Suppose wx, t is a solution of the Korteweg–de Vries equation.. , C4are arbitrary constants, is also a solution of the equation.. This is an equation of a steady laminar boundary layer o
Trang 1T9.4.2 Monge–Amp `ere Equations
1.
∂2w
∂x∂y
2
– ∂
2w
∂x2
∂2w
∂y2 = 0.
Homogeneous Monge–Amp`ere equation.
1◦ General solution in parametric form:
w = tx + ϕ(t)y + ψ(t),
x + ϕ (t)y + ψ (t) =0,
where t is the parameter, and ϕ = ϕ(t) and ψ = ψ(t) are arbitrary functions.
2◦ Solutions involving one arbitrary function:
w (x, y) = ϕ(C1x + C2y ) + C3x + C4y + C5,
w (x, y) = (C1x + C2y )ϕ
y x
+ C3x + C4y + C5,
w (x, y) = (C1x + C2y + C3)ϕ
C4x + C5y + C6
C1x + C2y + C3
+ C7x + C8y + C9,
where C1, , C9are arbitrary constants and ϕ = ϕ(z) is an arbitrary function.
2.
∂2w
∂x∂y
2
– ∂
2w
∂x2
∂2w
∂y2 = A.
Nonhomogeneous Monge–Amp`ere equation.
1◦ General solution in parametric form for A = a2 >0:
x= β – λ
2a , y= ψ (λ) – ϕ (β)
2a , w= (β + λ)[ψ (λ) – ϕ (β)] +2ϕ (β) –2ψ (λ)
where β and λ are the parameters, ϕ = ϕ(β) and ψ = ψ(λ) are arbitrary functions.
2◦ Solutions:
w (x, y) =
√ A
C2 x (C1x + C2y ) + ϕ(C1x + C2y ) + C3x + C4y,
w (x, y) = C1y2+ C
2xy+ 1
4C1(C
2
2– A)x2+ C3y + C4x + C5,
w (x, y) = 1
x + C1
C2y2+ C
3y+ C
2 3
4C2
12C2(x
3+3C1x2) + C
4y + C5x + C6,
w (x, y) = 2√ A
3C1C2(C1x – C
2
2y2+ C3)3 2+ C4x + C5y + C6,
where C1, , C6are arbitrary constants and ϕ = ϕ(z) is an arbitrary function.
Trang 2T9.5 Higher-Order Equations
T9.5.1 Third-Order Equations
∂t + ∂
3w
∂x3 – 6w ∂w
∂x = 0.
Korteweg–de Vries equation It is used in many sections of nonlinear mechanics and
physics
1◦ Suppose w(x, t) is a solution of the Korteweg–de Vries equation Then the function
w1= C12w (C1x+6C1C2t + C3, C13t + C4) + C2,
where C1, , C4are arbitrary constants, is also a solution of the equation
2◦ One-soliton solution:
2cosh21
2√
a (x – at – b) ,
where a and b are arbitrary constants.
3◦ Two-soliton solution:
w (x, t) = –2 ∂2
∂x2 ln 1+ B1e θ1+ B2e θ2 + AB1B2e θ1 +θ2
,
θ1= a1x – a31t, θ2 = a2x – a32t, A=
a1– a2
a1+ a2
2
,
where B1, B2, a1, and a2are arbitrary constants
4◦ N -soliton solution:
w (x, t) = –2 ∂2
∂x2
ln det
I + C(x, t)4
Here, I is the N×N identity matrix and C(x, t) the N ×N symmetric matrix with entries
C mn (x, t) =
√
ρ m (t)ρ n (t)
p m + p n exp
–(p m + p n )x
,
where the normalizing factors ρ n (t) are given by
ρ n (t) = ρ n(0) exp 8p3
n t
, n=1,2, , N
The solution involves2N arbitrary constants p n and ρ n(0)
The above solution can be represented, for t → ∞, as the sum of N single-soliton
solutions
5◦ “One soliton + one pole” solution:
w (x, t) = –2p2
cosh–2(pz)–(1+px)–2tanh2(pz)
1–(1+px)–1tanh(pz)–2
, z = x–4p2t –c,
where p and c are arbitrary constants.
Trang 36◦ Rational solutions (algebraic solitons):
w (x, t) = 6x (x3–24t)
(x3+12t)2 ,
w (x, t) = –2 ∂2
∂x2 ln(x6+60x3t–720t2).
7◦ There is a self-similar solution of the form w = t– 2 3U (z), where z = t– 1 3x.
8◦ Solution:
w (x, t) =2ϕ (z) +2C1t, z = x +6C1t2+ C
2t,
where C1and C2are arbitrary constants, and the function ϕ(z) is determined by the second-order ordinary differential equation ϕ zz =6ϕ2– C
2ϕ – C1z + C3
9◦ The Korteweg–de Vries equation is solved by the inverse scattering method Any
rapidly decreasing function F = F (x, y; t) as x → +∞ that simultaneously satisfies the two
linear equations
∂2F
∂x2 –
∂2F
∂y2 =0, ∂F
∂t +
∂
∂x + ∂
∂y
3
F =0
generates a solution of the Korteweg–de Vries equation in the form
w= –2 d
dx K (x, x; t), where K(x, y; t) is a solution of the linear Gel’fand–Levitan–Marchenko integral equation
K (x, y; t) + F (x, y; t) +
∞
x K (x, z; t)F (z, y; t) dz =0
Time t appears in this equation as a parameter.
∂t + ∂
3w
∂x3 – 6w ∂w
2t w= 0.
Cylindrical Korteweg–de Vries equation.
The transformation
w (x, t) = – x
12t – 1
2t u (z, τ ), x= z
τ, t= – 1
2τ2
leads to the Korteweg–de Vries equation T9.5.1.1:
∂u
∂τ + ∂3u
∂z3 –6u ∂u
∂z =0
∂t + ∂
3w
∂x3 + 6σw2∂w
∂x = 0.
Modified Korteweg–de Vries equation.
Trang 41◦ One-soliton solution for σ =1:
2
√
4a2+ k2 cosh z +2a, z = kx – (6a2k + k3)t + b,
where a, b, and k are arbitrary constants.
2◦ Two-soliton solution for σ =1:
w (x, t) =2 a1e θ1+ a2e θ2 + Aa2e2 1+θ2+ Aa1e θ1+2 2
1+ e2 1+ e2 2+2(1– A)e θ1 +θ2+ A2e2 (θ1 +θ2 ,
θ1= a1x – a31t + b1, θ2 = a2x – a32t + b2, A =
a1– a2
a1+ a2
2
,
where a1, a2, b1, and b2are arbitrary constants
3◦ Rational solutions (algebraic solitons) for σ =1:
w (x, t) = a – 4a
4a2z2+1, z = x –6a2t,
w (x, t) = a – 12a z4+ 3
2a–2z2– 163 a–4–24tz
4a2 z3+12t– 34a–2z 2
+3 z2+ 1
4a–2
2,
where a is an arbitrary constant.
4◦ There is a self-similar solution of the form w = t– 1 3U (z), where z = t– 1 3x.
5◦ The modified Korteweg–de Vries equation is solved by the inverse scattering method.
∂y
∂2w
∂x
∂2w
∂y2 = ν ∂
3w
∂y3
This is an equation of a steady laminar boundary layer on a flat plate (w is the stream
function)
1◦ Suppose w(x, y) is a solution of the equation in question Then the function
w1 = C1w C2x + C3, C1C2y + ϕ(x)
+ C4,
where ϕ(x) is an arbitrary function and C1, , C5are arbitrary constants, is also a solution
of the equation
2◦ Solutions involving arbitrary functions:
w (x, y) = C1y + ϕ(x),
w (x, y) = C1y2+ ϕ(x)y + 1
4C1ϕ
2(x) + C
2,
w (x, y) = 6νx + C1
y + ϕ(x) +
C2
[y + ϕ(x)]2 + C3,
w (x, y) = ϕ(x) exp(–C1y ) + νC1x + C2,
w (x, y) = C1exp
–C2y – C2ϕ (x)
+ C3y + C3ϕ (x) + νC2x + C4,
w (x, y) =6νC1x1 3tanh ξ + C
2, ξ = C1 y
x2 3 + ϕ(x),
w (x, y) = –6νC1x1 3tan ξ + C
2, ξ = C1 y
x2 3 + ϕ(x),
where C1, , C4are arbitrary constants and ϕ(x) is an arbitrary function The first two solutions are degenerate—they are independent of ν and correspond to inviscid fluid flows.
Trang 5TABLE T9.1 Invariant solutions to the hydrodynamic boundary layer equation (the additive constant is omitted)
No Solution structure Function F or equation for F Remarks
1 w = F (y) + νλx F (y) =
C1 exp(–λy) + C2y if λ≠ 0 ,
C1y2+ C2y if λ =0 λis any number
2 w = F (x)y–1 F (x) =6νx + C1
3 w = x λ+1F (z), z = x λ ( 2λ+ 1)(F z )2– (λ +1)F F zz = νF zzz λis any number
4 w = e λx F (z), z = e λx y 2λ (F z )2– λF F zz = νF zzz λis any number
5 w = F (z) + a ln|x|, z = y/x –(F z )2– aF zz = νF zzz ais any number
3◦ Table T9.1 lists invariant solutions to the hydrodynamic boundary layer equation.
Solution 1 is expressed in additive separable form, solution 2 is in multiplicative separable form, solution 3 is self-similar, and solution 4 is generalized self-similar Solution 5
degenerates at a =0into a self-similar solution (see solution 3 with λ = –1) Equations 3–5
for F are autonomous and generalized homogeneous; hence, their order can be reduced by
two
4◦ Generalized separable solution linear in x:
where the functions f = f (y) and g = g(y) are determined by the autonomous system of
ordinary differential equations
(f y )2– f f yy = νf yyy , (2)
f
y g y – f g yy = νg yyy. (3) Equation (2) has the following particular solutions:
f =6ν (y + C)–1,
f = Ce λy – λν, where C and λ are arbitrary constants.
Let f = f (y) be a solution of equation (2) (f const) Then the corresponding general solution of equation (3) can be written in the form
g (y) = C1+ C2f + C3
f
ψ dy–
f ψ dy
(f y )2 exp
–1
ν
f dy
∂y
∂2w
∂x
∂2w
∂y2 = ν ∂
3w
∂y3 + f (x).
This is a hydrodynamic boundary layer equation with pressure gradient
1◦ Suppose w(x, y) is a solution of the equation in question Then the functions
w1 = w(x, y + ϕ(x)) + C,
where ϕ(x) is an arbitrary function and C is an arbitrary constant, are also solutions of the
equation
Trang 6TABLE T9.2 Invariant solutions to the hydrodynamic boundary layer equation
with pressure gradient (a, k, m, and β are arbitrary constants)
No. Function f (x) Form of solution w = w(x, y) Function u or equation for u
1 f (x) =0 See equation 4 See equation 4
2 f (x) = ax m w = x m+43u (z), z = x m–41y m+21(u z)2–m+43uu zz = νu zzz + a
3 f (x) = ae βx w = e1βx u (z), z = e1βx y 12β (u z)2–14βuu zz = νu zzz + a
4 f (x) = a w = kx + u(y) u (y) =
C1exp –k ν y
–2a k y2+C2y if k≠ 0 , –6a ν y3+ C2y2+ C1y if k =0
5 f (x) = ax– 3 w = k ln|x|+ u(z), z = y/x –(u z)2– ku zz = νu zzz + a
2◦ Degenerate solutions (linear and quadratic in y) for arbitrary f (x):
w (x, y) = y
2
f (x) dx + C1
1 2
+ ϕ(x),
w (x, y) = C1y2+ ϕ(x)y + 1
4C1
ϕ2(x) –2 f (x) dx
+ C2,
where ϕ(x) is an arbitrary function and C1and C2are arbitrary constants These solutions
are independent of ν and correspond to inviscid fluid flows.
3◦ Table T9.2 lists invariant solutions to the boundary layer equation with pressure gradient.
4◦ Generalized separable solution for f (x) = ax + b:
w (x, y) = xF (y) + G(y), where the functions F = F (y) and G = G(y) are determined by the system of ordinary
differential equations
(F y )2– F F yy = νF yyy + a, F
y G y – F G yy = νG yyy + b.
5◦ Solutions for f (x) = –ax– 5 3:
w (x, y) = 6νx
y + ϕ(x)
√
3a
x1 3[y + ϕ(x)],
where ϕ(x) is an arbitrary function.
6◦ Solutions for f (x) = ax– 1 3– bx– 5 3:
w (x, y) = √
3b z + x2 3θ (z), z = yx–1 3,
where the function θ = θ(z) is determined by the ordinary differential equation 13(θ z)2–
2
3θθ zz = νθ zzz + a.
7◦ Generalized separable solution for f (x) = ae βx:
w (x, y) = ϕ(x)e λy– a
2βλ2ϕ (x) e βx–λy – νλx +
2νλ2
β y+ 2νλ
β ln|ϕ (x)|,
where ϕ(x) is an arbitrary function and λ is an arbitrary constant.
Trang 7T9.5.2 Fourth-Order Equations
1. ∂
2w
∂t2 + ∂
∂x
∂x
+ ∂
4w
∂x4 = 0.
Boussinesq equation This equation arises in hydrodynamics and some physical
applica-tions
1◦ Suppose w(x, t) is a solution of the equation in question Then the functions
w1= C12w (C1x + C2, C12t + C3),
where C1, C2, and C3are arbitrary constants, are also solutions of the equation
2◦ Solutions:
w (x, t) =2C1x–2C2
1t2+ C2t + C3,
w (x, t) = (C1t + C2)x – 1
12C2 1
(C1t + C2)4+ C3t + C4,
w (x, t) = – (x + C1)
2
(t + C2)2 +
C3
t + C2 + C4(t + C2)
2,
w (x, t) = – x
2
t2 + C1t3x–
C2 1
54t8+ C2t2+
C4
t ,
w (x, t) = – (x + C1)2
(t + C2)2 –
12
(x + C1)2,
w (x, t) = –3λ2cos– 21
2λ (x λt ) + C1
,
where C1, , C4and λ are arbitrary constants.
3◦ Traveling-wave solution (generalizes the last solution of Item1◦):
w (x, t) = w(ζ), ζ = x + λt, where the function w(ζ) is determined by the second-order ordinary differential equation
w
ζζ + w2+2λ2w + C
1ζ + C2=0
4◦ Self-similar solution:
w (x, t) = t–1u (z), z = xt–1 2,
where the function u = u(z) is determined by the ordinary differential equation u zzzz+
(uu z) z+ 14z2u
zz+ 74zu z+2u=0
5◦ There are exact solutions of the following forms:
w (x, t) = (x + C)2F (t) –12(x + C)–2;
w (x, t) = G(ξ) –4C2
1t2–4C1C2t, ξ = x – C1t2– C
2t;
w (x, t) = 1
t H (η) – 1
4
x
t + Ct
2
, η= √ x
t – 1
3Ct3 2;
w (x, t) = (a1t + a0)2U (ζ) –
a1x + b1
a1t + a0
2
, ζ = x(a1t + a0) + b1t + b0,
where C, C1, C2, a1, a0, b1, and b0are arbitrary constants