If y1x is the solution of the Mathieu equation satisfying the initial conditions y10 =1 and y 10 =0, the characteristic index can be determined from the relation cosh2πμ = y1π.. The gen
Trang 18. y
xx + (a – 2q cos 2x)y = 0.
Mathieu equation.
1◦ Given numbers a and q, there exists a general solution y(x) and a characteristic index μ
such that
y (x + π) = e2πμy (x).
For small values of q, an approximate value of μ can be found from the equation
cosh(πμ) =1+2sin2 12π √
a
2
(1– a) √
asin π √
a
+ O(q4)
If y1(x) is the solution of the Mathieu equation satisfying the initial conditions y1(0) =1
and y 1(0) =0, the characteristic index can be determined from the relation
cosh(2πμ ) = y1(π).
The solution y1(x), and hence μ, can be determined with any degree of accuracy by means
of numerical or approximate methods
The general solution differs depending on the value of y1(π) and can be expressed in terms of two auxiliary periodical functions ϕ1(x) and ϕ2(x) (see Table T5.3).
TABLE T5.3 The general solution of the Mathieu equation T5.2.2.8 expressed
in terms of auxiliary periodical functions ϕ1(x) and ϕ2(x)
Constraint General solution y = y(x) Period of
y1(π) >1 C1e2μx ϕ1(x) + C2e–2μx ϕ2(x) π μis a real number
y1(π) < –1 C1e2ρx ϕ1(x) + C2e–2ρx ϕ2(x) 2π μ = ρ +12i, i2= – 1 ,
ρ is the real part of μ
|y1(π)| < 1 (C1cos νx + C2sin νx)ϕ1(x)
+ (C1cos νx – C2sin νx)ϕ2(x) π
μ = iν is a pure imaginary number,
cos( 2πν) = y1(π)
2◦ In applications, of major interest are periodical solutions of the Mathieu equation that
exist for certain values of the parameters a and q (those values of a are referred to as
eigenvalues) The most important periodical solutions of the Mathieu equation have the form
ce2n(x, q) = ∞
m=0 A
2n 2mcos(2mx), ce2n+1(x, q) = ∞
m=0 A
2n+1 2m+1cos
(2m+1)x
;
se2n(x, q) = ∞
m=0 B
2n 2msin(2mx), se2n+1(x, q) = ∞
m=0 B
2n+1 2m+1sin
(2m+1)x
;
where A i j and B j i are constants determined by recurrence relations
The Mathieu functions ce2n(x, q) and se2n(x, q) are discussed in Section 18.16 in more
detail
Trang 29. y
xx + a tan x y x + by = 0.
1◦ The substitution ξ = sin x leads to a linear equation of the form T5.2.1.21: (ξ2–1)y ξξ+ (1– a)ξy ξ – by =0
2◦ Solution for a = –2:
y cos x =
C1sin(kx) + C2cos(kx) if b +1= k2 >0,
C1sinh(kx) + C2cosh(kx) if b +1= –k2 <0
3◦ Solution for a =2and b =3: y = C1cos3x + C2sin x (1+2cos2x)
T5.2.3 Equations Involving Arbitrary Functions
Notation: f = f(x) and g = g(x) are arbitrary functions; a, b, and λ are arbitrary
parameters
1. y
xx + f y x + a(f – a)y = 0.
Particular solution: y0= e–ax
2. y
xx + xf y x – f y = 0.
Particular solution: y0= x.
3. xy
xx + (xf + a)y x + (a – 1)f y = 0.
Particular solution: y0= x1–a
4. xy
xx + [(ax + 1)f + ax – 1]y x + a2xf y = 0.
Particular solution: y0= (ax +1)e–ax
5. xy
xx + [(ax2+ bx)f + 2]y x + bf y = 0.
Particular solution: y0= a + b/x.
6. x2y
xx + xf y x + a(f – a – 1)y = 0.
Particular solution: y0= x–a
7. y
xx + (f + ae λx )y x + ae λx (f + λ)y = 0.
Particular solution: y0= exp
–a
λ e
λx
8. y
xx – (f2+ f x )y = 0.
Particular solution: y0= exp
f dx
9. y
xx + 2f y x + (f2+ f x )y = 0.
Solution: y = (C2x + C1) exp
– f dx
Trang 3
10. y
xx + (1 – a)f y x – a(f2+ f x )y = 0.
Particular solution: y0= exp
a
f dx
11. y
xx + f y x + (f g – g2+ g x )y = 0.
Particular solution: y0= exp
– g dx
12. f y
xx – af x y x – bf2a+1 y= 0.
Solution: y = C1e u + C2e–u , where u = √
b
f a dx.
13. f2y
xx + f (f x + a)y x + by = 0.
The substitution ξ =
f–1dx leads to a constant coefficient linear equation: y
ξξ +ay ξ +by =0
14. y
xx – f x y x + a2e2f y= 0.
Solution: y = C1sin
a
e f dx
+ C2cos
a
e f dx
15. y
xx – f x y x – a2e2f y= 0.
Solution: y = C1exp
a
e f dx
+ C2exp
–a e f dx
T5.3 Second-Order Nonlinear Equations
T5.3.1 Equations of the Form yxx = f (x, y)
1. y
xx = f (y).
Autonomous equation.
Solution:
*
C1+2 f (y) dy
+–1 2
dy = C2 x
Particular solutions: y = A k , where A k are roots of the algebraic (transcendental)
equation f (A k) =0
2. y
xx = Ax n y m.
Emden–Fowler equation.
1◦ With m≠ 1, the Emden–Fowler equation has a particular solution:
y = λx1n+2–m, where λ = * (n +2)(n + m +1)
A (m –1)2
+ 1
m–1
2◦ The transformation z = x n+2 y m–1 , w = xy
x /y leads to a first-order (Abel) equation:
z [(m –1)w + n +2]w z = –w2+ w + Az.
3◦ The transformation y = w/t, x = 1/t leads to the Emden–Fowler equation with the
independent variable raised to a different power: w tt = At–n–m–3 w m.
4◦ The books by Zaitsev and Polyanin (1994) and Polyanin and Zaitsev (2003) present
28 solvable cases of the Emden–Fowler equation (corresponding to some pairs of n and m).
Trang 43. y
xx + f (x)y = ay–3 .
Ermakov’s equation Let w = w(x) be a nontrivial solution of the second-order linear
equation w xx + f (x)w =0 The transformation ξ =
dx
w2, z =
y
w leads to an autonomous
equation of the form T5.3.1: z ξξ = az–3
Solution: C1y2 = aw2+ w2
C2+ C1
dx
w2
2
Further on, f, g, h, and ψ are arbitrary composite functions of their arguments indicated
in parentheses after the function name (the arguments can depend on x, y, y x)
4. y
xx = f (ay + bx + c).
The substitution w = ay + bx + c leads to an equation of the form T5.3.1.1: w xx = af (w).
5. y
xx = f (y + ax2+ bx + c).
The substitution w = y+ax2+bx+c leads to an equation of the form T5.3.1.1: w xx = f (w)+2a
6. y
xx = x–1f (yx–1 ).
Homogeneous equation The transformation t = – ln|x|, z = y/x leads to an autonomous equation: z tt – z t = f (z).
7. y
xx = x–3f (yx–1 ).
The transformation ξ =1/x , w = y/x leads to the equation of the form T5.3.1.1: w ξξ = f (w).
8. y
xx = x–3/2 f (yx–1/2).
Having set w = yx–1 2, we obtain dx d (xw x)2 = 12ww
x+2f (w)w x Integrating the latter equation, we arrive at a separable equation
Solution: *
C1+ 14w2+2 f (w) dw+– 1 2
dw = C2 ln x.
9. y
xx = x k–2 f (x–k y).
Generalized homogeneous equation The transformation z = x–k y , w = xy
x /y leads to a
first-order equation: z(w – k)w z = z–1f (z) + w – w2
10. y
xx = yx–2f (x n y m).
Generalized homogeneous equation The transformation z = x n y m , w = xy
x /y leads to a
first-order equation: z(mw + n)w z = f (z) + w – w2
11. y
xx = y–3f
y
√
ax2+ bx + c
.
Setting u(x) = y(ax2+ bx + c)–1 2 and integrating the equation, we obtain a first-order separable equation:
(ax2+ bx + c)2(u x)2= (14b2– ac)u2+2 u–3f (u) du + C
1.
Trang 512. y
xx = e–ax f (e ax y).
The transformation z = e ax y , w = y x /y leads to a first-order equation: z(w + a)w z =
z–1f (z) – w2.
13. y
xx = yf (e ax y m).
The transformation z = e ax y m , w = y
x /y leads to a first-order equation: z(mw + a)w z =
f (z) – w2
14. y
xx = x–2f (x n e ay).
The transformation z = x n e ay , w = xy
x leads to a first-order equation: z(aw + n)w z =
f (z) + w.
15. y
xx =
ψ
xx
ψ y + ψ–3f y
ψ
The transformation ξ = dx
ψ2, w =
y
ψ leads to an equation of the form T5.3.1.1: w ξξ = f (w).
Solution:
*
C1+2 f (w) dw
+–1 2
dw = C2
dx
ψ2(x).
T5.3.2 Equations of the Form f (x, y)yxx = g(x, y, yx )
1. y
xx – y x = f (y).
Autonomous equation The substitution w(y) = y x leads to a first-order equation For solvable equations of the form in question, see the book by Polyanin and Zaitsev (2003)
2. y
xx + f (y)y x + g(y) = 0.
Lienard equation The substitution w(y) = y xleads to a first-order equation For solvable equations of the form in question, see the book by Polyanin and Zaitsev (2003)
3. y
xx + [ay + f (x)]y x + f x (x)y = 0.
Integrating yields a Riccati equation: y x + f (x)y + 12ay2= C.
4. y
xx + [2ay + f (x)]y x + af (x)y2= g(x).
On setting u = y x + ay2, we obtain a first-order linear equation: u x + f (x)u = g(x).
5. y
xx = ay x + e2ax f (y).
Solution: *
C1+2 f (y) dy+– 1 2
dy = C2 1
a e
ax.
6. y
xx = f (y)y x .
Solution:
F (y) + C1 = C2+ x, where F (y) =
f (y) dy.
Trang 67. y
xx =
e αx f (y) + α
y
x.
The substitution w(y) = e–αx y
x leads to a first-order separable equation: w y = f (y).
Solution:
F (y) + C1 = C2+
1
α e
αx , where F (y) = f (y) dy.
8. xy
xx = ny x + x2n+1 f (y).
1◦ Solution for n≠–1:
*
C1+2 f (y) dy+– 1 2
dy = x
n+1
n+1 + C2.
2◦ Solution for n = –1:
*
C1+2 f (y) dy+– 1 2
dy = ln|x|+ C2
9. xy
xx = f (y)y x.
The substitution w(y) = xy x /y leads to a first-order linear equation: yw y = –w +1+ f (y).
10. xy
xx =
x k f (y) + k – 1
y
x.
F (y) + C1 = C2+
1
k x
k , where F (y) = f (y) dy.
11. x2y
xx + xy x = f (y).
The substitution x = e t leads to an autonomous equation of the form T5.3.1.1: y tt = f (y).
12. (ax2+ b)y xx + axy x + f (y) = 0.
The substitution ξ = √ dx
ax2+ b leads to an autonomous equation of the form T5.3.1.1:
y
ξξ + f (y) =0
13. y
xx = f (y)y x + g(x).
Integrating yields a first-order equation: y x = f (y) dy + g (x) dx + C.
14. xy
xx + (n + 1)y x = x n–1 f (yx n).
The transformation ξ = x n , w = yx n leads to an autonomous equation of the form T5.3.1.1:
n2w
ξξ = f (w).
15. gy
xx+ 1 2g
x y x = f (y), g = g(x).
Integrating yields a first-order separable equation: g(x)(y x )2=2 f (y) dy + C1
Solution for g(x)≥ 0:
*
C1+2 f (y) dy+– 1 2
dy = C2 √ dx
g (x).
Trang 716. y
xx = –ay x + e ax f (ye ax).
The transformation ξ = e ax , w = ye ax leads to the equation w ξξ = a–2f (w), which is of the
form of T5.3.1.1
17. xy
xx = f (x n e ay )y x .
The transformation z = x n e ay , w = xy
x leads to the following first-order separable equation:
z (aw + n)w z = [f (z) +1]w.
18. x2y
xx + xy x = f (x n e ay).
The transformation z = x n e ay , w = xy
x leads to the following first-order separable equation:
z (aw + n)w z = f (z).
19. yy
xx + (y x) 2+ f (x)yy x + g(x) = 0.
The substitution u = y2 leads to a linear equation, u xx + f (x)u x+2g (x) =0, which can be
reduced by the change of variable w(x) = u xto a first-order linear equation
20. yy
xx – (y x) 2+ f (x)yy x + g(x)y2 = 0.
The substitution u = y x /y leads to a first-order linear equation: u x + f (x)u + g(x) =0
21. yy
xx – n(y x ) 2+ f (x)y2+ ay4n–2 = 0.
1◦ For n =1, this is an equation of the form T5.3.2.22
2◦ For n ≠ 1, the substitution w = y1–n leads to Ermakov’s equation T5.3.1.5: w
xx+
(1– n)f (x)w + a(1– n)w– 3=0
22. yy
xx – n(y x ) 2+ f (x)y2+ g(x)y n+1= 0.
The substitution w = y1–n leads to a nonhomogeneous linear equation: w xx +(1–n)f (x)w+
(1– n)g(x) =0
23. yy
xx + a(y x ) 2+ f (x)yy x + g(x)y2 = 0.
The substitution w = y a+1 leads to a linear equation: w xx + f (x)w x + (a +1)g(x)w =0
24. yy
xx = f (x)(y x) 2
The substitution w(x) = xy x /y leads to a Bernoulli equation T5.1.4: xw x = w+[f (x)–1]w2
25. y
xx – a(y x) 2+ f (x)e ay + g(x) = 0.
The substitution w = e–ay leads to a nonhomogeneous linear equation: w xx – ag(x)w =
af (x).
26. y
xx – a(y x) 2+ be4ay + f (x) = 0.
The substitution w = e–ay leads to Ermakov’s equation T5.3.1.5: w xx – af (x)w = abw–3
27. y
xx + a(y x) 2 – 1 2y
x = e x f (y).
The substitution w(y) = e–x (y x )2 leads to a first-order linear equation: w y+2aw=2f (y).