Unformatted text preview: V1.1 PHY501 Lecture IIPart 1 Boundary Value Problems Boundary Value Problems
Statement of the Problem:
Find the solution to Poisson's (Laplace's) Equation
2 = ( x)
0 (1) in a volume V bounded by a surface (or surfaces) S. The boundary conditions have two basic forms. Dirichlet Boundary Conditions (DBC): ( x ) specified for x on S (fixed voltages on surfaces)
Neumann Boundary Conditions (NBC): ( x) specified for x on S (fixed charges on surfaces) n S
Note that one can't in general arbitrarily specify both DBC's and NBC's at the same time, since this would overdetermine the solution to Eq. 1. Note also that the solution to Eq. 1 for a given set of boundary conditions is unique (to within an overall added constant for NBC). Proof of this assertion can be found in J1.9 Method of Images
In situations possessing a high degree of symmetry, it is sometimes possible to deduce the form of ( x ) by introducing fictitious "image charges," which, when acting in concert with the real charges of the problem, result in potentials that satisfy the BC's. z +q x' d y x d x'' q
Thus
z +q d A point charge +q at z=+d is placed above a conducting plane held at ground potential =0 . Intuitively we know that by placing q at z=d we can produce a potential such that (x, y,0) = 0 . DBC : =0 ( x) =
R q 4
0 1 x x' x 1 x '' ^ x '= +dz where ^ x ''= dz (2) r rcos x The resulting potential is easily expressed in spherical polar coordinates, using the Law of Cosines d q R' rsin d R = r 2 + d 2 2rd cos R' = r 2 + d 2 + 2rd cos Method of Images
Thus (r, ) = q 4 q 4
0 0 1 R 1 R' 1 1 2rd cos
=
2 = r +d 2 2 r + d + 2rd cos 2 2 (3) Note that for z = 0 and The charge induced on the conducting plane is given by ( r, 2 ) = 0
= Ez =
Thus z =+
0 1 r
2 (x, y)
0 (4) = 0 r z = 2 = q rd sin 4 r ( r 2 + d 2 2rd cos q d (4) 2 2 3/2 2 (r + d ) rd sin ) 3/2 (r 2 + d + 2rd cos 2 ) 3/2 = 2 = (r) r/d
4 3 2 1 1 2 3 Method of Images
The total charge induced on the conducting plane Q=
S qd (x, y)dA = 2 2 d
0 0 rdr (r 2 +d 2 3/2 ) = q (5) Formal Solution of the Boundary Value Problem
We start with Green's Theorem V ( 2 2 ) dV =
1 x x' S n n da (6) and take = G( x, x ') =
Using 1 and R = ( x) (7) '2 ( x ') =
we obtain ( x ')
0 and '2 G( x, x ') = 4
1
0 V (x x ') (8) 4
V ( x ') ( x x ')dV ' + G( x, x ') ( x ')dV ' ( x ') da' n' (9) =
S G( x, x ') ( x ') G( x, x ') n' Formal Solution of the Boundary Value Problem
Thus ( x) = 1 4 1 4
0 V G( x, x ') ( x ')dV ' G( x, x ') G( x, x ') n' ( x ') da' n' (10) ( x ')
S It appears that the solution to Poisson's Equation has been "reduced to quadrature." Life, however, is not so simple. The problem is that Eq. 10 contains both DBC's and NBC'si.e., we need both ( x ') and n'. The way out of this problem is to note that Eq. 10 was derived under the assumption that so that if we introduce '2 G( x, x ') = 4 (x x ')
(11) G'( x, x ') = G( x, x ') + F( x, x ') = 0, x ' on S
in which case Eq. 10 reduces to ( x) = 1 4
0 V G'( x, x ') ( x ')dV ' 1 4 ( x ')
S Note : '2 F = 0 in V G'( x, x ') da' n' (12) Note that we know ( x ') and determined. ( x ') so that ( x ) is Formal Solution of the Boundary Value Problem
For the NBC case we might guess that G' G'( x, x ') = G( x, x ') + F( x, x ') = is s.t. = 0 (13) n' S Since this would eliminate the ( x ') term in Eq. 10.
This can't be, however, since S G' da'= n' = ^ 'G' n ' da'=
S V ' 'G' dV '=
V '2 G' dV ' V ( '2 G+ '2 F ) dV '= 4 (14) So we settle instead for the surface area of S. 4 G' = n' s
1 (15) where s is With that Eq. 10 becomes ( x) = ( x) S + + 1 4 4
S G'( x, x ') ( x ')dV '
0 V G'( x, x ') n' da' (16) where ( x ') Sis the average of ( x ) over S (and is of little significance, since it's a constant). Formal Solution of the Boundary Value Problem
z r a
Example: Take the z=0 plane as an infinite conducting plane with (in cylindrical coordinates) y V r<a (r, ,0) = 0 r>a
Find x ( x ') above the plane. This is an example of a DBC problem, so we use Eq. 12. ( x) = 1 4
0 V G'( x, x ') ( x ')dV ' 1 4 ( x ')
S G'( x, x ') da' n' (12) where ( x ) = 0 and G'( x, x ') must be such that G'( x, x ') = 0 for x on S (17)
But we know of such a function from the imagecharge examplei.e., G'( x, x ') = 1 x x' x 1 x '' ^ ^ where x ''= x ' 2( x ' z ) z (18)
x ''= (x', y', z') If we take x = (x, y,z), x '= (x', y',z') Formal Solution of the Boundary Value Problem
Hence G'( x, x ') = 1 (x x') + ( y 2 y') + ( z z') 1 x') + ( y
2 2 2 2 (19) (x
G' = n' G' = z' y') + ( z + z') 2 To "complete" Eq. 12 we use (z z') [( x x') + ( y 2 y') + ( z z') (z + z')
2 2 2 3/2 [( x
= x') + ( y 2z x') + ( y
( x ')da'
2 2 y') + ( z + z') y') + z
2 2 2 3/2 (20) [( x
2 3/2 Combining Eqs. 12, 19, and 20, we arrive at z ( x) = 2 S [( x x') + ( y y') + z 2 2 3/2 (21) Formal Solution of the Boundary Value Problem
For the special case of x on the z axis, we get z ( x) = 2 2 a d '
0 0 V0 r' dr' The method just described is straightforward in principle (determination of the potential is reduced to an integral), but often difficult in practice since it depends on the availability of a suitable Green's function. We summarize some general properties. (If you aren't already comfortable with the concepts surrounding orthogonal functions, you should consult your favorite math text.) A set of functions defined for a < x < b are said to be orthogonal if b (r' 2 +z 2 3/2 ) = V0 1 z z +a
2 2 (22) Orthogonal Functions u* (x)un (x)dx = Cm n
a nm (23) If the Cm = 1, the functions are said to be orthonormal The utility of the un(x) is that we can "build" other functions out of themi.e., we can write
b f (x) =
n= 0 An un (x) (24) where An =
a f (x)u* (x)dx (25) n (Be sure that you know how to derive this result.) Orthogonal Functions
An amusing feature becomes apparent when we substitute Eq. 25 into Eq. 24.
b f (x) =
n= 0 b An un (x) =
n= 0 a f (x')u* (x')un (x)dx' (26) n and compare to f (x) =
a f (x') (x' x)dx'
n u* (x')un (x) = (x' x) (27) n which is known as closure. Perhaps the best known example of orthogonal functions is the Fourier Series. Separation of Variables
We start with the example Laplace's equation in cartesian coordinates
2 2 2 2 2 2 = x + y + z 2 =0 (28) We assume
2 (x, y,z) = X(x)Y (y)Z(z) so that d2X d 2Y d 2Z = YZ 2 + XZ 2 + XY 2 = 0 (29) dx dy dz Separation of Variables
Dividing through by (x, y,z) = X(x)Y (y)Z(z)
(30) 1 d 2 X 1 d 2Y 1 d 2 Z + + =0 2 2 2 Y dy Z dz X dx Since the terms in Eq. 30 are independent of one another, they must be separately constant and
2 = 2 = 2 =+ 2 = 2 + 2 (31) The values of conditions. , , and are determined by boundary Example: 2D box y = V1 = V2 b DBC : (x,0) = (x,b) = 0 (0, y) = V1 (32) (a, y) = V2
x a Separation of Variables
The 2D Laplacian leads to
2 2 2 2 = x + y 2 =0 (33) (34) 1 d 2Y 1 d2X = +k 2 and = k2 Y dy 2 X dx 2 We choose k2 for y since this is the easiest way to get = 0 for two y values. Y(y) = sin ky where the BC kb = n , n = 1, 2, ... (35) this leaves +k2 in the x direction, where we have X(x) = Ae
hence kx + Be kx (36) (x, y) =
At x=0
n ( Ane kn x + Bn e kn x ) sin k n y (37) (0, y) =
thus
n ( An + Bn ) sin kn y
n Cn sin k n y = V1 (38) 4V1 2 b n y n odd dy = n Cn = V1 sin b b 0 0 n even (39) Separation of Variables
At x=a (a, y) =
thus
n (A e
n ka + Bn e ka ) sin k n y C'n sin k n y = V2 (40)
n 4V2 2 b n y n odd dy = n C'n = V2 sin b 0 b 0 n even (41) Determination of the An's and Bn's is straightforward 4V1 n 4V2 ka ka An e + Bn e = C'n = n An + Bn = Cn =
Example 2: Polar Coordinates 4 V1 V2e kn a An = n 1 e 2kn a 4e Bn = n
kn a V2 V1e kn a 1 e 2kn a y 2 ( , )= 1 + 1
2 2 2 = 0 (41) Assume ( , ) = R( ) ( ) (42) x =0 Separation of Variables
If we multiply both sides of Eq. 41 by
2 R 1 d2 dR + d d 2 = 0 (43) We assume the first and second terms are 2 equal to + 2 and respectively. 1 d2 d 2 = 2 ( ) = Acos( ) + Bsin( ) (44) Note: in problems where is unrestricted, we would impose a periodic boundary conditioni.e., ( ) = ( + 2 ) In this case, however, 0 < But we have the DBC: < (45) (0) = 0 & ( )=0
2 A=0 &
v =n
v In the radial direction we have d Rd
Further R( dR = d R( ) = a +b (46) ) must remain finite as
an
n
n 0
n b = 0.
(47) ( , )= sin( n ), = n Separation of Variables
We can examine the fields E = E ^ + E ^ as 0. (48) ( , )=
Thus
n an n sin( n )
^ a1 sin E= = a1 ^ 1
1 (49) ^ + cos ^
0 0 sin For For < , > , 1 >0 1 <0 fields are finite as fields diverge as For =2 we have a lightning rod Separation of Variables in Spherical Polar Coordinates
The Laplacian is
2 1 1 = 2 r2 + 2 r r r r sin sin 1 + 2 2 r sin 2 2 (50) For now, we will assume azimuthal symmetry which means we drop the 3rd term in Eq. 50. Assume also that (r, ) = R(r)P( )
So Laplace's equation becomes (51) 1 d 2 dR d dP 1 =0 r + sin R dr dr sin P d d
= + ( + 1)
Yielding (52) =
2 ( + 1) d 2 dR r dr dr d 2R dR + 1) R = 0 r + 2r ( ( + 1)R = 0 (53) 2 dr dr B (54) Having solutions: R(r) = A r + +1 r 1 d dP + ( + 1) P = 0 (55) sin For P( ) we have: sin d d Legendre Polynomials
Defining x cos d d dx d = = sin then d dx d dx And Eq. 55 becomes d dP 1 x2) + ( dx dx ( + 1) P = 0 (56) We won't belabor the solution of Legendre's Equation see J3.2but rather will just list the first several solutions We state Rodriques's formula without proof: P0 = 1 P1 = x 1 (57) P2 = ( 3x 2 1) 2 1 P3 = (5x 3 3x ) 2 1 P4 = ( 35x 4 30x 2 + 3) 8 P ( x) =
1 d 2 ! dx 1 (x 2 1) (58) This can be used to show that I ' 2 P ' ( x ) P ( x ) dx = 2 +1 1 ' (59) Conducting Sphere Example
We consider the example of a conducting sphere in a uniform external electric field.
r z a +z ( ) E 0z = E 0 r cos In general =0
(r, ) = Ar + B P (cos +1 r ) (59) as r
E 0 r cos = E 0 rP1 (cos ) A = E 0, = 1 A = 0, 1 (60) at r = a
B1 = A1a
3 = A1a + B1 P1 (cos 2 a
E 0a3 = r2 )=0
r cos (61)
(62) What about the other Bl's? Why must they also all be zero? Potential Expansions
Note that once the Al's and Bl's are determined, the potential (r, ) is uniquely determined throughout space. It is often possible to fix the Al`s and Bl`s by considering only a limited region of space. For example consider the potential due to a charge q a distance r' away from the origin along the z axis.
z It is easy to determine at any point r along the z axisi.e., r r' y (r = z) = q r r' q
x q r' = 1+ + r' r r r1 r = q q r = 1+ + r r' r' r1 r' r' r r r' 2 +
2 q = r q = r' =0 r' r r r' r > r' (61) r < r' + =0 Potential Expansions
Or more compactly q (r = z) = = r r' r> q =0 r< r> =
=0 B A r + +1 P (1) r (61) Hence for r > r' r < r' r> = r & r< = r' r> = r' & r< = r
q r> r< r> Now if we go "off axis" we get A = 0 & B = qr' q A = ( +1) & B = 0 r' (r, ) = P (cos ) (62)
x x' x x' r= x =0 A useful byproduct is the result cos = 1 x x' = 1 r> =0 r< r> P (cos ) (63) r'= x ' ...
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 '11
 MARLOW
 Magnetism, Boundary value problem, Boundary conditions, Green's function, DBC

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