Power Series Solution of Nonlinear Functional Equations In some situations, a solution of a functional equation can be found in the form of power series expansion... The function w shoul
Trang 1920 DIFFERENCEEQUATIONS ANDOTHERFUNCTIONALEQUATIONS
Remark System (17.4.2.1), and therefore the original functional equation, may have several (even in-finitely many) solutions or no solutions at all.
Example Consider the nonlinear equation
y2(x) = f (x)y(a – x). (17 4 2 2 )
Replacing x by a – x in (17.4.2.2), we get
y2(a – x) = f (a – x)y(x). (17 4 2 3 )
Eliminating y(a – x) from (17.4.2.2)–(17.4.2.3), we obtain two solutions of the original equation:
y (x) = [f2(x)f (a – x)]1/3 and y (x)≡ 0
2◦ Consider a reciprocal equation of the form
F x , y(x), y(a/x)
=0
Replacing x by a/x, we obtain a similar equation with the unknown function having
the same arguments:
F a/x , y(a/x), y(x)
=0
Eliminating y(a/x) from this and the original equation, we come to the usual algebraic (or
transcendental) equation of the formΨ x , y(x)
=0
In other words, solutions of the original functional equation y = y(x) are defined in a
parametric manner by means of a system of two algebraic (or transcendental) equations:
F (x, y, t) =0, F (a/x, t, y) =0,
where t is a parameter.
17.4.2-2 Reciprocal (cyclic) functional equations of general form
Reciprocal functional equations have the form
F x , y(x), y(ϕ(x)), y(ϕ[2](x)), , y(ϕ[n–1](x))
Here we use the notation ϕ[n] (x) = ϕ(ϕ[n–1](x)) and ϕ(x) is a cyclic function satisfying the
condition
The value n is called the order of a cyclic (reciprocal) equation.
Successively replacing n times the argument x by ϕ(x) in the functional equation
(17.4.2.4), we obtain the following system (the original equation coincides with the first equation of this system):
F x , y0, y1, , y n–1
=0,
F ϕ (x), y1, y2, , y0
=0, ,
F ϕ[n–1 ](x), y n–1, y0, , y n–2
=0,
(17.4.2.6)
where we have set y0 = y(x), y1 = y(ϕ(x)), , y n–1 = y(ϕ[n–1](x)); condition (17.4.2.5) implies that y n = y0
Eliminating y1, y2, , y n–1from the system of nonlinear algebraic (or transcendental) equations (17.4.2.6), we obtain a solution of the functional equation (17.4.2.4) in implicit formΨ(x, y0) =0, where y0= y(x).
Trang 217.4.3 Nonlinear Functional Equations Reducible to Difference
Equations
17.4.3-1 Functional equations with arguments proportional to x.
Consider the equation
F x , y(a1x ), y(a2x ), , y(a n x)
The transformation
y (x) = w(z), z = ln x
reduces (17.4.3.1) to the difference equation
F e z , w(z + h
1), w(z + h2), , w(z + h n)
=0, h k = ln a k.
17.4.3-2 Functional equations with powers of x as arguments.
Consider the equation
F x , y(x n1), y(x n2), , y(x n m)
The transformation
y (x) = w(z), z = ln ln x
reduces (17.4.3.2) to the difference equation
F e e z
, w(z + h1), w(z + h2), , w(z + h n)
=0, h k = ln ln n k.
17.4.3-3 Functional equations with exponential functions of x as arguments.
Consider the equation
F x , y(e λ1x ), y(e λ2x ), , y(e λ n x)
=0
The transformation
y (x) = w(ln z), z = ln x
reduces this equation to the difference equation
F e z , w(z + h
1), w(z + h2), , w(z + h n)
=0, h k = ln λ k.
17.4.3-4 Equations containing iterations of the unknown function
Consider the equation
F x , y(x), y[2](x), , y[n] (x)
=0,
where y[2](x) = y y (x)
, , y[n] (x) = y y[n–1 ](x)
We seek its solution in parametric form
x = w(t), y = w(t +1)
Then the original equation is reduced to the following nth-order difference equation:
F w (t), w(t +1), w(t +2), , w(t + n) =0
Trang 3922 DIFFERENCEEQUATIONS ANDOTHERFUNCTIONALEQUATIONS
17.4.4 Power Series Solution of Nonlinear Functional Equations
In some situations, a solution of a functional equation can be found in the form of power series expansion
Consider the nonlinear functional equation
y (x) – F x , y(ϕ(x)
Suppose that the following conditions hold:
1) there exist a and b such that ϕ(a) = a, F (a, b) = b;
2) the function ϕ(x) is analytic in a neighborhood of a and|ϕ (a)|<1;
3) the function F is analytic in a neighborhood of the point (a, b) and|∂F
∂b (a, b)|<1 Then the formal power series solution of equation (17.4.4.1),
y (x) = b +
∞
k=1
has a positive radius of convergence
The formal solution is obtained by substituting the expansions
ϕ (x) = a +
∞
k=1
a k (x – a) k,
F (x, y) = b +
∞
i,j=1
b ij (x – a) i (y – b) j,
and (17.4.4.2) into equation (17.4.4.1) Gathering the terms with the same powers of the
difference ξ = (x – a) and then equating to zero the coefficients of different powers of ξ, we obtain a triangular system of algebraic equations for the coefficients c k
See also Section T12.2, which gives exact solutions of some nonlinear difference and functional equations with one independent variable
17.5 Functional Equations with Several Variables
17.5.1 Method of Differentiation in a Parameter
17.5.1-1 Classes of equations Description of the method
Consider linear functional equations of the form
w (x, t) = θ(x, t, a) w ϕ (x, t, a), ψ(x, t, a)
where x and t are independent variables, w = w(x, t) is the function to be found, θ = θ(x, t, a),
ϕ = ϕ(x, t, a), ψ = ψ(x, t, a) are given functions, and a is a free parameter that can take any value (on some interval) Assume that for a particular a = a0, we have
θ (x, t, a0) =1, ϕ (x, t, a0) = x, ψ (x, t, a0) = t, (17.5.1.2)
i.e., for a = a0the functional equation (17.5.1.1) turns into identity
Trang 4Let us expand (17.5.1.1) in powers of the small parameter a in a neighborhood of a0,
taking into account (17.5.1.2), and then divide the resulting equation by a – a0 and pass to
the limit for a → a0 As a result, we obtain a first-order linear partial differential equation
for the function w:
ϕ ◦
a (x, t) ∂w ∂x + ψ ◦ a (x, t) ∂w ∂t + θ ◦ a (x, t)w =0, (17.5.1.3) where we have used the following notation:
ϕ ◦
a (x, t) =
∂ϕ
∂a
a=a0, ψ ◦
a (x, t) =
∂ψ
∂a
a=a0, θ ◦
a (x, t) =
∂θ
∂a
a=a0
In order to solve equation (17.5.1.3), one should consider the corresponding system of equations for characteristics (see Subsection 13.1.1):
dx
ϕ ◦ a (x, t) =
dt
ψ ◦ a (x, t) = –
dw
Let
u1(x, t) = C1, u2(x, t, w) = C2 (17.5.1.5)
be independent integrals of the characteristic system (17.5.1.4) Then the general solution
of equation (17.5.1.3) has the form
u2(x, t, w) = F u1(x, t)
where F (z) is an arbitrary function The function w should be expressed from (17.5.1.6)
and substituted into the original equation (17.5.1.1) for verification [there is a possibility
of redundant solutions; it is also possible that a solution of the partial differential equation (17.5.1.3) is not a solution of the functional equation (17.5.1.1); see Example 3 below]
Remark 1. Equation (17.5.1.3) can be obtained from (17.5.1.1) by differentiation in the parameter a, after which one should take a = a0.
Remark 2. It is convenient to choose the second integral in (17.5.1.5) to be a linear function of w, i.e.,
u2(x, t, w) = ξ(x, t)w, and rewrite (17.5.1.6) as a relation resolved with respect to w.
17.5.1-2 Examples of solutions of some specific functional equations
Example 1 Self-similar solutions, which often occur in mathematical physics, may be defined as solutions
invariant with respect to the scaling transformation, i.e., solutions satisfying the functional equation
w (x, t) = a k w (a m x , a n t), (17 5 1 7 )
where k, m, n are given constants, and a >0 is an arbitrary constant.
Equation (17.5.1.7) turns into identity for a =1 Differentiating (17.5.1.7) in a and taking a =1 , we come
to the first-order partial differential equation
mx ∂w
∂x + nt ∂w
The first integrals of the corresponding characteristic system of ordinary differential equations
dx
mx = dt
nt = –dw
kw
can be written in the form
xt–m/n = C1 , t k/n w = C2 (n≠ 0 ).
Trang 5924 DIFFERENCEEQUATIONS ANDOTHERFUNCTIONALEQUATIONS
Therefore, the general solution of the partial differential equation (17.5.1.8) has the form
w (x, t) = t–k/n F (z), z = xt–m/n, (17 5 1 9 )
where F (z) is an arbitrary function Direct verification shows that expression (17.5.1.9) is a solution of the
functional equation (17.5.1.7).
Example 2 Consider the functional equation
w (x, t) = a k w (a m x , t + β ln a), (17 5 1 10 )
where k, m, β are given constants, a >0 is an arbitrary constant.
Equation (17.5.1.10) turns into identity for a =1 Differentiating (17.5.1.10) in a and taking a =1 , we come to the first-order partial differential equation
mx ∂w
∂x + β ∂w
The corresponding characteristic system of ordinary differential equations
dx
mx = dt
β = –dw
kw
admits the first integrals
x exp(–mt/β) = C1 , w exp(kt/β) = C2 Therefore, the general solution of the partial differential equation (17.5.1.11) has the form
w (x, t) = exp(–kt/β)F (z), z = x exp(–mt/β), (17 5 1 12 )
where F (z) is an arbitrary function Direct verification shows that (17.5.1.12) is a solution of the functional
equation (17.5.1.10).
Example 3 Now consider the functional equation
w (x, t) = a k w x+ ( 1– a)t, a n t
where a >0is arbitrary and n is a constant.
Equation (17.5.1.13) turns into identity for a =1 Differentiating (17.5.1.13) in a and taking a =1 , we come to the first-order partial differential equation
–t ∂w
∂x + nt ∂w
The corresponding characteristic system
–dx
t = dt
nt = –dw
kw
has the first integrals
t + nx = C1 , wt k/n = C2 Therefore, the general solution of the partial differential equation (17.5.1.14) has the form
w (x, t) = t–k/n F (nx + t), (17 5 1 15 )
where F (z) is an arbitrary function.
Substituting (17.5.1.15) into the original equation (17.5.1.13) and dividing the result by t–k/n, we obtain
F (nx + t) = F (nx + σt), σ= ( 1– a)n + a n (17 5 1 16 )
Hence, for F (z)≠const we have σ =1 or
( 1– a)n + a n= 1 (17 5 1 17 )
Since (17.5.1.16) must hold for all a >0, it follows that (17.5.1.17), too, must hold for all a >0 This can take place only if
n= 1
In this case, the solution of equation (17.5.1.13) is given by the following formula [see (17.5.1.15) for n =1 ]:
w (x, t) = t–k F (x + t), (17 5 1 18 )
where F (z) is an arbitrary function.
If n≠ 1, then equation (17.5.1.13) admits only a degenerate solution w(x, t) = Ct–k/n , where C is an arbitrary constant [the degenerate solution corresponds to F = const in (17.5.1.16)].
Trang 617.5.2 Method of Differentiation in Independent Variables
17.5.2-1 Preliminary remarks
1◦ In some situations, differentiation in independent variables can be used to eliminate
some arguments of the functional equation under consideration and reduce it to an ordinary differential equation (see Example 1 below) The solution obtained in this way should be then inserted into the original equation in order to get rid of redundant integration constants, which may appear due to the differentiation
2◦ In some situations, differentiation in independent variables should be combined with
the multiplication (division) of the equation and the results of its differentiation by suitable functions Sometimes it is useful to take logarithm of the equation or the results of its transformation (see Example 2 below)
3◦ In some situations, differentiation of a functional equation in independent variables
allows us to eliminate some arguments and reduce the equation to a simpler functional equation whose solution is known (see Subsection 17.5.5)
17.5.2-2 Examples of solutions of some specific functional equations
Example 1 Consider the Pexider equation
f (x) + g(y) = h(x + y), (17 5 2 1 )
where f (x), g(y), h(z) are the functions to be found.
Differentiating the functional equation (17.5.2.1) in x and y, we come to the ordinary differential equation
h zz (z) =0, where z = x + y Its solution is the linear function
Substituting this expression into (17.5.2.1), we obtain
f (x) + g(y) = ax + ay + b.
Separating the variables, we find the functions f and g:
f (x) = ax + b + c,
Thus, the solution of the Pexider equation (17.5.2.1) is given by the formulas (17.5.2.2), (17.5.2.3), where a, b,
care arbitrary constants.
Example 2 Consider the nonlinear functional equation
f (x + y) = f (x) + f (y) + af (x)f (y), a≠ 0 , (17 5 2 4 )
which occurs in the theory of probability with a = –1
Differentiating both sides of this equation in x and y, we get
f zz (z) = af x (x)f y (y), (17 5 2 5 )
where z = x + y Taking the logarithm of both sides of equation (17.5.2.5) and differentiating the resulting relation in x and y, we come to the ordinary differential equation
[ln f zz (z)] zz= 0 (17 5 2 6 )
Integrating (17.5.2.6) in z twice, we get
f zz (z) = C1exp(C2z), (17 5 2 7 )
where C1and C2are arbitrary constants Substituting (17.5.2.7) into (17.5.2.5), we obtain the equation
C1exp[C2(x + y)] = af x (x)f y (y),
which admits separation of variables Integration yields
f (x) = A exp(C2x ) + B, A= 1
C2
C1
a (17 5 2 8 )
Substituting (17.5.2.8) into the original equation (17.5.2.4), we find the values of the constants: A = –B =1/a
and C2 = β is an arbitrary constant As a result, we obtain the desired solution
f (x) = 1
a e
βx– 1 .
Trang 7926 DIFFERENCEEQUATIONS ANDOTHERFUNCTIONALEQUATIONS
17.5.3 Method of Substituting Particular Values of Independent
Arguments
In order to solve functional equations with several independent variables, one often uses the method of substituting particular values of independent arguments This procedure is aimed at reducing the problem for a functional equation with several independent variables
to a problem for a system of algebraic (transcendental) equations with a single independent variable
Basic ideas of this method can be demonstrated by the example of a fairly general functional equation of the form
Φ f (x), f (x + y), f (x – y), x, y
1◦ Taking y =0in this equation, we get
If the left-hand side of this relation does not vanish identically for all f (x), then this equation can be resolved with respect to f (x) Then the function f (x) obtained in this way should
be inserted into the original equation (17.5.3.1) and one should find conditions under which this function is its solution
Now, assume that the left-hand side of (17.5.3.2) is identically equal to zero for all f (x).
2◦ In equation (17.5.3.1), we consecutively take
x=0, y = t; x = t, y =2t; x = t, y = –2t
We obtain a system of algebraic (transcendental) equations
Φ a , f (t), f (–t),0, t =0,
Φ f (t), f (3 t ), f (–t), t,2t
=0,
Φ f (t), f (–t), f (3 t ), t, –2 t
=0
(17.5.3.3)
for the unknown quantities f (t), f (–t), f (3 t ), where a = f (0) Resolving system (17.5.3.3) with respect to f (t) [or with respect to f (–t) or f (3 x)], we obtain an admissible solution that should be inserted into the original equation for verification
3◦ In some cases the following trick can be used In equation (17.5.3.1), one consecutively
takes
where a is a free parameter We obtain the system of equations
Φ f(0), f (t), f (–t),0, t =0,
Φ f (t + a), f (t +2a ), f (t), t + a, a
=0,
Φ f (a), f (t +2a ), f (–t), a, t + a
=0
(17.5.3.5)
Letting f (0) = C1and f (a) = C2, where C1 and C2 are arbitrary constants, we eliminate
f (t) and f (t +2a) from system (17.5.3.5) (it is assumed that this is possible) As a result,
we come to the reciprocal equation
Ψ f (t + a), f (–t), t, a, C1, C2
In order to find a solution of equation (17.5.3.6), we replace t with –t – a We get
Ψ f (–t), f (t + a), –t – a, a, C1, C2
Further, eliminating f (t + a) from equations (17.5.3.6)–(17.5.3.7), we come to the algebraic (transcendental) equation for f (–t).