Partial Differential Equations, 2nd Edition, L.C.Evans The Calculus of Variations
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1 Partial Differential Equations, 2nd Edition, L.C.Evans Chapter 8 The Calculus of Variations Yung-Hsiang Huang Notation: denotes a bounded smooth, open subset of R n. All given functions are assumed smooth, unless otherwise stated. 1. This is called Rademacher s functions. Consult Brezis [1, Exercise 4.18]. (a) This is Riemann-Lebesgue lemma, which holds not only for L 2 but also for L p, (p [1, )), and weak-* convergence in L ; I put my comment on various proofs below: (1) I think the easy way to understand is through Bessel s inequality. But this method does not work on L p, except L 2. (2) The second way is through integration by parts, and one needs the density theorem (simple functions or Cc functions). This method is adapted to our claim, except for weak convergence in L 1. But for L 1 case, it s a simple consequence of squeeze theorem in freshman s Calculus. It s also easy to see f n 0 a.e. or in measure. Proof. (b)you can apply (2) s method, start with step functions χ B. 2. ( 1 ) L(p, z, x) = e φ(x) 2 p 2 f(x)z. A good intuition is explained in 3. The equation can be rewrite as 0 1 ɛ u t + 1 ɛ xu + u tt which is equivalent to 0 = (e t/ɛ u t ) t + 1 ɛ x(e t/ɛ u). Completed: Ex 1,2,3,7,9,10,15. Department of Math., National Taiwan niversity. d @ntu.edu.tw 1
2 So this motivates us to define L = 1 2 e t/ɛ ( u 2 t + 1 ɛ D xu 2 ), and then we calculate the first variation to conclude that, for each v C c ( (0, T )) 0 = T 0 v e t/ɛ (u t x u ɛu tt ). 4. Proof. (a) We use the results and notations in p without mentions: for k = 1, n, (L P l j (P, Z, x)) xj + L Z l(p, Z, x) (η(cofp ) k j ) xj + η Z k det P [ η Z l Z l x j (cofp ) k j + η(cofp ) k j,x j ] + η det P. Zk Given u C ( R n ; R n ), set Z = u, P = Du, then the Euler-Lagrange equations becomes η u l u l x j (cofdu) k j η η u l u l (cofdu) k j η x j u det P k η u l δl k det P + η det P = 0, uk (cofdu) k j,x j + η det Du uk where δ l k is the Kronecker delta. (b) Follows from the alternative characterization of null Lagrangians, that is, Theorem 1 in p.461, and the fact that the above computations are suitable to C 2 functions. 5. Proof. 6. Proof. 7. Proof. Expand L in coordinate form directly, we have L(P ) = i=1 k=1 P i kp k i P i i P k k Then we see for l = 1, n, (L P l j (P )) xj = 2 (P j l ) x j 2 i=1 (P i i ) xl. 2
3 For any u C ( R n ; R n ), we plug P j l = (u j ) xl into the above identity, and see (L P l j (Du)) xj = 2 ((u j ) xl ) xj 2 ((u i ) xi ) xl = 0. i=1 Remark 1. Does this problem have any useful application? Remark 2. See Giaquinta-Hildebrandt [3, Chapter 1] for more discussions on Null Lagrangians. 8. Explain why the methods in Section 8.2 will not work to prove the existence of a minimizer of the functional I[w] := (1 + Dw 2 ) 1 2 dx over A := {W 1,q () w = g on }, for any 1 q <. Remark 3. This is the minimal surface problem, many great textbooks treat it, e.g. Murrary, Giusti, or Colding-Minicozzi. Proof. The best coercive estimate one can get is I(w) Dw L 1 since lim s (1+ s 2 ) 1 2 s = 1. However, L 1 (or W 1,1 ) are not reflexive spaces. So the boundedness of minimizing sequence in L 1 does not imply existence of a weakly convergent sequence, e.g. approximation of the identity. If φ k (x) = kχ {(0, 1 )}(x) is assumed to be converge weakly in L1 up to a subsequence k φ kj and k j+1 k j as j. Then for the bounded test function g(x) = ( 1) j if x ( 1 k j+1, 1 k j ), φkj g 1 as odd j and φ kj g 1 as even j. 9. Proof. (a) Let u : R n R m be the minimizer of I[w] := L(Dw, w, x) dx, then for each v C c (; R m ) 0 d2 I[u + tv] = dt2 {L p ki p lj D iv k D j v l + 2L p ki u lvl D i v k + L z k z lvk v l } The above identity is true for all Lipschitz continuous R m -valued function v with compact support. In particular, we take v(x) = ɛρ( x ξ ɛ )ηζ(x), where ξ R n, η R m, ζ C c (; R) and ρ : R R is the same tent map in Section Thus, ρ = 1 a.e. and D i v k = ρ ( x ξ ɛ )ξ iη k ζ + O(ɛ), as ɛ 0. After substituting this into the first expression and sending ɛ 0, we obtain 0 L p k i p ξ iη k ξ l j j η l ζ 2 dx. 3
4 Since this holds for all ζ C c (), we deduce that 0 L p k i p l j ξ iη k ξ j η l. (b) Consider L(P ) = det P = p 1 1p 2 2 p 1 2p 2 1 on M 2 2. t (0, 1), But ( L t tl + (1 t) (1 t)l ) = det This function is not convex since for 1 t 0 1 t t > 0. (1) = 0. (2) Direct computation shows that L satisfies the Legendre-Hadamard condition, that is, for all P M 2 2, ξ R 2, η R 2, 2 2 i, k, 2 L(P ) η p k k η l ξ i ξ j = 2η 1 η 2 ξ 1 ξ 2 2η 1 η 2 ξ 2 ξ 1 = 0. i pl j Remark 4. Dacorogna [2, Theorem 5.3] says that for C 2 function L : R mn R { }, the Legendre-Hadamard condition rank one convexity, that is, L(λξ + (1 λ)η) λl(ξ) + (1 λ)l(η), for every λ [0, 1] and rank(ξ η) 1. Proof. ( ) Mean-Value theorem and note that m n matrix A is of rank 1 iff A = vw T for some v R m and w R n. ( ) Givenξ R m and η R n, consider φ(t) = f(p + t ξη T ) which is C 2 and convex, and hence φ (0) 0, which is exactly the Legendre-Hadamard condition. 10. se the methods of to show the existence of a nontrivial weak solution u H 1 0(), u 0, of u = u q 1 u in and u = 0 on Ω for 1 < q < n+2 n 2, n 3. Our method is not a direct application of theorems in Section Since the corresponding g does not satisfy the growth condition. Proof. First, we are going to show the existence of minimizer for the energy functional E on the admissible class A where and E(u) = 1 2 u 2 dx 1 u q+1 q + 1 A = {u H 1 0() : u L q+1 = 1}. 4
5 Then E is coercive on A since E A = 1 2 which is bounded in H 1 0(Ω) by coercivity. u 2 dx 1 q+1. Let u m be the minimizing sequence, Hence weak compactness theorem and Rellich s compactness theorem imply that, up to a subsequence, u m u in H 1 0(Ω) and u m u in L q+1 (Ω). Hence u L q+1 = 1, that is, u A. Therefore u is a minimizer since lim inf m E(u m) E(u) inf A E = lim m E(u m) Finally, a almost identically argument to Lagrange mulitiplier theorem to show this minimizer solves v = λ v q 1 v for some λ R with zero Dirichlet boundary condition weakly. If λ = 0, by uniqueness theorem for laplace equation, we see u 0. But this contradicts to u L q+1 = 1. Hence we see a nontrivial function ũ = λ 1 q 1 u solves the given boundary value problem weakly. We remark two details in the beginning of the proof of Lagrange multiplier theorem: (1.)Since u L q+1 = 1 0, u q 1 u is not identical zero a.e. (2.)For any v H 1 0(), the integral u q 1 uv dx is well-defined by Sobolev embedding theorem, the assumption that is a bounded domain, and Hölder s inequality to exponent pair ( q+1 q, q + 1). 11. If u C () and Multiply the equation with v C (), then we see (f, v) L 2 = u v dx v u ν ds = u v dx + vβ(u) ds For u H 1 (), since for any z, w R, β(z) β(w) b z w, trace theorem and standard approximation argument implies that we can define u H 1 () is a weak solution to our nonlinear boundary-value problem provided for each v H 1 (), (f, v) L 2 = u v dx + where T : H 1 () L 2 ( ) is the trace operator. T v β(t u) ds Proof. 12. Proof. 13. Proof. 14. Proof. 15. (Pointwise gradient constraint) 5
6 (a) Show tere exists a unique minimizer u A of 1 I[w] := 2 Dw 2 fw dx, where f L 2 () and A := {w H 1 0() : Dw 1 a.e.}. (b) Prove for all w A Du D(w u) dx f(w u) Proof. The proof is almost identical to the proof of Theorem 3 and 4 in Section 8.4.2, except we have to check A is weakly closed by Mazur s theorem, a corollary of the Hahn-Banach Theorem: Theorem 5. [1, Section 3.3] Let C be a convex set in a Banach space E, then C is weakly closed if and only if it is closed. One can see the convexity of A easily. The closedness of A is proved by the same argument as Step 1 of Theorem 3 I referred in the beginning. So we complete the proof. 16. Proof. 17. Proof. 18. Proof. 19. Proof. 20. Proof. References [1] Haim Brezis. Functional Analysis, Sobolev Spaces and Partial Differential Equations. Springer Science & Business Media, [2] Bernard Dacorogna. Direct methods in the calculus of variations, volume 78. Springer Science & Business Media, [3] Mariano Giaquinta and Stefan Hildebrandt. Calculus of Variations I, volume 310. Springer Science & Business Media,
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