advertisement

Math 567: Assignment 4 Solutions 1. Asymptotic stability in Hamiltonian systems (a) Show that a critical point ~x0 ∈ Rn , ∇H(~x0 ) = 0, of a finite-dimensional Hamiltonian system ~x0 = J∇H(~x), cannot be asymptotically stable. We may assume H is a continuous function. If H is constant in a neighbourhood of ~x0 , then there is a sequence of critical points ~x0 6= ~xj (∇H(~xj ) = 0) with ~xj → ~x0 , and hence a sequence of (constant) solutions with initial data converging to ~x0 , none of which → ~x0 as t → ∞. Thus ~x0 is not asymptotically stable. On the other hand, if H is not constant in a neighbourhood of ~x0 , there is a sequence ~xj → ~x0 with H(~xj ) 6= H(~x0 ). By conservation of the Hamiltonian, the solution ~yj (t) with initial data ~yj (0) = ~xj has H(~yj (t)) ≡ H(~xj ) 6= H(~x0 ), and so cannot satisfy ~y (t) → ~x0 as t → ∞. Thus again, ~x0 is not asymptotically stable. (b) On the other hand, show that every solution u(x, t) of the linear Schrödinger equation in RnR (which is Hamiltonian) with u(x, 0) ∈ L2 (Rn ) tends to zero locally in L2 : B |u(x, t)|2 dx → 0 as t → ∞ for any ball B ⊂ Rn (and so the zero solution is asymptotically stable in this sense). Let L2 ∩ L1 3 uj (x) → u(x, 0) in L2 (by density). By the dispersive decay estimate keit∆ φkL∞ ≤ ct−n/2 kφkL1 , the conservation of L2 -norm, the triangle and Hölder inequalities, ku(·, t)kL2 (B) ≤ keit∆ uj kL2 (B) + keit∆ (uj − u(·, 0))kL2 (B) ≤ [V ol(B)]1/2 keit∆ uj kL∞ + kuj − u(·, 0)kL2 ≤ Ct−n/2 kuj kL1 + kuj − u(·, 0)kL2 Given > 0, choose j large enough so that the second term is < /2, then choose t large enough so that the first term is < /2 to conclude that this quantity is < for large enough t. Hence it → 0 as t → ∞. 2. Euler-Lagrange equation for the best constant in the G-N-S inequality Derive the Euler-Lagrange equation satisfied by a maximizer w(x) of the G-N-S variational problem R 4 dx R2 |u(x)| R J(w) = max J(u), J(u) := R 2 2 06=u∈H 1 (R2 ) R2 |∇u(x)| dx R2 |u(x)| dx 1 (by differentiating J(w + ξ) with respect to , where ξ(x) is an arbitrary smooth, compactly supported function on R2 ) and show that a re-scaled version v(x) of such a minimizer (assuming it exists) satisfies the soliton equation −∆v − |v|2 v + v = 0. For a maximizer w(x) and a smooth compactly supported function ξ(x), d J(w + ξ)|=0 d R R R R R R R R |∇w|2 |w|2 4|w|2 Re(w̄ξ) − |w|4 |∇w|2 2Re(w̄ξ) + |w|2 2Re∇w̄ · ∇ξ R R = ( |∇w|2 )2 ( |w|2 )2 Z ¯ = Re α|w|2 w − βw + γ∆w ξ(x)dx 0= where we used an integration by parts (divergence theorem) and R 2 |w|4 4 2 R R R α= R , β=R , γ= R . 2 2 2 2 2 2 |∇w| |w| |∇w| ( |w| ) ( |∇w| )2 |w|2 Thus −γ∆w − α|w|2 w + βw = 0, p p and so v(x) := aw(bx), a := α/β, b := γ/β, solves −∆v − |v|2 v + v = 0. 3. Sharp threshold for global existence for mass cirtical focusing NLS: Consider the 2D cubic focusing NLS: iut + ∆u = −|u|2 u (N LS3− ). u(x, 0) = u0 (x) ∈ H 1 (R2 ) (a) Use the fact that the ground state soliton profile v(x) realizes the best constant in the G-N-S inequality above, together with the Pohozhaev relations for v, to show that solutions to (N LS3− ) with ku0 kL2 < kvkL2 are global. Recall v(x) solves −∆v − |v|2 v + v = 0 and so satisfies the Pohozhaev relations (from class) kvk4L4 = 2kvk2L2 , k∇vk2L2 = kvk2L2 , =⇒ max J(u) = J(v) = u 2 . kvk2L2 Global existence follows from an a priori bound on the H 1 norm. Since we already have conservation of ‘mass’ ku(·, t)kL2 ≡ ku0 kL2 , it suffices to bound k∇ukL2 which we do using the conserved energy E = 12 k∇uk2L2 − 41 kuk4L4 and the sharp form of the G-N-S inequality: 1 1 k∇u(·, t)k2L2 = 2E + kuk4L4 ≤ 2E + J(v)ku0 k2L2 k∇uk2L2 2 2 ku0 k2L2 = 2E + k∇uk2L2 . kvk2L2 2 Thus if ku0 kL2 < kvkL2 , we find k∇u(·, t)k2L2 ≤ 1− 2E ku0 k2 2 , L kvk2 2 L the desired a priori bound. (b) On the other hand, show that the pseudo-conformal transformation i|x|2 x 1 −1 − 4(1−t) u(x, t) 7→ ũ(x, t) := (1 − t) e u , 1−t 1−t maps smooth solutions of (N LS3− ) for t > 0 to solutions of (N LS3− ) for t < 1, and apply this transformation to the soliton solution u(x, t) = v(x)eit to show that there exist solutions of (N LS3− ) which blow up in finite time. Compute, with y := x 1−t , τ := |x|2 −3 −i 4(1−t) ũt = (1 − t) e 1 1−t , |x|2 (1 − t)u(y, τ ) − i u(y, τ ) + x · ∇y u(y, τ ) + uτ , 4 |x|2 h x i ∇x ũ = (1 − t) e −i u(y, τ ) + ∇y u(y, τ ) , 2 2 |x| |x|2 −i ∆x ũ = (1 − t)3 e 4(1−t) − u(y, τ ) − ix · ∇y u(y, τ ) − i(1 − t)u(y, τ ) + ∆y u , 4 −2 −i 4(1−t) |ũ|2 ũ = (1 − t)−3 e i|x|2 − 4(1−t) |u|2 u(y, τ ), so adding up, we find i|x|2 2 iũt + ∆x ũ + |ũ| ũ = (1 − t) −3 − 4(1−t) e iuτ + ∆y u + |u|2 u = 0. Applying this transform to the ground state soliton solution u(x, t) = eit v(x) produces a solution −1 ũ(x, t) = (1 − t) e i 1−t 1− |x|2 4 v x 1−t . Notice −2 ∇ũ = (1 − t) e i 1−t 1− |x|2 4 3 x ix x ∇v − v , 1−t 2 1−t and since v is real-valued, 2 −4 |∇ũ| = (1 − t) " 2 2 # 2 x |x| x ∇v + v , 1−t 4 1−t so by change of variable k∇ũ(·, t)k2L2 = (1 − t)−2 k∇vk2L2 + kxvk2L2 → ∞ as t → 1−, and the solution blows up at t = 1. 4. Non-compactness of the Strichartz ’embedding’ due to various ‘bubbles’: Let φ(x) be a fixed non-zero function in Ḣ 1 (R3 ). In each case, show that the given sequence uk (x) • is bounded in Ḣ 1 (R3 ) R • converges to 0 weakly in Ḣ 1 (R3 ) (that is, for any fixed f ∈ Ḣ 1 (R3 ), R3 ∇f (x) · ∇uk (x)dx → 0 as k → ∞; hint: if it helps, you can argue that it suffices to consider f smooth and compactly supported, by the density of such functions in Ḣ 1 ) 10 3 • eit∆ uk converges to 0 weakly in L10 t Lx (Rx × Rt ) 10 3 • has no subsequence for which eit∆ uk converges in L10 t Lx (R × R): (a) uk (x) = φ(x − xk ) with |xk | → ∞ Clearly k∇uk kL2 = k∇φkL2 is bounded. Next we show weak convergence to 0 in Ḣ 1 . RFirst suppose φ and f are both compactly supported functions in Ḣ 1 . Then ∇f · ∇uk → 0 since for large enough k the supports are disjoint. For general φ ∈ Ḣ 1 , given > 0, let ψ have compact support and k∇(ψ − φ)kL2 < (by density). So Z lim sup| ∇f · ∇uk | k→∞ Z Z ≤ lim sup k ∇f · ∇ψ(· − xk )| + lim sup | ∇f · ∇(ψ − φ)(· − xk )| k→∞ k→∞ ≤ k∇f kL2 R and since was arbitrary, ∇f · ∇uk → 0. Finally, an identical approximation argument allows us to draw the same conclusion for general f ∈ Ḣ 1 . By translation invariance of the linear Schrödinger equation, eit∆ uk = w(x − 10 3 xk , t) where w(x, t) = eit∆ φ. So the weak convergence to zero in L10 t Lx (R × R) 4 R R 10/9 10/9 (i.e. R R3 f (x, t)(eit∆ uk )(x)dx dt → 0 for any f ∈ Lt Lx (R3 × R)) follows from an argument very similar to the one above. Since the norm keit∆ uk kL10 10 = kwkL10 L10 is constant (non-zero) but the sex t Lx t quence converges weakly to zero, strong convergence of any subsequence is impossible. −1/2 (b) uk (x) = λk φ(x/λk ) with 0 < λk → 0 or → ∞ k∇uk k2L2 = Z 2 3 2 λ−3 k |∇φ(x/λk )| d(x/λk )λk = k∇φkL2 is constant, hence bounded. For the weak convergence, again first suppose φ and f have compact support. If λk → 0, suppose supp(φ) ⊂ {|x| ≤ R}, so that supp(uk ) ⊂ {|x| ≤ λk R} and Z Z | ∇uk · ∇f | = | ∇uk · ∇f | ≤ k∇uk kL2 k∇f kL2 (|x|≤λk R) |x|≤λk R = k∇φkL2 k∇f kL2 (|x|≤λk R) → 0 as k → ∞. If λk → ∞, change variables Z Z Z −3/2 3/2 ∇uk · ∇f = λk ∇φ(x/λk ) · ∇f (x)dx = ∇φ(y) · λk ∇f (λk y)dy, and repeat the above argument with roles of f and φ reversed. Finally, the convergence to zero for general φ and f in Ḣ 1 follows from an approximation argument just like the one above. −1/2 By the scale invariance of the linear Schrödinger equation, eit∆ uk = λk w(x/λk , t/λ2k ) 10 where w(x, t) = eit∆ φ, and so the weak convergence to zero in L10 t Lx follows from a very similar argument to the one just above. Again, keit∆ uk kL10 10 = kwkL10 L10 is a (non-zero) constant, and thus weak x t t Lx convergence to zero prohibits strong convergence of any subsequence. (c) uk (x) = eitk ∆ φ with tk → ±∞ (hint: the decay estimate keit∆ f kL∞ (R3 ) ≤ ct−3/2 kf kL1 (R3 ) in concert with an approximation argument might help) The norm is constant: by Plancherel, 2 k∇uk k2L2 = k∇eitk ∆ φk2L2 = kξe−itk |ξ| φ̂(ξ)k2L2 = kξ φ̂(ξ)k2L2 = k∇φk2L2 . Hence bounded. If ∇φ ∈ L1 ∩ L2 and ∇f ∈ L1 ∩ L2 , then by Hölder, Z | ∇f ·∇uk | ≤ k∇f kL1 keitk ∆ ∇φkL∞ ≤ ck∇f kL1 k∇φkL1 |tk |−3/2 → 0 as k → ∞. 5 The result for general φ, f ∈ Ḣ 1 then follows from an approximation argument, as above. We have eit∆ uk = eit∆ eitk ∆ φ = ei(t+tk )∆ φ = w(x, t + tk ) where w(x, t) = eit∆ φ (i,e, it is just time translation). So the weak convergence 10 to zero in L10 t Lx can be argues just as in part (a). Again the norm is constant: keit∆ uk kL10 10 = kwkL10 L10 and so the weak conx t Lx t vergence to zero prohibits strong convergence of any subsequence. 5. Asymptotic non-interaction of bubbles: Let φ, ψ ∈ Ḣ 1 (R3 ). In each case, show that lim k∇uk k2L2 (R3 ) = k∇φk2L2 (R3 ) + k∇ψk2L2 (R3 ) k→∞ (a) uk (x) = φ(x − xk ) + ψ(x − yk ), |xk − yk | → ∞ Since ∇uk (x) = ∇φ(x − xk ) + ∇ψ(x − yk ), by changing variables, Z 2 2 2 k∇uk kL2 = k∇φ(· − xk )kL2 + k∇ψ(· − yk )kL2 + 2Re ∇φ̄(x − xk ) · ∇ψ(x − yk )dx Z 2 2 = k∇φkL2 + k∇ψkL2 + 2Re ∇φ̄(x) · ∇ψ(x − [yk − xk ])dx and the last integral goes to zero as |yk − xk | → ∞ just as in 4(a) (i.e. weak convergence). −1/2 −1/2 µk λk (b) uk (x) = λk φ λxk + µk ψ µxk , µ k + λk → ∞ −3/2 −3/2 Since ∇uk (x) = λk k∇uk k2L2 ∇φ(x/λk ) + µk ∇ψ(x/µk ), by changing variables, Z Z −3 −3 2 = λk |∇φ(x/λk )| dx + µk |∇ψ(x/µk )|2 dx Z −3/2 −3/2 + 2Re λk ∇φ̄(x/λk ) · µk ∇ψ(x/µk )dx Z −3/2 = k∇φk2L2 + k∇ψk2L2 + 2Re ∇φ̄(x) · ηk ∇ψ(x/ηk )dx where ηk := µk /λk , and the last integral goes to zero as ηk → 0 or ∞ just as in 4(b) (i,e, weak convergence). −1/2 −1/2 |xk −yk |2 x−yk µk λk k + µ ψ , → ∞. (c) uk (x) = λk φ x−x k λk µk µk + λk + λk µk As in (a) and (b), k∇uk k2L2 = k∇φk2L2 + k∇ψk2L2 Z + 2Re 6 ∇φ̄ · −3/2 ηk ∇ψ y xk − yk + ηk µk dx with ηk := µk /λk , and it remains to show the last integral goes to zero (weak convergence again!). Suppose not. Then along some subsequence it is bounded away from zero. If along some further subsequence, ηk → ∞, then if ∇ψ were bounded and ∇φ were integrable, the integral would tend to zero (and the general case follows from approximation as in question 4), a contradiction. Similarly, if ηk → 0 among some further subsequence, change variables in the integral to put the scaling and translation on φ (rather than ψ) and argue similarly to get a contradiction. Hence ηk must be bounded away from 0 and ∞ along our subsequence. It follows then that (along our subsequence) |xk − yk | µk 2 = |xk − yk |2 1 → ∞, λk µk ηk and so if φ and ψ had compact support, the integral would tend to zero (the supports would become disjoint) – and the general case again follows from approximation – another contradiction. (d) uk (x) = eitk ∆ φ + eisk ∆ φ, |tk − sk | → ∞ Since ∇uk (x) = eitk ∆ ∇φ(x) + eisk ∆ ∇ψ(x), by the unitarity of eit∆ , k∇uk k2L2 = keitk ∆ ∇φk2L2 + keisk ∆ ∇ψk2L2 + 2Reheitk ∆ ∇φ, eisk ∆ ∇ψi = k∇φk2L2 + k∇ψk2L2 + 2Reh∇φ, ei(sk −tk )∆ ∇ψi, and the last integral goes to zero as |tk − sk | → ∞, just as in 4(c). 7