On a class of non-Linear differential equations with exact solutions

Vietnam Journal of Mechanics, VAST, Vol. 34, No. 1 (2012), pp. 7 – 17 ON A CLASS OF NON-LINEAR DIFFERENTIAL EQUATIONS WITH EXACT SOLUTIONS Dao Huy Bich1, Nguyen Dang Bich2 1Hanoi University of Science, VNU 2Institute for Building Science and Technology Abstract. The present paper deals with a class of non-linear ordinary second-order differential equations with exact solutions. A procedure for finding the general exact solution based on a known particular one is derived. For illustration solutions of some non-linear equations occured in many problems of solid mechanics are considered. Key words: Non-linear differential equation, general exact solution, varying coefficient iteration method, procedure for finding exact solution. 1. INTRODUCTION Generally for seeking an exact solution of a non-linear differential equation it is necessary to find an appropriate transformation deriving the non-linear equation to a linear one, but finding of such transformation is very complicated. In fact many problems of solid mechanics reduce to different types of non-linear differential equations, solutions of which can demonstrate specific effects, however through only exact solutions these effects can be observed profoundly. Hence finding exact solutions becomes very important in researching non-linear mechanical problems. The present paper introduces an idea and a procedure to find general exact solu- tions of a class of non-linear ordinary second-order differential equations based on known particular solutions. If a particular exact solution is known, then a general exact solution can be found, but for a received approximate particular one, a general solution may be obtained approximately with desired accuracy by the varying coefficient iteration method. 2. IDEA AND PROCEDURE FOR FINDING EXACT SOLUTION Consider a non-linear second-order differential equation d dt ( a1 x˙ x+ d + a2x+ a3 + b x+ d ) + ( a1 x˙ x + d + a2x+ a3 + b x+ d )2 + + d dt ( b1 x˙ x+ d + b2x+ b3 + b x+ d ) − ( b1 x˙ x+ d + b2x+ b3 + b x+ d )2 = 0. (1) where ai, bi(i = 1, 2, 3), b, d are constants, x(t) is unknown function of t. 8 Dao Huy Bich, Nguyen Dang Bich Using notation u˙ u = a1 x˙ x+ d + a2x+ a3 + b x+ d , (2) 1 2 v˙ v = b1 x˙ x+ d + b2x+ b3 + b x+ d , (3) equation (1) can be rewritten as follows: d dt ( u˙ u ) + ( u˙ u )2 + d dt ( 1 2 v˙ v ) − ( 1 2 v˙ v )2 = 0, (4) or in the other form d dt ( u˙ u + 1 2 v˙ v ) + ( u˙ u − 1 2 v˙ v )( u˙ u + 1 2 v˙ v ) = 0. (5) Eq. (1) has a particular solution when u˙ u + 1 2 v˙ v = 0, (6) or (a1 + b1) x˙ x+ d + (a2 + b2)x+ a3 + b3 + 2b x + d = 0. (7) The particular solution can be obtained from (7) as: (i). A function of t x+ d = −a3 − a2d+ b3 − b2d 2(a2 + b2) + √ δ 2(a2 + b2) th [ √ δt 2(a1 + b1) + ϕ ] , (8) where δ = (a3 − a2d+ b3 − b2d)2 − 8b(a2 + b2), ϕ - integral constant. (ii). A constant x+ d = −a3 − a2d+ b3 − b2d 2(a2 + b2) + √ δ 2(a2 + b2) , (9) x+ d = −a3 − a2d+ b3 − b2d 2(a2 + b2) − √ δ 2(a2 + b2) . (10) In order to represent funtion u through funtion v based on Eq. (5) it can be obtained [1] u˙ u = −1 2 v˙ v + C2v C1 + C2 t∫ 0 vdt , (11) thus u = 1 v1/2  C1 + C2 t∫ 0 vdt   , (12) where C1, C2 - integral constants. On a class of non-linear differential equations with exact solution 9 Subtracting side by side of Eqs.(2) and (3) and taking (11) into account yields (a1 − b1) x˙ x+ d + (a2 − b2)(x+ d) + a3 − a2d− (b3 − b2d) = C2v C1 +C2 t∫ 0 vdt − v˙ v . (13) Eq. (13) can lead to a linear differential equation with respect to unknown 1 x+ d d dt ( 1 x+ d ) + 1 a1 − b1   C2v C1 +C2 t∫ 0 vdt − v˙ v − (a3 − a2d) + (b3 − b2d)   1x+ d = a2 − b2a1 − b1 (14) Eq. (14) is derived from Eq.(1) by use of an equivalent transformation, consequently it is equivalent with Eq.(1), Eq.(14) is a linear differential equation with varying coefficient, that is in this case 1 a1 − b1   C2v C1 +C2 t∫ 0 vdt − v˙ v − (a3 − a2d) + (b3 − b2d)   . Following the idea of varying coefficient iteration method [2], at the first iteration this coefficient can be evaluated by particular solutions (8), (9) and (10). Denote the particular solution of Eq.(1) by x1, then Eq. (3) and Eq. (7) give: 2b1 x˙1 x1 + d + 2b2x1 + 2b3 + 2b x1 + d = v˙ v (15) (a1 + b1) x˙1 x1 + d + (a2 + b2)x1 + (a3 + b3) + 2b x1 + d = 0. (16) Subtracting side by side of two these equations yields: (a1 − b1) x˙1 x1 + d + (a2 − b2)x1 + a3 − b3 = − v˙ v , (17) from that v = (x1 + d) −(a1−b1) exp  −(a2 − b2) t∫ 0 (x1 + d)dt   exp [(−a3 + a2d+ b3 − b2d) t] . (18) Consequently at first iteration Eq. (14) has a general solution as follows 1 x+ d = 1 µ  a2 − b2 a1 − b1 t∫ 0 µdt+C   , (19) 10 Dao Huy Bich, Nguyen Dang Bich where C is an integral constant and µ = (x1 + d) exp  a2 − b2 a1 − b1 t∫ 0 (x1 + d)dt    C1 +C2 t∫ 0 vdt   1 a1 − b1 . (20) In order to verify the general solution (19), (20) satisfying exactly or not Eq.(14), i.e. the equivalent equation of Eq.(1), a procedure is established as following. Taking logarithm both sides of Eq.(19) and differentiating with respect to t yield d dt ( 1 x + d ) 1 x+ d = − µ˙ µ + a2 − b2 a1 − b1µ a2 − b2 a1 − b1 t∫ 0 µdt+ C = − µ˙ µ + a2 − b2 a1 − b1 (x+ d) In other side Eq.(14) reduces to d dt ( 1 x+ d ) 1 x+ d + 1 a1 − b1   C2v C1 +C2 t∫ 0 vdt − v˙ v − (a3 − a2d) + (b3 − b2d)   = a2 − b2a1 − b1 (x+ d). Eliminating d dt ( 1 x+ d ) 1 x + d from obtained equations results 1 a1 − b1   C2v C1 +C2 t∫ 0 vdt − v˙ v − (a3 − a2d) + (b3 − b2d)   = µ˙µ, (21) and from (20) it follows: µ˙ µ = x˙1 x1 + d + a2 − b2 a1 − b1 (x+ d) + C2v C1 +C2 t∫ 0 vdt 1 a1 − b1 . (22) Substitution of (22) into (21) reduces to Eq. (17) (a1 − b1) x˙1 x1 + d + (a2 − b2)(x1 + d) + a3 − a2d− (b3 − b2d) = − v˙ v . Summing up side by side of this equation and equation (15) we can get again Eq.(16). Indeed Eq.(16) is the equation for seeking particular solution to Eq.(1). In other side we can verify directly that, the general solution (19), (20) satisfies exactly the original equation (1). Remark: As seen that Eq.(14) is equivalent to Eq.(1), it has general exact solution (19), (20) when x1 is a particular solution of Eq.(1). So that at first iteration if x1 is On a class of non-linear differential equations with exact solution 11 a particular exact solution, then the expression (19), (20) is a general exact solution of Eq.(14). If x1 is a particular approximate solution, then (19), (20) is a general approximate solution of Eq. (14) at first iteration. For obtaining a general approximate solution with desired accuracy it need to fulfill successive iteration by the varying coefficient method. In the present paper we consider only the case when x1 is a particular exact solution. Consequently, the idea of the proposed method is to derive a non-linear differential equation to a linear differential equation with varying coefficient. This varying coefficient is determined by a particular solution of the original equation. If the particular solution is exact, the general exact solution can be obtained, if the particular solution is approximate then the approximated general solution can be evaluated with desired accuracy. Procedure can be established by following steps: Step 1: Seeking a particular solution of linear equation (16). Step 2: Calculate v according to (18). Step 3: Calculate µ according to (20). Step 4: Seeking a general solution according to (19). The solution of Eq.(14), i.e. the expressions (19), (20) consists of four integral con- stants and particular solution, which are chosen appropriately for each concrete problem. 3. PARTICULAR CASES OF EQ. (1) 3.1. Case 1: b1 = 0, a2 = b2 = 0, b = 0, d = 0 Eq. (1) is of the form: x¨+ 2a3x˙+ a23 − b23 a1 x = (1− a1) x˙ 2 x . (23) From Eq. (16) the particular solution is found directly x1 = exp [ −a3 + b3 a1 t ] , the integral constant is taken ϕ = 0. According to Eq.(18) function v is calculated as v = e2b3t and substituting the obtained value v into (20) yields µ = [ e−(a3+b3)t ( C1 + C2 2b3 e2b3t )]1/a1 , where C1, C2 are integral constants. The expression µ can be rewritten in other form µ = [ Ae−a3t ch (b3t+ γ) ] 1 a1 , A, γ − integral constants. By use of Eq. (19), the general solution of Eq. (23) is found x = B [ e−a3t ch (b3t+ γ) ] 1 a1 , (24) where B, γ are integral constants. 12 Dao Huy Bich, Nguyen Dang Bich If the coefficient of the Eq. (23) are denoted by a3 = ν, a23 − b23 a1 = ω2, i.e. b3 = √ ν2 − a1ω2, then with ν2 − a1ω2 < 0, the solution has form x = B [ e−νt cos ( t √ a1ω2 − ν2 + γ )] 1 a1 . (25) The general solution (25) for Eq. (23) was found in [3] by other method. 3.2. Case 2: a1 − b1 = −1, a2 = b2, b = 0, d = 0 Eq. (1) in this case has the form: x¨− 2 x˙ 2 x + 2(a1a3 − b1b3) a1 + b1 x˙+ 2a2(a3 − b3)x2 a1 + b1 + (a23 − b23)x a1 + b1 = 0. (26) According to Eqs. (8), (10) two particular solutions are obtained x11 = −a3 + b3 4a2 + a3 + b3 4a2 th [ a3 + b3 2(a1 + b1) t+ ϕ ] , (27) x12 = −a3 + b3 2a2 . (28) Based on Eqs.(14), (19) and (20) the general solution of Eq.(26) has the form: 1 x = C ( C1 + C2 t∫ 0 vdt ) x1 . (29) In the formulas (29) figure four integral constants C, C1, C2, ϕ and the particular solution x1, which need to be chosen appropriately. Such choice can be established as follows 1 x = CC1 x11 + CC2 t∫ 0 vdt x12 . (30) Carrying out some simple calculations in Eq.(27) yields 1 x11 = − 2a2 a3 + b3  1 + e2ϕe a3 + b3 a1 + b1 t   = − 2a2 a3 + b3 +Ae a3 + b3 a1 + b1 t , (31) where A = − 2a2 a3 + b3 e2ϕ is an integral constant. The function v is determined by use of Eq. (18) and (28) v = −a3 + b3 2a2 e−(a3−b3)t, On a class of non-linear differential equations with exact solution 13 and the second term in Eq.(30) leads to CC2 t∫ 0 vdt x12 = − CC2 a3 − b3 e −(a3−b3)t = Be−(a3−b3)t, (32) where B = − CC2 a3 − b3 is an integral constant. Since C1, C2 are integral constants, by choosing CC1 = 1, and substituting expres- sions (31), (32) into Eq.(30) we can write the general solution of Eq.(26) as follows 1 x = − 2a2 a3 + b3 + Ae a3 + b3 a1 + b1 t +Be−(a3−b3)t = − 2a2 a3 + b3 + e − a1a3 − b1b3 a1 + b1 t  Ae b1a3 − a1b3 a1 + b1 t +Be − b1a3 − a1b3 a1 + b1 t   , (33) or in other form: 1 x = − 2a2 a3 + b3 +De − a1a3 − b1b3 a1 + b1 t ch [ b1a3 − a1b3 a1 + b1 t+ γ ] , (34) where D, γ - integral constants. By use of notation β = a1a3 − b1b3 a1 + b1 , m + β2 = −a 2 3 − b23 a1 + b1 . Eq.(26) has the form x¨− 2 x˙ 2 x + 2βx˙+ 2a2(a3 − b3) a1 + b1 x2 − (1 + β2)x = 0, (35) general solution of which 1 x = − 2a2 a3 + b3 +De−βt ch ( t √−m+ γ) when m < 0, 1 x = − 2a2 a3 + b3 +De−βt cos ( t √ m+ γ ) when m > 0. (36) This solution was found in [4] by other method. 3.3. Case 3: a1 − b1 = 1 In this case Eq.(1) has the form: (a1 + b1)x¨+ (a2 + b2 + 2a1a2 − 2b1b2)xx˙ + (a2d+ b2d+ 2a1a3 − 2b1b3)x˙+ (a22 − b22)x3 + (a22d− b22d+ 2a2a3 − 2b2b3)x2 + (2a2a3d− 2b2b3d+ a23 − b23 + 2a2b− 2b2b)x+ a23d− b23d+ 2a3b− 2b3b = 0. (37) According to (10) the constant particular solution to Eq. (37) is defined x1 + d = − (a3 − a2d+ b3 − b2d) 2 (a2 + b2) + √ δ 2 (a2 + b2) , (38) 14 Dao Huy Bich, Nguyen Dang Bich where: δ = (a3 − a2d+ b3 − b2d)2 − 8b (a2 + b2). Based on (18) function v is calculated as: v = 1 x1 + d e−αt, (39) where α = (a2 − b2)(x1 + d) + a3 − a2d− (b3 − b2d). Function µ can be determined by (20) and (39): µ = (x1 + d)e −βt [ C1e αt − C2 (x1 + d)α ] , (40) where β = a3 − a2d− (b3 − b2d). According to Eq.(19) and using (40) the general solution to Eq.(37) can be obtained: x+ d = (x1 + d) [ Aeαt − 1 (x1 + d)α ] Aeαt +Beβt + α − β (x1 + d)αβ , (41) where A = C1 C2 , B = C C2 - integral constants. Differentiating both sides of Eq. (41) with respect to t yields x˙ = (α− β)(x1 + d)ABe(α+β)t + α β Aeαt + β α Beβt[ Aeαt + Beβt + α− β (x1 + d)αβ ]2 . (42) Suppose initial conditions x(t)|t=0 = x0, x˙(t)|t=0 = x˙0, (43) are satisfied, then A = 1 (x1 + d)α − x0 + d (α− β)(x1 − x0)(x0 + d)− x˙0(x1 + d) , B = 1 (x1 + d)β − x1 − x0 (α− β)(x1 − x0)(x0 + d)− x˙0(x1 + d) . (44) Now by use of mentioned above calculations we can seek general solutions of some non-linear equations, which usually have place in mechanical problems such as x¨+ (2ν + 3σx)x˙+ λx3 + 2qx2 + kx+ c = 0. (45) On a class of non-linear differential equations with exact solution 15 Eq. (45) coincides entirely with Eq. (37) when putting a2 + b2 + 2a1a2 − 2b1b2 = 3σ(a1 + b1), a2d+ b2d+ 2a1a3 − 2b1b3 = 2ν(a1 + b1), a22 − b22 = λ(a1 + b1), a22d− b22d+ 2a2a3 − 2b2b3 = 2q(a1 + b1), 2a2a3d− 2b2b3d+ a23 − b23 + 2a2b− 2b2b = k(a1 + b1), a23d− b23d+ 2a3b− 2b3b = c(a1 + b1). (46) The system of equations (46) consists of 6 equations with 6 unknowns a2, b2, a3, b3, b, d. So that 6 these unknowns can be determined through a1+b1 and 6 known parameters σ, v, λ, q, k, c, where a1 + b1 can be chosen arbitrarily such that a1 − b1 = 1. After that substituting just determined values into the expression of general solution (41) and integral constants (44) we can obtain the solution of Eq.(45) combined with initial conditions (43). Note that Eq.(45) with all coefficients has a similar form of Van der Pol equation, and when parameter σ = 0 has the form of Duffing equation, and when σ = 0, v = 0 - the form of KdV equation, respectively. Explicit solution of the system of equations (46) can be received, it may be expressed as real or complex values. This problem requires further investigation. 4. EXAMPLES 4.1. Case: a1 = 1, b1 = 0, b2 = 0, b = 0, d = 0 Eq (37) has the form x¨+ (2a3 + 3a2x)x˙+ a 2 2x 3 + 2a2a3x 2 + (a23 − b23)x = 0. (47) Using calculations in Section 3.3 results x1 + d = −a3 + b3 a2 , α = −2b3, β = a3 − b3, and the general solution is defined 1 x = a2 a23 − b23 [−a3 + b3 th (b3t+ ϕ)] + C0 e a3t ch (b3t+ ϕ) , (48) where C0, ϕ - integral constants. Solution (48) was found in [5] by other method. 4.2. Numerical case a1 = 1, b1 = 0, b2 = 0, b = 0, d = 0, a2 = 0.2, a3 = −0.5, b3 = 0.2, with initial conditions: x[0] = 2, x˙[0] = 0.1. a. Seeking solution by numerical method Numericalsolution = NDSolve[{x′′[t] + (2a3+3a2 ∗ x[t])∗ x′[t] + a2∧ 2 ∗ x[t]∧ 3+ 2a2 ∗ a3 ∗ x[t] ∧ 2 + (a3 ∧ 2− b3 ∧ 2) ∗ x[t], x[0] = 2, x′[0] = 0.1, x, {t, 0, 100}]. 16 Dao Huy Bich, Nguyen Dang Bich b. Seeking solution by analytical method δ = (a3 − a2d+ b3 − b2d)2 − 8b (a2 + b2) = 0.09, x1 + d = − (a3 − a2d+ b3 − b2d) 2 (a2 + b2) + √ δ 2 (a2 + b2) = 1.5, α = (a2 − b2)(x1 + d) + a3 − a2d− (b3 − b2d) = −0.4, β = a3 − a2d− (b3 − b2d) = −0.7, A = 1 (x1 + d)α − x0 + d (α− β)(x1 − x0)(x0 + d)− x˙0(x1 + d) = 2.77778, B = 1 (x1 + d)β − x1 − x0 (α− β)(x1 − x0)(x0 + d)− x˙0(x1 + d) = −0.15873, x+ d = (x1 + d) [ Aeαt − 1 (x1 + d)α ] Aeαt +Beβt + α − β (x1 + d)αβ . P lot[{x[t], Evaluate[x[t]/.numericalsolution}, {t, 0, 100},PlotStyle→ {Thick, Red, Blue}] (see Fig.1). Fig. 1. Graph of exact solution [{x(t), Evaluate[x[t]/.numericalsolution}, {t, 0, 100}] to Eq. (47) As can be observed, a very good agreement is obtained in this comparison study. On a class of non-linear differential equations with exact solution 17 5. CONCLUSION - Equation of the form (1) is non-linear differential one, general exact solutions of which can be found. They include many particular cases, where exact solutions were defined by other methods. - To obtain exact solutions of non-linear differential equations of the form (1) one needs to know important informations about particular solutions (8), (9), (10) and expres- sions (11), (12). - Proposed method for finding general exact solution of non-linear differential equa- tions of the form (1) is based upon a known particular solution and expressions (11), (12) in use of idea of the varying coefficient iteration method. - Verification of the proposed method is proven by obtaining exact solutions in many particular cases. These solutions coincide with exact solutions obtained by other analytical methods and are in good agreement with results using numerical methods. REFERENCES [1] Nguyen Dang Bich, Ngo Dinh Bao Nam, Conditions for the approximated analytical solution of a parametric oscillation problem described by the Mathieu equation, Journal of Science, Mathematics - Physics, VNU, 21(2) (2006) 9 - 16. [2] Dao Huy Bich, Plasticity theory and its applications, Publishing House of Civil Engineering, Hanoi, (2004). [3] Nguyen Dang Bich, Nguyen Vo Thong, Aerodynamical instability of cylindrical bar supported on elastic hold with viscous damping, Journal of Mechanics, 16(3) (1994) 1 - 4. [4] Dao Huy Bich, Nguyen Dang Bich, On the method for solving a class of non-linear differen- tial equations in mechanics, Proceedings of the sixth National Congress on Mechanics, 3 - 5 December, Hanoi, 3 (1997) 11 - 17. [5] Dao Huy Bich, Nguyen Dang Bich, Nguyen Vo Thong, The elastic aerodynamic problem with aerodynamical damping of turbulent characteristic, Proceedings of the ICCMS/IBST 2001, 28 - 29 March, Hanoi, Vietnam, 87 - 93. Received October 4, 2011