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This constitutes a ﬁrst necessary condition for this problem. Applying (2) to the present problem and using (33) of chapter 3 gives at point 2 1+y2 y √ =0 dx + (dy − y dx) √ y−α 1+y2 y−α (4) where y , y are values on the minimizing arc y12 at point 2 and dy, dx are values of the curve N at point 2. After multiplying and dividing by 1 + y 2 one obtains the condition dx + y dy = 0 39 (5) y=α 1 N y12 2 Figure 17: Path of quickest descent, y12 , from point 1 to the curve N which is the transversality condition for this problem.
2 y12 −4 N 3 L 5 y56 6 1 G Figure 16: Shortest arc from a ﬁxed point 1 to a curve N. G is the evolute Let τ2 be the parameter value deﬁning the intersection point 2 of N. Clearly the arc y12 is a straight-line segment. The length of the straight-line segment joining the point 1 with an arbitrary point (x(τ ) , y(τ )) of N is a function I(τ ) which must have a minimum at the value τ2 deﬁning the particular line y12 . The formula (3) of chapter 3 is applicable to the one-parameter family of straight lines joining 1 with N when in that formula we replace C by the point 1 and D by N.
Solve the Euler-Lagrange equation associated with b I = a y 2 − yy + (y ) 2 dx 5. What is the relevant Euler-Lagrange equation associated with I = 1 0 y 2 + 2xy + (y ) 2 dx 6. Investigate all possibilities with regard to tranversality for the problem b min a 1 − (y )2 dx 7. Determine the stationary functions associated with the integral 44 I = 1 0 2 (y ) − 2αyy − 2βy dx where α and β are constants, in each of the following situations: a. The end conditions y(0) = 0 and y(1) = 1 are preassigned.