# Maxwell Equations E,B Power Series

I am trying to solve Hughston Tod's Problem 2.13 and 2.14:

Suppose that $E_i$ and $B_i$ can each be developed into a power series in time: $E_i=\sum t^n E_i^n$, $B_i=\sum t^n B_i^n$ where the index $n$ (not a tensor index) runs from zero to infinity; $E_i^n$ and $B_i^n$ depend only on spatial coordinates. Under the assumption that the relevant series converge show that if $E_i^0$ and $B_i^0$ are specified, subject to $\nabla_i E_i^0=0$ and $\nabla_i B_i^0=0$, then Maxwell equations in vacuo determine $E_i$ and $B_i$ uniquely. Generalize the result to the case when charge and current are present.

So this is what I have now done: $$\epsilon_{ijk}\nabla_jE_k=-\dot{B}_i$$ Plugging in the power series expansion one gets: $$B_i^{m+1}=-\frac{\epsilon_{ijk}\nabla_jE_k^m}{m+1}$$ Similarly we yield for $E$: $$E_i^{m+1}=\frac{\epsilon_{ijk}\nabla_jB_k^m}{m+1}$$ Therefore: $$E_i^{m+2}=\frac{\epsilon_{ijk}\nabla_jB_k^{m+1}}{m+2}$$ And substituting for $B_k^{m+1}$ from the previous equation: $$E_i^{m+2}=\frac{\nabla_i \nabla_j E_j^m-\nabla_j\nabla_j E_i^m}{(m+1)(m+2)}$$ I am sure a similar expression can be found for $B$. But how to go on from here?

• If it's telling you explicitly to use the Maxwell equations, why are you using the (strictly weaker) wave equation instead? Dec 14 '16 at 17:16
• hmm, good point. I will try it with the MEs. Dec 14 '16 at 17:17
• There is a small mistake in your last equation. The second term in the numerator on the right should be $\nabla_j \nabla_j E_i^m$
– GRB
Dec 14 '16 at 22:27
• correct! thanks for bringing my awareness to that. Dec 14 '16 at 23:10
• You're welcome. By the way, even though it's useless (see my answer), you also have a divergence of the electric field ($\nabla_j E_j^m$) in your last equation and since you are dealing with the vacuum equations it should be equal to 0, simplifying the expression.
– GRB
Dec 14 '16 at 23:34

You already did most of the job, you just went past the solution.

Maxwell's equations in vacuum are: \begin{align} \vec{\nabla} \cdot \vec{E} &= 0 \quad & \vec{\nabla} \times \vec{E} & = -\frac{\partial \vec{B}}{\partial t}, \\ \vec{\nabla} \cdot \vec{B} &= 0 \quad & \vec{\nabla} \times \vec{B} & = \frac{1}{c^2} \frac{\partial \vec{E}}{\partial t} \end{align} where $\vec{E} = \sum \limits_{n=0}^{\infty} t^n \vec{E}_n$ and $\vec{B} = \sum \limits_{n=0}^{\infty} t^n \vec{B}_n$, and $\vec{E}_0$ and $\vec{B}_0$ are given.

Once you plug these two definitions in the curl equations, you obtain: $$\begin{cases} \vec{B}_{n+1} & = - \frac{1}{n+1} \vec{\nabla} \times \vec{E}_n \\ \vec{E}_{n+1} & = \frac{c^2}{n+1} \vec{\nabla} \times \vec{B}_n \end{cases}$$ You don't need to go any further now. You know the starting components $\vec{E}_0$ and $\vec{B}_0$, so you can recursively construct every other component and both $\vec{E}$ and $\vec{B}$ result to be completely determined. Seems a lot like the mathematically infamous "a solution exists", but that's exactly what the question asks. If you go further with the calculations, you end defining only half of the components of the fields.

For the general case, Maxwell's equations become: \begin{align} \vec{\nabla} \cdot \vec{E} &= \frac{\rho}{\varepsilon_0} \quad & \vec{\nabla} \times \vec{E} & = -\frac{\partial \vec{B}}{\partial t}, \\ \vec{\nabla} \cdot \vec{B} &= 0 \quad & \vec{\nabla} \times \vec{B} & = \frac{1}{c^2} \frac{\partial \vec{E}}{\partial t} + \mu_0 \vec{J} \end{align} The homogeneous curl equation remains the same: $$\vec{B}_{n+1} = - \frac{1}{n+1} \vec{\nabla} \times \vec{E}_n$$ while for the inhomogeneous we need to assume that $\vec{J} = \sum \limits_{n=0}^{\infty} t^n \vec{J}_n$ and thus: $$\vec{E}_{n+1} = \frac{c^2}{n+1} \vec{\nabla} \times \vec{B}_n + \mu_0 \vec{J}_n$$ Again, a solution exists.

As a side note, at first it appeared strange to me that I could solve the problem using only the curl equations in both cases, but I suppose it depends on the fact that part of the fields ($\vec{E}_0$ and $\vec{B}_0$) are already defined and the two divergence equations don't say anything new.

• thanks a lot! yes it is surprising that one can obtain the solutions using only half of the Maxwell equations. Dec 14 '16 at 23:41
• This is not true. The other half is hidden in the fact that the solution at $t=0$ is already known.
– zap
Dec 15 '16 at 0:18