Problem set 2: Solutions Math 207B, Winter 2016

Transcription

1 Problem set : Solutions Math 07B, Winter A particle of mass m with position x(t) at time t has potential energy V ( x) an kinetic energy T = 1 m x t. The action of the particle over times t t 1 is the time-integral of the ifference between the kinetic an potential energy: S( x) = t1 (T V ) t. (a) Show that an extremal x(t) of S satisfies Newton s secon law F = m a for motion in a conservative force fiel F = V. (b) Show that the total energy of the particle E = T + V is a constant inepenent of time. (a) The Euler-Lagrange equation for the action S( x) = t1 L( x, x t ) t, L( x, x t ) = 1 m x t V ( x) is given by t ( ) L x t + L x = 0. Using the summation convention, an the Kronecker-δ efine by { 1 if i = j, δ ij = 0 if i j, we have [ ] a a i = a i (a j a j ) = a j δ ij = a i = [ a] i, so F / x t = m x t, an F / x = V ( x).

2 The Euler-Lagrange equation is m x tt = V ( x), which is Newton s secon law. (b) Taking the scalar prouct of the ODE with x t, we get that m x t x tt = x t V ( x). It follows that so [ ] 1 t m x t x t + V ( x) = 0, 1 m x t + V ( x) = constant.

3 . Let R n be a boune region with smooth bounary (so the ivergence theorem hols) an f : R a smooth function. Derive the Euler-Lagrange equation an natural bounary conition that are satisfie by a smooth extremal u : R of the functional ( ) 1 J(u) = u fu x. For φ C (), we have ɛ J(u + ɛφ) ɛ=0 = [ ] 1 ( u + ɛ φ) ( u + ɛ φ) f(u + ɛφ) x ɛ = ( u φ fφ) x The ivergence theorem implies that ( u φ) x = [ (φ u) φ u] x = φ u n S φ u x. If φ = 0 on, then ɛ J(u + ɛφ) ɛ=0 = ( u + f) φ x. ɛ=0 The funamental lemma of the calculus of variations implies that an extremal u, with ɛ J(u + ɛφ) ɛ=0 = 0, for all φ satisfies the Euler-Lagrange equation u = f in.

4 For functions φ that o not necessarily vanish on the bounary, we have ɛ J(u + ɛφ) ɛ=0 = φ u n S. It follows that the natural bounary conition for an extremal u is the Neumann conition u n = 0 on.

5 3. The transverse isplacement at position x an time t of an elastic string vibrating in the (x, y)-plane is given by y = u(x, t), where a x b an t t 1. If the ensity of the string per unit length is ρ(x) an the tension in the string is a constant T, then (for small isplacements) the motion of the string is an extremum of the action t1 b ( 1 S(u) = ρu t 1 ) T u x xt. Derive the Euler-Lagrange equation for u(x, t). a For every φ C c ((a, b) (, t 1 )), we have t1 ɛ S(u + ɛφ) ɛ=0 = b a (ρu t φ t T u x φ x ) xt. Integrating by parts an using the fact that φ = 0 at x = a, b an t =, t 1, we get that t1 ɛ S(u + ɛφ) ɛ=0 = b a ( ρu tt + T u xx ) φ xt. The funamental lemma of the calculus of variations then implies that a smooth extremal u of S satisfies the wave equation ρu tt = T u xx.

6 4. The (n-imensional) area of a surface y = u(x) over a region R n is given by J(u) = 1 + u x. Fin the Euler-Lagrange equation (calle the minimal surface equation) that is satisfie by a smooth extremum of this functional. Using the ivergence theorem, we fin for every φ Cc () that ɛ J(u + ɛφ) ɛ=0 = 1 + u + ɛ φ x ɛ ɛ=0 1 = 1 + u ɛ u + ɛ φ ɛ=0 x u φ = x 1 + u = ( ) u φ x. 1 + u The Euler-Lagrange equation for smooth extremals of J is therefore ( ) u = u Using the summation convention, we can write this equation in component form as ( u xi 1 + uxj u xj )xi = 0.

7 5. Let X = {u C 1 ([ 1, 1]) : u( 1) = 1, u(1) = 1}, where C 1 ([a, b]) enotes the space of continuously ifferentiable functions on [a, b]. Define J : X R by (a) Show that J(u) = 1 1 x 4 (u ) x. inf J(u) = 0, u X but J(u) > 0 for every u X (so J oes not attain its infimum on X). (b) What happens when you try to solve the Euler-Lagrange equation for extremals of J? We have J(u) 0 for every u X, so inf J(u) 0. To make J(u) as small as possible, we want to make u small except near x = 0, where u can be large. In orer to o this, let f : R R be any continuously ifferentiable, o function such that f(x) 1 as x an 0 x 4 f (x) x = C <. For example, we coul choose f(x) = tanh x. If we were to inclue piecewise smooth functions in X, or functions in a Sobolev space such as H 1 ( 1, 1), we coul instea take 1 if < x 1, f(x) = x if 1 < x < 1, 1 if 1 x <. For ɛ > 0, efine u ɛ (x) = f(x/ɛ) f(1/ɛ). Then u ɛ ( 1) = 1 an u ɛ (1) = 1 so u ɛ X.

8 Making the change of variables x = ɛt, we have J(u ɛ ) = ɛ f(1/ɛ) = ɛ3 f(1/ɛ) Cɛ3 f(1/ɛ) /ɛ It follows that J(u ɛ ) 0 as ɛ 0 +, so 0 inf J(u) = 0. u X x 4 [ f ( x ɛ )] x t 4 f (t) t If J(u) = 0 an u C 1 ([ 1, 1]), then x 4 (u ) = 0, so u (x) = 0 except at x = 0, an then u (0) = 0 by continuity. It follows that u = constant, but no constant function satisfies both bounary conitions for functions in X, so J(u) > 0 for every u X. (b) The Euler-Lagrange equation for J(u) is ( x 4 u ) = 0. Integrating this ODE once, we get that x 4 u = c, where c is a constant of integration. Integrating again, we get that u(x) = c 1 + c x 3 where c 1, c are constants of integration. The bounary conitions imply that c 1 c = 1, c 1 + c = 1, so c 1 = 0, c = 1. The solution of the Euler-Lagrange equation is u(x) = 1/x 3, which is singular at x = 0, so there is no smooth solution. Remark. The Lagrangian for J(u) is F (x, u ) = x 4 (u ). When x 0, the Lagrangian is a strictly convex function of u (with F u u = x4 > 0), but strict convexity is lost at x = 0. The nonexistence of a minimizer of J(u) is associate with this loss of strict convexity.