Handbook of mathematics for engineers and scienteists part 72 pps

Reduction of the Abel equation of the second kind to the canonical form.1◦.. Therefore the function Rz in the right-hand side of the Abel equation 12.1.6.5 can be identified with the two

12.1.6-2 Reduction of the Abel equation of the second kind to the canonical form.

1◦ The substitution

w = (y + g)E, where E= exp



– f2dx



, (12.1.6.2) brings equation (12.1.6.1) to the simpler form

ww 

x = F1(x)w + F0(x), (12.1.6.3) where

F1= (f1–2f2g + g x  )E, F0 = (f0– f1g + f2g2)E2.

2◦ In turn, equation (12.1.6.3) can be reduced, by the introduction of the new independent

variable

z=



F1(x) dx, (12.1.6.4)

to the canonical form

ww 

z – w = R(z). (12.1.6.5)

Here, the function R(z) is defined parametrically (x is the parameter) by the relations

R= F0(x)

F1(x), z=



F1(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 ˆ w , z = aˆz + b brings (12.1.6.5) to a similar equation, ˆ w wˆz ˆ– ˆw=

a–1R (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–1R (az + b).

Remark 2. Any Abel equations of the second kind related by linear (in y) transformations 2x = ϕ1(x), 2y = ϕ2(x)y + ϕ3(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 + (f0– f1g + f2g2)u3+ (f

1–2f2g + g x  )u2+ f2u=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)

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 = t dx, 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.

12.1.7-4 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 = x dt + t dx + f  (t) dt; performing the substitution dy = t dx, 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 kare roots of the

equation f (t) – t =0 , may be lost These solutions can be particular or singular solutions of Lagrange’s equation.

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 ) and G(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 F U

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 + U F Y ,

admits the obvious first integral:

F (X, Y , U ) = C1, (12.1.8.5)

where C1is 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 

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 , C1) = C2.

Then the general representation of G = G(X, Y , U ) in transformation (12.1.8.1) is given

by

G=Ψ(F , 2Φ),

where Ψ(F , 2Φ) is an arbitrary function of two variables, F = F (X, Y , U) and 2Φ = Φ(X, Y , F ).

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 a Bernoulli 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 lny  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= 12(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 (x0) = y0 (initial condition) (12.1.9.2)

is reduced to the equivalent integral equation:

y (x) = y0+ x

x0f (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) = y0+ x

x0f (t, y n (t)) dt; n=0,1, 2, The initial approximation y0(x) can be chosen arbitrarily; the simplest way is to take y0

to be a number The iterative process converges as n → ∞, provided the conditions of the

theorems in Paragraph 12.1.1-3 are satisfied

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 – x0):

y (x) = y(x0) + y x  (x0)(x – x0) + y xx  (x0)

2! (x – x0)

2+· · · (12.1.9.4)

The first coefficient y(x0) in solution (12.1.9.4) is prescribed by the initial condition

(12.1.9.2) The values of the derivatives of y(x) at x = x0 are determined from equa-tion (12.1.9.1) and its derivative equaequa-tions (obtained by successive differentiaequa-tion), taking

into account the initial condition (12.1.9.2) In particular, setting x = x0 in (12.1.9.1) and substituting (12.1.9.2), one obtains the value of the first derivative:

y 

x (x0) = f (x0, y0). (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 = x0, as well as the initial condition (12.1.9.2) and the first deriva-tive (12.1.9.5), into the right-hand side of this equation, one calculates the value of the second derivative:

y 

xx (x0) = f x (x0, y0) + f (x0, y0)f y (x0, y0).

Likewise, one can determine the subsequent derivatives of y at x = x0

Solution (12.1.9.4) obtained by this method can normally be used in only some

suffi-ciently small neighborhood of the point x = x0

Example Consider the Cauchy problem for the equation

y  = e y + cos x with the initial condition y(0 ) = 0

Since x0 = 0, we will be constructing a series in powers of x If follows from the equation that y ( 0 ) =

e0+ cos 0 = 2 Differentiating the original equation yields y  = e y y  – sin x Using the initial condition and the condition y ( 0 ) = 2just obtained, we have y ( 0) = e0 × 2 – sin 0 = 2 Similarly, we find that

y  = e y y + e y (y )2– cos x, whence y ( 0) = e0× 2+ e0× 2 2 – cos 0 = 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 + x2+56x3+· · ·

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

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 

Y 

1 = g(x, Y0)Y1+ f1(x, Y0), g (x, y) = ∂f0

∂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):

Y0(x0) = y0, Y1(x0) =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 Y0

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(x0) = y0 Our aim is to construct an approximate solution

y = y(x) of this equation on an interval [x0, x ∗]

Let us split the interval [x0, x ∗ ] into n equal segments of length Δx = x ∗ – x0

seek approximate values y1, y2, , y n of the function y(x) at the partitioning points x1, x2,

, x n = x ∗

For a given initial value y0 = y(x0) and a sufficiently small Δx, the values of the

unknown function y k = y(x k ) at the other points x k = x0+k Δx 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 stepΔx).

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 specified by the recurrence formulas

y k+1= y k + f x k+ 12Δx, y k+ 12f k Δx)Δx,

y k+1= y k+ 12

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