Deriving a partial differential equation I was studying for my differential equations exam and came across this problem:
Given the equations:
$$\begin{gather} \vec{B} = \nabla \times(\hat{z}\Psi) \\ \frac{\partial \vec{E}}{\partial t} = \nabla \times \vec{B} \\ \vec{E} = \frac{\partial(\hat{z}\Psi)}{\partial t}
\end{gather}$$
with all function independent of $z$ and $\hat{z}$ being the unit vector in the $z$ direction. It asks to find which the partial differential equation which satisfies $\Psi(x,y,t)$. 
The solution says that $\frac{\partial^2 \Psi}{\partial x^2} + \frac{\partial^2 \Psi}{\partial y^2} =\frac{\partial^2 \Psi}{\partial t^2}$. I can see the obvious relations between the equations but I am not confident in working with the $\nabla$ operator. Could someone show me how this equation is derived? 
The next parts of the question asks, given that $\Psi(x,t)$ is independent of $y$ what is the form of the differential equation. I know that I have to look at the eigenvalues of the equation which are $-1$ and $1$ thus it is parabolic. Is this correct? The final bit asks to solve using separation of variables, which should not be an issue. 
 A: I assume you can derive this:
$$\frac{\partial^2(\hat{z}\Psi)}{\partial t^2}= \nabla \times \left(\nabla \times(\hat{z}\Psi)\right).$$
From here you just need to use the definition of curl: for a given function $F$, the curl is obtained by computing the following determinant:
$$\nabla \times F = \begin{vmatrix} \hat i & \hat j & \hat k \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ F_x & F_y & F_z \end{vmatrix}.$$
In your case, you have to apply it twice. Note that $\hat{z}\Psi = (0, 0, \Psi_z)$, so the computation is quite easier.
A: We have $\nabla = (\partial_x,\partial_y,\partial_z)$, so for $\vec{B}$, we get
\begin{align*}
\nabla \times (\hat{z} \Psi) =& \det \left(
\begin{matrix}
\hat{x} & \hat{y} & \hat{z} \\
\partial_x & \partial_y & \partial_z \\
0 & 0 & \Psi
\end{matrix} \right) \\
=& \left( \begin{matrix}
\Psi_y \\
-\Psi_x \\
0
\end{matrix} \right)
\end{align*}
For $\vec{E}$, we get
\begin{align*}
\vec{E} =& \partial_t \left( \begin{matrix}
0 \\
0 \\
\Psi
\end{matrix} \right) \\
=& \left( \begin{matrix}
0 \\
0 \\
\Psi_t
\end{matrix} \right) 
\end{align*}
Now, to get the formula we are after, we use the second one you gave.
\begin{align*}
\partial_t \vec{E} =& \nabla \times \vec{B} \\
\left( \begin{matrix}
0 \\
0 \\
\Psi_{tt}
\end{matrix} \right) =& \det \left( \begin{matrix}
\hat{x} & \hat{y} & \hat{z} \\
\partial_x & \partial_y & \partial_z \\
\Psi_y & -\Psi_x & 0
\end{matrix} \right) \\
\left( \begin{matrix}
0 \\
0 \\
\Psi_{tt}
\end{matrix} \right) =& \left( \begin{matrix}
\Psi_{xz} \\
\Psi_{yz} \\
-\Psi_{xx} - \Psi_{yy}
\end{matrix} \right)
\end{align*}
Since $\Psi$ is only dependent on $x$, $y$, and $t$, the first two entries work out. The last one gives our differential equation, although the sign is opposite to what you proposed. Are you sure you wrote it down correctly?
