Handbook of mathematics for engineers and scienteists part 189 potx

The signs at λ’s in the formula for w1are taken arbitrarily.. A fairly general method for constructing particular solutions involves the following.. Then the real and imaginary parts of

T8.2.7 Equations of the Form

∂2w

∂t2 +k ∂w

2∂2w

∂x2 +b ∂w

∂x +cw +Φ(x, t)

1. ∂

2w

∂t2 + k ∂w

∂t = a2∂

2w

∂x2 + bw.

For k >0and b <0, it is the telegraph equation

The substitution w(x, t) = exp –12kt

u (x, t) leads to the equation

∂2u

∂t2 = a2

∂2u

∂x2 + b+ 14k2

u, which is discussed in Subsection T8.2.4

2. ∂

2w

∂t2 + k ∂w

∂t = a2∂

2w

∂x2 + b ∂w

∂x + cw + Φ(x, t).

The substitution w(x, t) = exp –12a–2bx– 1

2kt

u (x, t) leads to the equation

∂2u

∂t2 = a2

∂2u

∂x2 + c+ 14k2– 14a–2b2

u+ exp 12a–2bx+ 1

2kt

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:

∂2w

∂x2 +

∂2w

∂y2 =0 in the Cartesian coordinate system, 1

r

∂

∂r



r ∂w

∂r

 + 1

r2

∂2w

∂ϕ2 =0 in the polar coordinate system,

where x = r cos ϕ, y = r sin ϕ, and r =

x2+ y2.

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(x2– y2) + Bxy,

w (x, y) = A(x3–3xy2) + B(3 x2y – y3),

w (x, y) = Ax + By

x2+ y2 + 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, and μ are arbitrary constants.

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, and D are arbitrary constants, and m =1,2,

3◦ If w(x, y) is a solution of the Laplace equation, then the functions

w1= Aw( λx + C1, λy + C2) + B,

w2= Aw(x cos β + y sin β, –x sin β + y cos β),

w3= Aw



x

x2+ y2,

y

x2+ y2



are also solutions everywhere they are defined; A, B, C1, C2, β, and λ are arbitrary constants The signs at λ’s in the formula for w1are 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; i2 = –1) Then the real and

imaginary parts of f both satisfy the two-dimensional Laplace equation,

Δ2u=0, Δ2v=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) at y=0

Solution:

w (x, y) = 1

π

 ∞

–∞

yf (ξ) dξ (x – ξ)2+ y2 =

1

π

 π/2

–π/2f (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) at y =0

Solution:

w (x, y) = 1

π

 ∞

–∞ f (ξ) ln

(x – ξ)2+ y2dξ + 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.

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 = f1(y) at x =0, w = f2(x) at y =0

Solution:

w (x, y) = 4

π xy

 ∞

0

f1(η)η dη

[x2+ (y – η)2][x2+ (y + η)2] +

4

π xy

 ∞

0

f2(ξ)ξ dξ

[(x – ξ)2+ y2][(x + ξ)2+ y2].

T8.3.1-5 Domain: –∞ < x < ∞, 0 ≤y≤a First boundary value problem

An infinite strip is considered Boundary conditions are prescribed:

w = f1(x) at y = 0, w = f2(x) at y = a.

Solution:

w (x, y) = 1

2asin



πy a

  ∞

–∞

f1(ξ) dξ cosh[π(x – ξ)/a] – cos(πy/a)

+ 1

2asin



πy a

  ∞

–∞

f2(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 = f1(x) at y = 0, ∂ y w = f2(x) at y = a.

Solution:

w (x, y) = 1

2π

 ∞

–∞ f1(ξ) ln

5

cosh[π(x – ξ)/a] – cos(πy/a)6

dξ

– 1

2π

 ∞

–∞ f2(ξ) ln

5

cosh[π(x – ξ)/a] + cos(πy/a)6

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 = f1(y) at x=0, w = f2(y) at x = a,

w = f3(x) at y=0, w = f4(x) at y = b.

Solution:

w (x, y) =

∞



n=1

A nsinh



nπ

b (a – x)

 sin



nπ

b y

 +

∞



n=1

B nsinh



nπ

b x

 sin



nπ

b y



+

∞



n=1

C nsin



nπ

a x

 sinh



nπ

a (b – y)

 +

∞



n=1

D nsin



nπ

a x

 sinh



nπ

a y

 ,

where the coefficients A n , B n , C n , and D nare expressed as

A n = 2

λ n

 b

0 f1(ξ) sin



nπξ b



dξ, B n = 2

λ n

 b

0 f2(ξ) sin



nπξ b



dξ, λ n = b sinh



nπa b



,

C n = 2

μ n

 a

0 f3(ξ) sin



nπξ a



dξ, D n= 2

μ n

 a

0 f4(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) at r = R.

Solution in the polar coordinates:

w (r, ϕ) = 1

2π

 2π

0 f (ψ)

R2– r2

r2–2Rr cos(ϕ – ψ) + R2 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) at r = R.

Solution in the polar coordinates:

w (r, ϕ) = R

2π

 2π

0 f (ψ) ln

r2–2Rr cos(ϕ – ψ) + R2

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 EquationΔw +Φ(x) = 0

The two-dimensional Poisson equation has the following form:

∂2w

∂x2 +

∂2w

∂y2 +Φ(x, y) =0 in the Cartesian coordinate system, 1

r

∂

∂r



r ∂w

∂r

 + 1

r2

∂2w

∂ϕ2 +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.

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) at y=0

Solution:

w (x, y) = 1

π

 ∞

–∞

yf (ξ) dξ (x – ξ)2+ y2 +

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 = f1(y) at x =0, w = f2(x) at y =0

Solution:

w (x, y) = 4

π xy

 ∞

0

f1(η)η dη

[x2+ (y – η)2][x2+ (y + η)2] +

4

π xy

 ∞

0

f2(ξ)ξ dξ

[(x – ξ)2+ y2][(x + ξ)2+ y2] + 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 = f1(y) at x =0, w = f2(y) at x = a,

w = f3(x) at y =0, w = f4(x) at y = b.

Solution:

w (x, y) =

 a

0

 b

0 Φ(ξ, η)G(x, y, ξ, η) dη dξ

+

 b

0 f1(η)



∂

∂ξ G (x, y, ξ, η)



ξ=0

dη–

 b

0 f2(η)



∂

∂ξ G (x, y, ξ, η)



ξ=a

dη

+

 a

0 f3(ξ)



∂

∂η G (x, y, ξ, η)



η=0dξ–

 a

0 f4(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

T8.3.2-5 Domain: 0 ≤r≤R First boundary value problem.

A circle is considered A boundary condition is prescribed:

w = f (ϕ) at r = R.

Solution in the polar coordinates:

w (r, ϕ) = 1

2π

 2π

0 f (η)

R2– r2

r2–2Rr cos(ϕ – η) + R2 dη+

 2π

0

 R

0 Φ(ξ, η)G(r, ϕ, ξ, η)ξ dξ dη,

where

G (r, ϕ, ξ, η) = 1

4π lnr

2ξ2–2R2rξ cos(ϕ – η) + R4

R2[r2–2rξ cos(ϕ – η) + ξ2] .

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:

∂2w

∂x2 +

∂2w

∂y2 + λw = – Φ(x, y) in the Cartesian coordinate system,

1

r

∂

∂r



r ∂w

∂r

 + 1

r2

∂2w

∂ϕ2 + 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 = μ2,

w = (Ax + B)(C cosh μy + D sinh μy), λ = –μ2,

w = (A cos μx + B sin μx)(Cy + D), λ = μ2,

w = (A cosh μx + B sinh μx)(Cy + D), λ = –μ2,

w = (A cos μ1x + B sin μ1x )(C cos μ2y + D sin μ2y), λ = μ21+ μ22,

w = (A cos μ1x + B sin μ1x )(C cosh μ2y + D sinh μ2y), λ = μ21– μ22,

w = (A cosh μ1x + B sinh μ1x )(C cos μ2y + D sin μ2y), λ = –μ21+ μ22,

w = (A cosh μ1x + B sinh μ1x )(C cosh μ2y + D sinh μ2y ), λ = –μ21– μ22,

where A, B, C, and D are arbitrary constants.

2◦ Particular solutions of the homogeneous equation in the polar coordinate system:

w = [AJ0(μr) + BY0(μr)](Cϕ + D), λ = μ2,

w = [AI0(μr) + BK0(μr)](Cϕ + D), λ = –μ2,

w = [AJ m (μr) + BY m (μr)](C cos mϕ + D sin mϕ), λ = μ2,

w = [AI m (μr) + BK m (μr)](C cos mϕ + D sin mϕ), λ = –μ2,

where m =1, 2, ; A, B, C, D are arbitrary constants; Jm (μ) and Y m (μ) are Bessel functions; and I m (μ) and K m (μ) are modified Bessel functions.

3◦ Suppose w = w(x, y) is a solution of the homogeneous Helmholtz equation Then the

functions

w1= w( x + C1, y + C2),

w2= w(x cos θ + y sin θ + C1, –x sin θ + y cos θ + C2),

where C1, C2, and θ are arbitrary constants, are also solutions of the equation The signs in the formula for w1are taken arbitrarily

T8.3.3-2 Domain: –∞ < x < ∞, –∞ < y < ∞.

1◦ Solution for λ = –s2<0:

w (x, y) = 1

2π

 ∞

–∞

 ∞

–∞ Φ(ξ, η)K0(s) dξ dη, =

(x – ξ)2+ (y – η)2

2◦ Solution for λ = k2>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) at y=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 = –s2 <0:

G (x, y, ξ, η) = 1

2π



K0(s1) – K0(s2)

,

1=

(x – ξ)2+ (y – η)2, 2=

(x – ξ)2+ (y + η)2

2◦ The Green’s function for λ = k2>0:

G (x, y, ξ, η) = – i

4



H(2 )

0 (k1) – H0(2)(k2)

Remark. The radiation Sommerfeld conditions at infinity were used to obtain the solution with λ >0 ; see Tikhonov and Samarskii (1990) and Polyanin (2002).

imaginary parts of f both satisfy... C2, β, and λ are arbitrary constants The signs at λ’s in the formula for w1are taken arbitrarily

4◦ A fairly general method for constructing particular... y2.

T8.3.1-1 Particular solutions and methods for their construction

1◦ Particular solutions in the Cartesian coordinate