Regularity of solutions to elliptic equations with Grushin's operator
aa r X i v : . [ m a t h . A P ] M a r Regularity of solutions to elliptic equations with Grushin’s operator
Xiaohuan Wang a,b and Jihui Zhang a a Institute of Mathematics, School of Mathematical ScienceNanjing Normal University, Nanjing 210023, China a School of Mathematics and Statistics, Henan UniversityKaifeng, Henan 475001, China [email protected]
March 21, 2019
Abstract
In this paper, we consider the regularity of solutions to elliptic equation with Grushin’soperator. By using the Feynman-Kac formula, we first get the expression of heat kernel, andthen by using the properties of heat kernel, the optimal regularity of solutions will be obtained.The novelty of this paper is that the Grushin’s operator is a degenerate operator.
Keywords : Grushin’s operator; Schauder estimate; L p -theory. AMS subject classifications (2010): 35J70, 35J08.
The regularity of solutions to second order elliptic equation has been extensively studied by manyauthors, see the book [2]. But for degenerate elliptic equation, there is few work about the regularityresults until now. In this paper, we focus on a special degenerate elliptic equation–Grushin’s ellipticequation. The main reason why we can deal with it is that we can get the expression of heat kernelby using the probability method.For Grushin’s operator, many authors studied it. Beckner [3] obtained the Sobolev estimates forthe Grushin’s operator in low dimensions by using hyperbolic symmetry and conformal geometry.Riesz transforms and multipliers for the Grushin’s operator was considered by Jotsaroop et al. [4].Tri [8] studied the generalized Grushin’s equation. L p -estimates for the wave equation associatedto the Grushin operator was studied by Jotsaroop-Thangavelu [5]. The fundamental solution fora degenerate parabolic pseudo-differential operator covering the Grushin’s operator was obtainedby Tsutsumi [9, 10]. Furthermore, Tsutsumi [11] constructed a left parametrix for a pesudo-differential operator. We remark that Tsutsumi did not give the exact expression of heat kernel.In the book [1], they gave the expression of heat kernel (see Page 191), but he expression is hardto use because they used the inverse Fourier transform. Similar degenerate elliptic equation wasstudied by Robinson-Sikora [7] and the Hardy inequalities for Grushin’s operator was consideredby [12].In this paper, in view of probability point, we give a new expression and then get the regularityof solution by using the properties of heat kernel.This paper is arranged as follows. In next section, some preliminaries are given and the mainresults will be proved in section 3. Throughout this paper, we write C as a general positive constantand C i , i = 1 , , · · · as a concrete positive constant.1 Consider the Grushin’s operator L = 12 ( ∂ x + x ∂ y ) , which is the generator of the diffusion process ( X t , Y t ), where ( X t , Y t ) satisfies dX t = dW t ,dY t = X t dW t ,X = µ , Y = µ . Here W it ( i = 1 ,
2) are standard i.i.d Brownian motion. It is easy to see that the process ( X t , Y t ) isa Gaussian stochastic process. Direct calculations show that E (cid:18) X t Y t (cid:19) = (cid:18) µ µ (cid:19) , Cov ( X t , Y t ) = (cid:18) t µ tµ t µ t + t (cid:19) . Therefore, we get the heat kernel of the operator L is K ( t, x, µ , y, µ ) = 12 πt / exp (cid:26) − ( x − µ ) t − [ µ ( x − µ ) − y + µ ] t (cid:27) , which yields that ∇ x K ( t, x, y ) = − xt K ( t, x, y ) , ∇ y K ( t, x, y ) = − yt K ( t, x, y ) . We first want to solve the following elliptic equation L λb := ( L − λ ) u + b · ∇ u = f. (2.1) Theorem 2.1
Assume that b ∈ C βb ( R ) . There exists a λ > such that for any f ∈ C βb ( R ) and λ > λ , there is a unique solution u ∈ C βb ( R ) to equation (2.1) such that λ δ k u k C βb ( R ) ≤ k f k C βb ( R ) , (2.2) where δ > is defined in Lemma 3.1.Assume further that ∇ y h ( x, · ) ∈ C βb ( R ) for any x ∈ R , where h = b or f . Then there is aunique solution u ∈ C βb ( R ) to equation (2.1) such that λ δ k u k C βb ( R ) ≤ k f k C βb ( R ; C β ( R )) . Let u ( x, y ) := ( λ − L ) − f ( x ) = Z ∞ e − λt K ( t, · ) ∗ f ( x, y ) dt. Next, we consider the L p -regularity of u ( x, y ). Theorem 2.2
Assume that f ( · , y ) ∈ L p ( R ) for every y ∈ R with and f ( x, · ) ∈ W s,q ( R , R ) forevery y ∈ R , < s < and q > . Then we have the following estimates: k u k L r ( R ) ≤ Cλ − − r + p + q k f k L p ( R ,L q ( R )) , k∇ x u k L r ( R ) ≤ Cλ − − r + p + q k f k L p ( R ,L q ( R )) , k∇ y u k L r ( R ) ≤ Cλ − s − r + p k f k L p ( R ,W s,q ( R )) . Moreover, if we take < s < and p, q, r > such that − r + p + q ≤ and − s − r + p + 1 ≤ ,we have k∇ x u k L r ( R ) ≤ C, k∇ y u k L r ( R ) ≤ C. Denote [ · ] β by the semi-norm of C β . Set A := (cid:26) f ( x, y ) : f ( · , y ) ∈ L ∞ ( R ) f or any y ∈ R , f ( x, · ) ∈ C βb ( R ) f or any x ∈ R with < β < (cid:27) Lemma 3.1
Assume that f ∈ A . Then k u k L ∞ ( R ) ≤ Cλ − k f k L ∞ ( R ) , k∇ x u k L ∞ ( R ) ≤ Cλ − k f k L ∞ ( R ) , k∇ y u k L ∞ ( R ) ≤ Cλ − β k f k L ∞ ( R ,C βb ( R )) , [ ∇ x u ] β ≤ Cλ − + β k f k L ∞ ( R ,C βb ( R )) . Moreover, if f ( · , y ) ∈ C βb ( R ) for any y ∈ R , it holds that [ ∇ y u ] β ≤ Cλ − δ k f k C β ( R ) , where < δ < ( β ∧ (1 − β )) . That is to say, ∇ u ∈ C βb ( R ) . Proof.
Simply calculations show that k u k L ∞ ( R ) = k Z ∞ e − λt K ( t, · ) ∗ f ( x, y ) dt k L ∞ ( R ) ≤ Z ∞ e − λt k K ( t, · ) k L ( R ) k f k L ∞ ( R ) dt ≤ Cλ − k f k L ∞ ( R ) , k∇ x u k L ∞ ( R ) = k Z ∞ e − λt ∇ x K ( t, · ) ∗ f ( x, y ) dt k L ∞ ( R ) ≤ Z ∞ e − λt k∇ x K ( t, · ) k L ( R ) k f k L ∞ ( R ) dt ≤ Cλ − k f k L ∞ ( R ) , and k∇ y u k L ∞ ( R ) = k Z ∞ e − λt ∇ y K ( t, · ) ∗ f ( x, y ) dt k L ∞ ( R ) = k Z ∞ e − λt Z R ∇ y K ( t, x − u, y − v )( f ( u, v ) − f ( u, y )) dudvdt k L ∞ ( R ) ≤ k f k L ∞ ( R ,C βb ( R )) k Z ∞ e − λt Z R |∇ y K ( t, x − u, y − v ) | · | y − v | β dudvdt k L ∞ ( R ) ≤ C k f k L ∞ ( R ,C βb ( R )) Z ∞ e − λt t − β dt ≤ Cλ − β k f k L ∞ ( R ,C βb ( R )) , where we used the fact that Z R ∇ y K ( t, x − u, y − v ) f ( u, y ) dudv = Z R (cid:18)Z R ∇ y K ( t, x − u, y − v ) dv (cid:19) f ( u, y ) du = 0 . We also remark that the meaning of k f k L ∞ ( R ,C βb ( R )) is that we take infinity norm for the firstvariable and take H¨older norm for the second variable.Recall the following interpolation inequality[ u ] σα +(1 − σ ) γ ≤ ([ u ] α ) σ ([ u ] γ ) − σ , ≤ α < γ ≤ , σ ∈ (0 , . Now, if 0 < β <
1, applying the above inequality with α = 0 , γ = 1 and β = γ (1 − σ ), we have[ ∇ x u ] β = (cid:20)Z ∞ e − λt ∇ x K ( t, · ) ∗ f ( x, y ) dt (cid:21) β ≤ Z ∞ e − λt Z R [ ∇ x K ( t, · , · )] β ( x, y ) dxdy k f k L ∞ ( R ) dt ≤ k f k L ∞ ( R ) Z ∞ t − − β e − λt dt ≤ Cλ − + β k f k L ∞ ( R ) . Next, we consider the derivative of the second variable.[ ∇ y u ] β = (cid:20)Z ∞ e − λt ∇ y K ( t, · ) ∗ f ( x, y ) dt (cid:21) β ≤ Z ∞ e − λt sup x, ˆ x,y, ˆ y ∈ R ,x =ˆ x,y =ˆ y | x − ˆ x | β + | y − ˆ y | β × (cid:12)(cid:12)(cid:12) Z R ∇ y K ( t, x − u, v )( f ( u, y − v ) − f ( u, ˆ y − v )) dudv + Z R ∇ y K ( t, u, ˆ y − v )( f ( x − u, v ) − f (ˆ x − u, v )) dudv (cid:12)(cid:12)(cid:12)(cid:19) dt = Z ∞ e − λt sup x, ˆ x,y, ˆ y ∈ R ,x =ˆ x,y =ˆ y | x − ˆ x | β + | y − ˆ y | β | I + I | ! dt. By dividing the real line into two parts, we have I = Z R ∇ y K ( t, x − u, y − v )( f ( u, v ) − f ( u, y ) dudv + Z R ∇ y K ( t, x − u, ˆ y − v )( f ( u, ˆ y ) − f ( u, v )) dudv = Z R ∇ y K ( t, u, y − v )( f ( x − u, v ) − f ( x − u, y )) dudv + Z R ∇ y K ( t, u, ˆ y − v )( f ( x − u, ˆ y ) − f ( x − u, v )) dudv = Z R Z | y − v |≤ | y − ˆ y | ∇ y K ( t, u, y − v )( f ( x − u, v ) − f ( x − u, y ) dudv + Z R Z | y − v |≤ | y − ˆ y | ∇ y K ( t, u, ˆ y − v )( f ( x − u, ˆ y ) − f ( x − u, v )) dudv + Z R Z | y − v | > | y − ˆ y | ( ∇ y K ( t, u, y − v ) − ∇ y K ( t, u, ˆ y − v ))( f ( x − u, v ) − f ( x − u, y )) dudv + Z R Z | y − v | > | y − ˆ y | ∇ y K ( t, u, ˆ y − v )( f ( x − u, ˆ y ) − f ( x − u, y )) dudv =: I + · · · + I . Let us estimate I - I . By using the form of heat kernel, we get | I | = (cid:12)(cid:12)(cid:12) Z R Z | y − v |≤ | y − ˆ y | ∇ y K ( t, u, y − v )( f ( x − u, v ) − f ( x − u, y − u ) dudv (cid:12)(cid:12)(cid:12) ≤ Ct − Z R e − u t Z | y − v |≤ | y − ˆ y | | y − v | t e − ( y − v )2 t | y − v | β dv ! du = Ct − Z R e − u t Z | z |≤ | y − ˆ y | | z | t e − z t | z | β dv ! du ≤ Ct − β | y − ˆ y | β . Similarly, we can obtain I ≤ Ct − β | y − ˆ y | β . Next, we consider I . | I | = (cid:12)(cid:12)(cid:12) Z R Z | y − v | > | y − ˆ y | ( ∇ y K ( t, u, y − v ) − ∇ y K ( t, u, ˆ y − v ))( f ( x − u, v ) − f ( x − u, y )) dudv (cid:12)(cid:12)(cid:12) ≤ C Z R Z | y − v | > | y − ˆ y | |∇ y K ( t, u, y − v ) − ∇ y K ( t, u, ˆ y − v ) | y − v | β dudv ≤ Ct − Z R e − u t Z | y − v | > | y − ˆ y | (cid:12)(cid:12)(cid:12) y − vt e − ( y − v )2 t − ˆ y − vt e − (ˆ y − v )2 t (cid:12)(cid:12)(cid:12) | y − v | β dudv. Notice that | y − v | > | y − ˆ y | . So for every ξ ∈ [ y, ˆ y ],12 | y − v | ≤ | ξ − v | ≤ | y − v | . We recall the following fractional mean value formula (see (4.4) of [6]) f ( x + h ) = f ( x ) + Γ − (1 + β ) h γ f ( γ ) ( x + θh ) , where 0 < γ < θ > h satisfyinglim h ↓ θ γ = Γ (1 + γ )Γ(1 + 2 γ ) . Denote ˜ K ( t, v ) = vt e − v t . By using the above fractional mean value formula with γ > β and the interpolation inequality inH¨older space | I | ≤ Ct − | y − ˆ y | β Z R e − u t Z | y − v | > | y − ˆ y | [ ˜ K ( t, y − v )] γ | y − ˆ y | γ − β | y − v | β dudv ≤ Ct − | y − ˆ y | β Z R e − u t du Z | y − v | > | y − ˆ y | [ ˜ K ( t, y − v )] γ | y − v | γ dv ≤ Ct − γ − β | y − ˆ y | β Z R e − u du Z ∞ | v | γ e − v dv ≤ Ct − γ − β | y − ˆ y | β . Lastly, by using the properties of heat kernel K , it is easy to see that Z | u − ( y − v ) | > | y − ˆ y | ∇ y K ( t, u, y − v ) dv = Z | v | > | y − ˆ y | ∇ y K ( t, u, u − v ) dv = (2 π ) − t − e u t e − v t (cid:12)(cid:12)(cid:12) v = − | y − ˆ y | v =2 | y − ˆ y | = 0 . Using the above equality and similar to the operation of I , we have | I | = (cid:12)(cid:12)(cid:12) Z R Z | y − v | > | y − ˆ y | ∇ y K ( t, u, ˆ y − v )( f ( u, ˆ y − u ) − f ( u, y − u )) dudv (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) Z R Z | y − v | > | y − ˆ y | ( ∇ y K ( t, u, ˆ y − v ) − ∇ y K ( t, u, y − v )) × ( f ( u, ˆ y − u ) − f ( u, y − u )) dudv (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) Z R ( f ( x − u, ˆ y ) − f ( x − u, y )) du Z | y − v | > | y − ˆ y | × ( ∇ y K ( t, u, ˆ y − v ) − ∇ y K ( t, u, y − v )) dv (cid:12)(cid:12)(cid:12) ≤ C k f k L ∞ ( R ,C βb ( R )) | y − ˆ y | β Z R du × Z | y − v | > | y − ˆ y | |∇ y K ( t, u, ˆ y − v ) − ∇ y K ( t, u, y − v ) | dv ≤ C k f k L ∞ ( R ,C βb ( R )) t − γ − β | y − ˆ y | β . Substituting I − I into I , we get Z ∞ e − λt sup x, ˆ x,y, ˆ y ∈ R ,x =ˆ x,y =ˆ y | x − ˆ x | β + | y − ˆ y | β | I | ! dt ≤ C Z ∞ e − λt (cid:16) t − β + t − γ − β (cid:17) dt ≤ Cλ − ( β ∧ ( γ − β )) . Similarly, we can prove that if f ( · , y ) ∈ C βb ( R ) for any y ∈ R , Z ∞ e − λt sup x, ˆ x,y, ˆ y ∈ R ,x =ˆ x,y =ˆ y | x − ˆ x | β + | y − ˆ y | β | I | ! dt ≤ Cλ − ( β ∧ ( γ − β )) . Summing the above discussion, we obtain[ ∇ y u ] β ≤ Cλ − ( β ∧ ( γ − β )) k f k C β ( R ) . Noting that the above inequality holds for 0 < γ <
1. The proof of this lemma is complete. (cid:3)
Remark 3.1
It is well known that if f ∈ C α ( R n ) , then the solution u of the following equation u t − ∆ u = f, u = 0 , belongs to C α ( R n ) , which is the Schauder theory. Noting that the heat kernel of above equationis Gauss heat kernel, that is K ( t, x ) = 1(2 πt ) n e − x t . It is easy to see that x ∼ √ t . But in our case, different axis has different scaling, that is, x ∼ √ t, y ∼ t. Thus when we take derivative for variable x , we can get t − , and take double derivative for variable x , we will get t − . But if we take derivative for variable y , we shall get t − , which is differentfrom the classical case. In Schauder theory, we can get C α ( R n ) estimates, but in our case the C α ( R n ) should be optimal. Like the classical case, we can get the C α -estimate for the x -axis if f ( · , y ) ∈ C βb ( R ) for any y ∈ R ,but we can not get the same estimate for y -axis. If we want to get the C α -estimate for the y -axis,we need more regularity about the second variable. In other words, we have the following results. Corollary 3.1
Assume that f ( · , y ) ∈ C βb ( R ) for any y ∈ R and ∇ y f ( x, · ) ∈ C βb ( R ) for any x ∈ R . Then [ ∇ u ] β ≤ Cλ − δ k f k C β ( R ; C β ( R )) , where < δ < ( β ∧ (1 − β )) . That is to say, ∇ u ∈ C βb ( R ) . Proof of Theorem 2.1.
We use Picard’s iteration to solve (2.1). Let u = 0 and define for n ∈ N , u n := ( λ − L ) − ( f − b · ∇ u n − ) . (3.1)It follows from Lemma 3.1 that λ δ k u n k C βb ( R ) ≤ k f − b · ∇ u n − k C βb ( R ) ≤ k f k C βb ( R ) + k b k L ∞ ( R ) · k∇ u n − k C βb ( R ) + k b k C βb ( R ) · k∇ u n − k L ∞ ( R ) , (3.2)and λ δ k u n − u m k C βb ( R ) ≤ k b k L ∞ ( R ) · k∇ u n − − ∇ u m − k C βb ( R ) + k b k C βb ( R ) · k∇ u n − − ∇ u m − k L ∞ ( R ) . (3.3)Choosing λ be large enough so that Cλ − δ k b k C βb ( R ) < / λ ≥ λ , we get k u n k C βb ( R ) ≤ λ − δ ( k f k C βb ( R ) + 2 k b k C βb ( R ) ) + 12 k u n − k C βb ( R ) and for all n ≥ m , k u n − u m k C βb ( R ) ≤ k u n − − u m − k C βb ( R ) . Substituting them into (3.2) and (3.3), we obtain λ − δ k u n k C βb ( R ) ≤ C k f k C βb ( R ) and for all n ≥ m , λ − δ k u n − u m k C βb ( R ) ≤ C m . Hence there is a u ∈ C βb ( R ) such that (2.2) holds and λ − δ k u − u m k C βb ( R ) ≤ C m , and u solves equation (2.1) by taking limits for (3.1). The second result can be obtained similarly.The proof is complete. (cid:3) Proof of Theorem 2.2.
For simplicity, we only consider a special case, that is, f ( x, y ) = f ( x ) f ( y ). Assume that f ∈ L p ( R ) and f ∈ L q ( R ). Denote K ( t, y ) = Z R K ( t, x − u, y ) f ( u ) du. By using the above inequality, Minkowski’s inequality and the properties of the heat kernel, wehave k u k L r ( R ) = k Z ∞ e − λt K ( t, · ) ∗ f ( x, y ) dt k L r ( R ) ≤ Z ∞ e − λt (cid:18)Z R | Z R K ( t, x, y − v ) f ( v ) dv | r dxdy (cid:19) r dt = Z ∞ e − λt (cid:18)Z R k K ( t, x, · ) ∗ f k rL r ( R ) dx (cid:19) r dt ≤ k f k L q ( R ) Z ∞ e − λt (cid:18)Z R k K ( t, x, · ) k rL m ( R ) dx (cid:19) r dt = k f k L q ( R ) Z ∞ e − λt (cid:18)Z R (cid:12)(cid:12)(cid:12) Z R | K ( t, x, y ) | m dy (cid:12)(cid:12)(cid:12) rm dx (cid:19) r dt ≤ k f k L q ( R ) Z ∞ e − λt (cid:18)Z R (cid:12)(cid:12)(cid:12) Z R | Z R K ( t, x − u, y ) f ( u ) du | r dx (cid:12)(cid:12)(cid:12) mr dy (cid:19) m dt = k f k L q ( R ) Z ∞ e − λt (cid:18)Z R (cid:12)(cid:12)(cid:12) Z R k K ( t, · , y ) ∗ f k mL r ( R ) dy (cid:19) m dt ≤ k f k L p ( R ) k f k L q ( R ) Z ∞ e − λt k K ( t, · , · ) k L n ( R ,L m ( R )) dt ≤ C Z ∞ e − λt t − + n + m dt ≤ Cλ − − r + p + q , where 1 + 1 r = 1 n + 1 p , r = 1 m + 1 q , (3.4) C = k f k L p ( R ) k f k L q ( R ) Z R (cid:18)Z R e − p ′ x − p ′ ( y − x ) dx (cid:19) q ′ p ′ dy q ′ . We remark that (cid:18)Z R |∇ x K ( t, x, y ) | r dx (cid:19) r ≤ Ct − + r , (cid:18)Z R |∇ y K ( t, x, y ) | r dy (cid:19) r ≤ Ct − r . Similarly, we obtain k∇ x u k L r ( R ) ≤ Cλ − − r + p + q . Furthermore, we can get k∇ x u k L r ( R ) ≤ Cλ − r + p + q . Thus if we take p = q = r , then we have k∇ x u k L r ( R ) ≤ C . However, if we deal with the secondvariable, it is difficult to get the decay estimate. More precisely, we have for r < p + q , k∇ y u k L r ( R ) ≤ Cλ − r + p + q → ∞ , as λ → ∞ . h ∈ W s,p ( R n ) with 0 < s <
1, then k h k W s,p ( R n ) = (cid:18) k h k pL p ( R n ) + Z R n Z R n | h ( x ) − h ( y ) | p | x − y | n + sp dxdy (cid:19) p . If we assume that f ∈ W s,q ( R ) and let K ( t, y ) = Z R ∇ y K ( t, x, y − v ) f ( v ) dv. we get k∇ y u k L r ( R ) = k Z ∞ e − λt ∇ y K ( t, · ) ∗ f ( x, y ) dt k L r ( R ) ≤ Z ∞ e − λt (cid:18)Z R | Z R K ( t, x − u, y ) f ( u ) du | r dxdy (cid:19) r dt ≤ k f k L p ( R ) Z ∞ e − λt (cid:18)Z R k K ( t, · , y ) k nL r ( R ) dy (cid:19) n dt = k f k L p ( R ) Z ∞ e − λt × Z R (cid:12)(cid:12)(cid:12) Z R ∇ y K ( t, x, v ) | v | s + q f ( y − v ) − f ( y ) | v | s + q dv (cid:12)(cid:12)(cid:12) r dx ! nr dy n dt ≤ k f k L p ( R ) k f k W s,q ( R ) Z ∞ e − λt (cid:18)Z R (cid:16) |∇ y K ( t, x, y ) | m | y | sm + mq dy (cid:17) nm dx (cid:19) n dt ≤ C Z ∞ e − λt t − + s + q + n + m dt ≤ Cλ − s − r + p , where m, n satisfy (3.4) and C = 2 k f k L p ( R ) k f k W s,q ( R ) Z R (cid:18)Z R | y − x | p | y | sp + pq e − p ′ x − p ′ ( y − x ) dx (cid:19) q ′ p ′ dy q ′ . Moreover, under the condition that f ∈ W s,p ( R ), we can similarly get k∇ y u k L r ( R ) ≤ Cλ − s − r + p +1 . Hence it is easy to see that we can take suitable s ∈ (0 , , p > r > − s − r + p +1 ≤
0. That is to say, we have k∇ y u k L r ( R ) ≤ C under the condition that f ∈ W s,p ( R ). The proof is complete. (cid:3) Remark 3.2
It is well known that if f ∈ L p ( R n ) , then the solution u of the following equation u t − ∆ u = f, u = 0 , belongs to W ,p ( R n ) , which is the L p -theory. Noting that the heat kernel of above equation isGauss heat kernel, and similar to Remark 3.1, it is easy to find the difference from the classicalLaplacian operator. Due to the singularity of the variable y , we must give two different assumptions.Comparing the classical L p -theory, in our case the regularity of Theorem 2.2 should be optimal. Acknowledgment
The first author was supported in part by NSFC of China grants 11571176 and11771123.
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