Handbook of mathematics for engineers and scienteists part 195 pptx

1326 NONLINEAR MATHEMATIC AL PHYSICS EQUATIONS T9.4.2. Monge–Amp ` ere Equations 1.  ∂ 2 w ∂x∂y  2 – ∂ 2 w ∂x 2 ∂ 2 w ∂y 2 =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)andψ = ψ(t) are arbitrary functions. 2 ◦ . Solutions involving one arbitrary function: w(x, y)=ϕ(C 1 x + C 2 y)+C 3 x + C 4 y + C 5 , w(x, y)=(C 1 x + C 2 y)ϕ  y x  + C 3 x + C 4 y + C 5 , w(x, y)=(C 1 x + C 2 y + C 3 )ϕ  C 4 x + C 5 y + C 6 C 1 x + C 2 y + C 3  + C 7 x + C 8 y + C 9 , where C 1 , , C 9 are arbitrary constants and ϕ = ϕ(z) is an arbitrary function. 2.  ∂ 2 w ∂x∂y  2 – ∂ 2 w ∂x 2 ∂ 2 w ∂y 2 = A. Nonhomogeneous Monge–Amp ` ere equation. 1 ◦ . General solution in parametric form for A = a 2 > 0: x = β – λ 2a , y = ψ  (λ)–ϕ  (β) 2a , w = (β + λ)[ψ  (λ)–ϕ  (β)] + 2ϕ(β)–2ψ(λ) 4a , where β and λ are the parameters, ϕ = ϕ(β)andψ = ψ(λ) are arbitrary functions. 2 ◦ . Solutions: w(x, y)= √ A C 2 x(C 1 x + C 2 y)+ϕ(C 1 x + C 2 y)+C 3 x + C 4 y, w(x, y)=C 1 y 2 + C 2 xy + 1 4C 1 (C 2 2 – A)x 2 + C 3 y + C 4 x + C 5 , w(x, y)= 1 x + C 1  C 2 y 2 + C 3 y + C 2 3 4C 2  – A 12C 2 (x 3 + 3C 1 x 2 )+C 4 y + C 5 x + C 6 , w(x, y)= 2 √ A 3C 1 C 2 (C 1 x – C 2 2 y 2 + C 3 ) 3/2 + C 4 x + C 5 y + C 6 , where C 1 , , C 6 are arbitrary constants and ϕ = ϕ(z) is an arbitrary function. T9.5. HIGHER-ORDER EQUATIONS 1327 T9.5. Higher-Order Equations T9.5.1. Third-Order Equations 1. ∂w ∂t + ∂ 3 w ∂x 3 –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 w 1 = C 2 1 w(C 1 x + 6C 1 C 2 t + C 3 , C 3 1 t + C 4 )+C 2 , where C 1 , , C 4 are arbitrary constants, is also a solution of the equation. 2 ◦ . One-soliton solution: w(x, t)=– a 2 cosh 2  1 2 √ a (x – at – b)  , where a and b are arbitrary constants. 3 ◦ . Two-soliton solution: w(x, t)=–2 ∂ 2 ∂x 2 ln  1 + B 1 e θ 1 + B 2 e θ 2 + AB 1 B 2 e θ 1 +θ 2  , θ 1 = a 1 x – a 3 1 t, θ 2 = a 2 x – a 3 2 t, A =  a 1 – a 2 a 1 + a 2  2 , where B 1 , B 2 , a 1 ,anda 2 are arbitrary constants. 4 ◦ . N-soliton solution: w(x, t)=–2 ∂ 2 ∂x 2  ln det  I + C(x, t)   . Here, I is the N × N identity matrix and C(x, t)theN × 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)aregivenby ρ n (t)=ρ n (0)exp  8p 3 n t  , n = 1, 2, , N. The solution involves 2N arbitrary constants p n and ρ n (0). The above solution can be represented, for t → ∞,asthesumofN single-soliton solutions. 5 ◦ . “One soliton + one pole” solution: w(x, t)=–2p 2  cosh –2 (pz)–(1+px) –2 tanh 2 (pz)  1–(1+px) –1 tanh(pz)  –2 , z = x–4p 2 t–c, where p and c are arbitrary constants. 1328 NONLINEAR MATHEMATIC AL PHYSICS EQUATIONS 6 ◦ . Rational solutions (algebraic solitons): w(x, t)= 6x(x 3 – 24t) (x 3 + 12t) 2 , w(x, t)=–2 ∂ 2 ∂x 2 ln(x 6 + 60x 3 t – 720t 2 ). 7 ◦ . There is a self-similar solution of the form w = t –2/3 U(z), where z = t –1/3 x. 8 ◦ . Solution: w(x, t)=2ϕ(z)+2C 1 t, z = x + 6C 1 t 2 + C 2 t, where C 1 and C 2 are arbitrary constants, and the function ϕ(z) is determined by the second- order ordinary differential equation ϕ  zz = 6ϕ 2 – C 2 ϕ – C 1 z + C 3 . 9 ◦ . The Korteweg–de Vries equation is solved by the inverse scattering method. Any rapidly decreasing function F = F (x, y; t)asx → +∞ that simultaneously satisfi es the two linear equations ∂ 2 F ∂x 2 – ∂ 2 F ∂y 2 = 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. 2. ∂w ∂t + ∂ 3 w ∂x 3 –6w ∂w ∂x + 1 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 2τ 2 leads to the Korteweg–de Vries equation T9.5.1.1: ∂u ∂τ + ∂ 3 u ∂z 3 – 6u ∂u ∂z = 0. 3. ∂w ∂t + ∂ 3 w ∂x 3 +6σw 2 ∂w ∂x =0. Modified Korteweg–de Vries equation. T9.5. HIGHER-ORDER EQUATIONS 1329 1 ◦ . One-soliton solution for σ = 1: w(x, t)=a + k 2 √ 4a 2 + k 2 cosh z + 2a , z = kx –(6a 2 k + k 3 )t + b, where a, b,andk are arbitrary constants. 2 ◦ . Two-soliton solution for σ = 1: w(x, t)=2 a 1 e θ 1 + a 2 e θ 2 + Aa 2 e 2θ 1 +θ 2 + Aa 1 e θ 1 +2θ 2 1 + e 2θ 1 + e 2θ 2 + 2(1 – A)e θ 1 +θ 2 + A 2 e 2(θ 1 +θ 2 ) , θ 1 = a 1 x – a 3 1 t + b 1 , θ 2 = a 2 x – a 3 2 t + b 2 , A =  a 1 – a 2 a 1 + a 2  2 , where a 1 , a 2 , b 1 ,andb 2 are arbitrary constants. 3 ◦ . Rational solutions (algebraic solitons) for σ = 1: w(x, t)=a – 4a 4a 2 z 2 + 1 , z = x – 6a 2 t, w(x, t)=a – 12a  z 4 + 3 2 a –2 z 2 – 3 16 a –4 – 24tz  4a 2  z 3 + 12t – 3 4 a –2 z  2 + 3  z 2 + 1 4 a –2  2 , where a is an arbitrary constant. 4 ◦ . There is a self-similar solution of the form w = t –1/3 U(z), where z = t –1/3 x. 5 ◦ . The modified Korteweg–de Vries equation is solved by the inverse scattering method. 4. ∂w ∂y ∂ 2 w ∂x∂y – ∂w ∂x ∂ 2 w ∂y 2 = ν ∂ 3 w ∂y 3 . 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 w 1 = C 1 w  C 2 x + C 3 , C 1 C 2 y + ϕ(x)  + C 4 , where ϕ(x) is an arbitrary function and C 1 , , C 5 are arbitrary constants, is also a solution of the equation. 2 ◦ . Solutions involving arbitrary functions: w(x, y)=C 1 y + ϕ(x), w(x, y)=C 1 y 2 + ϕ(x)y + 1 4C 1 ϕ 2 (x)+C 2 , w(x, y)= 6νx + C 1 y + ϕ(x) + C 2 [y + ϕ(x)] 2 + C 3 , w(x, y)=ϕ(x)exp(–C 1 y)+νC 1 x + C 2 , w(x, y)=C 1 exp  –C 2 y – C 2 ϕ(x)  + C 3 y + C 3 ϕ(x)+νC 2 x + C 4 , w(x, y)=6νC 1 x 1/3 tanh ξ + C 2 , ξ = C 1 y x 2/3 + ϕ(x), w(x, y)=–6νC 1 x 1/3 tan ξ + C 2 , ξ = C 1 y x 2/3 + ϕ(x), where C 1 , , C 4 are 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. 1330 NONLINEAR MATHEMATIC AL PHYSICS EQUATIONS TABLE 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)=  C 1 exp(–λy)+C 2 y if λ ≠ 0, C 1 y 2 + C 2 y if λ = 0 λ is any number 2 w = F (x)y –1 F (x)=6νx + C 1 3 w = x λ+1 F (z), z = x λ y (2λ + 1)(F  z ) 2 –(λ + 1)FF  zz = νF  zzz λ is any number 4 w = e λx F (z), z = e λx y 2λ(F  z ) 2 – λFF  zz = νF  zzz λ is any number 5 w = F (z)+a ln|x|, z = y/x –(F  z ) 2 – aF  zz = νF  zzz a is 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 = 0 into 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: w(x, y)=xf(y)+g(y), (1) where the functions f = f(y)andg = g(y) are determined by the autonomous system of ordinary differential equations (f  y ) 2 – ff  yy = νf  yyy ,(2) f  y g  y – fg  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)=C 1 + C 2 f + C 3  f  ψdy–  fψdy  ,whereψ = 1 (f  y ) 2 exp  – 1 ν  fdy  . 5. ∂w ∂y ∂ 2 w ∂x∂y – ∂w ∂x ∂ 2 w ∂y 2 = ν ∂ 3 w ∂y 3 + 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 w 1 = 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. T9.5. HIGHER-ORDER EQUATIONS 1331 TABLE T9.2 Invariant solutions to the hydrodynamic boundary layer equation with pressure gradient (a, k, m,andβ 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+3 4 u(z), z =x m–1 4 y m+1 2 (u  z ) 2 – m+3 4 uu  zz = νu  zzz + a 3 f(x)=ae βx w = e 1 4 βx u(z), z = e 1 4 βx y 1 2 β(u  z ) 2 – 1 4 βuu  zz = νu  zzz + a 4 f(x)=a w = kx + u(y) u(y)=  C 1 exp  – k ν y  – a 2k y 2 +C 2 y if k ≠ 0, – a 6ν y 3 + C 2 y 2 + C 1 y 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 + C 1  1/2 + ϕ(x), w(x, y)=C 1 y 2 + ϕ(x)y + 1 4C 1  ϕ 2 (x)–2  f(x) dx  + C 2 , where ϕ(x) is an arbitrary function and C 1 and C 2 are arbitrary constants. These solutions are independent of ν and correspond to inviscid fluid flows. 3 ◦ . TableT9.2 lists invariant solutions to theboundary 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)andG = G(y) are determined by the system of ordinary differential equations (F  y ) 2 – FF  yy = νF  yyy + a, F  y G  y – FG  yy = νG  yyy + b. 5 ◦ . Solutions for f(x)=–ax –5/3 : w(x, y)= 6νx y + ϕ(x) √ 3a x 1/3 [y + ϕ(x)], where ϕ(x) is an arbitrary function. 6 ◦ . Solutions for f(x)=ax –1/3 – bx –5/3 : w(x, y)= √ 3bz+ x 2/3 θ(z), z = yx –1/3 , where the function θ = θ(z) is determined by the ordinary differential equation 1 3 (θ  z ) 2 – 2 3 θθ  zz = νθ  zzz + a. 7 ◦ . Generalized separable solution for f(x)=ae βx : w(x, y)=ϕ(x)e λy – a 2βλ 2 ϕ(x) e βx–λy – νλx + 2νλ 2 β y + 2νλ β ln |ϕ(x)|, where ϕ(x) is an arbitrary function and λ is an arbitrary constant. 1332 NONLINEAR MATHEMATIC AL PHYSICS EQUATIONS T9.5.2. Fourth-Order Equations 1. ∂ 2 w ∂t 2 + ∂ ∂x  w ∂w ∂x  + ∂ 4 w ∂x 4 =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 w 1 = C 2 1 w(C 1 x + C 2 , C 2 1 t + C 3 ), where C 1 , C 2 ,andC 3 are arbitrary constants, are also solutions of the equation. 2 ◦ . Solutions: w(x, t)=2C 1 x – 2C 2 1 t 2 + C 2 t + C 3 , w(x, t)=(C 1 t + C 2 )x – 1 12C 2 1 (C 1 t + C 2 ) 4 + C 3 t + C 4 , w(x, t)=– (x + C 1 ) 2 (t + C 2 ) 2 + C 3 t + C 2 + C 4 (t + C 2 ) 2 , w(x, t)=– x 2 t 2 + C 1 t 3 x – C 2 1 54 t 8 + C 2 t 2 + C 4 t , w(x, t)=– (x + C 1 ) 2 (t + C 2 ) 2 – 12 (x + C 1 ) 2 , w(x, t)=–3λ 2 cos –2  1 2 λ(x λt)+C 1  , where C 1 , , C 4 and λ are arbitrary constants. 3 ◦ . Traveling-wave solution (generalizes the last solution of Item 1 ◦ ): w(x, t)=w(ζ), ζ = x + λt, where the function w(ζ) is determined by the second-order ordinary differential equation w  ζζ + w 2 + 2λ 2 w + C 1 ζ + C 2 = 0. 4 ◦ . Self-similar solution: w(x, t)=t –1 u(z), z = xt –1/2 , where the function u = u(z) is determined by the ordinary differential equation u  zzzz + (uu  z )  z + 1 4 z 2 u  zz + 7 4 zu  z + 2u = 0. 5 ◦ . There are exact solutions of the following forms: w(x, t)=(x + C) 2 F (t)–12(x + C) –2 ; w(x, t)=G(ξ)–4C 2 1 t 2 – 4C 1 C 2 t, ξ = x – C 1 t 2 – C 2 t; w(x, t)= 1 t H(η)– 1 4  x t + Ct  2 , η = x √ t – 1 3 Ct 3/2 ; w(x, t)=(a 1 t + a 0 ) 2 U(ζ)–  a 1 x + b 1 a 1 t + a 0  2 , ζ = x(a 1 t + a 0 )+b 1 t + b 0 , where C, C 1 , C 2 , a 1 , a 0 , b 1 ,andb 0 are arbitrary constants. . self-similar solution of the form w = t –2/3 U(z), where z = t –1/3 x. 8 ◦ . Solution: w(x, t)=2ϕ(z)+2C 1 t, z = x + 6C 1 t 2 + C 2 t, where C 1 and C 2 are arbitrary constants, and the function ϕ(z). 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,. 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