44 2. The First Variation Note that M(x) = ∂f ∂y − x a ∂f ∂y du (2.74) is continuous on [a, b] and that the assumptions of the lemma are satisfied. It now follows that ∂f ∂y = x a ∂f ∂y du + c (2.75) for all x ∈ [a, b]. This is known as the integrated form of the Euler– Lagrange equation. The right-hand side of equation (2.75) is differentiable. This, in turn, implies that the left-hand side of equation (2.75) is differentiable and that ˆ(x) satisfies the Euler–Lagrange equation, d dx ∂f ∂y = ∂f ∂y . (2.76) In other words, all solutions with continuous first derivatives, not just those with continuous second derivatives, must satisfy the Euler– Lagrange equation. The differentiability of fy (x, ˆ ˆ ) can also be used to prove (Whittemore, 1900–1901) the existence of the second derivative ˆ (x) for all values of x for which fy y (x, ˆ ˆ ) = 0. We will see later that we can weaken the conditions on ˆ(x) and η(x) even further, so that they are merely piecewise continuously dif- ferentiable. One can then show that the Euler–Lagrange equation is satisfied between corners of the solution and that additional condi- tions must be satisfied at the corners. Determining these conditions requires additional tools, and so we will defer the topic of corners until Chapter 10. For the time being, we will focus our attention on continuously differentiable solutions.

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