Workshop: A Nonconvex Region With Several Boundary Components

Why Every Boundary Component Matters

To find absolute extrema on a region with a hole, it is not enough to search the outer edge. The hole creates another boundary component, and the objective can have its own minimum or maximum there. A complete search must include interior critical points and every part of the boundary.

We will use one region throughout the worked examples. It is a disk of radius \(5\) with the open disk of radius \(1\), centered at \((1,0)\), removed. The result is nonconvex: the points \((0,0)\) and \((2,0)\) belong to the region, but their midpoint \((1,0)\) lies in the removed hole.

D=\{(x,y):x^2+y^2\leq 25,\ (x-1)^2+y^2\geq 1\}.

The boundary has two components: the outer circle \(x^2+y^2=25\) and the inner circle \((x-1)^2+y^2=1\). The inner circle is entirely inside the outer disk, since every point on it is at distance at most \(2\) from the origin, which is less than \(5\).

Definition: An absolute maximum of \(f\) on \(D\) is a value \(f(p)\) such that \(f(p)\geq f(q)\) for every \(q\in D\). An absolute minimum is defined by reversing the inequality. A candidate is a point that must be checked because it may attain one of these values.

A Complete Candidate Search

The region \(D\) is closed and bounded, hence compact. If \(f\) is continuous on \(D\), the extreme value theorem guarantees that \(f\) attains both an absolute minimum and an absolute maximum. If \(f\) is also differentiable in the interior, any interior point where an extremum occurs must have \(\nabla f=\mathbf{0}\), by Fermat’s theorem. Boundary extrema require a separate search on each boundary component.

For a smooth parametrization \(\mathbf{r}(t)=(x(t),y(t))\) of a boundary component, define the restricted function \(h(t)=f(x(t),y(t))\). A boundary extremum must be a stationary point of this one-variable function, so we solve \(h'(t)=0\). For closed curves such as circles, the parameter is periodic; there are no endpoints of the curve to check separately.

Why this method is complete: A point attaining an absolute extremum is either in the interior or on the boundary. An interior extremum of a differentiable function has zero gradient. A boundary extremum must be an extremum of the restriction to its boundary component, and therefore a stationary point of that restriction. Thus the interior critical points and the stationary points on every boundary component form a complete candidate set, provided the relevant searches find all such points.

The boundary step follows from the chain rule: \(h'(t)=\nabla f(\mathbf{r}(t))\cdot\mathbf{r}'(t)\). At a local extremum of \(h\), its derivative is zero. This does not mean the gradient of \(f\) must vanish at a boundary extremum; the necessary condition is that the derivative along the boundary vanishes.

Worked Examples on the Region

Consider the objective \(f(x,y)=x^2+(y-3)^2\). We will first check its interior candidate and then examine the outer and inner boundary restrictions separately.

Worked Example: Checking the Interior Candidate

Differentiate \(f\) with respect to each variable and set both partial derivatives equal to zero:

f_x=2x,\qquad f_y=2(y-3),

2x=0,\qquad 2(y-3)=0 \quad\Longrightarrow\quad (x,y)=(0,3).

Now check that this point is strictly inside \(D\). Its squared distance from the origin is \(0^2+3^2=9<25\), and its squared distance from the hole’s center is \((0-1)^2+3^2=10>1\). Thus \((0,3)\) is an interior point, not a point on either boundary. Its objective value is

$$ f(0,3)=0^2+(3-3)^2=0. $$

As a check, the gradient at this point is \(\nabla f(0,3)=(0,0)\), as required for an interior critical point. This establishes an interior candidate; the boundary still needs to be checked before deciding whether its value is an absolute extremum.

Worked Example: Searching the Outer Circle

Parametrize the outer circle by \(x=5\cos t\), \(y=5\sin t\), for \(0\leq t\leq 2\pi\). Substituting into the objective gives

h_{\mathrm{out}}(t) =(5\cos t)^2+(5\sin t-3)^2 =25\cos^2 t+25\sin^2 t-30\sin t+9 =34-30\sin t.

Differentiate and solve for the stationary parameter values:

h_{\mathrm{out}}'(t)=-30\cos t=0 \quad\Longrightarrow\quad t=\frac{\pi}{2},\ \frac{3\pi}{2} \quad (0\leq t<2\pi).

At \(t=\pi/2\), the point is \((0,5)\), and substitution into the original objective gives \(f(0,5)=0^2+(5-3)^2=4\). At \(t=3\pi/2\), the point is \((0,-5)\), and \(f(0,-5)=0^2+(-5-3)^2=64\). The restricted formula confirms these values: \(34-30(1)=4\) and \(34-30(-1)=64\). Since these are all the stationary points of the smooth, closed outer circle, its minimum is \(4\) and its maximum is \(64\).

Worked Example: Searching the Hole Boundary

Parametrize the inner circle by \(x=1+\cos t\), \(y=\sin t\), for \(0\leq t\leq 2\pi\). Substituting into \(f\) and using \(\cos^2t+\sin^2t=1\) gives

h_{\mathrm{in}}(t) =(1+\cos t)^2+(\sin t-3)^2 =11+2\cos t-6\sin t.

Its derivative is \(h_{\mathrm{in}}'(t)=-2\sin t-6\cos t\). The stationary points satisfy \(\sin t=-3\cos t\). Combining this relation with \(\cos^2t+\sin^2t=1\) gives \(\cos^2t=1/10\) and \(\sin^2t=9/10\). The two solutions on the unit circle are

(\cos t,\sin t) =\left(\frac{1}{\sqrt{10}},-\frac{3}{\sqrt{10}}\right) \quad\text{or}\quad \left(-\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right).

For the first pair, the boundary point is \(\left(1+\frac{1}{\sqrt{10}},-\frac{3}{\sqrt{10}}\right)\). Substituting into the restricted objective gives \(11+\frac{2}{\sqrt{10}}+\frac{18}{\sqrt{10}}=11+2\sqrt{10}\). For the second pair, the point is \(\left(1-\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right)\), and the value is \(11-\frac{2}{\sqrt{10}}-\frac{18}{\sqrt{10}}=11-2\sqrt{10}\).

As a check, each parameter pair satisfies the stationary equation: for the first, \(-2(-3/\sqrt{10})-6(1/\sqrt{10})=0\); for the second, \(-2(3/\sqrt{10})-6(-1/\sqrt{10})=0\). The two values are therefore the boundary candidates, and the smaller and larger are \(11-2\sqrt{10}\) and \(11+2\sqrt{10}\), respectively.

Assembling and Comparing the Candidates

The search is not complete until every candidate value is compared. The exact comparisons here are simple. Since \(7^2=49>40=(2\sqrt{10})^2\), we have \(11-2\sqrt{10}>4\). Also, \(\sqrt{10}<3\), so \(11+2\sqrt{10}<17<64\). Including the interior value and both outer-boundary values gives:

Location	Candidate point or points	Objective value or values
Interior	\((0,3)\)	\(0\)
Outer boundary	\((0,5)\), \((0,-5)\)	\(4,\ 64\)
Inner boundary	\(\left(1-\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right)\), \(\left(1+\frac{1}{\sqrt{10}},-\frac{3}{\sqrt{10}}\right)\)	\(11-2\sqrt{10},\ 11+2\sqrt{10}\)

The least of all the listed values is \(0\), attained at \((0,3)\), and the greatest is \(64\), attained at \((0,-5)\). Because \(f\) is continuous on the compact region and the candidate search includes the interior and both boundary components, these are the absolute minimum and maximum on \(D\).

A Common Omission: Ignoring the Hole

A tempting but incorrect shortcut is to check only the outer boundary and interior critical points. For example, consider \(g(x,y)=x^2+y^2\) on the same region. Its gradient is \((2x,2y)\), which vanishes at \((0,0)\). But this point is not an interior critical point: it lies on the hole boundary, since \((0-1)^2+0^2=1\). If the inner boundary is omitted, the search misses a point where \(g(0,0)=0\); checking only the outer boundary, where \(g=25\), would falsely suggest that the minimum is \(25\).

The correction is to include the hole boundary. There, \(x=1+\cos t\) and \(y=\sin t\), so

g(1+\cos t,\sin t) =(1+\cos t)^2+\sin^2t =2+2\cos t.

This expression has minimum \(0\) at \(t=\pi\), corresponding to \((0,0)\), and maximum \(4\) at \(t=0\), corresponding to \((2,0)\). On the outer circle \(g=25\). Thus the absolute minimum is \(0\) and the absolute maximum is \(25\). Checking the hole changes the minimum, exactly as a complete boundary search must allow.

Takeaway: On a region with a hole, the inner boundary is part of the boundary of the region. Search it just as deliberately as the outer boundary, then compare its candidates with all interior and outer-boundary candidates.

Check Your Understanding

Use the region \(D\) and the candidate-search method to answer these questions.

Why is \(D\) nonconvex, and what are its two boundary components?
For a differentiable objective, what condition must hold at an interior point where an absolute extremum occurs?
For a boundary parametrization \(\mathbf{r}(t)\), what one-variable function should be differentiated to find boundary candidates?
Why is it necessary to check the inner circle even if the outer boundary and interior have already been examined?
For \(f(x,y)=x^2+(y-3)^2\), what are the absolute minimum and maximum on \(D\), and where are they attained?