Handbook of mathematics for engineers and scienteists part 189 potx

1284 LINEAR EQUATIONS AND PROBLEMS OF MATHEM ATICAL PHYSICS T8.2.7. Equations of the Form ∂ 2 w ∂t 2 + k ∂w ∂t = a 2 ∂ 2 w ∂x 2 + b ∂w ∂x + cw + Φ(x, t) 1. ∂ 2 w ∂t 2 + k ∂w ∂t = a 2 ∂ 2 w ∂x 2 + bw. For k > 0 and b < 0,itisthetelegraph equation. The substitution w(x, t)=exp  – 1 2 kt  u(x, t) leads to the equation ∂ 2 u ∂t 2 = a 2 ∂ 2 u ∂x 2 +  b + 1 4 k 2  u, which is discussed in Subsection T8.2.4. 2. ∂ 2 w ∂t 2 + k ∂w ∂t = a 2 ∂ 2 w ∂x 2 + b ∂w ∂x + cw + Φ(x, t). The substitution w(x, t)=exp  – 1 2 a –2 bx – 1 2 kt  u(x, t) leads to the equation ∂ 2 u ∂t 2 = a 2 ∂ 2 u ∂x 2 +  c + 1 4 k 2 – 1 4 a –2 b 2  u +exp  1 2 a –2 bx + 1 2 kt  Φ(x, t), which is discussed in Subsection T8.2.5. T8.3. Elliptic Equations T8.3.1. Laplace Equation Δw =0 The Laplace equation is often encountered in heat and mass transfer theory, fluid mechanics, elasticity, electrostatics, and other areas of mechanics and physics. The two-dimensional Laplace equation has the following form: ∂ 2 w ∂x 2 + ∂ 2 w ∂y 2 = 0 in the Cartesian coordinate system, 1 r ∂ ∂r  r ∂w ∂r  + 1 r 2 ∂ 2 w ∂ϕ 2 = 0 in the polar coordinate system, where x = r cos ϕ, y = r sin ϕ,andr =  x 2 + y 2 . T8.3.1-1. Particular solutions and methods for their construction. 1 ◦ . Particular solutions in the Cartesian coordinate system: w(x, y)=Ax + By + C, w(x, y)=A(x 2 – y 2 )+Bxy, w(x, y)=A(x 3 – 3xy 2 )+B(3x 2 y – y 3 ), w(x, y)= Ax + By x 2 + y 2 + C, w(x, y)=exp( μx)(A cos μy + B sin μy), w(x, y)=(A cos μx + B sin μx)exp( μy), w(x, y)=(A sinh μx + B cosh μx)(C cos μy + D sin μy), w(x, y)=(A cos μx + B sin μx)(C sinh μy + D cosh μy), where A, B, C, D,andμ are arbitrary constants. T8.3. ELLIPTIC EQUATIONS 1285 2 ◦ . Particular solutions in the polar coordinate system: w(r)=A ln r + B, w(r, ϕ)=  Ar m + B r m  (C cos mϕ + D sin mϕ), where A, B, C,andD are arbitrary constants, and m = 1, 2, 3 ◦ .Ifw(x, y) is a solution of the Laplace equation, then the functions w 1 = Aw( λx + C 1 , λy + C 2 )+B, w 2 = Aw(x cos β + y sin β,–x sin β + y cos β), w 3 = Aw  x x 2 + y 2 , y x 2 + y 2  are also solutions everywhere they are defined; A, B, C 1 , C 2 , β,andλ are arbitrary constants. The signs at λ’s in the formula for w 1 are taken arbitrarily. 4 ◦ . A fairly general method for constructing particular solutions involves the following. Let f (z)=u(x, y)+iv(x, y) be any analytic function of the complex variable z = x + iy (u and v are real functions of the real variables x and y; i 2 =–1). Then the real and imaginary parts of f both satisfy the two-dimensional Laplace equation, Δ 2 u = 0, Δ 2 v = 0. Thus, by specifying analytic functions f(z) and taking their real and imaginary parts, one obtains various solutions of the two-dimensional Laplace equation. T8.3.1-2. Domain: –∞ < x < ∞, 0 ≤ y < ∞. First boundary value problem. ∗ A half-plane is considered. A boundary condition is prescribed: w = f(x)aty = 0. Solution: w(x, y)= 1 π  ∞ –∞ yf(ξ) dξ (x – ξ) 2 + y 2 = 1 π  π/2 –π/2 f(x + y tan θ) dθ. T8.3.1-3. Domain: –∞ < x < ∞, 0 ≤ y < ∞. Second boundary value problem. ∗ A half-plane is considered. A boundary condition is prescribed: ∂ y w = f(x)aty = 0. Solution: w(x, y)= 1 π  ∞ –∞ f(ξ)ln  (x – ξ) 2 + y 2 dξ + C, where C is an arbitrary constant. * For the Laplace equation and other elliptic equations, the first boundary value problem is often called the Dirichlet problem, and the second boundary value problem is called the Neumann problem. 1286 LINEAR EQUATIONS AND PROBLEMS OF MATHEMATIC AL PHYSICS T8.3.1-4. Domain: 0 ≤ x < ∞, 0 ≤ y < ∞. First boundary value problem. A quadrant of the plane is considered. Boundary conditions are prescribed: w = f 1 (y)atx = 0, w = f 2 (x)aty = 0. Solution: w(x, y)= 4 π xy  ∞ 0 f 1 (η)ηdη [x 2 +(y – η) 2 ][x 2 +(y + η) 2 ] + 4 π xy  ∞ 0 f 2 (ξ)ξdξ [(x –ξ) 2 + y 2 ][(x + ξ) 2 + y 2 ] . T8.3.1-5. Domain: –∞ < x < ∞, 0 ≤ y ≤ a. First boundary value problem. An infinite strip is considered. Boundary conditions are prescribed: w = f 1 (x)aty = 0, w = f 2 (x)aty = a. Solution: w(x, y)= 1 2a sin  πy a   ∞ –∞ f 1 (ξ) dξ cosh[π(x – ξ)/a]–cos(πy/a) + 1 2a sin  πy a   ∞ –∞ f 2 (ξ) dξ cosh[π(x – ξ)/a]+cos(πy/a) . T8.3.1-6. Domain: –∞ < x < ∞, 0 ≤ y ≤ a. Second boundary value problem. An infinite strip is considered. Boundary conditions are prescribed: ∂ y w = f 1 (x)aty = 0, ∂ y w = f 2 (x)aty = a. Solution: w(x, y)= 1 2π  ∞ –∞ f 1 (ξ)ln  cosh[π(x – ξ)/a]–cos(πy/a)  dξ – 1 2π  ∞ –∞ f 2 (ξ)ln  cosh[π(x – ξ)/a]+cos(πy/a)  dξ + C, where C is an arbitrary constant. T8.3.1-7. Domain: 0 ≤ x ≤ a, 0 ≤ y ≤ b. First boundary value problem. A rectangle is considered. Boundary conditions are prescribed: w = f 1 (y)atx = 0, w = f 2 (y)atx = a, w = f 3 (x)aty = 0, w = f 4 (x)aty = b. Solution: w(x, y)= ∞  n=1 A n sinh  nπ b (a – x)  sin  nπ b y  + ∞  n=1 B n sinh  nπ b x  sin  nπ b y  + ∞  n=1 C n sin  nπ a x  sinh  nπ a (b – y)  + ∞  n=1 D n sin  nπ a x  sinh  nπ a y  , T8.3. ELLIPTIC EQUATIONS 1287 where the coefficients A n , B n , C n ,andD n are expressed as A n = 2 λ n  b 0 f 1 (ξ)sin  nπξ b  dξ, B n = 2 λ n  b 0 f 2 (ξ)sin  nπξ b  dξ, λ n = b sinh  nπa b  , C n = 2 μ n  a 0 f 3 (ξ)sin  nπξ a  dξ, D n = 2 μ n  a 0 f 4 (ξ)sin  nπξ a  dξ, μ n = a sinh  nπb a  . T8.3.1-8. Domain: 0 ≤ r ≤ R. First boundary value problem. A circle is considered. A boundary condition is prescribed: w = f (ϕ)atr = R. Solution in the polar coordinates: w(r, ϕ)= 1 2π  2π 0 f(ψ) R 2 – r 2 r 2 – 2Rr cos(ϕ – ψ)+R 2 dψ. This formula is conventionally referred to as the Poisson integral. T8.3.1-9. Domain: 0 ≤ r ≤ R. Second boundary value problem. A circle is considered. A boundary condition is prescribed: ∂ r w = f (ϕ)atr = R. Solution in the polar coordinates: w(r, ϕ)= R 2π  2π 0 f(ψ)ln r 2 – 2Rr cos(ϕ – ψ)+R 2 R 2 dψ + C, where C is an arbitrary constant; this formula is known as the Dini integral. Remark. The function f (ϕ) must satisfy the solvability condition  2π 0 f(ϕ) dϕ = 0. T8.3.2. Poisson Equation Δw + Φ(x) =0 The two-dimensional Poisson equation has the following form: ∂ 2 w ∂x 2 + ∂ 2 w ∂y 2 + Φ(x, y)=0 in the Cartesian coordinate system, 1 r ∂ ∂r  r ∂w ∂r  + 1 r 2 ∂ 2 w ∂ϕ 2 + Φ(r, ϕ)=0 in the polar coordinate system. T8.3.2-1. Domain: –∞ < x < ∞,–∞ < y < ∞. Solution: w(x, y)= 1 2π  ∞ –∞  ∞ –∞ Φ(ξ,η)ln 1  (x – ξ) 2 +(y – η) 2 dξ dη. 1288 LINEAR EQUATIONS AND PROBLEMS OF MATHEMATIC AL PHYSICS T8.3.2-2. Domain: –∞ < x < ∞, 0 ≤ y < ∞. First boundary value problem. A half-plane is considered. A boundary condition is prescribed: w = f(x)aty = 0. Solution: w(x, y)= 1 π  ∞ –∞ yf(ξ) dξ (x – ξ) 2 + y 2 + 1 2π  ∞ 0  ∞ –∞ Φ(ξ,η)ln  (x – ξ) 2 +(y + η) 2  (x – ξ) 2 +(y – η) 2 dξ dη. T8.3.2-3. Domain: 0 ≤ x < ∞, 0 ≤ y < ∞. First boundary value problem. A quadrant of the plane is considered. Boundary conditions are prescribed: w = f 1 (y)atx = 0, w = f 2 (x)aty = 0. Solution: w(x, y)= 4 π xy  ∞ 0 f 1 (η)ηdη [x 2 +(y –η) 2 ][x 2 +(y +η) 2 ] + 4 π xy  ∞ 0 f 2 (ξ)ξdξ [(x – ξ) 2 +y 2 ][(x + ξ) 2 +y 2 ] + 1 2π  ∞ 0  ∞ 0 Φ(ξ,η)ln  (x – ξ) 2 +(y +η) 2  (x + ξ) 2 +(y –η) 2  (x – ξ) 2 +(y –η) 2  (x + ξ) 2 +(y +η) 2 dξ dη. T8.3.2-4. Domain: 0 ≤ x ≤ a, 0 ≤ y ≤ b. First boundary value problem. A rectangle is considered. Boundary conditions are prescribed: w = f 1 (y)atx = 0, w = f 2 (y)atx = a, w = f 3 (x)aty = 0, w = f 4 (x)aty = b. Solution: w(x, y)=  a 0  b 0 Φ(ξ,η)G(x, y, ξ, η) dη dξ +  b 0 f 1 (η)  ∂ ∂ξ G(x, y, ξ, η)  ξ=0 dη –  b 0 f 2 (η)  ∂ ∂ξ G(x, y, ξ, η)  ξ=a dη +  a 0 f 3 (ξ)  ∂ ∂η G(x, y, ξ, η)  η=0 dξ –  a 0 f 4 (ξ)  ∂ ∂η G(x, y, ξ, η)  η=b dξ. Two forms of representation of the Green’s function: G(x, y, ξ, η)= 2 a ∞  n=1 sin(p n x)sin(p n ξ) p n sinh(p n b) H n (y, η)= 2 b ∞  m=1 sin(q m y)sin(q m η) q m sinh(q m a) Q m (x, ξ), where p n = πn a , H n (y, η)=  sinh(p n η)sinh[p n (b – y)] for b ≥ y > η ≥ 0, sinh(p n y)sinh[p n (b – η)] for b ≥ η > y ≥ 0; q m = πm b , Q m (x, ξ)=  sinh(q m ξ)sinh[q m (a – x)] for a ≥ x > ξ ≥ 0, sinh(q m x)sinh[q m (a – ξ)] for a ≥ ξ > x ≥ 0. T8.3. ELLIPTIC EQUATIONS 1289 T8.3.2-5. Domain: 0 ≤ r ≤ R. First boundary value problem. A circle is considered. A boundary condition is prescribed: w = f (ϕ)atr = R. Solution in the polar coordinates: w(r, ϕ)= 1 2π  2π 0 f(η) R 2 – r 2 r 2 – 2Rr cos(ϕ – η)+R 2 dη +  2π 0  R 0 Φ(ξ,η)G(r, ϕ, ξ, η)ξdξdη, where G(r, ϕ, ξ, η)= 1 4π ln r 2 ξ 2 – 2R 2 rξ cos(ϕ – η)+R 4 R 2 [r 2 – 2rξ cos(ϕ – η)+ξ 2 ] . T8.3.3. Helmholtz Equation Δw + λw =–Φ(x) Many problems related to steady-state oscillations (mechanical, acoustical, thermal, elec- tromagnetic) lead to the two-dimensional Helmholtz equation. For λ < 0, this equation describes mass transfer processes with volume chemical reactions of the first order. The two-dimensional Helmholtz equation has the following form: ∂ 2 w ∂x 2 + ∂ 2 w ∂y 2 + λw =–Φ(x, y) in the Cartesian coordinate system, 1 r ∂ ∂r  r ∂w ∂r  + 1 r 2 ∂ 2 w ∂ϕ 2 + λw =–Φ(r, ϕ) in the polar coordinate system. T8.3.3-1. Particular solutions of the homogeneous Helmholtz equation with Φ ≡ 0. 1 ◦ . Particular solutions of the homogeneous equation in the Cartesian coordinate system: w =(Ax + B)(C cos μy + D sin μy), λ = μ 2 , w =(Ax + B)(C cosh μy + D sinh μy), λ =–μ 2 , w =(A cos μx + B sin μx)(Cy + D), λ = μ 2 , w =(A cosh μx + B sinh μx)(Cy + D), λ =–μ 2 , w =(A cos μ 1 x + B sin μ 1 x)(C cos μ 2 y + D sin μ 2 y), λ = μ 2 1 + μ 2 2 , w =(A cos μ 1 x + B sin μ 1 x)(C cosh μ 2 y + D sinh μ 2 y), λ = μ 2 1 – μ 2 2 , w =(A cosh μ 1 x + B sinh μ 1 x)(C cos μ 2 y + D sin μ 2 y), λ =–μ 2 1 + μ 2 2 , w =(A cosh μ 1 x + B sinh μ 1 x)(C cosh μ 2 y + D sinh μ 2 y), λ =–μ 2 1 – μ 2 2 , where A, B, C,andD are arbitrary constants. 2 ◦ . Particular solutions of the homogeneous equation in the polar coordinate system: w =[AJ 0 (μr)+BY 0 (μr)](Cϕ + D), λ = μ 2 , w =[AI 0 (μr)+BK 0 (μr)](Cϕ + D), λ =–μ 2 , w =[AJ m (μr)+BY m (μr)](C cos mϕ + D sin mϕ), λ = μ 2 , w =[AI m (μr)+BK m (μr)](C cos mϕ + D sin mϕ), λ =–μ 2 , where m = 1, 2, ; A, B, C, D are arbitrary constants; J m (μ)andY m (μ) are Bessel functions; and I m (μ)andK m (μ) are modified Bessel functions. 1290 LINEAR EQUATIONS AND PROBLEMS OF MATHEMATIC AL PHYSICS 3 ◦ . Suppose w = w(x, y) is a solution of the homogeneous Helmholtz equation. Then the functions w 1 = w( x + C 1 , y + C 2 ), w 2 = w(x cos θ + y sin θ + C 1 ,–x sin θ + y cos θ + C 2 ), where C 1 , C 2 ,andθ are arbitrary constants, are also solutions of the equation. The signs in the formula for w 1 are taken arbitrarily. T8.3.3-2. Domain: –∞ < x < ∞,–∞ < y < ∞. 1 ◦ . Solution for λ =–s 2 < 0: w(x, y)= 1 2π  ∞ –∞  ∞ –∞ Φ(ξ,η)K 0 (s) dξ dη,  =  (x – ξ) 2 +(y – η) 2 . 2 ◦ . Solution for λ = k 2 > 0: w(x, y)=– i 4  ∞ –∞  ∞ –∞ Φ(ξ,η)H (2) 0 (k) dξ dη,  =  (x – ξ) 2 +(y – η) 2 . Remark. The radiation Sommerfeld conditions at infinity were used to obtain the solution with λ > 0;see Tikhonov and Samarskii (1990) and Polyanin (2002). T8.3.3-3. Domain: –∞ < x < ∞, 0 ≤ y < ∞. First boundary value problem. A half-plane is considered. A boundary condition is prescribed: w = f(x)aty = 0. Solution: w(x, y)=  ∞ –∞ f(ξ)  ∂ ∂η G(x, y, ξ, η)  η=0 dξ +  ∞ 0  ∞ –∞ Φ(ξ,η)G(x, y, ξ, η) dξ dη. 1 ◦ . The Green’s function for λ =–s 2 < 0: G(x, y, ξ, η)= 1 2π  K 0 (s 1 )–K 0 (s 2 )  ,  1 =  (x – ξ) 2 +(y – η) 2 ,  2 =  (x – ξ) 2 +(y + η) 2 . 2 ◦ . The Green’s function for λ = k 2 > 0: G(x, y, ξ, η)=– i 4  H (2) 0 (k 1 )–H (2) 0 (k 2 )  . Remark. The radiation Sommerfeld conditions at infinity were used to obtain the solution with λ > 0;see Tikhonov and Samarskii (1990) and Polyanin (2002). . y) be any analytic function of the complex variable z = x + iy (u and v are real functions of the real variables x and y; i 2 =–1). Then the real and imaginary parts of f both satisfy the two-dimensional. J m (μ)andY m (μ) are Bessel functions; and I m (μ)andK m (μ) are modified Bessel functions. 1290 LINEAR EQUATIONS AND PROBLEMS OF MATHEMATIC AL PHYSICS 3 ◦ . Suppose w = w(x, y) is a solution of. defined; A, B, C 1 , C 2 , β ,and are arbitrary constants. The signs at λ’s in the formula for w 1 are taken arbitrarily. 4 ◦ . A fairly general method for constructing particular solutions involves