By Kenneth S. Miller.

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Du + ∂x ∂u ∂u • The variation of a function F (x, u, u ) is δF = ∂F ∂F ∂F δx + δu + δu . ∂x ∂u ∂u 33 34 2 Calculus of Variations However, because x is an independent variable, which does not vary, δx = 0 and δF = ∂F ∂F δu , δu + ∂u ∂u which is the variation of F from function to function. The stationary point of a function f (x, y, z) is the point (x, y, z) where the • differential vanishes, that is, df = 0. • The stationary function of a functional I[u] is the function u(x) that causes the variation to vanish, that is, δI = 0.

Observe that for F (x, u, u ) ∂F ∂F du ∂F d2 u dF . = + + dx ∂x ∂u dx ∂u dx2 Thus, ∂F du ∂F d2 u dF ∂F = + − . 18) = 0, and multiply by du/dx to obtain d ∂F du − ∂u dx dx ∂F du ∂F d2 u + = 0. 18) for the ﬁrst and third terms leads to the following form of the Euler equation d ∂F du dF ∂F − = 0. − ∂x dx ∂u dx dx In special case II, in which F(u, u ) does not explicitly depend on the independent variable x, ∂F /∂x = 0, and we have dF d − dx dx ∂F du ∂u dx = 0. 19) where c is the constant of integration.

8 Adjoint and Self-Adjoint Differential Operators 23 where the inner product is taken with respect to the weight function r(x). 12). 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.