## PHY454H1S Continuum Mechanics. Lecture 20: Asymptotic solutions of ill conditioned equations. Taught by Prof. K. Das.

Posted by peeterjoot on March 30, 2012

# Disclaimer.

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.

### Exact solution.

Our homogeneous equation is

with solution

Looking for a solution of the form

we find

and integrate to find

Application of the boundary value constraints give us

This is plotted in figure (1).

### Limiting cases.

We want to consider the limiting case where

and we let . If , then we have

or just

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

If and

so

### Approximate solution in the inner region.

When define a new scale

so that our LDE takes the form

When we have

We have solution

or

**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.

### Exact solution.

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

with roots

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

or

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

so that

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 [1].

# References

[1] EJ Hinch. *Perturbation methods*, volume 6. Cambridge Univ Pr, 1991.

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