Asymptotic solution to ODE

Our adventure starts with the following ordinary differential equation:

y' = y = \frac{1}{x}

Analytic solution

We can solve this equation in closed-form, depending on your definition of closed-form, by multiplying by an integrating factor.

e^x y' + e^x y = \frac{e^x}{x}

The left side factors to

(e^x y)'

and so

y = e^{-x} \left( \int^x \frac{e^t}{t}\, dt + c\right)

The indefinite integral above cannot be evaluated in elementary terms, though it can be evaluated in terms of special functions.

We didn’t specify where the integration starts or the constant c. You can make the lower limit of integration whatever you want, and the value of c will be whatever is necessary to satisfy initial conditions.

The initial conditions hardly matter for large x because the constant c is multiplied by a negative exponential. As we’ll see below, the solution y decays like 1/x while the effect of the initial condition decays exponentially.

Asymptotic series solution

I wrote a few months ago about power series solutions for differential equations, but that approach won’t work here, not over a very large range. If y has a Taylor series expansion, so does it’s derivative, and so y‘ + y has a Taylor series. But the right side of our differential equation has a singularity at 0.

What if instead of assuming y has a Taylor series expansion, we assume it has a Laurant series expansion? That is, instead of assuming y is a weighted sum of positive powers of x, we assume it’s a weighted sum of negative powers of x:

y = \sum_{n=1}^\infty a_nx^{-n}

Then formally solving for the coefficients we find that

y = \frac{1}{x} + \frac{1}{x^2} + \frac{2}{x^3} + \cdots + \frac{(n-1)!}{x^n} + \cdots

 

There’s one problem with this approach: The series diverges for all x because n! grows faster than xn.

And yet the series gives a useful approximation! With a convergent series, we fix x and let n go to infinity. The divergent series above is an asymptotic series, and so we fix n and let x go to infinity.

To see how well the asymptotic solution compares to the analytic solution, we’ll pick the lower limit of integration to be 1 and pick c = 0 in the analytic solution.

The plot above shows the analytic solution, and the first and second order asymptotic solutions. (The nth order solution takes the first n terms of the asymptotic series.) Note that the three curves differ substantially at first but quickly converge together.

Now let’s look at the relative error.

That’s interesting. Eventually the second order approximation is more accurate than the first order, but not at first. In both cases the relative error hits a local minimum, bounces back up, then slowly decreases.

Does this pattern continue if we move on to a third order approximation. Yes, as illustrated below.

Superasymptotic series

The graphs above suggests that higher order approximations are more accurate, eventually, but we might do better than any particular order by letting the order vary, picking the optimal order for each x. That’s the idea behind superasymptotics. For each x, the superasymptotic series sums up terms in the asymptotic series until the terms start getting larger.

When this approach works, it can produce a series that converges exponentially fast, even though each truncated asymptotic series only converges polynomially fast.

Update: To get an idea how accuracy varies jointly with the argument x and the asymptotic approximation order n, consider the following graph. Note that the horizontal axis is now n, not x. The different curves correspond to different values of x, generally moving down as x increases.

Relative error as a function of asymptotic order

Every curve is non-monotone because for every value of x, increasing the order only helps to a point. Then the error gets worse as the series diverges. But as you can see from the graph, you can do quite well by picking the optimal approximation order for a given x.

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