Appendix A Solutions to the Exercises For many exercises extensive calculations are necessary that are difficult to perform by pencil and paper. It is recommended to apply the userfunctions provided on the website www.alltypes.de in those cases; a short description is given in Appendix E. Chapter 1 00 0 1.1. For y C a1y C a2y D 0 the answer is ˇ ˇ ˇ ˇ ˇ ˇ 1 ˇ y y ˇ 1 ˇ y0 y0 ˇ ˇ y y ˇ a D ˇ 1 2 ˇ ;aD ˇ 1 2 ˇ W .2/ D ˇ 1 2 ˇ : 1 .2/ ˇ 00 00 ˇ 2 .2/ ˇ 00 00 ˇ where ˇ 0 0 ˇ W y1 y2 W y1 y2 y1 y2 000 00 0 For y C a1y C a2y C a1y D 0 the answer is ˇ ˇ ˇ ˇ ˇ y y y ˇ ˇ y y y ˇ 1 ˇ 1 2 3 ˇ 1 ˇ 1 2 3 ˇ a D ˇ 0 0 0 ˇ ;aD ˇ 00 00 00 ˇ ; 1 .3/ ˇ y1 y2 y3 ˇ 2 .3/ ˇ y1 y2 y3 ˇ W ˇ 000 000 000 ˇ W ˇ 000 000 000 ˇ y1 y2 y3 y1 y2 y3 ˇ ˇ ˇ ˇ ˇ y0 y0 y0 ˇ ˇ y y y ˇ 1 ˇ 1 2 3 ˇ ˇ 1 2 3 ˇ a D ˇ 00 00 00 ˇ where W .3/ D ˇ 0 0 0 ˇ : 3 .3/ ˇ y1 y2 y3 ˇ ˇ y1 y2 y3 ˇ W ˇ 000 000 000 ˇ ˇ 00 00 00 ˇ y1 y2 y3 y1 y2 y3 1.2. Go to the ALLTYPES user interface and define e1:=Df(z 1,x)-z 2; e2:=Df(z 2,x)-z 3+a 1*z 2+a 2*z 1; e3:=Df(z 3,x)-a 3*z 1+a 1*z 3; T==|LDFMOD(DFRATF(Q,fa 1,a 2,a 3g,fxg,GRLEX), fz 3,z 2,z 1g,fxg,LEX)|; JanetBasis(fe1,e2,e3g|T|); The result of Example 1.1 is returned. The term orders T==|LDFMOD(DFRATF(Q,fa 1,a 2,a 3g,fxg,GRLEX), fz 1,z 2,z 3g,fxg,LEX)|; JanetBasis(fe1,e2,e3g|T|); F. Schwarz, Loewy Decomposition of Linear Differential Equations, Texts & Monographs 181 in Symbolic Computation, DOI 10.1007/978-3-7091-1286-1, © Springer-Verlag/Wien 2012 182 A Solutions to the Exercises T==|LDFMOD(DFRATF(Q,fa 1,a 2,a 3g,fxg,GRLEX), fz 3,z 1,z 2g,fxg,LEX)|; JanetBasis(fe1,e2,e3g|T|); yield the associated equations for z3 and z2 respectively. 1.3. Let the given fourth-order equation be 0000 000 00 0 y C a1y C a2y C a3y C a4y D 0: (A.1) In addition to the functions z1, z2,andz3 defined in Example 1.1 the functions ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ 0 0 ˇ ˇ 00 00 ˇ ˇ y1 y2 ˇ ˇ y1 y2 ˇ ˇ y1 y2 ˇ z4 D ˇ 000 000 ˇ ; z5 D ˇ 000 000 ˇ ; z6 D ˇ 000 000 ˇ : y1 y2 y1 y2 y1 y2 are required. They obey the system 0 0 0 0 z1 D z2; z2 D z3 C z4; z3 D z5; z4 D z5 a1z4 a2z2 C a4z1; 0 0 z5 D z6 a1z5 a2z3 C a4z1; z6 Da1z6 C a3a3 C a4z2: A Janet basis in lex term order with z1 ::: z6 has the form X5 X5 .VI/ .k/ 0 .k/ z1 C rk.a1;:::;a4/z1 ; z2 D z1; and zi D fi;k.a1;:::;a4/z1 kD0 kD0 for i D 3;:::;6.Therk and fi;k are differential functions of the coefficients a1;:::;a4. In order to determine the coefficients of a second-order factor (compare the discussion in Example 1.3) it suffices to find a solution with rational logarithmic derivative of the first equation for z1. 1.4. Substituting y D z into the given equation yields Á Á 0 00 0 z00 C 2 C p z0 C C p C q z D 0: 0 R 0 The coefficient of z vanishes if D1 p; hence D C exp 1 pdx , C a 2 2 constant, is the most general transformation with this property. Substitution into the 1 1 2 coefficient of z leads to r D2 4 p C q. 1.5. The first-order right factors of a L 2 type decomposition have the form 0 3 .1/ r l .C / D D p r C C ,wherep and r originate from the solution of the Riccati equation (notation as in (B.1)), and C is a constant. The Lclm for two operators of this form is 00 Á 00 r r L Á Lclm l.1/.C /; l.1/.C / D D2 C 2p D C p p0 C p2: 1 2 r0 r0 By division it is shown that any operator l.1/.C / is a divisor of L, i.e. it is contained in the left intersection ideal generated by it. A Solutions to the Exercises 183 L 2 1.6. For 1 , the first solution y1 is obtained from .D C a1/y D 0, the second from .D C a1/y DNy2 with yN2 a solution of .D C a2/y D 0. L 2 For 2 the two solutions are obtained from .D C ai /y D 0. Linear dependence over the base field would imply a relation q1y1 Cq2y2 D 0 withR q1;q2 from the base q1 field. Substituting the solutions this would entail Dexp .a2 a1/dx.Dueto q2 the non-equivalence of a1 and a2, its difference is not a logarithmic derivative; thus the right hand side cannot be rational. L 2 For 3 the equation .D Ca.C //y D 0 has to be solved, then C is specialized to N r0 r0 CN and CN . Substituting a1 D Cp and a2 D Cp in the above quotient, r C CN r C CN N q1 r C CN the integration may be performed with the result q D which is rational. p 2 r C CN 1.7. Define Á A2 4B. Two cases are distinguished. If ¤ 0,two 1 1 first-order right factors are l1;2 D D C 2 A ˙ 2 ; L has the decomposition L D Lclm.l ;l / of type L 2; a fundamental system is y D exp . 1 A ˙ 1 /x . 1 2 2 1;2 2 2 If D 0, the type L 2 decomposition is L D Lclm D C 1 A 1 , C a 3 2 x C C 1 1 constant; it yields the fundamental system y1 D exp 2 A , y2 D x exp 2 A .This result shows: A second-order lode with constant coefficients is always completely reducible. 1.8. According to Lemma 1.1, case .i/, the coefficient a of a first-order factor D C a has to satisfy 1 1 2 2 a00 3a01 C 1 a0 C a3 1 a2 C x a D 0: x x x x2 This second-order Riccati equation does not have a rational solution; thus a first- order right factor does not exist. According to case .ii/ ofthesamelemma,the coefficient b of a second-order factor D2 C bD C c follows from 2 2 4 2 2 2 b00 3b0b C 2 b0 C b3 2 b2 C x C 1 C b x D D 0: x x x x2 x x2 1 Its single rational solution b D 1 leads to c D x x and yields the second-order factor given in Example 1.3. 1.9. Let y1, y2,andy3 be a fundamental system for the homogeneous equation and W its Wronskian. The general solution may be written as Z r 0 0 y D C1y1 C C2y2 C C3y3 C y1 .y2y3 y2y3/dx Z W Z r r y .y y0 y0 y /dx C y .y y0 y0 y /dx 2 W 1 3 1 3 3 W 1 2 1 2 000 where C1, C2 and C3 are constants. For y D r with the fundamental system 2 y1 D 1, y2 D x and y3 D x there follows 184 A Solutions to the Exercises Z Z Z 1 y D C C C x C C x2 C 1 x2rdx x xrdx C x2 rdx: 1 2 3 2 2 1.10. A simple calculation shows that the commutator between l1 D D C a1 and 0 0 l2 D D C a2 vanishes if a1 a2 D 0,i.e.ifa1 and a2 differ by a constant. If this 2 0 is true, the representation (1.17) simplifies to L D D C .a1 C a2/D C a1a2 C a1; furthermore L D l1l2 D l2l1. An example of this case is D2 4xD C 4x2 3 D Lclm.D 2x C 1; D 2x 1/: An example of non-commutative first-order factors is 1 1 1 D2 1 C D C D Lclm D 1; D 1 C : x x x C 1 Chapter 2 2.1. The product of l1 Á @x C a1@y C b1 and l2 Á @x C a1@y C b2 is l1l2 D @xx C .a1 C a2/@xy C a1a2@yy C .b1 C b2/@x C.a2;x C a1a2;y C a1b2 C a2b1/@y C b2;x C a1b2;y C b1b2: The product l2l1 is obtained from this expression by interchange of all indices 1 and 2.