Laplace's equation

**Laplace's equation** is a second-order partial differential equation. It is written as

∇
2

# V
0

{\displaystyle \nabla^2 V = 0}

where V is a scalar field that satisfies Laplace's equation.

## Specific Applications

### Electric Potential

One application of solutions to Laplace's equation is when calculating the electric potential of a charge distribution in the electrostatic approximation.

In electrostatics,

∇
⋅

E
→

=

ρ

ε
0

{\displaystyle \nabla \cdot \vec{E} = \frac{\rho}{\varepsilon\_0}}

 (Gauss' law)

∇
×

E
→

=
0

{\displaystyle \nabla \times \vec{E} = 0}

 (Faraday's law)

E
→

=
−
∇
V

{\displaystyle \vec{E} = -\nabla V}

 (Definition of electric potential)
Substituting the definition of electric potential into Gauss' law gives

∇
2

# V
−

ρ

ε
0

{\displaystyle \nabla^2V = -\frac{\rho}{\varepsilon\_0}}

Setting the charge density ρ to 0 - as is the case outside a localized charge distribution - gives Laplace's equation

∇
2

# V
0

{\displaystyle \nabla^2 V = 0}

The specific form of the potential depends on the boundary conditions - the values of the potential, or of the electric field, on surfaces surrounding the volume of zero charge density.

### Magnetic Vector Potential

In magnetostatics,

∇
×

B
→

=

μ
0

J
→

{\displaystyle \nabla \times \vec{B} = \mu\_0 \vec{J}}

 (Ampere's law)

∇
.

B
→

=
0

{\displaystyle \nabla . \vec{B} = 0}

 (Gauss' law for magnetism)

B
→

=
∇
×

A
→

{\displaystyle \vec B = \nabla \times \vec A}

 (Definition of magnetic vector potential)
Substituting the definition of magnetic vector potential into Ampere's law gives

∇
×
∇
×

A
→

=

μ
0

J
→

{\displaystyle \nabla \times \nabla \times \vec{A} = \mu\_0 \vec{J}}

For the case where the divergence of **A** is zero, this is equivalent to

∇
2

A
→

=
−

μ
0

J
→

{\displaystyle \nabla^2 \vec{A} = -\mu\_0 \vec{J}}

Setting the current density **J** to 0, such as outside of a localised current distribution, gives Laplace's equation

∇
2

A
→

=
0

{\displaystyle \nabla^2 \vec{A} = 0}

### Magnetic Scalar Potential

In magnetostatics, and in regions where there is no current,

∇
⋅

B
→

=
0

{\displaystyle \nabla \cdot \vec{B} = 0}

 (Gauss' law for magnetism)

∇
×

B
→

=
0

{\displaystyle \nabla \times \vec{B} = 0}

 (Ampere's law)

B
→

=
−
∇
Ψ

{\displaystyle \vec{B} = - \nabla \Psi}

 (Definition of magnetic scalar potential)
Substituting the definition of magnetic scalar potential into Gauss' law for magnetism gives Laplace's equation

∇
2

# Ψ
0

{\displaystyle \nabla^2 \Psi = 0}

## Solving by Separation of Variables

Often, solutions of Laplace's equation can be found by separating the variables - writing the function V of position as the product of functions along various axes.

### Cylindrical Coordinates, no Z dependence (V = RΘ)

If the potential is defined in cylindrical coordinates, and only depends on the radius r and the the angle θ, and is separable, then we can define

V
(
r
,
θ
# )
R
(
r
)
Θ
(
θ
)

{\displaystyle V(r, \theta) = R(r) \Theta (\theta)}

In cylindrical coordinates,

∇
2

# V

1
r

∂

∂
r

(

r

∂
V

∂
r

)

1

r
2

∂
2

V

∂

θ
2

∂
2

V

∂

z
2

=
0

{\displaystyle \nabla^2 V = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial V}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 V}{\partial \theta^2} + \frac{\partial^2 V}{\partial z^2}= 0}

Substituting in the expression for V gives

Θ
r

∂

∂
r

(

r

∂
R

∂
r

)

R

r
2

∂
2

Θ

∂

θ
2

=
0

{\displaystyle \frac{\Theta}{r} \frac{\partial}{\partial r} \left( r \frac{\partial R}{\partial r} \right) + \frac{R}{r^2} \frac{\partial^2 \Theta}{\partial \theta^2} = 0}

Rearranging to get one term dependent only on r and one term dependent only on θ gives

r
R

∂

∂
r

(

r

∂
R

∂
r

)

1
Θ

∂
2

Θ

∂

θ
2

=
0

{\displaystyle \frac{r}{R} \frac{\partial}{\partial r} \left( r \frac{\partial R}{\partial r} \right) + \frac{1}{\Theta} \frac{\partial^2 \Theta}{\partial \theta^2} = 0}

Since the first term depends only on r, and the second term depends only on θ, they must be both constant and of opposite sign. Calling this constant ℓ2, we have

r
R

∂

∂
r

(

r

∂
R

∂
r

)

=

ℓ
2

{\displaystyle \frac{r}{R} \frac{\partial}{\partial r} \left( r \frac{\partial R}{\partial r} \right) = \ell^2}

1
Θ

∂
2

Θ

∂

θ
2

=
−

ℓ
2

{\displaystyle \frac{1}{\Theta} \frac{\partial^2 \Theta}{\partial \theta^2} = -\ell^2}

The differential equation for R can be expanded to give

r
2

d
2

R

d

r
2

r

d
R

d
r

−
R

ℓ
2

=
0

{\displaystyle r^2 \frac{d^2 R}{d r^2 } + r \frac{dR}{dr} -R\ell^2 = 0}

This is an Euler-Cauchy equation. When ℓ is nonzero, this has power law solutions

R
(
r
# )

A
ℓ

r
ℓ

B
ℓ

r

−
ℓ

{\displaystyle R(r) = A\_\ell r^\ell + B\_\ell r^{-\ell}}

When ℓ is zero, this has logarithmic solutions

R
(
r
# )

A
0

ln
⁡
r

B
0

{\displaystyle R(r) = A\_0 \ln r + B\_0 }

The differential equation for Θ can be rearranged to give

d
2

Θ

d

θ
2

Θ

ℓ
2

=
0

{\displaystyle \frac{d^2 \Theta}{d \theta^2} + \Theta \ell^2 = 0}

When ℓ is nonzero, this has sinusoidal solutions

Θ
(
θ
# )

C
ℓ

sin
⁡
ℓ
θ

D
ℓ

cos
⁡
ℓ
θ

{\displaystyle \Theta(\theta) = C\_\ell \sin \ell \theta + D\_\ell \cos \ell \theta}

When ℓ is zero, this has linear solutions

Θ
(
θ
# )

C
0

θ

D
0

{\displaystyle \Theta(\theta) = C\_0 \theta + D\_0 }

However, because θ is periodic, requiring Θ(0) = Θ(2π), some of these solutions cannot be valid. These invalid solutions are the ones for which ℓ is not an integer or for which D0 is nonzero.

Since Laplace's equation is linear, the general solution is the sum of all valid solutions, namely

# V

(

A
0

ln
⁡
r

B
0

)

D
0

∑

# ℓ
1

∞

(

A
ℓ

r
ℓ

B
ℓ

r

−
ℓ

)

(

C
ℓ

sin
⁡
ℓ
θ

D
ℓ

cos
⁡
ℓ
θ

)

{\displaystyle V = \left(A\_0 \ln r + B\_0 \right)D\_0 + \sum\_{\ell=1}^{\infty} \left( A\_\ell r^\ell + B\_\ell r^{-\ell} \right) \left( C\_\ell \sin \ell \theta + D\_\ell \cos \ell \theta \right) }

The coefficients can be determined by including boundary conditions.