Potential theory

If the function has a singularity within the stadium S, i.e., close to the real line, then we have to do more analysis and the effect of point distribution also has to be considered in the analysis. Define

then

If $\gamma_n(x,t) > 1$, then we get exponential convergence as $n \to \infty$. Define

If $\alpha_n \ge \alpha > 1$ for all sufficiently large $n$, and if $f$ is analytic in the region bounded by Γ, then $p(x) \to f(x)$ at the rate $O(\alpha^{-n})$.

We can give a geometric interpretation of the condition $\alpha_n > 1$; the geometric mean distance of every $t \in \Gamma$ from $\{ x_j \}$ is strictly greater than the geometric mean distance of every $x \in X$ to $\{ x_j \}$.

1Potential theory¶

We want to study the behaviour of the function $\gamma_n(x,t)$. Taking logarithm

Define the potential function

$u_n$ is a harmonic function in the complex plane away from $\{x_j\}$. We may think of each $x_j$ as a point charge of strength $\frac{1}{n+1}$, like an electron, and of $u_n$ as the potential induced by all the charges, whose gradient defines an electric field.

In terms of the potential

and hence

We require $u_n(t)-u_n(x) > 0$; in particular if

then

and the interpolants converge exponentially, $\norm{f-p} = O(\alpha^{-n})$. We can write the condition as

Hence convergence depends on the difference of values taken by the potential function on the set of points $X$ where the interpolant is to be evaluated and on a contour Γ inside which $f$ is analytic. If $f$ is analytic everywhere, we can take take Γ far out in the complex plane and we easily satisfy the above condition. But if there is a singularity close to the real line, Γ cannot be too far away, and then we have to analyze the condition more carefully.

2From discrete to continuous potentials¶

We can write the potential $u_n$ as a Lebesque-Stieltjes integral

where $\mu_n$ is a measure consisting of a sum of Dirac delta functions, each of strength $1/(n+1)$

What is the limiting measure as $n \to \infty$ ? This limit depends on the grid point distribution. Note that

The second property can also be written as

2.1Uniform point distribution¶

The number of nodes in an interval of length $\Delta\tau$ is proportional to $\Delta\tau$ so that

and using the condition $\int_{-1}^1\mu(\tau)\ud\tau=1$ we get $C=\half$ and

The limiting potential is

At the end-point and middle, it takes the values

The equipotential curve $u(s) = -1 + \log(2)$ is shown in red in figure and it cuts the imaginary axis at $\pm 0.52552491457\ii$ and passes through the end points $x=\pm 1$.

If $f$ has a singularity outside the red curve, then we can take Γ to be an equipotential curve $u(s) = u_0$ with $u_0 > -1 + \log(2)$. Then

and we have exponential convergence. But if the singularity is inside the red curve, then we cannot choose a curve Γ that satisfies the above condition.

2.2Chebyshev points¶

The Chebyshev points are uniformly distributed with respect to the variable $\theta \in [0,\pi]$ where $x = -\cos\theta$, so that

and using the condition $\int_{-1}^1\mu(\tau)\ud\tau=1$, we get

The limiting potential is

For $s \in [-1,+1]$, $u(s) = -\log 2$, i.e., a constant. Thus $X = [-1,+1]$ is an equipotential curve of the potential function $u(s)$. For $u_0 > -\log 2$, the equipotential curve $u(s) = u_0$ is the Bernstein ellipse $E_\rho$ with $\rho = 2 \ee^{u_0}$ which encloses the set $X$.

No matter how close the singularity of $f$ is to $X$, we can always find an equipotential curve $u(s) = \rho$ with $\rho > -\log(2)$, which encloses $X$ and inside which $f$ is analytic. If we take Γ to be this equipotential curve then

and we obtain exponential convergence.