By John Gregory Ph.D., Cantian Lin Ph.D. (auth.)

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30b). 29b). (f) + Ja f(x, y + fz, y' + Ez')dx. Using Leibnitz's rule as before we have 0= F' (0) = I' (y, z) = gx(b, y(b)) . 29b) leads to + gy(b,y(b))y'(b) + f(b,Y(b)'Y'(b))l¢'(b;~)Y'(b) + gy (b, y(b ))z(b) + z(b)fy, (b, y(b), y' (b)) = 0 [gx(b,y(b)) or gx + gyY' + f + [gy + fy, H¢' - y'li (b,y(b),y'(b)) = o. 31). 32) f + fy, [¢' - y'll (b,y(b),y'(b)) = O. 33) gx + gy¢' + f + fy, [¢' - y'll (b,y(b),y'(b)) = O. Thus, we have the definition of critical point solutions, below. 48 Chapter 2. 32). 33).

4 above. To avoid being tedious, we assume that (a, y( a)) is given in both these situations and leave the remaining cases to the reader. 26) 1 43 b minimize I(y) = such that y( a) = A, f(x, y, y')dx b is given. 4, which motivates our results, is shown above. 21) above, except that we consider a fixed variation z E Za = {y E Y : y(a) = O}. 21) and all subsequent results must hold. 26). 2 and leave all subsequent results (including the corner conditions) to the reader. 21) except for at most a finite number of interior points.

3 with J 2(x, y) = yT Ax where A is a symmetric n x n matrix. Show that J2(x) = h(x, x) has a strict global minimum at x = 0 if A is positive definite. 32 Chapter 2. 4 is negative then J2(X) has no local minimum by considering the Taylor series expansion for J(x + EY) where y =f 0 is an eigenvector corresponding to a negative eigenvalue. Some important ideas of quadratic forms are best illustrated in this context by two simple examples. 12). In this case, J 1 (y} = J 1 (y,z) = J 1 (y + EZ) 1 b (y'2+ y2)dx, ~J~(Y,z) = Io b (ylZI +yz)dx and = J1 (y) + 2EJ1 (y, z) + E2 J1 (z).