# A partial differential equation for a vector function and resulting constraint between its components

If we have an algebraic equation connecting 3-variables $x,y,z$, such as $x^2+y^2+z^2=2$, we can immediately conclude that all the 3 variabes are not independent. Now, consider the following Maxwell's (partial differential) equation in vacuum for the electric field $\textbf{E}$: $$\nabla\cdot \textbf{E}(\textbf{r},t)=0.$$ Can we similarly conclude from this that the components of the electric field i.e., $E_x(\textbf{r},t),E_y(\textbf{r},t),E_z(\textbf{r},t)$ are not all independent?

Here is my attempt: In the Fourier space, $$\textbf{E}(\textbf{r},t)=\int\frac{d^3\textbf{k}}{(2\pi)^{3/2}}e^{i\textbf{k}\cdot\textbf{r}}\tilde{\textbf{E}}(\textbf{k},t)$$ and therefore, $$\nabla\cdot \textbf{E}(\textbf{r},t)=0\Rightarrow \int\frac{d^3\textbf{k}}{(2\pi)^{3/2}}e^{i\textbf{k}\cdot\textbf{r}}(i\textbf{k})\cdot\tilde{\textbf{E}}(\textbf{k},t)=0\Rightarrow \textbf{k}\cdot\tilde{\textbf{E}}=0.$$ Therefore, $$k_x\tilde{E}_x(\textbf{k},t)+k_y\tilde{E}_y(\textbf{k},t)+k_z\tilde{E}_z(\textbf{k},t)=0$$ Therefore, for a given $\textbf{k}$, the components of $\tilde{\textbf{E}}(\textbf{k},t)$ are algebraically related. But I could not derive any algebraic relation between the components of $\textbf{E}(\textbf{r},t)$. However, it is not immediately clear to me, whether such a relation can act as a constraint between the components $E_x(\textbf{r},t), E_y(\textbf{r},t)$ and $E_z(\textbf{r},t)$ [in the sense that specification of the values of $E_y(\textbf{r},t)$ and $E_z(\textbf{r},t)$ will uniquely determine $E_x(\textbf{r},t)$].

• – Qmechanic Dec 3 '16 at 14:33

There is no such algebraic equation: you can choose electric field with arbitrary values of the components in one point and zero divergence everywhere.

• @ akhmeteli Do you mean there is no constraint? – SRS Dec 3 '16 at 14:26
• @SRS: I mean there is no algebraic constraint, as far as I can see. – akhmeteli Dec 3 '16 at 14:38

You can consider the divergence as a differential constraint: if $\nabla\cdot \textbf{E}(\textbf{r},t)=0$, knowing $E_y(\textbf{r},t)$ and $E_z(\textbf{r},t)$ everywhere imposes a constraint on $\partial_x E_x(\textbf{r},t)$, which means that for each $y$ and $z$, $E_x(\textbf{r},t)$ is known up to a constant. Hence the 3 components are not independant, although the constraint is not strong enough to determine $E_x(\textbf{r},t)$ uniquely.

Addendum: Regarding uniqueness, if another field $E'$ satisfies the same constraint (divergence free and same $y$ and $z$ components), then it follows from $\partial_x E_x(\textbf{r},t)-\partial_x E'_x(\textbf{r},t)=0$ that $$E'_x(\textbf{r},t)=E_x(\textbf{r},t)+g(y,z)$$ where $g$ is an arbitrary scalar function that does not depend on $x$. That gives the full picture of the consequence of the constraint.

All that you've done in deriving eq (3) is a Fourier expansion. You have illustrated essentially why Fourier expansions are nice when working with differential equations, it's because $e^{kx}$ is an eigenfunction of $\frac{\partial}{\partial x}$. When you use the differential operator, $\nabla$, on it's eigenfunctions, it acts by scaling the function (in this case by $k_i$).

That is the reason you were able to derive an algebraic relationship when considering the Fourier expansion of $\textbf{E}$, but only (as pointed out by Claude Chuber) a differential relationship when considering $\textbf{E}(\textbf{x})$.

## protected by Qmechanic♦Dec 3 '16 at 20:33

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