Ch3.1 - Boundary values problems in electrostatics The...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Boundary values problems in electrostatics The uniqueness theorem If we specified the potentials at all the boundaries (i.e. surfaces) of a charge-free region of space, and want to know the potentials at every points in the region, we have a boundary value problem . Question: Given the boundary condition, does it uniquely specify the physical situation? In other words, is the potential function found from this specification unique ? It is easy to show that merely specifying the boundary values does not guarantee a unique solution. Example: suppose the boundaries values are fixed by : f(0) = 0 , f(1) = 0 , we can immediately find at least two possibilities for f : f(x) = x 2 – x or f(x) = sin 2 π x For an electrostatic boundary value problem, however, we can safely say that the problem is indeed uniquely specified. This is a consequent of the fact that the potential has to satisfy the Laplace’s equation V 2 = 0. An important property of the solutions to the Laplace’s equation: V( r ) cannot have a local maximum or minimum (i.e. extrema) within the volume of the region, the extrema can only occur at the boundary . We first illustrate this by the simple case of 1-D Laplace’s equation : 2 2 dx V d = 0. Clearly the solution is V(x) = ax + b , where a and b are constants. A straight line either keeps on increasing, or decreasing, depending on the sign of a , obviously V(x) cannot have a local maximum or minimum, except at the boundary point.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
In 3-D, it can be shown that if V is a solution to the Laplace’s equation, then V at a point r is equal to the average of the value of V on any spherical surface centred at r , i.e. V(x) = sphere 2 da V R 4 1 π (2.1) The proof of (2.1) is given in Section 3.1.4 of Griffiths. With (2.1), it is easy to deduce that, just like the 1-D case, V cannot have a local maximum or minimum, except at the boundary. Reason: if V has a local maximum at a point r , then for a suitably chosen spherical surface, the value of V at every points on the sphere would be smaller than V( r ) . This however, will contradict (2.1). We can now prove the Uniqueness theorem : The solution to Laplace’s equation, V( r ) , in a region is uniquely determined if the values of V at the boundary surfaces of the region are specified. Proof : Suppose we have two distinct solutions V 1 and V 2 to the Laplace’s equation, i.e. 1 2 V = 0 and 2 2 V = 0 at every point in the region, and V 1 and V 2 satisfy the same boundary conditions. Define a function V 3 by V 3 = V 1 – V 2 . Clearly, V 3 is also a solution to Laplace’s equation. Therefore, the extrema of V 3 must occur only at the boundary. But the boundary condition for V 3 now become V 3 = 0 , since V 1 = V 2 at the boundary. With both the maximum and minimum equal to zero, V 3 must equal zero everywhere in the region. Thus, V 1 must equal V 2 everywhere in the region and could not be distinct.
Background image of page 2
Important consequent of the uniqueness theorem : For a given boundary value problem, if we can somehow guess a solution that satisfies the boundary condition, then this solution must be the only solution to the problem. Hence, we don’t necessarily
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 40

Ch3.1 - Boundary values problems in electrostatics The...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online