PHY454H1S Continuum Mechanics. Lecture 20: Asymptotic solutions of ill conditioned equations. Taught by Prof. K. Das.
Posted by peeterjoot on March 30, 2012
Peeter’s lecture notes from class. May not be entirely coherent.
Two ill conditioned LDEs.
We’ll consider two cases, both ones that we can solve exactly
- With , and letting , we’ll look at solutions of the ill conditioned LDE
- With , , and we’ll look at the second order ill conditioned LDE
The first order LDE.
Our homogeneous equation is
Looking for a solution of the form
and integrate to find
Application of the boundary value constraints give us
This is plotted in figure (1).
We want to consider the limiting case where
and we let . If , then we have
However, if then we have to be more careful constructing an approximation. When is very small, but is also of the same order of smallness we have
Approximate solution in the inner region.
When define a new scale
so that our LDE takes the form
When we have
We have solution
Question: Couldn’t we just Laplace transform.
Answer given: We’d still get into trouble when we take . My comment: I don’t think that’s strictly true. In an example like this where we have an exact solution, a Laplace transform technique should also yield that solution. I think the real trouble will come when we attempt to incorporate the non-linear inertial terms of the Navier-Stokes equation.
Second order example.
We saw above in the first order system that our specific solution was polynomial. While that was found by the method of variation of parameters, it seems obvious in retrospect. Let’s start by looking for such a solution, starting with a first order polynomial
Application of our LDE operator on this produces
Now let’s move on to find a solution to the homogeneous equation
As usual, we look for the characteristic equation by assuming a solution of the form . This gives us
So our homogeneous equation has the form
and our full solution is
with the constants and to be determined from our boundary value conditions. We find
We’ve got and by subtracting
So the exact solution is
This is plotted in figure (2).
Solution in the regular region.
For small relative to our LDE is approximately
which has solution
Our boundary value constraint gives us
Our solution in the regular region where and is therefore just
Solution in the ill conditioned region.
Now let’s consider the inner region. We’ll see below that when , and we allow both and tend to zero independently, we have approximately
We’ll now show this. We start with a helpful change of variables as we did in the first order case
When and we have
This puts the LDE into a non ill conditioned form, and allows us to let . We have approximately
We’ve solved this in our exact solution work above (in a slightly more general form), and thus in this case we have just
at we have
and we find for the inner region
Taking these independent solutions for the inner and outer regions and putting them together into a coherent form (called matched asymptotic expansion) is a rich and tricky field. For info on that we’ve been referred to .
 EJ Hinch. Perturbation methods, volume 6. Cambridge Univ Pr, 1991.