Scherk Surface
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High Quality Content by WIKIPEDIA articles! The Scherk surface given by the graph of u(x, y) = log ( cos(x) / cos(y) ) for x and y between - /2 and /2. Consider the following minimal surface problem on a square in the Euclidean plane: for a natural number n, find a minimal surface n as the graph of some function u_{n} : left( - frac{pi}{2}, + frac{pi}{2} right) times left( - frac{pi}{2}, + frac{pi}{2} right) to mathbb{R} such that lim_{y to pm pi / 2} u_{n} left( x, y right) = + n mbox{ for } - frac{pi}{2} x + frac{pi}{2}, lim_{x to pm pi / 2} u_{n} left( x, y right) = - n mbox{ for } - frac{p...
High Quality Content by WIKIPEDIA articles! The Scherk surface given by the graph of u(x, y) = log ( cos(x) / cos(y) ) for x and y between - /2 and /2. Consider the following minimal surface problem on a square in the Euclidean plane: for a natural number n, find a minimal surface n as the graph of some function u_{n} : left( - frac{pi}{2}, + frac{pi}{2} right) times left( - frac{pi}{2}, + frac{pi}{2} right) to mathbb{R} such that lim_{y to pm pi / 2} u_{n} left( x, y right) = + n mbox{ for } - frac{pi}{2} x + frac{pi}{2}, lim_{x to pm pi / 2} u_{n} left( x, y right) = - n mbox{ for } - frac{pi}{2} y + frac{pi}{2}. That is, un satisfies the minimal surface equation mathrm{div} left( frac{nabla u_{n} (x, y)}{sqrt{1 + nabla u_{n} (x, y) ^{2}}} right) equiv 0 and Sigma_{n} = left{ (x, y, u_{n}(x, y)) in mathbb{R}^{3} left - frac{pi}{2} x, y + frac{pi}{2} right. right}.
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