Uniqueness of ground state solutions of non-local equations in R N Rupert L. Frank Department of Mathematics Princeton University Joint work with Enno Lenzmann and Luis Silvestre Uniqueness and non-degeneracy of ground states for ( ) s Q + Q Q α+1 = 0 in R Acta Math., to appear. Preprint: arxiv:1009.4042 Institute for Advanced Study, Princeton, October 19, 2012 R. Frank Uniqueness of ground states October 19, 2012 # 1
Introduction The classical result If α > 0 is subcritical (i.e., 0 < α < in N = 1, 2 and 0 < α < 4/(N 2) = 2 2 in N 3), then the equation Q Q α+1 = Q in R N has a unique (up to translations) positive solution in H 1 (R N ). Key points: Existence (Strauss, Lieb,...): minimize the ratio u θ 2 u 1 θ 2 u α+2, θ = α 2(α + 2) Symmetry (Alexandrov, Serrin, Gidas Ni Nirenberg,...) any positive solution is radial Uniqueness (Coffman, McLeod Serrin, Kwong) the radial, positive solution is unique Side remark: If α = 4/(N 2) = 2 2, the positive solutions of Q Q α+1 = 0 are known explicitly (and unique up to translation and dilation) R. Frank Uniqueness of ground states October 19, 2012 # 2
Goal A similar result for the fractional Laplacian What about uniqueness of positive solutions for ( ) s Q Q α+1 = Q in R N? Here 0 < s < 1, and α > 0 is subcritical, i.e., α < 2 2 = restriction if s N 2 ) Existence and radial symmetry are known 4s N 2s if s < N 2 (no Fractional Laplacian (( )s Q)(ξ) = ξ 2s Q(ξ) defined via Fourier transform Q(ξ) = (2π) N/2 Q(x)e iξ x dx Previous results: Only one special case N = 1, s = 1 2, α = 1 (Amick Toland), where explicit solutions are known Side remark: In the critical case α = 2 2, s < N 2, the solutions of ( )s Q Q α+1 = 0 are known explicitly (and unique up to translation and dilation) (Y.Y. Li, Chen Li Ou, Lieb) R. Frank Uniqueness of ground states October 19, 2012 # 3
Motivation for this question Solitary wave solutions for non-linear dispersive model equations generalized Benjamin-Ono u t + u x (( ) s u) x + u α u x = 0, generalized Benjamin-Bona-Mahony u t + u x + (( ) s u) t + u α u x = 0, fractional NLS iu t ( ) s u + u α u = 0. Solutions: (c(α+1)) 1 1 1 α Q(c 2s (x (1+c)t)), (c(α+1)) α Q(( c 1 + c ) 1 2s (x (1+c)t)), e iωt ω 1 1 α Q(ω Uniqueness and non-degeneracy crucial for stability and blow-up analysis Uniqueness conjectured by Weinstein (1987) and Kenig Martel Robbiano (2010); (cf. also Lieb Yau) Where does uniqueness come from? Relation to maximum principle? ( ) s has positive heat kernel for 0 < s 1 2s x) R. Frank Uniqueness of ground states October 19, 2012 # 4
Main result The Morse index of a non-negative solution Q of ( ) s Q Q α+1 = Q is the number of (strictly) negative eigenvalues of the linearization L + = ( ) s (α + 1)Q α + 1 in L 2 (R N ). Theorem 1. Let N 1, 0 < s < 1 and assume that 0 < α < 4s N 2s if 0 < s < N 2 and that 0 < α < if N 2 s < 1. Then there is a unique (up to translations) non-negative solution of ( ) s Q Q α+1 = Q with Morse index one. Moreover, L + is non-degenerate, i.e., ker L + = span{ x1 Q,..., xn Q}. The restriction on α is optimal. It is an open problem whether every positive solution has Morse index one. Corollary 2. For N, s and α as before, the infimum inf u ( ) s/2 u θ 2 u 1 θ 2 u α+2, θ = Nα 2s(α + 2), is attained by a unique (modulo symmetries) function. R. Frank Uniqueness of ground states October 19, 2012 # 5
Proof of the corollary Corollary 2. For N, s and α as before, the infimum inf u ( ) s/2 u θ 2 u 1 θ 2 u α+2, θ = Nα 2s(α + 2), is attained by a unique (modulo symmetries) function. More precisely, there is a positive, radially decreasing function Q such that any minimizer is of the form aq(b(x c)) for some a C \ {0}, b > 0 and c R N. Proof. Up to rescaling and multiplication by a constant, any minimizer Q solves the Euler Lagrange equation ( ) s Q Q α+1 = Q. Since Q, L + Q = α Q α+2 < 0, the Morse index of any non-negative solution Q 0 is 1. The condition d2 dϵ 2 ϵ=0 J[Q + ϵϕ] 0 for every ϕ (local stability) implies that L + is non-negative on a subspace of co-dimension one. Thus, the Morse index of any non-negative local minimizer is 1. R. Frank Uniqueness of ground states October 19, 2012 # 6
Main tool of our proof Non-degeneracy Let Q be a positive (and hence radially decreasing) solution of ( ) s Q Q α+1 = Q. The linearization L + = ( ) s (α + 1)Q α + 1 commutes with angular momentum. Since L + xj Q = 0 and xj Q = ( r Q)(x j /r) with r Q < 0, we have ker L + ang.mom.=1 = span{ x1 Q,..., xn Q} by Perron-Frobenius. Moreover, by monotonicity with respect to l, ker L + ang.mom.=l = {0} for l 2. What about ker L + radial? Since Q, L + Q = α Q α+2 < 0, we have number of strictly negative eigenvalues of L + radial = N (L + radial ) 1. Proposition 3. If N (L + radial ) = 1, then ker L + radial = {0}. This proposition immediately proves the non-degenercay part of the main theorem. As we shall see, it also implies the uniqueness part! R. Frank Uniqueness of ground states October 19, 2012 # 7
Strategy of our proof I The branch Let s 0, α be as in the theorem and let Q be a non-negative (and hence radially decreasing) solution of ( ) s 0 Q Q α+1 = Q with Morse index one. By nondegeneracy, ker L +,s0 radial = {0}. Construct branch Q s, s [s 0, s 0 + δ) with Q s0 = Q such that ( ) s Q s Q α+1 s = Q s. Implicit function theorem in L 2 (R N ) L α+2 (R N ) All Q s are positive and radially decreasing. Key fact: If N (L +,s0 radial ) = 1 at s 0, then ker L +,s radial = {0} for all s (consequence of non-degeneracy) Branch extends continuously in L 2 (R N ) L α+2 (R N ) up to s = 1. I.e. with Q 1 0. Q 1 Q α+1 1 = Q 1 R. Frank Uniqueness of ground states October 19, 2012 # 8
Strategy of the proof II Analysis at s = 1 Q 1 Q α+1 1 = Q 1 By uniqueness at s = 1 (Kwong s theorem) we have Q 1 (x) = P (x) for a universal function P (depending only on α and N). Proof of the main theorem: If Q and Q are two Morse index one solutions at s 0, they can be continuously connected to Q 1 = Q 1. By non-degeneracy at s = 1 we have Q s = Q s for s (1 δ, 1]. By non-degeneracy at s 0 s < 1, we have Q s = Q s for s [s 0, 1]. R. Frank Uniqueness of ground states October 19, 2012 # 9
Why does the branch extend all the way up to s = 1? s = sup { s 1 (s 0, 1] : s Q s is C 1 ((s 0, s 1 ), L 2 L α+2 ) } Key technical point: upper and lower bounds on ( ) s/2 Q s 2, Q s α+2 and Q s 2, uniform in s [s 0, s ) Use energy and Pohozaev identities plus Sobolev inequalities to get lower bounds Most difficult: upper bound on Q s 2, say One can deduce that Q s extends continuously to s = s The eigenvalues of L +,s are continuous functions of s up to s By non-degeneracy, λ 2 (L +,s radial ) can never cross zero if it is initially above zero As long as λ 2 (L +,s radial ) > 0 and s < 1, the branch can be continued. Conclusion: s = 1 R. Frank Uniqueness of ground states October 19, 2012 # 10
And now for something completely different... Zeroes of eigenfunctions of fractional Schrödinger operators Theorem 4. Let N 1, 0 < s < 1 and assume that V is a radial, non-decreasing function. Let λ 2 < inf ess spec(( ) s +V ) be the second eigenvalue of ( ) s +V rad. Then λ 2 is simple and the corresponding ψ 2 changes sign exactly once on (0, ). We have (presumably non-sharp) results for higher eigenvalues More precise results for ( ) s on an interval (Banuelos et al., Kwasnicki) The fact that ψ 2 changes sign at most twice on (0, ) follows from the local variational characterization λ k = inf {c s u 2 y 1 2s dx dy + V (x) u(x, 0) 2 dx : R R 2 + u(x, 0) 2 dx = 1, u(, 0) M k 1 } R, M k 1 = span{ψ 1,..., ψ k }, plus Courant argument. (For N = 1 we even conclude that ψ 2 changes sign once.) R. Frank Uniqueness of ground states October 19, 2012 # 11
How to rule out two sign changes in N 2 Lemma 5. Let N 2, 0 < s < 1 and assume that V is a radial, non-decreasing function. Assume that ψ k corresponds to the k-th radial eigenvalue of ( ) s + V. Then ψ k (0) 0. This immediately implies simplicity of all radial eigenvalue. With some more work (a deformation argument) it also implies that ψ 2 changes sign only once. Proof (s = 1/2). The harmonic extension u(x, t) = (exp( t )ψ k )(x) satisfies u = 0 in R N+1 + and t u + V u = λ k u on R N {0}. Following Cabré Sire we consider H(r) = 0 ( r u 2 t u 2) dt (V (r) λ k ) u(r, 0) 2. This is decreasing: H (r) = 2(N 1)r 1 0 r u 2 dt ( r V ) u(r, 0) 2 < 0. The assertion follows from H(0) > H( ) = 0. R. Frank Uniqueness of ground states October 19, 2012 # 12
Back to non-degeneracy Let Q be a positive (and hence radially decreasing) solution of ( ) s Q Q α+1 = Q and let L + = ( ) s (α + 1)Q α + 1. Proposition 6. If N (L + radial ) = 1, then ker L + radial = {0}. Proof (following Chang, Gustafson et al.). Assume 0 ψ ker L + radial. Then ψq dx = 0 and ψq α+1 dx = 0 (orthogonality conditions) (because of equation for Q and dilation invariance ), In particular, ψq(q α Q α (R)) dx = 0 for any R > 0. On the other hand, by corollary, ψ(r) 0 for 0 r r and ψ(r) 0 for r r <. Hence ψq(q α Q α (r )) dr > 0 by monotonicity of Q. This is a contradiction. R. Frank Uniqueness of ground states October 19, 2012 # 13
THANK YOU FOR YOUR ATTENTION! R. Frank Uniqueness of ground states October 19, 2012 # 14