If follows from the first equation of 15.11.2.5 that Substituting this into the second and third equations of 15.11.2.5, after some rearrangements we obtain ξ t + w1ξ – νξ xx= 0, It is o
Trang 1752 NONLINEARPARTIALDIFFERENTIALEQUATIONS
where it has been taken into account that ξ x 0.
If follows from the first equation of (15.11.2.5) that
Substituting this into the second and third equations of (15.11.2.5), after some rearrangements we obtain
ξ t + w1ξ – νξ xx= 0,
It is obvious that if the first equation of (15.11.2.7) is valid, then the second is satisfied identically The last equation in (15.11.2.5) is the Burgers equation.
Hence, formula (15.11.2.4) in view of (15.11.2.6) can be rewritten as
w= –2ν ∂
where the functions w and w1satisfy the Burgers equation and the function ξ is described by the first equation
of (15.11.2.7) Given a solution w1of the Burgers equation, formula (15.11.2.8) allows obtaining other solutions
of it by solving the first equation in (15.11.2.7), which is linear in ξ.
Taking into account that w1 = 0 is a particular solution of the Burgers equation, let us substitute it into (15.11.2.7) and (15.11.2.8) This results in the Cole–Hopf transformation
w= –2νξ
ξ This transformation allows constructing solutions of the nonlinear Burgers equation (15.11.2.1) via solutions
of the linear heat equation
ξ t = νξ xx.
Example 2 Consider the Korteweg–de Vries equation
∂w
∂t + w ∂w
∂x + ∂
3w
First step Let us substitute the leading term of the expansion (15.11.1.1) into equation (15.11.2.9) and
then multiply the resulting relation by ξ p+3(the product ξ p+3w xxxgives a zeroth order quantity) to obtain
w0 ξ3+ pw0x 0ξ2– pw2ξ2–p – p(p +1)(p + 2)w 0 = 0,
where ξ = x – x0, x0= x0(t), and w0= w0(t) From the balance of the highest-order terms (only the last two
terms are taken into account) it follows that
p= 2, w0= –12 (m =0).
Since p is a positive integer, the first necessary condition of the Painlev´e test is satisfied.
Second step To fine the Fuchs indices (resonances), we substitute the binomial
w= –12ξ – 2+ w m ξ m–2
into the leading terms of equation (15.11.2.9), where the second and the third term are taken into account.
Isolating the term proportional to w m, we have
(m +1)(m – 4)(m – 6)wm ξ m–5+· · · =0.
Equating (m +1)(m – 4)(m – 6) to zero gives the Fuchs indices
m1= 4, m2= 6.
Since they are both positive integers, the second necessary condition of the Painlev´e test is satisfied.
Third step We substitute the expansion (15.11.1.1), while considering, according to the second step, the
terms up to number m =6 inclusive,
w= –12ξ – 2+ w1ξ–1+ w2+ w3ξ + w4ξ2+ w5ξ3+ w6ξ4+· · · , ξ = x – x0(t) (15.11.2.10)
Trang 2into equation (15.11.2.9) Then we collect the terms of equal powers in ξ and equate the coefficients of the different powers of ξ to zero to arrive at a system of equations for the w m:
ξ–4: 2w 1 = 0,
ξ–3: 24w 2 – 24x
0– w2 = 0,
ξ–2: 12w 3+ w1x 0– w1w2= 0,
ξ–1: 0 ×w4+ w1= 0, 1: –6w 5+ w 2– w3x 0+ w1w4+ w2w3 = 0,
ξ: 0 ×w6+ w3– 2w 4x 0+ w2+ 2w 1w5+ 2w 2w4= 0.
(15.11.2.11)
Simple computations show that the equations with resonances corresponding to the powers ξ– 1 and ξ are
satisfied identically Hence, the Korteweg–de Vries equation (15.11.2.9) passes the Painlev´e test.
The solution of equation (15.11.2.11) results in the following expansion coefficients in (15.11.2.10):
w1= 0, w 2= x 0(t), w3 = 0, w 4= w4(t), w5 = 16x 0(t), w6= w6(t), where x0(t), w4(t), and w6(t) are arbitrary functions.
Truncated series expansion and the B¨acklund transformation For further analysis, let us use a truncated
expansion of the general form (15.11.1.6) with p =2:
w= w0
ξ2 +w1
Substituting (15.11.2.12) into (15.11.2.9) and equating the functional coefficients of the different powers of ξ
to zero, in the same way as in Example 2, we arrive at the B¨acklund transformation
w= 12(ln ξ)xx + w2 ,
ξ t ξ + w2ξ x2+ 4ξ ξ xxx– 3ξ 2
xx= 0,
ξ xt + w2ξ xx + ξ xxx= 0,
(w2)t + w2(w2)x + (w2)xxx= 0.
(15.11.2.13)
It relates the solutions w and w2of the Korteweg–de Vries equations Eliminating w2from the second and
third equations in (15.11.2.13), one can derive an equation for ξ, which can further be reduced, via suitable
transformations, to a system of linear equations.
15.11.3 Construction of Solutions of Nonlinear Equations That Fail
the Painlev ´e Test, Using Truncated Expansions
In some cases truncated expansions of the form (15.11.1.6) may be effective in finding exact solutions to nonlinear equations of mathematical physics that fail the Painlev´e test
In such cases, the expansion parameter p must be a positive integer; it is determined in the
same way as at the first step of performing the Painlev´e test We illustrate this by a specific example below
Example Consider the nonlinear diffusion equation with a cubic source
First step Let us substitute the leading term of the expansion (15.11.1.1) into equation (15.11.3.1) and
then multiply the resulting relation by ξ p+2to obtain
w 0ξ2+ pw0x 0ξ = p(p +1)w 0 – 2w 3ξ2 – 2p,
where ξ = x – x0, x0= x0(t), and w0= w0(t) From the balance of the highest-order terms (both terms on the
right-hand side are taken into account) it follows that
Since p is a positive integer, the first necessary condition of the Painlev´e test is satisfied.
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Second step The equation is invariant under the substitution of –w for w Hence, it suffices to consider
only the positive value of w0 in (15.11.3.2) Therefore, in order to find resonances, we substitute the binomial
w = ξ–1+ w m ξ m–1
in the leading terms w xx and bw3of equation (15.11.3.1) Collecting the terms proportional to w m, we get
(m +1)(m – 4)wm ξ m–3+· · · =0.
Equating (m +1)(m – 4) to zero gives the Fuchs index m 1 = 4 Since it is integer and positive, the second necessary condition of the Painlev´e test is satisfied.
Third step We substitute the expansion (15.11.1.1) into equation (15.11.3.1); according to the second
step, the terms up to number m =4 inclusive must be taken into account It can be shown that the consistency condition (15.11.1.5) is not satisfied, and therefore the equation in question fails the Painlev´e test.
Using a truncated expansion for finding exact solutions For further analysis, we use a truncated expansion
of the general form (15.11.1.6) with p =1, which from the first step This results in formula (15.11.2.4).
Substituting (15.11.2.4) into the diffusion equation (15.11.3.1) and collecting the terms of equal powers in ξ,
we obtain
ξ–3 2w 0ξ x2– 2w 3
+ ξ–2
w0ξ t– 2(w 0 )x x – w0ξ xx– 6w 2w
1
+ ξ–1
–(w0 )t + (w0 )xx– 6w 0w2
– (w1 )t + (w1 )xx– 2w 3 = 0.
Equating the coefficients of like powers of ξ to zero, we arrive at the system of equations
w0(ξ x2– w2) = 0,
w0ξ t– 2(w 0 )x x – w0ξ xx– 6w 2w
1 = 0,
–(w0)t + (w0)xx– 6w 0w2= 0,
(w1)t – (w1)xx+ 2w 3 = 0.
(15.11.3.3)
From the first equation in (15.11.3.3) we have
The other solution differs in sign only and gives rise to the same result, and therefore is not considered Substituting (15.11.3.4) into the second and third equations of (15.11.3.3) and canceling by nonzero factors,
we obtain
ξ t– 3ξxx– 6w1ξ = 0,
The latter equation in (15.11.3.3), which coincides with the original equation (15.11.3.1), is satisfied if
On inserting (15.11.3.4) and (15.11.3.6) in (15.11.2.4), we get the following representation for solutions:
w= ξ
where the function ξ is determined by an overdetermined linear system of equations resulting from the
substi-tution of (15.11.3.6) into (15.11.3.5):
ξ t– 3ξxx= 0,
Differentiate the first equation with respect to x and then eliminate the mixed derivative w xtusing the second
equation to obtain ξ xxx= 0 It follows that
ξ = ϕ2(t)x2+ ϕ1(t)x + ϕ0(t). (15.11.3.9)
In order to determine the functions ϕ k (t), let us substitute (15.11.3.9) into equations (15.11.3.8) to obtain
ϕ 2x2+ ϕ 1x + ϕ 0– 6ϕ 2 = 0,
–ϕ 2x – ϕ 1= 0.
Equating the functional coefficients of the different powers of x to zero and integrating the resulting equations,
we get
ϕ2= C2 , ϕ1= C1 , ϕ0 = 6C 2t + C0 , (15.11.3.10)
where C0, C1, and C2are arbitrary constants Substituting (15.11.3.9) into (15.11.3.7) and taking into account (15.11.3.10), we find an exact solution of equation (15.11.3.1) in the form
w= 2C 2x + C1
C2x2+ C1x+ 6C 2t + C0
Trang 415.12 Methods of the Inverse Scattering Problem
(Soliton Theory)
Preliminary remarks The methods of the inverse scattering problem rely on “implicit”
linearization of equations Main idea: Instead of the original nonlinear equation in the
unknown w, one considers an auxiliary overdetermined linear system of equation for a
(vector) function ϕ, with the coefficients of this system generally dependent on w and the derivatives of w with respect to the independent variables The linear system for ϕ
is chosen so that the compatibility condition for its equations gives rise to the original
nonlinear equation for w.
15.12.1 Method Based on Using Lax Pairs
15.12.1-1 Method description Consistency condition Lax pairs
We will be studying a nonlinear evolution equation of the form
∂w
where the right-hand side F(w) depends on w and its derivatives with respect to x.
Consider two auxiliary linear differential equations, one corresponding to an eigenvalue
problem and involving derivatives with respect to the space variable x only,
and the other describing the evolution of the eigenfunction in time,
∂ϕ
The coefficients of the linear differential operators L and M in equations (15.12.1.2)
and (15.12.1.3) depend on w and its derivatives with respect to x.
Since system (15.12.1.2)–(15.12.1.3) is overdetermined (there are two equations for ϕ),
the operators L and M cannot be arbitrary—they must satisfy a compatibility condition In
order to find this condition, let us first differentiate (15.12.1.2) with respect to t Assuming that the eigenvalues λ are independent of time t, we have
Lt ϕ + Lϕ t = λϕ t
Replacing ϕ there by the right-hand side (15.12.1.3), we get
Lt ϕ – LMϕ = –λMϕ.
Taking into account the relations λMϕ = M(λϕ) and λϕ = Lϕ, we arrive at the compatibility
condition
Lt ϕ = LMϕ – MLϕ,
which can be rewritten in the form of an operator equation:
∂L
The linear operators L and M [or the linear equations (15.12.1.2) and (15.12.1.3)] are
said to form a Lax pair for the nonlinear equation (15.12.1.1) if the compatibility condition
(15.12.1.4) coincides with equation (15.12.1.1) The right-hand side of equation (15.12.1.4)
represents the commutator of the operators L and M, which is denoted by [L, M] = LM–ML
for short
Thus, if a suitable Lax pair is found, the analysis of the nonlinear equation (15.12.1.1) can be reduced to that of two simpler, linear equations, (15.12.1.2) and (15.12.1.3)
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Remark The operator M in equations (15.12.1.3) and (15.12.1.3) is defined to an additive function of time; it can be changed according to the rule
M=⇒ M + p(t),
where p(t) is an arbitrary function This function is found in solving a Cauchy problem for equation (15.12.1.1);
see Paragraph 15.12.3-2.
15.12.1-2 Examples of Lax pairs for nonlinear equations of mathematical physics
Example 1 Let us show that a Lax pair for the Korteweg–de Vries equation
∂w
∂t +∂
3w
∂x3 – 6w∂w
is formed by the operators
L = w – ∂
2
∂x2, M = 4 ∂3
∂x3 – 6w ∂
∂x – 3∂w
which generate the linear equations
ϕ xx + (λ – w)ϕ =0,
ϕ t+ 4ϕxxx– 6wϕx– 3wx ϕ + p(t)ϕ =0. (15.12.1.7)
Here, p(t) is an arbitrary function.
It is not difficult to verify that the following formulas hold:
LM(ϕ) = –4ϕxxxxx+ 10wϕxxx+ [15wx – p(t)]ϕ xx+ (12wxx– 6w 2)ϕ x
+ [3wxxx– 3wwx + wp(t)]ϕ, ML(ϕ) = –4ϕxxxxx+ 10wϕxxx+ [15wx – p(t)]ϕ xx+ (12wxx– 6w 2)ϕ x
+ [4wxxx– 9wwx + wp(t)]ϕ, where ϕ(x, t) is an arbitrary function It follows that
From (15.12.1.6) and (15.12.1.8) we obtain
Lt = w t, LM – ML = –w xxx+ 6wwx.
On inserting these expressions into (15.12.1.4), we arrive at the Korteweg–de Vries equation (15.12.1.5) Remark A procedure for solving the Cauchy problem for equation (15.12.1.5) is outlined in Subsec-tion 15.12.3.
The linear equations (15.12.1.2) and (15.12.1.3) for the auxiliary function ϕ, which
form a Lax pair, can be vector; in this case, the linear operators L and M are represented
by matrices In other words, the individual equations (15.12.1.2) and (15.12.1.3) may be replaced by appropriate systems of linear equations
Example 2 The sinh-Gordon equation
w xt= 4asinh w
can be represented as a vector Lax pair
Lϕ = λϕ,
ϕ t = –Mϕ,
where
ϕ1
ϕ2
, L =
0
∂ x+12w x
∂ x–12w x 0
, M = a
λ
0
e w
e–w 0
. The determination of a Lax pair for a given nonlinear equation is a very complex problem that is basically solvable for isolated equations only Therefore, the “implicit” linearization
of equations is usually realized using a simpler method, described in Subsection 15.12.2
Trang 615.12.2 Method Based on a Compatibility Condition for Systems of
Linear Equations
15.12.2-1 General scheme Compatibility condition Systems of two equations Consider two systems of linear equations
where ϕ is an n-dimensional vector and A and B are n×nmatrices The right-hand sides of systems (15.12.2.1) and (15.12.2.2) cannot be arbitrary—they must satisfy a compatibility condition To find this condition, let us differentiate systems (15.12.2.1) and (15.12.2.2)
with respect to t and x, respectively, and eliminate the mixed derivative ϕ xt from the
resulting equations Then replacing the derivatives ϕ x and ϕ tby the right-hand sides of (15.12.2.1) and (15.12.2.2), we obtain
where [A, B] = AB – BA It turns out that, given a matrix A, there is a simple deductive procedure for finding B as a result of which the compatibility condition (15.12.2.3) becomes
a nonlinear evolution equation
Let us dwell on the special case where the vector function ϕ two components, ϕ = ϕ1
ϕ2
We choose a linear system of equations (15.12.2.1) in the form
(ϕ1)x = –iλϕ1+ f ϕ2,
where λ is the spectral parameter, f and g are some (generally complex valued) functions
of two variables x and t, and i2= –1 As system (15.12.2.2) we take the most general linear
system involving the derivatives with respect to t:
(ϕ1) = Aϕ1+ Bϕ2,
where A, B, C, and D are some functions (dependent on the variables x, t and the parame-ter λ) to be deparame-termined in the subsequent analysis Differentiating equations (15.12.2.4) with respect to t and equations (15.12.2.5) with respect to x and assuming that (ϕ1,2)xt = (ϕ1,2)tx,
we obtain compatibility conditions in the form
A x = Cf – Bg,
B x+2iλB = f t – (A – D)f ,
C x–2iλC = g t + (A – D)g, –D x = Cf – Bg.
(15.12.2.6)
For simplicity, we set
D = –A.
In this case, the first and last equations in (15.12.2.6) coincide, so that (15.12.2.6) turns into
a system of three determining equations:
A x = Cf – Bg,
B x+2iλB = f t–2Af,
C x–2iλC = g t+2Ag
(15.12.2.7)
The functions A, B, and C must be expressed in terms of f and g The general solution
of system (15.12.2.7) for arbitrary functions f and g cannot be found So let us look for
particular solutions in the form finite expansions in positive and negative powers of the
parameter λ.
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15.12.2-2 Solution of the determining equations in the form of polynomials in λ.
The simplest polynomial representations of the unknown functions that give rise to nontrivial
results are quadratic in the spectral parameter λ:
A = A2λ2+ A
1λ + A0,
B = B2λ2+ B
1λ + B0,
C = C2λ2+ C
1λ + C0
(15.12.2.8)
Let us substitute (15.12.2.8) into (15.12.2.7) and collect the terms with equal powers in λ
to obtain
λ2(A
2x – C2f + B2g ) + λ(A1x – C1f + B1g ) + A0x – C0f + B0g=0, (15.12.2.9)
2iλ3B
2+ λ2(B2x+2iB1+2A2f)
+ λ(B1x+2iB0+2A1f ) + B0x+2A0f – f t=0, (15.12.2.10) –2iλ3C
2+ λ2(C2x–2iC1–2A2g)
+ λ(C1x–2iC0–2A1g ) + C0x–2A0g – g t=0 (15.12.2.11)
Let us equate the coefficients of like powers of λ to zero starting from the highest power Setting the coefficients of λ3to zero gives
B2= C2=0 (15.12.2.12)
Equating the coefficients of λ2to zero and taking into account (15.12.2.12), we find that
A2= a = const, B1= iaf , C1= iag. (15.12.2.13)
Setting the coefficient of λ in (15.12.2.9) to zero and then replacing B1and C1in accordance
with (15.12.2.13), we have A1x =0, whence A1= a1 = const For simplicity, we dwell on
the special case a1 =0(arbitrary a1gives rise to more general results), so that
By equating to zero the coefficients of λ in the equations obtained from (15.12.2.10)
and (15.12.2.11) and taking into account (15.12.2.13) and (15.12.2.14), we get
B0= –12af x, C0 = 12ag x. (15.12.2.15)
Setting the coefficient of λ0 in (15.12.2.9) to zero and then integrating, we find that A0 =
1
2af g + a0, where a0is an arbitrary constant As before, we set a0=0for simplicity, which results in
A0 = 12af g (15.12.2.16) Then equations (15.12.2.10) and (15.12.2.11) in view of (15.12.2.15) and (15.12.2.16) become
f t= –12af xx + af2g,
g t= 12ag xx – af g2.
(15.12.2.17) Substituting (15.12.2.8) and (15.12.2.12)–(15.12.2.17) into (15.12.2.5) yields
(ϕ1) = a(λ2+ 12f g )ϕ1+ a(iλf – 12f x )ϕ2,
(ϕ2) = a(iλg + 12g x )ϕ1– a(λ2+ 12f g )ϕ2 (15.12.2.18)