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out to be incorrect.
Hence, there is a unique solution to Laplace’s equation satisfied given boundary conditions.
3.2 Direct Integration
Note: The potential field is a function of only one variable.
g, n S
where g is a continuous function.
3. Mixed boundary conditions
The Uniqueness Theorem
The uniqueness theorem states that there is only one (unique) solution to Poisson’s or Laplace’s equation satisfied given boundary conditions.
The integral is equal to zero since 3 0 on the surface S.
Thus,
S33 d S 0 V 3 2 dV 0
The integral can be zero is if 3 is a constant
3 1 2 constant
We know that 3 0 on the surface S, so we get
The types of boundary conditions
Consider a region V bounded by a surface S.
1. Dirichlet boundary conditions specify the potential
function on the boundary.
Example 3.2.1 The inner conductor of radius a of a coaxial cable is held at a potential of U while the outer conductor of radius b is grounded. Determine (a) the potential distribution between the conductors, (b) the surface charge density on the inner conductor, and (c) the capacitance per unit length. Solution
f, S
where f is a continuous function.
1＝2
1
1
n
2
2
n
S
2. Neumann boundary conditions specify the normal
derivative of the potential function on the boundary.
21 0
and
22 0
Let
3 1 2
then
23 0
Applying Green’s first identity
u2vdV u v dV uv d S
V
V
S
We have
V
323
3
2
dV
V 3 2 dV
S 33
dS
where volume V bounded by the enclosed surface S.
Poisson’s equation
2＝－ V
Laplace’s equation 2 0
Boundary conditions
f, S
g,
n S
Prove (proof by contradiction)
Consider a volume V bounded by some surface S. Suppose that we are given the charge density
n
2
2
n
S
Equations
dV dq
2＝－ V
or 2＝0
V
2.The calculation methods of boundary value problems
(1)Analytical method
(2)Numerical method
3.1 Types of Boundary Conditions and Uniqueness Theorem
3 0
That is,
1 2
Throughout V and on surface S. Our initial assumption
that 1 and 2 are two different solutions of Laplace’s
equations, satisfying the same boundary conditions, turns
Since the two conductors of radii a and b form
equipotential surfaces, the potential must be a functionቤተ መጻሕፍቲ ባይዱof only.
Thus, Laplace’s equation reduces to
1 d d 0 (a b)
d d
Integrating twice, we obtain
3 Boundary Value Problems
1.The types of problems in electrostatic field (1) Distribution problems (2) Boundary value problems
Boundary conditions
1＝2
1
1
V 0 throughout V and the value of the scalar
potential on surface S.
Assume that there exist two solutions 1 and 2 of
Laplace’s equation subject to the same boundary conditions. Then,
Hence, there is a unique solution to Laplace’s equation satisfied given boundary conditions.
3.2 Direct Integration
Note: The potential field is a function of only one variable.
g, n S
where g is a continuous function.
3. Mixed boundary conditions
The Uniqueness Theorem
The uniqueness theorem states that there is only one (unique) solution to Poisson’s or Laplace’s equation satisfied given boundary conditions.
The integral is equal to zero since 3 0 on the surface S.
Thus,
S33 d S 0 V 3 2 dV 0
The integral can be zero is if 3 is a constant
3 1 2 constant
We know that 3 0 on the surface S, so we get
The types of boundary conditions
Consider a region V bounded by a surface S.
1. Dirichlet boundary conditions specify the potential
function on the boundary.
Example 3.2.1 The inner conductor of radius a of a coaxial cable is held at a potential of U while the outer conductor of radius b is grounded. Determine (a) the potential distribution between the conductors, (b) the surface charge density on the inner conductor, and (c) the capacitance per unit length. Solution
f, S
where f is a continuous function.
1＝2
1
1
n
2
2
n
S
2. Neumann boundary conditions specify the normal
derivative of the potential function on the boundary.
21 0
and
22 0
Let
3 1 2
then
23 0
Applying Green’s first identity
u2vdV u v dV uv d S
V
V
S
We have
V
323
3
2
dV
V 3 2 dV
S 33
dS
where volume V bounded by the enclosed surface S.
Poisson’s equation
2＝－ V
Laplace’s equation 2 0
Boundary conditions
f, S
g,
n S
Prove (proof by contradiction)
Consider a volume V bounded by some surface S. Suppose that we are given the charge density
n
2
2
n
S
Equations
dV dq
2＝－ V
or 2＝0
V
2.The calculation methods of boundary value problems
(1)Analytical method
(2)Numerical method
3.1 Types of Boundary Conditions and Uniqueness Theorem
3 0
That is,
1 2
Throughout V and on surface S. Our initial assumption
that 1 and 2 are two different solutions of Laplace’s
equations, satisfying the same boundary conditions, turns
Since the two conductors of radii a and b form
equipotential surfaces, the potential must be a functionቤተ መጻሕፍቲ ባይዱof only.
Thus, Laplace’s equation reduces to
1 d d 0 (a b)
d d
Integrating twice, we obtain
3 Boundary Value Problems
1.The types of problems in electrostatic field (1) Distribution problems (2) Boundary value problems
Boundary conditions
1＝2
1
1
V 0 throughout V and the value of the scalar
potential on surface S.
Assume that there exist two solutions 1 and 2 of
Laplace’s equation subject to the same boundary conditions. Then,