# Do we need to specify additional conditions in order to find the unique potential satisfying Laplace's equation?

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

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?

• You have specified boundary conditions for the space $1\le r\le\infty$. Sep 5, 2019 at 19:43
• If you have charge, you no longer have Laplace's equation. Sep 5, 2019 at 19:59
• @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 $\phi(0,0,0)$? Sep 5, 2019 at 20:08
• The boundary of the ball $r<1$ is the sphere $r=1$ so specifying the potential on the bounding sphere should be sufficient. Sep 5, 2019 at 20:20
• @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. Sep 5, 2019 at 20:31

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.