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](http://www-math.mit.edu/~stevenj/18.369/pml.pdf)


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 $\partial ^2 /\partial x^2$?



Application of coordinate-stretching transformation for Perfectly Matched Layer
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. 


