# Existence of solution to Poisson's equation

1. Jan 5, 2007

### Manchot

I'm reteaching electrodynamics to myself on a more rigorous footing, and I'm trying to prove to myself that setting the divergence of the vector potential is justified using a gauge shift. I could use the Helmholtz theorem to do this, but the problem with this from my perspective is that I haven't actually justified the full version of the theorem, only the weaker version which requires that a vector function decay to zero faster than 1/r at infinity. This isn't a problem for the field quantities (since all physical fields decay like 1/r^2); nevertheless, they do pose a problem for the potential quantities (which in general will not even decay). Basically, given a vector potential $$\vec{A}$$, I want to show that fixing the divergence of the gauge-shifted potential $$\vec{A}\prime$$ to some scalar function $$D$$ is equivalent to adding the gradient of some some scalar function $$\phi$$.

$$\nabla \cdot \vec{A}' = D$$
$$\nabla \cdot (\vec{A} + \nabla \phi) = D$$
$${\nabla}^2 \phi = D - \nabla \cdot \vec{A}$$

Since the right-hand side is just some function of position, proving that the divergence can be adjusted by adding the gradient of a scalar amounts to proving that Poisson's equation has a solution for an arbitrary source term. No boundary conditions are specified, so I would expect that there are actually an infinite number of solutions; however, I cannot prove this. Does anyone have any insights? Thanks.

Last edited: Jan 5, 2007
2. Jan 6, 2007

### Hurkyl

Staff Emeritus
Can we solve this system of equations?

$$\nabla \cdot \mathbf{X} = \theta$$
$$\nabla \times \mathbf{X} = 0$$

If we know $\nabla \times \mathbf{V} = 0$, can we solve this equation?

$$\nabla \varphi = \mathbf{V}$$

If so, then we can chain these results to solve the Poisson equation.

$$\nabla^2 \varphi = \theta$$

If you know a counterexample to one of the above questions, I suspect you could use it to construct a Poisson equation without a solution.

3. Jan 7, 2007

### Manchot

Thanks for the response, Hurkyl. I've been thinking about it some more, and I've been able to informally construct a solution to the equation, but I'm still not quite satisfied. Since there are no boundary conditions specified, and I'm just trying to construct any solution, I arbitrarily specified that

$$\phi (x=0) = \frac{\partial \phi}{\partial x} = 0$$

It would then seem to me that if you numerically integrated with a small enough dx, you could construct phi as follows:

$$\phi (x+dx,y,z) = \phi (x,y,z) + \frac{\partial \phi}{\partial x} (x,y,z) \cdot dx$$

$$\frac{\partial \phi}{\partial x} (x+dx,y,z) = \frac{\partial \phi}{\partial x} (x,y,z) + \frac{{\partial}^2\phi}{\partial {x}^2} (x,y,z) \cdot dx$$

$$= \frac{\partial \phi}{\partial x} (x,y,z) + (f(x,y,z) - \frac{{\partial}^2\phi}{\partial {y}^2}- \frac{{\partial}^2\phi}{\partial {z}^2}) \cdot dx$$

(f(x,y,z) is the source term.) Obviously, this method fails for some source terms, like the pathological sin(1/x) or any function which blows up, but other than those cases, it seems to be pretty solid. I've tried to "un-discretize" the construction to form integrals, but the expression gets pretty messy quickly. Could anyone explain how to to it a little more formally, since I'm still not convinced?

Last edited: Jan 7, 2007
4. Jan 7, 2007

### Hurkyl

Staff Emeritus
5. Jan 7, 2007

### Manchot

Yeah, but AFAIK, the problem with that solution is that the integral will only converge if the source term decays faster than 1/r^2, which is not in general the case when dealing with potentials.