Handbook of mathematics for engineers and scienteists part 209 doc

On the interval xa, b, there exists a decreasing solution of the original equation involving an arbitrary function: y x =ϕ x for x a, c], ϕ–1x for x c, b, where c is an arbitrary point b

2◦ On the interval x

(a, b), there exists a decreasing solution of the original equation

involving an arbitrary function:

y (x) =ϕ (x) for x

(a, c],

ϕ–1(x) for x

(c, b), where c is an arbitrary point belonging to the interval (a, b), and ϕ(x) is an arbitrary continuous decreasing function on (a, c] such that

lim

x→a+0ϕ (x) = b, ϕ (c) = c.

3◦ Solution in parametric form:

x=Θ2t, y =Θt+21, whereΘ(t) = Θ(t +1) is an arbitrary periodic function with period 1

4◦ Solution in parametric form:

x=Θ1(t) +Θ2(t) sin(πt),

y=Θ1(t) –Θ2(t) sin(πt),

whereΘ1(t) andΘ2(t) are arbitrary periodic functions with period 1.

5◦ The original functional equation has a single increasing solution: y(x) = x.

6◦ Particular solutions of the equation may be represented in implicit form using the

algebraic (or transcendental) equation

Φ(x, y) =0, whereΦ(x, y) = Φ(y, x) is some symmetric function with two arguments.

8. y y(x)

+ ay(x) + bx = 0.

1◦ General solution in parametric form:

x=Θ1(t)λ t1+Θ2(t)λ t2,

y =Θ1(t)λ t+11+Θ2(t)λ t+2 1,

where λ1and λ2are roots of the quadratic equation

λ2+ aλ + b =0 andΘ1=Θ1(t) andΘ2=Θ2(t) are arbitrary periodic functions with period 1.

2◦ ForΘ1(t) = C1= const andΘ2(t) = C2= const, we have a particular solution in implicit form

λ2x – y(x)

λ2– λ1 = C1



λ1x – y(x)

C2(λ1– λ2)

k

, k= ln λ1

ln λ2.

9. y y(y(x))

– x = 0.

A special case of equation T12.1.2.21

1◦ Particular solutions:

y1(x) = – C

2

C + x, y2(x) = C –

C2

x , y3(x) = C1– (C1+ C2)

2

C2+ x ,

where C, C1, C2are arbitrary constants

2◦ Solution in parametric form:

x=Θ3t, y =Θt+31, whereΘ(t) = Θ(t +1) is an arbitrary periodic function with period 1

T12.1.2-3 Equations involving unknown function with three different arguments.

10. Ay(ax) + By(bx) + y(x) = 0.

This functional equation has particular solutions of the form y(x) = Cx β , where C is an arbitrary constant and β is a root of the transcendental equation Aa β + Bb β+1=0

11. Ay(x a ) + By(x b ) + y(x) = 0.

This functional equation has particular solutions of the form y(x) = C| ln x| p , where C is an arbitrary constant and p is a root of the transcendental equation A| a|p + B| b|p+1=0

12. Ay(x) + By  ax – β

x + b



+ Cy  bx + β

a – x



= f (x), β = a2+ ab + b2

Let us substitute x in the equation first by ax – β

x + b and then by

bx + β

a – x to obtain two more

equations So we get the system (the original equation is given first)

Ay (x) + By(u) + Cy(w) = f (x),

Ay (u) + By(w) + Cy(x) = f (u),

Ay (w) + By(x) + Cy(u) = f (w),

(1)

where u = ax – β

x + b and w =

bx + β

a – x . Eliminating y(u) and y(w) from the system of linear algebraic equations (1) yields the

solution of the original functional equation

13. f1(x)y(x) + f2(x)y  ax – β

x + b



+ f3(x)y  bx + β

a – x



= g(x), β = a2+ ab + b2

Let us substitute x in the equation first by ax – β

x + b and then by

bx + β

a – x to obtain two more

equations So we get the system (the original equation is given first)

f1(x)y(x) + f2(x)y(u) + f3(x)y(w) = g(x),

f1(u)y(u) + f2(u)y(w) + f3(u)y(x) = g(u),

f1(w)y(w) + f2(w)y(x) + f3(w)y(u) = g(w),

(1)

where

u= ax – β

x + b , w=

bx + β

a – x . Eliminating y(u) and y(w) from the system of linear algebraic equations (1) yields the solution y = y(x) of the original functional equation.

T12.1.2-4 Higher-order linear difference equations

14. y n+m + a m–1 y n+m–1+· · · + a1y n+1 + a0y n= 0.

A homogeneous mth-order linear difference equation defined on a discrete set of points

x=0,1, 2, Notation adopted: yn = y(n).

Let λ1, λ2, , λ mbe roots of the characteristic equation

P (λ)≡λ m + a

m–1λ m–1+· · · + a1λ + a0=0 (1)

If all the roots of the characteristic equation (1) are distinct, then the general solution of the original difference equation has the form

y n=

m–1



i=0

y i

m–i–1

j=0

a i+j+1

m



k=1

λ n+1

k

where the prime denotes a derivative

Formula (2) involves the initial values y0, y1, , y m They can be set arbitrarily

In the case of complex conjugate roots, one should separate the real and imaginary parts

in solution (2)

15. y n+m + a m–1 y n+m–1+· · · + a1y n+1 + a0y n = f n.

A nonhomogeneous mth-order linear difference equation defined on a discrete set of points

x=0,1, 2, Notation adopted: yn = y(n).

The general solution of the difference equation has the form y(x) = Y (x) + ¯y(x), where

Y (x) is the general solution of the corresponding homogeneous equation (with f n≡ 0) and

¯y(x) is any particular solution of the nonhomogeneous equation.

Let λ1, λ2, , λ mbe roots of the characteristic equation

P (λ)≡λ m + a

m–1λ m–1+· · · + a1λ + a0=0 (1)

If all the roots of the characteristic equation (1) are distinct, then the general solution of the original difference equation has the form

y n=

m–1



i=0

y i

m–i–1

j=0

a i+j+1

m



k=1

λ n+1

k

P  (λ k) +

n



ν=m

f n–ν

m



k=1

λ ν–1

k

P  (λ k), (2) where the prime denotes a derivative

Formula (2) involves the initial values y0, y1, , y m They can be set arbitrarily

In the case of complex conjugate roots, one should separate the real and imaginary parts

in solution (2)

16. y(x + n) + a n–1 y(x + n – 1) + · · · + a1y(x + 1) + a0y(x)= 0.

A homogeneous nth-order constant-coefficient linear difference equation.

Let us write out the characteristic equation:

λ n + a n–1λ n–1+· · · + a1λ + a0=0 (1) Consider the following cases

1◦ All roots λ1, λ2, , λ

nof equation (1) are real and distinct Then the general solution

of the original finite-difference equation has the form

y (x) =Θ1(x)λ x1 +Θ2(x)λ x2+· · · + Θ n (x)λ x n, (2) whereΘ1(x),Θ2(x), ,Θn (x) are arbitrary periodic functions with period 1, which means

thatΘk (x) =Θk (x +1), k =1, 2, , n

ForΘk (x)≡C k, formula (2) gives a particular solution

y (x) = C1λ x

1+ C2λ x

2 +· · · + C n λ x

n, where C1, C2, , C nare arbitrary constants

2◦ There are m equal real roots, λ1 = λ2 = · · · = λ m (m≤n), the other roots real and distinct In this case, the general solution of the functional equation is expressed as

y=

Θ1(x)+xΘ2(x)+ · · ·+x m–1Θm (x)

λ x

1+Θm+1(x)λ x m+1+Θm+2(x)λ x m+2+· · ·+Θ n (x)λ x n.

3◦ There are m equal complex conjugate roots, λ = ρ(cos β i sin β) (2 m≤n), the other roots real and distinct Then the original functional equation has a solution corresponding

toΘn (x)≡constk:

y = ρ x cos(βx)(A1+ A2x+· · · + A m x m–1)

+ ρ x sin(βx)(B1+ B2x+· · · + B m x m–1)

+ C m+1λ x

m+1+ C m+2λ x

m+2+· · · + C n λ x

n, where A1, , A m , B1, , B m , C2m+1, , C nare arbitrary constants

17. y(x + n) + a n–1 y(x + n – 1) + · · · + a1y(x + 1) + a0y(x) = f (x).

A nonhomogeneous nth-order constant-coefficient linear difference equation.

1◦ Solution:

y (x) = Y (x) + ¯y(x),

where Y (x) is the general solution of the corresponding homogeneous equation

Y (x + n) + a n–1Y (x + n –1) +· · · + a1Y (x +1) + a0Y (x) =0

(see the previous equation), and ¯y(x) is any particular solution of the nonhomogeneous

equation

2◦ For f (x) =n

k=0A k x

n, the nonhomogeneous equation has a particular solution of the form

¯y(x) = n

k=0B k x

n ; the constants B

kare found by the method of undetermined coefficients.

3◦ For f (x) =n

k=1A k exp(λ k x), the nonhomogeneous equation has a particular solution of

the form¯y(x) =n

k=1B k exp(λ k x ); the constants B kare found by the method of undetermined

coefficients

4◦ For f (x) = n

k=1A k cos(λ k x), the nonhomogeneous equation has a particular solution of

the form¯y(x) = n

k=1B k cos(λ k x) +

n



k=1D k sin(λ k x ); the constants B k and D k are found by

the method of undetermined coefficients

5◦ For f (x) = n

k=1A k sin(λ k x), the nonhomogeneous equation has a particular solution of

the form¯y(x) = n

k=1B k cos(λ k x) +

n



k=1D k sin(λ k x ); the constants B k and D k are found by

the method of undetermined coefficients

18. y(x + b n ) + a n–1 y(x + b n–1 ) + · · · + a1y(x + b1) + a0y(x)= 0.

There are particular solutions of the form y(x) = λ x k , where λ kare roots of the transcendental (or algebraic) equation

λ b n + a n–1λ b n–1 +· · · + a1λ b1 + a0 =0

T12.1.2-5 Equations involving unknown function with many different arguments.

19. y(a n x) + b n–1 y(a n–1 x)+· · · + b1y(a1x) + b0y(x)= 0.

This functional equation has particular solutions of the form y = Cx β , where C is an arbitrary constant and β is a root of the transcendental equation

a β

n + b n–1a β

n–1+· · · + b1a β

1 + b0 =0

20. y x a n

+ b n–1 y x a n–1

+· · · + b1y x a1

+ b0y(x)= 0.

This functional equation has particular solutions of the form y(x) = C| ln x| p , where C is an arbitrary constant and p is a root of the transcendental equation

|a n|p + b

n–1|a n–1|p+· · · + b1|a1|p + b0=0

21. y[n] (x) + a n–1 y[n–1] (x) + · · · + a1y(x) + a0x= 0.

Notation used: y[2](x) = y y (x)

, , y[n] (x) = y y[n–1 ](x)

1◦ Solutions are sought in the parametric form

x = w(t), y = w(t +1)

With it, the original equation is reduced to an nth-order linear finite-difference equation

(see equation 16 above):

w (t + n) + a n–1w (t + n –1) +· · · + a1w (t +1) + a0w (t) =0

2◦ In the special case a n–1 = = a1 =0and a0 = –1, we have the following solution in parametric form:

x=Θt

n



, y =Θt+1

n



, whereΘ(t) = Θ(t +1) is an arbitrary periodic function with period 1

T12.2 Nonlinear Functional Equations in One

Independent Variable

T12.2.1 Functional Equations with Quadratic Nonlinearity

T12.2.1-1 Difference equations

1. y n y n+1 = a n y n+1 + b n y n + c n.

Riccati difference equation Here, n =0, 1, and the constants an , b n , c n satisfy the

condition a n b n + c n≠ 0

1◦ The substitution

y n= u u n+1

n + a n

leads to the linear second-order difference equation of the form T12.1.2.1:

u n+2+ (a n+1– b n )u n+1– (a n b n + c n )u n=0

2◦ Let y ∗

nbe a particular solution of the Riccati difference equation Then the substitution

z n= 1

y n – y n ∗, n=0, 1, reduces this equation to the nonhomogeneous first-order linear difference equation

z n+1+ (y

∗

n – a n)2

a n b n + c n z n+

y ∗

n – a n

a n b n + c n =0

About this equation, see Paragraph 17.1.1-2

2. y(x + 1) – ay2(x) = f(x).

A special case of equation T12.2.3.1

3. y(x)y(x + 1) + a[y(x + 1) – y(x)] = 0.

Solution:

y (x) = a

x+Θ(x),

whereΘ(x) = Θ(x +1) is an arbitrary periodic function with period 1

4. y(x)y(x + 1) = a(x)y(x + 1) + b(x)y(x) + c(x).

Riccati difference equation Here, the functions a(x), b(x), c(x) satisfy the condition

a (x)b(x) + c(x)0

1◦ The substitution

y (x) = u (x +1)

u (x) + a(x)

leads to the linear second-order difference equation

u (x +2) + [a(x +1) – b(x)]u(x +1) – [a(x)b(x) + c(x)]u(x) =0

2◦ Let y0(x) be a particular solution of Riccati difference equation Then the substitution

z (x) = 1

y (x) – y0(x)

reduces this equation to the nonhomogeneous first-order linear difference equation

z (x +1) + [y0(x) – a(x)]2

a (x)b(x) + c(x) z (x) +

y0(x) – a(x)

a (x)b(x) + c(x) =0

T12.2.1-2 Functional equations involving y(x) and y(a – x).

5. y(x)y(a – x) = b2

Solutions:

y (x) = b exp

Φ(x, a – x), whereΦ(x, z) = –Φ(z, x) is any antisymmetric function with two arguments.

On settingΦ(x, z) = C(x – z), we arrive at particular solutions of the form

y (x) = be C(2x–a),

where C is an arbitrary constant.

6. y(x)y(a – x) = –b2

Discontinuous solutions:

y (x) =

⎧

⎨

⎩

bϕ (x) if x≥a/2,

ϕ (a – x) if x < a/2, where ϕ(x) is an arbitrary function There are no continuous solutions.

7. y(x)y(a – x) = f2(x).

The function f (x) must satisfy the condition f (x) = f (a – x).

1◦ The change of variable y(x) = f (x)u(x) leads to one of the equations of the form

T12.2.1.5 or T12.2.1.6

2◦ For f (x) = f (a – x), there are solutions of the form

y (x) = f (x) exp

Φ(x, a – x), whereΦ(x, z) = –Φ(z, x) is any antisymmetric function with two arguments.

On settingΦ(x, z) = C(x – z), we arrive at particular solutions

y (x) = e C(2x–a) f (x),

where C is an arbitrary constant.

8. y2(x) + y2(a – x) = b2

1◦ Solutions:

y (x) =

 1

2b2+Φ(x, a – x) ,

whereΦ(x, z) = –Φ(z, x) is any antisymmetric function of two arguments.

2◦ Particular solutions:

y1 , 2(x) = √ b

2, y3,4(x) = b sin

πx

2a



, y5 , 6(x) = b cos

πx

2a



9. y2(x) + Ay(x)y(a – x) + By2(a – x) + Cy(x) + Dy(a – x) = f(x).

A special case of equation T12.2.3.2

Solution in parametric form (w is a parameter):

y2+ Ayw + Bw2+ Cy + Dw = f (x),

w2+ Ayw + By2+ Cw + Dy = f (a – x).

Eliminating w gives the solutions in implicit form.