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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.

Harry Peter
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user135626
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3 Answers3

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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 $\nabla\cdot H = 0$.

Winther
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  • I usually wait a week before awarding a bounty, but, after reading this answer it seems quite unlikely it will be improved upon, so thanks. I should have had enough sense to tinker at the level of $\nabla$ notation. – James S. Cook Jul 11 '14 at 15:24
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ 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. $\pars{1}$: $$ -\pars{k^{2} - \mu}\tilde{H}\pars{\vec{k}} =-\vec{k}\ \vec{k}\cdot\tilde{H}\pars{\vec{k}} $$
Felix Marin
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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: $$\bar H(x,y) = F_x(x)G_x(y) \hat x + F_y(x)G_y(y) \hat y$$

Klaas van Aarsen
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  • When is seperation of variable a good assumption? Does it have to do with the symmetry of the boundary values or their independence on $x$ and $y$? What if the boundary is a curve of the shape $y=f(x)$? – user135626 Jul 06 '14 at 22:06
  • It depends indeed on your boundary conditions which separation you should use. The fact that you specified the problem in cartesian coordinates suggests the boundary conditions are rectangular. If they are not, you should probably switch to a different coordinate system and do the separation there. In polar coordinates you would use $F(r)G(\theta)$ with the additional boundary condition that $G(\theta)=G(2\pi + \theta)$ for any $\theta$. With a shape $y=f(x)$ you should use coordinates that align with that shape. – Klaas van Aarsen Jul 07 '14 at 18:39
  • What if I cannot find a known conformal mapping to transform the coordinated to alight with the shape? Is there any method to proceed then to solve for that kind of boundary? – user135626 Jul 11 '14 at 01:09