Handbook of mathematics for engineers and scienteists part 190 potx

T8.3. ELLIPTIC EQUATIONS 1291
T8.3.3-4. 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)=–

∞
–∞
f(ξ)G(x, y, ξ, 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).
T8.3.3-5. 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)=

∞

0
f
1
(η)

∂
∂ξ
G(x, y, ξ, η)

ξ=0
dη +

∞
0
f
2
(ξ)

∂
∂η
G(x, y, ξ, η)

η=0
dξ
+

∞
0

∞

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
)–K
0
(s
3
)+K
0
(s
4
)

,


1
=

(x – ξ)
2
+(y – η)
2
, 
2
=

(x – ξ)
2
+(y + η)
2
,

3
=

(x + ξ)
2
+(y – η)
2
, 
4
=

(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
)–H
(2)
0
(k
3
)+H
(2)

0
(k
4
)

.
1292 LINEAR EQUATIONS AND PROBLEMS OF MATH EMATIC AL PHYSICS
T8.3.3-6. Domain: 0 ≤ x < ∞, 0 ≤ y < ∞. Second boundary value problem.
A quadrant of the plane is considered. Boundary conditions are prescribed:
∂
x
w = f
1
(y)atx = 0, ∂
y
w = f
2
(x)aty = 0.
Solution:
w(x, y)=–

∞
0
f
1
(η)G(x, y, 0, η) dη –

∞
0
f

2
(ξ)G(x, y, ξ, 0) dξ
+

∞
0

∞
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
)+K
0
(s

3
)+K
0
(s
4
)

,

1
=

(x – ξ)
2
+(y – η)
2
, 
2
=

(x – ξ)
2
+(y + η)
2
,

3
=

(x + ξ)

2
+(y – η)
2
, 
4
=

(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
)+H
(2)
0
(k
3
)+H
(2)
0
(k
4
)

.
T8.3.3-7. 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.
1
◦
. Eigenvalues of the homogeneous problem with Φ ≡ 0 (it is convenient to label them
with a double subscript):

λ
nm
= π
2

n
2
a
2
+
m
2
b
2

; n = 1, 2, ; m = 1, 2,
Eigenfunctions and the norm squared:
w
nm
=sin

nπx
a

sin

mπy
b

, w

nm

2
=
ab
4
.
2
◦
. Solution for λ ≠ λ
nm
:
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.
T8.3. ELLIPTIC EQUATIONS 1293
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)
β
n
sinh(β

n
b)
H
n
(y, η)=
2
b
∞

m=1
sin(q
m
y)sin(q
m
η)
μ
m
sinh(μ
m
a)
Q
m
(x, ξ),
where
p
n
=
πn
a
, β

n
=

p
2
n
–λ, H
n
(y, η)=

sinh(β
n
η)sinh[β
n
(b–y)] for b ≥ y > η ≥ 0,
sinh(β
n
y)sinh[β
n
(b–η)] for b ≥ η > y ≥ 0;
q
m
=
πm
b
, μ
m
=

q

2
m
–λ, Q
m
(x, ξ)=

sinh(μ
m
ξ)sinh[μ
m
(a–x)] for a ≥ x >ξ ≥ 0,
sinh(μ
m
x)sinh[μ
m
(a–ξ)] for a ≥ ξ > x ≥ 0.
T8.3.3-8. Domain: 0 ≤ r ≤ R. First boundary value problem.
A circle is considered. A boundary condition is prescribed:
w = 0 at r = R.
Eigenvalues of the homogeneous boundary value problem with Φ ≡ 0:
λ
nm
=
μ
2
nm
R
2
; n = 0, 1, 2, ; m = 1, 2, 3,
Here, μ

nm
are positive zeros of the Bessel functions, J
n
(μ)=0.
Eigenfunctions:
w
(1)
nm
= J
n

r

λ
nm

cos nϕ, w
(2)
nm
= J
n

r

λ
nm

sin nϕ.
Eigenfunctions possessing the axial symmetry property: w
(1)

0m
= J
0

r
√
λ
0m

.
T8.3.3-9. Domain: 0 ≤ r ≤ R. Second boundary value problem.
A circle is considered. A boundary condition is prescribed:
∂
r
w = 0 at r = R.
Eigenvalues of the homogeneous boundary value problem with Φ ≡ 0:
λ
nm
=
μ
2
nm
R
2
,
where μ
nm
are roots of the quadratic equation J

n

(μ)=0.
Eigenfunctions:
w
(1)
nm
= J
n
(r

λ
nm
)cosnϕ, w
(2)
nm
= J
n
(r

λ
nm
)sinnϕ.
Here, n = 0, 1, 2, ;forn ≠ 0, the parameter m assumes the values m = 1, 2, 3, ;for
n = 0, a root μ
00
= 0 (the corresponding eigenfunction is w
00
= 1).
Eigenfunctions possessing the axial symmetry property: w
(1)
0m

= J
0

r
√
λ
0m

.
1294 LINEAR EQUATIONS AND PROBLEMS OF MATH EMATIC AL PHYSICS
T8.4. Fourth-Order Linear Equations
T8.4.1. Equation of the Form
∂
2
w
∂t
2
+ a
2
∂
4
w
∂x
4
=0
This equation is encountered in studying transverse vibration of elastic rods.
T8.4.1-1. Particular solutions.
w(x, t)=(Ax
3
+ Bx

2
+ Cx+ D)t + A
1
x
3
+ B
1
x
2
+ C
1
x + D
1
,
w(x, t)=

A sin(λx)+B cos(λx)+C sinh(λx)+D cos(λx)

sin(λ
2
at),
w(x, t)=

A sin(λx)+B cos(λx)+C sinh(λx)+D cos(λx)

cos(λ
2
at),
where A, B, C, D, A
1

, B
1
, C
1
, D
1
,andλ are arbitrary constants.
T8.4.1-2. Domain: –∞ < x < ∞. Cauchy problem.
Initial conditions are prescribed:
w = f(x)att = 0, ∂
t
w = ag

(x)att = 0.
Boussinesq solution:
w(x, t)=
1
√
2π

∞
–∞
f

x – 2ξ
√
at

cos ξ
2

+sinξ
2

dξ
+
1
a
√
2π

∞
–∞
g

x – 2ξ
√
at

cos ξ
2
–sinξ
2

dξ.
T8.4.1-3. Domain: 0 ≤ x < ∞. Free vibration of a semi-infinite rod.
The following conditions are prescribed:
w = 0 at t = 0, ∂
t
w = 0 at t = 0 (initial conditions),
w = f(t)atx = 0, ∂

xx
w = 0 at x = 0 (boundary conditions).
Boussinesq solution:
w(x, t)=
1
√
π

∞
x/
√
2at
f

t –
x
2
2aξ
2

sin
ξ
2
2
+cos
ξ
2
2

dξ.

T8.4.1-4. Domain: 0 ≤ x ≤ l. Boundary value problems.
For solutions of various boundary value problems, see Subsection T8.4.2 for Φ ≡ 0.
T8.4. FOURTH-ORDER LINEAR EQUATIONS 1295
T8.4.2. Equation of the Form
∂
2
w
∂t
2
+ a
2
∂
4
w
∂x
4
= Φ(x, t)
T8.4.2-1. Domain: 0 ≤ x ≤ l. Solution in terms of the Green’s function.
We consider boundary value problems on an interval 0 ≤ x ≤ l with the general initial
condition
w = f(x)att = 0, ∂
t
w = g(x)att = 0
and various homogeneous boundary conditions. The solution can be represented in terms
of the Green’s function as
w(x, t)=
∂
∂t

l

0
f(ξ)G(x, ξ, t) dξ +

l
0
g(ξ)G(x, ξ, t) dξ +

t
0

l
0
Φ(ξ, τ)G(x, ξ, t – τ) dξ dτ .
Paragraphs T8.4.2-2 through T8.4.2-6 present the Green’s functions for various types
of homogeneous boundary conditions.
T8.4.2-2. Both ends of the rod are clamped.
Boundary conditions are prescribed:
w = ∂
x
w = 0 at x = 0, w = ∂
x
w = 0 at x = l.
Green’s function:
G(x, ξ, t)=
4
al
∞

n=1
λ

2
n

ϕ

n
(l)

2
ϕ
n
(x)ϕ
n
(ξ)sin(λ
2
n
at),
where
ϕ
n
(x)=

sinh(λ
n
l)–sin(λ
n
l)

cosh(λ
n

x)–cos(λ
n
x)

–

cosh(λ
n
l)–cos(λ
n
l)

sinh(λ
n
x)–sin(λ
n
x)

;
λ
n
are positive roots of the transcendental equation cosh(λl)cos(λl)=1. The numerical
values of the roots can be calculated from the formulas
λ
n
=
μ
n
l
,whereμ

1
= 1.875, μ
2
= 4.694, μ
n
=
π
2
(2n – 1)forn ≥ 3.
T8.4.2-3. Both ends of the rod are hinged.
Boundary conditions are prescribed:
w = ∂
xx
w = 0 at x = 0, w = ∂
xx
w = 0 at x = l.
Green’s function:
G(x, ξ, t)=
2l
aπ
2
∞

n=1
1
n
2
sin(λ
n
x)sin(λ

n
ξ)sin(λ
2
n
at), λ
n
=
πn
l
.
1296 LINEAR EQUATIONS AND PROBLEMS OF MATH EM ATI CA L PHYSICS
T8.4.2-4. One end of the rod is clamped and the other is hinged.
Boundary conditions are prescribed:
w = ∂
x
w = 0 at x = 0, w = ∂
xx
w = 0 at x = l.
Green’s function:
G(x, ξ, t)=
2
al
∞

n=1
λ
2
n
ϕ
n

(x)ϕ
n
(ξ)
|ϕ

n
(l)ϕ

n
(l)|
sin(λ
2
n
at),
where
ϕ
n
(x)=

sinh(λ
n
l)–sin(λ
n
l)

cosh(λ
n
x)–cos(λ
n
x)


–

cosh(λ
n
l)–cos(λ
n
l)

sinh(λ
n
x)–sin(λ
n
x)

;
λ
n
are positive roots of the transcendental equation tan(λl)–tanh(λl)=0.
T8.4.2-5. One end of the rod is clamped and the other is free.
Boundary conditions are prescribed:
w = ∂
x
w = 0 at x = 0, ∂
xx
w = ∂
xxx
w = 0 at x = l.
Green’s function:
G(x, ξ, t)=

4
al
∞

n=1
ϕ
n
(x)ϕ
n
(ξ)
λ
2
n
ϕ
2
n
(l)
sin(λ
2
n
at),
where
ϕ
n
(x)=

sinh(λ
n
l)+sin(λ
n

l)

cosh(λ
n
x)–cos(λ
n
x)

–

cosh(λ
n
l)+cos(λ
n
l)

sinh(λ
n
x)–sin(λ
n
x)

;
λ
n
are positive roots of the transcendental equation cosh(λl)cos(λl)=–1.
T8.4.2-6. One end of the rod is hinged and the other is free.
Boundary conditions are prescribed:
w = ∂
xx

w = 0 at x = 0, ∂
xx
w = ∂
xxx
w = 0 at x = l.
Green’s function:
G(x, ξ, t)=
4
al
∞

n=1
ϕ
n
(x)ϕ
n
(ξ)
λ
2
n
ϕ
2
n
(l)
sin(λ
2
n
at),
where
ϕ

n
(x)=sin(λ
n
l)sinh(λ
n
x)+sinh(λ
n
l)sin(λ
n
x);
λ
n
are positive roots of the transcendental equation tan(λl)–tanh(λl)=0.
T8.4. FOURTH-ORDER LINEAR EQUATIONS 1297
T8.4.3. Biharmonic Equation ΔΔw =0
The biharmonic equation is encountered in plane problems of elasticity (w is the Airy stress
function). It is also used to describe slow flows of viscous incompressible fluids (w is the
stream function).
In the rectangular Cartesian system of coordinates, the biharmonic operator has the form
ΔΔ ≡ Δ
2
=
∂
4
∂x
4
+ 2
∂
4
∂x

2
∂y
2
+
∂
4
∂y
4
.
T8.4.3-1. Particular solutions.
w(x, y)=(A cosh βx + B sinh βx + Cxcosh βx + Dxsinh βx)(a cos βy + b sin βy),
w(x, y)=(A cos βx + B sin βx + Cxcos βx + Dx sin βx)(a cosh βy + b sinh βy),
w(x, y)=Ar
2
ln r + Br
2
+ C lnr + D, r =

(x – a)
2
+(y – b)
2
,
where A, B, C, D, E, a, b, c,andβ are arbitrary constants.
T8.4.3-2. Various representations of the general solution.
1
◦
. Various representations of the general solution in terms of harmonic functions:
w(x, y)=xu
1

(x, y)+u
2
(x, y),
w(x, y)=yu
1
(x, y)+u
2
(x, y),
w(x, y)=(x
2
+ y
2
)u
1
(x, y)+u
2
(x, y),
where u
1
and u
2
are arbitrary functions satisfying the Laplace equation Δu
k
= 0 (k = 1, 2).
2
◦
. Complex form of representation of the general solution:
w(x, y)=Re

zf(z)+g(z)


,
where f (z)andg(z) are arbitrary analytic functions of the complex variable z = x + iy;
z = x – iy, i
2
=–1. The symbol Re[A] stands for the real part of the complex quantity A.
T8.4.3-3. Boundary value problems for the upper half-plane.
1
◦
. Domain: –∞ < x < ∞, 0 ≤ y < ∞. The desired function and its derivative along the
normal are prescribed at the boundary:
w = 0 at y = 0, ∂
y
w = f(x)aty = 0.
Solution:
w(x, y)=

∞
–∞
f(ξ)G(x – ξ, y) dξ, G(x, y)=
1
π
y
2
x
2
+ y
2
.
2

◦
. Domain: –∞< x < ∞, 0 ≤ y < ∞. The derivatives of the desired function are prescribed
at the boundary:
∂
x
w = f(x)aty = 0, ∂
y
w = g(x)aty = 0.
Solution:
w(x, y)=
1
π

∞
–∞
f(ξ)

arctan

x – ξ
y

+
y(x – ξ)
(x – ξ)
2
+ y
2

dξ +

y
2
π

∞
–∞
g(ξ) dξ
(x – ξ)
2
+ y
2
+ C,
where C is an arbitrary constant.