Preliminary inspection of extremal metrics yields a constant value of ρ(x,y)=1ρ(x,y)=1 as a function which satisfies the constraints and has metric area 5. However, ρ≡1ρ≡1 is not extremal for the Swiss Cross. This can be shown by the function in Figure 2.

Figure 2

Theorem 3 The function in Figure 2 satisfies the constraints of the extremal length problem, and has a smaller metric area than ρ≡1ρ≡1.

Proof Suppose that we define ρρ as in Figure 2. Then we can calculate the metric area as

Now it is only necessary to show that the constraint is satisfied. Let γ∈Γγ∈Γ be a curve with

Without loss of generality, γγ moves from the left edge of the Swiss Cross to the right. All loops in γγ can be cut at their intersection point and the loop itself discarded to create

Any curve of γ'γ' which lies completely inside one constant-weighted region can be replaced by a curve from the entry point to the exit point which minimizes the length. If a line would leave the Swiss Cross, then draw two lines, one from the entry point to the corner and another from the corner to the exit point since this is a special case of lying inside the circle. Furthermore, we know that ∃∃ a geodesic between the 2 points. In a more general sense, we are replacing the earlier curve with a geodesic one, which may or may not travel along regional boundaries. Call this new curve γ''γ''.

Diagram 1

Now suppose γ''γ'' cuts back, i.e. it leaves a region going towards the right, then returns to the same x-value at some point along γ''γ''. Then cut γ''γ'' at the largest x-value before the turn back and at the smallest x-value after the turn back. One of these sections of the curve has smaller integral length than the other (if they are the same length, arbitrarily pick one), and we can translate γ''γ'' up or down either on the right or left in order that the "longer" path becomes a loop, while maintaining the same line length in each region. Then this γ'''γ''' can perform step 1 to get rid of the loops. We now have αα running completely from left to right. By construction if αα travels through one of the 4 circles,

Diagram 2

Also by construction, if αα travels all of the way around the circumference of a circle to get from one 1-ϵ2ϵ2 region to another, then

Also, any curve αα that does not enter the top or bottom 1-ϵ2ϵ2 regions while traveling will have

Thus, suppose αα travels along in the 1+ϵϵ region before breaking off to a 1-ϵ2ϵ2 region. Then shift αα up or down so that it is tangent to one of the circles if necessary, and make sure the first section and last section of αα are completely horizontal to minimize Euclidean distance and create a new αα. Then,

Now if we were somehow able to travel only across the 1-ϵ2ϵ2 region, then a minimal path would add 1-ϵ2ϵ2 in length. Traveling along the boundary of the circle in the 1+ϵϵ region we would pick up θθ**ϵϵ3434(1+ϵϵ) = θθϵϵ3434 + θθϵϵ7474. As ϵϵ gets very small, this will be much greater than the advantage given by entering the 1-ϵ2ϵ2 region. Thus

□ □

There is clearly a great deal of symmetry in the Swiss Cross, and it would be convenient to use that. Theorem 4 allows us to do so.

Theorem 4 The extremal metric on the Swiss Cross will be symmetric about all eight axes of symmetry.

Proof Consider a line of symmetry about the Swiss Cross. Then we can take opposite points and average their weights. If the original weights are aa and bb, then this yields by the Cauchy-Schwarz Inequality:

which implies that:

□ □

Also, this averaging will not affect the integral length of the function from side to side.

Another useful result is a bound on the metric at the center of the Swiss Cross.

Consider a discrete metric on the Swiss Cross. Then create a square in the center of the Swiss Cross with side length ϵϵ, and other ϵϵ squares at the center of the boundary of each outer square. If the weight in the center is wcenterwcenter, and the weight on the sides is wsidewside then we want to minimize:

Diagram 3

subject to:

We can use Lagrange multipliers to solve this equation. So:

When we solve for the ratio wcenterwsidewcenterwside, we get:

In order to find a lower bound for the metric area of the Swiss Cross, we can consider the simplified case where Γ={(StraightlinesconnectingAandB)∪(StraightlinesconnectingCandD)}Γ={(StraightlinesconnectingAandB)∪(StraightlinesconnectingCandD)}. It is easy to verify that the extremal metric is as shown in Figure 3. This function has a metric area of 4.5, so the minimum metric area for the entire set of curves ΓΓ will be greater than or equal to 4.5.

Figure 3

In order to visualize a solution, we used a recursive Matlab program to create an approximation of the extremal metric. Using a modified version of Dijkstra's algorithm we were able to find the shortest path between two sides, and progressively lower the metric area. In order to find a global minimum instead of a local, we used the method of gradient descent. Briefly, by selectively raising certain areas and rerunning the recursion, we wound up with a better metric area.

Diagram 4: A 36x36 matrix of a discrete approximation of the extremal metric on the Swiss Cross

Again, a prime candidate for extremal metric, ρ≡1ρ≡1 seems reasonable which gives a metric area of ππ. However, it is not extremal as is demonstrated in Figure 4. The metric area of Figure 4 is about 2.95, and it satisfies the constraints.

Figure 4

Using a symmetry argument similar to the Swiss Cross, it is evident that the metric on the disk will be a radial function, and hence in polar coordinates will have no θθ dependence.

At this point it became useful to find a lower bound for the metric area of the circle. However, the straight line case is degenerate as every line passes through one point[1]. Thus using other methods it is possible to show that

Thus the extremal metric will have metric area between 8π≈8π≈ 2.54 and 2.95.