Calculus of Variations: The Direct Approach

Transcription

1 Clculus of Vritions: The Direct Approch Lecture by Andrejs Treibergs, Notes by Bryn Wilson June 7, 2010 The originl lecture slides re vilble online t: An Exmple of Vritionl Problem: Curve of Shortest Length Suppose we wish to find the shortest curve from (, y 1 ) to (b, y 2 ) in the Eucliden plne (we of course lredy know how to do this, but simply wish to introduce rigorous nottion for future exmples). We then look for curve γ(t) = (t, u(t)) with endpoints s described bove, or in other words, u(t) belonging to the dmissible set A = {w C 1 ([, b]) : w() = y 1, w(b) = y 2 } It must minimize the length integrl, which is written L(u) = b 1 + u2 (t) dt. The minimizing curve must stisfy the Euler eqution, which is in this cse ( ) d u = 0 dt 1 + u 2 nd the solution is stright line u(t) = c 1 t + c 2, s we expected. Let us now dd n dditionl constrint to the sme problem - tht the re under the curve be fixed number j 0. Our dmissible set is now w() = y 1, A = w C1 : w(b) = y 2, b u(t)dt = j 0 1

2 This is clled the Isoperimetric Problem. As shown in the proof outline below, we will see tht the ssocited Euler-Lgrnge equtions demnd the minimum-length curve hve constnt curvture, thus must be n rc of circle. Note tht in some cses there my be no minimum. We my solve more generl problem by letting Ω R n be bounded domin with smooth boundry (to ct s the intervl [, b] from before) nd letting φ C 1 (Ω) be the boundry conditions. We then look for functions u(t) tht mximize some I(u) = f(x, u(x), Du(x)) dx subject to the constrint tht J(u) = Ω Ω g(x, u(x), Du(x)) dx = j 0 In our previous problem, I(u) ws the negtive of rc length nd J(u) ws the re under the curve u(t). Look t two-prmeter vritions U(x, ε 1, ε 2 ) = u(x) + ε 1 η 1 (x) + ε 2 η 2 (x), where η 1 (z) = η 2 (z) = 0 re smooth functions which vnish on the boundry of Ω. Thus U is in our dmissble set. Define the following: I(ε 1, ε 2 ) = J(ε 1, ε 2 ) = b b f(t, U, DU) dx g(t, U, DU) dx Note tht I is t mximum when ε 1 = ε 2 = 0 Using Lgrnge Multipliers, there is constnt λ so tht the solution is the criticl point of the Lgrnge function where L(ε 1, ε 2 ) = I(ε 1, ε 2 ) + λj(ε 1, ε 2 ) = b h(t, U, DU) dt h(t, U, DU) = f(t, U, DU) + λg(t, U, DU). We my use this to find the wek Euler-Lgrnge Equtions L b { } h ε i = ε1 =ε 2 =0 u η i + D p h D x η i dx = 0, Integrting by prts (which we ssume is OK) yields b { } h η i u div (D ph) dt = 0. 2

3 And remembering tht η i re rbitrry, this cn only hold if h u div (D ph) = 0. Returning to the originl isometric problem we were interested in nd plugging in vlues, we hve h = f + λg = 1 + u 2 + λu. Inserting this into the Euler-Lgrnge Eqution nd with some lgebric mnipultion, we get tht 0 = λ κ, or in other words, tht the curvture κ is constnt not dependnt on t. This mens our solution is n rc of circle. Dirichlet s Principle nd Hilbert Dirichlet s Principle is derived from n electrosttics problem: If two electric bttery poles re ttched t points to thin conducting sheet, the potentil cross the sheet is the solution of boundry vlue problem, nd one cn look for solution which produces miniml het. Generlizing, Dirichlet s Principle. Let G R 2 (or in smooth surfce) be compct domin nd φ C( G). Then there is u C 1 (G) C(G) tht stisfies u = φ on G nd minimizes the Dirichlet Integrl D[u] = Du 2 da. Moreover, u = 0 on G. G Dirichlet ws mistken in ssuming minimizer must exist. Weierstrss found the flw nd Hilbert finlly proved the principle rigorously (but ssuming pproprite smoothness). Hilbert lter (in his list of fmous problems to ttck in the 20th century) suggested the following venues be explored in Clculus of Vritions: Are solutions of regulr vrition problems lwys nlytic? Does solution lwys exist, nd if not, when re we sure tht solution must exist? Also, cn we lwys modify the definition of solution in meningful wy for ny problem? Due to these questions, much progress ws mde in the 20th century. The Direct Method The direct method for solution to minimiztion problem on functionl F(u) is s follows: Step 1: Find sequence of functions such tht F(u n ) inf A F(u) Step 2: Choose convergent subsequence u n which converges to some limit u 0. This is the cndidte for the minimizer. Step 3: Exchnge Limits: 3

4 ( ) F(u 0 ) = F lim u n n = lim F(u n n ) = I. There re obviously issues with ssuming some of the bove steps re possible, for exmple: 1) There my not be lower bound. 2) The set A of dmissible functions my not be compct. 3) Only llowed to exchnge limits if F is lower-semicontinuous With nice enough spces nd functions, though, the direct method ssures existence of minimizing solution. Illustrtion with Poisson Minimiztion Problem The Poisson minimiztion problem uses the following functionl: Here, ψ L 2 (Ω) nd φ C 1 (Ω). The Euler Eqution is F(u) = Ω 1 2 Du 2 + ψu dx. 0 = ψ div(du) This is usully written u = φ on Ω, u = ψ in Ω nd is Poisson s Eqution. We will see using the direct method tht the following theorem holds true: Poisson s Minimiztion Problem. Let Ω R n be bounded, connected domin with smooth boundry. Let φ, ψ C (Ω). For u C 1 (Ω), let F(u) = 1 Ω 2 Du 2 + ψu dx. Then there is unique u 0 C (Ω) with u 0 = φ on Ω such tht F(u 0 ) = inf u A F(u) where A = { u C(Ω) C 1 (Ω) : u = φ on Ω.) }. Also, u 0 = ψ in Ω. Note: We enlrge the spce of dmissible functions using the Hilbert Spce: { } ll distributionl derivtives H 1 (Ω) := u L 2 (Ω) : exist nd u x i L 2 (Ω) for ll i. This mkes sense becuse to prove n inequlity, you only need to prove it on dense set, nd H 1 (Ω) is the completion of C (Ω). Similrly denote by H 1 0(Ω) the completion of C 0 (Ω). Denote by A 1 the extension of A to this Hilbert spce, s follows: A 1 := { u H 1 (Ω) : u φ H 1 0(Ω) }. 4

5 Coercivity. We need to ddress the three issues listed bove to mke sure the direct method is going to work. Our first tsk is to prove F is bounded below (coercive). Lemm: There re constnts c 1, c 2 > 0 depending on ψ nd Ω so tht for ll u A 1, F(u) c 1 u 2 H 1 c 2. It follows esily tht F is bounded below by c 2 nd I = inf v A 1 F(v) exists nd is finite. The proof is bit involved nd is not included here. We now know tht we my choose minimizing sequence u n A 1 so tht lim F(u n) = I. n Compctness. We cn ssume F(u n ) < I + 1 for ll n by renumbering, so u n H 1 I c 2 c 1. FACT: In ny Hilbert Spce, e.g. in H 1, ny bounded sequence {u n } is wekly sequentilly compct: there is subsequence {u n } tht wekly converges in H 1 to u 0 H 1. Tht is, for ny v H 1, u n, v H 1 u 0, v H 1 s n. FACT: The embedding H 1 (Ω) L 2 (Ω) is compct. i.e., by going to sub-subsequence if necessry, we my ssume u n u 0 in L 2 (Ω). FACT: A 1 is closed subspce of H 1 (Ω). If ll u n belong to closed subspce nd {u n } converges wekly to u 0 in H 1, then u 0 lso belongs to the closed subspce. i.e., u 0 A 1. u 0 is the cndidte to be the minimizer of the vritionl problem. Lower Semi-Continuity. We need this in order to switch the order of F nd the limit in Step 3 of the Direct Method. Lemm: Let u n be minimizing sequence for F(u) such tht u n u 0 strongly in L 2 (Ω) nd wekly in H 1 (Ω). Then F(u 0 ) lim inf F(u n). n Proof. Since u n u 0 in L 2 (Ω), Ω ψu n Ω ψu 0 nd u n L 2 u 0 L 2. 5

6 In ny Hilbert Spce the norm is SWLSC: u 0 H 1 lim inf n u n H 1. F(u 0 ) = 1 Du L + ψu 2 0 = 1 u H 1 u L + ψu 2 0 { 1 u 2 n 2 H 1 u 1 2 n 2 L + 2 lim inf n ψu n } = lim inf n F(u n) = I. Uniqueness of Solution. Uniqueness follows from the convexity of the functionl F(u), which we must define. F C 1 is convex on A 1 H 1 if F(u + w) F(u) DF(u)[w] whenever u, u + w A 1. F is strictly convex if = holds iff w = 0. We hve not yet proven F C 1 so this is the first step in the rgument. By definition, this is true if F is differentible nd DF : H 1 (H 1 ) is continuous. We prove them here. Thus F is differentible: F(u + v) F(u) DF(u)[v] = 1 2 DF is continuous: For u, v, w H 1, DF(u)[w] DF(v)[w] = s u v H 1 0. Ω Dv 2 dx 1 2 v 2 H 1 = o ( v H 1) s v H 1 0. Ω (Du Dv) Dw dx D(u v) L 2 Dw L 2 u v H 1 w H 1. DF(u) DF(v) (H 1 ) = sup DF(u)[w] DF(v)[w] w 0 w H 1 It is esily checked tht our prticulr F in the Poisson problem is convex, so it follows with some work tht the solution is unique. Wht hve we shown? Using the Direct Method, we showed the existence of unique wek solution u 0 H 1 of the Dirichlet problem for Poisson s Eqution. If the coefficients re known to be smoother (smooth boundry, smooth ψ nd φ) then the solution u will lso hve more regulrity. 0 6