Deriving a partial differential equation

Question

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.

Answer 1

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?

Answer 2

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.