By Singer I.

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Consequently, the curves y = y{m) that give extrema for a problem with moving boundaries should be extremals.

32 CALCULUS OF VARIATIONS EXAMPLE 3. 21 v(y(z)) = J y dz, 2 Y(Zo) =Yo, y(z1) =Yi· Zo The Euler equation is F 11 = 0 or y = 0. The extremal y = 0 passes through the end points only for Yo = 0 and y 1 = 0 (Fig. 10). Consequently, if y 0 = 0 and y 1 = 0, then the function y = 0 minimizes Z1 the functional v = f y2 M;, for v(y {z)) > 0, and v = 0 when y = 0. long the z-axis close to the point (0, z 1 ), but again going upwards to (zi, y 1 ) just before reaching the limit point z 1 (Fig. 11). It is evident that the values of the functional are arbitrarily small along the curves of such a sequence, and therefore the greatest lower bound of the functional is zero.

The Euler-Poisson equation is yiv_ y = O. The general solution of this equation is y = 0 1 eZ+02 e-z+ 0 3 cosz+ 0 4 sinz. From boundary conditions we have 0 1 = 0, 0 2 = 0, 0 3 = I, 0 4 = O. Hence an extremum can be taken on only along the curve y = cosz. 46 CALCULUS OF VARIATIONS EXAMPLE 3. iy" + ey)dz, 2 _·z satisfying the boundary conditio·ns y ( - Z) = y' ( - Z) = 0, 0, y (Z) = y' (Z) 0, = O• This variational problem arises in the study of the buckled axis of an elastic cylindrical beam with both ends fixed.

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