ISSN: [Engida* et al., 6(4): April, 2017] Impact Factor: 4.116

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1 IJESRT INTERNATIONAL JOURNAL OF ENGINEERING SCIENCES & RESEARCH TECHNOLOGY ELLIPTICITY, HYPOELLIPTICITY AND PARTIAL HYPOELLIPTICITY OF DIFFERENTIAL OPERATORS Temesgen Engida*, Dr.Vishwajeet S. Goswami * Vice Dean, College of Natural and Computational Sciences, Dilla University Ethiopia Assistant Professor of Mathematics, Department of Mathematics, College of Natural and Computational Sciences, Dilla University Ethiopia DOI: /zenodo ABSTRACT The conditions for ellipticity and hypoellipticity of operator P(D) in terms of fundamental solutions were described by HÄormander []. But the relationship and the difference of condition of ellipticity, hypoellipticity and partial hypoellipticity of operator P(D) in terms of fundamental solutions was not established. We look at the case when the solution u of the equation P(D)u = f is conditionally smooth whenever f is a smooth function. INTRODUCTION Distributions (also known as a generalized functions) are widely used to formulate generalized solutions of partial differential equation. Where a classical solution may not exist or be very difficult to establish, a distribution solution to a differential equation is often much easier. One of the most useful aspects of the distribution theory is that discontinues functions can be handled as easily as continuous or differentiable functions. This paper aimed to describe the relationship and the difference of condition of ellipticity, hypoellipticity and partial hypoellipticity of differential operator P(D) in terms of its Fundamental Solutions. For the first time some remarkable results about infinitely differentiability and analyticity of solutions of elliptic differential operators were obtained by Bernstein at the beginning of the 0th century. Later on Petrovskii proved that all classical solutions of equation P(D)u = f with analytic right hand side are analytic if and only if operator P(D) is elliptic. The conditions for ellipticity and hypoellipticity of operator P(D) in terms of fundamental solutions were described by HÄormander []. But the relationship and the difference of condition of ellipticity, hypoellipticity and partial hypoellipticity of operator P(D) in terms of fundamental solutions was not established. We look at the case when the solution u of the equation P(D)u = f is conditionally smooth whenever f is a smooth function. WE DESCRIBE THE PARTIAL HYPOELLIPTICITY OF DIFFERENTIAL OPERATORS WITH CONSTANT COEFFICIENTS IN TERMS OF FUNDAMENTAL SOLUTIONS. Fundamental solution of linear differential equation Definition:.1.1 Let P be a non-zero differential operators with Constant coefficients a α (x) = a α such that P (D) = K ad. (3.1) Where K is a finite set in N o n the space of multi-indices; i.e. N o n = N o. N o (n copies) with N o = N {0}. for α N o n =α 1 + α + + α n Max = m; a α k are constants; D α = D 1 α 1. D α... D n α n and Dj α j = 1 i α j x j α j [44]

2 Aim: To describe the condition of partial hypoellipticity of differential operator (3.1) in terms of its Fundamental Solutions. A distribution ε(x) D (R n ) that satisfies the equation P (D) ε(x) = δ(x) (3.) in R n is said to be a fundamental solution of the operator P(D) Remark.1.1: A fundamental solution of the operator P (D) generally is not unique. Proof; suppose ε(x) D (R n ) is fundamental solution of the operator P(D) Then; P (D) ε(x) = δ(x) (*) Let ε 0 (x) D (R n ) be arbitrary solution of the homogeneous equation, P(D)ε 0 (x) = 0 (**) From (*) and (**) we have P (D)[ε(x) + ε 0 (x)] = P (D) ε(x) + P(D)ε 0 (x) = δ(x) This implies that ε(x) + ε 0 (x) is also the fundamental solution of the operator P (D) Lemma.1.1: A distribution εs (R n )is said to be a fundamental solution of the operator P (D) if and only if the Fourier transform F[ε] satisfy the equation P(i ξ)f[ε]=(π) n, Where, P (D) = k a Proof: Let εs be a fundamental solution of the operator P (D) P (D) ε(x) = δ(x) F[P(D)ε] = F[δ] = (π) n F [ m ad ] =(π) n ma F[Dα ε] = (π) n m a(i ) F[ε] = (π) n P(i ξ)f[ε] = (π) n By using the notation of fundamental solution ε(x) of the operator P(D), we can construct a solution of equation P(D)u(x) =f(x),..(3.3) With an arbitrary function f. Theorem.1.1: Let f D such that ε f exists in D. then the equation (3.3) has a solution in D that is given by the formula U = ε f.. (3.4) This solution is unique in the class of distributions belongs to D for which a convolution with ε exist. Proof: Form (.9) and (3.) we have P(D) (ε f) = ma Dα (ε f ) = ( a m Dα ε) f = P(D)ε f = δ f = f Hence u= ε f is a solution of equation (3.3) Uniqueness: It suffice to show that the corresponding homogeneous equation P(D)u=0 has only a zero solution in the class of generalized function beloges to D whose convolution with ε exists in D. u = u δ = u P(D)ε = P(D)u ε = 0 Hence u= ε f is a unique solution ind Example: Find a fundamental solution of the heat conduction operator differential equation L t a Solution: Let ε be a fundamental solution of the heat conduction operator differential equation. Then ε satisfy ε(x,t) t a ε(x, t) = δ(x, t).. ( ) [45]

3 Now, applying the Fourier transform F x to ( ), we obtain F x [ ε(x,t) a ε(x, t)] = F t x [δ(x, t)] But, ε(x, t) F x [ ] = t t F x [ε(x, t)] F x [ ε(x, t)] = ε(x, t) t e i(ξ,x) dx = ε(x, t) e i(ξ,x) t dx = ε(x, t)(iξ) e i(ξ,x) dx = i (ξ 1 + ξ + + ξ n ) ε(x, t)e i(ξ,x) dx = ξ F x [ε(x, t)] (F x [δ(x, t)], φ) = (δ(x, t), F ξ [φ]) = F ξ [φ(x, t)](0) = φ(ξ, 0) = 1(ξ)δ(t) Therefore, F t x [ε(x, t)] + a ξ F x [ε(x, t)] = 1(ξ)δ(t) Let E(ξ, t) = F x [ε(x, t)], then E(ξ, t) + t a ξ E(ξ, t) = 1(ξ)δ(t) Let L E(t) = E(t) + t a ξ E(t) is a linear differential operator with constant coefficients. The solution is a generalized function θ(t)z(t) where z(t) is the solution of initial value problem L z(t) = 0, z(0) = 1 Hence, z(t) + t a ξ z(t) = 0. this implies that z(t) = e a ξ t Therefore, E(ξ, t) = θ(t)z(t) = θ(t)e a ξ t and applying inverse Fourier transform F 1 x, we obtain ε(x, t) = F ξ 1 [θ(t)e a ξ t ] = θ(t)f ξ 1 [e a ξ t ] = θ(t) π e a ξ t e i(ξ,x) dξ x = θ(t)e 4a t π a t e t dt = θ(t) π e a t( ξ + = θ(t) π t( ξ + e a = θ(t)e π = θ(t)e π = π x 4a t x 4a t x = θ(t)e 4a t a πt i x ξ a t ) dξ i x a t ) x 4a tdξ e a t( ξ + i x a t ) dξ i x (a t( ξ + a e t dξ )) Description of the ellipticity, hypoellipticity and partial hypoellipticity in terms of fundamental solutions Definition..1: A distribution ε is called a fundamental solution for differential operator (3.1) if, and only,if P (D) ε(x) = δ(x). (3.5) Where δ is the Dirac delta function. One of the basic and the most profound results is the proof of the existence of a fundamental solution ε in D of any differential operator of the type (3.1) Equation (3.5) in the class S is equivalent to the algebraic equation P(i ξ)f[ε] = (π) n.(3.6) The problem of seeking a fundamental solution of slow growth turns out to be a special case of the more general problem of dividing a generalized function of slow growth by a polynomial, that is, of the problem of finding a solution u in S of the equation P(ξ)u = f.. (3.7) [46]

4 Where P is a nonzero polynomial and f is a specified generalized function in S. Elliptic and hypoelliptic operators Definition..: Let Ω R n be an open set and let u D (Ω) a Differential operator (3.1) is called elliptic, if P(D)u A(Ω), then u A(Ω). Definition..3: Let Ω R n be an open set and let u D (Ω) a differential operator (3.1) is called hypo elliptic, if P(D)u C then u C Note that every elliptic operator is hypo elliptic. Example: -The Laplace and Cauchy-Riemann operators are elliptic - The heat conducting operator is hypo elliptic The conditions for ellipticity and hypoellipticity of operator P(D) in terms of fundamental solutions were described by HÄormander. But the Condition for partial hypoellipticity of operator P(D) in terms of fundamental solutions was not established. We look at the case when the solution u of the equation P(D)u = f is conditionally smooth whenever f is a smooth function. That is, we want to study the smoothness of u with respect to some variables. Lemma..1: Let Ω R n be an open set, u D (Ω) and ψ D(R n ). Then the set Ω ψ = (Ω + y) y spp(ψ ) is open. Moreover, the convolution u ψ exists in Ω ψ. That is for all ψ D( Ω ψ ) (u ψ,φ)(x) = (u(x)(ψ(y), φ(x + y))) Lemma..: Let Ω R n be an open set, m, n N 0 are such that 0 m < n and x = (x, x n m ).Let for. Let ψ D(R n ), ψ m =ψ(x ) δ(x ): supp(ψ m ) = supp(ψ) {0}, Where δ is a Dirac delta function Then the set Ω ψ = (Ω + y) y spp(ψ ) is open. Moreover the convolution u ψ m exists in Ω ψm for Ω R n be an open set and u D ( Ω),that is ψ D(Ω ψm ) (u ψ m, ψ )(x) := (u(x), (ψ(y ), φ(x + y ))) Definition..4: Let Ω R n be an open set and u D ( Ω) a differential operator P(D) is called partially hypoelliptic with respect to the plane x = 0, if P(D)u C, then for any function ψ D(R m ) the convolution u ψ m C (Ω ψm ). Remark..1: If m = 0 in Definition (3), then m=0, then ψ o= δ and in this case we obtain definition (). In this paper we prove the following theorem in which a description of the partial hypoellipticity of a differential operator (3.1) in terms of its fundamental solutions is given. Theorem..1: For the differential operator (3.1) to be partially hypoelliptic with respect to the plane x = 0, it is necessary and sufficient that the existence of a fundamental solution ε of operator (3.1) such that for any function ψ m D(R n )the convolution ε ψ m C (Ω ψm supp(ψ m )) Remark..: From the existence of such fundamental solution follows that every fundamental solution satisfies ε ψ m C (Ω ψm supp(ψ m )) [47]

5 Preliminaries In this section we will describe the basic tools we need in the proof of our theorem and we prove them for the sake of completeness. For sets A and B we will use the following notations, inf A-B ={x y: x A and y B } and the distance between A and B, ρ(a;b) = x y x A,x B Theorem..: Let Ω R n be an open set, u D (Ω) and ψ D(R n ). Then the set Ω ψ = (Ω + y) is open y spp(ψ ). Moreover, the convolution u ψ exists in Ω ψ. That is for all ψ D( Ω ψ ), (u ψ,φ)(x) = (u(x), (ψ(y), φ(x + y))) Note that for the convolution u ψ m to be defined for all φ D( Ω ψ ) it is important to show that the set Ω ψm = (Ω + y) y spp(ψ m ) is open. Proposition..1 If Ω R n and x, y R n, then ρ(x y, Ω)= ρ(x, Ω + y) Proposition.. If Ω R n and x, y, z R n, then ρ(x, Ω + y) ρ(x, Ω + z) y z Proof: ρ(x, Ω + y) ρ(x, Ω + z) = ρ(x y, Ω) ρ(x z, Ω) And ρ(x y, Ω) ρ(x z, Ω) x y (x z) y z by Proposition 3..1 Proposition..3. Let Ω α R n ; where α A R. Then for any x R n the condition is equivalent to ρ (x, Ω α ) > 0 (1) inf α A ρ(x, Ω α) > 0 ( ) Proof: Condition () is equivalent to the existence of c > 0, such that ρ(x, Ω α ) > c, for all α A. Then c B(x, c) Ω α for all α A and this means B(x, c) Ω α c = ( Ω α ) c ρ(x, Ω α ) > 0 Conversely, let condition (1) be satisfied and suppose that condition () be not satisfied. If A is a finite set, then there exists α 0 A such that ρ(x, Ω αo ) = 0 Hence That is, ρ (x, Ω α ) ρ(x, Ω αo ) = 0 ρ (x, Ω α ) = 0 If A is an infinite set, then for each k N there exists α k A such that ρ(x, Ω αk ) 1/k Consequently, [48]

6 and taking the limit we get ρ (x, Ω α ) ρ(x, Ω αk ) 1/k ρ (x, Ω α ) = 0 In both cases we arrived at a contradiction to condition (1) and this completes the proof Lemma..3; Let Ω R n be an open set and F compact, then the set G = (Ω + y).proof: That G is open means that ρ(x, G c ) > 0 for all x G. That is, for all x G, is open ρ (x, [ (Ω + y) ] c ) = ρ (x, (Ω + y) c ) By proposition that is equivalent to = ρ (x, (Ω c + y) ) > 0 inf y F ρ(x, (Ωc + y)) > 0 for all x G. Assume to the contrary that there exists x G such that inf y F ρ(x, (Ωc + y)) = 0 This implies that, for every k N their exists y k F such that ρ(x, (Ω c + y k )) 1/k We can select a convergent subsequence {y ks } such that lim y ks = y o, for y o F s Then, ρ(x, (Ω c + y o )) = lim ρ(x, (Ω c + y ks )) = 0 s since (Ω c + y o ) is closed by the assumption, x (Ω c + y o ) (Ω c + y) = (Ω + y) c = [ (Ω + y) ] c = G c This contradicts the fact that x G. Consequently, G = (Ω + y) is open. From the Lemma..3 we can conclude that for ψ D(R n ) and open set Ω R n the set [49]

7 Ω ψ = (Ω + y) is open y supp(ψ). Remark..3: (i) For any u D ( Ω) and ψ D (R n ) the convolution is well defined onω ψ. ii) Suppose Ω R n is an open set m, n Z are such that 0 m < n, and for any ψ D (R m ) define ψ m (x) = ψ(x ) δ(x ), where x = (x, x ),wher x R m, x R n m. Then supp (ψ m ) = supp(ψ) {0} and the set Ω ψm = y supp(ψ m ) (Ω + y) is open. Definition..5: Let Ω R n be open, u D ( Ω) and ψ D (R m ). Then for D( Ω ψm ), we define (u ψ m, φ) = (u(x), ψ(y ) supp(ψ) φ(x + y )dy ) Remark..4: For any u D ( Ω) and ψ D (R m ) the convolution u ψ m is well define on Ω ψm. Lemma..4: Let Ω R n be an open set,m, n Z be such that 0 m < n, and ψ D (R m ). Let φ, φ k D( Ω ψm ) for k N and φ k φ in D( Ω ψm ) as k. Then, (ψ(y ), φ k (x + y )) (ψ(y), φ(x + y )) in D( Ω ψ ) Proof: φ k φ in D( Ω ψm ) means φ k φ 0 in D( Ω ψm ). Hence, (ψ(y ), φ k (x + y )) (ψ(y ), φ(x + y )) = (ψ(y ), φ k (x + y ) φ(x + y))) (ψ(y ), 0) = 0 That is, (ψ(y ), φ k (x + y )) (ψ(y), φ(x + y )) in D( Ω ψ ). Lemma..5: Let Ω R n be an open set,m, n Z be such that 0 m < n, and ψ D(R m ). Then u ψ m D ( Ω ψm ) for any u D ( Ω) Proof: From Remark 3..4 it follows that u ψ m is defined on Ω ψm and from lemma 3..4 it follows that the functional is continuous. Since it is linear on D( Ω ψm ) we conclude that u ψ m D ( Ω ψm ) Remark..5 If Ω C (Ω) then the convolution u ψ m ( C Ω ψm ). Description of the partial hypoellipticity in term of fundamental solutions. All classical solutions of equations p(d)u = f with analytic right hand side are analytic if and only if operator (1) is elliptic.in other words if Ω R n is an open set,then for any u C (Ω) such that P(D) is an analytic function in Ω, u is an analytic function in Ω if and only if operator (3.1)is elliptic. In terms of the polynomial p this is equivalent to p(ξ) 0 ξ 0 Where, p(d) = a α D α αεk: α =1 is the principal part of operator(3.1). The corresponding description for hypoelliptic operators was obtained by HÄormander. That is, a differential operator (3.1) is hypoelliptic if and only if for all multi-indices α 0, P (α) (ξ) p(ξ) 0 as ξ The following is Hormander s theorem about description of the hypoellipticity in terms fundamental solutions. Theorem..3 For an operator (3.1) to be hypoelliptic it is necessary and sufficient that there exists a fundamental solution ε of (3.1) such that ε C (R n \{0}) Remark..6. From the existence of such fundamental solution if follows that any fundamental solution ε C (R n \{0} ) Proof of Theorem..3: Necessity Let Ω R n be open and u D ( Ω) be a solution of p(d)u = f Where, f C (R n ) Since ε is fundamental solution of operator (3.1), One has p(d)ε = δ where δ C (R n \{0}). Consequently by definition 3..3, [50]

8 ε C (R n \{0} ) Sufficiency: Let ε C (R n \{0}) be a fundamental solution of operator (3.1) and u D ( Ω) be a solution to the equation P(D)u = f with f C (R n ) Let Ω be an arbitrary open set such that Ω Ω and let η D( Ω) with η(x) = 1 for all x Ω. The generalized function ηu is compactly supported in Ω and satisfies the equation p(d)(ηu) = ηf + f 1 where ηf D( Ω) andf 1 D ( Ω) such that supp(f 1 ) supp(η) \Ω. Therefore, ηu= δ ηu = p(d)ε ηu = ε p(d)ηu = ε (ηf + f 1 ) = ε (ηf) + ε f 1. Since ηf D( R n ), ε ηf C (R n ) and ε f 1 C (R n \ supp(f 1 )), we have ε f 1 C (R n (Ω\ Ω ) ), that is, ε f 1 C (Ω). And since ηu= u in Ω and Ω Ω is arbitrary, u C (Ω). It is known that operator (3.1) is partially hypoelliptic with respect to the plane x = 0 if and only if for all α 0 P (α) (ξ,ξ ) p(ξ,ξ ) 0 as ξ and ξ remains bounded. Lemma..5 Let G R n be a closed set, f D ( R n ) C (R n \G) and g ε (R n ). Then f g C (R n \ supp(g) + G). Moreover, for all γ > 0, (f g) = (g(y), η(y)f(x y)) on R n \((supp(g) )+G), γ where η D( R n ), η = 1 in some neighborhood of supp(g) and supp(η) (supp(g)) γ. Remark..6 If Ω = G c then from Lemma..5 and Remark..5 we observe that (f g) = (g(y), η(y)f(x y)) C (Ω g ) where, Ω g = (G c + y) = ( (G + y) c y supp(g) y supp(g) = (G + supp(g)) c = R n \ (supp(g) + G). Remark..7 If G = {0} in Lemma 3..5, then it can be easily shown that (f g) = (g(y), η(y)f(x y)) C (R n \ (supp(g))) Remark..8 If G = R n is a closed set, f D ( R n ) C (R n \G) (f g 1,, g n )(x) exists in D (R n ) and we have, (f g 1,, g n )(x) C (R n \(G + supp(g 1 ) + + supp(g n ). and g 1,, g n ε (R n ) then Proof of Theorem..1 Necessity: Let differential operator (3.1) be partially hypoelliptic with respect to the plane x = 0 and ε be its fundamental solution, i.e. p(d)ε = δ. Then using Definition 3..4, and Remarks 3..5, 3..6 and 3..7, we deduce that for any ψ D( R n ) and ψ m (x) = ψ(x ) δ(x ), We have (ε ψ m )(x) C (Ω ψm supp(ψ m )). Sufficiency: Let ε be a fundamental solution of operator (3.1) and that (ε ψ m )(x) C (Ω ψm supp(ψ m )). Moreover, let u D (Ω) be a solution of the equation P(D)u = f with f C (Ω). Then, ηu ψ m C ( Ω ψm ) where η D(Ω), η(x) = 1 x Ω Ω Indeed, ηu ψ m = ηu (ψ m δ) = ηu (ψ m P(D)ε) = ηu P(D)( (ψ m ε) = P(D)(ηu (ε ψ m )) = P(D)(ηu) (ε ψ m ) = (ηf + f 1 ) (ε ψ m ) Where ηf and f 1 are as in the proof of Theorem 3..3 Since, ηf D( Ω) the convolution ηf (ε ψ m ) C (R n ) By Lemma 3..5, f 1 (ε ψ m ) = (ε ψ m ) f 1 C (R n \ (supp(ψ m ) + supp(f 1 )) [51]

9 And since R n \ ((Ω ) c + supp(ψ m )) R n \ (supp(ψ m ) + supp(f 1 )) it follows that (ε ψ m ) f 1 C (R n \ ((Ω ) c + supp(ψ m ))). But, (Ω ) c + supp(ψ m ) = y supp(ψ (Ω ) c m ) + y) = y supp(ψ (Ω + y) c m ) = ( y supp(ψ (Ω m ) + y) ) c =(Ω ) c ψ m Consequently, (ε ψ m ) f 1 C (Ω ψm ) and ηu ψ m C ( Ω ψm ) Since Ω Ω is arbitrary, ηu ψ m C ( Ω ψm ). Next, let x o Ω ψm.then there exists δ > 0 such that B(x o, δ) Ω ψm, Ω x o = B(x o, δ) supp(ψ m ) and Ω x o Ω. Let η x0 D(Ω) be such that η x0 (x) = 1 x Ω x o. Observe that η x0 u ψ m = u ψ m on B(x o, δ) That is, (η x0 u ψ m, φ) = (u ψ m, φ) φ D(B(x o, δ)). Indeed, (η x0 u ψ m, φ) = (η x0 (x)u(x), (ψ m (y), φ(x + y)) = (η x0 (x)u(x), (ψ(y ) δ(y ), φ(x + y, x + y ))) = (η x0 (x)u(x), (ψ(y ), φ(x + y, x ))) = (u(x), ((η x0 (x)ψ(y ), φ(x + y ))) = (u(x), η x0 (x)ψ(y )φ(x + y )dy ) R m Since (x, x ) = (x + y, x ) (y, 0) B(x o, δ) supp(ψ m ) for all y supp(ψ m ). Then for all x with x + y supp(φ) B(x o, δ), we have η x0 (x) = 1 Consequently, (η x0 u ψ m, φ) = (u(x), φ(x + y )ψ(y )dy ) R m = (u ψ m, φ) for all φ D(B(x o, δ)) Hence, ε ψ m C (Ω ψm supp(ψ m )) and we are done SUMMARY AND RESULTS Definition Differential Operator (3.1) is elliptic if and only if u D (Ω), P(D)u A(Ω) u A(Ω) Differential Operator (3.1) is hypoelliptic if and only if u D (Ω), P(D)u C (Ω) u C (Ω) Differential Operator (3.1) is partial hypoellipticity if and only if u D (Ω), P(D)u C (Ω), ψ D(Ω) ε ψ m C (Ω ψm ) In terms polynomial. For ξ 0 The principal part P m (ξ) 0 P (α) (ξ) p(ξ) P (α) (ξ,ξ ) p(ξ,ξ ) 0 as (ξ ) ξ remains bounded. 0 as (ξ ) And In terms of fundamental solution. Any Fundamental solution ε is such that ε A([{0}] c ) Any Fundamental solution ε is such that ε C ([{0}] c ) For m, n N 0 : 0 m < n and ψ D(R m ) ε ψ m C ([supp(ψ m )] c ) REFERENCES [1] Vladimirov V. S., Generalized Functions in Mathematical Physics. Mir publisher. Moscow, [] Hormander L., Linear Partial Differential Operators. New York, [3] F.G.Friedlander, Introduction to the theory of distributions, Cambridge university press, London, 198. [4] Zemanian, A. Distribution theory and transform analysis, McGraw Hill, New York, 1965 [5]