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

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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.

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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.
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

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

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