Find the expansions of the solutions of x^2 + (4+epsilon) x + 4 - epsilon = 0 around epsilon = 0.© BrainMass Inc. brainmass.com March 5, 2021, 12:11 am ad1c9bdddf
We want to obtain the solutions of the equation:
x^2 + (4+epsilon) x + 4 - epsilon = 0
using perturbation theory, where epsilon is our small parameter. For epsilon = 0 we obtain the unperturbed solution x0 = -2. There are thus two roots that collapse into one when epsilon = 0. In such cases, perturbation theory yields a non-regular series (one with non integral powers of epsilon). This is actually part of what is called "singular perturbation theory", a more typical example of that is when the small parameter multiplies the highest power of x. Then, the unperturbed equation has lower degree and you have alo have less solutions, and somehow the perturbation series has to yield more solution. In that case, there is a series that starts with negative fractional power of epsilon (so, in the limit of epsilon to zero, one solution moves to infinity).
To find the perturbation series, we can put x = -2 + u and expand the equation in powers of u. We then find:
u^2 + epsilon u - 3 epsilon = 0
Then u is a function of epsilon, and for epsilon = 0, u = 0. Now, we need to choose for u some appropriate power series with undetermined coefficients for u, and then solve for those coefficients. A regular power series with integral powers of epsilon doesn't work. Putting
u = a epsilon + b epsilon^2 + ...
(note that the constant term is absent, because for epsilon = 0, u is zero). Then the term linear in epsilon in the equation is -3 epsilon, and -3 is not equal to zero, we ...
We explain singular perturbation theory and apply it to solve the problem.