By Alan L. T. Paterson

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6) in variational notation is δI = δ x1 F x, u, u dx = x0 x1 δF x, u, u dx = 0. x0 We want to ﬁnd the function u(x) for which the functional I[u(x)] is stationary, in which case the variation of I is zero, that is, δI = 0. 4. In taking the variation inside the integral, it has been assumed that the end points x0 and x1 do not vary with x. We say that a quantity varies when it changes from function to function. Therefore, I[u(x)], u(x), and u (x) vary, but x does not. Using this variational notation, note the analogies between differentials and variations of functions and functionals: • The total differential of a function f (x, y, z) is df = ∂f ∂f ∂f dx + dy + dz, ∂x ∂y ∂z which is the change in f (x, y, z) along the curve from point to point.

6. Integrating the second term by parts gives b b b a1 vu dx = a1 vu − a a u(a1 v) dx, a where pdq = pq − qdp with p = va1 , q = u, dp = (va1 ) dx, dq = u dx. 14) by parts twice results in b a b a0 vu dx = a0 vu b − a b u (a0 v) dx = a0 vu − (a0 v) u b + a a (1) (2) u(a0 v) dx, a (1) p = va0 , q=u, dp = (va0 ) dx, dq = u dx, p = (va0 ) , q = u, dp = (va0 ) dx, dq = u dx. 14) yields b v, Lu = a0 vu − (a0 v) u + a1 vu b + a a 1 r(x)u(x) (a0 v) − (a1 v) + a2 v r(x) dx, where the expression in {} is L∗ v, and the integral is u, L∗ v .

5) It is the rate of change of the functional I with respect to the function u(x) on which it depends. Intuitively, we might expect that for u(x) to be a stationary function of I, the variational derivative should vanish according to δI ∂F d = − δu ∂u dx ∂F ∂u = 0. This is indeed the case, which will be established more formally in the next section. 2 Derivation of Euler’s Equation Although we are in a position to address the question of what function u(x) is a stationary function of the functional I[u(x)], let us ﬁrst consider the nuances of our developments in the previous section in order to provide a more formal proof and introduce some of the arguments to be used later.