Boundary Parameterization Method

The boundary parameterization method is a strategy for finding the absolute maximum and minimum values of a function of two variables over a closed and bounded region.

When interior critical points either don’t exist or aren’t sufficient to determine extrema, we need to analyze the function along the boundary of the domain.

However, we can't apply the standard critical point method (based on partial derivatives) along the boundary, since those points lie outside the function’s interior domain.

That’s where parameterization comes in.

What’s the idea?

Suppose we’re working with a function $f(x, y)$ defined on a domain $A$ whose boundary is curved or piecewise linear. To analyze $f$ along this boundary, we follow these steps:

1. Break the boundary into simpler segments.
2. Express each segment parametrically as a curve or line: $ (x(t), y(t)) $ with $ t \in [a, b] $. This reduces the problem to a one-variable function in terms of $t$.
3. Find the local extrema of $f$ along each segment by analyzing the resulting one-variable function.
4. Evaluate $f$ at all critical points identified, and determine which yield the global maximum and minimum.

For instance, to analyze the function $f(x, y) = x^2 + y^2$ along the boundary of the circle $x^2 + y^2 = 1$, we can parameterize the circle as: $$ x(t) = \cos t,\quad t \in [0, 2\pi] $$ $$ y(t) = \sin t,\quad t \in [0, 2\pi] $$ Composing $f$ with the parameterization gives: $$ g(t) = f(\cos t, \sin t) = \cos^2 t + \sin^2 t = 1 $$

A Concrete Example

Let’s consider the function

$$ f(x,y) = x^2 + 2y $$

on the closed, bounded region

$$ A = \{(x,y) \in \mathbb{R}^2 \mid x^2 - 1 \le y \le 3 \} $$

We want to find its absolute maximum and minimum values.

Since $A$ is closed and bounded, the Weierstrass Theorem guarantees that absolute extrema exist.

We'll analyze the function both inside the domain and along its boundary.

1] Interior Critical Points

Interior critical points include stationary points - where the gradient is zero - and singular points - where the gradient doesn't exist.

We compute the partial derivatives:

$$ f_x = 2x $$

$$ f_y = 2 $$

Solving the system:

$$
\begin{cases}
f_x = 0 \Rightarrow x = 0 \\
f_y = 0 \Rightarrow 2 = 0
\end{cases}
$$

This system has no solution, so there are no stationary points in the interior.

Also, the function is differentiable everywhere, so there are no singular points either.

2] Boundary Analysis

The boundary consists of two parts: a straight segment and a parabola.

• Segment 1: The top segment is the line $y = 3$ for $x \in [-2, 2]$. We can parameterize it as $(t, 3)$ with $t \in [-2, 2]$. Substituting into $f$ gives: $$ g(t) = f(t, 3) = t^2 + 6 $$ The minimum occurs at $$t = 0 \Rightarrow f(0, 3) = 6$$ and the maximum at $$t = \pm 2 \Rightarrow f(\pm 2, 3) = 10$$ This segment gives three candidate points: $(0, 3)$, $(-2, 3)$, and $(2, 3)$.
  
• Segment 2: The second boundary component is a parabola: $y = x^2 - 1$ for $x \in [-2, 2]$. We parameterize it as $(t, t^2 - 1)$ with $t \in [-2, 2]$. Substituting gives: $$ g(t) = f(t, t^2 - 1) = t^2 + 2(t^2 - 1) = 3t^2 - 2 $$ The minimum occurs at $$t = 0 \Rightarrow f(0, -1) = -2$$ and the maximum at $$t = \pm 2 \Rightarrow f(\pm 2, 3) = 10$$ This segment gives three more candidates: $(0, -1)$, $(-2, 3)$, and $(2, 3)$.

We now evaluate the function at all the candidate points:

$$
\begin{align*}
f(-2, 3) &= 4 + 6 = 10 \\
f(2, 3) &= 4 + 6 = 10 \\
f(0, 3) &= 0 + 6 = 6 \\
f(0, -1) &= 0 - 2 = -2
\end{align*}
$$

So the absolute maximum of $f$ is 10, occurring at both $(-2, 3)$ and $(2, 3)$.

The absolute minimum is $-2$, which occurs at $(0, -1)$.

How to Parameterize Boundary Segments

The right parameterization depends on the geometry of the boundary segment.

• Circle
  To parameterize a circle of radius $r$ centered at the origin: $$ x(t) = r \cos t,\quad y(t) = r \sin t,\quad t \in [0, 2\pi] $$ If the center is $(a, b)$: $$ x(t) = a + r \cos t,\quad y(t) = b + r \sin t $$
• Ellipse
  For an ellipse centered at the origin: $$ x(t) = a \cos t,\quad y(t) = b \sin t,\quad t \in [0, 2\pi] $$ If shifted to $(x_0, y_0)$: $$ x(t) = x_0 + a \cos t,\quad y(t) = y_0 + b \sin t $$
• Line segments
  To parameterize a segment from $P = (x_1, y_1)$ to $Q = (x_2, y_2)$: $$ x(t) = (1 - t)x_1 + t x_2,\quad y(t) = (1 - t)y_1 + t y_2,\quad t \in [0, 1] $$
• Hyperbola
  For a hyperbola like $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: $$ x(t) = a \cosh t,\quad y(t) = b \sinh t,\quad t \in \mathbb{R} $$
• Spirals
  For the Archimedean spiral: $$ x(t) = a t \cos t,\quad y(t) = a t \sin t,\quad t \in [0, 2\pi] \text{ or beyond} $$
• Special curves
  For arcs of curves like $y = \sqrt{x}$, just pick $t$ in the right interval and substitute: $x = t$, $y = \sqrt{t}$.

The general idea is always the same: express the path you want to follow as $x(t), y(t)$ by letting $t$ vary over the appropriate interval.

And so on.