pendicular to the straight line joining their centers. The question presents itself whether these are the only minimal surfaces bounded by the two circles. The answer is in the affirmative, and we therefore have explicit examples of curves in space bounding more than one minimal surface and all of them known. More generally, we shall also consider the case when the two bounding circles lie in parallel planes but do not necessarily have a common axis of symmetry. It will be shown that any minimal surface bounded by them has the property that its intersection by a plane parallel to the planes of the two bounding circles is again a circle. This property allows the minimal surfaces bounded by the two circles to be explicitly determined and indeed this determination has already been made by B. Riemann under this assumption [6, pp. 305-352, especially pp. 341-347]. In precise terms, let S be a minimal surface in space bounded by two plane curves IN, r2 lying in parallel planes. The following theorems will be proved: THEOREM 1. If Fi, F2 are circles, then the intersection of S by a plane parallel to the planes of F1, F2 is again a circle. THEOREM 2. If ri, r2 are convex curves, the the intersection of S by a plane parallel to the planes of rF, r2 is again a convex curve. With reference to Theorem 1, it easily follows that when the two circles have a common axis of symmetry, then all the minimal surfaces bounded by ri, r2 are rotationally symmetric and are therefore catenoids. This completes a classical example, and furnishes an example of boundary curves bounding more than one minimal surface all of which are known. In this connection, mention should be made of the question posed by T. Rado in the Princeton Bicentennial Conference in 1946: to estimate the number of minimal surfaces bounded by a given curve or curves. 2. Let S be a minimal surface given in terms of conformal parameters (u, v). It is given by the equations x = x(u, v), y = y(u, v), z = z(u, v), where the functions x(u, v), y(u, v), z(u, v) are harmonic functions of (u, v) and x? + yu + zu = xv + y? + zV, +YYv + zuZv = O.
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