Pair of PDEs to be solved together I have the following pair of equations to be solved together to find the functions $H_{x}$ and $H_{y}$, which are the components of a vector $\bar{H}(x,y)=H_{x}(x,y)\hat{x}+H_{y}(x,y)\hat{y}$ in planar Cartesian coordinates $(x,y)$ and with $\mu$ being a constant:
\begin{eqnarray}
\frac{\partial^{2}H_{y}}{\partial x \partial y}&=&\frac{\partial^{2}H_{x}}{\partial y^{2}}+\mu H_{x}\\
\frac{\partial^{2}H_{x}}{\partial x \partial y}&=&\frac{\partial^{2}H_{y}}{\partial x^{2}}+\mu H_{y}
\end{eqnarray}
How do I proceed to solve this set of equations for $H_{x}$ and $H_{y}$?
NOTE: Combining this set into vector format can be written as one equation: $\nabla\times\nabla\times\bar{H}(x,y)=\mu\bar{H}(x,y)$. This can be an alternative way to express the above equations, but still, I am not sure how to tackle it and find $\bar{H}$.
Thanks for any help.
 A: Taking $\partial_x$ on the first equation and $\partial_y$ on the second equation we get
$${\partial^3 H_y \over \partial x^2 \partial y} = {\partial^3 H_x \over \partial x\partial y^2} + \mu {\partial H_x \over \partial x}$$
$${\partial^3 H_x \over \partial x \partial y^2} = {\partial^3 H_y \over \partial x^2\partial y} + \mu {\partial H_y \over \partial y}$$
which implies
$$\mu\left[ {\partial H_x \over \partial x} +  {\partial H_y \over \partial y}\right] = 0$$
or in vector notation $\mu(\nabla \cdot H) = 0$. This can also be more easily derived from $\nabla\times\nabla\times H = \mu H$ and the fact that $\nabla\cdot (\nabla\times A) = 0$. We now use this to simplify the original equation set
$$-{\partial^2 H_x \over \partial x^2} = {\partial^2 H_x \over \partial y^2} + \mu H_x$$
$$-{\partial^2 H_y \over \partial y^2} = {\partial^2 H_y \over \partial x^2} + \mu H_y$$
which can be more compactly written
$$\nabla^2 H_x + \mu H_x = 0$$
$$\nabla^2 H_y + \mu H_y = 0$$
or in vector notation $\nabla^2 H + \mu H = 0$ given $\mu(\nabla\cdot H) = 0$. This is a spatial Helmholtz equation.
To solve it you can apply separation of variables $H_x = A(x)B(y)$ and similar for $H_y$. This will give you equations
$$\frac{\partial_{xx}A}{A} = E$$
$$\frac{\partial_{yy}B}{B} = -\mu-E$$
The allowed values for $E$ follows from the boundary conditions. More details are given here and here. The only thing left to do after solving the two equations for $H_x$ and $H_y$ is to enforce the condition $\nabla\cdot H = 0$.
A: $\newcommand{\+}{^{\dagger}}
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With the identity $\ds{\nabla\times\nabla = \nabla\nabla\cdot\ -\ \nabla^{2}}$
the equation for $\ds{\tilde{H}\pars{x,y}}$ becomes:

\begin{align}
\pars{\nabla^{2} + \mu}\tilde{H}\pars{x,y}
=\nabla\bracks{\nabla\cdot\tilde{H}\pars{x,y}}\tag{1}
\end{align}



*
*I suspect $\ds{\tilde{H}\pars{x,y}}$ is a Magnetic Field $\ds{\tt\pars{~\mbox{Is't true ?}~}}$ which satisfies
$\nabla\cdot\tilde{H}\pars{x,y} = 0$. In that case the equation becomes a Helmholtz one.

*Otherwise, you can Fourier transform Eq. $\pars{1}$:
$$
-\pars{k^{2} - \mu}\tilde{H}\pars{\vec{k}}
=-\vec{k}\ \vec{k}\cdot\tilde{H}\pars{\vec{k}}
$$


A: This looks like Poisson's equation. The standard approach to solve it is by separation of variables.
In your case we would assume that $\bar H(x,y)$ can be written as:
$$\bar H(x,y) = F_x(x)G_x(y) \hat x + F_y(x)G_y(y) \hat y$$
