Pair of PDEs to be solved together

Question

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.

Answer 1

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: $$\bar H(x,y) = F_x(x)G_x(y) \hat x + F_y(x)G_y(y) \hat y$$

Answer 2

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 $\nabla\cdot H = 0$.

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ 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}

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