Equations with ß Space Variables
8.5. Equations with ß Space Variables
8.5.1. Laplace Equation à á â =0
The ➴ -dimensional Laplace equation in the rectangular Cartesian system of coordinates ã 1 , Ñ✘Ñ✘Ñ , ã q
has the form
2 q = 0.
For ➴ = 2 and ➴ = 3, see Subsections 7.1.1 and 8.1.1.
A regular solution of the Laplace equation is called a harmonic function. In what follows we use the notation: x = { ã 1 , Ñ✘Ñ✘Ñ , ã q } and |x| = ä ã 2 1 + ➶✘➶✘➶ + ã 2 q .
8.5.1-1. Particular solutions.
1 ➪ . Fundamental solution: å å
2 ➪ . Solution containing arbitrary functions of ➴ − 1 variables:
2 q r +1
( ã 1 , Ñ✘Ñ✘Ñ ã q
, r )= (−1) Ø ( ã 1 , Ñ✘Ñ✘Ñ , ã q
( ì ã 1 , Ñ✘Ñ✘Ñ , ã q −1 ) and ( ã 1 , Ñ✘Ñ✘Ñ , ã q −1 ) are arbitrary infinitely differentiable functions.
. Let ( ã 1 , Ñ✘Ñ✘Ñ , ã q ) be a harmonic function. Then the functions Ö Ö
( î ➥ ã 1 + ï 1 , Ñ✘Ñ✘Ñ , î ð ã q + ï q ),
2 , Ñ✘Ñ✘Ñ , 2 ò | , x| | x|
| q x| −2
are also harmonic functions everywhere they are defined; Ö Ò , ï 1 , Ñ✘Ñ✘Ñ , ï q , and ð are arbitrary constants. The signs at ó★ô ð in the expression of 1 can be taken independently of one another.
References : A. V. Bitsadze and D. F. Kalinichenko (1985), R. Courant and D. Hilbert (1989).
8.5.1-2. Domain: − õ < ã 1 < õ , ö✘ö✘ö ,− õ < ã q −1 < õ ,0≤ ã q < õ .
The first boundary value problem for an ÷ -dimensional half-space is considered. A boundary condition is prescribed:
= Ø ( ã 1 , ö✘ö✘ö , ã q −1 ) at ã q = 0.
( û 1 , ö✘ö✘ö , û q −1 ) û 1 ö✘ö✘ö✁ û q −1 ,
where ó★ô ✂ ( ✄ ) is the gamma function. ù ù
Reference : A. V. Bitsadze and D. F. Kalinichenko (1985).
8.5.1-3. Domain: |x| ≤ 1. First boundary value problem.
A sphere of unit radius in the ÷ -dimensional space is considered. A boundary condition is prescribed:
= ÿ (x) for | x| = 1.
Solution (Poisson integral):
Reference : A. V. Bitsadze and D. F. Kalinichenko (1985).
8.5.2. Other Equations
This is the Poisson equation in ÷ independent variables. For ÷ = 2 and ÷ = 3, see Sections 7.2 and 8.2.
−2 . ( ü 1 − û 1 ) 2 + ✓✔✓✔✓ +( ü q − û q ) 2 ✕ 2
Reference : S. G. Krein (1972).
2 ✍ . Domain: 0 ≤ ü r ≤ ✖ r ; ✗ = 1, ö✘ö✘ö , ÷ . First boundary value problem.
A rectangular parallelepiped is considered. Boundary conditions are prescribed:
= ÿ r ( ü 1 , ö✘ö✘ö , ü r −1 , ü r +1 , ö✘ö✘ö , ü q ) at ü r = 0,
= ✘ r ( ü 1 , ö✘ö✘ö , ü r −1 , ü r +1 , ö✘ö✘ö , ü q ) at ü r = ✖ r . Green’s function:
) sin( r ) ( 1 ,
sin( ✛ r 1 ü 1 ) sin( ✛ r 1 û 1 ) ö✘ö✘ö sin( ✛ r ✏ ü q
1 ✛ =1 r ✏ =1 r 1 + ✓✔✓✔✓ + ✛ r ✏
, ö✘ö✘ö , ✛ r ✏ =
3 ✍ . Domain: 0 ≤ ü r ≤ ✖ r ; ✗ = 1, ö✘ö✘ö , ✜ . Mixed boundary value problem.
A rectangular parallelepiped is considered. Boundary conditions are prescribed:
, −1 , ü r +1 , ö✘ö✘ö , ü q ) at ü r = 0,
= ✘ r ( ü 1 , ö✘ö✘ö , ü r −1 , ü r +1 , ö✘ö✘ö , ü q ) at ü r = ✖ r . Green’s function:
sin( ✛ r 1 ü 1 ) sin( ✛ r 1 û 1 ) ö✘ö✘ö sin( ✛ r ✏ ü q ) sin( ✛ r ✏ û q
( ü 1 , ö✘ö✘ö , ü q , û 1 , ö✘ö✘ö , û q
1 r 1 =0
=0 ✛ r 2 1 + ✓✔✓✔✓ + ✛ 2 r ✏
This is the Helmholtz equation in ✜ independent variables. For ✜ = 2 and ✜ = 3, see Sections 7.3 and 8.3.
1 ✍ . Fundamental solution for ★ = ✗ 2 > 0:
for even ✜ ,
where ✮ ✯ ( ✄ ) and ✫ ✯ ( ✄ ) are the Bessel functions.
2 ✍ . Domain: 0 ≤ ü r ≤ ✖ r ; ✗ = 1, ö✘ö✘ö , ✜ . First boundary value problem.
A rectangular parallelepiped is considered. Boundary conditions are prescribed:
= ÿ r ( ü 1 , ö✘ö✘ö , ü r −1 , ü r +1 , ö✘ö✘ö , ü q ) at ü r = 0,
= ✘ r ( ü 1 , ö✘ö✘ö , ü r −1 , ü r +1 , ö✘ö✘ö , ü q ) at ü r = ✖ r . Green’s function:
sin( ✛ r 1 ü 1 ) sin( ✛ r 1 û 1 ) ö✘ö✘ö sin( ✛ r ✏ ü q ) sin( ✛ r ✏ û q ) ( 1 ,
3 ✍ . Domain: 0 ≤ ü r ≤ ✖ r ; ✗ = 1, ö✘ö✘ö , ✜ . Second boundary value problem.
A rectangular parallelepiped is considered. Boundary conditions are prescribed:
= ÿ r ( ü 1 , ö✘ö✘ö , ü r −1 , ü r +1 , ö✘ö✘ö , ü q ) at ü r = 0,
= ✘ r ( ü 1 , ö✘ö✘ö , ü r −1 , ü r +1 , ö✘ö✘ö , ü q ) at ü r = ✖ r . Green’s function:
cos( 1 1 ) cos( 1 û 1 ) ö✘ö✘ö cos( ✛ r ✏ ü q ) cos( ✛ r ✏ û q ) ( 1 ,
, ✖ ✛ r 2 = ✖ , ö✘ö✘ö , ✛ r ✏ =
1 for = 0,
1 r 2 ✱✲✱✲✱ r
✖ ö✘ö✘ö✥✖ q
2 for ✶ ≠ 0.
4 ✍ . Domain: 0 ≤ ü r ≤ ✖ r ; ✗ = 1, ö✘ö✘ö , ✜ . Mixed boundary value problem.
A rectangular parallelepiped is considered. Boundary conditions are prescribed:
= ÿ r ( ü 1 , ö✘ö✘ö , ü r −1 , ü r +1 , ö✘ö✘ö , ü q ) at ü r = 0,
= ✘ r ( ü 1 , ö✘ö✘ö , ü r −1 , ü r +1 , ö✘ö✘ö , ü q ) at ü r = ✖ r . Green’s function:
2 sin( r 1 ü 1 ) sin( ✛ r 1 û 1 ) ö✘ö✘ö sin( ✛ r ✏ ü q ) sin( ✛ r ✏ û q ) ( 1 ,
Reference : V. M. Babich, M. B. Kapilevich, S. G. Mikhlin, et al. (1964).
1 ✽ , ö✘ö✘ö It is assumed that for any real numbers ✽ û , û q the relation ✼
holds, where
is some positive constant. Fundamental solution:
1 , ö✘ö✘ö , ü q , û 1 , ö✘ö✘ö û q
✷★ô where is the determinant of the matrix A = { ✾ } and the ✾ are the entries of the inverse of A.
Reference : V. M. Babich, M. B. Kapilevich, S. G. Mikhlin, et al. (1964). s
= 0 is set on the entire boundary of the domain. Eigenvalues and eigenfunctions:
1 ✍ . Case 0 < ❊
< 1. First boundary value problem. The condition ☎
where ❋ ✯ is the ✶
th positive root of the equation ✮ ✯ ( ) = 0, 1− ❊
2 ❚ . Case 1 ≤ ❊ < 2. Boundary conditions: the solution must be bounded at P ■ = 0, and the condition
= 0 must hold on the rest of the boundary of the domain. The eigenvalues and eigenfunctions of this problem are given by the relations of Item 1 ❚ with
Reference : M. M. Smirnov (1975).
Chapter 9
Higher-Order Partial Differential Equations