Stability of the saddle solutions for the Allen-Cahn equation
aa r X i v : . [ m a t h . A P ] J a n STABILITY OF THE SADDLE SOLUTIONS FOR THE ALLEN-CAHNEQUATION
YONG LIU, KELEI WANG, AND JUNCHENG WEI
Abstract.
We are concerned with the saddle solutions of the Allen-Cahn equation con-structed by Cabr´e and Terra [5, 6] in R m = R m × R m . These solutions vanish precisely onthe Simons cone. The existence and uniqueness of saddle solution are shown in [5, 6, 7].Regarding the stability, Schatzman [31] proved that the saddle solution is unstable for m = 1 , Cabr´e [7] showed the instability for m = 2 , m ≥
7. This hasleft open the case of m = 4 , ,
6. In this paper we show that the saddle solutions arestable when m = 4 , ,
6, thereby confirming Cabr´e’s conjecture in [7]. The conjecturethat saddle solutions in dimensions 2 m ≥ AMS 2010 Classification: primary 35B08, secondary 35B06, 35B51, 35J15.1.
Introduction and statement of the main result
Allen-Cahn type equation is a model arising from the phase transition theory. In thispaper, we will investigate the stability of the saddle solutions to the following Allen-Cahnequation:(1) − ∆ u = u − u , in R n . Properties of solutions for this equation have delicate dependence on the dimension n . Inthe simplest case n = 1 , we know all the solutions, thanks to the phase plane analysistechnique. In this case, (1) has a heteroclinic solution H ( x ) = tanh (cid:16) x √ (cid:17) . It is monotoneincreasing and plays an important role in the De Giorgi conjecture. As we will see, thisfunction also plays a role in our later analysis on the stability in higher dimensions. Recallthat the De Giorgi conjecture states that monotone bounded solutions of (1) have tobe one-dimensional if n ≤ . This conjecture has been proved to be true in dimension n = 2(Ghoussoub-Gui [19]), n = 3 (Ambrosio-Cabr´e [3]). In dimension 4 ≤ n ≤ , Savin[30] proved it under an additional limiting condition:lim x n →±∞ u ( x ′ , x n ) = ± . Counter examples in dimension n ≥ the minimal surface theory. To explain this, let us recall some basic facts from the minimalsurface theory. In R , there is a famous minimal cone with one singularity at the originwhich minimizes the area, called Simons cone. It is given explicitly by: (cid:8) x + ... + x = x + ... + x (cid:9) . The minimality of this cone is proved in [4] and this property is related to the regularitytheory of minimal surfaces. More generally, if we consider the so-called Lawson’s cone(2 ≤ i ≤ j ) C i,j := (cid:26) ( x, y ) ∈ R i ⊕ R j : | x | = i − j − | y | (cid:27) , then for i + j ≤ , C i,j is unstable minimal cone(Simons [33]). For i + j ≥ , and( i, j ) = (2 , , C i,j are area minimizing, and C , is not area minimizing but it is one-sidedminimizer. (See [2], [10], [24], [27] and [28]).There are analogous objects as the cone C m,m in the theory of Allen-Cahn equation.They are the so-called saddle-shaped solutions, which are solutions in R m of (1) vanishingexactly on the cone C m,m (See [8, 20, 31] for discussion on the dimension 2 case, andCabr´e-Terra [5, 6] and Cabr´e [7] for higher dimension case). We denote them by U m . Inthis paper, we will simply call it saddle solution. It has been proved in [7] that thesesolutions are unique in the class of symmetric functions. Furthermore in [5, 6] it is provedthat for 2 ≤ m ≤ , the saddle solution is unstable, while for m ≥ , they are stable [7].It is conjectured in [7] that for m ≥ , U m should be stable. In this paper, we confirm thisconjecture and prove the following Theorem 1.
The saddle solution U m is stable for m = 4 , , . As a corollary, Theorem 1 together with the result of Cabr´e tells us that U m is stablefor m ≥ m ≤ . We remark that actually U m is conjectured to be a minimizer of the corresponding energyfunctional for all m ≥ . But this seems to be difficult to prove at this moment.Let us now briefly explain the main idea of the proof. We focus on the case of m = 4 . That is, saddle solution in R . Suppose u depends only on the variables s := p x + ... + x m and t := p x m +1 + ... + x m . Then (1) reduces to(2) − ∂ s u − ∂ t u − m − s ∂ s u − m − t ∂ t u = u − u . Throughout the paper, we use the notationΩ := { ( s, t ) : s > t > } , Ω ∗ := { ( s, t ) : s > | t | > } . Then the saddle solution satisfies U ( s, t ) = − U ( t, s ) and U > . We will use L todenote the linearized Allen-Cahn operator around U : Lη := ∆ η − (cid:0) U − (cid:1) η. By definition, U is stable if and only if: Z R ( η · Lη ) ≤ , for any η ∈ C ∞ (cid:0) R (cid:1) . TABILITY OF SADDLE SOLUTIONS 3
To prove Theorem 1, we would like to construct a positive function Φ satisfying(3) L Φ ≤ R . It is known that the existence of such a supersolution implies the stability of U. We define f := tanh (cid:16) st (cid:17) √ s √ s + t + 14 . (cid:0) − e − s t (cid:1)! ( s + t ) − . ,h := − tanh (cid:18) ts (cid:19) √ t √ s + t + 14 . (cid:16) − e − t s (cid:17)! ( s + t ) − . . Then we set Φ := f U s + hU t + 0 . (cid:16) s − . e − t + t − . e − s (cid:17) . Here U s , U t are the derivatives of U with respect to s and t. We will prove in Section 3that Φ satisfies (3) . The choice of f is governed by the Jacobi fields of the Simons cone,which are of the form c ( s + t ) − + c ( s + t ) − . Note that 2 . ∈ (2 , . We also point outthat the admissable constants choosen here are not unique.The key ingredients of our proof are some estimates of the first and second derivatives of U, obtained in the next section. These estimates are partly inspired by the explicit saddlesolution in the plane of the elliptic sine-Gordon equation(4) − ∆ u = sin u. The double well potential of this equation is 1 + cos u. It can be checked(see [25]) that thefunction 4 arctan cosh (cid:16) y √ (cid:17) cosh (cid:16) x √ (cid:17) − π is a saddle solution to (4) . However, in dimension 2 m with m > , (we believe)the saddlesolution of (4) does not have explicit formula. More generally, one may conjecture thatthe saddle solution in dimension 8 is stable for general Allen-Cahn type equations of theform ∆ u = F ′ ( u ) , where F is a double well potential. However, as we shall see later on in this paper, itseems that the stability of the saddle solution will also depend on the nonlinearity F .(Atleast, our computations have used the explicitly formula of the one dimensional heteroclinicsolution)This paper is organized as follows. In Section 2, we obtain some point-wise estimatesfor the derivatives of U in dimension 8 . The key will be the estimate of u s + u t . In Section3, we use these estimates to show that Φ is a supersolution of the linearized operator indimension 8 . In Section 4, we briefly discuss the case of dimensions 10 and 12 . Acknowledgement
Y. Liu is partially supported by “The Fundamental ResearchFunds for the Central Universities WK3470000014,” and NSFC no. 11971026. K. Wangis partially supported by NSFC no. 11871381. J. Wei is partially supported by NSERC ofCanada.
Y. LIU, K. WANG, AND J. WEI Estimates for the saddle solution and its derivatives in R . In this section, we analyze the saddle solution in dimension 8 . That of dimension 10 and12 follows from straightforward modifications.One of the main difficulties in the proof of the stability stems from the fact that wedon’t have an explicit formula for the saddle solution . Hence we need to estimate U andits derivatives. This will be the main aim of this section.To begin with, we would like to control U from below and above. For this purpose, weshall construct suitable sub and super solutions.Recall that U is the saddle solution in dimension 2 . Let s = q x + ... + x , t = q x + ... + x . Then the function U ( s, t ) satisfies(5) − ∂ s U − ∂ t U = U − U . In the rest of the paper, ∆ will represent the Laplacian operator in dimension 8 . That is,in the ( s, t ) coordinate, ∆ := ∂ s + ∂ t + 3 s ∂ s + 3 t ∂ t . Let H ( x ) = tanh (cid:16) x √ (cid:17) be the one dimensional heteroclinic solution: − H ′′ = H − H . It has the following expansion: H ( x ) = 1 − e −√ x + O (cid:16) e − √ x (cid:17) , for x large.Moreover, √ H ′ = 1 − H .For simplicity, in the rest of this section, we will also write U as u. Recall that boundedsolutions of Allen-Cahn equation satisfy the Modica estimate:(6) 12 |∇ u | ≤ F ( u ) := (1 − u ) . This inequality will be used frequently later on in our analysis. Note that it providesan upper bound for the gradient. The lower bound of the gradient turns out to be muchmore delicate. Nevertheless, we will prove in this section that(7) u s u + u ss ≥ . Let d ( s, t ) := u + u s . Then inequality (7) implies ∂ s d ≥ d ( s, t ) ≥ d ( t, t ) , in Ω . This inequality will give us a lower bound of u s , provided we have some information of d ( t, t ) . The proof of (7) is quite nontrivial and requires many delicate estimates. It willbe one of the main contents in this section.
TABILITY OF SADDLE SOLUTIONS 5
Following Cabr´e [7], we introduce the new variables y = s + t √ , z = s − t √ . Then the Allen-Cahn equation has the form:(8) − ∂ y u − ∂ z u − y − z ( y∂ y u − z∂ z u ) = u − u . The estimates obtained in this paper rely crucially on the following maximum principle,due to Cabr´e [7].
Theorem 2 (Proposition 2.2 of [7]) . Suppose c ≥ in Ω . Then the maximum principleholds for the operator L − c. It is known that u s > u t < . Moreover, based on this maximum principle,it is proved in [7] that u st ≥ , u tt ≤ . But u ss will change sign in Ω . Indeed, u ss is positive near the origin and y axis. But we don’t know the precise region where u ss ispositive. Here we point out that the estimate of the upper bound of | u tt | near the s axisis the most difficult one.Differentiating equation (2) with respect to s and t, we obtain(9) Lu s = 3 s u s , Lu t = 3 t u t . Lemma 3. In Ω , U satisfies t∂ s U + s∂ t U ≤ . Proof.
Consider the linearized operator around U : P φ := ∂ s φ + ∂ t φ + (cid:0) − U (cid:1) φ. We compute P ( ∂ s U ) = 0 , P ( ∂ t U ) = 0 . Moreover, P ( t∂ s U ) = 2 ∂ s ∂ t U , and P ( s∂ t U ) = 2 ∂ s ∂ t U . Hence, using the fact that ∂ s ∂ t U ≥ , we get P ( t∂ s U + s∂ t U ) ≥ . By the maximum principle(Theorem 2), and the boundary condition: t∂ s U + s∂ t U = 0 , on ∂ Ω , we get t∂ s U + s∂ t U ≤ , in Ω . This completes the proof. (cid:3)
The next result is more or less standard.
Lemma 4.
The functions H ( y ) H ( z ) and U ( s, t ) are super solutions of U in R . Con-sequently, U ( s, t ) ≤ U ( s, t ) ≤ H ( y ) H ( z ) in Ω . Y. LIU, K. WANG, AND J. WEI
Proof.
We first compute − ∆ ( H ( y ) H ( z )) + ( H ( y ) H ( z )) − H ( y ) H ( z )= − H ′′ ( y ) H ( z ) − H ( y ) H ′′ ( z ) + ( H ( y ) H ( z )) − H ( y ) H ( z ) − yH ′ ( y ) H ( z ) − zH ( y ) H ′ ( z ) y − z = H ( y ) H ( z ) (cid:0) − H ( y ) (cid:1) (cid:0) − H ( z ) (cid:1) − yH ′ ( y ) H ( z ) − zH ( y ) H ′ ( z ) y − z . Note that when y > , the function yH ′ ( y ) H ( y ) is monotone decreasing. It follows that yH ′ ( y ) H ( z ) − zH ( y ) H ′ ( z ) ≤ , if y ≥ z. This in turn implies that − ∆ ( H ( y ) H ( z )) + ( H ( y ) H ( z )) − H ( y ) H ( z ) ≥ . Therefore H ( y ) H ( z ) is a supersolution.Next, by Lemma 3, we have − ∆ ( U ( s, t )) − U ( s, t ) + U ( s, t )= − s ∂ s U − t ∂ t U ≥ . Hence U is also a supersolution of U , in Ω . Indeed, the fact that U ( y, z ) ≤ H ( y ) H ( z )has already been proved in [31]. (cid:3) Note that although U is a supersolution, we still don’t have explicit formula for U . Onthe other hand, using H ( y ) H ( z ) , the upper bound near the origin can be improved byiterating the solution once. Indeed, after some tedious computation, we can show that u isbounded from above near the origin by 0 . yz. Note that for y, z small, the supersolution H ( y ) H ( z ) ∼ . yz. We remark that in Ω , the saddle solution is not concave. However, we conjecture thatits level lines should be convex. But we don’t know how to prove it at this moment.Next, we want to find (explicit) subsolutions of u. In R , it is known[31] that the function H (cid:16) y √ (cid:17) H (cid:16) z √ (cid:17) is a subsolution of U . In higher dimensions, the construction of (explicit)subsolutions are more delicate. We have the following
Lemma 5.
For a ∈ (0 , . , the function H ( ay ) H ( az ) is a subsolution of u. Proof.
Let us denote H ( ay ) H ( az ) by η and write ˜ y = ay, ˜ z = az. Then − ∆ η − η + η isequal to H (˜ y ) H (˜ z ) (cid:0) a − − a H (˜ y ) − a H (˜ z ) + H (˜ y ) H (˜ z ) (cid:1) − a √ y − ˜ z (cid:16) ˜ yH (˜ z ) − ˜ z ˜ H (˜ y ) + H (˜ y ) H (˜ z ) (˜ zH (˜ z ) − ˜ yH (˜ y )) (cid:17) . TABILITY OF SADDLE SOLUTIONS 7
This is an explicit function of the variables ˜ y and ˜ z. One can verify directly that it isnegative when a ∈ (0 , . . (cid:3) We remark that this subsolution is not optimal, especially regarding the decaying rateaway from the Simons cone. In the sequel, we shall write the supersolution H ( y ) H ( z ) as u ∗ . We also set φ = u ∗ − u ≥ . To estimate φ, we introduce the function ρ ( z ) = H ′ ( z ) Z z (cid:18) H ′− Z + ∞ s H ′ (cid:19) ds. Note that the function ρ can be explicitly written down and it satisfies ρ ′′ − (cid:0) H − (cid:1) ρ = − H ′ . Lemma 6. In Ω , we have: φ ≤ H ( y ) ( H ( z ) + zH ′ ( z )) y − z . Moreover, (10) φ ≤ (cid:18) t − s (cid:19) H ( y ) ρ ( z ) , for z > . Proof.
The function φ satisfies − ∆ φ + (cid:0) u − (cid:1) φ = − uφ − φ − ∆ u ∗ + u ∗ − u ∗ . Recall that u ∗ is a supersolution and we have − ∆ u ∗ + u ∗ − u ∗ = H ( y ) H ( z ) (cid:0) − H ( y ) (cid:1) (cid:0) − H ( z ) (cid:1) − yH ′ ( y ) H ( z ) y − z + 6 zH ( y ) H ′ ( z ) y − z . (11)Let g ( z ) = ( H ( z ) + zH ′ ( z )) . We compute∆ (cid:18) g ( z ) y − z (cid:19) = (cid:18) y + 2 z ( y − z ) − yy − z y ( y − z ) (cid:19) g + (cid:18) z + 2 y ( y − z ) − zy − z zy − z (cid:19) g + 2 2 zy − z y − z g ′ − zy − z y − z g ′ + g ′′ y − z . The left hand side is equal to − y − z ( y − z ) g − z ( y − z ) g ′ + g ′′ y − z . Y. LIU, K. WANG, AND J. WEI
It follows that∆ (cid:18) H ( y ) y − z g ( z ) (cid:19) = − y − z ( y − z ) H ( y ) g − z ( y − z ) H ( y ) g ′ − H ( y ) H ( z ) y − z + (cid:18) H ′′ ( y ) + 6 yH ′ ( y ) y − z (cid:19) gy − z − y ( y − z ) H ′ ( y ) g = − y − z ( y − z ) H ( y ) − √ H ( y ) H ′ ( y ) y − z + 2 yH ′ ( y )( y − z ) ! g − zy − z H ( y ) g ′ + H ( y ) g ′′ y − z . Let a be a constant to be determined later on, we have L (cid:18) aH ( y ) gy − z − φ (cid:19) ≤ − aH ( y ) H ( z ) y − z − ∆ u ∗ − u ∗ + u ∗ + 3 a ( u ∗ − u ) y − z H ( y ) g + − y − z ( y − z ) H ( y ) − √ H ( y ) H ′ ( y ) y − z + 2 yH ′ ( y )( y − z ) ! ag − azH ( y )( y − z ) g ′ − uφ − φ . Let us denote aH ( y ) gy − z − φ by η. Note that for z close to 0 ,H ( y ) H ( z ) (cid:0) − H ( y ) (cid:1)(cid:0) − H ( z ) (cid:1) ≤ yH ′ ( y ) H ( z ) y − z . If we choose a = 8 , then Lη ≤ , in the region where η ≤ . Hence by maximum principle,(12) φ ≤ H ( y ) ( H ( z ) + zH ′ ( z )) y − z in Ω . TABILITY OF SADDLE SOLUTIONS 9
It remains to prove (10) . We first observe that due to (12) , the inequality (10) is truefor z = 1 . Let ˜ η = (cid:0) t − s (cid:1) H ( y ) ρ ( z ) . We compute L ˜ η + 3 (cid:0) u − u ∗ (cid:1) ˜ η = (cid:18) s − t (cid:19) H ( y ) ρ + √ s − √ t ! H ′ ( y ) ρ + √ s + √ t ! H ( y ) ρ ′ + (cid:18) t − s (cid:19) − H ( y ) + 3 √ (cid:18) s + 1 t (cid:19) H ′ ( y ) ! ρ + (cid:18) t − s (cid:19) √ (cid:18) s − t (cid:19) ρ ′ − H ′ ( z ) ! H ( y )= (cid:18) s − t (cid:19) H ( y ) ρ + √ (cid:18) t − s (cid:19) H ′ ( y ) ρ + √ t + 1 s − (cid:18) t − s (cid:19) ! H ( y ) ρ ′ + (cid:18) t − s (cid:19) (cid:0) − H ( y ) ρ − H ′ ( z ) H ( y ) (cid:1) . We can verify L (cid:0) ˜ η − φ (cid:1) is negative in the region where ˜ η < φ. Hence by maximumprinciple, φ ≤ ˜ η. This finishes the proof. (cid:3)
With the estimate of φ at hand, we see from Modica estimate that u y ≤ |∇ u |√ ≤ (cid:0) − u (cid:1) ≤
12 (1 + H ( y ) H ( z )) (1 − H ( y ) H ( z ) + φ ) ≤ (cid:0) − H ( y ) H ( z ) (cid:1) + 32 H ( y ) (1 + H ( y ) H ( z )) ( H ( z ) + zH ′ ( z )) y − z . In particular, as y → + ∞ , u y decays at least like O ( y − ) . (Note that u y decays exponen-tially fast away from the Simons cone). However, we expect that u y decays like O ( y − ) . Let us consider ρ ( z ) := H ′ ( z ) Z z (cid:18) H ′− Z + ∞ s tH ′ ( t ) dt (cid:19) ds. It satisfies − ρ ′′ + (cid:0) H − (cid:1) ρ = zH ′ ( z ) . Then intuitively, near the Simons cone, for y large, u y ∼ y ρ ( z ) . However, it turns out to be quite delicate and difficult to get an explicit global bound of u y with this decay rate. The estimate of u y will be one of our main aim in this section. Lemma 7. In Ω , tu s + su t ≤ . Proof.
By monotonicity, we know that u s + u t ≥ . Let us define η := t k u s + s k u t . We have Lη = u s t k (cid:18) s + k + 2 kt (cid:19) + u t s k (cid:18) t + k + 2 ks (cid:19) + 2 ku st (cid:0) s k − + t k − (cid:1) . We write this equation as Lη − (cid:18) t + k + 2 ks (cid:19) g = u s t k (cid:18) s + k + 2 kt − (cid:18) t + k + 2 ks (cid:19)(cid:19) + 2 ku st (cid:0) s k − + t k − (cid:1) . Suppose k ≥ . Using the fact that u s ≥ u st ≥ , we obtain Lη − (cid:18) t + k + 2 ks (cid:19) η ≥ . Hence from the maximum principle, t k u s + s k u t ≤ , if k ≥ . (cid:3) The above lemma in particular gives us a lower bound of | u t | in terms of u s . That is, | u t | ≥ ts u s in Ω . We also note that this inequality implies that u s + u t has the following decaying property:(13) u s + u t ≤ zy + z u s . Lemma 8. In Ω , we have u s ≤ (cid:18) e . t + 4 . √ t (cid:19) e − . s . Proof. u s satisfies Lu s − u s s = 0 . Let a, b be parameters to be determined and η = (cid:16) e at + b √ t (cid:17) e − as . We have Lη − ηs = (cid:18) a + 3 at − as + 1 − u − s (cid:19) e a ( t − s ) − b t e − as + b √ t (cid:18) a − as (cid:19) e − as + (cid:18) − u − s (cid:19) b √ t e − as . TABILITY OF SADDLE SOLUTIONS 11
Note that in the region z > , u is close to 1 . More precisely, u ≥ max { H (0 . y ) H (0 . z ) , u ∗ − φ } . Let Γ := Ω ∩ { z > } . Choose a = . , b = 4 . . We then have Lη − ηs ≤ , in Γ . Note that on ∂ Γ , u s ≤ η. The desired estimate then follows from the maximum principle. (cid:3)
Lemma 9. u s s − u ss ≥ in Ω ∗ . Proof.
Let η = u s s − − u ss . Then by (22) , η ≥ ∂ Ω ∗ . We compute Lη = 8 s u s − u ss s − u s u. Hence Lη − s η ≤ . By maximum principle, η ≥ . (cid:3) Lemma 10. In Ω ∗ , we have u s s + u t t − u ss − u tt ≥ . Proof.
Let η = u s s + u t t − u ss − u tt . Then η = 0 on ∂ Ω ∗ . We have Lη = 8 u s s + 8 u t t − u ss s − u tt t − u (cid:0) u s + u t (cid:1) = 8 t η + (cid:18) t − s (cid:19) (cid:16) u ss − u s s (cid:17) − u (cid:0) u s + u t (cid:1) . Using the fact that u ss − u s s ≤ , we find that Lη − t η ≤ . Then by maximum principle, η ≥ . (cid:3) We know that for y large, u y decays like O ( y − ) . However, to estimate the secondderivatives of u, we need to have some explicit global estimate of u s + u t in Ω . We shallprove that u s + u t can be bounded by functions of the form (cid:18) t − s (cid:19) ( au z + bu αz ) , with suitable constants a, b, α. We would like to prove the following(non-optimal)
Proposition 11. In Ω , we have (14) u s + u t ≤ (cid:18) t − s (cid:19) (cid:0) u s − u t ) + √ u s − u t (cid:1) . Proof. u s + u t satisfies L ( u s + u t ) = 3 u s s + 3 u t t . We write it in the form: L ( u s + u t ) − (cid:18) s + 1 t (cid:19) ( u s + u t ) = 32 (cid:18) s − t (cid:19) ( u s − u t ) . Let α ∈ [0 ,
1] be a parameter. Define the function η α = (cid:18) t − s (cid:19) ( u s − u t ) α . Using the fact that ∆ ( t − ) = ∆ ( s − ) = 0 , we get Lη α = α ( u s − u t ) α − (cid:18) s ( u ss − u st ) − t ( u st − u tt ) (cid:19) + (cid:18) t − s (cid:19) L (( u s − u t ) α ) . (15)We compute, L (( u s − u t ) α ) = ( u s − u t ) α − (cid:18) αu s s − αu t t + (1 − α ) (cid:0) − u (cid:1) ( u s − u t ) (cid:19) + α ( α −
1) ( u s − u t ) α − |∇ ( u s − u t ) | . Note that the term αu s s − αu t t is positive. However, for α ∈ [0 , , we also have,3 αu s s − αu t t − (cid:18) s + 1 t (cid:19) ( u s − u t )= 3 ( α − (cid:18) t + 1 s (cid:19) ( u s − u t ) − α (cid:18) t − s (cid:19) ( u s + u t ); ≤ − α (cid:18) t − s (cid:19) ( u s + u t ) ≤ . ¿From this, we know that intuitively, for α = 1 , Lη α − (cid:0) s + t (cid:1) η α is negative near theSimons cone; while for 0 < α < , it is negative away from the Simons cone. In view ofthis, we consider a combination of η and η α , where α = (this choice may not be optimal).That is, h := η + 12 η = (cid:18) t − s (cid:19) (cid:18) u s − u t + 12 ( u s − u t ) α (cid:19) . We then set h ∗ := h − ( u s + u t ) . Observe that h ∗ ≥ ∂ Ω . We would like to use themaximum principle to show h ∗ ≥ . TABILITY OF SADDLE SOLUTIONS 13
Using (15) , we compute Lh ∗ − (cid:18) s + 1 t (cid:19) h ∗ = (cid:18) α ( u s − u t ) α − (cid:19) (cid:18) s ( u ss − u st ) − t ( u st − u tt ) (cid:19) + (cid:18) t − s (cid:19) L (cid:18) u s − u t + 12 ( u s − u t ) α (cid:19) − (cid:18) t + 1 s (cid:19) h + 32 (cid:18) t − s (cid:19) ( u s − u t ) . We first observe that (cid:18) t − s (cid:19) ( u s − u t ) α − (cid:18) αu s s − αu t t (cid:19) − (cid:18) t + 1 s (cid:19) (cid:18) t − s (cid:19) ( u s − u t ) α = 3 (cid:18) t − s (cid:19) ( u s − u t ) α − (cid:18) αu s s − αu t t − (cid:18) t + 1 s (cid:19) ( u s − u t ) (cid:19) = 3 (cid:18) t − s (cid:19) ( u s − u t ) α − (cid:18) ( α − (cid:16) u s s − u t t (cid:17) + 12 (cid:18) s − t (cid:19) ( u s + u t ) (cid:19) . Let us denote the right hand side by I. We know that I ≤ . It follows that Lh ∗ − (cid:18) s + 1 t (cid:19) h ∗ ≤ (cid:0) α ( u s − u t ) α − (cid:1) (cid:18) s ( u ss − u st ) − t ( u st − u tt ) (cid:19) + (1 − α ) (cid:18) t − s (cid:19) (cid:0) ( u s − u t ) α (cid:0) − u (cid:1) − α ( u s − u t ) α − |∇ ( u s − u t ) | (cid:1) + 32 (cid:18) t − s (cid:19) ( u s − u t ) + 12 I. (16)Suppose to the contrary that the inequality (14) was not true. Consider the region Γ where u s + u t > h. Now we would like to show that in Γ , Lh ∗ − (cid:0) s + t (cid:1) h ∗ ≤ . We have1 s ( u ss − u st ) − t ( u st − u tt )= u ss s + u tt t − (cid:18) s + 1 t (cid:19) u st = u − us − s (cid:16) u s s + u t t (cid:17) + (cid:18) t − s (cid:19) u tt − (cid:18) s + 1 t (cid:19) u st . In particular, since u tt < u st > , (17) 1 s ( u ss − u st ) − t ( u st − u tt ) ≤ u − us − s (cid:16) u s s + u t t (cid:17) . On the other hand, |∇ ( u s − u t ) | = ( u ss − u st ) + ( u st − u tt ) ≥
12 ( u ss + u tt − u st ) ≥ (cid:16) u − u − (cid:16) u s s + u t t (cid:17)(cid:17) . (18)Inserting estimates (17) , (18) into (16) , and using the fact that1 − u ≥ u s − u t , we conclude that Lh ∗ − (cid:18) s + 1 t (cid:19) h ∗ ≤ , in Γ . By maximum principle, h ∗ ≥ . The proof is then completed. (cid:3)
We remark that the estimates in the previous Proposition is not optimal, since we havenot used the information of u st , which is, intuitively, of the order − ( u ss + u tt ). Lemma 12. u satisfies (19) u − u + u ss ≥ in Ω . Proof.
This inequality follows directly from the equation u ss + u tt + 3 s u s + 3 t u t + u − u = 0 , and the fact that in Ω , s u s + 3 t u t ≤ , u tt ≤ . Here we give another proof using the maximum principle. We have∆ u σ = σu σ − ∆ u + σ ( σ − u σ − |∇ u | , which implies Lu σ = σu σ − (cid:0) u − u (cid:1) + (cid:0) − u (cid:1) u σ + σ ( σ − u σ − |∇ u | = (1 − σ ) u σ + ( σ − u σ +2 + σ ( σ − u σ − |∇ u | . In particular, Lu = − u , Lu = − u + 6 u |∇ u | . Hence(20) L (cid:0) u − u (cid:1) = − u |∇ u | . Define η = u − u + u ss . Applying (20) and (24) , we get Lη = − u |∇ u | + 6 s u ss − s u s + 6 u s u. Therefore, in Ω ∗ , Lη − s η = − s (cid:0) u − u (cid:1) − s u s − u t u ≤ . TABILITY OF SADDLE SOLUTIONS 15
By the maximum principle, η ≥ . This finishes the proof. (cid:3)
It turns out that the estimate of Lemma 12 is also not optimal. Indeed, u ss can beestimated by u s . This is the following
Lemma 13. u satisfies √ u s u + u ss ≥ , in Ω . Proof.
We compute L ( u s u ) = 3 s u s u + 2 u ss u s + 2 u st u t + u s (cid:0) u − u (cid:1) . Let η = au s u + u ss , where a > Lη = 3 as u s u + 2 au ss u s + 2 au st u t + au s (cid:0) u − u (cid:1) + 6 s u ss − s u s + 6 u s u. We can write it in the form Lη = (cid:18) au s + 6 s (cid:19) η + 3 as u s u + 2 au st u t + au s (cid:0) u − u (cid:1) − s u s + 6 u s u − (cid:18) au s + 6 s (cid:19) ( au s u ) . That is, Lη − (cid:18) au s + 6 s (cid:19) η = − as u s u + 2 au st u t − s u s + (cid:0) − a (cid:1) u s u + au s (cid:0) u − u (cid:1) . By Modica estimate, 1 − u ≥ √ |∇ u | ≥ √ u s . Hence Lη − (cid:18) au s + 6 s (cid:19) η ≤ (cid:0) − a (cid:1) u s u − √ au s u = (cid:16) − a − √ a (cid:17) u s u. In particular, if we choose a to be √ , then Lη − (cid:0) au s + s (cid:1) η ≤ . In this case, by themaximum principle, η ≥ . Hence √ u s u + u ss ≥ . (cid:3) We remark that this estimate together with Modica estimate implies (19) . Lemma 14.
We have √ u t u + u st ≤ in Ω . Proof.
Let η = √ u t u + u st . Then Lη = 3 √ t u t u + 2 √ u st u s + 2 √ u tt u t + √ u t (cid:0) u − u (cid:1) + (cid:18) t + 3 s (cid:19) u st + 6 u s u t u. This can be written as Lη − (cid:18) √ u s + 3 t + 3 s (cid:19) η = 3 √ t u t u + 2 √ u tt u t + √ u t (cid:0) u − u (cid:1) + 6 u s u t u − √ u t u (cid:18) √ u s + 3 t + 3 s (cid:19) . Since u tt ≤ u t ≤ , we get Lη − (cid:18) √ u s + 3 t + 3 s (cid:19) η ≥ , in Ω . By the maximum principle, √ u t u + u st ≤ . (cid:3) Next we prove the following non-optimal estimate:
Lemma 15. In Ω , we have u s + u t ) + u st + u ss ≥ . Proof.
Let η = 2 ( u s + u t ) + u st + u ss . First of all, η ≥ ∂ Ω . We compute Lη = 6 s u s + 6 t u t + (cid:18) s + 3 t (cid:19) u st + 6 u s u t u + 6 s u ss − s u s + 6 u s uLη − s η = − s ( u s + u t ) + (cid:18) t − s (cid:19) (2 u t + u st ) − s u s + 6 u s u ( u s + u t ) . Using Lemma 14, we obtain 2 u t + u st ≤ (cid:16) − √ u (cid:17) u t . Hence Lη − s η ≤ (cid:18) u s u − s (cid:19) ( u s + u t ) − s u s + (cid:16) − √ u (cid:17) (cid:18) t − s (cid:19) u t . TABILITY OF SADDLE SOLUTIONS 17
Applying the estimate of u s + u t (Proposition 11), we see that Lη − s η ≤ . It then followsfrom the maximum principle that η ≥ . (cid:3) Lemma 16. In Ω , we have u s + u t ) − u st − u tt ≥ . Proof.
Let η = 2 ( u s + u t ) − u st − u tt . First of all, since u tt ≤ , we have η ≥ ∂ Ω . We compute Lη = 6 u s s + 6 u t t − (cid:18) s + 3 t (cid:19) u st − u s u t u − t u tt + 6 t u t − u t u. Therefore, Lη − t η = (cid:18) s − t (cid:19) u s − u t t − (cid:18) s − t (cid:19) u st + 6 t u t − u t u ( u s + u t )= (cid:18) s − t (cid:19) ( u s + u t ) + (cid:18) t − s (cid:19) (2 u t + u st )+ 6 t u t − u t u ( u s + u t ) ≤ (cid:18) s − t − u t u (cid:19) ( u s + u t )+ 6 t u t + (cid:18) t − s (cid:19) (cid:16) − √ u (cid:17) u t ≤ . It then follows from the maximum principle that η ≥ . (cid:3) It follows immediately from these lemmas that4 ( u s + u t ) + u ss − u tt ≥ . We conjecture that u ( u s + u t ) + u ss + u st ≥ ,u ( u s + u t ) − u st − u tt ≥ . However, we are not able to prove them in this paper. The main difficulty here is thefollowing: In L ( u ( u s + u t )) , we have u tt term. Hence one needs to handle terms like u st + u tt (In particular in the region t < L ( u tt ) = t u tt − t u t + 6 u t u and u t t blows up as t → . Lemma 17. uy + uz − u y − u z ≥ , in Ω ∗ . Proof.
Let η = uy + uz − u y − u z . We first observe that η = 0 on ∂ Ω ∗ . We have Lη = − u y − u y y + u (cid:18) y − y ( y − z ) (cid:19) − u z − u z z + u (cid:18) z + 6 z ( y − z ) (cid:19) − y + 6 z ( y − z ) u y + 12 yz ( y − z ) u z + 12 yz ( y − z ) u y − y + 6 z ( y − z ) u z . That is, Lη − y + z ) η = − u y − u y y + u (cid:18) y − y ( y − z ) (cid:19) − u z − u z z + u (cid:18) z + 6 z ( y − z ) (cid:19) − y + z ) (cid:18) uy + uz (cid:19) = − u (cid:18) y + 1 z (cid:19) − u y y − u z z + 2 uy + 2 uz . On the other hand, we know that in Ω ,yu y − zu z = 12 ( s + t ) ( u s + u t ) −
12 ( s − t ) ( u s − u t )= su t + tu s < , while in Ω r , yu y − zu z > . Suppose at some point p, uy + uz − u y − u z < . Then due to symmetry, we can assume p ∈ Ω . Since u − yu y > u − zu z , we have, at p,uz − u z < . We then write Lη − y + z ) η − y η ≤ (cid:18) z − y (cid:19) (cid:16) uz − u z (cid:17) ≤ . This contradicts with the maximum principle. Hence uy + uz − u y − u z ≥ . (cid:3) TABILITY OF SADDLE SOLUTIONS 19
The above lemma in particular implies that(21) u ≥ yu y in Ω . On the other hand, we conjecture that u yy < ∗ . If this is true, then we will also have(21) . Taking the z derivative in (21), we find that(22) u z − yu yz ≥
0, if z = 0 . With this estimates at hand, we want to prove the following
Lemma 18. − u t t + u st + u tt ≥ , in Ω . Proof.
Let a > η = − au t t + u st + u tt . We compute Lη = − at u t + 2 at u tt + (cid:18) s + 3 t (cid:19) u st + 6 u s u t u + 6 t u tt − t u t + 6 u t u = 6 + 2 at η + (cid:18) s + − − at (cid:19) u st + 6 u t u ( u s + u t ) . It follows that for a ≥ , Lη − at η ≤ . Let us choose a = 1 . It remains to verify that η ≥ ∂ Ω . If z = 0 , then u st = 0 and u tt = − u yz . Then − u t + tu tt = u z √ y √ − u yz ) = 1 √ u z − yu yz ) . By (22) , we know that − u t + tu tt ≥ (cid:3) Lemma 19.
We have the following estimate(not optimal): (cid:18) t − s (cid:19) u s + u ss + 2 u st ≥ , in Ω . Proof.
Let a, d be two parameters. Define η = a (cid:18) t − s (cid:19) u s + u ss + du st . We compute Lη = 3 as (cid:18) t − s (cid:19) u s − at u st + 2 as u ss + a (cid:18) s − t (cid:19) u s + 6 s u ss − s u s + 6 u s u + d (cid:18) s + 3 t (cid:19) u st + 6 du s u t u. Then Lη − as η = (cid:18) − as a (cid:18) t − s (cid:19) + 3 as (cid:18) t − s (cid:19) + a (cid:18) s − t (cid:19) − s (cid:19) u s + (cid:18) − as d − at + (cid:18) s + 3 t (cid:19) d (cid:19) u st + 6 u s ( u s + du t ) u. The right hand side is equal to (cid:18)(cid:18) t − s (cid:19) − a − a s + a (cid:18) s − t (cid:19) − s (cid:19) u s + (cid:18) − as d − a − dt (cid:19) u st + 6 u s ( u s + du t ) u. Let us take d = 2 , a = 3 . Since | u t | ≥ ts u s , we have u s + 2 u t ≤ , if s ≤ t. Then applying Proposition 11 in the region { s > t } , we get Lη − s η ≤ . By maximum principle, η ≥ . The proof is finished. (cid:3)
Next we want to estimate u st in terms of u ss + u tt . This is the content of the following
Lemma 20. (cid:18) t − s (cid:19) ( u s − u t ) + 2 u st + u ss + u tt ≥ , in Ω . Proof.
Let η = a (cid:0) t − s (cid:1) ( u s − u t ) + 2 u st + u ss + u tt . Then Lη = − a (cid:18) t − s (cid:19) ( u s − u t ) + 2 at ( u tt − u st )+ 2 as ( u ss − u st ) + a (cid:18) t − s (cid:19) (cid:18) u s s − u t t (cid:19) + 6 u ss s − u s s + 6 u tt t − u t t + (cid:18) s + 6 t (cid:19) u st + 6 ( u s + u t ) u. TABILITY OF SADDLE SOLUTIONS 21
Let us denote h = u ss + u st , h = u st + u tt , and h = h + h . Then the terms in Lη involvingsecond order derivatives can be written as6 + 2 as h + 6 + 2 at h − (cid:18) as + 4 at (cid:19) u st = (3 + a ) (cid:18) t + 1 s (cid:19) h + (3 + a ) (cid:18) t − s (cid:19) ( u tt − u ss ) − (cid:18) at + 4 as (cid:19) u st . On the other hand, the terms in Lη involving u s , u t can be written as (cid:18) − a (cid:18) t − s (cid:19) + a (cid:18) t − s (cid:19) (cid:18) s + 32 t (cid:19) − s + 3 t (cid:19) ( u s − u t )+ (cid:18) a (cid:18) t − s (cid:19) (cid:18) s − t (cid:19) − s − t (cid:19) ( u s + u t ) + 6 ( u s + u t ) u Hence Lη − (3 + a ) (cid:0) t + s (cid:1) η is equal to (cid:18) − a (cid:18) t − s (cid:19) + a (cid:18) t − s (cid:19) (cid:18) s + 32 t (cid:19) − s + 3 t − a (3 + a ) (cid:18) t − s (cid:19) (cid:18) t + 1 s (cid:19)(cid:19) ( u s − u t )+ (3 + a ) (cid:18) t − s (cid:19) ( u tt − u ss ) − (cid:18) at + 4 as (cid:19) u st + (cid:18) a (cid:18) t − s (cid:19) (cid:18) s − t (cid:19) − s − t (cid:19) ( u s + u t ) + 6 ( u s + u t ) u. The coefficient before u s − u t , divided by t − s is(3 − a ) (cid:18) t + 1 st + 1 s (cid:19) + a (cid:18) s + 32 t (cid:19) − a (3 + a ) (cid:18) t + 1 s (cid:19) = (cid:18) − a + 3 a − a (3 + a ) (cid:19) (cid:18) s + 1 t (cid:19) + 3 − ast = (cid:18) − a − a (cid:19) (cid:18) s + 1 t (cid:19) + 3 − ast . Let us choose a = 3 . Using Lemma 19, we obtain(3 + a ) (cid:18) t − s (cid:19) ( u tt − u ss ) − (cid:18) at + 4 as (cid:19) u st ≤ (cid:18) t − s (cid:19) ( − u ss − u st ) ≤ (cid:18) t − s (cid:19) (cid:18) t − s (cid:19) u s . Then Applying the estimate of u s + u t , we get Lη − (cid:18) t + 1 s (cid:19) η ≤ . By maximum principle, η ≥ . This completes the proof. (cid:3)
Proposition 21. In Ω , (23) u st + u ss + (cid:18) t − s (cid:19) (cid:0) u s − u t ) + √ u s − u t (cid:1) ≥ . Proof.
The functions u ss and u st satisfy(24) Lu ss = 6 s u ss − s u s + 6 u s u, (25) Lu st = (cid:18) t + 3 s (cid:19) u st + 6 u s u t u. Hence L ( u ss + u st ) − s ( u ss + u st )= (cid:18) t − s (cid:19) u st − u s s + 6 u s u ( u s + u t ) . Let η = (cid:0) t − s (cid:1) ( u s − u t + ( u s − u t ) α ) . Then same computation as before yields L ( η + u st + u ss ) − s ( η + u ss + u st )= (cid:18) α ( u s − u t ) α − (cid:19) (cid:18) s ( u ss − u st ) − t ( u st − u tt ) (cid:19) + (cid:18) t − s (cid:19) L (cid:18) u s − u t + 12 ( u s − u t ) α (cid:19) + (cid:18) t − s (cid:19) u st − u s s + 6 u s u ( u s + u t ) − s η. In the region Γ where (23) is not true, we have1 s ( u ss − u st ) − t ( u st − u tt ) ≤ − (cid:18) s + 1 t (cid:19) u st + 1 t u tt − s (cid:18) t − s (cid:19) (cid:0) u s − u t + √ u s − u t (cid:1) . Denote J := 1 s (cid:18) t − s (cid:19) (cid:18) u s − u t + 12 √ u s − u t (cid:19) . TABILITY OF SADDLE SOLUTIONS 23
Using the fact that u tt ≤ , we also have |∇ ( u s − u t ) | = ( u ss − u st ) + ( u st − u tt ) ≥ (2 u st + J ) + u st = 5 u st + 4 u st J + J . Then applying the decay of u s , we get L ( η + u st + u ss ) − s ( η + u ss + u st ) ≤ . This implies that η + u st + u ss ≥ . (cid:3) Lemma 22. u s u + u ss ≥ , in Ω . Proof.
Let η = u s u + u ss . The computation in Lemma 13 tells us that Lη − (cid:18) u s + 6 s (cid:19) η = − s u s u + 2 u st u t − s u s + 4 u s u + u s (cid:0) u − u (cid:1) . This can be written as Lη − (cid:18) u s − u t + 6 s (cid:19) η = − s u s u + 2 ( u st + u ss ) u t + 2 ( u s + u t ) u s u − s u s + 2 u s u + u s (cid:0) u − u (cid:1) . Note that by Modica estimate,1 − u ≥ q u s + u t ) . = 2 u s + q u s + u t ) − u s = 2 u s + 2 ( u t − u s ) p u s + u t ) + 2 u s . We then deduce Lη ≤ − s u s u + 2 ( u st + u ss ) u t + 2 ( u s + u t ) u s u − s u s + u s u u s − u t ) p u s + u t ) + 2 u s . Now inserting the estimate of u st obtained in Lemma 20 into the proof of Proposition 11,we can bootstrap the estimate of u s + u t to(26) u s + u t ≤ (cid:18) t − s (cid:19) (cid:18) u s − u t + 12 √ u s − u t (cid:19) . Applying these estimates, we obtain Lη ≤ . Hence η ≥ . (cid:3) Similarly, we have the following
Lemma 23. − u t u − u st ≥ , in Ω . The proof of this lemma is almost identical to that of Lemma 22. We omit the details.Next we would like to prove the following
Lemma 24. In Ω , we have (cid:18) t − s (cid:19) u s + u ss + u st ≥ . Proof.
Let η = a (cid:18) t − s (cid:19) u s + u ss + u st . We compute Lη = 3 as (cid:18) t − s (cid:19) u s − at u st + 2 as u ss + a (cid:18) s − t (cid:19) u s + 6 s u ss − s u s + 6 u s u + (cid:18) s + 3 t (cid:19) u st + 6 u s u t u. Then Lη − as η = (cid:18) − as a (cid:18) t − s (cid:19) + 3 as (cid:18) t − s (cid:19) + a (cid:18) s − t (cid:19) − s (cid:19) u s + (cid:18) − as − at + 3 s + 3 t (cid:19) u st + 6 u s ( u s + u t ) u. The right hand side is equal to (cid:18)(cid:18) t − s (cid:19) − a − a s + a (cid:18) s − t (cid:19) − s (cid:19) u s + (cid:18) − as − at + 3 s + 3 t (cid:19) u st + 6 u s ( u s + u t ) u = (cid:18)(cid:18) t − s (cid:19) − a − a s + a (cid:18) s − t (cid:19) − s (cid:19) u s + (cid:18) − − as + 3 − at (cid:19) u st + 6 u s ( u s + u t ) u. TABILITY OF SADDLE SOLUTIONS 25
Hence applying Lemma 20 and (26) , we get Lη − s η ≤ . By maximum principle, η ≥ . The proof is finished. (cid:3)
Similarly, we have
Lemma 25. (cid:18) s − t (cid:19) u t − u st − u tt ≥ . Proof.
Let η = a (cid:18) s − t (cid:19) u t − u st − u tt . We compute Lη = 3 at (cid:18) s − t (cid:19) u t + 2 at u tt − as u st − a (cid:18) s − t (cid:19) u t − (cid:18) s + 3 t (cid:19) u st − u s u t u − t u tt + 6 t u t − u t u. Then Lη − − at η = (cid:18) − − at a (cid:18) s − t (cid:19) + 3 at (cid:18) s − t (cid:19) − a (cid:18) s − t (cid:19) + 6 t (cid:19) u t + (cid:18) − at − as − s − t (cid:19) u st − u t ( u s + u t ) u. The left hand side is equal to (cid:18) a (6 − a ) − a + 6 t + − a (6 − a ) + 3 ast − as (cid:19) u t + (cid:18) − − as + 3 − at (cid:19) u st − u t ( u s + u t ) u. If we choose a = , then this is equal to (cid:18) a (6 − a ) − a + 6 t + − a (6 − a ) + 3 ast − as (cid:19) u t − s u st − u t ( u s + u t ) u = (cid:18) t − s (cid:19) u t − s u st − u t ( u s + u t ) u. This will be negative. Hence η ≥ . (cid:3) Proposition 26. In Ω , u s + u t − u st − u tt ≥ . Proof.
Let η = u s + u t − u st − u tt . Then Lη = 3 u s s + 3 u t t − (cid:18) s + 3 t (cid:19) u st − u s u t u − t u tt + 6 t u t − u t u = 6 t η − t ( u s + u t − u st ) + 3 u s s + 3 u t t − (cid:18) s + 3 t (cid:19) u st − u s u t u − t u tt + 6 t u t − u t u = 6 t η + (cid:18) − t + 3 s (cid:19) u s + (cid:18) − t (cid:19) u t + (cid:18) t − s (cid:19) u st + 6 t u t − u t u ( u s + u t ) . We write it as Lη − t η = (cid:18) t − s (cid:19) ( u st + u t ) + (cid:18) − t + 3 s (cid:19) ( u s + u t )+ 6 t u t − u t u ( u s + u t ) . It then follows from the estimate of u s + u t that Lη − t η ≤ , which implies η ≥ . (cid:3) Similarly, we have
Proposition 27. In Ω , we have u s + u t + u st + u ss ≥ . The proof of Proposition 27 is similar to the proof of Proposition 26. That is, denoting η = u s + u t + u st + u ss . We can show that Lη − s η ≤ , which then implies that η ≥ . We sketch the proof below.
Proof.
Let η = u s + u t + u st + u ss . First of all, since u s u + u ss ≥ , we obtain η ≥ ∂ Ω . We compute Lη = 3 s u s + 3 t u t + (cid:18) s + 3 t (cid:19) u st + 6 u s u t u + 6 s u ss − s u s + 6 u s u. Then we get Lη − s η = − s ( u s + u t ) + (cid:18) t − s (cid:19) ( u st + u t ) − s u s + 6 u s u ( u s + u t ) . Using the fact that u t u + u st ≤ u s + u t , we see that Lη − s η ≤ . It then follows from the maximum principle that η ≥ . (cid:3) TABILITY OF SADDLE SOLUTIONS 27
An immediate consequence of Proposition 27 and estimate (26) is the following
Corollary 28. In Ω , we have (cid:18) t − s (cid:19) (cid:18) u s − u t + 12 √ u s − u t (cid:19) + u st + u ss ≥ . Now we would like to establish a lower bound on u s . Let us define the function E := u + 2 u s . By Lemma 22 and Lemma 23, we have(27) u s u − u t u + u ss − u st ≥ . This implies ∂ z E ≥ . We will slightly abuse the notation and still write the function u, u s in ( y, z ) variables as u ( y, z ) , u s ( y, z ) . Recall that u = H ( y ) H ( z ) − φ. From (27) and thefact that u ≤ H ( y ) H ( z ) , we get the following estimate in Ω :2 u s ( y, z ) ≥ u s ( y,
0) + u ( y, − u ( y, z ) ≥ u s ( y, − ( H ( y ) H ( z )) = 2 ∂ s ( H ( y ) H ( z ) − φ ) | ( z =0) − ( H ( y ) H ( z )) . (28)We have mentioned that the estimate of u tt is most delicate. To conclude this section,let us derive certain upper bound on | u tt | . Note that so far we have good control on | u ss | and | u st | , in terms of u s and u t respectively. We first recall that u ss + u tt + 3 s u s + 3 t u t = − u (cid:0) − u (cid:1) . Since 2 u s ( y, z ) ≥ u s ( y, − u ( y, z ) , there holds 1 − u ( y, z ) ≤ u s ( y, z ) + 1 − u s ( y, . Hence(29) u ss + u tt + 3 s u s + 3 t u t ≥ − u (2 u s ( y, z ) + 1 − u s ( y, . This together with the fact that u tt is negative, clearly gives us an upper bound on | u tt | . Construction of supersolution in dimension φ of the form f u s + hu t . Wehave Lφ = u s ∆ f + 2 u ss f s + 2 u st f t + 3 s u s f + u t ∆ h + 2 u st h s + 2 u tt h t + 3 t u t h. Hence Lφ = (cid:18) ∆ f + 3 s f (cid:19) u s + (cid:18) ∆ h + 3 t h (cid:19) u t + 2 u ss f s + (2 f t + 2 h s ) u st + 2 u tt h t . (30) Let us write it as Lφ = C s u s + C st u st + C ss u ss + C tt u tt + C t u t , where C s := ∆ f + 3 s f, C st := 2 f t + 2 h s ,C ss := 2 f s , C tt := 2 h t , C t := ∆ h + 3 t h. Recall that in dimension n ≥ , Cabr´e [7] made the choice f = t − α , h = − s α , for suitableconstant α > . In our case, to construct a supersolution, we choose f ( s, t ) := tanh (cid:16) st (cid:17) √ s √ s + t + 14 . (cid:0) − e − s t (cid:1)! ( s + t ) − . ,h ( s, t ) := − tanh (cid:18) ts (cid:19) √ t √ s + t + 14 . (cid:16) − e − t s (cid:17)! ( s + t ) − . . We now define Φ := 0 . (cid:16) s − . e − t + t − . e − s (cid:17) and Φ = f u s + hu t . Then we set(31) Φ := Φ + Φ . Note that Φ > s, t ) = Φ ( t, s ) . The reason that we choose this specific f, h, insteadof the more natural choice of µ := ( s + t ) − . ( u s − u t ) , is the following: Although near theSimons cone, the function µ is well behaved, it does not satisfies Lµ ≤ s, t ) = (3 , (cid:0) st (cid:1) s √ s + t . Next, we add a small perturbation term Φ , because the function f u s + hu t essentially decays as e t − s and is not a good supersolution when s − t is verylarge, where the linearized operator looks like − ∆ + 2 . Finally, the term . (cid:0) − e − s t (cid:1) isused to control the sign of L Φ near the point (5 , . . We would like to show that Φ is a supersolution of the linearized operator L . Due tosymmetry, in the sequel, we only need to consider the problem in Ω . Our first observation is the following fact: C s < , C st < , C ss < , in Ω . We emphasize that C t and C tt may change sign. As a matter of fact, C t changes sign nearthe Simons cone, and in most part of Ω , C t is negative. When t > , C tt is negative in theregion(approximately desribed by) 2 s/ < t < s/ . Moreover, in this region, | C tt | is smallcompared to | C ss | . (See Figure 11). It follows that C s u s and C st u st are negative, which canbe regarded as “good” terms. The “bad” terms are C ss u ss and C t u t . The main idea of our proof is to control the other positive terms using C s u s + C st u st and C ss u ss , based on the estimates obtained in the previous section. We have the following Proposition 29.
Let Φ be the function defined by (31) . For all ( s, t ) , we have L Φ ≤ . Proof.
Since Φ is even with respect to the Simons cone, it will be suffice to prove thisinequality in Ω . TABILITY OF SADDLE SOLUTIONS 29
Lemma 24 tells us that(32) 32 (cid:18) t − s (cid:19) u s + u st + u ss ≥ . For notational convenience, we will also write the coefficient as λ. That is λ := 32 (cid:18) t − s (cid:19) . It plays an important role in our analysis, since it measures how close is u st to u ss . Note that by Proposition 27 and the fact that | u t | ≥ ts u s , we have(33) s − ts u s + u st + u ss ≥ u s + u t + u st + u ss ≥ . Hence when t < , the inequality (32) is weaker than (33) . Moreover, (33) has the followingsimple consequence: If | u t | = au s for some constant a < , then(34) u ss + u st + (1 − a ) u s ≥ . This estimate is useful, because a priori, we don’t know the precise value of | u t /u s | , although it is always bounded from below by t/s. At this stage, it will be crucial tohave some information on the ratio C t /C s . We have C t /C s < . , if s/ < t < s. See Figure 6 for detailed information on the function C t /C s . This tells us that C s u s + C t u t < , for s/ < t < s. Moreover, it turns out that C s u s + C t u t + C st u st + C tt u tt can beused to control the term C ss u ss . Indeed, first of all, we have(See Figure 7), in the region s/ < t < s, C ss C st − C tt < . Then in this region, we can use (34) to estimate u st + u ss , and applying Proposition 26 todeduce(35) 2 ( u s + u t ) + u ss − u tt ≥ . It follows that if C tt > , then C s u s + C st u st + C ss u ss + C tt u t + C t u t ≤ ( C s − (1 − a ) C ss − aC t + (1 − a ) max { C st − C ss , } ) u s , (36)where a can be choosen to be | u t | /u s or λ. With the help of these estimates, let us consider the subregion E := (cid:26) ( s, t ) : 0 . s < t < s and t > (cid:27) . Consider the function T ( r ) := (1 − r ) C ss C s +(1 − r ) max { C st − C ss , }− rC t . It turns out that0 < (1 − r ) C ss C s + (1 − r ) max { C st − C ss , } − rC t < E , for all r ∈ [1 − λ, . See Figure 8 on the picture of this function in the case of r = 1 − λ. We conclude that L Φ < , in E . We also observe(37) L (cid:16) s − . e − t (cid:17) = − . s − . e − t + s − . e − t (cid:18) − t + 109 − u (cid:19) . Hence L Φ will be positive only in the region close to the Simons cone, where − t + − u > . Note that we always have u ≥ H (0 . y ) H (0 . z ) . In E , we then verify that L (Φ + Φ ) ≤ , see Figure 9. Hence we conclude that L Φ ≤ E . Next we consider the region E := { ( s, t ) : 1 / < t < . s } . In this region, L Φ may bepositive. However, we already know that, if C tt > , then (36) holds. Moreover, if C tt < , then by(Lemma 18)(38) − t u t + u st + u tt ≥ , we get C s u s + C st u st + C ss u ss + C tt u t + C t u t ≤ (cid:18) C s − (1 − a ) C ss − aC t + 1 t | C tt | (cid:19) u s , (39)We emphasize that in the region where C tt < , actually | C tt /C ss | is small. On the otherhand, by Lemma 8, u s has the following decay(40) u s ≤ (cid:18) e . t + 4 . √ t (cid:19) e − . s . Note that here we can also use the Modica estimate u s ≤ − u √ to estimate u s . In particular,(40) implies that u s decays at least like e − . t along the line t = 0 . s. We also should keepin mind that (40) does not mean that u s blows up as t → . Indeed, by u st > , we knowthat u s ( s, < u s ( s, t ) in Ω . Now we recall that Φ decays like s − . e − t . In particular,along the line s = 0 . t, u s decays faster than L Φ . From (36) , (39) , (40) and (37) , we canindeed verify that L Φ + L Φ ≤ , in E . Now we would like to consider E := { ( s, t ) : t < / } . Here | C t | could be large comparedto | C s | . Hence the above arguments does not work. However, as t → , we know that u t /t → u tt . This implies that for t small, u t is of the order tu tt . More precisely, we have u t ( s, t ) = Z t u tt ( s, r ) dr. On the other hand, we can estimate u tt using the fact that(Lemma 9) u s /s ≥ u ss and theAllen-Cahn equation u ss + u tt + 3 s u s + 3 t u t + u − u = 0 . TABILITY OF SADDLE SOLUTIONS 31
Indeed, since u s /s + t/u t <
0, we have | u ss + u tt | < u − u . Let u ∗ tt = min r ∈ (0 ,t ) u tt ( s, r ) . It turns out that | C s − C t ) tu ∗ tt | < L Φ . Hence L Φ ≤ E . See Figure 10.Combing the above analysis in E , E , E , we get the desired inequality. The proof isthus completed. (cid:3) Proposition 29 tells us that Φ is a supersolution and it follows from standard argu-ments(see, for instance, [7, 29]) that the saddle solution u is stable in dimension 8 . Thisfinishes the proof of Theorem 1.4.
Stability in dimension and n = 10 and n = 12 . We construct supersolution in the form Φ + Φ , where Φ = f u s + hu t , with f ( s, t ) := tanh (cid:16) st (cid:17) s √ s + t ( s + t ) − n − ,h ( s, t ) := − tanh (cid:18) ts (cid:19) t √ s + t ( s + t ) − n − . In principle, the cases n = 10 and 12 are easier than the dimension 8 case. Observethat in the definiton of f, we don’t need the term 1 − e − s t . In the previous section, thisterm is used to control the behaviour of the supersolution near the point ( s, t ) = (5 , . . The reason that we choose ( s + t ) − n − is as follows. Let us consider the function φ :=( s + t ) α ( u s − u t ) . Then Lφ is equal to( s + t ) α − (cid:16) α ( α −
1) + (cid:16) n − (cid:17) α (cid:0) s − + t − (cid:1) ( s + t ) + (cid:16) n − (cid:17) s − ( s + t ) (cid:17) u s − ( s + t ) α − (cid:16) α ( α −
1) + (cid:16) n − (cid:17) α (cid:0) s − + t − (cid:1) ( s + t ) + (cid:16) n − (cid:17) t − ( s + t ) (cid:17) u t + 2 α ( u ss − u tt ) ( s + t ) α − . When s = t, as t → + ∞ , Lφ asymptotically looks like2 ( α ( α −
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Y. Liu: Department of Mathematics, University of Science and Technology of China,Hefei, China
E-mail address : [email protected] K. Wang: School of Mathematics and Statistics, Wuhan University, Wuhan, Hubei,China
E-mail address : [email protected] J. Wei: Department of Mathematics, University of British Columbia, Vancouver, BCV6T 1Z2, Canada
E-mail address : [email protected]@math.ubc.ca