Disk Optimization by Circular Parametrization

Why a Disk Requires Two Checks

A disk has both an interior and a curved boundary. A differentiable function can attain its largest or smallest value at an interior critical point, but it can also attain an extreme value on the circle. Finding extrema on the whole disk therefore requires investigating both parts.

The circle has a convenient one-variable parametrization: $x=r\cos t$, $y=r\sin t$. Substituting these expressions into the function turns the boundary search into a problem in one variable. The interior and boundary searches produce candidates that must then be compared using the original function values.

Definition: For $r>0$, the closed disk of radius $r$ centered at the origin is $$ D=\{(x,y)\in\mathbb{R}^2:x^2+y^2\le r^2\}. $$ Its interior is given by $x^2+y^2<r^2$, and its boundary circle by $x^2+y^2=r^2$.

Assumptions and the Candidate Principle

Suppose $f$ is continuous on the entire closed disk $D$. Because $D$ is closed and bounded, it is compact, so the extreme value theorem guarantees that $f$ attains both an absolute maximum and an absolute minimum there. To find them by the procedure in this tutorial, also assume that $f$ is differentiable at every point in the interior of $D$, and that its restriction to the circle is differentiable as a function of the parameter $t$.

These are separate requirements. In particular, differentiability of the boundary restriction does not by itself guarantee that the partial derivatives of $f$ exist at the boundary. The boundary search below uses the derivative of the one-variable restriction, not a formula involving boundary partial derivatives.

Theorem: Under the assumptions above, every absolute extremum of $f$ on $D$ occurs either at an interior point where $f_x=f_y=0$, or at a boundary point whose parameter $t$ is a stationary point of $g(t)=f(r\cos t,r\sin t)$. Comparing the values of $f$ at all such candidates gives the absolute maximum and minimum.

If the boundary restriction has points where it is not differentiable, those points must also be included as boundary candidates. For the differentiable-boundary procedure here, stationary means $g'(t)=0$.

Why the Candidate Principle Works

The argument divides according to where an absolute extremum occurs. The existence of an absolute maximum and minimum follows from continuity on the compact disk. Consider either one of those extremizing points.

If the point is in the interior.
It is a local extremum in an open region where $f$ is differentiable. Fermat’s theorem in two variables gives $f_x=0$ and $f_y=0$.

If the point is on the circle.
The parametrization $(r\cos t,r\sin t)$ covers the entire circle. At a parameter $t_0$ representing the extremizing point, the periodic function $g(t)=f(r\cos t,r\sin t)$ has a local extremum at $t_0$. Since $g$ is differentiable, the one-variable Fermat theorem gives $g'(t_0)=0$.

Compare the resulting values.
Every absolute extremum is among these candidates, and every candidate lies in the disk. Thus the largest and smallest candidate values are the absolute maximum and minimum.

The parameter can be taken modulo $2\pi$. The values $t=0$ and $t=2\pi$ describe the same point. Since $g$ is periodic, an extremum at that point is still a local extremum of the periodic function on the real line; it is not an exception to the derivative test.

A Practical Procedure

Write down the disk $x^2+y^2\le r^2$ and confirm that the function is continuous on it.
Find interior critical points by solving $f_x(x,y)=0$ and $f_y(x,y)=0$, retaining only solutions with $x^2+y^2<r^2$.
Parametrize the circle by $x=r\cos t$, $y=r\sin t$, and form $g(t)=f(r\cos t,r\sin t)$.
Find the boundary stationary parameters by solving $g'(t)=0$ over one full period, such as $0\le t<2\pi$.
Evaluate $f$ at every interior candidate and at the boundary points generated by those parameters. Compare all the values.

The strict inequality in the interior test matters: a solution of the partial-derivative equations that lies on the circle belongs to the boundary search, not the interior search. The boundary parametrization already includes it.

Worked Examples

Worked Example: A Linear Boundary Restriction with an Interior Candidate

Find the absolute extrema of $f(x,y)=x^2+y^2+x$ on the unit disk $x^2+y^2\le1$.

Interior candidates. The partial derivatives are

f_x=2x+1,\qquad f_y=2y.

Setting both equal to zero gives $2x+1=0$, so $x=-\frac12$, and $2y=0$, so $y=0$. This point is interior because $(-\frac12)^2+0^2=\frac14<1$. Its function value is

f\left(-\frac12,0\right) =\left(-\frac12\right)^2+0^2-\frac12 =\frac14-\frac12 =-\frac14.

Boundary candidates. Here $r=1$, so $x=\cos t$, $y=\sin t$. The boundary restriction and its derivative are

g(t)=\cos^2t+\sin^2t+\cos t=1+\cos t, \qquad g'(t)=-\sin t.

Solving $-\sin t=0$ over $0\le t<2\pi$ gives $t=0$ and $t=\pi$. At these parameters,

g(0)=1+\cos 0=2, \qquad g(\pi)=1+\cos\pi=0.

The candidate values are therefore $-\frac14$, $2$, and $0$. The absolute minimum is $-\frac14$, attained at $(-\frac12,0)$; the absolute maximum is $2$, attained at $(1,0)$.

Check: The boundary point $(1,0)$ gives $1^2+0^2+1=2$. The interior point gives $\frac14-\frac12=-\frac14$, agreeing with the computed values.

Worked Example: A Boundary Minimum and Maximum from a Saddle-Shaped Function

Find the absolute extrema of $f(x,y)=x^2-y^2$ on the disk $x^2+y^2\le4$.

Interior candidates. We have $f_x=2x$ and $f_y=-2y$. The equations $2x=0$ and $-2y=0$ give $(x,y)=(0,0)$, which is interior. Its value is $f(0,0)=0$.

Boundary candidates. The radius is $2$, so $x=2\cos t$ and $y=2\sin t$. Substitution gives

g(t)=(2\cos t)^2-(2\sin t)^2 =4\cos^2t-4\sin^2t =4\cos(2t).

Consequently,

g'(t)=-8\sin(2t).

The stationary condition is $\sin(2t)=0$, which means $2t=k\pi$ for an integer $k$. In the range $0\le t<2\pi$, the distinct solutions are $0,\frac{\pi}{2},\pi,\frac{3\pi}{2}$. Their values are

g(0)=4,\quad g\left(\frac{\pi}{2}\right)=-4,\quad g(\pi)=4,\quad g\left(\frac{3\pi}{2}\right)=-4.

Comparing these with the interior value $0$, the absolute maximum is $4$, attained at $(2,0)$ and $(-2,0)$. The absolute minimum is $-4$, attained at $(0,2)$ and $(0,-2)$.

Check: At $(2,0)$, $x^2-y^2=4-0=4$. At $(0,2)$, $x^2-y^2=0-4=-4$. The other listed points give the same respective values.

Worked Example: No Interior Critical Point and Several Boundary Candidates

Find the absolute extrema of $f(x,y)=x^2+2y$ on the disk $x^2+y^2\le4$.

Interior candidates. The partial derivatives are $f_x=2x$ and $f_y=2$. The system would require $2x=0$ and $2=0$, which is impossible. Therefore there are no interior critical points.

Boundary candidates. Use $x=2\cos t$ and $y=2\sin t$. The boundary restriction is

g(t)=(2\cos t)^2+2(2\sin t) =4\cos^2t+4\sin t.

Differentiating each term gives

g'(t)=-8\sin t\cos t+4\cos t =4\cos t(1-2\sin t).

Thus $g'(t)=0$ when $\cos t=0$ or $\sin t=\frac12$. In $0\le t<2\pi$, these give $t=\frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{6},\frac{5\pi}{6}$. Evaluate each:

g\left(\frac{\pi}{2}\right)=4(0)^2+4(1)=4, \qquad g\left(\frac{3\pi}{2}\right)=4(0)^2+4(-1)=-4,

g\left(\frac{\pi}{6}\right) =4\left(\frac{\sqrt3}{2}\right)^2+4\left(\frac12\right) =4\left(\frac34\right)+2 =5,

g\left(\frac{5\pi}{6}\right) =4\left(-\frac{\sqrt3}{2}\right)^2+4\left(\frac12\right) =4\left(\frac34\right)+2 =5.

There are no interior candidates, so the absolute maximum is $5$, attained at $(\frac{\sqrt3}{1},1)$ and $(-\frac{\sqrt3}{1},1)$, and the absolute minimum is $-4$, attained at $(0,-2)$. Here $\frac{\sqrt3}{1}=\sqrt3$.

Check: The points $(\sqrt3,1)$ and $(-\sqrt3,1)$ lie on the circle because $3+1=4$, and each gives $x^2+2y=3+2=5$. The point $(0,-2)$ also lies on the circle and gives $0+2(-2)=-4$.

A Common Mistake: Checking Only the Interior

A frequent error is to find interior critical points and treat that list as the complete candidate set. For example, consider $f(x,y)=x$ on the unit disk. Its partial derivatives are $f_x=1$ and $f_y=0$, so there is no interior point with both partial derivatives equal to zero. The incorrect conclusion would be that the function has no absolute extrema.

That conclusion fails because the disk includes its boundary. On the circle, $g(t)=\cos t$, so $g'(t)=-\sin t$. The stationary parameters are $t=0,\pi$, with values $1$ and $-1$. Thus the absolute maximum is $1$ at $(1,0)$, and the absolute minimum is $-1$ at $(-1,0)$. The correction is to complete both searches and compare their values.

Use the Boundary Derivative That the Assumptions Permit

For the boundary method, define $g(t)$ by substitution and differentiate that one-variable function when it is differentiable. Do not automatically write a chain-rule expression involving $f_x$ and $f_y$ at boundary points: the assumptions used here ensure differentiability in the interior and differentiability of $g$, but do not ensure that those partial derivatives exist at the boundary.

If $f$ is differentiable on an open set containing the disk, then the chain rule does apply on the boundary and gives a formula for $g'(t)$ in terms of the boundary partial derivatives. That is a stronger assumption than this procedure needs. Keeping the boundary restriction and its differentiability explicit makes the candidate test valid under the stated assumptions.

Check Your Understanding

Use the interior-and-boundary procedure to answer these questions.

What condition must an interior candidate satisfy when $f$ is differentiable there?
Write the circular parametrization for a disk of radius $3$.
Why can an absolute extremum on the boundary be found by solving $g'(t)=0$, provided the boundary restriction is differentiable?
For $f(x,y)=x+y$ on the unit disk, what is the boundary restriction $g(t)$?
Why is continuity on the closed disk useful before beginning the candidate search?