Example of Second-order Singular Perturbation Theory
Consider the following equation for the unknown variable :
For the initial problem with, the solution is . For small the lowest-order approximation may be found by inserting the ansatz
into the equation and demanding the equation to be fulfilled up to terms that involve powers of higher than the first. This yields . In the same way, the higher orders may be found. However, even in this simple example it may be observed that for (arbitrarily) small there are four other solutions to the equation (with very large magnitude). The reason we don't find these solutions in the above perturbation method is because these solutions diverge when while the ansatz assumes regular behavior in this limit.
The four additional solutions can be found using the methods of singular perturbation theory. In this case this works as follows. Since the four solutions diverge at, it makes sense to rescale . We put
such that in terms of the solutions stay finite. This means that we need to choose the exponent to match the rate at which the solutions diverge. In terms of the equation reads:
The 'right' value for is obtained when the exponent of in the prefactor of the term proportional to is equal to the exponent of in the prefactor of the term proportional to, i.e. when . This is called 'significant degeneration'. If we choose larger, then the four solutions will collapse to zero in terms of and they will become degenerate with the solution we found above. If we choose smaller, then the four solutions will still diverge to infinity.
Putting in the above equation yields:
This equation can be solved using ordinary perturbation theory in the same way as regular expansion for was obtained. Since the expansion parameter is now we put:
There are 5 solutions for : 0, 1, -1, i and -i. We must disregard the solution . The case corresponds to the original regular solution which appears to be at zero for, because in the limit we are rescaling by an infinite amount. The next term is . In terms of the four solutions are thus given as:
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