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Can the Electric field and/or electric potential be determined (uniquely) given only boundary conditions? For example I was wondering if we have coaxial cylinders with inner radius a and outer radius b with a<b. The voltage difference between a and b is V. Is this sufficient to determine (uniquely) an electric field? Or are there many electric fields that satisfy that the potential between the coaxial cylinders' radius a and b is V. If yes. How much information do we need to do so?

dfphysicsi
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  • Yes. In fact boundary conditions are necessary to obtain unique solutions. Usually the process is proving that there are circumstances (i.e., conditions) where uniqueness is met. In that case whatever solution you calculate you can be sure it’s unique (and that way you don’t even need to prove EXISTENCE of a solution to your problem). – Newbie Jan 23 '22 at 19:17
  • Not specifically related. But also just knowing the location of charges and currents, also does not uniquely determine the E and B field. The initial de/dt, db/dt needs to be known aswell ( which commonly we just set to be zero, but this may not be the case) – jensen paull Jan 23 '22 at 19:22

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Assuming electrostatics here for simplicity as it is hard to define unique potentials otherwise.

There is a uniqueness theorem that guarantees the solution of these sorts of boundary value problems is unique in the sense that if a solution to Laplace's equation $\nabla^2 V=0$ can be found that satisfies the boundary conditions, then the solution is unique.

I haven't been able to find a proof succinct enough to include as an answer here (but I am interested to learn of one.).

There is one here How do I show that the Laplace equation has a unique solution under the Dirichlet closed-surface boundary condition? but I don't think it covers the general case. There's also one at Wikipedia: Uniqueness theorem for Poisson's equation. I can transcribe that one here, but I think it makes more sense as a reference.

GrapefruitIsAwesome
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    I remember seeing succinct proofs in Griffith’s electrodynamics book, not sure though. – Newbie Jan 28 '22 at 02:32
  • @Newbie I don't have a copy of Griffith's handy, but I I find something I'll add it. I'd also welcome someone who does have it to summarize the proof as an answer as well, that would be more authoritative than this response. – GrapefruitIsAwesome Jan 28 '22 at 03:10