12.1. FIRST-ORDER DIFFERENTIAL EQUATIONS 465 12.1.6-2. Reduction of the Abel equation of the second kind to the canonical form. 1 ◦ . The substitution w =(y + g)E,whereE =exp – f 2 dx ,(12.1.6.2) brings equation (12.1.6.1) to the simpler form ww x = F 1 (x)w + F 0 (x), (12.1.6.3) where F 1 =(f 1 – 2f 2 g + g x )E, F 0 =(f 0 – f 1 g + f 2 g 2 )E 2 . 2 ◦ . In turn, equation (12.1.6.3) can be reduced, by the introduction of the new independent variable z = F 1 (x) dx,(12.1.6.4) to the canonical form ww z – w = R(z). (12.1.6.5) Here, the function R(z)isdefined parametrically (x is the parameter) by the relations R = F 0 (x) F 1 (x) , z = F 1 (x) dx. Substitutions (12.1.6.2) and (12.1.6.4), which take the Abel equation to the canonical form, are called canonical. Remark 1. The transformation w = a ˆw, z = aˆz + b brings (12.1.6.5) to a similar equation, ˆw ˆw ˆz –ˆw = a –1 R(aˆz + b). Therefore the function R(z) in the right-hand side of the Abel equation (12.1.6.5) can be identified with the two-parameter family of functions a –1 R(az + b). Remark 2. Any Abel equations of the second kind related by linear (in y) transformations x = ϕ 1 (x), y = ϕ 2 (x)y + ϕ 3 (x) have identical canonical forms (up to the two-parameter family of functions specified in Remark 1). 12.1.6-3. Reduction to an Abel equation of the first kind. The substitution y + g = 1/u leads to an Abel equation of the first kind: u x +(f 0 – f 1 g + f 2 g 2 )u 3 +(f 1 – 2f 2 g + g x )u 2 + f 2 u = 0. For equations of this type, see Subsection 12.1.5. 12.1.7. Equations Not Solved for the Derivative 12.1.7-1. Method of “integration by differentiation.” In the general case, a first-order equation not solved for the derivative, F (x, y, y x )=0,(12.1.7.1) can be rewritten in the equivalent form F (x, y, t)=0, t = y x .(12.1.7.2) 466 ORDINARY DIFFERENTIAL EQUATIONS We look for a solution in parametric form: x = x(t), y = y(t). In accordance with the first relation in (12.1.7.2), the differential of F is given by F x dx + F y dy + F t dt = 0.(12.1.7.3) Using the relation dy = tdx, we eliminate successively dy and dx from (12.1.7.3). As a result, we obtain the system of two first-order ordinary differential equations: dx dt =– F t F x + tF y , dy dt =– tF t F x + tF y .(12.1.7.4) By finding a solution of this system, one thereby obtains a solution of the original equa- tion (12.1.7.1) in parametric form, x = x(t), y = y(t). Remark 1. The application of the above method may lead to loss of individual solutions (satisfying the condition F x + tF y = 0); this issue should be additionally investigated. Remark 2. One of the differential equations of system (12.1.7.4) can be replaced by the algebraic equation F (x,y, t)=0; see equation (12.1.7.2). This technique is used subsequently in Paragraphs 12.1.7-2, 12.1.7-3, and 12.1.7-5. 12.1.7-2. Equations of the form y = f(y x ). This equation is a special case of equation (12.1.7.1), with F (x, y, t)=y – f(t). The procedure described in Paragraph 12.1.7-1 yields dx dt = f (t) t , y = f(t). (12.1.7.5) Here, the original equation is used instead of the second equation in system (12.1.7.4); this is valid because the first equation in (12.1.7.4) does not depend on y explicitly. Integrating the first equation in (12.1.7.5) yields the solution in parametric form, x = f (t) t dt + C, y = f (t). 12.1.7-3. Equations of the form x = f(y x ). This equation is a special case of equation (12.1.7.1), with F (x, y, t)=x – f(t). The procedure described in Paragraph 12.1.7-1 yields x = f(t), dy dt = tf (t). (12.1.7.6) Here, the original equation is used instead of the first equation in system (12.1.7.4); this is valid because the second equation in (12.1.7.4) does not depend on x explicitly. Integrating the second equation in (12.1.7.5) yields the solution in parametric form, x = f (t), y = tf (t) dt + C. 12.1. FIRST-ORDER DIFFERENTIAL EQUATIONS 467 12.1.7-4. Clairaut’s equation y = xy x + f(y x ). Clairaut’s equation is a special case of equation (12.1.7.1), with F (x, y, t)=y – xt – f(t). It can be rewritten as y = xt + f(t), t = y x .(12.1.7.7) This equation corresponds to the degenerate case F x + tF y ≡ 0, where system (12.1.7.4) cannot be obtained. One should proceed in the following way: the first relation in (12.1.7.7) gives dy = xdt+ tdx+ f (t) dt; performing the substitution dy = tdx, which follows from the second relation in (12.1.7.7), one obtains [x + f (t)] dt = 0. This equation splits into dt = 0 and x + f (t)=0. The solution of the first equation is obvious: t = C; it gives the general solution of Clairaut’s equation, y = Cx + f(C), (12.1.7.8) which is a family of straight lines. The second equation generates a solution in parametric form, x =–f (t), y =–tf (t)+f (t), (12.1.7.9) which is a singular solution and is the envelope of the family of lines (12.1.7.8). Remark. There are also “compound” solutions of Clairaut’s equation; they consist of part of curve (12.1.7.9) joined with the tangents at finite points; these tangents are defined by formula (12.1.7.8). 12.1.7-5. Lagrange’s equation y = xf (y x )+g(y x ). Lagrange’s equation is a special case of equation (12.1.7.1), with F(x, y, t)=y–xf (t)–g(t). In the special case f(t) ≡ t, it coincides with Clairaut’s equation; see Paragraph 12.1.7-4. The procedure described in Paragraph 12.1.7-1 yields dx dt + f (t) f(t)–t x = g (t) t – f(t) , y = xf(t)+g(t). (12.1.7.10) Here, the original equation is used instead of the second equation in system (12.1.7.4); this is valid because the first equation in (12.1.7.4) does not depend on y explicitly. The first equation of system (12.1.7.10) is linear. Its general solution has the form x = ϕ(t)C + ψ(t); the functions ϕ and ψ are defined in Paragraph 12.1.2-5. Substituting this solution into the second equation in (12.1.7.10), we obtain the general solution of Lagrange’s equation in parametric form, x = ϕ(t)C + ψ(t), y = ϕ(t)C + ψ(t) f(t)+g(t). Remark. With the above method, solutions of the form y = t k x + g(t k ), where the t k are roots of the equation f(t)–t = 0, may be lost. These solutions can be particular or singular solutions of Lagrange’s equation. 468 ORDINARY DIFFERENTIAL EQUATIONS 12.1.8. Contact Transformations 12.1.8-1. General form of contact transformations. A contact transformation has the form x = F (X, Y , Y X ), y = G(X, Y , Y X ), (12.1.8.1) where the functions F(X, Y , U)andG(X, Y , U) are chosen so that the derivative y x does not depend on Y XX : y x = y X x X = G X + G Y Y X + G U Y XX F X + F Y Y X + F U Y XX = H(X, Y , Y X ). (12.1.8.2) The subscripts X, Y ,and U after F and G denote the respective partial derivatives (it is assumed that F U 0 and G U 0). It follows from (12.1.8.2) that the relation ∂G ∂U ∂F ∂X + U ∂F ∂Y – ∂F ∂U ∂G ∂X + U ∂G ∂Y = 0 (12.1.8.3) holds; the derivative is calculated by y x = G U F U ,(12.1.8.4) where G U /F U const. The application of contact transformations preserves the order of differential equations. The inverse of a contact transformation can be obtained by solving system (12.1.8.1) and (12.1.8.4) for X, Y , Y X . 12.1.8-2. Method for the construction of contact transformations. Suppose the function F = F (X, Y , U) in the contact transformation (12.1.8.1) is specified. Then relation (12.1.8.3) can be viewed as a linear partial differential equation for the second function G. The corresponding characteristic system of ordinary differential equations (see Subsection 13.1.1), dX 1 = dY U =– F U dU F X + UF Y , admits the obvious first integral: F (X, Y , U)=C 1 ,(12.1.8.5) where C 1 is an arbitrary constant. It follows that, to obtain the general representation of the function G = G(X, Y , U), one has to deal with the ordinary differential equation Y X = U ,(12.1.8.6) whose right-hand side is defined in implicit form by (12.1.8.5). Let the first integral of equation (12.1.8.6) have the form Φ(X, Y , C 1 )=C 2 . Then the general representation of G = G(X, Y , U) in transformation (12.1.8.1) is given by G = Ψ(F , Φ), where Ψ(F , Φ) is an arbitrary function of two variables, F = F (X, Y , U)and Φ = Φ(X, Y , F ). 12.1. FIRST-ORDER DIFFERENTIAL EQUATIONS 469 12.1.8-3. Examples of contact transformations. Example 1. Legendre transformation: x = Y X , y = XY X – Y , y x = X (direct transformation); X = y x , Y = xy x – y, Y X = x (inverse transformation). This transformation is used for solving some equations. In particular, the nonlinear equation (xy x – y) a f(y x )+yg(y x )+xh(y x )=0 can be reduced by the Legendre transformation to aBernoulli equation: [Xg(X)+h(X)]Y X =g(X)Y –f(X)Y a (see Paragraph 12.1.2-6). Example 2. Contact transformation (a ≠ 0): x = Y X + aY , y = be aX Y X , y x = be aX (direct transformation); X = 1 a ln y x b , Y = 1 a x – y y x , Y X = y y x (inverse transformation). Example 3. Contact transformation (a ≠ 0): x = Y X + aX, y = 1 2 (Y X ) 2 + aY , y x = Y X (direct transformation); X = 1 a x – y x , Y = 1 2a 2y –(y x ) 2 , Y X = y x (inverse transformation). 12.1.9. Approximate Analytic Methods for Solution of Equations 12.1.9-1. Method of successive approximations (Picard method). The method of successive approximations consists of two stages. At the first stage, the Cauchy problem y x = f (x, y) (equation), (12.1.9.1) y(x 0 )=y 0 (initial condition) (12.1.9.2) is reduced to the equivalent integral equation: y(x)=y 0 + x x 0 f(t, y(t)) dt.(12.1.9.3) Then a solution of equation (12.1.9.3) is sought using the formula of successive approxi- mations: y n+1 (x)=y 0 + x x 0 f(t, y n (t)) dt; n = 0, 1, 2, The initial approximation y 0 (x) can be chosen arbitrarily; the simplest way is to take y 0 to be a number. The iterative process converges as n →∞, provided the conditions of the theorems in Paragraph 12.1.1-3 are satisfied. 470 ORDINARY DIFFERENTIAL EQUATIONS 12.1.9-2. Method of Taylor series expansion in the independent variable. A solution of the Cauchy problem (12.1.9.1)–(12.1.9.2) can be sought in the form of the Taylor series in powers of (x – x 0 ): y(x)=y(x 0 )+y x (x 0 )(x – x 0 )+ y xx (x 0 ) 2! (x – x 0 ) 2 + ··· .(12.1.9.4) The first coefficient y(x 0 ) in solution (12.1.9.4) is prescribed by the initial condition (12.1.9.2). The values of the derivatives of y(x)atx = x 0 are determined from equa- tion (12.1.9.1) and its derivative equations (obtained by successive differentiation), taking into account the initial condition (12.1.9.2). In particular, setting x = x 0 in (12.1.9.1) and substituting (12.1.9.2), one obtains the value of the first derivative: y x (x 0 )=f(x 0 , y 0 ). (12.1.9.5) Further, differentiating equation (12.1.9.1) yields y xx = f x (x, y)+f y (x, y)y x .(12.1.9.6) On substituting x = x 0 , as well as the initial condition (12.1.9.2) and the fi rst deriva- tive (12.1.9.5), into the right-hand side of this equation, one calculates the value of the second derivative: y xx (x 0 )=f x (x 0 , y 0 )+f(x 0 , y 0 )f y (x 0 , y 0 ). Likewise, one can determine the subsequent derivatives of y at x = x 0 . Solution (12.1.9.4) obtained by this method can normally be used in only some suffi- ciently small neighborhood of the point x = x 0 . Example. Consider the Cauchy problem for the equation y = e y +cosx with the initial condition y(0)=0. Since x 0 = 0, we will be constructing a series in powers of x. If follows from the equation that y (0)= e 0 +cos0 = 2. Differentiating the original equation yields y = e y y –sinx. Using the initial condition and the condition y (0)=2 just obtained, we have y (0)=e 0 ×2–sin0 = 2. Similarly, we find that y = e y y + e y (y ) 2 –cosx, whence y (0)=e 0 ×2+ e 0 ×2 2 –cos0 = 5. Substituting the values of the derivatives at x = 0 into series (12.1.9.4), we obtain the desired series representation of the solution: y = 2x + x 2 + 5 6 x 3 + ···. 12.1.9-3. Method of regular expansion in the small parameter. Consider a general first-order ordinary differential equation with a small parameter ε: y x = f (x, y, ε). (12.1.9.7) Suppose the function f is representable as a series in powers of ε: f(x, y, ε)= ∞ n=0 ε n f n (x, y). (12.1.9.8) One looks for a solution of the Cauchy problem for equation (12.1.9.7) with the initial condition (12.1.9.2) as ε → 0 in the form of a regular expansion in powers of the small parameter: y = ∞ n=0 ε n Y n (x). (12.1.9.9) 12.1. FIRST-ORDER DIFFERENTIAL EQUATIONS 471 Relation (12.1.9.9) is substituted in equation (12.1.9.7) taking into account (12.1.9.8). Then one expands the functions f n into a power series in ε and matches the coefficients of like powers of ε to obtain a system of equations for Y n (x): Y 0 = f 0 (x, Y 0 ), (12.1.9.10) Y 1 = g(x, Y 0 )Y 1 + f 1 (x, Y 0 ), g(x, y)= ∂f 0 ∂y .(12.1.9.11) Only the first two equations are written out here. The prime denotes differentiation with respect to x. The initial conditions for Y n can be obtained from (12.1.9.2) taking into account (12.1.9.9): Y 0 (x 0 )=y 0 , Y 1 (x 0 )=0. Success in the application of this method is primarily determined by the possibility of constructing a solution of equation (12.1.9.10) for the leading term in the expansion of Y 0 . It is significant that the remaining terms of the expansion, Y n with n ≥ 1, are governed by linear equations with homogeneous initial conditions. Remark 1. Paragraph 12.3.5-2 gives an example of solving a Cauchy problem by the method of regular expansion for a second-order equation and also discusses characteristic features of the method. Remark 2. The methods of scaled coordinates, two-scale expansions, and matched asymptotic expansions are also used to solve problems defined by first-order differential equations with a small parameter. The basic ideas of these methods are given in Subsection 12.3.5. 12.1.10. Numerical Integration of Differential Equations 12.1.10-1. Method of Euler polygonal lines. Consider the Cauchy problem for the first-order differential equation y x = f (x, y) with the initial condition y(x 0 )=y 0 . Our aim is to construct an approximate solution y = y(x) of this equation on an interval [x 0 , x ∗ ]. Let us split the interval [x 0 , x ∗ ]inton equal segments of length Δx = x ∗ – x 0 n .We seek approximate values y 1 , y 2 , , y n of the function y(x) at the partitioning points x 1 , x 2 , , x n = x ∗ . For a given initial value y 0 = y(x 0 ) and a sufficiently small Δx, the values of the unknown function y k = y(x k ) at the other points x k = x 0 +kΔx are calculated successively by the formula y k+1 = y k + f(x k , y k )Δx (Euler polygonal line), where k = 0, 1, , n – 1. The Euler method is a single-step method of the first-order approximation (with respect to the step Δx). 12.1.10-2. Single-step methods of the second-order approximation. Two single-step methods for solving the Cauchy problem in the second-order approximation are specifi ed by the recurrence formulas y k+1 = y k + f x k + 1 2 Δx, y k + 1 2 f k Δx)Δx, y k+1 = y k + 1 2 f k + f(x k+1 , y k + f k Δx) Δx, where f k = f (x k , y k ); k = 0, 1, , n – 1. . is a singular solution and is the envelope of the family of lines (12.1.7.8). Remark. There are also “compound” solutions of Clairaut’s equation; they consist of part of curve (12.1.7.9) joined. application of contact transformations preserves the order of differential equations. The inverse of a contact transformation can be obtained by solving system (12.1.8.1) and (12.1.8.4) for X, Y. solutions can be particular or singular solutions of Lagrange’s equation. 468 ORDINARY DIFFERENTIAL EQUATIONS 12.1.8. Contact Transformations 12.1.8-1. General form of contact transformations. A