By Russak

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**Sample text**

1) x1 and a class of admissible arcs y where superscript bar designates a vector arc, with components y : yi(x) x1 ≤ x ≤ x2 i = 1, · · · , N (2) on which the integral (1) has a well deﬁned value. e. arcs that satisfy yi(x1 ) = yi1 yi(x2 ) = yi2 (3) where y 1 has coordinates (y11 , · · · , yN 1 ) and y 2 has coordinates (y12 , · · · , yN 2). Analogous to the proof used in obtaining the ﬁrst Euler equation in chapter 3 for the two-dimensional problem we could obtain the following condition along a minimizing arc x Fyi (x) = x1 Fyi dx + ci i = 1, · · · , N (4) where ci are constants.

We next continue using the general theory results to develop two auxiliary formulas for the Brachistochrone problem which are the analogues of (3), (4) for the shortest distance problem. Two Important Auxiliary Formulas If a segment y34 of a cycloid varies so that its endpoints describe two curves C and D, as shown in Figure 15 then it is possible to ﬁnd a formula for the diﬀerential of the value of the integral I taken along the moving segment, and a formula expressing the diﬀerence of the values of I at two positions of the segment.

The path y12 must furthermore be cut at right angles by the curve N at their intersection point 2. Example: Minimize the integral π/4 I = 0 y 2 − (y )2 dx with left end point ﬁxed y(0) = 1 and the right end point is along the curve x = π . 4 Since F = y 2 − (y )2 , then the Euler equation becomes y + y = 0. The solution is y(x) = A cos x + B sin x Using the condition at x = 0, we get y = cos x + B sin x 40 Now for the transversality condition F + (φ − y )Fy x= π 4 = 0 where φ is the curve on the right end.