Solutions of wave equation but not Maxwell equations

[crick]

Maxwell equation in absence of charges and currents are

$$\nabla \cdot \bf{E} = 0 \\ \nabla\cdot B=0 \\ \nabla \times E=-\frac{\partial B}{\partial t} \\\nabla \times B=\mu \epsilon \frac{\partial E}{\partial t}$$

Wave equation is $$\nabla ^2 \bf{E}=\mu \epsilon \frac{\partial^2 E}{\partial t^2}\tag{1}$$

How can I prove that, given a solution $\bf{E}$ that satisfies both Maxwell equations and $(1)$, then the vector $\bf{E+E'}$ where $\bf{}{}E'$ is a vector such that $\nabla \times \bf{}{}E'=0$ (and $\nabla \cdot E \bf{}$) is a solution of $(1)$?

As I found explained here https://www.photonics.ethz.ch/fileadmin/user_upload/Courses/EM_FieldsAndWaves/WaveEquation.pdf,

Although the field $E(r, t)$ fulfills the wave equation it is not yet a rigorous solution of Maxwell’s equations. We still have to require that the fields are divergence free, i.e. $∇·E(r, t) = 0$. This condition restricts the k-vector to directions perpendicular to the electric field vector $(k·E_0 = 0)$.

Can anyone provide a proof of the fact that a non divergence free (but irrotational) vector can be solution of wave equation without being solution of maxwell equation?

 

[I like Serena]

Maxwell equation in absence of charges and currents are

$$\nabla \cdot \bf{E} = 0 \\ \nabla\cdot B=0 \\ \nabla \times E=-\frac{\partial B}{\partial t} \\\nabla \times B=\mu \epsilon \frac{\partial E}{\partial t}$$

Wave equation is $$\nabla ^2 \bf{E}=\mu \epsilon \frac{\partial^2 E}{\partial t^2}\tag{1}$$

How can I prove that, given a solution $\bf{E}$ that satisfies both Maxwell equations and $(1)$, then the vector $\bf{E+E'}$ where $\bf{}{}E'$ is a vector such that $\nabla \times \bf{}{}E'=0$ (and $\nabla \cdot E \bf{}$) is a solution of $(1)$?

As I found explained here https://www.photonics.ethz.ch/fileadmin/user_upload/Courses/EM_FieldsAndWaves/WaveEquation.pdf,



Can anyone provide a proof of the fact that a non divergence free (but irrotational) vector can be solution of wave equation without being solution of maxwell equation?

Hi crick,

Perhaps we can use the following calculation rule?
$$\nabla\times(\nabla\times f)=\nabla(\nabla\cdot f)-\nabla^2 f$$

And how about:
$$E=(x^2+c^2t^2)\mathbf{\hat x}$$