# 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.

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.
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?