# What is meant when we say "any solution is *the* solution" due to the uniqueness theorem?

I understand the proofs for the uniqueness theorems in electrostatics, but I'm having trouble understanding their application to problem solving.

A classic example is a system of concentric shells of charge: When we seek to find the potential of the smaller sphere $$\phi_2$$, we say $$\phi_2 = Q_2/R_2 + \phi_1(R_2)$$ where $$\phi_1(x)$$ is the potential at $$x$$ due to the bigger sphere. We know that $$\phi_1(R_1) = Q_1/R_1$$, but what about $$\phi_1(R_2)$$? Well, we have the boundary condition $$\phi_1(R_1) = Q_1/R_1$$, and we need $$\nabla^2\phi_1 = 0$$ inside the boundary. Let's try $$\phi_1(x) = Q_1/R_1$$: this satisfies the boundary condition and Laplace's equation, so it must be the solution to this configuration. From this we can conclude that $$\phi_2 = Q_2/R_2 + \phi_B(R_2) = Q_2/R_2 + Q_1/R_1$$

Now my question is: if we legitimately guess a solution that has no foundation in any physical deduction, and it just so happens to satisfy the Laplace equation and fulfill the boundary conditions, is it the solution? I know there are no other equations $$f_n(x)$$ that satisfy Laplace's equation and take on the value $$Q_1/R_1$$ when $$x=R_1$$ (there's the canonical proof by contradiction), but this confuses me since it suggests that I can always "guess" that $$\phi(x)=\phi_{\text{boundary}} \, \forall x$$ and that this is correct because it satisfies the Laplace equation. But surely this wouldn't work. Why not?

• "it just so happens to be Laplacian" - presumably you meant "it just so happens to be a solution of the Laplace equation", i.e. $\nabla ^2 \phi=0$. I'm not sure where you got the term, but it's not a broadly-used description for that purpose. Dec 16 '18 at 1:01
• Please clarify something: Is $Q_1$ the charge on sphere #1? Dec 16 '18 at 1:05
• Uniqueness theorem tells us that if we find a solution that obeys the PDE and the boundary conditions by any means it is THE solution. You are showing us the method of images, a very powerful technique for solving PDE with BC problems. The U.T. allows us to make an image source config that matches the B.C.
– user196418
Dec 16 '18 at 1:11
• @DanielSank Correct. Dec 16 '18 at 1:39
• While all mathematical formulations are correct, you are losing physical insights with your statement: "if we legitimately guess a solution that has no foundation in any physical deduction..". In this particular case there is indeed good physical foundation. For example you deduct from Gauss Law that field inside R1 due to Q1 must be zero hence potential must be constant and therefore must keep the value of the boundary. Dec 16 '18 at 10:16

if we legitimately guess a solution that has no foundation in any physical deduction, and it just so happens to satisfy the Laplace equation and fulfill the boundary conditions, is it the solution?

Yes. In my university it was known as the Method of Divine Inspiration. It works...

I know there are no other equations $$f_n(x)$$ that satisfy Laplace's equation and take on the value $$Q_1/R_1$$ when $$x=R_1$$ (there's the canonical proof by contradiction)

...and this is precisely why.

this confuses me since it suggests that I can always "guess" that $$\phi(x)=\phi_{\text{boundary}} \, \forall x$$ and that this is correct because it satisfies the Laplace equation. But surely this wouldn't work. Why not?

If your boundary conditions are that the potential must be held at the same value at every boundary, then yes, this will indeed be the solution.

On the other hand, if your boundary conditions require different potentials on different components of the boundary, then this obviously won't work. Similarly, if your problem statement includes a slightly different spin (such as a fixed total charge per component of the boundary, as in your example), and your constant-$$\phi$$ solution does not fit to those expectations, then it is obviously not the solution that you're looking for. It's still the solution, but it's the solution to a different problem.

• So when finding a solution for $\phi$, we need to verify that (1) $\nabla^2 \phi = 0$ (2) $\phi$ is the correct value at all boundaries and (3) the charge distribution implied by our choice of $\phi$ is correct? Dec 16 '18 at 1:46
• @TiwaAina (1) and (2) are sufficient; (3) is not necessary, as there are no charges in the region in which $\nabla^2\phi=0$. Using imaginary charges to inspire a choice for $\phi$ is perfectly valid, but of course those charges can't exist in the domain of interest, and must be imagined as being outside its boundary. Dec 16 '18 at 7:05
• @TiwaAina It depends on the details of the problem you're solving. If what's fixed is the charge on each surface, then you do need to check that that condition is satisfied. (and, frankly, you should completely disregard J. Murray's comment.) Dec 16 '18 at 13:25