Real and imaginary part of the solution to the Laplace Equation violates uniqueness?

Question

I am trying to solve for the magnetic vector potential on $\mathbb{R}^2$. I have used the phasor formulation of Maxwell's equations and therefore I believe I am solving the equation on $\mathbb{C}^2$. The equation I have derived is $$\nabla^2\hat{A}(r,\phi)=-\frac{\mu_0}{r}\delta(r-r')\delta(\phi-\phi')$$ where $\hat{A}(r,\phi)$ is the complex magnitude of the solution $A(r,\phi)=\Re({\hat{A}(r,\phi)e^{j \omega t}})$.

I believe that the transformation to phasor form is mostly a mathematical convenience and that it should have no effect on my final solution. However, when I have a solution $\hat{A}(r,\phi)$ that satisfies the Poisson equation and Dirichlet boundary conditions ($\hat{A}(r\to \infty,\phi)=0$), I will then have two real solutions namely $A_1(r,\phi)=\frac{\hat{A}e^{j \omega t}+\hat{A}^*e^{-j \omega t}}{2}=\Re({\hat{A}(r,\phi)e^{j \omega t}})$ and $A_2(r,\phi)=\frac{\hat{A}e^{j \omega t}-\hat{A}^*e^{-j \omega t}}{2 j}=\Im({\hat{A}(r,\phi)e^{j \omega t}})$.

My difficulty is in choosing which one of these is the "correct" solution (i.e. why I should choose one and disregard the other). And also why this doesn't violate the uniqueness of the Laplace equation with Dirichet boundary conditions. My understanding is that if I solve the Poisson equation on $\mathbb{R}^2$ with Dirichlet boundary conditions then I am guaranteed a unique solution (See Wiki: Uniqueness Theorem for Poisson Equation).

Answer 1

You're presumably looking to solve the (real) equation $$ \nabla^2 A(r,\phi,t)=-\mu_0 J(r,\phi,t) $$ and have chosen a source of the form $J(r, \phi, t) = \frac{1}{r}\delta(r-r')\delta(\phi-\phi') \cos( \omega t)$. When you go to the complex domain, you have implicitly replaced $\Re (e^{i \omega t})$ in the "real source" with $e^{i \omega t}$ as a "complex source"; and so when you solve the problem and want to find the "real potential" from the "complex potential", you have to undo this step. In other words, if $J(r, \phi)e^{i \omega t}$ really stands for $\Re (J(r, \phi)e^{i \omega t} )$, then $\hat{A}(r, \phi)e^{i \omega t}$ really stands for $\Re (\hat{A}(r, \phi)e^{i \omega t} )$.

To see this more formally, consider the solutions to two real equations: \begin{align} \nabla^2 A_r(r,\phi,t)&=-\mu_0 J(r,\phi) \cos \omega t \tag{1} \\ \nabla^2 A_i(r,\phi,t)&=-\mu_0 J(r,\phi) \sin \omega t \tag{2} \end{align} The complex function $A_c(r, \phi, t) \equiv A_r + i A_i$ will then satisfy (by linearity) $$ \nabla^2 A_c(r,\phi,t)=-\mu_0 J(r,\phi) e^{i \omega t} \tag{3} $$ So one way to solve both (1) and (2) above is to solve (3) for $A_c$, and then take the real and imaginary parts of $A_c$. The real part of $A_c$ gives the response to the real part of the source, and the imaginary part of $A_c$ gives the response to the imaginary part of the source. (In typical situations we're not interested in this latter response, but it's still a valid solution to Poisson's equation for that source.)

See this excellent answer from Emilio Pisanty for more perspective on this last paragraph.