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A Perfectly Matched Layer (PML) is an absorbing boundary condition (ABC) which can be used to approximate free-field conditions for the numerical solution of wave equation problems.

PML note

The PML is normally applied to a PDE using the following transformation:

$ \frac{\partial }{{\partial x}} \to \frac{1}{{1 + i\frac{{\sigma (x)}}{\omega }}}\frac{\partial }{{\partial x}} $

In the above, $i = \sqrt { - 1}$ and $\sigma(x)$ is a function of position in the ABC. Now apparently

$ \frac{{\partial ^2 }}{{\partial x^2 }} \to \frac{1}{s}\frac{\partial }{{\partial x}}\left( {\frac{1}{s}\frac{\partial }{{\partial x}}} \right) $

$ s = 1 + i\frac{{\sigma (x)}}{\omega } $

But is it possible to apply a coordinate stretching in the following fashion: $ \frac{{\partial ^2 }}{{\partial x^2 }} \to \frac{1}{u}\frac{{\partial ^2 }}{{\partial x^2 }} $

The coordinate stretching performed in this fashion would be similar in function to the stretching performed by the transformation applied to $\partial /\partial x$.

Essentially what I would like to do is to apply the coordinate stretching directly to $\partial ^2 /\partial x^2$. I've looked in the PML literature for a very long time, and it seems that most interest is in the application of the coordinate stretching directly to $\partial /\partial x$.

Moreover, I can imagine the coordinate-stretching occurring in a similar fashion to stretching a rubber sheet. If the stretching is being done to the coordinates of $\partial /\partial x$, then what is happening to $\partial ^2 /\partial x^2$?

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@Willie: I think that you are right; I'll edit the question above in an attempt to be a little more clear. Thank you for suggesting this. –  Nicholas Kinar Jul 25 '10 at 20:15

2 Answers 2

up vote 1 down vote accepted

Now, I am not familiar with the PML method to categorically say that it is generally impossible to have an absorbing boundary layer with the second order transform that you seek. But I can say confidently that if you are looking from the point of view of the complex coordinate transformation, then you cannot have a transformation that substitutes $$\frac{\partial^2}{\partial x^2}\to \frac{1}{u}\frac{\partial^2}{\partial x^2} $$ This is because the second coordinate derivative is not tensorial in nature.

Let $y_i$ and $x_i$ be two difference coordinate systems for the same domain, such that $y_i = y_i(x_1,\ldots,x_n)$ can be written as functions of $x_j$s. The first derivative of any function is tensorial, in the sense that the coefficients $$ \frac{\partial f}{\partial x_i} = \sum_{j} \frac{\partial y_j}{\partial x_i} \frac{\partial f}{\partial y_j} $$ the matrix $(\partial y_j / \partial x_i)$ is sometimes called the Jacobian matrix for the coordinate transformation.

However, the second derivative of functions are, in general, not tensorial. So usually we have $$ \frac{\partial^2 f}{\partial x_i \partial x_j} \neq \sum_{kl} \frac{\partial y_k}{\partial x_i} \frac{\partial y_l}{\partial x_j} \frac{\partial^2 f}{\partial y_k \partial y_l}$$ The only exception is when $f$ has a critical point: if $\frac{\partial f}{\partial y_k} = 0$ for each $k$, then the above expression is actually an equality. In general the action of a coordinate transformation on any higher order (order $>1$) partial differential operator will introduce lower order terms. The best way to explain this is in terms of connections on the tangent bundle in the language of differential geometry.

Another way to illustrate this is to consider the case where the underlying domain is one-dimensional, just the real line $\mathbb{R}$. Now let $v$ be any non-vanishing continuous vector field on $\mathbb{R}$, then $v = \phi(x) \partial_x$ for some continuous, non-vanishing coefficient function $\phi$. For any such $v$ we can integrate it to obtain a coordinate function $y$ such that $\partial_y = v$. Now, $$ \partial_y^2 f = v(v(f)) = \phi \partial_x(\phi \partial_x f) = \phi^2 \partial_x^2 f + \phi \partial_x\phi \partial_x f$$ And you see that unless $\phi$ is the constant function (in which case $y$ is just a dilation of $x$), in general the second derivative relative to the $y$ coordinate system will depend on first derivatives in the $x$ coordinate system.

This all just goes to say that in general, if you are looking for a transformation that directly substitutes $$ \frac{\partial^2}{\partial x^2} \to \frac{1}{u}\frac{\partial^2}{\partial x^2} $$ it cannot be obtained by any sort of coordinate transformation.

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Thank you for pointing this out; this allows me to better understand what I should do next and how I should structure my further research into problems involving the PML. –  Nicholas Kinar Jul 26 '10 at 1:23

The complex contour $\tilde x$ is given by

$ \tilde x(x) = x + if(x) $

In the above, $x$ is the real part of the contour, $i = \sqrt { - 1}$, and $\frac{{\partial f}}{{\partial x}} = \frac{{\sigma (x)}}{\omega }$

Then

$ \frac{{\partial \tilde x(x)}}{{\partial x}} = 1 + i\frac{{\partial f(x)}}{{\partial x}} $

$ \frac{{\partial ^2 \tilde x(x)}}{{\partial x^2 }} = i\frac{{\partial ^2 f(x)}}{{\partial x^2 }} $

Now applying this to the differential along the deformed contour, the transformation becomes:

$ \frac{{\partial ^2 }}{{\partial x^2 }} \to \frac{1}{{i\frac{{\partial ^2 f(x)}}{{\partial x^2 }}}}\frac{{\partial ^2 }}{{\partial x^2 }} $

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that is not the correct transformation rule for second derivatives. Under a coordinate transformation $y = y(x)$, you have $\partial_x \to \partial_x y \cdot \partial_y$, but the second derivative transforms like $\partial_x^2 \to (\partial_x y)^2 \partial_y^2 + l.o.t.$, where $l.o.t.$ stands for first order derivatives. –  Willie Wong Jul 25 '10 at 21:31
    
@Willie: Okay, then could you follow a similar transformation and post an alternate answer? –  Nicholas Kinar Jul 25 '10 at 21:44

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