LAPLACE'S AND POISSON'S EQUATIONS
∇ 2 Φ = −4πρ
Poisson's equation
In regions of no charges the equation turns into:
∇2Φ = 0
Laplace's equation
Solutions to Laplace's equation are called Harmonic Functions.
Properties of harmonic functions
1)
Principle of superposition holds
2)
A function Φ(r) that satisfies Laplace's equation in an enclosed volume
and satisfies one of the following type of boundary conditions on the
enclosing boundary is unique.
(a) the value of the function is specified on the whole boundary (Dirichlet
condition).
(b) the value of the normal derivative, n ⋅ gradΦ, is specified on the whole
boundary (Neumann condition).
(c) the Φ is specified on part of the boundary and n ⋅ gradΦ on the rest.
If Φ(r) satisfies Laplace's equation in a region V, bounded by the surface
S, Φ can attain neither a maximum nor a minimum within V.
Extreme values occur only at the surface
Discussion: If the potential is constant on a boundary of a volume not
containing any charges the potential has the same constant value within the
whole volume.
3)
Bo E. Sernelius
4:2
SOLUTION OF LAPLACE'S EQUATION WITH SEPARATION OF
VARIABLES.
There are eleven different coordinate systems in which the Laplace equation is
separable. We will here treat the most important ones: the rectangular or
cartesian; the spherical; the cylindrical.
The geometry of a typical electrostatic problem is a region free of charges
bounded by a surface of a given geometry. It can be of rectangular box type,
spherical, cylindrical or of some other type. The standard method then is to
choose a coordinate system in which the boundary surface coincides with the
surface where one of the coordinates is constant.
In the special case of a 2D configuration, where the bounding surface and the
boundary conditions on the surface only depend on two variables, one may use
conformal mapping to go from one geometrical shape to another.
Other situations are too complicated to solve by these methods. Then one has
to rely on purely numerical methods, like solving finite-difference versions of
the Laplace's equation, finite element methods (FEM) or some other method.
Bo E. Sernelius
4:3
RECTANGULAR COORDINATES
∂ 2Φ ∂ 2Φ ∂ 2Φ
+ 2 + 2 =0
∂x 2
∂y
∂z
Assume we may write
Φ( x, y, z ) = X ( x )Y ( y) Z ( z )
YZ
d2X
dx 2
+ XZ
d 2Y
dy 2
+ XY
d2Z
dz 2
=0
Note that the derivatives are no longer partial.
1 d 2 X 1 d 2Y 1 d 2 Z
+
+
=0
X dx 2 Y dy 2 Z dz 2
The first term depends on x only, the second on y only and the third on z
only. The equation can only be valid if each of the terms is a constant:
1 d2X
= α' 2
2
X dx
1 d 2Y
= β' 2
2
Y dy
1 d2Z
= γ'2
2
Z dz
α ' 2 + β' 2 +γ ' 2 = 0
Since we are considering the electrostatic potential it is real valued. This means
that all these squares are real valued, but the last relation shows that the
constants themselves cannot all be real valued, neither can they all be
imaginary.
Bo E. Sernelius
4:4
We can only have the following cases
a)
b)
c)
d)
two real, one imaginary
one real, two imaginary
one real, one imaginary, one zero
three zero
An imaginary separation constant leads to an oscillatory solution while a real
valued leads to an exponential.
Let us arbitrarily let α' and β' be imaginary:
α ' 2 ≡ −α 2
β' 2 ≡ − β 2
γ'2 ≡ γ 2
α, β and γ are all real valued.
d2X
dx
2
d 2Y
dy
2
d2Z
dz
2
+α2X = 0
+ β 2Y = 0
− γ 2Z = 0
γ 2 = α2 + β2 ; γ = α2 + β2
X ( x ) = Ae iαx + Be − iαx
Y ( y) = Ce iβy + De − iβy
Z ( z ) = Eeγz + Fe −γz
Bo E. Sernelius
4:5
The complete solution is
Φ( x, y, z ) = X ( x )Y ( y) Z ( z )
∞
∑ ( Ar eiα r x + Br e −iα r x )(Cs eiβs y + Ds e −iβs y )
r ,s =1
(
⋅ Ers eγ rs z + Frs e −γ rs z
)
Short hand notation:
Φ( x, y, z ) ~ e ± iαx e ± iβy e ±γz
All the constants will be determined from the boundary conditions of the
problem.
Bo E. Sernelius
4:6
SPHERICAL COORDINATES
1 ∂  2 ∂Φ 
1
∂ 
∂Φ 
1
∂ 2Φ
∇ Φ= 2
r
+
+
=0
sin θ
∂θ  r 2 sin 2 θ ∂ϕ 2
r ∂r  ∂r  r 2 sin θ ∂θ 
2
Φ(r, θ , ϕ ) = R(r ) P(θ )Q(ϕ )
1
1
d  2 dR 
d 
dP 
d 2Q
r
+
sin θ
+
=0
dθ  r 2 Q sin 2 θ dϕ 2
r 2 R dr  dr  r 2 P sin θ dθ 
1
multiply with r 2 sin 2 θ :
sin 2 θ d  2 dR  sin θ d 
1 d 2Q
dP 
sin θ
r
+
=−
R dr  dr 
P dθ 
dθ 
Q dϕ 2
The left-hand side depends only on r and θ, while the right-hand side depends
only on φ. Thus the two sides must be a constant, m2.
d 2Q
dϕ
2
+ m 2 Q = 0 ; Q(ϕ ) ~ e ± imϕ ; m = 0,1, 2 …
Note: If the physical problem limits φ to a restricted range m can be a noninteger.
Now we return to the left-hand side and rearrange the terms:
1 d  2 dR 
1
d 
dP 
m2
r
=−
sin θ
+
R dr  dr 
P sin θ dθ 
dθ  sin 2 θ
The new left-hand side depends only on r and the right-hand side on only θ.
Thus, they must be a constant, l(l+1).
Bo E. Sernelius
4:7
We get
d  2 dR 
r
− l(l + 1) R = 0
dr  dr 
and
1 d 
dP  
m2 
+ l(l + 1) − 2  P = 0
sin θ
dθ  
sin θ dθ 
sin θ 
To solve the first, we make the ansatz: R = Ar α and obtain the two solutions
rl and r-(l+1). The general solution is then
1
Rl (r ) = Al r l + Bl l +1
r
For the polar-angle function P(θ) it is customary to make the substitution
cos θ → x ; −
1 d
d
→
sin θ dθ
dx
This gives
d 
m2 
2 dP  
+ l(l + 1) −
1− x
P = 0
dx 
dx  
1 − x 2 
(
)
We will first limit ourselves to axial or azimuthal symmetry.
Bo E. Sernelius
4:8
Axial symmetry
(
) dx
2
dP
2 d P
1− x
−
2
x
+ l(l + 1) P = 0
2
dx
Legendre's equation
Note that if x=±1 are excluded from the problem l may be non-integer.
The solution is the Legendre polynomial of order l: Pl (cos θ )
Thus we have the general solution to Laplace's equation in spherical
coordinates for the special case of axial symmetry as:
Φ( r , θ ) =
∞

1 
∑  Al r l + Bl r l +1  Pl (cosθ )
l =0
The Legendre polynomials can be obtained from
1 dl
(
)
Pl ( x ) = l
x2 − 1
l
2 l! dx
l
Rodrigues' formula
or from the generating function
F( x, µ ) =
1
(1 − 2 xµ + µ )
2 12
=
∞
∑ µ l Pl ( x )
l =0
or from recursion relations such as:
(l + 1) Pl +1 ( x ) = (2l + 1) xPl ( x ) − lPl −1 ( x )
or
(1 − x 2 ) dPdxl = −lxPl ( x) + lPl−1( x)
Bo E. Sernelius
4:9
The polynomials form a complete, orthogonal set of functions in the domain
-1≤x≤1 (0≤θ≤π)
f ( x) =
∞
∑ Al Pl ( x )
l =0
2l + 1
Al =
2
1
∫ f ( x )Pl ( x )dx
−1
Bo E. Sernelius
4:10
General case, no axial symmetry.
In this case we have in general a non-zero m value and the differential equation
for P is more elaborate. The Legendre polynomials are replaced by the
associated Legendre polynomials, Plm (cos θ ) . For a given l-value there are
2l+1 possible m-values: m = 0, ±1, ±2,, ±3, ...
There is a more general Rodrigues' formula for these functions:
( −1)
Plm ( x ) = l
m
2 l!
l+m
l
2 m2 d
2
;
1− x
x
−
1
l+m
(
)
dx
(
)
( −l ≤ m ≤ +l )
For any given m the functions Plm (cos θ ) and Pl'm (cos θ ) are orthogonal and
the associated Legendre polynomials for a fixed m form a complete set of
functions in the variable x.
The product of Plm ( x ) and e imϕ forms a complete set for the expansion of an
arbitrary function on the surface of a sphere. These functions are called
spherical harmonics.
2l + 1 (l − m)! m
Pl (cos θ )e imϕ
4π (l + m)!
Ylm (θ , ϕ ) =
They are orthonormal
m
m'
∫4π Yl (θ , ϕ )Yl' * (θ ,ϕ )dΩ
2π
π
dϕ sin θdθYlm (θ , ϕ )Yl'm' * (θ , ϕ ) = δ ll' δ mm'
0
0
=∫
∫
Bo E. Sernelius
f (θ , ϕ ) =
∞
4:11
l
∑ ∑ Clm Ylm (θ ,ϕ )
l = 0 m =− l
and
Clm = ∫
4π
f (θ , ϕ )Ylm * (θ , ϕ )dΩ
The general solution to Laplace's equation in terms of spherical harmonics is
Φ( r , θ , ϕ ) =
∞
l
 m l
m 1  m
A
r
+
B
l l +1 Yl (θ , ϕ )
 l
r 
l = 0 m =− l
∑ ∑
Bo E. Sernelius
4:12
CYLINDRICAL COORDINATES
1 ∂  ∂Φ  1 ∂ 2 Φ ∂ 2 Φ
∇ Φ=
r
+
+ 2 =0
r ∂r  ∂r  r 2 ∂θ 2
∂z
2
Φ(r, θ , z ) = R(r )Q(θ ) Z ( z )
1 d  dR(r ) 
1 d 2 Q(θ )
1 d 2 Z(z)
+
+
=0
r
rR(r ) dr  dr  r 2 Q(θ ) dθ 2
Z ( z ) dz 2
r d  dR(r ) 
r 2 d 2 Z(z)
1 d 2 Q(θ )
2
+
r
=
−
=
n
R(r ) dr  dr  Z ( z ) dz 2
Q(θ ) dθ 2
d 2Q
dθ
2
+ n2Q = 0
Q(θ ) ~ e ± inθ ; n = 0,1, 2,… (n may sometimes be non-integer)
1 d  dR  n 2
1 d2Z
r
− 2 =−
= −k 2
2
rR dr  dr  r
Z dz
d2Z
dz
2
− k2Z = 0
Z ( z ) ~ e ± kz
r
(
)
d  dR 
r
+ k 2r 2 − n2 R = 0


dr dr
Bo E. Sernelius
Cylindrical symmetry and Cylindrical Harmonics
Then we may let k vanish and
r
d  dR 
r
− n2 R = 0


dr dr
The n = 0 term has to be treated separately
 A0 + B0 ln r, (n = 0)

Rn (r ) = 
1
n
A
r
B
+
n n , (n = 1, 2, 3…)
 n
r

C0 [ + D0θ ], (n = 0)
Qn (θ ) = 
Cn cos nθ + Dn sin nθ , (n = 1, 2, 3…)
General solution in cylindrical coordinates with no z-dependence.
∞
1
Φ(r, θ ) = A0 + B0 ln r + ∑  An r n + Bn n [Cn cos nθ + Dn sin nθ ]
r 
n =1 
The terms are called cylindrical harmonics.
4:13
Bo E. Sernelius
4:14
No cylindrical symmetry and Bessel functions.
Now, we have to keep the constant k in the differential equation for R.
r
(
)
d  dR 
r
+ k 2r 2 − n2 R = 0
dr  dr 
To solve this one usually makes the substitution
u = kr ;
d
d
=k
dr
du
This leads to Bessel's equation:
(
)
2
dR
2
2
2d R
u
+
u
+
u
−
n
R=0
2
du
du
The solution to this equation is the so-called Bessel function of order n, Jn(u).
J-n(u) is also a solution. These are linearly dependent for integer orders but not
for non-integer orders.
One usually introduces another function instead of J -n (u), the so-called
Neumann function or Bessel function of the second kind, Nn(u).
J (u) cos nπ − J − n (u)
N n (u ) = n
sin nπ
General solution to Bessel's equation may be written as
Rn (kr ) = An J n (kr ) + Bn N n (kr )
Jn(u) is regular at origin and at infinity.
Nn(u) is not regular at origin but at infinity.
Bo E. Sernelius
4:15
The general solution to Laplace's equation in cylindrical coordinates can be
written as the Fourier-Bessel expansion:
Φ( r , θ , z ) ~
∑ [ Amn Jn (km r ) + Bmn Nn (km r )]e ±inθ e ± km z
m, n
Other useful properties of the Bessel function
Let kmρ be the mth root of Jn(kr), i.e., Jn(kmρ) = 0.
Then Jn(kmr) form a complete orthogonal set for the expansion of a function
of r in the interval 0≤r≤ρ.
f (r ) =
∞
∑ Dmn Jn (km r )
m =1
(for any n)
∞
f (r ) J n (km r )rdr
0
2
Dmn = 2 2
∫
ρ J n +1 (km ρ )
analogous to the Fourier transform.
Fourier-Bessel series
Bo E. Sernelius
4:16
Discussion: If we had chosen +k2 instead of -k2:
1 d  dR  n 2
1 d2Z
r
− 2 =−
= +k 2
2
rR dr  dr  r
Z dz
The z- dependence had been plane waves instead of exponentials and the r
dependence had been found as solutions to the modified Bessel equation:
(
)
2
dR
2
2
2d R
u
+
u
−
u
+
n
R=0
2
du
du
with the modified Bessel functions In(u) and K n(u) as solutions. The first is
bounded for small arguments and the second for large.
Thus, an alternative expression for the general solution is
Φ( r , θ , z ) ~
∑ [ Amn In (km r ) + Bmn Kn (km r )]e ±inθ e ±ikm z
m.n