Uniqueness of excited states to -Δu+u-u^3=0 in three dimensions
UUNIQUENESS OF EXCITED STATES TO − ∆ u + u − u = 0 IN THREEDIMENSIONS
ALEX COHEN, ZHENHAO LI, WILHELM SCHLAG
Abstract.
We prove the uniqueness of several excited states to the ODE ¨ y ( t )+ t ˙ y ( t )+ f ( y ( t )) = 0, y (0) = b , and ˙ y (0) = 0 for the model nonlinearity f ( y ) = y − y . The n -th excited state is a solutionwith exactly n zeros and which tends to 0 as t → ∞ . These represent all smooth radial nonzerosolutions to the PDE ∆ u + f ( u ) = 0 in H . We interpret the ODE as a damped oscillator governedby a double-well potential, and the result is proved via rigorous numerical analysis of the energy andvariation of the solutions. More specifically, the problem of uniqueness can be formulated entirelyin terms of inequalities on the solutions and their variation, and these inequalities can be verifiednumerically. Contents
1. Introduction 12. Overview of approach 42.1. Toy example: finding zeros of a function 42.2. Finding and isolating excited states 42.3. Approximating solutions via interval arithmetic 53. Analytical description of the damped oscillator dynamics 53.1. Basic properties of the ODE 53.2. The equation of variation 63.3. Picard approximation 63.4. The equation at infinity 93.5. Limit sets and convergence theorems 103.6. Passing over the saddle 114. Proof of Theorem 1 134.1. Outline of proof 134.2. Planning section 154.3. Proving section 165. Using
VNODE-LP and the data 16References 181.
Introduction
Consider the ODE ¨ y ( t ) + 2 t ˙ y ( t ) + f ( y ( t )) = 0 , (1.1) y (0) = b, ˙ y (0) = 0 . (1.2)In this paper, we will only consider the model case f ( y ) = y − y , but it will be convenient to usethe more general notation for the nonlinearity. Smooth solutions to this ODE exist for all t ≥ W. Schlag was partially supported by NSF grant DMS-1902691. a r X i v : . [ m a t h . C A ] J a n ALEX COHEN, ZHENHAO LI, WILHELM SCHLAG and any b ∈ R , and they are unique. We denote them by y b , or simply y . The singular coefficientat t = 0 can be dealt with by a power series ansatz, or by Picard iteration. Solutions to this ODEcorrespond to radial smooth solutions of the PDE∆ u + f ( u ) = 0 (1.3)in three dimensions, under the identification t = r , the radial variable. Dynamics of (1.1) resemblesparticle motion in a double well as in Figure 1, with time varying friction. The qualitative behavior Figure 1.
Potential function V ( y )exhibits a trichotomy: we either have • ( y b ( t ) , ˙ y b ( t )) → (1 , • ( y b ( t ) , ˙ y b ( t )) → ( − , • ( y b ( t ) , ˙ y b ( t )) → (0 , bound state is a nonzero solution with y ( t ) →
0. Only these solutions giverise to nontrivial solutions u ∈ H ( R ) of the elliptic PDE (1.3). A ground state is a positive boundstate, and an n th excited state is a bound state with precisely n zero crossings. The 0 th boundstate is the ground state. Existence and uniqueness of these bound states have been investigatedsince the 1960s. In 1967, using a variational characterization, Ryder [20, Theorem II] showed theexistence of both ground and excited states with any finite number of zero crossings. In 1972, Coff-man [6, Section 6] related Ryder’s characteristic values to degree theory in infinite dimensions andLyusternik-Schnirelman techniques. Most importantly, [6] also established uniqueness of the groundstate for the cubic case. For more general nonlinearities, ground state uniqueness was then shownby Peletier, Serrin [19, 18], McLeod, Serrin [13], Zhang [24], Kwong [11], and finally in greatestgenerality by McLeod in 1992 [12]. Clemons, Jones [5] gave a different proof of McLeod’s theorembased on the Emden-Folwer transformation and unstable manifold theory. In 1983, Berestycki andLions [1, 2] solved the existence problem of radial bound states for (1.3) for all H subcriticalnonlinearities f ( u ) in all dimensions, see also the earlier work by Strauss [21].However, uniqueness of exited states in the radial class, i.e., for the ODE (1.1), remained openfor most nonlinearities. In fact, in their 2012 text, Hastings, McLeod [10, Chapter 19] list thisproblem as one of three major open problems in nonlinear ODEs. We note that there has beensome uniqueness results for specific nonlinearities; in 2005, Troy [23] proved the uniqueness ofthe first excited state for a piecewise linear nonlinearity by analyzing the explicit solutions, andin 2009, Cort´azar, Garc´ıa-Huidobro, and Yarur [7] proved uniqueness of the first excited statewith restrictions on yf (cid:48) ( y ) /f ( y ). However, neither cover the cubic nonlinearity. In this paper,we provide a rigorous computer-assisted proof of the uniqueness of the first excited state for thecubic nonlinearity. The proof technique combines analytical dynamics with the rigorous ODE solver NIQUENESS OF EXCITED STATES 3 y (0) = b − . − . . . . y ( T ) a s T →∞ Figure 2.
Limiting position y b ( T ) as T → ∞ , plotted as a function of b up to the20th excited state. This graph holds due to Theorem 1 and the rigorous numericalwork done in this paper. VNODE-LP , see Section 2.3 and [16]. The latter works with interval arithmetic and therefore does notcompute precise solutions (which is impossible), but rather intervals containing the solution at anygiven time. These inclusions accommodate all errors incurred through floating point arithmetic,and are therefore themselves free of errors.
Theorem 1.
The first twenty excited states of ODE (1.1-1.2) are unique for f ( y ) = − y + y . The method is robust, and extends to both more general nonlinearities, as well as other dimen-sions. But we will leave the verification of this claim for another paper. The code involved in theproof is publicly available, see the GitHub repository https://github.com/alexander-cohen/NLKG-Uniqueness-Prover . The readers can verify uniqueness of higher excited states beyond the20 th using the arguments of this paper. Computation time is the main obstacle to going furtherthan the twentieth one, to which the authors chose to limit themselves. See Figure 2 for a graphof the limiting position of y b ( T ) as a function of b , up to the twentieth excited state. The rigorousnumerical work done in this paper proves that this graph holds.The uniqueness property of the ground state soliton is of fundamental importance to the clas-sification of its long-term evolution under the nonlinear cubic Schr¨odinger or Klein-Gordon flows.See for example [4] and [14]. The uniqueness property of the excited states should therefore alsobe seen as a bridge to dynamical results. As a first step, one needs to determine the spectrum ofthe linearized operator H = − ∆ + 1 − φ in the radial subspace of L ( R ). Here φ is any radial bound state solution of the PDE (1.3). If φ is the ground state, then it is known that the spectrum over the radial functions contains a uniquenegative eigenvalue, and no other discrete spectrum up to and including 0 energy (nonradially, dueto the translation symmetry, 0 is an eigenvalue of multiplicity 3); see [14]. A particularly delicatequestion pertains to the shape of the spectrum in the interval (0 , ,
1] is a spectral gap, including the threshold, which is nota resonance. Due to the absence of an explicit expression of the ground state soliton, the methodof [8] depended on an approximation of this special solution. Note that without uniqueness such anapproximation has no meaning. The authors intend to investigate the spectral problem of excitedstates in another publication using the methods of [9], which essentially require the uniquenessproperty of the special solution φ (in the case of [9] φ is the so-called Skyrmion). ALEX COHEN, ZHENHAO LI, WILHELM SCHLAG Overview of approach
Toy example: finding zeros of a function.
Suppose we wish to find the number of zerosof the function f as shown in the figure. Numerical computations make it clear that f has exactly3 zeros – how can we use a computer to prove this rigorously?A first approach might be to find the approximate location of those zeros with reasonably highprecision, using floating point arithmetic. Say they lie at approximately y , y , y . Then we can useinterval arithmetic to show rigorously that f is bounded away from zero everywhere but the threesmall intervals ( y − / , y + 1 / y − / , y + 1 / y − / , y + 1 / f has at least one zero ineach of these intervals. Finally, to show that each of them contains at most one zero, we can applythe mean value theorem. If we prove rigorously, using interval arithmetic, that f (cid:48) is bounded awayfrom zero in those intervals, then uniqueness follows. Notice that if f has infinitely many zeroswith a limit point, then f (cid:48) must be zero at that limit point. As expected, this method would breakdown in such a scenario.2.2. Finding and isolating excited states.
We apply a similar idea to the ODE (1.1). We nowoutline the approach by means of the ground state. Suppose we find numerically that the uniqueground state should be at height b ≈ . b ∈ (1 , b − . y b ( t ) > T , and E ( T ) <
0. This will imply by analyticalarguments, see the next section, that y b ( t ) → y b ( t ) is positive, so it is not a ground state.Similarly, we can show that for b ∈ ( b + 0 . , y = 0, and thus is nota ground state. It follows from a connectedness argument that there is some ground state in theinterval ( b − . , b + 0 . exactly one ground state in that interval, wefind some large time T such that δ b ( T ) , ˙ δ b ( T ) < b ∈ ( b − . , b +0 . δ b = ∂ b y b .This means that if b ∗ is the actual ground state (rather than an approximation), for any b > b ∗ inour interval, y b ( T ) < y b ∗ ( T ) and ˙ y b ( T ) < ˙ y b ∗ ( T ) by the mean value theorem. One can then proveassuming this condition that y b ( T ) crosses over zero and lands in the second well, see Lemma 7.This will show that there is at most one ground state in the interval ( b − . , b + 0 . , ∞ ). To this end, we rescalethe ODE (1.1) so that it takes the form ¨ w + t ˙ w + w − b − w = 0, w (0) = 1, ˙ w (0) = 0. Again using VNODE-LP we then show that the solution of this equation exhibits more than any given number ofzeros provided b is sufficiently large. This then implies the same for y b .The same approach works just as well for excited states as it does for ground states. NIQUENESS OF EXCITED STATES 5
Approximating solutions via interval arithmetic.
We now outline our computationalapproach. Our main tool is the
VNODE-LP package for rigorous ODE solving. The supportingwebsite is at [15], and the documentation is available at [17]. VNODE-LP uses exact interval arithmetic, a toolset which allows for rigorous numerical computa-tions. Rather than computing with floating point numbers as usual, interval arithmetic treats allvalues as intervals of real numbers, of the form a = [ a , a ] where a , a are machine representablefloating point numbers. All mathematical operations are rounded properly so that any input withinthe original interval ends up within the output interval. The VNODE-LP package combines intervalarithmetic with ODE solving: given an initial value problem ˙ y = f ( y, t ) with initial values in aninterval b , a starting time interval t , and an ending time interval t , the package outputs aninterval y such that for any b ∈ b , t ∈ t , t ∈ t , y b,t ( t ) ∈ y .A difficulty in applying VNODE-LP to our problem is that ODE (1.1) is singular at t = 0. To dealwith this, we approximate y b ( t ) near t = 0 by Picard iteration. We explicitly bound the error termsin this approximation so that we can rigorously obtain an interval containing y b ( t ) , ˙ y b ( t ) for t small. Then VNODE-LP can be applied to this desingularized initial value problem, and all in all wewe will have rigorous bounds on our solutions and quantities defined in terms of the solutions.Section 5 explains in detail how we use this software, and provides links to websites containingthe code and all supporting data needed in the proof of our theorem. This will hopefully allow thereader to implement the methods of this paper in other related settings.3.
Analytical description of the damped oscillator dynamics
Basic properties of the ODE.
It is an elementary property that smooth solutions of (1.1),(1.2) exist for all times t ≥
0; in fact, we will reestablish this fact below in passing. Taking it forgranted, we note that the energy E ( t ) := 12 ˙ y ( t ) + V ( y ( t )) = 12 ˙ y ( t ) + 14 y ( t ) − y ( t ) satisfies ˙ E ( t ) = − t ˙ y ( t )and thus E ( t ) ≤ E (0) = V ( b ) for all times. In fact, E ( t ) is strictly decreasing unless it is a constantand that can only happen for the unique stationary solutions ( y, ˙ y ) equal (0 ,
0) or ( ± , V ( b ) ≤
0, then E ( t ) < t > y ( t ) = 0 is a constant. We will seebelow that this implies that ( y ( t ) , ˙ y ( t )) → (1 ,
0) as t → ∞ (recall that we are assuming b > y, ˙ y )( t ) approaches the minimum of the potential well on the right of Figure 1 andso inf t> y b ( t ) >
0. The range of b here is 0 < b ≤ √ b > √
2, then V ( y ( t )) ≤ E ( t ) < E (0) = V ( b ) for all t > y ( t ) ( y ( t ) − ≤ b ( b − | y ( t ) | ≤ b for all t ≥
0. We will assume from now on that b > √
2. Rewriting the initialvalue problem (1.1), (1.2) in the form ddt ( t ˙ y ( t )) + t f ( y ( t )) = 0 , In the online version of this paper, click on
VNODE-LP In the online version click on the hyperlink documentation
ALEX COHEN, ZHENHAO LI, WILHELM SCHLAG where f ( y ) = y − y throughout, we arrive at the integral equations y ( t ) = b + (cid:90) t ˙ y ( s ) ds, ˙ y ( t ) = − (cid:90) t s t f ( y ( s )) ds = (cid:90) t s t y ( s )(1 − y ( s ) ) ds (3.1)For short times, we obtain a unique solution by the contraction mapping principle which is smoothnear t = 0. Picard iteration gives better constants, which is important for starting VNODE atsome positive time. We shall determine the quantitative bounds in Section 3.3. But first we recallthe equation of variation of (1.1) relative to the initial height b .3.2. The equation of variation.
We let δ b ( t ) := ∂∂b y b ( t ). Then differentiating (1.1), δ b ( t ) satisfiesthe ODE ¨ δ + 2 t ˙ δ + f (cid:48) ( y ) δ = 0 . with initial conditions δ (0) = 1 and ˙ δ (0) = 0. Notice that the ODE for δ depends on the solution y b ( t ). Altogether, we can make one ODE in four variables that includes y and δ : ddt yv y δv δ = v y − t v y − f ( y ) v δ − t v δ − f (cid:48) ( y ) δ . with initial vector yv y δv δ (0) = b Switching again to the variables t y ( t ), respectively, t δ ( t ), and writing the resulting ODE inintegral form, this is equivalent to Z ( t ) := yv y δv δ ( t ) = b + (cid:90) t v y ( s ) − t − s f ( y ( s )) v δ ( s ) − t − s f (cid:48) ( y ( s )) δ ( s ) ds = b + (cid:90) t v y ( s ) t − s y ( s )(1 − y ( s ) ) v δ ( s ) t − s (1 − y ( s ) ) δ ( s ) ds (3.2)The first three Picard iterates of this system are Z ( t ) = b , Z ( t ) = b − t f ( b )1 − t f (cid:48) ( b ) , Z ( t ) = b − t f ( b ) − t f ( b )1 − t f (cid:48) ( b ) − t f (cid:48) ( b ) (3.3)3.3. Picard approximation.
The purpose of this section is to compare the actual solution Z ( t )in (3.2) to the second Picard iterate Z ( t ) in (3.3) which we denote in the form Z ( t ) =: ˜ y ( t )˙˜ y ( t )˜ δ ( t )˙˜ δ ( t ) = ˜ y ( t )˜ v y ( t )˜ δ ( t )˜ v δ ( t ) (3.4) NIQUENESS OF EXCITED STATES 7
In fact, we will prove the following inequalities on each of the four entries of this vector.
Lemma 1.
Suppose b ≥ √ . Then for all times t ≥ , ˜ y ( t ) ≤ y ( t ) ≤ ˜ y ( t ) + f ( b ) f (cid:48) ( b )120 t ≤ ˜ y ( t ) + b t , ˙˜ y ( t ) ≤ ˙ y ( t ) ≤ ˜ y ( t ) + f ( b ) f (cid:48) ( b )30 t ≤ ˙˜ y ( t ) + b t (3.5) For all ≤ t ≤ t ∗ with t ∗ := min (cid:16)(cid:115) √ b − √ b ( b − , log 4 √ b (cid:17) (3.6) one has ˜ δ ( t ) ≤ δ ( t ) ≤ ˜ δ ( t ) + b t , ˙˜ δ ( t ) ≤ ˙ δ ( t ) ≤ ˙˜ δ ( t ) + b t (3.7) Proof.
Since energy is decreasing and b ≥ √ | y ( t ) | ≤ b for all t ≥
0. Note that f ( b ) ≥ f ( √ > √ , which is the absolute value of the local minima and maxima, so | f ( y ) | ≤ f ( b ) for all | y | ≤ b .Therefore f ( y ( t )) ≤ f ( b ) for all t ≥
0. Substituting this bound into (3.1) yields | ˙ y ( t ) | ≤ t f ( b ) , (3.8)0 ≤ b − y ( t ) ≤ t f ( b ) . (3.9)for all times t ≥
0. Leveraging these bounds, we now compare the actual solution to its secondPicard iterates as in (3.3). In view of (3.4),˜ y ( t ) = b − t f ( b ) , ˙˜ y ( t ) = − t f ( b )and we obtain via the mean value theorem that0 ≤ ˙ y ( t ) − ˙˜ y ( t ) = (cid:90) t s t (cid:2) f ( b ) − f ( y ( s )) (cid:3) ds ≤ f ( b ) f (cid:48) ( b )30 t , ≤ y ( t ) − ˜ y ( t ) ≤ f ( b ) f (cid:48) ( b )120 t , where we used that | f (cid:48) ( y ) | ≤ f (cid:48) ( b ) = 3 b − | y | ≤ b .The last two rows of (3.2) imply that | δ ( t ) − | ≤ (cid:90) t | ˙ δ ( s ) | ds | ˙ δ ( t ) | ≤ t − f (cid:48) ( b ) (cid:90) t s | δ ( s ) | ds ≤ t f (cid:48) ( b ) + t − f (cid:48) ( b ) (cid:90) t s | δ ( s ) − | ds ≤ tb + 3 b (cid:90) t | δ ( s ) − | ds ALEX COHEN, ZHENHAO LI, WILHELM SCHLAG whence h ( t ) := | ˙ δ ( t ) | + µ | δ ( t ) − | with µ := √ b satisfies h ( t ) ≤ tb + µ (cid:90) t h ( s ) dsh ( t ) ≤ b µ ( e µt −
1) (3.10)We infer from the last two rows of (3.2) and (3.4) that δ ( t ) − ˜ δ ( t ) = (cid:90) t ( v δ ( s ) − ˜ v δ ( s )) dsv δ ( t ) − ˜ v δ ( t ) = t − (cid:90) t s ( f (cid:48) ( b ) − f (cid:48) ( y ( s )) δ ( s )) ds = t − (cid:90) t s (cid:2) f (cid:48) ( b ) − f (cid:48) ( y ( s )) + f (cid:48) ( y ( s ))(1 − δ ( s )) (cid:3) ds (3.11)as well as 1 − δ ( t ) = − (cid:90) t v δ ( s ) ds − v δ ( t ) = t − (cid:90) t s f (cid:48) ( y ( s )) δ ( s ) ds (3.12)Let t ∗ > f (cid:48) ( y ( t )) ≥ δ ( t ) ≥ ≤ t ≤ t ∗ . Then by (3.12), δ ( t ) ≤ δ ( t ) − ˜ δ ( t ) ≥ v δ ( t ) − ˜ v δ ( t ) ≥ ≤ t ≤ t ∗ . By (3.10), we have δ ( t ) ≥ e µt ≤ , t ≤ log 4 √ b Moreover, f (cid:48) ( y ( t )) ≥ √ y ( t ) ≥ √ y ( t ) ≥ , t ≤ (cid:115) √ b − √ b ( b − − v δ ( t ) ≤ tb , − δ ( t ) ≤ t b b − y ( t ) ≤ t b Inserting these bounds into (3.11) yields by the mean value theorem v δ ( t ) − ˜ v δ ( t ) ≤ t − (cid:90) t s ( b s + 3 b s / ds ≤ b t δ ( t ) − ˜ δ ( t ) = (cid:90) t ( v δ ( s ) − ˜ v δ ( s )) ds ≤ b t as claimed. (cid:3) The equation at infinity.
As explained in Section 2.2, to prove uniqueness of the firstexcited state we will need to show that all y b have at least two crossings for all sufficiently large b .For the second excited state, we need to do the same with three crossings, and so on. This will beaccomplished by means of the following lemma. Lemma 2.
Let y ( t ) be a solution to ODE (1.1-1.2). Let w ( s ) = b y ( s/b ) . Then w satisfies ¨ w + 2 s ˙ w + w − β w = 0; (3.13) w (0) = 1 , ˙ w (0) = 0 . (3.14) where β := b .Proof. Immediate by scaling. (cid:3)
We will analyse this initial value problem with
VNODE-LP , but as before can only start at positivetimes rather than at t = 0. The analogue of Lemma 1 is the following. We only need to approximatethe ODE in (3.13). Indeed, since the initial condition is fixed, the equation of variations does notarise. Lemma 3.
Suppose < β ≤ , and let ˜ w ( t ) := 1 − − β t , and ˙˜ w ( t ) := − β t . Then for all times t ≥ , ˜ w ( t ) ≤ w ( t ) ≤ ˜ w ( t ) + t , ˙˜ w ( t ) ≤ ˙ w ( t ) ≤ ˙˜ w ( t ) + t
10 (3.15)
Proof.
We write the equation (3.13) in the form ddt ( t ˙ w ( t )) = − t f β ( w ( t ))with f β ( w ) := w − β w = w ( w − β )Solutions are global, and the energy takes the form E β ( t ) = 12 ˙ w ( t ) + V β ( w ( t )) , V β ( w ) = 14 w − β w which is nonincreasing as before. Thus, V β ( w ( t )) ≤ E β ( t ) ≤ V β (1) = − β whence | w ( t ) | ≤ w is of the form w ( t ) = 1 + (cid:90) t ˙ w ( s ) ds, ˙ w ( t ) = − (cid:90) t s t f β ( w ( s )) ds = (cid:90) t s t w ( s )( β − w ( s ) ) ds (3.16)Inserting w = 1 into the right-hand side of the second row of (3.16) gives ˙˜ w ( t ) := − f β (1)3 t , and ˜ w isobtained by inserting this expression into the right-hand side of the first row of (3.16). These areprecisely the approximate solutions appearing in the formulation of the lemma. The stated boundsare now obtained as in Lemma 1 and we leave the details to the reader. (cid:3) Limit sets and convergence theorems.
As we have already noted, an important quantityassociated with the equation (1.1) is the energy E ( y, ˙ y ) = ˙ y + V ( y ), where V ( y ) = (cid:82) y f ( y ) is thepotential energy. Explicitly, V ( y ) = y − y resembles a double well as in the first figure above.Were we to modify our ODE to ¨ y + f ( y ) = 0, then the energy would be preserved. The term t ˙ y adds a time dependent frictional force, so energy decreases monotonically:˙ E ( t ) = ˙ y ¨ y ( t ) + f ( y ( t )) ˙ y ( t ) = − y ( t ) t . The interpretation of the radial form of the PDE (1.3) as a damped oscillator with the role of timebeing played by the radial variable is of essential importance in this section. Tao [22] emphasizedthis already in his exposition of ground state uniqueness, but here we will rely on this interpretationeven more heavily. In particular, the proof of the long-term trichotomy given by the solution vectorof the main ODE approaching one of the three critical points of the potential follows the dynamicalargument in the damped oscillator paper [3].The following lemma determines the ω -limit set of every trajectory in phase space. The lemmacombined with the monotonicty of the energy will help us determine the desired long-term tri-chotomy. Lemma 4. If y ( t ) = y b ( t ) is the global solution to the initial value problem (1.1) , (1.2) , then thereexists an increasing unbounded sequence { t j } such that ( y ( t j ) , ˙ y ( t j )) → (0 , or ( ± , as j → ∞ .Proof. From boundedness of the energy, we see that ∞ (cid:88) n =1 n I n ≤ (cid:90) ∞ ˙ y ( t ) t dt < ∞ , I n := (cid:90) nn − ˙ y ( t ) dt. Therefore, I n j → j → ∞ for some subsequence. We can pick t j ∈ ( n j − , n j ) so that ˙ y ( t j ) → E ( t ) and y ( t ) are bounded, ¨ y is bounded, and differentiating (1.1), we then see that ... y is alsobounded. Therefore I n j → y ( t j ) →
0. This implies that | f ( y ( t j )) | ≤ | ¨ y ( t j ) | + 2 t | ˙ y ( t j ) | → , so there must be a subsequence of y ( t n j ) that converges either to 0 or 1. (cid:3) Next, we establish that each trajectory must converge to the point in its limit set, cf. theconvergence theorems in [3].
Lemma 5.
Either y b ( t ) → − , or y b ( t ) → , or y b ( t ) → as t → ∞ . In all cases ˙ y ( t ) → .Proof. Since E ( t ) is monotonically decreasing, the limit lim t →∞ E ( t ) exists as a real number,which is either negative or non-negative. In the former case, there must be a sequence t j sothat ( y ( t j ) , ˙ y ( t j )) → ( ± ,
0) by Lemma 4. Monotonicity of the energy then implies that E ( t ) = E ( y b ( t ) , ˙ y b ( t )) tends toward the global minimum value of the potential energy, which means that( y b ( t ) , ˙ y b ( t )) → ( ± , E ( t ) → t → ∞ . Suppose y ( t ) doesnot converge to 0. Let τ j denote the j -th time at which ˙ y ( τ j ) = 0, and if y ( t ) does not tend to 0,then { τ j } is an infinite sequence. We will show that this leads to a contradiction since too muchenergy will be lost in each oscillation. To do so, we first upper bound τ j +1 − τ j . Assume withoutloss of generality that ˙ y > τ j and τ j +1 . We have V ( y ) ≤ − y for y ∈ ( − , τ j < t < t < τ j +1 so that y ( t ) = − y ( t ) = 1. In particular, the portion of the trajectorybetween t and t is the part of the trajectory going over the hill in the potential, which should be NIQUENESS OF EXCITED STATES 11 the most time-consuming part of the trajectory, and indeed, (cid:90) t t dt = (cid:90) − y (cid:48) dy = (cid:90) − (cid:112) E ( y ) − V ( y )) dy ≤ (cid:90) (cid:112) E ( t ) + y / dy ≤ (cid:90) √ E ( t )0 (cid:112) E ( t ) dy + 2 √ (cid:90) √ E ( t ) y dy (cid:46) − log E ( t ) , (3.17)assuming that τ j is sufficiently large so that 0 < E ( t ) (cid:28)
1. Note that from the energy, ˙ y ( t ) canonly reverse sign if | y ( t ) | > √
2. Since the energy is always positive, E ( t ) ≥ (cid:90) τ j +1 t ˙ y ( t ) t dt ≥ τ j +1 (cid:90) √ (cid:112) E ( y ) − V ( y )) dy (cid:38) τ j +1 . Substituting this into (3.17), we find that t − t (cid:46) log τ j +1 (3.18)Finally, we show that for small energy, the time spent by y ( t ) in one oscillation outside the interval( − ,
1) is uniformly bounded by some constant. Fix some 0 < ε (cid:28) √ − | f ( y ) | ≥ α > y ∈ B ε ( ±√ ε -neighborhoods of ±√
2. Then there exists a sufficiently large T so that √ < | y ( τ j ) | < √ ε for all τ j > T and that | y ( t ) | /t < α/ t > T . This means that if τ j > T and ˙ y ( t ) > t ∈ ( τ j , τ j +1 ), then¨ y ( t ) = − f ( y ( t )) − t ˙ y ( t ) ≥ α y ( t ) ∈ B ε ( −√ , ¨ y ( t ) = − f ( y ( t )) − t ˙ y ( t ) ≤ − α when y ( t ) ∈ B ε (+ √ . In other words, between τ j and τ j +1 , the initial acceleration and final deceleration are both uni-formly lower bounded. Then there is a uniform constant bounding the time spent by the part of thetrajectory in B ε ( ±√ B ε ( ±√
2) and the interval ( − , τ j +1 − τ j (cid:46) log τ j +1 , whence τ j (cid:46) j log j. The cummulative loss in energy starting from some sufficiently large time τ N is therefore (cid:90) ∞ τ N ˙ y ( t ) t dt ≥ ∞ (cid:88) j = N τ j +1 (cid:90) τ j +1 τ j ˙ y ( t ) dt (cid:38) ∞ (cid:88) j = N j log j , which is not finite, a contradiction. (cid:3) Passing over the saddle.
We now turn to a lemma which establishes the following naturalproperty: consider the value 0 < y ( T ) = ε (cid:28) T so large that y ( t ) > t > T . Then any other y b with y b ( T ) ∈ (0 , ε ) and ˙ y b ( T ) < ˙ y ( T ) needs to cross 0after time T . For simplicity, we prove the lemma for f ( y ) = y − y , but it is easy to see that itworks for many nonlinearities via the same argument. Lemma 6.
Suppose b ∗ ∈ (0 , ∞ ) is a bound state, and without loss of generality assume y b ∗ ( t ) approaches from the right. That is, ˙ y b ∗ ( t ) < eventually. Let T be large enough that y b ∗ hasno more zero crossings after time T , and < y b ∗ ( T ) ≤ / √ . If y b ( t ) is another solution with < y b ( T ) < y b ∗ ( T ) and ˙ y b ( T ) < ˙ y b ∗ ( T ) , then y b ( T ) has a zero crossing after time T . Proof.
Let s ( t ) = y b ∗ ( t ) − y b ( t ). Then¨ s ( t ) + 2 t ˙ s ( t ) = f ( y b ( t )) − f ( y b ∗ ( t )) . (3.19)At t = T , s ( T ) > s ( T ) >
0. If y b ( t ) does not cross zero for any t > T , y b ( t ) → s ( t ) → −
1. In either case, s ( t ) must reach a maximum after t = T , so there existsa t ∗ > T so that ˙ s ( t ∗ ) = 0, s ( t ∗ ) > s ( t ∗ ) ≤
0. Then by (3.19), f ( y b ( t ∗ )) − f ( y b ∗ ( t ∗ )) ≤ f ( y ) is strictly decreasing for y ∈ (0 , / √ s ( t ∗ ) >
0, (3.20) leads to acontradiction with the assumption y b ∗ ( T ) ≤ / √ (cid:3) Note that the only property of f we used is that f (cid:48) (0) <
0, which holds for all nonlinearitiesassociated with a double-well potential.The next lemma provides a sufficient condition under which the trajectory will pass over thehill and be trapped in the following well. The underlying mechanism is the consumption of energydue to a necessary oscillation around the left well. If this amount exceeds the energy present atthe pass over the saddle at y = 0, then the remaining energy is negative, ensuring trapping. Thelemma will ensure that if y b ∗ ( t ) is a bound state, then for initial values b ∈ ( b ∗ , b ∗ + ε ) for somesmall (cid:15) , y b ( t ) will necessarily fall into the following potential well. Lemma 7.
Suppose y ( t ) is a solution of (1.1), (1.2) such that for some T > , ≤ y ( T ) < , ˙ y ( T ) < , < E ( T ) < ,E ( T ) (cid:18) T − E ( T ) + 32 (cid:19) < . Then if y ( t ) has a zero after (or at) time T , it must proceed to fall into the left well. That is, y ( t ) → − , and y ( t ) has no further zero crossings.Proof. Suppose y ( t ) has another zero crossing, say the minimal time t > T with this propertyis t > T . Then ˙ y ( t ) < y ( t ) until after y ( t ) haspassed −
1. So we can define T > T to be the first time after T at which y ( T ) = − / y ( t ) does not fall into the left well; in this situation, E ( T ) = α >
0. Then we must have E ( T ) > α . Let t ( s ), − / < s < /
2, be the nearest time after/before T such that y ( t ( s )) = s .Then we have (recall 0 < α < ) T − T < (cid:90) / − / | ˙ y ( t ( s )) | ds = (cid:90) / − / (cid:112) E ( t ( s )) − V ( s )) ds< (cid:90) / − / (cid:112) α + s − s / ds < (cid:90) / − / (cid:112) α + s / ds< (cid:90) √ α √ α ds + 2 √ (cid:90) / √ α s ds = √ − − α ) < − α ) . Thus T < T − α ). Next, we observe that if the conditions of the lemma are satisfied, then morethan α energy is lost in going from y = − / y = −
1. Letting t ( s ) be as before and assuming NIQUENESS OF EXCITED STATES 13 y ( t ) does not fall into the left well, E ( t ( s )) > − < s < − /
2. Using dEds = y dEdt = | ˙ y | t , wehave ∆ E = 2 (cid:90) − / − | ˙ y ( t ( s )) | t ( s ) ds = 2 (cid:90) − / − (cid:112) E ( t ( s )) − V ( t ( s ))) t ( s ) ds> (cid:90) − / − (cid:112) s − s / t ( s ) ds > T (cid:90) − / − | s | ds = 38 T where T > T is the first time at which y = −
1. Now we have T − T = (cid:90) − / − | ˙ y ( t ( s )) | ds = (cid:90) − / − (cid:112) E ( t ( s )) − V ( s )) ds< (cid:90) − / − (cid:112) y − y / dy< (cid:90) − / − | y | dy < . Altogether, we have ∆
E >
38 1 T − α ) + , and if ∆ E > α , then E ( T ) <
0, and the particle falls into the left well. This occurs when α (cid:18) T − α ) + 32 (cid:19) < α < E ( T ) and α ln( α ) is monotone decreasing for 0 < α < /
4, in the situation of thelemma, if
T E ( T ) − E ( T ) ln E ( T ) + 32 E ( T ) < T . (cid:3) Proof of Theorem 1
Outline of proof.
Theorem 1 is proved by running a
C++ computer program which combinesthe rigorous numerics of
VNODE-LP with the analytical lemmas of the preceding section. This codeis divided into two parts: a planning section, and a proving section. The planning section of thecode creates a plan for proving the first several bound states are unique, and the proving sectionexecutes this plan and outputs a rigorous proof of uniqueness. Separating these two sections isadvantageous because only the proving section must be mathematically rigorous, so only that partof the code needs to be checked for correctness. The planning section can be modified without fearof compromising the rigour of the code.In what follows we treat
VNODE-LP as a black box that takes in an input interval b = ( b , b ) anda time interval t = ( t , t ), and outputs an interval y b ( t ) = ( y , y ) × ( ˙ y , ˙ y ) × ( δ , δ ) × ( ˙ δ , ˙ δ )which contains y b ( t ) for any b ∈ ( b , b ) and t ∈ ( t , t ). We can also integrate the equation atinfinity (3.13) rigorously. To implement this functionality, we use the the explicit error boundsgiven in Lemma 1 to move past the singularity at t = 0. For instance, we may pick t = (0 . , . y a vector of four intervals which contain any y b ( t ), y ( T ) Position over time with VNODE intervals (scaled up x100)
Figure 3.
VNODE-LP numerical integration with solution intervals scaled up × y -uncertainty occur when δ = y b ( t ) ∼
0, andalthough it is not shown here, the ˙ y -uncertainty is larger at these points. b ∈ ( b , b ), and t ∈ (0 . , . y , t directlyinto VNODE-LP , which rigorously integrates to the desired ending time. See Figure 3 for a depictionof
VNODE-LP integration with solution intervals.We now describe our procedure for the ground state and first excited state, before describing theplanning and proving sections in detail. Bound states can only occur in the range b ∈ ( √ , ∞ ). Toprove the ground state is unique, we split this range into four intersecting intervals, I , I , I , I .For instance, we can take: I = (1 . , . , I = (4 . , . , I = (4 . , . , I = (6 . , ∞ ) . Now, numerical evidence shows that the ground state occurs in the range I . So, in the range I , the solution will eventually fall into the right energy well. We use VNODE-LP to prove this bysplitting I into smaller chunks, and verifying that in each of these chunks the energy of the solutioneventually falls below zero. We deal with the range I in the same way, by showing that for all b ∈ I , y b ( t ) eventually has negative energy. In the interval I , the solution should always have atleast one zero crossing. We prove this using the equation at infinity (3.16) as discussed in § b ∈ I corresponds to the finite range β ∈ (0 , . w ( t ) is eventually negative, we prove that y b ( t )eventually crosses zero for all b ∈ I . Notice that we could have replaced the interval I by I ∪ I ,and it would still be true that in this range there is at least one zero crossing. We handle with I separately because the numerics of ODE (3.16) are delicate near bound states, so I acts as a buffer interval .The only range left is I , which actually contains the ground state. We must show that I contains at most one ground state, and that it contains no first or higher excited states. To thisend, we use Lemmas 6 and 7 respectively. At some time T >
0, say T = 6, any y b ( T ) for b ∈ I willbe positive, moving in the negative direction, and small in magnitude. We use VNODE-LP to provethat δ b ( T ) , ˙ δ b ( T ) < b ∈ I . Let b ∈ I be a ground state. Then for any b > b , the mean valuetheorem implies that y b ( T ) < y b ( T ) and ˙ y b ( T ) < ˙ y b ( T ). By Lemma 6, y b ( T ) must cross zero. It NIQUENESS OF EXCITED STATES 15 follows that there is at most one ground state in I . Next, we check that the conditions of Lemma7 are satisfied for all y b ( T ), b ∈ I . This implies that if any solution y b ( t ) does cross zero, it mustfall into the left energy well, and cannot be a higher excited state. Altogether this shows that thereis at most one ground state for the ODE (3.1), as desired. It also follows from this analysis thatthe ground state exists, so we have successfully shown the ground state exists and is unique.We note a subtlety in this argument. A pathological issue would be that by time T = 6, y b ( T )crossed all the way to the left well, come back to the right well, and then started to approach y = 0from the right. This would kill our later attempt to prove that the second excited state is unique,because we would miss a second excited state in I . To deal with this, we use energy considerationsto bound | ˙ y b ( t ) | , and we make small enough time steps with VNODE-LP so that the solution cannotcross zero twice in between time steps. Then, we can be sure that the number of zero crossingsobserved by
VNODE-LP up to some time T is the actual number of zero crossings for all solutions inour initial interval b , up to time T .To prove the ground state is unique, we split the range ( √ , ∞ ) into subintervals to which weapplied three different proof methods. The method for I and I was FALL : we proved that thesolution eventually has negative energy and thus cannot be a bound state. The method for I was INFTY CROSSES MANY , we used the equation at infinity to show that there are sufficiently manyzero crossings and thus no ground states. The method for I was BOUND STATE GOOD , we used theanalytical Lemmas 6, 7 to show that there was at most one ground state and no other bound states.These are the same methods we use to deal with higher excited states. We have used the samenotation here as is used in the code, for ease of verifying that the code follows the mathematicalargument.Let us extend our procedure to prove the first excited state is unique. We split up ( √ , ∞ ) intosix pieces: I = (1 . , . , I = (4 . , . , I = (4 . , . ,I = (14 . , . , I = (14 . , . , I = (16 . , ∞ ) . We apply the
FALL method to I , I , I , we apply the BOUND STATE GOOD method to I , I , and weapply the INFTY CROSSES MANY method to I . For the interval I , our careful stepping procedure asdescribed above lets us find a time T >
0, for example T = 8, such that y b ( T ) crosses zero exactlyonce by time T for all b ∈ I . We can also verify that ˙ y b ( T ) > δ b ( T ) >
0, ˙ δ b ( T ) >
0, so that theconditions of Lemma 6 are satisfied and there is at most one first excited state in the interval I .Next we verify that the conditions of Lemma 7 are satisfied uniformly for b ∈ I at time T , so thatthere are no second or higher excited states in I . Altogether this shows that the first excited stateexists and is unique, and sets us up to prove subsequent excited states are unique as well.4.2. Planning section.
We now describe the planning section of the code. Given a value N ≥ I , I , . . . , I k , along with which method is to be used in eachinterval. The proving section will use this plan to verify that all bound states up to the N th (that is,all bound states with ≤ N zero crossings, or in other words the first N excited states) are unique.Let b , b , . . . , b N denote the locations of the first N excited states (assuming for now that theyare unique). We find their locations numerically with a binary search. To find b k , we keep trackof a lower bound l < b k and an upper bound u > b k , and at each iteration, check how many times y m ( t ) crosses zero, m = ( l + u ) /
2. If y m ( t ) crosses zero more than k times, we set u := m , and ifnot we set l := m . We iterate until we have a small enough interval ( l, m ) containing b k .Next, we find small enough intervals around each bound state so that the BOUND STATE GOOD method can run successfully for each bound state. We start with a large interval around b k , width0 .
5, and then keep on dividing the width by two until
BOUND STATE GOOD succeeds.Third, we fill in the space between the bound states with
FALL intervals. We include a bufferinterval above the last bound state so as to make
INFTY CROSSES MANY run faster.
Finally, we create an interval βββ = (0 , β ) corresponding to the infinite interval (1 /β, ∞ ) where wewill show the ODE crosses zero at least N + 1 times.4.3. Proving section.
The proving section receives a list of intervals and methods from the plan-ning section, and outputs a rigorous proof that the first N excited states are unique. The first stepis to verify that subsequent intervals intersect each other, so that every real number in the range( √ , ∞ ) is covered by some interval. Next, the different methods are implemented as follows.The FALL method receives an interval, e.g. (1 . , . b in that interval y b ( t ) eventually has negative energy. It begins by attempting to integrate with that potentiallyvery large input interval for b . Of course, VNODE-LP will likely fail to integrate with such a largeinput interval. If this happens, we bisect the interval into two halves, an upper and lower half,and recursively apply the
FALL method to each half. Once the starting intervals are small enough,
VNODE-LP will successfully integrate and prove that the energy is eventually negative. This bisectionmethod allows us to use larger intervals away from the bound states and smaller intervals closer tothe bound states, where the computations are more delicate.The
BOUND STATE GOOD method receives an interval I which supposedly contains an n th boundstate. It must prove that there is at most one n th bound state, no lower bound states, and nohigher bound states in I . We use a careful stepping procedure to find some time T > b ∈ I , y b ( t ) crosses zero exactly n times by time T . This already shows that I doesn’t containany lower bound states. Increasing T if necessary, we also verify that ˙ y b ( T ), δ b ( T ), and ˙ δ b ( T ) allhave the opposite sign as y b ( T ) uniformly in I . As discussed earlier, Lemma 6 then implies thatthere is at most one n th bound state in I . Finally, we verify that the conditions in Lemma 7 applyat time T , so there are no higher bound states in I . Throughout we use interval arithmetic, neverfloating point arithmetic.The INFTY CROSSES MANY method works similarly to the
FALL method. We bisect the interval(0 , β ) into smaller pieces, and in each of these small pieces we prove that w ( t ) has at least N + 1crossings.Altogether, these methods show that if it exists, the n th bound state (counting from n = 0)must be unique and lie in the n th BOUND STATE GOOD interval. This proves that all bound states upto the N th are unique, as desired. In fact, the code may also be used to show these bound statesexist by counting crossing numbers, but this is already known by synthetic methods [10].5. Using
VNODE-LP and the data
The code and full output logs from the proof procedure can be found at https://github.com/alexander-cohen/NLKG-Uniqueness-Prover , the most recent commit at the time of writ-ing is 9cf63c06ca1838e64dd35fe11ca4fdfd45591714. The code is contained in the single
C++ file“nlkg uniqueness prover.cc”, and output logs are titled “uniqueness output N=*.txt”. The codeproved the first 20 excited states are unique in ∼
4h running on a MacBook Pro 2017, 2.5 GHz.Time is the main limiting factor to proving uniqueness of more excited states. We summarize theoutput of the proof for N = 3 excited states. Intervals are rounded for space, in actuality theyhave a nonempty intersection. See Figure 4 for a visual representation of the same information.Interval Method Details[1.414, 4.266] FALL [4.266, 4.433]
BOUND STATE GOOD
Bound state 0, used T = 1 . y ( T ) ∈ [0 . , . y ( T ) ∈ [ − . , − . δ ∈ [ − . , − . δ ∈ [ − . , − . NIQUENESS OF EXCITED STATES 17 [4.433, 14.095]
FALL [14.085, 14.115]
BOUND STATE GOOD
Bound state 1, used T = 2 . y ( T ) ∈ − . , y ( T ) ∈ . , δ ∈ . , δ ∈ . , FALL [29.090, 29.174]
BOUND STATE GOOD
Bound state 2, used T = 4 . y ( T ) ∈ . , y ( T ) ∈ − . , δ ∈ − . , δ ∈ − . , FALL [49.339, 49.381]
BOUND STATE GOOD
Bound state 3, used T = 5 . y ( T ) ∈ − . , y ( T ) ∈ [0 . , . δ ∈ . , δ ∈ . , FALL [0.000, 0.019]
INFTY CROSSES MANY
FALLBOUND_STATE_GOODINFTY_CROSSES_MANY
Figure 4.
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Department of Mathematics, Yale University, New Haven, CT 06511, USA
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