In this case, cos a =0and the function f t + a is “dropped” from the equations; summing up the first two equations term by term and subtracting the third equation from the resulting rela
Trang 1Remark For the sake of analysis, it is sometimes convenient to choose suitable values of the parameter
ain order to simplify system (17.5.3.5).
Example Consider the equation
f(x + y) + f (x – y) =2f(x) cos y. (17 5 3 8 )
This equation holds as identity for y =0and any f (x).
Substituting (17.5.3.4) into (17.5.3.8), we get
f(t) + f (–t) =2C1cos t, f(t +2a) + f (t) =2f(t + a) cos a, f(t +2a) + f (–t) =2C2cos(t + a),
(17 5 3 9 )
where C1= f (0), C2= f (a).
System (17.5.3.9) becomes much simpler for a = π/2 In this case, cos a =0and the function f (t + a)
is “dropped” from the equations; summing up the first two equations term by term and subtracting the third equation from the resulting relation, we immediately find a solution of the functional equation (17.5.3.8) in the form
f(t) = C1cos t + C2sin t. (17 5 3 10 ) Verification shows that the function (17.5.3.10) is indeed a solution of the functional equation (17.5.3.8).
4◦ Now consider a functional equation more general than (17.5.3.1),
Φ f (x), f (y), f (x + y), f (x – y), x, y
=0 (17.5.3.11)
Letting y =0, we get
Φ f (x), a, f (x), f (x), x,0 =0, (17.5.3.12)
where a = f (0) If the left-hand side of this relation does not vanish identically for all f (x), then it can be resolved with respect to f (x) Then, inserting an admissible solution f (x) into the original equation (17.5.3.11), we find possible values of the parameter a (there are
cases in which the equation has no solutions)
5◦ If the left-hand side of (17.5.3.11) identically vanishes for all f (x) and a, the following
approach can be used In (17.5.3.11), we consecutively take
x=0, y = t; x = t, y=2t; x=2t, y = t; x = t, y = t.
We get
Φ a , f (t), f (t), f (–t),0, t
=0,
Φ f (t), f (2t ), f (3t ), f (–t), t,2t
=0,
Φ f(2t ), f (t), f (3t ), f (t),2t , t
=0,
Φ f (t), f (t), f (2t ), a, t, t
=0,
(17.5.3.13)
where a = f (0) Eliminating f (–t), f (2t ), and f (3t) from system (17.5.3.13), we come
to an equation for f (t) The solution obtained in this manner should be inserted into the
original equation (17.5.3.11)
Trang 217.5.4 Method of Argument Elimination by Test Functions
17.5.4-1 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)
, (17.5.4.1)
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, which can take
any value (on some interval)
Instead of equation (17.5.4.1), consider an auxiliary more general functional equation
w (x, t) = θ(x, t, ξ) w ϕ (x, t, ξ), ψ(x, t, ξ)
, (17.5.4.2)
where ξ = ξ(x, t) is an arbitrary function.
Basic idea: If an exact solution of equation (17.5.4.2) can somehow be obtained, this
function will also be a solution of the original equation (17.5.4.1) [since (17.5.4.1) is a
special case of equation (17.5.4.2) with ξ = a].
In view that the function ξ = ξ(x, t) can be arbitrary, let us first take a test function so
that it satisfies the condition
ψ (x, t, ξ) = b, (17.5.4.3)
where b is a constant (usually, it is convenient to take b =1or b =0) Resolving (17.5.4.3)
with respect to ξ and substituting the test function ξ = ξ(x, t) thus obtained into (17.5.4.2),
we have
w (x, t) = θ(x, t, ξ(x, t))Φ ϕ (x, t, ξ(x, t))
, (17.5.4.4) whereΦ(ϕ)≡w (ϕ, b).
Expression (17.5.4.4) is crucial for the construction of an exact solution of the original functional equation: this expression should be substituted into (17.5.4.2) and one should find out for which functionsΦ(ϕ) it is indeed a solution of the equation for arbitrary ξ = ξ(x, t)
(in this connection, some constraints on the structure of the determining functions θ, ϕ, ψ
may appear)
Remark 1 Expression (17.5.4.4) may be substituted directly into the original equation (17.5.4.1).
Remark 2 Condition (17.5.4.3) corresponds to the elimination of the second argument (it is replaced by
a constant) in the right-hand side of equation (17.5.4.2).
Remark 3. Instead of (17.5.4.3), a similar condition ϕ(x, t, ξ) = b can be used for choosing the test function ξ = ξ(x, t).
17.5.4-2 Examples of solutions of specific functional equations
Example 1 Consider the functional equation
w(x, t) = a k w(a m x, a n t), (17 5 4 5 )
(k, m, n are given constants, a >0 is an arbitrary constant), which is a special case of equation (17.5.4.1) for
θ(x, t, a) = a k , ϕ(x, t, a) = a m x, ψ(x, t, a) = a n t.
Following the scheme described in Paragraph 17.5.4-1, let us use the auxiliary equation
w(x, t) = ξ k w(ξ m x, ξ n t) (17 5 4 6 )
and the test function ξ defined, according to (17.5.4.3), from the condition
ξ n t= 1 (b =1 ) (17 5 4 7 )
Hence, we find that ξ = t– 1/n Substituting this expression into (17.5.4.6), we get
w(x, t) = t–k/n Φ(t–m/n x), (17 5 4 8 ) whereΦ(ϕ)≡w(ϕ,1 ).
It is easy to show by direct verification that (17.5.4.8) is a solution of the functional equation (17.5.4.5) for
an arbitrary function Φ and coincides (to within notation) with solution (17.5.1.9) obtained by the method of differentiation in a parameter.
Trang 3Remark Instead of 1on the right-hand side of (17.5.4.7) we can take any constant b≠ 0 and obtain the same result (to within notation of the arbitrary function Φ).
Example 2 Consider the functional equation
w(x, t) = a k w(a m x, t + β ln a), (17 5 4 9 )
(k, m, β are given constants, a >0 is an arbitrary constant), which is a special case of equation (17.5.4.1) for
θ(x, t, a) = a k , ϕ(x, t, a) = a m x, ψ(x, t, a) = t + β ln a.
Following the above scheme, consider a more general auxiliary equation
w(x, t) = ξ k w(ξ m x, t + β ln ξ). (17 5 4 10 )
The test function ξ is found from the condition
t + β ln ξ =0 (b =0 ).
We have ξ = exp(–t/β) Substituting this expression into (17.5.4.10), we get
w(x, t) = e–kt/β Φ(xe–mt/β), (17 5 4 11 ) whereΦ(ϕ)≡ w(ϕ,0 ) Direct verification shows that (17.5.4.11) is a solution of the functional equation (17.5.4.9) for an arbitrary function Φ and coincides with solution (17.5.1.12) obtained by the method of differentiation in a parameter.
Example 3 Now consider the functional equation
w(x, t) = a k w x+ ( 1– a)t, a n t
(a >0is arbitrary, n is a constant), which is a special case of equation (17.5.4.1) for θ(x, t, a) = a k , ϕ(x, t, a) =
x+ ( 1– a)t, ψ(x, t, a) = a n t.
Following the scheme described above, consider the auxiliary equation
w(x, t) = ξ k w x+ ( 1– ξ)t, ξ n t
(17 5 4 13 )
and define the test function ξ from the condition (17.5.4.7), according to (17.5.4.3) We have ξ = t– 1/n Substituting this expression into (17.5.4.13), we get
w(x, t) = t–k/n Φ(z), z = x + t – t(n–1)/n, (17 5 4 14 ) whereΦ(ϕ)≡w(ϕ,1 ).
Substituting (17.5.4.14) into the original equation (17.5.4.12) and dividing the result by t–k/n, we find that
Φ x + t – t(n–1)/n
= Φ x+ ( 1– a + a n )t – a n–1 t(n–1)/n
(17 5 4 15 )
Since this relation must hold for all a >0 , there are two possibilities:
1 ) nis arbitrary, Φ = C = const;
2 ) n= 1 , Φ is arbitrary. (17.5.4.16)
In the second case, which corresponds to n =1 in the functional equation (17.5.4.12), its solution can be written
in the form
w(x, t) = t–k F (x + t), (17 5 4 17 )
where F (z) is an arbitrary function, F (z) = Φ(z –1 ) We see that expression (17.5.4.17) coincides with solution (17.5.1.18) obtained by the method of differentiation in a parameter.
Remark 1 The results of solving specific functional equations obtained in Subsection 17.5.4 by the elimination of an argument coincide with those obtained for the same equations in Subsection 17.5.1 by the method of differentiation in a parameter However, it should be observed that the intermediate results, when solving equation (17.5.4.12) by these methods, may not coincide [cf (17.5.4.15) and (17.5.1.16)].
Remark 2 The method of elimination of an argument is much simpler than that of differentiation in a parameter, since the former only requires to solve algebraic (transcendental) equations of the form (17.5.4.3)
with respect to ξ and does not require solutions of the corresponding first-order partial differential equations
(see Subsection 17.5.1).
Trang 417.5.5 Bilinear Functional Equations and Nonlinear Functional
Equations Reducible to Bilinear Equations
17.5.5-1 Bilinear functional equations
1◦ A binomial bilinear functional equation has the form
f1(x)g1(y) + f2(x)g2(y) =0, (17.5.5.1)
where f n = f n (x) and g n = g n (y) (n =1,2) are unknown functions of different arguments
In this section, it is assumed that f n0, g n0
Separating the variables in (17.5.5.1), we find the solution:
f1= Af2, g2= –Ag1, (17.5.5.2)
where A is an arbitrary constant The functions on the right-hand sides in (17.5.5.2) are
assumed arbitrary
2◦ The trinomial bilinear functional equation
f1(x)g1(y) + f2(x)g2(y) + f3(x)g3(y) =0, (17.5.5.3)
where f n = f n (x) and g n = g n (y) (n =1,2,3) are unknown functions, has two solutions:
f1 = A1f3, f2= A2f3, g3= –A1g1– A2g2;
g1 = A1g3, g2 = A2g3, f3= –A1f1– A2f2, (17.5.5.4)
where A1, and A2 are arbitrary constants The functions on the right-hand sides of the equations in (17.5.5.4) are assumed arbitrary
3◦ The quadrinomial functional equation
f1(x)g1(y) + f2(x)g2(y) + f3(x)g3(y) + f4(x)g4(y) =0, (17.5.5.5)
where all f i are functions of the same argument and all g iare functions of another argument, has a solution
f1= A1f3+ A2f4, f2= A3f3+ A4f4,
g3= –A1g1– A3g2, g4= –A2g1– A4g2 (17.5.5.6)
depending on four arbitrary constants A1, , A4 The functions on the right-hand sides
of the solutions in (17.5.5.6) are assumed arbitrary
Equation (17.5.5.5) has two other solutions:
f1 = A1f4, f2= A2f4, f3= A3f4, g4= –A1g1– A2g2– A3g3;
g1 = A1g4, g2= A2g4, g3 = A3g4, f4= –A1f1– A2f2– A3f3 (17.5.5.7) involving three arbitrary constants
4◦ Consider a bilinear functional equation of the general form
f1(x)g1(y) + f2(x)g2(y) + · · · + f k (x)g k (y) =0, (17.5.5.8)
where f n = f n (x) and g n = g n (y) are unknown functions (n =1, , k).
Trang 5It can be shown that the bilinear functional equation (17.5.5.8) has k –1 different solutions:
f i (x) = A i,1f m+1(x) + A i,2f m+2(x) + · · · + A i,k–m f k (x), i=1, , m;
g m+j (y) = –A1 ,j g1(y) – A2,j g2(y) – · · · – A m,j g m (y), j =1, , k – m;
m=1,2, , k –1,
(17.5.5.9)
where the A i,j are arbitrary constants The functions f m+1(x), , f k (x), g1(y), , g m (y)
on the right-hand sides of solutions (17.5.5.9) can be chosen arbitrarily It is obvious that
for a fixed m, solution (17.5.5.9) contains m(k – m) arbitrary constants.
For a fixed m, solution (17.5.5.9) contains m(k – m) arbitrary constants A i,j Given k,
the solutions having the maximum number of arbitrary constants are defined by
Solution number Number of arbitrary constants Conditions on k
m= 12(k 1) 14(k2–1) if k is odd.
Remark 1. Formulas (17.5.5.9) imply that equation (17.5.5.8) may hold only if the functions f n (and g n) are linearly dependent.
Remark 2 The bilinear functional equation (17.5.5.8) and its solutions (17.5.5.9) play an important role
in the methods of generalized and functional separation of variables for nonlinear PDEs (see Section 15.5).
17.5.5-2 Functional-differential equations reducible to a bilinear equation
Consider a nonlinear functional-differential equation of the form
f1(x)g1(y) + f2(x)g2(y) + · · · + f k (x)g k (y) =0, (17.5.5.10)
where f i (x) and g i (x) are given function of the form
f i (x)≡F i x , ϕ1, ϕ 1, ϕ 1, , ϕ n , ϕ n , ϕ n
, ϕ p = ϕ p (x);
g i (y)≡G i y , ψ1, ψ 1, ψ1 , , ψ m , ψ m , ψ m
, ψ q = ψ q (y).
(17.5.5.11)
The problem is to find the functions ϕ p = ϕ p (x) and ψ q = ψ q (y) depending on different
variables Here, for simplicity, we consider an equation that contains only second-order derivatives; in the general case, the right-hand sides of (17.5.5.11) may contain higher-order
derivatives of ϕ p = ϕ p (x) and ψ q = ψ q (y).
The functional-differential equation (17.5.5.10)–(17.5.5.11) is solved by the method of splitting On the first stage, we treat (17.5.5.10) as a purely functional equation that depends
on two variables x and y, where f i = f i (x) and g i = g i (y) are unknown quantities The
solutions of this equation are described by (17.5.5.9) On the second stage, we successively
substitute the functions f i (x) and g i (y) from (17.5.5.11) into all solutions (17.5.5.9) to obtain systems of ordinary differential equations for the unknown functions ϕ p (x) and ψ q (y).
Solving these systems, we get solutions of the functional-differential equation (17.5.5.10)– (17.5.5.11)
Remark The method of splitting will be used in Paragraph 17.5.5-3 for the construction of solutions of some nonlinear functional equations.
Trang 617.5.5-3 Nonlinear equations containing the complex argument z = ϕ(x) + ψ(t).
Here, we discuss some nonlinear functional equations with two variables Such equations often arise when the method of functional separation of variables is used for finding solutions
of nonlinear equations of mathematical physics
1◦ Consider a functional equation of the form
f (t) + g(x) + h(x)Q(z) + R(z) =0, where z = ϕ(x) + ψ(t). (17.5.5.12)
Here, one of the two functions f (t) and ψ(t) is prescribed and the other is assumed unknown; also one of the functions g(x) and ϕ(x) is prescribed and the other is unknown, and the functions h(x), Q(z), and R(z) are assumed unknown.*
Differentiating equation (17.5.5.12) with respect to x, we obtain the two-argument
equation
g
x + h x Q + hϕ x Q z + ϕ x R z =0 (17.5.5.13) Such equations were discussed in Paragraph 17.5.5-2; their solutions are found with the help
of (17.5.5.6) and (17.5.5.7) Hence, we obtain the following system of ordinary differential equations [see formulas (17.5.5.6)]:
g
x = A1hϕ
x + A2ϕ
x,
h
x = A3hϕ
x + A4ϕ
x,
Q
z = –A1– A3Q,
R
z = –A2– A4Q,
(17.5.5.14)
where A1, , A4are arbitrary constants Integrating the system of ODEs (17.5.5.14) and substituting the resulting solutions into the original functional equation, one obtains the following results
Case 1 If A3 = 0 in (17.5.5.14), then the corresponding solution of the functional equation is given by
f = –12A1A4ψ2+ (A
1B1+ A2+ A4B3)ψ – B2– B1B3– B4,
g= 12A1A4ϕ2+ (A
1B1+ A2)ϕ + B2,
h = A4ϕ + B1,
Q = –A1z + B3,
R= 12A1A4z2– (A
2+ A4B3)z + B4,
(17.5.5.15)
where A k and B k are arbitrary constants and ϕ = ϕ(x) and ψ = ψ(t) are arbitrary functions Case 2 If A3 ≠ 0 in (17.5.5.14), then the corresponding solution of the functional
* In similar equations with a composite argument, it is assumed that ϕ(x) const and ψ(y) const.
Trang 7equation is
f = –B1B3e–A3ψ+
A2– A1A4
A3
ψ – B2– B4– A1A4
A2 3 ,
g= A1B1
A3 e
A3ϕ+
A2– A A1A4
3
ϕ + B2,
h = B1e A3ϕ– A4
A3,
Q = B3e–A3z– A1
A3,
R= A4B3
A3 e
–A3z+
A1A4
A3 – A2
z + B4,
(17.5.5.16)
where A k and B k are arbitrary constants and ϕ = ϕ(x) and ψ = ψ(t) are arbitrary functions Case 3 In addition, the functional equation has two degenerate solutions [formulas
(17.5.5.7) are used]:
f = A1ψ + B1, g = A1ϕ + B2, h = A2, R = –A1z – A2Q – B1– B2, (17.5.5.17)
where ϕ = ϕ(x), ψ = ψ(t), and Q = Q(z) are arbitrary functions, A1, A2, B1, and B2 are arbitrary constants, and
f = A1ψ + B1, g = A1ϕ + A2h + B2, Q = –A2, R = –A1z – B1– B2, (17.5.5.18)
where ϕ = ϕ(x), ψ = ψ(t), and h = h(x) are arbitrary functions, A1, A2, B1, and B2 are arbitrary constants
2◦ Consider a functional equation of the form
f (t) + g(x)Q(z) + h(x)R(z) =0, where z = ϕ(x) + ψ(t). (17.5.5.19)
Differentiating (17.5.5.19) in x, we get the two-argument functional-differential
equa-tion
g
x Q + gϕ x Q z + h x R + hϕ x R z=0, (17.5.5.20) which coincides with equation (17.5.5.5), up to notation
Nondegenerate case Equation (17.5.5.20) can be solved with the help of
formu-las (17.5.5.6)–(17.5.5.7) In this way, we arrive at the system of ordinary differential equations
g
x = (A1g + A2h )ϕ x,
h
x = (A3g + A4h )ϕ x,
Q
z = –A1Q – A3R,
R
z = –A2Q – A4R,
(17.5.5.21)
where A1, , A4are arbitrary constants
The solution of equation (17.5.5.21) is given by
g (x) = A2B1e k1ϕ + A2B2e k2ϕ,
h (x) = (k1– A1)B1e k1ϕ + (k2– A1)B2e k2ϕ,
Q (z) = A3B3e–k1z + A3B4e–k2z,
R (z) = (k1– A1)B3e–k1z + (k
2– A1)B4e–k2z,
(17.5.5.22)