Heat kernels of some Schrödinger operators

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1 Heat kernels of some Schrödinger operators Alexander Grigor yan Tsinghua University 28 September 2016

2 Consider an elliptic Schrödinger operator H = Δ + Φ, where Δ = n 2 i=1 is the Laplace operator in R n and Φ is a smooth x 2 i function on R n. Denote by p Φ t (x, y) the heat kernel of H that is, by definition, the integral kernel of the heat semigroup { e th} using a selfadjoint extension of H in L 2 (R n ). In other words, for any f C0 (R n ), t 0 the function u (x, t) = p Φ t (x, y) f (y) dy R n solves the Cauchy problem { t u = Δu Φu in R n (0, ) u (x, 0) = f (x) Recall that for Φ 0 we have the classical heat kernel of the Laplace operator: ( ) p 0 1 x y 2 t (x, y) = exp. n/2 (4πt) 4t 1

3 The main purpose of this talk is to state some estimates for p Φ t (x, y) for some non-trivial potentials Φ, in particular, for the potentials with quadratic decay: Φ (x) = b, x > 1. x 2 Weighted Laplace operator and h-transform. Consider so far an arbitrary smooth Φ and assume in the sequel that the equation Δh + Φh = 0 has a solution h (x) > 0 in R n. Define the weighted Laplace operator Δ := 1 h 2 div ( h 2 ). This operator is symmetric with respect to measure dμ = h 2 dx and has a symmetric positive heat kernel p t (x, y) with respect to measure μ. That is, the function u (x, t) = p t (x, y) f (y) dμ (y) R n solves the Cauchy problem { t u = Δu in R n (0, ) u (x, 0) = f (x) 2

4 The identity Δh + Φh = 0 implies, for any smooth function v (x) ( Δv+Φv = Δv + 2 h ) h v + Δh h v = 1 Δ (hv) (hδv + 2 h v + Δh v) = h h whence Δv = 1 (Δ (hv) Φhv) h and Δ = 1 (Δ Φ) h. h The mapping v hv is an isometry from L 2 (h 2 dx) onto L 2 (dx). Hence, the operators Δ in L 2 (h 2 dx) and Δ Φ in L 2 (dx) are conjugate, which implies p Φ t (x, y) = p t (x, y) h (x) h (y). (1) Hence, in order to estimate p Φ t the following two problems: by means of this identity, one must solve 1. Prove the existence of a positive solution to Δh + Φh = 0 and estimate h (x). 2. Estimate the heat kernel p t (x, y) of Δ. 3

5 Small potentials. We first show a simple example of application of this method. We say that a function Φ on R n is Green bounded if either Φ 0 or sup g (x y) Φ (y) dy <, x R n R n where g is the positive Green function of Δ, that is, g (x) = { cn x 2 n, n > 2 +, n 2. For example, in R 1 and R 2 only zero function is Green bounded, whereas in R n with n > 2 any function with the estimate Φ (x) C x (2+ε) is Green bounded, provided ε > 0. The following result shows that Green bounded potentials are small perturbations for the Laplacian. 4

6 Proposition 1 If Φ 0 and Φ is Green bounded then there exists a positive solution h to Δh Φh = 0 such that a h b for some a, b > 0. Consequently, the operator Δ = 1 div (h 2 ) is uniformly elliptic and, h 2 hence, p (x, y) and p Φ t (x, y) admit Aronson s estimate: p Φ t (x, y) with some positive constants C, c. C exp tn/2 ( c x y 2 t The sign means that both and hold, but possibly with different values of the constants C, c. Of course, Proposition 1 gives a non-trivial result only in R n with n 3, where there are plenty of Green bounded potentials. For potentials that are not Green bounded, like Φ (x) = C x 2, the function h is normally unbounded so that the operator Δ = h 2 div (h 2 ) is not uniformly elliptic. Estimating p t in this case requires new tools that will be discussed below. 5 ),

7 How to estimate the heat kernel p t. Fix a smooth positive function h (x) in R n (so far not necessarily a solution to the Schrödinger equation), consider measure dμ = h 2 dx and set V (x, r) := μ (B (x, r)) = h 2 (y) dy, B(x,r) where B (x, r) is a ball in R n. Consider also the function h (r) := sup h (x) x =r that is a radial version of h. For two positive functions f and g, we write f g (f is comparable to g) if a f b for some positive constants a, b, in the specified range of the g arguments of f and g. The next theorem is a highly non-trivial result allowing to estimate the heat kernel p t (x, y) of the operator Δ = 1 div (h 2 ) for a class of h 2 unbounded functions h in R n. In fact, the result is true if R n is replaced by a manifold with non-negative Ricci curvature. 6

8 Theorem 2 Assume that h (x) h (r) if x = r, h (r 1 ) h (r 2 ) if r 1 r 2 2r 1, and, for large enough R, R 1 h (r) 2 r n 1 dr Ch (R) 2 R n. (i) (ii) (iii) If n 2 then, for all x, y R n and t > 0, the heat kernel of Δ satisfies the estimate ( ) C p t (x, y) V ( x, t ) exp x y 2, (2) ct Moreover, if h satisfies (i) and (ii) then condition (iii) is not only sufficient but also necessary for p t to satisfy (2). 7

9 Condition (i) means that h is nearly radial. Condition (ii) is a doubling property of h. Condition (iii) is trivially satisfied if h (r) is an increasing function since in this case h (r) h (R). If h (r) r β for r > 1 then (iii) is equivalent to R 1 r 2β r n 1 dr CR 2β+n, which is true if and only if 2β + n > 0, that is, if β > n/2. For any function h satisfying (i) (iii), the function V (x, r) admits the estimate V (x, r) Cr n h 2 ( x + r). Substituting this into (2) and symmetrizing it in x, y, we obtain ) C exp ( x y 2 ct p t (x, y) t n/2 h ( x + t ) h ( y + ). (3) t 8

10 Compactly supported potentials in R 2. Consider a function Φ C0 (R 2 ) such that Φ 0 and Φ 0. Then it is possible to show that the equation Δh Φh = 0 has a positive solution h on R 2 such that h (x) log x where x = 2 + x. This function obviously satisfies (i) (iii) (indeed, (iii) holds because log x is monotone increasing in x ). By (1) and (3) we obtain p Φ t (x, y) = p t (x, y) h (x) h (y) C log x log y ( t log ( x + t ) log ( y + t ) exp x y 2 ct In particular, in the most interesting long time range t x 2 + y 2, we have the following estimate ). p Φ t (x, y) log x log y t log 2 t. Recall that, by Proposition 1, if Φ C0 (R n ) with n 3 then, in the same range of t, x, y, we have p Φ t (x, y) 1. t n/2 9

11 Potential Φ (x) = b x 2 in R n (n 2). function in R n such that Let h (x) be a smooth positive h (x) = x β for x > 1, where β R is to be specified later. Define function Φ by so that h satisfies the equation Φ (x) := Δh h for all x Rn, Δh Φh = 0. Computing explicitly Δ x β, we obtain Φ (x) = b 2 for x > 1, x where b = β 2 + (n 2) β. 10

12 Consider now b as given, we solve this quadratic equation for β and obtain provided β = n (n 2 1 ) 2 + b, (4) b ( n 2 1 ) 2. (5) Assuming in the sequel that b satisfies (5), we see that β n + 1, in 2 particular, β > n/2. Hence, the function h satisfies all the conditions (i), (ii), (iii) of Theorem 2. Using the relation (1), that is, p Φ t (x, y) = p t (x, y) h (x) h (y) and the estimate (3) of p t from Theorem 2, we obtain: Theorem 3 Under the above conditions, we have p Φ t (x, y) C ( ) β ( ) β ) t t x y exp (. (6) t n/2 x y ct In particular, in the long time range t x 2 + y 2, we obtain p Φ t (x, y) C x β y β. (7) tn/2+β 11

13 The Harnack inequality. Let us explain some ideas used in the proof of Theorem 2. Recall that we are given a positive smooth function h in R n, n 2, that satisfies the conditions (i) (iii), and need to show that the heat kernel p t (x, y) of the operator Δ = 1 div (h 2 ) satisfies the h 2 estimate (2) where V (x, r) = μ (B (x, r)) and dμ = h 2 dx. The proof uses the fact that the required estimate (2) of the heat kernel p t is equivalent to the uniform parabolic Harnack inequality (P HI) for the operator Δ which means the following: there exists a constant C H such that, for any ball B(z, r) in R n, any positive solution u of the heat equation t u = Δu in the cylinder C := (0, r 2 ) B(z, r) satisfies the inequality where sup u C H inf u, C C + C = ( 1 4 r2, 1 2 r2 ) B(z, 1 2 r), C + = ( 3 4 r2, r 2 ) B(z, 1 2 r). 12

14 r 2 3 / 4r / 2r 2 1 / 4r 2-0 B(z, 1 / 2r) B(z, r) sup u C H inf u C C + 13

15 It is known that (P HI) is equivalent to the conjunction of the following two conditions that should be satisfied for all balls B (x, r): the volume doubling property: V (x, 2r) CV (x, r) ; (V D) the Poincaré inequality: for all f C 1 (B (x, r)) (f f) 2 dμ Cr 2 f 2 dμ, (P I) B(x,r/2) B(x,r) where f = fdμ. B(x,r/2) Hence, the estimate (2) will be proved if we show that (V D) and (P I) are satisfied. For example, if h 1 then both (V D) and (P I) are trivially satisfied because measure μ in this case is comparable to the Lebesgue measure, and for the Lebesgue measure both (V D) and (P I) are known to be true. However, in Theorem 2 measure μ is generally not comparable to the Lebesgue measure, which requires new tools for the proof. 14

16 Approach to the proof of Theorem 2. A ball B (x, r) R n is called remote if r < 1 x, and central if x = o (=the origin in 2 Rn ). non-remote ball remote balls o central ball x x r It is easy to prove that if the conditions (V D),(P I) hold for all central and remote balls then they hold for all balls. The following lemma is the most interesting and difficult part of the proof of Theorem 2. 15

17 Lemma 4 The conditions (V D), (P I) hold for all balls if and only if (V D),(P I) hold for remote balls and the following volume comparison condition is satisfied: V (o, r) CV for all r > 0 and x R n with x = r. ( x, ) r, (V C) o r= x B(o,r) x B(x, 1 /100r) 16

18 The condition (V C) implies (V D) for central balls. However, it is also used to prove the Poincaré inequality in central balls. This is done by splitting a ball B (o, R) into a sequence of annuli and considering them as vertices in a linear weighted graph. R o 2 -k R 17

19 Condition (V C) is used to prove a certain discrete Poincaré inequality on this graph, which is then merged with the Poincaré inequality in each annulus. Theorem 2 is now proved as follows. By conditions (i),(ii), function h (x) is nearly constant on any remote ball, which implies that (V D),(P I) for measure μ hold in remote balls. It is not difficult to show that condition (iii) is equivalent to (V C) for measure μ. Hence, by Lemma 4, we obtain that (V D),(P I) for measure μ hold in all balls. 18

20 Potential Φ (x) = b x 2 in R 1. As we have seen, Theorem 2 implies the following heat kernel bound for the potential Φ (x) = b x 2 in R n with n 2: p Φ t (x, y) C ( ) β ( ) β ) t t x y exp ( t n/2 x y ct that was stated in Theorem 3. Neither Theorem 2 nor Theorem 3 holds in the case n = 1! For the both theorems, the obstacle is the following topological property of R 1 : it has two ends, R + and R, that are separated by a compact set {0}. In R 1 the heat kernel of the potential Φ (x) = b x 2 has a different estimate as is stated in the following theorem. 19

21 Theorem 5 Assuming b 0, set β = b. Then the heat kernel p Φ t (x, y) of the operator Δ + Φ satisfies the following estimate: for all x 1, y 1, t 1, ( ) p Φ t (x, y) C ( x 1 β y β + x β y 1 β) x y 2 exp c. t 1/2+β t In particular, in the long time range t > x 2 + y 2 we obtain ( ) p Φ t (x, y) x β y β 1 t 1/2+β x β 1 y 2β 1 For comparison, in R n with n 2 we had the following estimate in the same range of the variables: p Φ t (x, y) x β y β t n/2+β. 20

22 Potentials Φ (x) = b x (2 ε). long range potential For comparison, consider a so called Φ (x) = b x γ, x > 1, where 0 < γ < 2 and b > 0. Theorem 6 For this potential, we have p Φ t (o, o) C exp ( ct 2 γ 2+γ ), for t > 1. Hence, in the case γ < 2 the heat kernel decays superpolynomially in t as t, in contrast to a polynomial decay in the cases γ 2. At present, no estimates are known for p Φ t (x, y) for all x, y. 21

23 References A.Grigor yan & L.Saloff-Coste, Stability results for Harnack inequalities, Ann. Inst. Fourier, Grenoble, 55 (2005), (Theorem 2) A.Grigor yan, Heat kernels on weighted manifolds and applications, Cont. Math. 398 (2006) (Theorems 3 and 6) A.Grigor yan & L.Saloff-Coste, Heat kernel on manifolds with ends, Ann. Inst. Fourier, Grenoble, 59 (2009) (Theorem 5) All papers can be found on my web page: 22

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