The General Solutions of Linear ODE and Riccati Equation

The General Solutions of Linear ODE and Riccati Equation by Inte gral Series ✩ Y imin Y an Abstract This paper giv es ou t the genera l solutions of variable coe ffi cients Linear ODE and Riccati equation by way of integral series E ( X ) and F ( X ). Such kind s of integral series are the ge neralized form of expon ential f unction, and keep the proper ties of conver gent and reversible. K eywor d s: Linear ODE,Riccati equation,integral s eries, gen eral solution,variable coe ffi cients 1. Introduction It is a classical problem to solve the n-th order Linear ODE : d n d x n u + a 1 ( x ) d n − 1 d x n − 1 u + a 2 ( x ) d n − 2 d x n − 2 u + · · · + a n ( x ) u = f ( x ) (1) which is equiv alen t to d d x U = AU + F (2) with                                                U = h d n − 1 d x n − 1 u d n − 2 d x n − 2 u · · · u i T F = h f ( x ) 0 · · · 0 i T A ( x ) =                               − a 1 − a 2 − a 3 · · · − a n 1 0 0 · · · 0 0 1 0 · · · 0 · · · · · · · · · · · · · · · 0 0 · · · 1 0                               (3) As we all known, 1. if  a n ( x )  are all constants, Eq.( 1 ) could be solved by metho d of eigen value ( Eu ler), or by exponential fun ction in matrix form U = e A · x · C + e A · x · Z x 0 e − A · s · F ( s ) d s wher e C i s a n × 1 co nstant matrix . ✩ Many thank s to Prof.Qiyan Shi’ s guidance. Email addr ess: yanyimin@ foxmail.co m (Y imin Y an ) October 26, 2018 2. if  a n ( x )  are some variable coe ffi cients, such as some special functions [ W ang , P337,2 06] d 2 y d x 2 + 1 x d y d x +  1 − n 2 x 2  y = 0 (Bessel Equa tion) (1 − x 2 ) d 2 y d x 2 − 2 x d y d x + n ( n + 1 ) y = 0 (Legendre Equation) special function theory answers them. But wh en it com es to the gen eral cir cumstances, the e x isting m ethods meet di ffi culties in dealin g with Eq.( 2 ) , because of the variable coe ffi cients. I n order to overcome it, two function s are invited : 1.1. Definition                E  X ( x )  = I + Z x 0 X ( t ) d t + Z x 0 X ( t ) Z t 0 X ( s ) d sd t + Z x 0 X ( t ) Z t 0 X ( s ) Z s 0 X ( ξ ) d ξ d sd t + · · · F  X ( x )  = I + Z x 0 X ( t ) d t + Z x 0 Z t 0 X ( s ) d s X ( t ) d t + Z x 0 Z t 0 Z s 0 X ( ξ ) d ξ X ( s ) d s X ( t ) d t + · · · (4) It will be seen that su ch d efinition is reason able and nec essary . Clearly , when X ( x ) an d R x 0 X ( t ) d t ar e exchangeab le, then E  X ( x )  = e R x 0 X ( t ) d t = F  X ( x )  Besides, E ( X ) and F ( X ) extend s ome main pr operties of the expon ential f unctions, s uch as con vergent , rev ersible and deter minant (see Theor em 3.1 ). In ad dition, a n × m m atrix A ( x ) =  a i j ( x )  nm is b ounded an d integral in [0, b] means that all its element a i j ( x ) are bounded and integral in [0,b]. 2. Main Results Theorem 2.1. the general s olution of the Linear ODE ( 2 ) is: U = E  A ( x )  · C + E  A ( x )  · Z x 0 F  − A ( s )  · F ( s ) d s (5) wher e C is a n × 1 constant matrix . Theorem 2.2. F or the boun ded and inte grable matrix , A ( x ) = ( a i j ) nn , B ( x ) = ( b i j ) mm , P ( x ) = ( p i j ) mn , Q ( x ) = ( q i j ) nm , in [0,b], the general solution of Riccati equ ation d d x W + W PW + W B − AW − Q = 0 (6) is W = W 1 · W − 1 2 (7) wher e        W 1 W 2        = E         A Q P B         ·        W | x = 0 I        (8) or the other equivalent form: W = U − 1 2 · U 1 (9) wher e h U 1 U 2 i = h I W | x = 0 i · F         − B P Q − A         (10) 2 3. Solutio ns of Linear ODE 3.1. Pr operties of E ( X ) and F ( X ) From the Definition( 4 ), it holds that              d d x E  X ( x )  = X · E  X ( x )  d d x F  X ( x )  = F  X ( x )  · X (11) Now , we will see more explicit properties of E ( X ) and F ( X ). Theorem 3.1 (Pr operties of E ( X ) and F ( X ) ). I f X ( x ) is bound ed and integr able, it holds that 1. E ( X ) and F ( X ) ar e conver gent; 2. det E ( X ) = det F ( X ) = d et e R x 0 X ( t ) d t = e R x 0 tr X ( t ) dt = e tr R x 0 X ( t ) d t (12) 3. E ( X ) and F ( X ) ar e r eversible , and F ( X ) E ( − X ) = E ( − X ) F ( X ) = I (13) P r o of . 1. Firstly , E ( A ) is convergent,since  a k ( x )  n k = 1 are bound ed i n [0,b] : ∃ M > 0, s.t. | a k ( x ) | < M , ∀ x ∈ [0 , b ], k = 1 , 2 , · · · , n So (a)    Z x 0 A ( t ) d t    = ma x      Z x 0 a k ( t ) d t      < M | x | (b)    R x 0 A ( t ) R t 0 A ( s ) d sd t    = ma x     P i R x 0 a k ( t ) R t 0 a i ( s ) d sd t     < n M 2     R x 0 R t 0 1 d sdt     < n 2! ( M | x | ) 2 (c)    Z x 0 A ( t ) Z t 0 A ( s ) Z s 0 A ( ξ ) d ξ d sd t    = ma x         X i , j Z x 0 a k ( t ) Z t 0 a i ( s ) Z s 0 a j ( ξ ) d ξ d sd t         < n 2 M 3       Z x 0 Z t 0 Z s 0 d ξ d sdt       < n 2 3! ( M | x | ) 3 (d) · · · · · · It follows that kE ( A ) k < 1 + 1 n  n M | x | + 1 2! ( n M x ) 2 + 1 3! ( n M x ) 3 + · · ·  = 1 + 1 n e n M | x | Clearly , E ( A ) is con vergent. Similarly , F ( X ) is also convergent. 3 2. ∀ n × n matrix A ( x ) , if tr A ( x ) is bou nded an d inte gral , then det E ( A ( x )) = e R x 0 tr A ( t ) d t = e tr R x 0 A ( t ) d t (14) which is a special case of Abel’ s form ula[ Chen ]: If W an d B ar e n × n ma trixes , s.t. d d x W = BW (15) then, det W = e tr B (16) Here we just take 2 × 2 matr ix for verification: Let Y ( x ) = E  A ( x )  =        y 1 , 1 y 1 , 2 y 2 , 1 y 2 , 2        , A ( x ) =        a 1 , 1 a 1 , 2 a 2 , 1 a 2 , 2        so d d x Y = A · Y means that d d x        y 1 , 1 y 1 , 2 y 2 , 1 y 2 , 2        =        a 1 , 1 a 1 , 2 a 2 , 1 a 2 , 2        ·        y 1 , 1 y 1 , 2 y 2 , 1 y 2 , 2        (17) it follows d d x (det Y ) = d et        d d x y 1 , 1 d d x y 1 , 2 y 2 , 1 y 2 , 2        + d et        y 1 , 1 y 1 , 2 d d x y 2 , 1 d d x y 2 , 2        = d et        a 1 , 1 y 1 , 1 + a 1 , 2 y 2 , 1 a 1 , 1 y 1 , 2 + a 1 , 2 y 2 , 2 y 2 , 1 y 2 , 2        + det        y 1 , 1 y 1 , 2 a 2 , 1 y 1 , 1 + a 2 , 2 y 2 , 1 a 2 , 1 y 1 , 2 + a 2 , 2 y 2 , 2        = a 1 , 1 det        y 1 , 1 y 1 , 2 y 2 , 1 y 2 , 2        + a 2 , 2 det        y 1 , 1 y 1 , 2 y 2 , 1 y 2 , 2        =  a 1 , 1 + a 2 , 2  d etY = t r A · d et Y Thus, Abel’ s formu la holds and E ( A ( x )) is rev ersible. By the times: det F ( X ) = e R x 0 tr X ( t ) dt = e tr R x 0 X ( t ) d t (18) so, all we need to proo f is det e R x 0 X ( t ) d t = e R x 0 tr X ( t ) dt (19) Because e R x 0 X ( t ) d t no lon ger satisfies Abel’ s formu la (one reason is X and R x 0 X ( t ) d t ar e un necessarily exchang e- able ) , we seek the other appro ach: ∀ n × n matrix A, ∃ n × n reversible matrix P , s.t. P − 1 A P = d iag { J 1 , J 2 , · · · , J s } : = J J is A ’ s J or dan matrix, J i is the J or dan block with eigen value λ i ( x ) . It follows that e J i = e λ i ( x )                               1 1 1 2! 1 3! · · · · · · 0 1 1 1 2! · · · · · · 0 0 1 1 · · · · · · · · · · · · · · · · · · · · · · · · 0 0 0 0 · · · 1                               (20) 4 So, P − 1 e A P = e P − 1 AP = e J = d iag { e J 1 , e J 2 , · · · , e J s } Therefo re det e A = det e J = e tr J = e tr A which yields det e R x 0 X ( t ) d t = e R x 0 tr X ( t ) dt 3. Notice that ∀ n × n matrix A, ther e exists a companio n matrix A ∗ ,s.t. A · A ∗ = A ∗ · A = det A · I (21) so, if det A , 0 , A is invertible. Therefo re, E ( X ) and F ( X ) are in vertible. Furthermo re, it holds that F ( X ) E ( − X ) = E ( − X ) F ( X ) = I (22) Because: (a) d d x  F ( X ) E ( − X )  = d d x F ( X ) · E ( − X ) + F ( X ) · d d x E ( − X ) = F ( X ) X · E ( − X ) − F ( X ) · X E ( − X ) = 0 So, F ( X ) E ( − X ) = con st . =  F ( X ) E ( − X )     x = 0 = I (b) Due to the special prop erty( 21 ) of ma trix, Eq.( 22 ) is obtained.  3.2. Pr oof o f Theor em 2 .1 P r o of . Accord ing to Definition( 4 ) and Theorem 3.1 , it follo ws              d d x E  A ( x )  = A ( x ) · E  A ( x )  d d x G ( x ) = A ( x ) · G ( x ) + F (23) wher e G ( x ) = E  A ( x )  · Z x 0 F  − A ( s )  · F ( s ) d s because d d x G ( x ) = d d x E  A ( x )  · Z x 0 F  − A ( s )  · F ( s ) d s + E  A ( x )  · F  − A ( x )  · F ( x ) = A ( x ) · E  A ( x )  · Z x 0 F  − A ( s )  · F ( s ) d s + F Clearly U ( x ) = E  A ( x )  · C + E  A ( x )  · R x 0 F  − A ( s )  · F ( s ) d s is co n vergent. Moreover , since E ( A ) is reversible, U ( x ) is the general solution of Eq.( 2 ).  5 Theorem 3.2. Assume tha t A ( x ) = ( a i j ) n × n , B ( x ) = ( b i j ) m × m , P ( x ) = ( p i j ) n × m ar e bo unded and in te grable matrixes , and U ( x ) is the desir ed n × m matrix. The Lin ear ODE : d d x U = A ( x ) U + U B ( x ) + P ( x ) (24) has general solution s U ( x ) = E ( A )  Z x 0 F  − A ( t )  P ( t ) E  − B ( t )  d t + C  F ( B ) (25) where C is n × m constant matrix. P r o of . Let U = E ( A ) · W · F ( B ), then d d x U = A ( x ) U + U B ( x ) + E ( A ) d d x W · F ( B ) (26) So Eq.( 24 ) could be reduced to E ( A ) d d x W · F ( B ) = P (27) or , d d x W = F ( − A ) · P · E ( − B ) (28) It’ s obviously that W ( x ) = Z x 0 F  − A ( t )  P ( t ) · E  − B ( t )  d t + C (29) C is n × m constant matrix .  4. Solutio ns of Riccati equation In mathematical investigation of the dynam ics o f a system, the in troductio n o f a nonlinearity always leads to some form of the Riccati equation [ W atkins ]: d d x y + a ( x ) y 2 + b ( x ) y + c ( x ) = 0 (30) But it is usually the case that not even one solution o f the Riccati equation is known. In th e fo llowing te xt, we try to giv e out solutions of Riccati equation in matrix form: d d x W + W PW + W B − AW − Q = 0 (31) wher e A ( x ) = ( a i j ) nn , B ( x ) = ( b i j ) mm , P ( x ) = ( p i j ) mn , Q ( x ) = ( q i j ) nm . 4.1. Pr oof of Theorem. 2.2 P r o of . 1. Firstly , defin e[ Polyanin , Ch 0.1.4] W 2 : = E ( PW + B ) (32) so W 2 is rev ersible, if PW + B is bounded; meanwhile, d d x W 2 = ( PW + B ) W 2 (33) Secondly , let W 1 : = W W 2 , so d d x W 1 = d d x W · W 2 + W · d d x W 2 = d d x W · W 2 + W ·  PW + B  W 2 =  d d x W + W PW + W B  W 2 (34) 6 so, with Eq.( 3 1 ) and Definition ( 32 ), it holds d d x W 1 = AW 1 + QW 2 (35) T ake the relation ship ( 33 ) and ( 35 ) into consideratio n, d d x        W 1 W 2        =        A Q P B        ·        W 1 W 2        (36) we can solve W 1 and W 2 . On the other hand, accordin g to Definition ( 32 ) , it’ s obviously that W 2 | x = 0 = E ( PW + B ) | x = 0 = I (37) so it goes without saying that W 1 | x = 0 = ( W W 2 ) | x = 0 = W | x = 0 (38) W e im mediately obtain        W 1 W 2        = E         A Q P B         ·        W | x = 0 I        (39) Therefo re W = W 1 · W − 1 2 is the solution of Eq.( 31 ). 2. Similarly , we can get d d x h U 1 U 2 i = h I W | x = 0 i ·        − B P Q − A        (40) so, W = U − 1 2 · U 1 is also the solution of Eq.( 31 ). 3. But th e two solutions are equi valence! That is, W 1 · W − 1 2 = U − 1 2 U 1 (41) or U 2 · W 1 − U 1 · W 2 = 0 (42) Because, according to Eq.( 36 ) and Eq.( 40 ) d d x  U 2 · W 1 − U 1 · W 2  = d d x U 2 · W 1 + U 2 · d d x W 1 − d d x U 1 · W 2 − U 1 · d d x W 2 =  U 1 P − U 2 A  · W 1 + U 2 ·  AW 1 + QW 2  −  U 2 Q − U 1 B  · W 2 − U 1 ·  PW 1 + BW 2  = 0 (43) As a result, U 2 · W 1 − U 1 · W 2 = con st . =  U 2 · W 1 − U 1 · W 2     x = 0 = 0 (44) which implied that two solutions are equiv a lence. 7 4. Uniq ueness. If Eq.( 31 ) has mor e th an one solution,such as X ( x ) , Y ( x ), u nder the same in itial con dition,i.e. X ( 0) = Y (0). Let W ( x ) = X ( x ) − Y ( x ). So it is clear that what we need to prove is equ itant to sho w          d d x W + W PW + W B − AW = 0 W | x = 0 = 0 (45) has uniqu eness solutio n W ( x ) = 0. T ake advantage the proof s teps we have es tablished: acc ording to step( 39 ) and ( 35 ), Any solution of Eq.( 45 ), such as W ( x ) , it is r e asonable to define W 2 = E ( PW + B ) , W 1 = W · W 2 It follows that W 2 is bound ed , d d x W 1 = AW 1 (46) and W 1 = E [ A ] · W 1 | x = 0 = E [ A ] · W | x = 0 = 0 (47) Ther efor e, W ( x ) = W 1 · W − 1 2 = 0  4.2. Simplify solutions of Riccati equation by particular solution In the research of Riccati equ ation, particular solutio n plays crucial i mpor tant ro le. T oo much of w orks have been done. The fir st im portant r esult in the analysis of the Riccati equation is that if one solutio n is known then a wh ole family of solutions can be found [ W atkins ]. Theorem 4.1. The same condition s as theor em 2.2 , Riccati equa tion d d x W + W PW + W B − AW − Q = 0 (48) has the unique solution W = Y + E  A − Y P  ·  W | x = 0  ·  I + Z x 0 R ( t ) ( t ) d t · ( W | x = 0 )  − 1 ·F  − [ B + PY ]  (49) wher e      Y is solution of Eq.( 48 ) when W | x = 0 = 0 , i.e. Y | x = 0 = 0 R : = F  − [ B + PY ]  · P · E  A − Y P  (50) P r o of . 1. Acco rding to Theorem 2.2 , Eq.( 48 ) has solutions. T ake an y one of it, such as Y , an d let V = W − Y (51) It follows that V PV = ( W − Y ) P ( W − Y ) =  W PW − Y PY  − ( W − Y ) PY − Y P ( W − Y ) =  W PW − Y PY  − V PY − Y PV Eq . ( 48 ) =  [ − d d x W − W B + AW + Q ] − [ − d d x Y − Y B + AY + Q ]  − V PY − Y PV =  − d d x V + AV − V B  − V PY − Y PV = − d d x V + ( A − Y P ) V − V ( B + PY ) (52) That is, d d x V + V PV + V ( B + PY ) − ( A − Y P ) V = 0 ( 53) 8 2. Obviou sly , E  A − Y P  and F  − [ B + PY ]  are reversible , we may let V = E  A − Y P  · U · F  − [ B + PY ]  (54) Now Eq .( 53 ) could be transforme d into  E  A − Y P  · d d x U · F  − [ B + PY ]  + ( A − Y P ) V − V ( B + PY )  + V PV + V ( B + PY ) − ( A − Y P ) V = 0 (5 5) or , d d x U + U  F  − [ B + PY ]  · P · E  A − Y P  ·  U = 0 (56) 3. Let R : = F  − [ B + PY ]  · P · E  A − Y P  (57) According to Theor em 2.2 , U h as solution U = W 1 · W − 1 2 (58) where        W 1 W 2        = E         0 0 R 0         ·        U | x = 0 I        =  I + Z x 0        0 0 R 0        d t  ·        U | x = 0 I        =         U | x = 0 I + R x 0 R ( t ) ( t ) d t · U | x = 0         (59) Now , Let’ s consider how to choose Y , so that both W and U | x = 0 are as simple as possible. It’ s clear that when Y | x = 0 = 0, U | x = 0 = Y | x = 0 = W | x = 0 In this case, U = W | x = 0 ·  I + Z x 0 R ( t ) ( t ) d t · ( W | x = 0 )  − 1 (60) It should be noticed that  I + R x 0 R ( t ) ( t ) d t · ( W | x = 0 )  is rev ersible, otherwise I + Z x 0 R ( t ) ( t ) d t · ( W | x = 0 ) ≡ 0 (61) which is clearly impossible. According to transfor mation( 54 ), the solu tion of Eq.( 48 ) is W = Y + V = Y + E  A − Y P  · W | x = 0 ·  I + Z x 0 R ( t ) ( t ) d t · ( W | x = 0 )  − 1 ·F  − [ B + PY ]  (62) wher e Y ( x ) ≡ 0 , if and only if Q ( x ) ≡ 0 .  5. Acknowledgments Thanks Pro f.Qiyan Shi’ s enthusiastic instru ction and pr ecious adv ice on the thesis . The work is also sup ported by Prof.Y oudo ng Zeng; thanks for his many h elpful discussions and sug gestions on this p aper . Besides, than ks Prof.Guowei Chen for many v aluable personal commun ications and guidance concerning the school work. References [W ang] Z huxi W ang & Dunre n Guo, An Introduction to Special F unctio ns, Peking Univ ersity Press. [Polyani n] Andrei D. Pol yanin a nd V alen tin F . Zait sev , Handbook of Exact Solutions for Ordina ry Di ff e rential Equa tions, Chapma n and Hall / CRC , 2nd Edition(0.1.4), 2002. [Chen] Gongning Chen, The Theory and Applic ation of Matrix, Science Press(Beijing), 2007. [W atkins] Thayer W atkins, Silicon V alley & T ornado Alle y , The Sol ution of the Riccati E quation , applet -magic.com . 9