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The essence of the statement of the uniqueness theorem is that if we know the conditions the limit that needs to be met by the potential of the system, then we find the solution of the system , then that solution is the only solution that exists and is not other solutions may be found. If we know potential solutions of a system, can we determine the type of system that generate this potential? If so, prove the statement! If no, give an example of a case that breaks the statement!

The essence of the statement of the uniqueness theorem is that if we know the conditions the limit that needs to be met by the potential of the system, then we find the solution of the system , then that solution is the only solution that exists and is not other solutions may be found. If we know potential solutions of a system, can we determine the type of system that generate this potential? If so, prove the statement! If no, give an example of a case that breaks the statement!

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The essence of the statement of the uniqueness theorem is that if we know the conditions the limit that needs to be met by the potential of the system, then we find the solution of the system , then that solution is the only solution that exists and is not other solutions may be found. If we know potential solutions of a system, can we determine the type of system that generate this potential? If so, prove the statement! If no, give an example of a case that breaks the statement!

5.1.1 Uniqueness Theorem Let us consider two proposed solutions V1 and V2, each satisfying the boundary conditions and V²V = 0. The difference between these solutions, & = Vị – V2, then also satisfies V? = 0 and in addition satisfies = 0 or d4/an = 0 according to the boundary conditions specified. In either case, the product of & and (V4), must vanish on the boundary. Thus, with the aid of the divergence theorem and making use of the vanishing of the Laplacian of 4, we have SP (5–1) As (V4)? is nowhere negative, the integral can vanish only if (V&)² is identically zero. We conclude then that V = 0, implying & is constant. In the case that o was zero on the boundary (Dirichlet condition), it must be zero everywhere, implying V = V2. For the case that the normal derivative vanished (Neumann condition), V and V2 can differ only by a constant. This minor freedom in the latter case is not surprising; because we have essentially specified only the field at the boundary and we know that the field cannot determine the potential to better than an additive constant, we should have fully expected this inconsequential ambiguity in the solution. %3D The theorem we have just proved also proves incidently the useful observation that a charge-free region of space enclosed by a surface of constant potential has a constant potential and consequently a vanishing electric field within that boundary. We proceed now to explore a number of techniques for solving Laplace's equa- tion. Generally it will be best, if possible, to choose a coordinate system in which coordinate surfaces coincide with the boundary, as this makes the application of boundary conditions considerably easier. To clearly demonstrate the techniques we systematically explore one-, two- and three-dimensional solutions in a number of different coordinate systems.
Transcribed Image Text:5.1.1 Uniqueness Theorem Let us consider two proposed solutions V1 and V2, each satisfying the boundary conditions and V²V = 0. The difference between these solutions, & = Vị – V2, then also satisfies V? = 0 and in addition satisfies = 0 or d4/an = 0 according to the boundary conditions specified. In either case, the product of & and (V4), must vanish on the boundary. Thus, with the aid of the divergence theorem and making use of the vanishing of the Laplacian of 4, we have SP (5–1) As (V4)? is nowhere negative, the integral can vanish only if (V&)² is identically zero. We conclude then that V = 0, implying & is constant. In the case that o was zero on the boundary (Dirichlet condition), it must be zero everywhere, implying V = V2. For the case that the normal derivative vanished (Neumann condition), V and V2 can differ only by a constant. This minor freedom in the latter case is not surprising; because we have essentially specified only the field at the boundary and we know that the field cannot determine the potential to better than an additive constant, we should have fully expected this inconsequential ambiguity in the solution. %3D The theorem we have just proved also proves incidently the useful observation that a charge-free region of space enclosed by a surface of constant potential has a constant potential and consequently a vanishing electric field within that boundary. We proceed now to explore a number of techniques for solving Laplace's equa- tion. Generally it will be best, if possible, to choose a coordinate system in which coordinate surfaces coincide with the boundary, as this makes the application of boundary conditions considerably easier. To clearly demonstrate the techniques we systematically explore one-, two- and three-dimensional solutions in a number of different coordinate systems.
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