By Lamb H.

Sir Horace Lamb (1849-1934) the British mathematician, wrote a few influential works in classical physics. A scholar of Stokes and Clerk Maxwell, he taught for ten years because the first professor of arithmetic on the college of Adelaide prior to returning to Britain to soak up the put up of professor of physics on the Victoria college of Manchester (where he had first studied arithmetic at Owens College). As a instructor and author his said goal used to be readability: 'somehow to make those dry bones live'. the 1st variation of this paintings was once released in 1897, the 3rd revised version in 1919, and an extra corrected model previous to his demise. This variation, reissued right here, remained in print until eventually the Fifties. As with Lamb's different textbooks, each one part is via examples.

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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.

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. • The laws of variations of sums and products are analogous to those for differentials.

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.