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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.

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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.

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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?

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  • $\begingroup$ The solution was given by another student, so it may be incorrect. I now understand how to solve this type of question. Thank you $\endgroup$
    – MathsBBB
    Commented Jan 8, 2020 at 11:49
  • $\begingroup$ No problem, you're welcome :) $\endgroup$ Commented Jan 8, 2020 at 11:59

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