SECOND-ORDER DIFFERENTIAL EQUATIONS: CONDITIONS OF COMPLETE INTEGRABILITY 1. from Lectures on Integration of Differential Equations

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1 Applicable Analsis and Discrete Mathematics available online at Appl. Anal. Discrete Math. 008, doi:0.98/aadm0803e SECOND-ORDER DIFFERENTIAL EQUATIONS: CONDITIONS OF COMPLETE INTEGRABILITY Vasilij Petrovich Ermakov from Lectures on Integration of Differential Equations Linear second-order equations with variable coefficients can be completel integrated onl in ver rare cases. We consider the most important of them. To begin we prove that, if a particular integral of the equation + Ad + B = 0 is known, then the determination of a complete integral is reduced to a quadrature. Let u be a particular integral of this equation, ie d u + Adu + Bu = 0. We eliminate B from and to obtain 3 u d d u + A u d du = Mathematics Subject Classification. 34A05, 34A34. Kewords and Phrases. Ermakov, integral, complete integrabilit. Translated from Russian b A. O. Harin, under redaction b P. G. L. Leach 3

2 4 Vasilij Petrovich Ermakov A first integral of 3 is u d du = C exp A. When we integrate again, we obtain the complete integral = C u exp A u + C u. An linear differential equation of the second order, videlicet + A d + B = 0, can alwas be reduced b a transformation of the dependent variable to a form in which the first derivative is absent. Explicitl we set = z exp A to obtain d z = 4 A + da B z. We see below that this form makes it eas to discover conditions of integrabilit for differential equations. 3 The majorit of differential equations for which it is possible to find conditions of integrabilit reduce to the form 4 Ax + Bx + C d + Dx + E + F = 0, where the uppercase coefficients are parameters independent of the variables, x and. When we take the nth derivative of 4 and set we obtain z = dn n, 5 Ax + Bx + C d z D+Anx+E+Bx dz + F + An + Dn + An z = 0. Editor s Note: The integral for the second solution is known as Abel s formula.

3 Second-order differential equations: Conditions of complete integrabilit 5 Thus, if an integral of 4 is known, we can find an integral of 5 when n is a natural number. An integral of 4 can alwas be found in the case that F = 0, ie the equation has the form 6 Ax + Bx + C d + Dx + E = 0. We write Dx + E Ax = ϕx. + Bx + C Then we determine an integral of 5 of the form = α exp ϕx + β, where α and β are the arbitrar constants of integration. Consequentl we have proven that, if n is a natural number, then a particular integral of the equation Ax + Bx + C d z dz D + Anx + E + Bx + n D + A + An z = 0 is given b the formula z = dn n exp ϕx. 4 We pass to a more thorough investigation of particular cases. From what we proved above it is evident that the differential equation, 7 x+ax+b d z + n λx+b+n µx+a dz +nn λ µz = 0, is completel integrable if n is a natural number. In the present case λx + b + µx + a ϕx = = λlogx + a + µ logx + b. x + ax + b Hence a particular integral of the equation is given b the formula z = dn n x + a λ x + b µ. We appl the transformation of to 7. When we set z = x + a λ n/ x + b µ n/,

4 6 Vasilij Petrovich Ermakov we reduce 7 to the form x + ax + b d z = n λ When we compare this equation with 4 x + ax + b d = b a x + a + n µ 4 α x + a + a b x + b + λ + µ + z. 4 β x + b + γ, we obtain three algebraic equations the solutions of which are n = ± + 4α b a ± + 4β a b ± + 4γ, λ = ± + 4α b a + 4β a b 8 + 4γ, µ = + 4α b a ± + 4β a b + 4γ. Before each of the roots in these equations either upper or lower sign can be taken so that altogether there are eight solutions which leads to the following result. To find the complete integrabilit conditions of the differential equation 9 decompose the fraction into partial fractions according to = Ax + Bx + C x + a x + b Ax + Bx + C x + ax + b Ax + Bx + C x + ax + b = α x + a + β x + b + γ. The equation can be completel integrated if one of the eight expressions + 4α b a ± + 4β a b ± + 4γ is an odd integer. If this condition be satisfied, a particular integral of 9 is 0 = x + a n λ/ c + b n µ/ dn n a + a λ x + b µ,

5 Second-order differential equations: Conditions of complete integrabilit 7 where n, λ and µ are given b the formulæ 8 in which, naturall, signs should be chosen so that n be a positive integer. If two of the numbers in 0 are odd integers, we can find two independent particular integrals and consequentl the complete integral without the use of quadratures. 5 The conditions of integrabilit found above are not unique. We demonstrate the existence of other conditions. It is eas to verif that the complete integral of the equation t + b a d z dt + t dz dt δ z = 0 is given b the formula C t + δ t + b a + C t δ t + b a. As was proven in 3, the nth derivative of is the complete integral of the equation 3 t + b a d z dz + n + t dt dt + n δ z = 0. The nth derivative of is nothing but the coefficient of u n in the expansion of the expression, 4 C t + u + δ t + b a + C t + u δ t + b a, in increasing powers of u. Under the change of independent variable t x + a equation 3 becomes 5 x + ax + b d z + dz x + b + n + x + a + n δ z = 0. 4 When we substitute for t into 4, we find that the complete integral of 5 is the coefficient of u n in the expansion of the expression 6 C x + a + u + x + b + u x + a + u δ in increasing powers of u. + C x + a + u x + b + u x + a + u δ

6 8 Vasilij Petrovich Ermakov When we take the mth derivative of 5 and set s = d m z/ m, we obtain 7 x + ax + b d s + ds m + x + b + m + n + x + a + m + n δ z = 0. 4 The complete integral of 8 is the mth derivative of the coefficient of u n in the expansion of 6 in increasing powers of u. we obtain We appl the transformation of to 8. When we set s = x + a m+/4 x + b m+n+/4, 8 x + ax + b d = m 4 6 b a x + a + m + n 4 6 a b x + b + δ. 4 If we compare 8 with 9 x + ax + b d α = x + a + β x + b + γ, we make the identifications m = + + 4α b a, m + n = + + 4β a b and δ = + 4γ. Thus 9 is completel integrable if + + 4α b a and + + 4β a b are integers. In this case we obtain the complete integral through multipling x + a m+/4 x + b m n+/4 b the mth derivative of the coefficient of u n in the expansion of 6 in increasing powers of u. We have found that the equation 0 x + ax + b d α = x + a + β x + b + γ, 6

7 Second-order differential equations: Conditions of complete integrabilit 9 can be completel integrated if + + 4α b a and + + 4β a b are whole numbers. To find further conditions for integrabilit we transform the variables of 0 according to x + a b a t + a and z t + a to obtain t + at + b d z dt = γb a t + a + β t + b + α z. b a This equation is of the same form as 0 and hence is completel integrable if + + 4γ and + + 4β a b are whole numbers. With this condition 0 is also integrable. In the same wa we can prove that 0 is also integrable in the case that + + 4γ and are integers. Thus we obtain the result that + + 4α a b The equation x + ax + b d α = x + a + β x + b + γ, in addition to the cases indicated in 4, is completel integrable if two among the three numbers, + + 4α b a, + + 4β and + 4γ, a b are whole numbers. The equation x + ax + bx + c d α = x + a + 7 β x + b + γ x + c

8 30 Vasilij Petrovich Ermakov can be transformed to the form of the equation, 0, examined above b a transformation of variables. When we make the change c ac b 3 x + c = t + a + b c, = z t + a + b c, we obtain 4 t + at + b d z dt = α c at + a + β c bt + b + γ z. c ac b The integrabilit conditions for this equation can be found following the rules given in 4 and 6. Thus we obtain the following result. To find the integrabilit conditions for the differential equation decompose the fraction into partial fractions according to = Ax + Bx + C x + a x + b x + c Ax + Bx + C x + a x + b x + c Ax + Bx + C x + a x + b x + c = α x + a + The equation is completel integrable if 4α + a ba c ± + β x + b + 4β b ab c ± + γ x + c. 4γ c ac b is an odd integer. The equation is completel integrable also in the case when two of the three numbers, + + are whole numbers. 4α a ba c, + + 4β b ab c and + 4γ + c ac b, We now pass to a new form of the equation 5 x + a d z + n µ λx a dz λnz = 0. 8

9 Second-order differential equations: Conditions of complete integrabilit 3 This equation, as was shown in 3, is completel integrable if n is a positive integer. In the present case 6 ϕx = λ + µ = λx + µ logx + a. x + a Therefore a particular integral of 5 is given b the formula 7 z = dn n x + a µ e λx. We appl the transformation of to 5. When we set 8 z = x + a µ n e λx/, 5 becomes 9 d λ = λn + µ + 4 x + a + n µ 4x + a. Comparing this equation with the equation 30 α + = β x + a + we make the identifications n = ± + 4γ ± 3 µ = + 4γ ± λ = ± α. Thus we obtain the result that is completel integrable if = α + β x + a + + 4γ ± γ x + a β α, β α and γ x + a β α is an integer. Given that this condition holds, a particular integral of the equation is 3 = x + a n µ/ λx/ dn e n x + aµ e λx, where n, λ and µ are given b 3.

10 3 Vasilij Petrovich Ermakov 9 The integrabilit condition found in 8 is not unique. Moreover it does not hold in the case that α = 0, ie when the equation has the form d β 33 = x + a + γ x + a. To find the integrabilit condition for this equation we appl the transformation 34 x + a = t + a, = z t + a. Then 33 becomes 35 d z dt = β γ 4t + a z. Equation 35, hence 33, is completel integrable if + 4γ is an odd integer. Thus we obtain = α + β x + a + γ x + a is completel integrable if α = 0 and + 4γ is an odd integer. In addition to the two cases indicated the equation is also completel integrable when α = β = 0, ie, the equation is 36 = γ x + a. The solution of 36 is 37 = C x + a µ + C x + a µ, where µ = + + 4γ. 0 We demonstrate the determination of the integrabilit condition and solution given in 9. It is eas to show that 38 sx = C exp δ x + a + C exp δ x + a

11 Second-order differential equations: Conditions of complete integrabilit 33 is the solution of the equation 39 x + a d s + ds δ s = 0. We differentiate 39 n times and set Then the equation becomes z = dn s n. 40 x + a d z + n + dz δ z = 0. The solution of 40 is given b the formula 4 zx = dn n C exp δ x + a + C exp δ x + a. We appl the transformation of to 4. When we set 4 z = x + a +n/4, we obtain 43 d δ = x + a + n 4 6x + a. On comparison of 43 with 44 d β = x + a + γ x + a we obtain δ = β and n = + + 4γ. For n a positive integer the solution of 43 is +n/4 dn 45 x = x + a n C exp δ x + a + C exp δ x + a. 46 x + b d = α x + b + β x + ax + b + γ x + a

12 34 Vasilij Petrovich Ermakov can be transformed b a change of variables to the form of the equation examined in the three previous sections. When we appl the transformation 47 x + b = a b t + a and = z t + a to 46, we obtain 48 a b d z dt = α + β t + a + γ t + a. The integrabilit conditions for 48 can be found according to the rules developed in the three previous sections. Thus we obtain x + b d = α x + b + β x + ax + b + γ x + a is completel integrable in the following three cases:. if + 4γ a b ± β a b α is an odd integer,. if α = 0 and + + 4γ a b is an integer and 3. if α = β = d = α x + a 4 + β x + a 3 + γ x + a can be transformed b a change of variables to the form of the equation examined in 8, 9 and 0. If we set x + a = t + a and = z t + a,

13 Second-order differential equations: Conditions of complete integrabilit becomes 50 Thus we obtain = d z dt = α + β t + a + γ t + a z. α x + a 4 + β x + a 3 + γ x + a is completel integrable in the following three cases:. if + 4γ ± β α is an odd integer,. if α = 0 and + 4γ is an odd integer and 3. if α = β = 0. 3 The first of the integrabilit conditions given in and the solution of 50 can also be obtained as below. As was shown in 3, the differential equation 5 x + a d z + n λx + a + µ dz + nn λz = 0 is completel integrable if n is an integer. In the present case λ 5 ϕx = x + a µ x + a = λlogx + a + µ x + a. A particular integral of the equation is expressed b the formula 53 zx = dn µ n x + a λ exp. x + a When we appl the transformation of, namel z = x + a λ n/ exp we obtain 54 d = µ x + a, µ µn λ + 4x + a 4 x + a 3 + λ + 4x + a.

14 36 Vasilij Petrovich Ermakov On comparison with 49, namel we obtain n = d = α x + a 4 + β x + a 3 + γ x + a, ± + 4γ ± β, λ = ± + 4γ and µ = ± α. α If n is found to be a positive integer, a particular integral of the equation ma be written as 55 = x + a n λ/ µ d n µ exp x + a n x + a λ exp. x + a 56 4 = αx + βx + γ can be transformed b a change of variables to a particular form of the equation examined in 8. If we set we obtain 57 x + β α = t + a and = zt + a /4, d z α dt = 4αγ β + 4 6αt + a 3 6t + a z. As has been proved in 8, we obtain is completel integrable if is an odd integer. = αx + βx + γ ± 4αγ β 4α α

15 Second-order differential equations: Conditions of complete integrabilit 37 5 The integrabilit condition given in the previous section and the solution can be found as follows. 58 As has been shown in 3, the differential equation d z dz λ + µ λnz = 0 is completel integrable if n is a positive integer. In the case of 58 ϕx = λx + µ = λx + µx. Therefore a particular integral of the equation is given b z = dn n exp λx + µx. When we appl the transformation of, videlicet z = exp λx + µ, we obtain 59 d λ = x + λµx + µ 4 λ + λn. When we compare 59 with 56, videlicet we find that λ = ± α, = αx + βx + γ, µ = ± β and n = ± 4αγ β α 4α. α If n is a positive integer, the solution of 56 is 60 x = exp d n λx + µ expλx n + µx. 6 6 d = α x + a 6 + β x + a 5 + γ x + a 4

16 38 Vasilij Petrovich Ermakov can be transformed to 56 b the change of variables x + a = t and = z t. Specificall we obtain d z dt = αt + βt + γ z. As a consequence of the result of 5 we have d = α x + a 6 + β x + a 5 + γ x + a 4 can be completel integrated if ± 4αγ β 4α α is an integer. 7 All of the differential equations examined thus far can be expressed in terms of a single general formula, videlicet 6 = Ax + Bx + C Dx 3 + Ex + Fx + G. The methods of integration of 6 and its integrabilit conditions essentiall depend upon the roots of the equation 63 Dx 3 + Ex + Fx + G = 0. The sections in which we examined the different particular cases are. 7 when all roots of 63 are different,. when 63 has two roots equal, 3. 6 when 63 has three roots equal, 4. 4 and 6 when D = 0 and the roots of the equation 64 Ex + Fx + G = 0 are unequal,

17 Second-order differential equations: Conditions of complete integrabilit and 3 when D = 0 and the roots of 64 are equal, 6. 8, 9 and 0 when D = E = 0 and 7. 4 and 5 when D = E = F = We can transform the differential equation, Ex + Fx + G d + Hx + K + L = 0, to the standard form b the application of the transformation of, videlicet = z exp Hx + K Ex + Fx + G. Equation 65 becomes 66 d z = Ax + Bx + C Ex + Fx + G which is a particular case of 6 with D = 0. 8 There are man differential equations which can be reduced to the equations examined above b a change of variables. We consider some of them. 67 is transformed to 68 b the transformation = aex + be x + c αe x + β d z dt = 4t + at + bt + c t z αt + β x = log t and = z t. The integrabilit conditions for this equation can be found according to the rules developed in 4, 6, 8, 9, 0, and x d = alog x + b logx + c α log x + β

18 40 Vasilij Petrovich Ermakov is transformed to 70 b the transformation d z dt = 4 + at + bt + c z αt + β x = e t and = z expt/. The integrabilit conditions for 69 can be found according to the rules developed in 8, 9, 0, 4 and 5. 7 cos x d = a sin x + b sinx + c is transformed to 7 b the transformation t d z dt = a t + bt + c z 4 sin x = t and = z t /4. According to the rule developed in 4 7, hence 7, is completel integrable if 4 + a b + c ± 4 + a + b + c ± a is an odd integer. According to the rule developed in 6 this equation is also completel integrable in the case when two of the three numbers a b + c, a + b + c and + a are integers. 73 is transformed to 74 b the transformation = a tan x + b tanx + c t + d z dt = at + bt + c z tan x = t and = z + t /.

19 Second-order differential equations: Conditions of complete integrabilit 4 According to the rule developed in 4 74, hence 73, is completel integrable if + 4a ± a c + ib ± a c ib is an odd integer. 75 Pfaff s equation, ax δ + b x d + cx δ + e x d + fx δ + g = 0, can be transformed to the form 76 ax δ + b x d z = αx δ + βx δ + γ z b using the transformation of. Under the change of variables 77 x = t /δ and z = st δ/δ equation 76 becomes 78 δ d s dt = δ 4t + αt + βt + γ t at + b s. The integrabilit conditions for 78 can be found according to the rules developed in 4, 6, 8, 9, 0, and 3. When we set a = β = γ = 0 and b = in 76, we obtain 79 d z = αxδ z, which is known as Riccati s equation. When we appl the transformation 77, equation 79 becomes δ d s α dt = + δ 4t s. According to the rule developed in 8 we find that Riccati s equation is completel integrable if /δ is an odd integer. 9 Some nonlinear differential equations of the first and second orders are reduced to linear form b a transformation of the dependent variable. The equation 80 d = + A + B is reduced to the linear second-order equation 8 d z A dz + Bz = 0

20 4 Vasilij Petrovich Ermakov under the transformation The more general equation = z dz. 8 d = A + B + C, where A, B and C ma be functions of x, can be reduced to 80 b a change of the independent variable. We set = z A to obtain dz = z + B + da z + AC. A As has been shown above, this equation can be reduced to a linear second-order differential equation. 83 d d + A + B d + C = 0 can be transformed to the form of 8 b a change of the dependent variable. If we set 84 = exp z, we obtain dz = A + z Bz + C. As we have proven above, this equation can be reduced to the linear second-order differential equation. The particular case of 83 with A =, videlicet 85 d d + B d + C = 0, deserves special attention. Equation 85 can alwas be completel integrated since it reduces to the linear first-order equation under the transformation 84. dz + Bz = C

21 Second-order differential equations: Conditions of complete integrabilit 43 0 Some nonlinear differential equations are related to linear second-order differential equations. If an integral of the equation 86 = M is known, one can find an integral of the equation 87 where α is some constant. 88 d z = Mz + α z 3, We eliminate M from 86 and 87 to obtain d dz z d = α z 3. When we multipl both sides of 88 b dz z d, 88 becomes We integrate this to obtain 89 d dz z d = α z dz z d = C α z. d. z If and are two particular integrals of 86, we obtain two first integrals of 87 when we substitute them for in 89. The integrals are dz z d = C α z and dz z d = C α z. On the elimination of dz/ from these two first integrals we obtain the solution of 87. The solution of 87 can also be obtained as follows. From 89 we obtain = dz z d. C α z

22 44 Vasilij Petrovich Ermakov When we divide both sides b, this becomes z = d z. C z α We multipl b C and integrate both sides to obtain 90 C + C 0 = C z α, where C 0 is the second constant of integration. In 90 we have the solution of 87. Instead of it is sufficient to take an particular solution 3 of 86. Conversel, if a particular solution of 87 is known, we can find the complete solution of 86. Since it is sufficient to find particular integrals of 86, we can set C = 0 in 89. Thus we obtain dz z d = ± α z which is variables separable and can be written as d = dz z ± α z. On integration this becomes from which it follows that log = log z ± α z = z exp ± α z. We take the upper and then the lower sign to obtain two particular integrals of 86. The theorem proven in 0 can be generalised as below. If p is some known function of x and f is some other given function, then the solution of the equation 9 p d d p = p f p 3 Editor s Note: In 90 one has a generalisation of Abel s formula for a second solution of a linear second-order differential equation given a particular solution.

23 Second-order differential equations: Conditions of complete integrabilit 45 can be determined b means of quadratures. The multiplication of 9 b p d dp leads to the differential form d p d dp = f d p. p We integrate this to obtain p d dp = ϕ + C, p where ϕz = fzdz. This is the expression for a first integral of 9. We solve for, namel p d dp =, ϕ + C p divide b p and integrate to obtain 9 p + C 0 = d p, ϕ + C p where C 0 is the second constant of integration. Equation 9 represents the complete integral of the equation. In the particular case for which p = x equation 9 takes the form x 3 d = f. x REFERENCES. V. P. Ermakov: Second order differential equations. Conditions of complete integrabilit. Universita Izvestia Kiev, Series III Translator: A. O. Harin Received Jul 9, 008 Redactor: P. G. L. Leach School of Mathematical Sciences, Universit of KwaZulu-Natal, Private Bag X5400 Durban 4000, Republic of South Africa E mail: leachp@ukzn.ac.za and leach@math.aegean.gr