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Let us say we have a simple boundary value problem (BVP) in spherical coordinates : $$\Delta \phi = 0$$

along with $\phi=1$ at $\,r=1$, and $\,\phi =0\,$ at $\,\infty$.

A surface sphere with radius $R_{1}=1$ and charge $Q=1$ on it defines a potential $\phi_{1}$ that satisfies the BVP above. Another sphere with radius $R_{2}=0.5$ and charge $Q=1$ defines a potential $\phi_{2}$ that too satisfies the BVP.

According to the uniqueness theorem, the solution to the BVP is unique, however, both $\phi_{1}$ and $\phi_{2}$ are solutions, but these two solutions are different in the region $r<1$ since

$$\phi_{1}(r)=1 ~~~~~~~~\text{for}~~ r<1$$ whereas

$$\phi_{2}(r)=\begin {cases} 2 & r<0.5 \\ 1/r & 0.5<r<1 \end{cases}$$

So, besides the value of the potential on some surfaces, there must additional conditions that need to be specified in order to find the unique potential, or am I missing something?

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  • $\begingroup$ You have specified boundary conditions for the space $1\le r\le\infty$. $\endgroup$
    – G. Smith
    Commented Sep 5, 2019 at 19:43
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    $\begingroup$ If you have charge, you no longer have Laplace's equation. $\endgroup$
    – Javier
    Commented Sep 5, 2019 at 19:59
  • $\begingroup$ @G.Smith so the solution is unique within $1\le r \le \infty$, but how can we exactly specify the boundary conditions for the space $r<1$? Do we simply need to specify $\phi(0,0,0)$? $\endgroup$
    – Hilbert
    Commented Sep 5, 2019 at 20:08
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    $\begingroup$ The boundary of the ball $r<1$ is the sphere $r=1$ so specifying the potential on the bounding sphere should be sufficient. $\endgroup$
    – G. Smith
    Commented Sep 5, 2019 at 20:20
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    $\begingroup$ @G.Smith " specifying the potential on the bounding sphere should be sufficient." but this is what we did when we set $\phi$ to $1$ at $r=1$, so we specified the boundary condition for the space $r<1$, not $1\le r \le \infty$, as suggested in the first comment. $\endgroup$
    – Hilbert
    Commented Sep 5, 2019 at 20:31

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Your two functions $\phi_1$ and $\phi_2$ satisfy $$\phi_1 = \phi_2 = \frac1r, ~~~~~r > 1,$$ and thus the uniqueness theorem has indeed worked out.

Recall that the uniqueness theorem says that if you give the values of the potential on a boundary of some space $V$ then the potential inside that space $V$ is uniquely specified, if the equation $\Delta \phi = 0$ holds throughout that space. In this case your space is $$V = \mathbb R^3 - B_3(1)$$ where $B_n(r)$ is the closed $n$-ball of radius $r$ centered on the origin, whose boundary is the $(n-1)$-sphere, $S_{n-1}(r)$. This space can be specified very cleanly in spherical coordinates as $r > 1$. The boundary of this space is accordingly $S_2(1) \cup \infty,$ which you have specified the conditions on perfectly.

You appear to be confused about the two functions being different inside that ball $B_3(1).$ The theorem doesn’t say anything about what is happening outside of that volume where $\Delta \phi = 0$ is happening. That assumption is a crucial part of the theorem.

One of your counterexamples breaks this assumption with $\Delta \phi \ne 0$ on the sphere $S_2(1)$, the other breaks this assumption on the sphere $S_2(1/2).$ They both agree within $V$ because that is what the uniqueness theorem guarantees; they both disagree within $\mathbb R^3 - V$ because the precondition for the theorem, $\Delta \phi = 0,$ does not hold in that space.

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