Uniqueness theorem, boundary conditions, boundary value problems

Uniqueness theorem

Problem:

Prove, carefully explaining your reasoning, that the solution of ∇∙E = ρ/ε₀, ∇×E = 0, for E is unique.  You may assume that the sources of E are bounded in space and that therefore the field vanishes at sufficiently large distances from the sources.

Solution:

• Concepts:
  The uniqueness theorem, Green's first identity
  ∫_V (Φ∇²ψ + ∇Φ∙∇ψ) dV = ∮_S Φ(∂ψ/∂n) dS
  Green's first identity is derived starting with Green's theorem:
  ∫_V∇∙A dV = ∮_S A∙n dS
  Set A = Φ∇ψ, with Φ, ψ arbitrary scalar fields.
  ∇∙A = ∇∙(Φ∇ψ) =  Φ∇²ψ  +  ∇Φ∙∇ψ
  A∙n = Φ∇ψ∙n = Φ(∂ψ/∂n)
• Reasoning:
  We are asked to prove the uniqueness theorem.
• Details of the calculation:
  ∇×E = 0 <==> E = -∇Φ, E is the gradient of some scalar function.
  ∇·E = ρ/ε₀ <==> ∇²Φ = -ρ/ε₀.

  Consider the volume V bounded by surfaces at infinity.  Assume we know ρ(r) everywhere inside V and we know that Φ = 0 at the boundaries.
  Assume that two solutions for Φ exist that satisfy the same boundary conditions.
  ∇²Φ₁ = -ρ/ε₀,  ∇²Φ₂ = -ρ/ε₀.
  Let U = Φ₁ - Φ₂, then ∇²U = 0 and U = 0 at the boundaries.
  Apply Green's first identity with Φ = ψ = U.
  ∫_V (U∇²U + ∇U∙∇U) dV = ∮_S U(∂U/∂n) dS
  The right hand side is zero.  ∇²U = 0.  Therefore 
  ∫_V| ∇U|² dV = 0.
  Since |∇U|² is a non-negative everywhere, this implies ∇U = 0 everywhere inside V and U is constant.  Φ₁ and Φ₂ differ at most by a constant.  This constant is zero since Φ₁ = Φ₂ at the boundaries.  Φ is unique, E = -∇Φ is unique.

Boundary value problems, azimuthal symmetry

Problem:

A sphere of radius R carries a surface charge density σ = σ₀cosθ.
(a)  Derive the exact potential everywhere (both inside and outside the sphere) assuming that it vanishes as r goes to infinity.
(b)  Calculate the dipole moment of the charge distribution and deduce the approximate form of the potential at points far from the sphere (r >> R).
(c)  Compare (a) and (b).  What can you conclude about the higher multipoles?

Solution:

• Concepts:
  Boundary value problems,  Φ(r,θ) = ∑_n=0^∞[A_nrⁿ + B_n/rⁿ⁺¹]P_n(cosθ) is the general solution to Laplace's equation for problems with azimuthal symmetry.
• Reasoning:
  Φ(r,θ) = ∑_n=0^∞[A_nrⁿ + B_n/rⁿ⁺¹]P_n(cosθ) is the most general solution inside and outside of the sphere, since ρ = 0 inside and outside of the sphere.  To find the specific solution we apply boundary conditions.
• Details of the calculation:
  Assume Φ₁(r,θ) = ArP₁(cosθ) = Arcosθ inside the sphere and Φ₂(r,θ) = (B/r²)P₁(cosθ) = (B/r²)cosθ outside the sphere.  The symmetry of the situation suggests this form of the solution.  If we can satisfy the boundary conditions, the uniqueness theorem guaranties that we have found the only solution.  The potential vanishes as r goes to infinity.
  Boundary conditions:
  Φ is continuous across the boundary.  Φ₁(R,θ) = Φ₂(R,θ).  A = B/R³.
  (E₂ - E₁)·n₂ = σ/ε₀, dΦ₂/dr - dΦ₁/dr|_R = -σ/ε₀.  -2(B/R³)cosθ - Acosθ = -σ₀cosθ/ε₀.
  -2A -A = -σ₀/ε₀.  A = σ₀/3ε₀.
  Φ₁(r,θ) = (σ₀/3ε₀) r cosθ inside the sphere and Φ₂(r,θ) = (σ₀R³/3ε₀r²)cosθ outside the sphere. 
  (b)  dipole moment: p = ∫σ(r)r dA = k σ₀∫cosθ Rcosθ 2πR²sinθdθ = k σ₀(4π/3)R³.
  dipole potential: Φ_dip = (1/(4πε₀))p∙r/r³ = (σ₀R³/3ε₀r²)cosθ.
  The potential outside the sphere is a pure dipole potential.  The higher multipoles are zero.

Problem:

A hollow ball of radius R has a surface charge distribution which produces a potential on the surface of V(R,θ) = k(cos²θ + 1), where k is a constant and θ is the usual polar angle relative to the z-axis.
(a)  Find the potential at all points in space.
(b)  Show that the charge distribution on the ball is σ(R,θ) = (kε₀/R)(5cos²θ - (1/3)).

Solution:

• Concepts:
  General solution to Laplace's equation in situations with azimuthal symmetry, boundary value problems
• Reasoning:
  Φ(r,θ) = ∑_n=0^∞[A_nrⁿ + B_n/rⁿ⁺¹]P_n(cosθ) is the most general solution inside and outside of the sphere, since ρ = 0 inside and outside of the sphere.  To find the specific solution we apply boundary conditions.  The potential on the surface of the ball is given.
• Details of the calculation:
  (a)  Assume
  Φ₁(r,θ) = A₀ + A₂r²P₂(cosθ) inside the sphere and
  Φ₂(r,θ) = (B₀/r) + (B₂/r³)P₂(cosθ) outside the sphere.
  The symmetry of the situation suggests this form of the solution.  If we can satisfy the boundary conditions, the uniqueness theorem guaranties that we have found the only solution.  The potential vanishes as r goes to infinity.
  Boundary conditions:
  Φ₁(r,θ)  is finite at r = 0 and Φ₂(r,θ)  goes to zero as r goes to infinity.
  Φ is continuous across the boundary at r = R.
  A₀ = B₀/R,  A₂ = B₂/R⁵. 
  Φ(R,θ) = A₀ + A₂R²P₂(cosθ) = A₀ + A₂R²(3cos²θ -1)/2 = k(cos²θ + 1).
  3A₂R²/2 = k, A₀ - A₂R²/2 = k,  A₂ = 2k/(3R²),  A₀ = 4k/3.
  Φ₁(r,θ) = 4k/3 + (2k/(3R²))r²P₂(cosθ),  Φ₂(r,θ) = 4kR/(3r) + (2kR³/(3r³))P₂(cosθ) .
  
  (b)  (E₂ - E₁)·n₂ = σ/ε₀, ∂Φ₂/∂r - ∂Φ₁/∂r |_R = -σ/ε₀.
  ∂Φ₂/∂r = -4kR/(3r²) - (2kR³/r⁴)P₂(cosθ),  ∂Φ₁/∂r = (4k/(3R²))rP₂(cosθ).
  σ(R,θ)/ε₀ = 4k/(3R) + (2k/R)P₂(cosθ) + (4k/(3R))P₂(cosθ) 
  = 4k/(3R) + 10k/(3R))P₂(cosθ).
  σ(R,θ) = (kε₀/R)(4/3 + (10/3)(3cos²θ - 1)/2) = (kε₀/R)(5cos²θ - (1/3)).

Problem:

Consider a hollow sphere of radius R with a surface potential Φ(R,θ) = k sin²θ, where k is a constant and θ is the usual polar angle relative to the z-axis.
Find the potential everywhere inside the sphere.
Hint:
P₀(x) = 1
P₁(x) = x
P₂(x) = (3x² - 1)/2
P₃(x) = (5x³ - 3x)/2

Solution:

• Concepts:
  General solution to Laplace's equation in situations with azimuthal symmetry, boundary value problems
• Reasoning:
  Φ(r,θ) = ∑_n=0^∞[A_nrⁿ + B_n/rⁿ⁺¹]P_n(cosθ)
  is the most general solution inside and outside of the sphere, since ρ = 0 inside and outside of the sphere.  To find the specific solution we apply boundary conditions.  The potential on the surface of the sphere is given.
• Details of the calculation:
  Φ(R,θ) = k sin²θ = k(1 - cos²θ) = (2k/3)(P₀ - P₂(cosθ)).
  Assume
  Φ₁(r,θ) = A₀ + A₂r²P₂(cosθ) inside the sphere.
  The symmetry of the situation suggests this form of the solution.  If we can satisfy the boundary conditions, the uniqueness theorem guaranties that we have found the only solution. 
  Boundary conditions:
  Φ₁(r,θ) is finite at r = 0 .
  Φ is continuous across the boundary at r = R.
  A₀ = 2k/3,  A₂R² = -2k/3.  
  Φ₁(r,θ) = 2k/3 - (2k/3)(r²/R²)P₂(cosθ) = 2k/3 - (2k/3)(r²/R²)( 1 - (3/2)sin²θ)
  is the potential inside the sphere.

Problem:

A conducting sphere of radius a is located in an electric field that is uniform at infinity, i.e. E = E₀ k at infinity.   Put the origin of the coordinate system at the center of a sphere and set the potential Φ(0) = 0.
Solve for the electric potential  and the electric field everywhere by boundary value methods.

Solution:

• Concepts:
  Most general solution to Laplace's equation, boundary conditions
• Reasoning:
  Φ₁ = 0, E₁ = 0 inside the sphere since the interior of a conductor in electrostatics is field-free.  ∇²Φ₂ = 0 and outside the sphere since ρ = 0 outside the sphere.  We are looking for a solution for ∇²Φ₂ = 0, subject to the boundary conditions.
• Details of the calculation:
  We have azimuthal symmetry.
  Φ(r,θ) = ∑_n=0^∞[A_nrⁿ + B_n/rⁿ⁺¹]P_n(cosθ) is the most general solution of ∇²Φ = 0 for problems with azimuthal symmetry.
  Assume Φ₁ = 0 inside, Φ₂ = A₀' + A₁'rcosθ + B₀'/r + (B₁'/r²)cosθ outside.
  We assume all higher-order coefficients are zero.  If we find a solution with this assumption, it is the only solution.
  Boundary conditions:
  (i)  Φ is continuous at r = a .  0 = A₁'a + B₁'/a².
  (ii)  E₂ = E₀ at r >> a.
  -A₁'cosθ = E₀cosθ,  A₁'sinθ = -E₀sinθ,  A₁' = -E₀,  B₁' = E₀a³.
  Φ₂ = -E₀rcosθ  + (E₀a³/r²)cosθ.
  E₂ = -∇Φ₂.  E_2r = E₀cosθ  + 2(E₀a³/r³)cosθ,  E_2θ = -E₀'sinθ + (E₀a³/r³)sinθ.
  E₂ = E₀ + 2(E₀a³/r³)cosθ (r/r) + (E₀a³/r³)sinθ (θ/θ).
  E₂ = external field + field due to the polarized sphere.

Problem:

Two thin wires, each of length L, lie on the z-axis.  Wire 1 has a positive charge per unit length +λ and extends from (0, 0, L) to (0,0,0).  Wire 2 has a negative charge per unit length -λ and extends from (0,0,0) to (0,0,-L).

(a)  Find the potential Φ on the z-axis at z > L.
(b)  The potential Φ(r,θ) at a arbitrary points in space with r > L can be expanded in terms of Legendre polynomials,  Φ(r,θ) = ∑_n=0^∞[A_nrⁿ + B_n/rⁿ⁺¹]P_n(cosθ). 
Find the expansion coefficients.
Note: ln(1 + x) = ∑_n=1^∞(-1)ⁿ⁺¹xⁿ/n

Solution:

• Concepts:
  Symmetry, Φ(r,θ) = ∑_n=0^∞[A_nrⁿ + B_n/rⁿ⁺¹]P_n(cosθ) is the general solution to Laplace's equation for problems with azimuthal symmetry.
• Reasoning:
  Φ(r) = [1/(4πε₀)]∫ρ(r')dV'/|r - r'|.  We can evaluate this integral for r = rk. 
  For r > L, Φ(r) satisfies Laplace's equation, ∇²Φ(r) = 0.  The general solution is
  Φ(r,θ) = ∑_n=0^∞[A_nrⁿ + B_n/r^n-1]P_n(cosθ). 
  We can find the coefficients by comparing the general solution to the solution found by integration on the z-axis.
• Details of the calculation:
  (a)  Φ(z) = [1/(4πε₀)]∫₀^Lλdz'/(z - z') - [1/(4πε₀)] ∫_-L⁰λdz'/(z - z')
  = [1/(4πε₀)][-λ∫_z^z-Ldx/x + λ∫_z+L^zdx/x] = [1/(4πε₀)] [-λln((z - L)/z) + λln(z/(z + L))]
  =  -[λ/(4πε₀)]ln[(z² - L²)/z²].
  Φ(z) = -[λ/(4πε₀)]ln(1 - u), with u = L²/z².
  (b)  For r > L  ∑_n=0^∞(B_n/rⁿ⁺¹)P_n(cosθ).  Only the odd parity terms contribute and all A_n are zero since Φ = 0 at infinity. 
  We can therefore write  Φ(r,θ) = ∑_n=1^∞(B_2n-1/r²ⁿ)P_2n-1(cos(θ)). 
  On the positive z-axis  Φ(z) = ∑_n=1^∞(B_2n-1/z²ⁿ), since P_n(1) = 1.
  Expanding ln(1 - u) = ∑_n=1^∞(-1)ⁿ⁺¹(-u)ⁿ/n we have, equating the two expressions for Φ(z),
  ∑_n=1^∞(B_2n-1/z²ⁿ) = -[λ/(4πε₀)] ∑_n=1^∞(-1)ⁿ⁺¹(-u)ⁿ/n = [λ/(4πε₀)] ∑_n=1^∞(-1)²ⁿ⁺²uⁿ/n
  = [λ/(4πε₀)] ∑_n=1^∞L²ⁿ/(nz²ⁿ).
  Equating term with equal powers of z we have B_2n-1/z²ⁿ = [λ/(4πε₀)] L²ⁿ/(nz²ⁿ).
  B_2n-1 = [λ/(4πε₀)] L²ⁿ/n. 
  Φ(r,θ) = ∑_n=1^∞([λ/(4πε₀)] L²ⁿ/n)(1/r²ⁿ)P_2n-1(cos(θ)).

Problem:

A charge array consists of two charges, each of magnitude +q, located on the z-axis at (0, 0, ±a).
(a)  Find the potential V(0,0,z) at an arbitrary point z > a; then expand V(0, 0,z) in a power series in z.
(b)  Using this series as a "boundary condition", find the potential V(r,θ,φ) at an arbitrary location (r,θ,φ) with r > a.  An infinite series is acceptable.
(c)  Characterize the first three terms in 1/rⁿ.

Solution:

• Concepts:
  The electrostatic potential
• Reasoning:
  In regions with charge density ρ = 0 we have ∇²V = 0.  So in the region r > a we have
  ∇²V = 0.  For a problem with azimuthal  symmetry, the most general solution for ∇²V = 0 is
  V(r,θ) = Σ_n=0^∞(A_nrⁿ + B_nr^-(n+1))P_n(cosθ).
  On the positive z-axis P_n(cosθ) = 1 and V(z) = Σ_n=0^∞(A_nzⁿ + B_nz^-(n+1)).
  If we know the potential on the z-axis, we can solve for the expansion coefficients by equating term of equal power in z.
• Details of the calculation:
  (a)  V(z) = kq/(z - a) + kq/(z + a) = (kq/z)[(1 - a/z)^-1 + (1 + a/z)^-1].
  Assume a/z < 1.
  For x < 1 we have (1 ± x)^-1 = 1 ∓ x + x² ∓ x³ + ... .
  V(z) = (2kq/z)(1 + (a/z)² + (a/z) ⁴ + ...) = 2kq(1/z + a²/z³ + a⁴/z ⁵ + ...).
  
  (b)  V(z) = 2kq(1/z + a²/z³ + a⁴/z ⁵ + ...) = Σ_n=0^∞(A_nzⁿ + B_nz^-(n+1)).
  A_n = 0 for all n.  B_n = 0 for n = odd,  B_n = 2kqaⁿ for n = even.
  V(r,θ) = 2kqΣ_n=even aⁿr^-(n+1)P_n(cosθ).
  
  (c)  First term: Monopole term  ∝  1/r = 2kq/r. 
  Second term:  Dipole term  ∝  1/r² = 0.
  Third term: Quadrupole term  ∝  1/r³ = 2kqa²P₂(cosθ)/r³.

Boundary value problems, other symmetry

Problem:

A rectangular box occupying the region 0 ≤ x ≤ A, 0 ≤ y ≤ B, 0 ≤ z ≤ C, has an electrostatic potential V(x,y,z) = 0 on all faces except the face z  = C, where the potential is V₀.  Starting from Laplace's equation, find the potential V everywhere inside.

Solution:

• Concepts:
  General solution to Laplace's equation, boundary conditions
• Reasoning:
  ρ = 0 inside the box, ∇²V = 0 inside the box.
• Details of the calculation:
  In Cartesian coordinates ∇²V = ∂²V/∂x² + ∂²V/∂y² + ∂²V/∂z² = 0.
  Separation of variables:  V(x,y,z) = F₁(x)F₂(y)F₃(z).
  (1/F₁)∂²F₁/∂x² + (1/F₂)∂²F₂/∂y² + (1/F₃)∂²F₃/∂z² = 0.
  Since each term in the sum is a function of a different independent variable, each term must be equal to a constant and the constants have to sum to zero.
  Let (1/F₁)∂²F₁/∂x² = -a²,  ∂²F₁/∂x² + a²F₁ = 0.
  (1/F₂)∂²F₂/∂y² = -b²,  ∂²F₂/∂y² + b²F₂ = 0.  
  (1/F₃)∂²F₃/∂z² = (a² + b²),  ∂²F₃/∂z² - (a² + b²)F₃ = 0.
  Solutions that can fulfill the boundary conditions on all faces require
  F₁(x) = sin(ax), a = nπ/A,  F₂(y) = sin(by), b = mπ/B, n, m = 1. 2, 3, ... .
  F₃(z) = sinh(cz)  c = π(n²/A² + m²/B²)^½.
  The solution is
  V(x,y,z) = ∑_n=1^∞∑_m=1^∞D_nmsinh(c_nmz) sin(nπx/A) sin(mπy/B),
  subject to the boundary condition V(x,y,C) = V₀.
  If p, q = odd then ∫₀^Adx∫₀^Bdy V₀sin(pπx/A) sin(qπy/B) = [4V₀AB/(pqπ²)]
  [V₀AB/(pqπ²)] = D_pqsinh(c_pqC) ¼AB.
  Therefore D_nm = 16V₀/(nmπ²sinh(c_nmC)), n, m = odd.

Problem:

A long cylinder of radius a and relative permittivity ε carries a free surface charge density σ_f = σ₀cos²(φ).
(a)  Find the potential Φ inside and outside of the cylinder.
(b)  Find the electric field E inside and outside the cylinder.
(c)  Find the polarization surface charge density of the cylinder.

Let Φ =  Φ(ρ,φ),  0 ≤ θ ≤ 2π  in cylindrical coordinates.
Then the most general solution for ∇²Φ = 0 is
Φ(ρ,φ) = ∑_n=1^∞(A_ncos(nφ) +  B_nsin(nφ))ρⁿ + ∑_n=1^∞(A'_ncos(nφ) +  B'_nsin(nφ))ρ^-n + a₀ + b₀lnρ.

Solution:

• Concepts:
  Boundary value problems, 
  Φ(ρ,φ) = ∑_n=1^∞(A_ncos(nφ) +  B_nsin(nφ))ρⁿ + ∑_n=1^∞(A'_ncos(nφ) +  B'_nsin(nφ))ρ^-n + a₀ + b₀lnρ
• Reasoning:
  Φ = Φ(ρ,φ)  (0 ≤ φ  ≤ 2π) independent of z.  The problem involves the whole azimuthal range.
• Details of the calculation:
  (a)  σ = σ₀cos²(φ) = σ₀(½ + ½cos(2φ)).
  Assume Φ₁(ρ,φ) = A₂cos(2φ)ρ² inside,
  Φ₂(ρ,φ) = A'₂cos(2φ)ρ^-2 + a₀ + b₀lnρ outside.
  We assume all other coefficients are zero.  If we find a solution with this assumption, it is the only solution.
  Boundary conditions:
  ρ < a:  Φ is finite at the origin.  We choose Φ(0) = 0.
  ρ = a:  Φ is continuous, A₂cos(2φ)a² =  A'₂cos(2φ)a^-2 + a₀ + b₀lna.
  A₂a⁴ =  A'₂,  a₀ = -b₀lna.
  (D₂ - D₁)·n₂ = σ_f at ρ = a.
  (D₂ - D₁)∙(ρ/ρ) =  σ₀(½ + ½cos(2φ)) at ρ = a.
  E_ρ1 = -∂Φ₁/∂ρ = -2A₂cos(2φ)ρ,  E_ρ2 = -∂Φ₂/∂ρ = 2A₂cos(2φ)ρ^-3 - b₀/ρ.
  D_ρ1 =  -ε2A₂cos(2φ)ρ,  D_ρ2 = ε₀2A₂a⁴cos(2φ)ρ^-3 - ε₀b₀/ρ.
  (ε₀ + ε)2A₂cos(2φ)a - ε₀b₀/a =  σ₀(½ + ½cos(2φ)).
  A₂ = σ₀/(4a(ε₀ + ε)),  b₀ = -aσ₀/2ε_0.
  Φ₁(ρ,φ) = σ₀/cos(2φ)ρ²/(4a(ε₀ + ε)).
  Φ₂(ρ,φ) = a⁴σ₀cos(2φ)ρ^-2/(4a(ε₀ + ε)) - (½aσ₀/ε₀)ln(ρ/a).
  
  (b)   E = -∇Φ.  E_ρ = ∂Φ/∂ρ,  E_φ = (1/ρ)∂Φ/∂φ.
  E_in = -2A₂cos(2φ)ρ [ρ/ρ] + 2A₂sin(2φ)ρ² [φ/φ]
  = -σ₀cos(2φ)ρ/(2a(ε₀ + ε)) [ρ/ρ] + σ₀sin(2φ)ρ²/(2a(ε₀ + ε)) [φ/φ].
  E_out = (2A₂a⁴cos(2φ)ρ^-3 - b₀/ρ)[ρ/ρ] + 2A₂a⁴sin(2φ)ρ^-2 [φ/φ]
  = (σ₀a⁴cos(2φ)ρ^-3/(2a(ε₀ + ε)) + aσ₀/(2ε₀ρ) [ρ/ρ] + σ₀a⁴sin(2φ)ρ^-2/(2a(ε₀ + ε)) [φ/φ].
  
  (c)  (E₂ - E₁)·[ρ/ρ] = (σ_f + σ_p)/ε₀  at ρ = a.
  σ₀cos(2φ)/(2(ε₀ + ε)) + σ₀/(2ε₀) + σ₀cos(2φ)/(2(ε₀ + ε))
  = σ₀cos(2φ)/(ε₀ + ε) + σ₀/(2ε₀) = (σ_f + σ_p)/ε₀ .
  σ_p = ε₀σ₀cos(2φ)/(ε₀ + ε) - σ₀½cos(2φ) =  ½(ε₀ - ε)σ₀cos(2φ)/(ε₀ + ε).
  
  Or = σ_p = P·[ρ/ρ] evaluated at ρ = a.
  P = ε₀χ_eE_in = (ε₀ - ε₀)E_in,
  σ_p = ½(ε₀ - ε)σ₀cos(2φ)/(ε₀ + ε).