Gibbs Phenomenon

Brown University, Applied Mathematics

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Gibbs Phenomenon

Let $ F_N (x) $ be the finite Fourier sum with N+1 terms:

\[ F_N (x) = \frac{a_0}{2} + \sum_{k=1}^N \left( a_k \cos \frac{k\pi x}{\ell} + b_k \sin \frac{k\pi x}{\ell} \right) \]

If a function f(x) has a discontinuity at the point $ x_0 $ of amount $ f(x_0 +0) - f(x_0 -0) $ then finite Fourier sums experience overshoot and undershoot in a neighborhood of this point. These undershoots or overshoots cannot be eliminated by increasing the the number of terms in the finite Fourier sum and they approach with $ N\mapsto \infty $ the value

\[ 1.1789797444721675\ldots \left\vert f(x_0 +0) - f(x_0 -0) \right\vert . \]

Such behavior of Fourier series is usually referred to as the Gibbs phenomenon, which was first noticed and analyzed by the English mathematician Henry Wilbraham (1825--1883) in 1848. The term "Gibbs phenomenon" was introduced by the American mathematician Maxime Bocher in 1906. The 'magic' number is the approximation of

\[ \frac{2}{\pi} \,\mbox{Si} (\pi ) \approx 1.1789797444721675\ldots \qquad\mbox{where } \mbox{Si} (x) = \int_0^x \frac{\sin t}{t} \,{\text d}t \]

is sine integral (special function), which is build-in MuPad as Si(x).

Essentially, now that we have a function in terms of lambda, we want to get it back to a function of t to obtain our solution.
$ F(\lambda) => Inverse Laplace => f(t)H(t) $
When we find our function of time, we will want to multiply by the Heaviside function to define it for t > 0.
The inverse Laplace Tranform is denoted with the following noation:
$\mathcal{L}^{-1}[F(\lambda)]$

Finding the Inverse Laplace Transform

We will want to make sure our function of lambda as a fraction of 2 functions:
$F(\lambda) = \frac{P(\lambda)}{Q(\lambda)}$ where degree of Q > degree of P and a shifted function has the following property:
$F(\lambda)e^{-a\lambda} = \frac{P(\lambda)}{Q(\lambda)}e^{-a\lambda}$
This property suggests the following:
$\mathcal{L}^{-1}[\frac{P(\lambda)}{Q(\lambda)}] = f(t)H(t)$ and a shifted function yields: $\mathcal{L}^{-1}[\frac{P(\lambda)}{Q(\lambda)}e^{-a\lambda}] = f(t)H(t-a)$, showing how important multiplying by the Heaviside function is.

Solving Using Partial Fraction Decomposition

1. First you will split your fraction into a sum of simple fractions by using Partial Fraction Decomposition.
2. Next you will look up the Inverse Laplace for a given function of lambda in a table provided to write the answer (if you are working by hand).
.
Example 1:

Example 2:

Example 4:

Working In MuPAD

The syntax is exactly the same, only this time, we will be ussing "ilaplace" instead of "laplace".

ilaplace::addpattern(pat, s, t, res)

ilaplace(1/`λ`^5,`λ`,t)

ilaplace(1/(`λ`^2+1),`λ`,t)

Gibbs Phenomenon

Brown University, Applied Mathematics