Lagrange Multipliers in the Calculus of Variations

Francis J. Narcowich, January 2020

The problem1that we wish to address is the following: Consider the func- R b 0 R b 0 tionals J(y) = a f(x, y, y )dx and K(y) = a g(x, y, y )dx. Extremize (max- imize/minimize) J over all y such that y ∈ C1[a, b], y(a) = A, y(b) = B, and K(y) = L, where L is a constant.

Theorem 0.1. If y = y extremizes J(y), subject to the conditions above, then there exists a constant λ such that y is an unconstrained extremal for R b D(y) = a (f − λg)dx, y(a) = A, y(b) = B.

∂f 0 R x ∂f 0 Proof. Let F (x) = ∂y0 (x, y,¯ y¯ ) − a ∂y (t, y¯(t), y¯ (t))dt − c1, where c1 is de- R b ∂g R x ∂g termined by a F (x)dx = 0. Similarly, let G(x) = ∂y0 − a ∂y dt − c2, where R b a G(x)dx = 0. (Both F and G are evaluated with y = y, so they are fixed functions of x.) Next, consider the function

Z x Z x ye := y + ε F (t)dt + δ G(t)dt. a a This function satisfies the boundary conditions at x = a and x = b, because

ye(a) = y(a) +  · 0 + δ · 0 = y(a) = A, and Z b Z b ye(b) = y(b) + ε F (x)dx +δ G(x)dx = B. a a | {z } | {z } 0 0 We can require it to satisfy the constraint, too. The quantity K(ye) = R b g(x, y, y0)dx depends only on the parameters  and δ. The reason is that a e e y˜ depends on known functions, y, F (x) and G(x). These are integrated out when K(ye) is computed. The only variables left are ε and δ. The same is true when J(ye) is computed. Thus, we may let φ(ε, δ) := J(ye) and ψ(ε, δ) := K(ye). The constraint is satisfied by requiring ψ(ε, δ) = L, which implicitly defines a curve relating δ and ε.

1The proof presented here may be found in Akhiezer’s book on the Calculus of Varia- tions. See the list of references for a full citation.

1 The extremum for J occurs at ε = δ = 0, because φ(0, 0) = J(y), subject to the constraint ψ(ε, δ) = L. This is equivalent to a 2D Lagrange multiplier problem: Extremize φ(ε, δ) − λ(ψ(ε, δ) − L), with no constraint. Making use of the extremum occurring at (0, 0) gives us ∂φ ∂ψ ∂φ ∂ψ (0, 0) = λ (0, 0) and (0, 0) = λ (0, 0). (0.1) ∂ε ∂ε ∂δ ∂δ We need to compute the various derivatives in the equation above. For ∂φ ∂ε (0, 0), we have ∂φ ∂ Z b ∂f Z x ∂f  (0, 0) = J(y)| = F (t)dt + F (x) dx. e (0,0) 0 ∂ε ∂ε a ∂y a ∂y Integrating the first term on the right by parts then yields ∂φ Z b  ∂f Z x ∂f  (0, 0) = 0 − dt F (x)dx ∂ε a ∂y a ∂y | {z } F (x)+c1 Z b Z b 2 = F (x) dx + c1 F (x)dx = hF,F i a a | {z } 0 Similar calculations result in ∂ψ ∂ψ ∂φ (0, 0) = hF,Gi, (0, 0) = hG, Gi, and (0, 0) = hF,Gi. ∂ε ∂δ ∂δ Using these in (0.1) and putting the resulting equations in matrix form, we see that hF,F i hF,Gi  1  0 = hF,Gi hG, Gi −λ 0 By multiplying on the left by (1 − λ) and collecting terms, we arrive at kF − λGk2 = 0, which implies that F − λG = 0; consequently, ∂(f − λg) Z x ∂(f − λg) 0 − dt = c, ∂y a ∂y which is the du Bois-Reymond form of the Euler-Lagrange equations for D(y). These hold for y = y and do not involve the constraint, as required.

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