2.3. Lagrange’s approach 39 Fundamental lemma of the calculus of variations: If M(x) C[a, b] and if b a M(x) η(x) dx = 0 (2.47) for every η(x) C1 [a, b] such that η(a) = η(b) = 0 , (2.48) then M(x) = 0 (2.49) for all x [a, b]. Proof. The proof is by contradiction. Suppose (without loss of gen- erality) that M(x) is positive at some point in (a, b). M(x) must then, by continuity, be positive in some interval [x1,x2] within [a, b]. Now (see Figure 2.5), let η(x) = (x x1)2 (x x2)2 , x [x1,x2] , 0 , x [x1,x2] . (2.50) Clearly, η(x) C1[a, b]. With this choice of η(x), b a M(x) η(x) dx = x2 x1 M(x)(x x1)2(x x2)2 dx . (2.51) Since the integrand is nonnegative, b a M(x) η(x) dx 0 . (2.52) This in contrary to our original hypothesis and it now follows that M(x) = 0 , x (a, b) . (2.53) The continuity of M(x), in turn, guarantees that M(x) also vanishes at the endpoints of the interval.
Previous Page Next Page