 43 h 4 9 8 1 2 3 4 9 4 1 43 % 2X286 10 4 X EXAMPLE 1.34 Find the ®rst three successive approximations for solution of the system y H zY y00 z H Ày 2 Y z0 1 2 Y & D X À1 x 1 À1 y 1 À1 z 1X V b ` b X m Solution The functions f x Y yY z z and gxY yY z Ày 2 are continuous in D. max D jf xY yY z j max D jz j1; max D jgxY yY zj max D jÀy 2 j1, that is, M 1, df dy 0 A max df dy 0Y df dz 1 A max df dz 1 A function f satisfies the Lipschitz condition with L f 1Y dg dy À2y A dg dy jÀ2yj 2Y dg dz 0 A max dg dy 0 A function g satisfies the Lipschitz condition with L g 2X 1. Differential Equations 769 Differential Equations Then, h min aY b 1 M Y b 2 M 1X There is a unique solution y yx, z zx, x PÀ1Y 1 of the initial system. Writing y 0 x0, z 0 x 1 2 , then yields y n xy 0 x 0 f tY y nÀ1 tY z nÀ1 tdt x 0 z nÀ1 tdt z n xz 0 x 0 gtY y nÀ1 tY z nÀ1 tdt 1 2 x 0 Ày 2 nÀ1 tdt y 1 x x 0 z 0 tdt x 0 1 2 dt x 2 Y z 1 x 1 2 x 0 0dt 1 2 y 2 x x 0 z 1 tdt x 0 1 2 dt x 2 Y z 2 x 1 2 x 0 À t 2 4 dt 1 2 À x 3 12 y 3 x x 0 z 2 tdt x 0 1 2 À t 3 12 dt x 2 À x 4 48 Y z 3 x 1 2 x 0 À t 2 4 dt 1 2 À x 3 12 X 1.3.2 EXISTENCE AND UNIQUENESS THEOREM FOR DIFFERENTIAL EQUATIONS NOT SOLVED FOR DERIVATIVE In this section, let us consider equations of the form F xY yY y H 0X 1X126 It is obvious that generally, for such equations, not one but several integral curves pass through some point (x 0 Y y 0 ), since as a rule, when solving the equation F x Y yY y H 0 for y H , we ®nd several (not one) real values y H f i xY yY i 1Y 2Y FFF, and if each of the equations y H f i xY y in the neighborhood of point (x 0 Y y 0 ) satis®es the conditions of the theorem of existence and uniqueness, then for each one of these equations there will be a unique solution satisfying the condition yx 0 y 0 . THEOREM 1.6 There is a unique solution y yx , x 0 À h x x 0 h (with h 0 suf®- ciently small) of Eq. (1.126) that satis®es the condition yx 0 y 0 , for which y H x 0 y H 0 , where y H 0 is one of the real roots of the equation F x 0 Y y 0 Y y H 0 if in a closed neighborhood of the point (x 0 Y y 0 Y y H ), the function F xY yY y H satis®es the following conditions: 1. FxY yY y H is continuous with respect to all arguments. 2. The derivative dFady H exists and is nonzero in (x 0 Y y 0 Y y H . 3. There exists the derivative dFady bounded in absolute value dFady . m 770 Appendix: Differential Equations and Systems of Differential Equations Differential Equations Remark 1.19 The uniqueness property for the solution of equation F x Y yY y H 0, which satis®es condition yx 0 y 0 , is usually understood in the sense that not more than one integral curve of equation F xY yY y H 0 passes through a given point (x 0 Y y 0 ) in a given direction. EXAMPLE 1.35 Let us consider the problem xy H2 À 2yy H À x 0Y y10. (a) Study the application of Theorem 1.6. (b) Solve the problem. m Solution (a) The function F x Y yY y H xy H2 À 2yy H À x is continuous with respect to all arguments and x 0 1, y 0 0. F x 0 Y y 0 Y y H 0 A F 1Y 0Y y H 0, that is, y H2 À 1 0 A y H 01 1 and y H 02 À1. dF ady 2xy H À 2y; dF ady H 1Y 0Y 1 2 T 0; dF ady H 1Y 0Y À1À2 T 0. Hence, the derivative exists and is non- zero. dF ady À2y H ; dF ady jÀ2y H j M 1 . The considered problem has a unique solution y yx Y 1 À h x 1 hX (b) Solving the equation for y H (see Example 1.20), we obtain two equations y H y y 2 x 2 p x Y y H y À y 2 x 2 p x with the solutions y 1 2 cx 2 À1ac and y 1 2 c Àx 2 ac.Fory H 01 1, x 0 1, y 0 0, we obtain 0 1 2 c À1ac A c 1 and y 1 2 x 2 À 1. Setting y H 02 À1, x 0 1, y 0 0 yields y 1 2 1 À x 2 X EXAMPLE 1.36 For the equation 2y H2 xy H À y 0, y2À 1 2 : (a) Study the application of Theorem 1.6. (b) Solve the problem. m Solution (a) The function F xY yY y H 2y H2 xy H À y is continuous and dF ady À1 A dF ady 1 is bounded. dF ady H 4y H x, x 0 2, y 0 À 1 2 , F x 0 Y y 0 Y y H 0 A 2y H2 2y H 1 2 0 A y H 0 À 1 2 . This yields dF ady H x 0 Y y 0 Y y H dF ady H 2Y À 1 2 Y À 1 2 0. The derivative dF ady H exists, but is zero. The uniqueness condition is not ful®lled. (b) The equation 2y H2 xy H À y 0 has the general solution y cx 2c 2 and the singular solution y Àx 2 a8. Condition y2À 1 2 1. Differential Equations 771 Differential Equations gives À 1 2 2c 2c 2 A 4c 2 4c 1 0 A c À 1 2 ; y À 1 2 x 1 2 is a solu- tion, and y Àx 2 a8 is another solution. 1.3.3 SINGULAR SOLUTIONS OF DIFFERENTIAL EQUATIONS The set of points (x Y y) at which the uniqueness of solutions for equation F xY yY y H 0 1X127 is violated is called a singular set. If the conditions (1) and (3) of Theorem 1.6 are ful®lled, then in the points of a singular set, the equations F xY yY y H 0 and dF dy 0 1X128 must be satis®ed simultaneously. Eliminating y H from these equations, we ®nd the equations FxY y0Y 1X129 which must be satis®ed by the points of the singular set. However, the uniqueness of solution of Eq. (1.127) is not necessarily violated at every point that satis®es Eq. (1.129), because the conditions of Theorem 1.6 are only suf®cient for the uniqueness of solutions, but are not necessary, and hence the violation of a condition of theorem does not imply the violation of uniqueness. The curve determined by Eq. (1.129) is called a p-discriminant curve (PDC), since Eqs. (1.128) are most frequently written in the form F xY yY p0 and dF adp 0. If a branch y jx of the curve FxY y0 belongs to the singular set and at the same time is an integral curve, it is called a singular integral curve, and the function y jx is called a singular solution. Thus, in order to ®nd the singular solution of Eq. (1.127) it is necessary to ®nd the PDC de®ned by the equations F xY yY p0, dF adp 0, to ®nd out [by direct substitutions into Eq. (1.127)] whether there are integral curves among the branches of the PDC and, if there are such curves, to verify whether uniqueness is violated in the points of these curves or not. If the uniqueness is violated, then such a branch of the PDC is a singular integral curve. The envelope of the family of curves FxY yY c0 1X130 is the curve that in each of its points is tangent to some curve of the family (1.130), and each segment is tangent to an in®nite set of curves of this family. If Eq. (1.130) is the general integral of Eq. (1.127), then the envelope of the family (1.130) (if it exists) will be a singular integral curve of Eq. (1.127). The 772 Appendix: Differential Equations and Systems of Differential Equations Differential Equations envelope forms a part of the c-discriminant curves (CDCs) determined by the system FxY yY c0Y dF dc 0X 1X131 For some branch of a CDC de®ned by an envelope, it is suf®cient to satisfy on it the following: 1. There exist the bounded partial derivatives dF dx N 1 Y dF dy N 2 Y 1X132 where N 1 , N 2 are constants. 2. One of the following conditions is satis®ed: dF dx T 0or dF dy T 0X 1X133 Remark 1.20 Note that these conditions are only suf®cient; thus, curves involving a violation of one of conditions (1) or (2) can also be envelopes. EXAMPLE 1.37 Find the singular solutions of the differential equation 2y H2 xy H À y 0. m Solution The following p-discriminant curves are found: F xY yY p2p 2 xp À y 0 dF dp 4p x 0 A p À x 4 X V b ` b X Substituting p Àx a4 into the ®rst equation yields y Àx 2 a8. Substitut- ing y Àx 2 a8 and y H Àxa4 in the given equation 2y H2 xy H À y 0, we observe that y Àx 2 a8 is a solution of this equation. Let us test if the solution y Àx 2 a8 is singular. Version I The general solution is y cx 2c 2 X Let us write the conditions for the tangency of the curves y y 1 x and y y 2 x in the point with abscissa x x 0 : y 1 x 0 y 2 x 0 , y H 1 x 0 y H 2 x 0 . The ®rst equation shows the ordinate coincidence of curves and, second, the slope coincidence of tangents to those curves at the point with abscissa x x 0 . Setting y 1 xÀx 2 a8, y 2 xcx 2c 2 , then Àx 2 0 a8 cx 0 2c 2 , Àx 0 a4c. Substituting c Àx 0 a4 into the ®rst equation, we 1. Differential Equations 773 Differential Equations ®nd that Àx 0 a8Àx 0 a8, that is, for c Àx 0 a4 the ®rst equation is identically satis®ed, since x 0 is the abscissa of an arbitrary point. It can be concluded that at each of its points, the curve y Àx 2 a8 is touched by some other curves of the family y cx 2c 2 (see Example 1.27). Hence, y Àx 2 a8 is a singular solution for the given equation. Version II The c-discriminant curves are FxY yY ccx 2c 2 À y 0 dF dc xY yY cx 4c 0Y hence c À x 4 X @ Substituting c Àxa4 in the ®rst equation yields y Àx 2 a8. This is the c- discriminant curve. Making a direct substitution, we observe that a solution for the given equation is found. Then, dFady À1 T 0, so that one of the conditions (1.133) is ful®lled. The curve y Àx 2 a8 is the envelope of the family y cx 2c 2 ; hence y Àx 2 a8 is a singular solution for the equation 2y H2 xy H À y 0. 1.4 Linear Differential Equations 1.4.1 HOMOGENEOUS LINEAR EQUATIONS: DEFINITIONS AND GENERAL PROPERTIES DEFINITION 1.1 An n-order differential equation is called linear if it is of the ®rst degree in the unknown function y and its derivatives y H Y y H Y FFFY y n , or is of the form a 0 y n a 1 y nÀ1 ÁÁÁa n y f xY 1X134 where a 0 Y a 1 Y FFFY a n and f are given continuous functions of x and a 0 T 0 (assume a 0 1) for all the values of x in the domain in which Eq. (1.134) is considered. The function f x is called the right-hand member of the equation. If f xT0, then the equation is called a nonhomogeneous linear equation. But if f x0, then the equation has the form y n a 1 y nÀ1 ÁÁÁa n y 0 1X135 and is called a homogeneous linear equation. m Remark 1.21 If the coef®cients a i x are continuous on the interval a x b, then, in the neighborhood of any initial values yx 0 y 0 Y y H x 0 y H 0 Y FFFY y nÀ1 x 0 y nÀ1 0 Y 1X136 where x 0 PaY b, the conditions of the theorem of existence and uniqueness are satis®ed. 774 Appendix: Differential Equations and Systems of Differential Equations Differential Equations DEFINITION 1.2 The functions y 1 xY y 2 xY FFFY y n x are said to be linearly dependent in the interval (aY b) if there exist constants a 1 Y a 2 Y FFFY a n , not all equal to zero, such that a 1 y 1 xa 2 y 2 xÁÁÁa n y n x0 is valid. If, however, the identity holds only for a 1 a 2 ÁÁÁa n 0, then the functions y 1 xY y 2 xY FFFY y n x are said to be linearly independent in the interval (aY b). m EXAMPLE 1.38 Show that the system of functions 1Y xY x 2 Y FFFY x n 1X137 is linearly independent in the interval (ÀIY I). m Solution The identity a 0 1 a 1 x ÁÁÁa n x n 0 may hold only for a 0 a 1 ÁÁÁ a n 0. EXAMPLE 1.39 Show that the system of functions e k 1 x Y e k 2 x Y FFFY e k n x Y 1X138 where k 1 Y k 2 Y FFFY k n are different in pairs, is linearly independent in the interval ÀI x I. m Solution Suppose the contrary, that is, that the given system is linearly dependent in this interval. Then, a 1 e k 1 x a 2 e k 2 x ÁÁÁa n e k n x 0 on the interval (ÀIY I), and at least one of the members a 1 Y a 2 Y FFFY a n is nonzero, for example a n T 0. Dividing this identity by e k 1 x yields a 1 a 2 e k 2 Àk 1 x a 3 e k 3 Àk 1 x ÁÁÁa n e k n Àk 1 x 0. Differentiating this identity gives a 2 k 2 À k 1 e k 2 Àk 1 x a 3 k 3 À k 1 e k 3 Àk 1 x ÁÁÁa n k n À k 1 e k n Àk 1 x 0X After dividing this identity by e k 2 Àk 1 x , we ®nd a 2 k 2 À k 1 a 3 k 3 À k 1 e k 3 Àk 2 x ÁÁÁa n k n À k 1 e k n Àk 2 x 0X Differentiating yields a 3 k 3 À k 1 k 3 À k 2 e k 3 Àk 2 x ÁÁÁa n k n À k 1 k n À k 2 e k n Àk 2 x 0Y and then a n k n À k 1 k n À k 2 ÁÁÁk n À k nÀ1 e k n Àk nÀ1 x 0Y which is false, since a n T 0Y k n T k 1 Y k n T k 2 Y FFFY k n T k nÀ1 according to the condition, and e k n Àk nÀ1 x T 0. 1. Differential Equations 775 Differential Equations EXAMPLE 1.40 Prove that the system of functions e ax sin bxY e ax cos bxY b T 0 1X139 is linearly independent in the interval (ÀIY I). m Solution Consider the identity a 1 e ax sin bx a 2 e ax cos bx 0. If we divide by e ax T 0, then a 1 sin bx a 2 cos bx 0. Then we substitute in this identity the value of x 0 to get a 1 0 and hence a 1 sin bx 0; but the function sin bx is not identically equal to zero, so a 1 0. The initial identity holds only when a 1 a 2 0, that is, the given functions are linearly independent in the interval ÀI ` x ` I. DEFINITION 1.3 Consider the functions y 1 xY y 2 xY FFFY y n x that have derivatives of order n À 1. The determinant W y 1 Y y 2 Y FFFY y n y 1 x y 2 x ÁÁÁ ÁÁÁ y n x y H 1 x y H 2 x ÁÁÁ ÁÁÁ y H n x ÁÁÁ ÁÁÁ ÁÁÁ ÁÁÁ ÁÁÁ y nÀ1 1 x y nÀ1 2 x ÁÁÁ ÁÁÁ y nÀ1 n x 1X140 is called the Wronskian determinant for these functions. Let y 1 xY y 2 xY FFFY y n x be a system of functions given in the interval [aY b]. m DEFINITION 1.4 Let us set hy i Y y j i b a y i xy j xdxY iY j 1Y 2Y FFFY nX 1X141 The determinant Gy 1 Y y 2 Y FFFY y n hy 1 Y y 1 ihy 1 Y y 2 i ÁÁÁ ÁÁÁ hy 1 Y y n i hy 2 Y y 1 ihy 2 Y y 2 i ÁÁÁ ÁÁÁ hy 2 Y y n i ÁÁÁ ÁÁÁ ÁÁÁ ÁÁÁ ÁÁÁ hy n Y y 1 ihy n Y y 2 i ÁÁÁ ÁÁÁ hy n Y y n i 1X142 is called the Grammian of the system of functions fy k xg. m THEOREM 1.7 If a system of functions y 1 xY y 2 xY FFFY y n x is linearly dependent in the interval [aY b], then its Wronskian is identically equal to zero in this interval. m 776 Appendix: Differential Equations and Systems of Differential Equations Differential Equations THEOREM 1.8 Liouville's Formula Consider that y 1 xY y 2 xY FFFY y n x are solutions of the homogeneous linear Equations (1.135), and W xW y 1 xY y 2 xY FFFY y n x is the Wronskian determinant. Then W xW x 0 e À x x 0 a 1 xdx X 1X143 Equation (1.143) is called Liouville's formula. m THEOREM 1.9 If the Wronskian W xW y 1 xY y 2 xY FFFY y n x, formed for the solu- tions y 1 Y y 2 Y FFFY y n of the homogeneous linear Eqs. (1.135), is not zero for some value x x 0 on the interval [aY b], where the coef®cients of the equation are continuous, then it does not vanish for any value of x on that interval. m THEOREM 1.10 If the solutions y 1 Y y 2 Y FFFY y n of Eq. (1.135) are linearly independent on an interval [aY b], then the Wronskian W formed for these solutions does not vanish at any point of the given interval. m THEOREM 1.11 For a system of functions y 1 xY y 2 xY FFFY y n x to be linearly dependent, it is necessary and suf®cient that the Grammian be zero. Let us write the homogeneous linear Eq. (1.135) as Ly0Y 1X144 where Lyy n a 1 xy nÀ1 ÁÁÁa n xyX 1X145 Then Ly will be termed a linear differential operator. m THEOREM 1.12 The linear differential operator L has the following two basic properties: 1. LcycLy, c constant. 2. Ly 1 y 2 Ly 1 Ly 2 . m Remark 1.22 The linearity and homogeneity of Eq. (1.135) are retained for any transfor- mation of the independent variable x jt, where jt is an arbitrary n-times differentiable function, with j H tT0Y t PaY b. 1. Differential Equations 777 Differential Equations Remark 1.23 The linearity and homogeneity are also retained in a homogeneous linear transformation of the unknown function yxaxz x. THEOREM 1.13 The totality of all solutions of Eq. (1.135) is an n-dimensional linear space. A basis in such a space is a fundamental system of solutions, that is, any family of n linearly independent solutions of Eq. (1.135). m COROLLARY 1.1 (for Theorem 1.13) The general solution for a x b of the homogeneous linear equation (1.135) with the coef®cients a i xY i 1Y 2Y FFFY n continuous on the interval a x b is the linear combination y n i1 c i y i 1X146 of n linearly independent (on the same interval) partial solutions y i i 1Y 2Y FFFY n with arbitrary constant coef®cients. m THEOREM 1.14 If a homogeneous linear equation Ly0 with real coef®cients a i x has a complex solution yx uxivx, then the real part of this solution ux and its imaginary part vx are separately solutions of that homogeneous equation. m Remark 1.24 Knowing one nontrivial particular solution y 1 of the homogeneous linear Eq. (1.135), it is possible with the substitution y y 1 udxY u y y 1 H 1X147 to reduce the order of the equation and retain its linearity and homogeneity. Knowing k linearly independent (on the interval a x b) solutions y 1 Y y 2 Y FFFY y k of a homogeneous linear equation, it is possible to reduce the order of the equation to (n À k) on the same interval a x b. Remark 1.25 If two equations have the form y n a 1 xy nÀ1 ÁÁÁa n xy 0 1X148 y n b 1 xy nÀ1 ÁÁÁb n xy 0Y 1X149 where the functions a i x and b i xY i 1Y 2Y FFFY n are continuous on the interval a x b and have a common fundamental system of solutions, y 1 Y y 2 Y FFFY y n , then the equations coincide. This means that a i xb i x, i 1Y 2Y FFFY n on the interval a x b. 778 Appendix: Differential Equations and Systems of Differential Equations Differential Equations [...]... Equations is t2 À t1 p X w 1X213 2 Comparison equations: Consider the equation Differential Equations x HH Q t x 0Y 0`m Q t MY t P aY bY 1X 214 and the comparison equations & HH y my 0 z HH Mz 0X If x is a solution of Eq (1. 214) and t1 , t2 are two consecutive zeros of this equation, then p p p t2 À t1 p X m 1X215 m M EXAMPLE 1.67 Consider the Bessel equation x 2 y... X If a0 x T 0 for the interval of variation of x , then, after division by a0 x , we ®nd y n p1 x y nÀ1 Á Á Á pn x y f x X 1X155 This equation is written brie¯y as (see Eq (1 .145 )) Ly f x X 1X156 If, for a x b all the coef®cients pi x in Eq (1.155) and f x are continuous, then it has a unique solution that satis®es the conditions k y k x0 y0 where a x... 16l2 0 has the roots l1 l2 0; l3 l4 2i; l5 l6 À2i The general solution is ygYh x c1 c2 x c3 cos 2x c4 sin 2x c5 x sin 2x c6 x cos 2x X EXAMPLE 1.54 Differential Equation of Mechanical Vibrations Find the general solution of the equation y HH x py H x qy x 0Y p b 0Y q b 0X m Solution The characteristic equation l2 pl q 0 has the roots r r... 4i 3ax 2 2bx À 4 ax 3 bx 2 e 2ix 4 zpYn 48ai À 24 6ax 2b À 32i 3ax 2 3bx 16 ax 3 bx 2 e 2ix X Substituting into the equation and reducing e 2ix from both sides yields 1 1 14 6ax 2b 48ai x , whence a À 96, b À 64 i, so that 1 i 1 3 x 3 ix 2 cos 2x i sin 2x zpYn À x 3 À x 2 e 2ix À 96 64 96 2 1 3 2 1 3 2 3 3 x cos 2x À x sin 2x À i x... and the Liouville's formula (1.188) has the form x pH s p x0 W x0 p x0 À ds X W x W x0 e x0 p s W x0 e ln p x p x 1X206 The order of Eq (1.184) can be reduced (see Eq (1. 114) ), using the substitution yH zX y 1X207 z H Àz 2 À p x z À q x X 1X208 The result is the Riccati's equation EXAMPLE 1.61 Find the general solution of the equation 1 À x 2 y HH À 2xy H ... the particular solutions y1 e x , y2 x with the Wronskian x e x e x 1 À x T 0Y x P 2Y IX W y1 Y y2 x e 1 The solution y1 , y2 is linearly independent on [2Y I) Using Eq (1 .147 ) gives H y H y1 udx y1 u e x udx e x u HH H y HH y1 2y1 u y1 u H e x udx 2e x u e x u H HHH HH H y HHH y1 3y1 u 3y1 u H y1 u HH e x udx 3e x u 3e x u H e . FFFY y n x is the Wronskian determinant. Then W xW x 0 e À x x 0 a 1 xdx X 1X143 Equation (1 .143 ) is called Liouville's formula. m THEOREM 1.9 If the Wronskian W xW y 1 xY. zero. Let us write the homogeneous linear Eq. (1.135) as Ly0Y 1X144 where Lyy n a 1 xy nÀ1 ÁÁÁa n xyX 1X145 Then Ly will be termed a linear differential operator. m THEOREM. combination y n i1 c i y i 1X146 of n linearly independent (on the same interval) partial solutions y i i 1Y 2Y FFFY n with arbitrary constant coef®cients. m THEOREM 1 .14 If a homogeneous linear