1.3 The Lemma of DuBois-Reymond

Transcription

1 28 CHAPTER 1. INDIRECT METHODS 1.3 The Lemm of DuBois-Reymond We needed extr regulrity to integrte by prts nd obtin the Euler- Lgrnge eqution. The following result shows tht, t lest sometimes, the extr regulrity in such sitution need not be ssumed. Lemm 3 (cf. Lemm 1.8 in BGH). (The lemm of DuBois-Reymond) If f C 0 (, b) nd b f(x)η (x) dx = 0 for every η C c (, b). (1.16) then f c (constnt). Proof: Let ζ Cc (, b) be rbitrry nd tke µ Cc (, b) with µ = 1. Consider ( ) φ = ζ ζ µ = ζ cµ where c = ζ. Note tht φ C c (, b). Also, η(x) = hs η (x) = φ(x) nd (for ǫ > 0 smll) η(+ǫ) = +ǫ x φ(ξ) dξ φ(ξ) dξ = 0 nd η(b ǫ) = b (ζ cµ) dx = c c b µ dx = 0. Thus, η Cc (, b) with φ = η. According to (1.16) we hve ( 0 = fφ = f(ζ cµ) = fζ c fµ = fζ ) ζ c 1 where c 1 = fµ. Therefore, 0 = (f c 1 )ζ for every ζ C c (, b). The fundmentl lemm implies f c 1.

2 1.3. THE LEMMA OF DUBOIS-REYMOND Figure 1.5: For the lemm of DuBois-Reymond, we mollify piecewise smooth function. Another proof of the lemm of DuBois-Rymond Agin the uthors of BGH give different rgument nd more generl result. Lemm 4 (Lemm 1.8 in BGH). If f L 1 loc (, b) nd fη = 0 for every η Cc (, b). (1.17) (,b) then f c (constnt), i.e., there is some constnt c such tht f(x) = c for lmost every x. Proof: Let x nd x be Lebesgue points for f [f 0 ]. We might s well ssume < x < x < b. As suggested in BGH, let us lso tke δ > 0 smll nd fixed so tht η(ξ) = 2χ I (ξ) + [1 + 1δ ] (ξ x) χ T (ξ) [1 1δ ] (ξ x) χ T(ξ) with I = (x + δ, x δ), T = (x δ, x + δ) nd T = ( x δ, x + δ) gives the function with grph indicted in Figure 1.5. The resoning of Exercise 8 shows µ ǫ η Cc (, b) with { 0, ξ (, b)\(x δ ǫ, x + δ + ǫ) µ ǫ η(ξ) = 2, ξ [x + δ + ǫ, x δ ǫ] where µ ǫ is stndrd mollifier with ǫ < δ. Also, d dx (µ ǫ η) = µ ǫ η = 0

3 30 CHAPTER 1. INDIRECT METHODS on the interiors (, b)\[x δ ǫ, x + δ + ǫ] nd (x + δ + ǫ, x δ ǫ) with 1 δ d dx (µ ǫ η) 1 for ll x (, b). δ Let us compute wht hppens in the portions (x δ + ǫ, x + δ ǫ) nd ( x δ + ǫ, x + δ ǫ) of the trnsition intervls T nd T. For ξ in the first intervl µ ǫ η(ξ) = µ ǫ (t) η(ξ t) t = µ ǫ (t) [1 + 1δ ] (ξ t x) t = δ (ξ x) 1 tµ ǫ (t) δ where m(ǫ) = 1 δ t tµ ǫ (t) = 1 δ = η(ξ) m(ǫ) ǫ ǫ t tµ ǫ (t) dt stisfies m(ǫ) < ǫ δ. Similrly for ξ in the intervl round x we hve µ ǫ η(ξ) = η(ξ) + m(ǫ). The hypothesis (1.17) clerly pplies to give f(µ ǫ η) = 0. We wish to tke limit nd conclude f η = 0. (1.18) To this end, let us estimte f(µ ǫ η) f η = f [(µ ǫ η) η ] f (µ ǫ η) η = f (µ ǫ η) η. (1.19) (x δ ǫ, x+δ+ǫ)

4 1.3. THE LEMMA OF DUBOIS-REYMOND 31 In this cse, we do not know f L 2, so the Cuchy-Schwrz inequlity does not help us. We do know, howevever, tht for ǫ smll the intervl J = (x δ ǫ, x+δ+ǫ) (, b) so f L 1 (J). Furthermore, we hve shown the function (µ ǫ η) (ξ) η (ξ) is bounded on ll of (, b) nd stisfies lim (µ ǫ η) (ξ) η (ξ) = 0 ǫ 0 for every ξ (, b). It follows tht the integrnd in (1.19) is bounded independent of ǫ in L 1 (J) nd limits pointwise to zero lmost everywhere. By the Lebesgue dominted convergence theorem, the expression in (1.19) tends to zero, nd we hve estblished (1.18). Since η (ξ) = (1/δ)χ T (ξ) + (1/δ)χ T(ξ) lmost everywhere (i.e., except t the four corner points), we cn rewrite (1.18) s 1 f 1 f = 0. δ (x δ,x+δ) δ (x δ,x+δ) Tking the limit s δ ց 0 nd reclling tht x nd x were Lebesgue points, we get 2f(x) 2f( x) = 0. Tht is, f(x) = f( x), nd it follows tht f is constnt, tking single vlue on its Lebesgue points. Exercises Exercise 11. Show tht if the mollifier µ ǫ is chosen to be even, then the quntity m(ǫ) = tµ ǫ (t) vnishes. Show why this quntity need not be zero when µ ǫ is not even. Exercise 12. Show tht when g hs compct support in (, b), then for ǫ smll enough µ ǫ g lso hs compct support in (, b) nd my therefore be defined on ll of R. Exercise 13. Show directly tht t d dx (µ ǫ g)(x) = µ ǫ g (x) for ll x R

5 32 CHAPTER 1. INDIRECT METHODS when g 1 c (, b) is ny piecewise C1 function with compct support. Use this clcultion to give (new nd different) direct proof tht f η = 0 where f stisfies the hypothesis (1.17) or (1.16) nd η is the function defined in the proof of the DuBois-Reymond lemm from BGH. 1.4 The Euler-Lgrnge Eqution (revisited) Theorem 4 (Corollry 1.10 in BGH). (The Euler-Lgrnge Eqution for wek extremls) If u C 1 (, b) is wek extreml for the functionl b F(x, u(x), u (x)) dx with Lgrngin F C 1 ((, b) R R), then d dx F p(x, u, u ) F z (x, u, u ) = 0 on (, b). (1.20) Proof: This result follows, essentilly, from integrting by prts in the condition for wek extremls (1.1) in the reverse direction: By the fundmentl theorem of clculus ψ(x) = x F z (t, u(t), u (t)) dt is C 1 function with derivtive F z (x, u(x), u (x)). Thus, b F z (x, u, u )φ dx = ψφ b b ψφ dx = b ψφ dx. Combining this expression with the other integrl from (1.1), we get b [F p (x, u, u ) ψ]φ dx = 0 for ll φ C c (, b).

6 1.4. THE EULER-LAGRANGE EQUATION (REVISITED) 33 By the lemm of DuBois-Rymond, there is some constnt c such tht Tht is, F p (x, u, u ) = F p (x, u, u ) ψ = c. x F z (t, u(t), u (t)) dt + c. (1.21) While it my not be true tht F p hs ny higher prtil derivtives nd it my not be true tht u hs ny higher prtil derivtives, we hve shown tht the composition F p (x, u, u ) does hve derivtive: d dx F p(x, u, u ) = F z (x, u, u ). It is importnt to relize tht the Euler-Lgrnge eqution, under these hypotheses, my not llow expnsion of the left side by the chin rule. The eqution (1.21) is clled the DuBois-Rymond eqution or the Euler-Lgrnge eqution in integrted form. The following exmple (Exmple 4 on pge 14 of BGH) shows the weker regulrity llowed by Theorem 4 is sometimes needed. Consider the functionl on 1 1 u 2 (2x u ) 2 dx A = {u C 1 [ 1, 1] : u( 1) = 0, u(1) = 1}. Notice tht F is non-negtive nd F[u 0 ] = 0 where u 0 C 1 [ 1, 1]\C 2 ( 1, 1) is given by { 0, 1 x 0 u 0 (x) = x 2, 0 x 1. We wish to show u 0 is the unique minimizer in A. If u A is ny minimizer, then we must hve 1 1 u 2 (2x u ) 2 dx = 0. This mens tht on ny intervl where u 0, we must hve u = 2x nd u(x) = x 2 + c for some constnt c. In prticulr, integrting from x = 1, we must hve u(x) = 1 + x 1 (2ξ) dξ = x 2 for 0 x 1.

7 34 CHAPTER 1. INDIRECT METHODS If we ssume there is some x 0 with 1 < x 0 < 0 for which u(x 0 ) 0, then there is mximl intervl (, b) with 1 < x 0 < b 0 such tht u(x) = u(x 0 ) + x 2 x for < x < b, but u() = u(b) = 0. Evluting u(x) t x = nd x = b we conclude 2 = b 2 = x 2 0 u(x 0). This contrdicts the fct tht < b 0. Consequently, there is no such point x 0, we hve u(x) 0 for 1 x 0, nd u u 0. Exercise 14. () Find the Euler-Lgrnge eqution for 1 1 u 2 (2x u ) 2 dx, nd show tht u 0 given bove is solution of the eqution. (b) Assume u C 2 [ 1, 1] is clssicl extreml for F, nd use the chin rule (product rule etc.) to write the Euler-Lgrnge eqution s second order qusiliner ODE. Is u 0 lso solution of this eqution? (c) Wht cn you sy bout C 2 [ 1, 1] clssicl extremls for this functionl? 1.5 Exmples We now return to some exmples from the introduction nd write down the ssocited Euler-Lgrnge equtions. We lso mke some elementry observtions bout those exmples nd introduce some dditionl exmples Dirichlet energy Recll tht D[u] = 0 if u c (constnt), nd these re bsolute minimizers in C 1 [, b], but they my not be dmissible. For the Dirichlet energy, the Lgrngin is F(p) = p 2 nd the Euler- Lgrnge eqution is u = 0. (1.22) Notice the rgument of DuBois-Rymond now implies dded regulrity: Any C 1 (wek) extreml must be C 2 (clssicl) extreml. Given the dmissible

8 1.5. EXAMPLES 35 clss A 0 = {u C 1 [, b] : u() = u, u(b) = u b }, it is esy to integrte (1.22) to obtin the unique dmissible extreml: u 0 (x) = u b b (x ) + u. Let s try to show u 0 is the minimizer. Sy u is some other dmissible competitor. Then b { } D[u] D[u 0 ] = [u 0 + (u u 0)] 2 u 02 dx = 2 b b = 2u 0 { (u u 0 )2 + 2u 0 (u u 0 )} dx b u 0 (u u 0 ) dx (u u 0) dx = 2u 0 [u b u b (u u )] = 0. This shows D[u] D[u 0 ] with equlity if nd only if u u 0, which mens u u 0. Tht is, u 0 is the unique minimizer. This pproch suggests the following simple rephrsing of the condition for minimizers: Lemm 5 (Lemm 2.1 in Troutmn). If the universl set U is liner spce (e.g., C 1 [, b] or 1 [, b], etc.) nd u 0 A stisfies F[u 0 + v] F[u 0 ] 0 whenver v U nd u 0 + v A, (1.23) then u 0 is minimizer. Conversely, if u 0 A is minimizer, then (1.23) holds. Furthermore, if equlity holds in (1.23) only for v = 0, then u 0 is the unique minimizer (nd conversely, if u 0 is the unique minimzer, then equlity in (1.23) cn only hold when v = 0.) Poisson s functionl Here we consider 1 0 (u + u 2 ) dx

9 36 CHAPTER 1. INDIRECT METHODS on A 0 = {u C 1 [, b] : u(0) = 0 = u(1)}. Here the function u 0 is dmissible nd gives 0, but this is not the minimizer. The Lgrngin is F(z, p) = z + p 2 nd the Euler-Lgrnge eqution is 2u 1 = 0. Agin, we hve extr regulrity for extremls, nd the unique extreml is u 0 = 1 x(x 1). 4 Computing F[u 0 ] we get [ x2 1 4 x + 1 ( x 1 ) ] 2 dx = 1 ( ) = Thus, F[u 0 ] < 0. See Exercise Arclength functionl Here is n instnce where C 2 regulrity is not immedite from the Euler- Lgrnge eqution. The Lgrngin is 1 + p 2, nd the Euler-Lgrnge eqution is ( ) d u = 0. dx 1 + u 2 With dditionl regulrity, the differentition my be crried out so tht the left side is immeditely recognizble s the curvture of the grph of u: k = u (1 + u 2 ) 3/2. Of course, if the curvture vnishes, the grph is stright line. See Exercise Potentil eqution An ordinry differentil eqution (ODE) centrl to the study of mny physicl phenomen is the second order liner potentil eqution u + c(x)u = 0

10 1.6. EXERCISES 37 ssocited with the vritionl integrl b [ u 2 + c(x)u 2] dx. The function c = c(x) is clled the potentil. See Exercise 19. This is Exmple 2 on pge 14 of BGH. 1.6 Exercises Exercise 15. Compute the Dirichlet energy of the sequence of functions u j A 0 = {u C 1 [0, 1] : u(0) = 0, u(1) = 1} which converge in L 1 (0, 1) to the constnt zero function. Exercise 16. Show u 0 = x(x 1)/4 is the unique minimizer of Poisson s functionl on 1 0 (u + u 2 ) dx A 0 = {u C 1 [0, 0] : u(0) = 0 = u(1)}. Exercise 17. Explicitly integrte the Euler-Lgrnge eqution for the rclength functionl to verify ll extreml solutions must be ffine, hving the form u(x) = αx + β for some constnts α nd β. Exercise 18. Find the extremls for the totl vrition functionl. Exercise 19. Show the potentil eqution is the Euler-Lgrnge eqution of the functionl b [ u 2 + c(x)u 2] dx. Wht cn you sy bout the regulrity of solutions in C 1 [, b]? Exercise 20 (project exercise). Consider cylindricl continer (modeled by) {(x, y, z) : x 2 + y 2 = 1, z 0} {(x, y, 0) : x 2 + y 2 1}. Assume the continer contins volume V = π of liquid in grvity field so tht the volume initilly occupies the spce {(x, y, z) : x 2 + y 2 < 1, 0 < z < 1}.

11 38 CHAPTER 1. INDIRECT METHODS (Ignore surfce tension nd wetting energy.) Assume, more generlly, tht the liquid is bounded by the grph of n even function z = u(r) for 1 < r < 1 (in polr coordintes) in the dmissible clss A = { u C 1 [0, 1] : u (0) = 0, 2π 1 0 } ru(r) dr = π. If the continer (long with the liquid it contins) is rotting t constnt ngulr velocity ω, clculte the physicl energy of the system two wys: () The physicl energy is the kinetic energy of rottion. (b) With respect to rotting frme fixed to the continer, the physicl energy is the potentil energy with respect to the pprent field. Find (nd solve) the Euler-Lgrnge eqution for the interfce. Wht hppens if the continer hs some other shpe (but still rottes with given volume of liquid round some xis)? Solutions Solution 1 (Exercise 8). Let f(x) = χ ( δ,δ) ( x x0 δ nd consider the convolution integrl µ ǫ f(x) = Show the following: () µ ǫ f = f µ ǫ. (b) µ ǫ f C c (R). (c) 0 µ ǫ f(x) 1 for ll x R. ξ R ) µ(ξ)f(ξ x). (d) When ǫ < δ µ ǫ f(x) = { 0, x δ + ǫ 1, x δ ǫ.