The solutions of these equations are expressed in terms of elliptic functions.. Below, we discuss several types of three-argument functional equations of the form 15.6.4.2 that arise mos
Trang 1Let us dwell on the case (15.6.3.22b) According to (15.6.3.23),
h = A1(z + C1)2+ A2
where A1= –C3/3and A2= –C2are any numbers Substituting (15.6.3.22b) and (15.6.3.25) into (15.6.3.24) yields
w = B1ln|z + C1|+ B2, F = A1 B1(z + C1) + A2B1
(z + C1)2.
Eliminating z, we arrive at the explicit form of the right-hand side of equation (15.6.3.16):
F(w) = A1 B1e u + A2B1e–2u, where u= w – B2
For simplicity, we set C1=0, B1=1, and B2=0and denote A1= a and A2= b Thus, we have
w (z) = ln|z|, F(w) = ae w + be–2w, g (z) = –1 /z, h (z) = az2+ b/z. (15.6.3.27)
It remains to determine ψ(t) and ϕ(x) We substitute (15.6.3.27) into the functional differential
equa-tion (15.6.3.18) Taking into account (15.6.3.17), we find
[ψ tt – (ψ t)2– aψ3– b] – [ϕ xx ϕ – (ϕ x)2+ aϕ3] + (ψ tt –3aψ2)ϕ – ψ(ϕ xx+3aϕ2) =0 (15.6.3.28)
Differentiating (15.6.3.28) with respect to t and x yields the separable equation*
(ψ ttt–6aψψ t )ϕ x – (ϕ xxx+6aϕϕ x )ψ t=0, whose solution is determined by the ordinary differential equations
ψ ttt–6aψψ t = Aψ t,
ϕ xxx+6aϕϕ x = Aϕ x,
where A is the separation constant Each equation can be integrated twice, thus resulting in
(ψ t)2=2aψ3+ Aψ2+ C1ψ+ C2,
(ϕ x)2= –2aϕ3+ Aϕ2+ C3ϕ + C4, (15.6.3.29)
where C1, , C4are arbitrary constants Eliminating the derivatives from (15.6.3.28) using (15.6.3.29), we
find that the arbitrary constants are related by C3= –C1and C4= C2+ b So, the functions ψ(t) and ϕ(x) are
determined by the first-order nonlinear autonomous equations
(ψ t )2=2aψ3+ Aψ2+ C1ψ+ C2,
(ϕ x)2= –2aϕ3+ Aϕ2– C1ϕ + C2+ b.
The solutions of these equations are expressed in terms of elliptic functions
For the other cases in (15.6.3.22), the analysis is performed in a similar way Table 15.6 presents the final results for the cases (15.6.3.22a)–(15.6.3.22e)
Case 2 Integrating the third and fourth equations in (15.6.3.21) yields
B – C t + D2 if A=0;
4A (At + D1)2– B
A , ϕ = – 1
4A (Ax + D2)2+B – C
where D1and D2are arbitrary constants In both cases, the functionF(w) in equation (15.6.3.16) is arbitrary.
The first row in (15.6.3.30) corresponds to the traveling-wave solution w = w(kx + λt) The second row leads
to a solution of the form w = w(x2– t2)
* To solve equation (15.6.3.28), one can use the solution of functional equation (15.5.4.4) [see (15.5.4.5)]
Trang 2TABLE 15.6
Nonlinear Klein–Gordon equations ∂ tt w – ∂ xx w=F(w) admitting functional separable solutions of the form
w = w(z), z = ϕ(x) + ψ(t) Notation: A, C1, and C2are arbitrary constants; σ =1for z >0and σ = –1 for z <0
No Right-hand sideF(w) Solution w(z) Equations for ψ(t) and ϕ(x)
1 aw ln w + bw e z (ψ t)2= C1e–2ψ + aψ –12a + b + A,
(ϕ x)2= C2e–2ϕ – aϕ +12a + A
2=2aψ3+ Aψ2+ C1ψ+ C2,
(ϕ x)2= –2aϕ3+ Aϕ2– C1ϕ + C2+ b
3 a sin w + b
sin w ln tan w
4 +2sin
w
4
4arctan e z (ψ
t)2= C1e2ψ + C2e–2ψ + bψ + a + A, (ϕ x)2= –C2e2ϕ – C1e–2ϕ – bϕ + A
4 a sinh w + b
sinh w ln tanh w
4 +2sinh
w
2
2lncothz
2 (ψ t)2= C1e2ψ + C2e–2ψ – σbψ + a + A, (ϕ x)2= C2e2ϕ + C1e–2ϕ + σbϕ + A
5 a sinh w +2b
sinh w arctan e w/2+ cosh w
2
2lntanz
2 (ψ t)2= C1sin2ψ + C2cos2ψ + σbψ + a + A, (ϕ x)2= –C1sin2ϕ + C2cos2ϕ – σbϕ + A
15.6.4 Splitting Method Solutions of Some Nonlinear Functional
Equations and Their Applications
15.6.4-1 Three-argument functional equations of special form Splitting method The substitution of the expression
w = F (z), z = ϕ(x) + ψ(t) (15 6 4 1 ) into a nonlinear partial differential equation sometimes leads to functional differential equations of the form
f (t) + Φ1(x) Ψ1(z) + · · · + Φk(x) Ψk(z) = 0 , (15 6 4 2 ) where Φj(x) and Ψj(z) are functionals dependent on the variables x and z, respectively,
Φj(x) ≡ Φj x , ϕ, ϕ x, , ϕ(x n)
,
Ψj(z) ≡ Ψj F , Fz , , Fz(n)
It is reasonable to solve equation (15.6.4.2) by the splitting method At the first stage,
we treat (15.6.4.2) as a purely functional equation, thus disregarding (15.6.4.3)
Differenti-ating (15.6.4.2) with respect to x yields the standard bilinear functional differential equation
in two independent variables x and z:
Φ1(x) Ψ1(z)+ · · ·+Φ k(x) Ψk(z)+ Φ1(x)ϕ(x) Ψ1(z)+ · · ·+ϕ(x) Φk(x) Ψ k(z) = 0 , (15 6 4 4 ) which can be solved using the results of Subsections 15.5.3–15.5.5 Then, substituting the solutions Φm(x) and Ψm(z) into (15.6.4.2) and taking into account the second relation in (15.6.4.1), we find the function f (t) Further, substituting the functionals (15.6.4.3) into the
solutions of the functional equation (15.6.4.2), we obtain determining systems of ordinary
differential equations for F (z), ϕ(x), and ψ(t).
Trang 3Below, we discuss several types of three-argument functional equations of the form (15.6.4.2) that arise most frequently in the functional separation of variables in nonlinear equations of mathematical physics The results are used for constructing exact solutions for some classes of nonlinear heat and wave equations.
15.6.4-2 Functional equation f (t) + g(x) = Q(z), with z = ϕ(x) + ψ(t).
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 ψ(x) is prescribed and the other is unknown, and the function Q(z) is assumed unknown.*
Differentiating the equation with respect to x and t yields Q zz = 0 Consequently, the solution is given by
f (t) = Aψ(x) + B, g (x) = Aϕ(x) – B + C, Q (z) = Az + C, (15 6 4 5 )
where A, B, and C are arbitrary constants.
15.6.4-3 Functional equation f (t) + g(x) + h(x)Q(z) + R(z) = 0 , with z = ϕ(x) + ψ(t) Differentiating the equation with respect to x yields the two-argument equation
g
x+ h xQ + hϕ xQ z+ ϕ xR z = 0 (15 6 4 6 ) Such equations were discussed in Subsections 15.5.3 and 15.5.4 Hence, the following relations hold [see formulas (15.5.4.4) and (15.5.4.5)]:
g
x = A1hϕ
x+ A2ϕ
x,
h
x = A3hϕ
x+ A4ϕ
x,
Q
z = –A1– A3Q ,
R
z = –A2– A4Q ,
(15 6 4 7 )
where A1, , A4are arbitrary constants By integrating system (15.6.4.7) and substituting the resulting solutions into the original functional equation, one obtains the results given below.
Case 1 If A3= 0 in (15.6.4.7), 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,
(15 6 4 8 )
where Akand Bkare arbitrary constants and ϕ = ϕ(x) and ψ = ψ(t) are arbitrary functions.
* In similar equations with a composite argument, it is assumed that ϕ(x) const and ψ(t) const
Trang 4Case 2 If A3≠ 0 in (15.6.4.7), the corresponding solution of the functional equation is
f = –B1B3e–A3ψ+
A2– A A1A4
3
ψ – B2– B4– A1A4
A2 3
,
g = A1B1
A3 e
A3ϕ+
A2– A1A4
A3
ϕ + B2,
h = B1eA3ϕ– A4
A3,
Q = B3e–A3z– A1
A3,
R = A4B3
A3 e
–A3z+
A1A4
A3 – A2
z + B4,
(15 6 4 9 )
where Akand Bkare arbitrary constants and ϕ = ϕ(x) and ψ = ψ(t) are arbitrary functions.
Case 3 In addition, the functional equation has two degenerate solutions:
f = A1ψ + B1, g = A1ϕ + B2, h = A2, R = –A1z – A2Q – B1– B2, (15.6.4.10a)
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, (15.6.4.10b)
where ϕ = ϕ(x), ψ = ψ(t), and h = h(x) are arbitrary functions; and A1, A2, B1, and B2are arbitrary constants The degenerate solutions (15.6.4.10a) and (15.6.4.10b) can be obtained directly from the original equation or its consequence (15.6.4.6) using formulas (15.5.4.6).
Example 1 Consider the nonstationary heat equation with a nonlinear source
∂w
2w
We look for exact solutions of the form
Substituting (15.6.4.12) into (15.6.4.11) and dividing by w z yields the functional differential equation
ψ t = ϕ xx + (ϕ x)2w
zz
w z +F(w(z))
w z Let us solve it by the splitting method To this end, we represent this equation as the functional equation
f (t) + g(x) + h(x)Q(z) + R(z) =0, where
f (t) = –ψ t , g (x) = ϕ xx, h (x) = (ϕ x)2, Q (z) = w zz /w z , R (z) = f (w(z))/w z (15.6.4.13)
On substituting the expressions of g and h of (15.6.4.13) into (15.6.4.8)–(15.6.4.10), we arrive at overde-termined systems of equations for ϕ = ϕ(x).
Case 1 The system
ϕ xx= 12A1A4ϕ2+ (A1B1+ A2)ϕ + B2,
(ϕ x)2= A4ϕ+ B1
following from (15.6.4.8) and corresponding to A3=0in (15.6.4.7) is consistent in the cases
ϕ = C1x+ C2 for A2= –A1C2, A4= B2=0, B1= C2,
ϕ=14A4x2+ C1x+ C2 for A1= A2=0, B1= C2– A4C2, B2 = 12A4, (15.6.4.14)
where C1and C2are arbitrary constants
Trang 5The first solution in (15.6.4.14) with A1≠ 0leads to a right-hand side of equation (15.6.4.11) containing the inverse of the error function [the form of the right-hand side is identified from the last two relations in (15.6.4.8) and (15.6.4.13)] The second solution in (15.6.4.14) corresponds to the right-hand side ofF(w) = k1 w ln w+k2w
in (15.6.4.11) In both cases, the first relation in (15.6.4.8) is, taking into account that f = –ψ t , a first-order linear solution with constant coefficients, whose solution is an exponential plus a constant
Case 2 The system
ϕ xx= A1B1
A3ϕ+
A2–A1A4
A3
ϕ + B2,
(ϕ x)2= B1e A3ϕ– A4
A3,
following from (15.6.4.9) and corresponding to A3≠ 0in (15.6.4.7), is consistent in the following cases:
–A4/A3x + C1 for A2= A1A4/A3, B1= B2=0,
A3 ln|x|+ C1 for A1= 12A2, A2= A4= B2=0, B1=4A–2e–A3C1,
A3 lncos 12√
A3A4 x + C1 + C2 for A1= 1
2A2, A2= 12A3A4 , B2=0, A3A4>0,
A3 lnsinh 1
2
√
–A3A4x + C1 + C2 for A1= 1
2A2, A2= 12A3A4, B2=0, A3A4<0,
A3 lncosh 12√
–A3A4x + C1 + C2 for A1= 12A2, A2= 12A3A4, B2=0, A3A4<0,
where C1and C2are arbitrary constants The right-hand sides of equation (15.6.4.11) corresponding to these solutions are represented in parametric form
Case 3 Traveling-wave solutions of the nonlinear heat equation (15.6.4.11) and solutions of the linear
equa-tion (15.6.4.11) withF
w= const correspond to the degenerate solutions of the functional equation (15.6.4.10)
15.6.4-4 Functional equation f (t) + g(x)Q(z) + h(x)R(z) = 0 , with z = ϕ(x) + ψ(t) Differentiating with respect to x yields the two-argument functional differential equation
g
xQ + gϕ xQ z+ h xR + hϕ xR z= 0 , (15 6 4 15 ) which coincides with equation (15.5.4.4), up to notation.
Nondegenerate case Equation (15.6.4.15) can be solved using formulas (15.5.4.5) 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 ,
(15 6 4 16 )
where A1, , A4are arbitrary constants.
The solution of equation (15.6.4.16) is given by
g (x) = A2B1ek1ϕ+ A
2B2ek2ϕ,
h (x) = (k1– A1)B1ek1ϕ+ (k
2– A1)B2ek2ϕ,
Q (z) = A3B3e–k1z+ A
3B4e–k2z,
R (z) = (k1– A1)B3e–k1z+ (k
2– A1)B4e–k2z,
(15 6 4 17 )
where B1, , B4are arbitrary constants and k1and k2are roots of the quadratic equation
(k – A1)(k – A4) – A2A3= 0 (15 6 4 18 )
Trang 6In the degenerate case k1 = k2, the terms ek2ϕ and e–k2z in (15.6.4.17) must be replaced
by ϕek1ϕ and ze–k1z, respectively In the case of purely imaginary or complex roots, one
should extract the real (or imaginary) part of the roots in solution (15.6.4.17).
On substituting (15.6.4.17) into the original functional equation, one obtains conditions
that must be met by the free coefficients and identifies the function f (t), specifically,
B2= B4= 0 =⇒ f(t) = [A2A3+ (k1– A1)2]B1B3e–k1ψ,
B1= B3= 0 =⇒ f(t) = [A2A3+ (k2– A1)2]B2B4e–k2ψ,
A1= 0 =⇒ f(t) = (A2A3+ k12)B1B3e–k1ψ+ (A
2A3+ k22)B2B4e–k2ψ.
(15 6 4 19 )
Solution (15.6.4.17), (15.6.4.19) involves arbitrary functions ϕ = ϕ(x) and ψ = ψ(t).
Degenerate case In addition, the functional equation has two degenerate solutions,
f = B1B2eA1ψ, g = A
2B1e–A1ϕ, h = B
1e–A1ϕ, R = –B
2eA1z– A
2Q ,
where ϕ = ϕ(x), ψ = ψ(t), and Q = Q(z) are arbitrary functions; A1, A2, B1, and B2 are arbitrary constants; and
f = B1B2eA1ψ, h = –B1e–A1ϕ– A2g , Q = A2B2eA1z, R = B2eA1z,
where ϕ = ϕ(x), ψ = ψ(t), and g = g(x) are arbitrary functions; and A1, A2, B1, and B2
are arbitrary constants The degenerate solutions can be obtained immediately from the original equation or its consequence (15.6.4.15) using formulas (15.5.4.6).
Example 2 For the nonlinear heat equation (15.6.3.1), searching for exact solutions in the form w = w(z),
with z = ϕ(x) + ψ(t), leads to the functional equation (15.6.3.3), which coincides with the equation
f (t) + g(x)Q(z) + h(x)R(z) =0if
f (t) = –ψ t, g (x) = ϕ xx, h (x) = (ϕ x)2, Q (z) = F(w), R(z) = [F(w)w z] z
w z , w = w(z).
15.7 Direct Method of Symmetry Reductions of
Nonlinear Equations
15.7.1 Clarkson–Kruskal Direct Method
15.7.1-1 Simplified scheme Examples of constructing exact solutions.
The basic idea of the simplified scheme is as follows: for an equation with the unknown
function w = w(x, t), an exact solution is sought in the form
w = f (t)u(z) + g(x, t), z = ϕ(t)x + ψ(t). (15 7 1 1 )
The functions f (t), g(x, t), ϕ(t), and ψ(t) are found in the subsequent analysis and are chosen in such a way that, ultimately, the function u(z) would satisfy a single ordinary
differential equation.
Below we consider some cases in which it is possible to construct exact solutions of nonlinear equations of the form (15.7.1.1).
Trang 7Example 1 Consider the generalized Burgers–Korteweg–de Vries equation
∂w
We seek its exact solution in the form (15.7.1.1) Inserting (15.7.1.1) into (15.7.1.2), we obtain
af ϕ n u(z n) + bf2ϕuu z + f (bgϕ – ϕ t x – ψ t )u z + (bf g x – f t )u + ag x(n) + bgg x – g t=0 (15.7.1.3)
Equating the functional coefficients of u(z n) and uu zin (15.7.1.3), we get
Further, equating the coefficient of u zto zero, we obtain
Inserting the expressions (15.7.1.4) and (15.7.1.5) into (15.7.1.3), we arrive at the relation
ϕ2n–1(au(z n) + buu z) + (2– n)ϕ n–2ϕ t u+ 1
bϕ2
(2ϕ2– ϕϕ tt )x +2ϕ t ψ t – ϕψ tt
=0
Dividing each term by ϕ2n–1and then eliminating x with the help of the relation x = (z – ψ)/ϕ, we obtain
au(z n) + buu z+ (2– n)ϕ–n–1ϕ t u+1
b ϕ
– 2n–2(2ϕ2– ϕϕ tt )z + 1
b ϕ
– 2n–2(ϕψϕ tt – ϕ2ψ +2ϕϕ t ψ t–2ψϕ2) =0
(15.7.1.6)
Let us require that the functional coefficient of u and the last term be constant,
ϕ–n–1ϕ t = –A, ϕ–2n–2(ϕψϕ tt – ϕ2ψ +2ϕϕ t ψ t–2ψϕ2t ) = B, where A and B are arbitrary As a result, we arrive at the following system of ordinary differential equations for ϕ and ψ:
ϕ t = –Aϕ n+1,
ψ +2Aϕ n ψ t + A2(1– n)ϕ2n ψ = –Bϕ2n (15.7.1.7)
Using (15.7.1.6) and (15.7.1.7), we obtain an equation for u(z),
au(z n) + buu z + A(n –2)u + A2
b (1– n)z + B
For A≠ 0, the general solution of equations (15.7.1.7) has the form
ϕ (t) = (Ant + C1)–n1,
ψ (t) = C2(Ant + C1)n– n1 + C3(Ant + C1)–n1 + B
A2(n –1),
(15.7.1.9)
where C1, C2, and C3are arbitrary constants
Formulas (15.7.1.1), (15.7.1.4), (15.7.1.5), and (15.7.1.9), together with equation (15.7.1.8), describe an exact solution of the generalized Burgers–Korteweg–de Vries equation (15.7.1.2)
Example 2 Consider the Boussinesq equation
∂2w
∂t2 + ∂
∂x
∂x
+ a ∂
4w
Just as in Example 1, we seek its solutions in the form (15.7.1.1), where the functions f (t), g(x, t), ϕ(t), and
ψ (t) are found in the subsequent analysis Substituting (15.7.1.1) into (15.7.1.10) yields
af ϕ4u + f2ϕ2uu + f (z2+ gϕ2)u + f2ϕ2(u )2+ (f z tt+2f g x ϕ+2f t z t )u
+ (f g xx + f tt )u + g tt + gg xx + g x2+ ag x(4)=0 (15.7.1.11)
Equating the functional coefficients of u and uu , we get