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1 Degenerated operator equations of higher order L. Tepoyan November 14, 1 Introduction The main object of the present paper is the dierential{operator equation L u (;1) m D m t (t D m t u)+ad m;1 t (t ;1 D m t u)+t ;m Pu= f(t) (1.1) where m is a natural number, t ( b), b < 1,, 6= 1 3 ::: m ; 1, D t d=dt, f L ; (( b) H) H, A and P are operators acting in some Hilbert space H, commuting with D t and possessing a complete system of eigenfunctions f' k g 1 that forms a Riesz basis in H. We are interested in the character of boundary conditions for t = b, which guarantee the existence and uniqueness of solution of (1.1) for every f H. These conditions depend on and on properties of the operators A and P. In the case when m =1,A is the operation of multiplication by a constant, P Dx is an operator on a closed interval and 1, the dependence on the sign of A of the character of the conditions with respect to t was rst observed by Keldis [6]. Later, the corresponding eect was studied by Visik [13]. For m this problem has been considered in [], [1], [11]. For the case >m the factors t ;1 and t ;m in (1.1) are essential, and instead of the usual L ( b) we consider the weighted space L ; ( b). Our approach is close to that of Dezin [] and is based on the case A and P are the operators of multiplication by numbers a and p. We describe the spectrum of one{dimensional operator and prove embedding theorems for weighted Sobolev spaces. The results of this paper have been obtained during my visit of the research group \Partielle Dierentialgleichungen und Komplexe Analysis" of the Institute of Mathematics at the University ofpotsdam. Yerevan State University, supported by DAAD 1

2 The one{dimensional case.1 The space W m For simplicity, in this section we assume the function u(t) to be real{valued. Let C _ m be the set of m times continuously dierentiable functions on [ b] satisfying the conditions u (k) (t) = u (k) (t) t= t=b = k = 1 ::: m ; 1: (.1) Denote by _ W m the completion of _C m in the norm generated by the inner product fu vg =(u (m) v (m) ) where ( ) stands for the inner product in L ( b). Moreover, let W m of _W m in the norm Dene the inner product in W m ju W m j = by fu vg =(t u (m) v (m) ): denote the completion t ju (m) j dt: (.) It is clear that the embedding W _ m W m is proper and that the space W _ m coincides with the class of (m ; ) times continuously dierentiable functions u(t) for which u (m;1) (t) is absolutely continuous and (.1) is fullled. It is easy to check that the norms of _W m and W m are equivalent on [ b], >. Hence it is enough to study the properties of functions from W m for t in a neighbourhood of. For the proofofthefollowing proposition we refer to [11]. Proposition.1 For every u W m where ju (k) (t)j C k t m;k;1; ju W m j (.3) 6= n +1 n = 1 ::: m ; 1 k = 1 ::: m ; 1: For = n +1, n = 1 ::: m ; 1 in (.3) the factor t m;k;1; t m;k;n; j ln tj, k = 1 ::: m ; 1. should be replaced by The inequality (.3) implies that for < 1 (weak degeneracy) the conditions (.1) are \retained", while for 1(strongdegeneracy) not all boundary conditions are \retained". For instance, for 1 < 3, u (m;1) (t)j t= can become innite, while for m ; 1 all u (k) (t)j t= may be innite for k = 1 ::: m; 1. Proposition. For every, 6= 1 3 ::: m ; 1, we have the embedding W m L ;m : (.4)

3 Proof. Since _ C m is dense in W m, it is enough to prove (.4) in the case where u _ C m. First, for m =1have This implies t ; u (t) dt = = 1 ; 1 ; 1 u (t) dt ;1 t =;1 u (t) t = u(t) dt Z 4 b t ( ; (u (t)) dt 1) Z 4 b t ; u (t) dt t ( ; (u (t)) dt: 1) Then, repeating this procedure m times we obtain t ; u (t) dt: t ;m u (t) dt This completes the proof of Proposition.. Z 4 m b ( ; 1) ( ; 3) ( ; (m ; 1)) Remark.3 The embedding (.4) breaks down for =1 3 ::: m ; 1. t (u (m) ) dt: (.5) We prove this assertion for = 1 and m =1. Consider the function u(t) =j ln tj, t ( a), a<min(1 b). In L ;1 ( a) its norm is nite for <;1=, while in W 1 1 ( a) it is nite for <1=. Therefore, for ;1= <1= theembedding (.4) breaks down. Remark.4 The embedding (.4) is not compact. For simplicityweverify this assertion for m = 1 and =. Consider the bounded in W 1 ( a) sequence of functions u n (t) = n ;1= t ;1= j ln tj ;1=;1=n, where a < min(1 b). It is easy to check that there is no subsequence of u n (t) convergent in the metric of L ( a) (see [3]). Remark.5 For, 6= 1 3 ::: m ; 1 in the space W m that is equivalent to (.). kuk = t = u (m) we can dene the norm dt (.6) The proof of equivalence of the norms (.) and (.6) follows the same lines as those of Proposition. and the inequality (.5) (see [4]). Therefore, we can write W m = t ;= _W m : 3

4 This means that u _ W m implies v = t ;= u W m. Observe that for =1 3 ::: m ; 1 the norms (.) and (.6) are not equivalent. Denote by W ~ m the completion of _W m by the norm (.6). The inequality (.3) implies the embedding With the help of Mellin transform (see [9]) Mu(z) = ~ W m W m : Z 1 t z;1 u(t) dt a norm equivalent to(.)can be introduced. Let v v 1 C 1 [ b], v 1 =1; v be cut{o functions with v (x) =(v (x) = 1) in some neighbourhood of x = b (x =). Remark.6 The norm given by the formula Z ; jjujj m = 1+jzj m j(mv u)(z)j jdzj ;1=+=;m + Z ;1=;m ; 1+jzj m j(mrv 1 u)(z)j jdzj where ; = fz C : Re z = g, Rv(x) =v(b ; x) and <1, is equivalent to (.). For the proof we referto[4]. Denote by L the weight space. L = fu(t) : ju L j = t ju(t)j dt < 1g Proposition.7 For every the embedding W m L (.7) is compact. Proof. First consider the case when 6= 1 3 ::: m ; 1. It follows from (.3) that ju(t)j C k t m;1; ju W m j : This implies ju L jc 1 ju W m j: (.8) 4

5 To prove that the embedding (.7) is compact we use (.3) for k = 1 and write ju(t + h) ; u(t) L j = = t ju(t + h) ; u(t)j dt t Z t+h t c 1 ju W m j u () d t = c ju W m j jhj c jhj ju W m j dt t Z t+h (m;3;)= d t m;3; dt dt where [t t + h]. Therefore, ju(t + h) ; u(t) L jc jhjju W m j: (.9) The result now follows from compactness criterion in L (see [], [1]). For =1 3 ::: m ; 1 the proof is similar. Djarov in [3] has proved, that the embedding W m L is compact for every >; and m =1. For us it is sucient the case =.. Self{adjoint Equation We consider the one{dimensional version of equation (1.1) for A = : Bu (;1) m D m t (t D m t u)+pt ;m u = f(t) (.1) where, 6= 1 3 ::: m ; 1, f(t) L ;,andp is a constant. Denition.8 A function u W m is called a generalized solution of equation (.1), if for every v W m fu vg + p(t ;m u v) =(f v): (.11) We set d(m ) =4 ;m ( ; 1) ( ; 3) ( ; (m ; 1)) : (.1) Theorem.9 Assume that p + d(m ) >, and 6= 1 3 ::: m ; 1. Then equation (.1) has a unique generalized solution for every f L ;. 5

6 Proof. Uniqueness of the generalized solution of (.1) follows from (.5) and (.11) with f =andv = u. To prove the existence we consider the functional l f (v) (f v), f L ; over the space W m. Using (.8) we write jl f (v)j = = Z b f(t) v(t) dt t ;= f(t) t = v(t) dt t ; jf(t)j dt c jf L ; j jv W m j : t jv(t)j dt Therefore, l f (v) is a linear bounded functional over the space W m. The result now follows from Riesz's lemma on the representation of such functionals (see []). Theorem.9 is proved. Remark.1 Observe that the generalized solution u(t) of equation (.1) belongs to W m ( b; ) for every >. Hence in each interval ( b;) the generalized solution u(t) coincides with the usual solution of (.1). An element f L ; can be represented in the form It is clear that f 1 L and f(t) =t f 1 (t) (.13) jf L ; j = jf 1 L j: Now, using Denition.8 we dene an operator B : W m L! L. Denition.11 We say that a function u(t) W m belongs to the domain D(B ) of an operator B if (.11) is fullled for some f L ;. In this case we will write B u = f 1,where the function f 1 is specied by (.13). It follows from Denition.11 that the operator B acts by the formula B u t ; f(;1) m D m t (t D m t u)+pt ;m ug and for every u D(B ), v W m and f L ; (B u v) =(f v) where ( ) stands for the inner product in L. Theorem.1 Under the assumptions of Theorem.1 the operator B is positive and self{ adjoint in L. Moreover, B ;1 : L! L is a compact operator. 6

7 Proof. The symmetry and positivity of B is an immediate consequence of Denition.11. The self{adjointness of the symmetric operator B follows from the fact that, according to Theorem.9, for every f L ; the equation (1.1) is solvable, that is, for every f 1 L of the form (.13) the equation B u = f 1 is solvable. Using (.5), (.11) with v = u and the embedding L ;m L we can write Therefore, (d(m ) +p) c ju L j (d(m )+p) ju L ;m j ju W m j + p ju L ;m j jf L ; jju L j = jf 1 L jju L j: jb ;1 f 1 L jc 1 jf 1 L j: This implies that B ;1 is bounded. To prove thatb ;1 is compact it remains to observe that, according to Proposition., the embedding D(B ) W m L is compact. Theorem.1 is proved. Applying standard properties of the spectra of self{adjoint compact operators we get the following corollary (see [5]). Corollary.13 The operator B has a pure point spectrum, and the system of corresponding eigenfunctions is dense in L. Observe that if is an eigenvalue of B, and u(t) is the corresponding eigenfunction, then according to Denition.11, (;1) m D m t (t D m t u)+pt ;m u = t u: Remark.14 Note that, if p =, = m and f L ( b), then the spectrum of the operator B is pure continuous and coincides with the ray [d(m m) +1) (when p 6=, then we can take ; p instead of the spectral parameter ). For the proof we refer to [11] and note that it is in fact a consequence of the embedding (.4). Now we consider equation (.1), as above, with p = and f L m; Qu (;1) m D m t (t D m t u) =f(t) f L m; : (.14) We can dene (as for equation (.1)) generalized solutions for equation (.14) and prove that for every f L m; a generalized solution exists and is unique. Let f = t ;m f 1. It is clear that f 1 L ;m. Then we can dene an operator Q : L ;m! L ;m as in Denition.11. Theorem.15 The operator Q has a pure continuous spectrum which coincides with the ray [d(m ) +1). Theorem.15 can be proved similarly to [11] with the help of embedding (.4). 7

8 .3 Non{selfadjoint Equation Now we consider the one{dimensional version of equation (1.1) Su (;1) m D m t (t D m t u)+ad m;1 t (t ;1 D m t u)+pt ;m u = f(t) (.15) where, 6= 1 3 ::: m ; 1, f(t) L ;, a and p are constant. Denition.16 A function u W m every v W m is called generalized solution of equation (.15), if for fu vg + a(;1) m;1 (t ;1 D m t u D m;1 t u)+p(t ;m u v) =(f v): (.16) Theorem.17 Let the following condition be fullled a( ; 1)(;1) m > = d(m )+a=( ; 1)(;1) m d(m ; 1 ; ) + p> (.17) where d(m ) is dened in (.1). Then equation (.15) has a unique generalized solution for every f(t) L ;. Proof. Uniqueness. For the proof of uniqueness in equality (.16) we set f = and u = v. Let >1 (in the case <1 the proof is similar and we use that [see Proposition.1]). Then integrating by parts we get t ;1 u (m) u (m;1) = t ;1 u (m) (t)u (m;1) (t) dt = ; 1 t ;1 u (m;1) (t) It follows from (.3) that From this inequality and (.5) we get ; ; 1 t ;1; u (m;1) (t) j t= = j t= t ; u (m;1) (t) dt: t ;1; u (m;1) (t j t= is nite. Using (.5) we can write t u ; (m;1) (t) dt d(m ; 1 ; ) t ;m (u(t)) dt: =fu ug + a(;1) m;1 (t ;1 u (m) u (m;1) )+p(t ;m u u) + 1 a(;1)m t ;m (u(t)) dt t ;1 u (m;1) (t) j t= : 8

9 Now uniqueness of the generalized solution immediately follows from the condition (.17). Existence. First note that the functional l f (v) (f v) can be represented in the form (f v) = fu vg, where u W m (see the proof of Theorem.9). The last two terms in the left-hand side of equality (.16) also can be regarded as a continuous linear functional relative to u and represented in the form fu Kvg, wherekv W m. Indeed, using inequality (.5)wecan write ja(;1) m;1 (t ;1 Dt m u Dt m;1 u) + p(t ;m u v)j ja(t = Dt m u t =;1 D m;1 u)j + jp(t =;m u t =;m v)j Now from (.16) we obtain c 1 ju W m j t t ; v (m;1) (t) + c ju L ;m jjv L ;m j dt 1= c 1 j ; 1j ju W m jjv W m j + c 3 ju W m jjv W m j = cju W m jjv W m j: fu (I + K)vg = fu vg (.18) for every v W m. Note that the image of the operator I + K is dense in W m there exists a u W m such that fu (I + K)vg =. Indeed, if for every v W m,wegetu =, since we have already proved uniqueness of the generalized solution for equation (.15). Let <d(m ). Then we can write Finally, weget fu (I + K)ug fu ug +[(1; )d(m ) + a=( ; 1)(;1) m d(m ; 1 ; ) + p] = fu ug +( ; d(m )) fu ug: From (.19) it follows that (I +K) ;1 is dened on W m t ;m u (t) dt t ;m u (t) dt fu (I + K)ugfu ug: (.19) I + K and (I + K ) ;1 =((I + K) ;1 ) : Then from (.18) we obtain u =(I + K ) ;1 u : and is bounded therefore, there exist Theorem.17 is proved. Let f = t f 1. As in the self-adjoint case we can dene an operator S, according to Denition.11. 9

10 Denition.18 We say that u(t) D(S) if (.16) is fullled for some f L ;, and then we will write Su = f 1. Proposition.19 Under the assumptions of Theorem.17 the operator S ;1 : L! D(S) L is compact. Proof. For the proof we rst note that Indeed, setting v = u in (.16) we obtain ju L j(d(m )) ;1 jf 1 L j: (.) d(m )ju L j d(m )jl ;m j j(f u)j jf L ; jju L j = jf 1 L jju L j: Now to complete the proof of Proposition.19 it is enough to apply the compactness of the embedding (.7). Proposition.19 is proved. For the case a( ; 1)(;1) m < we consider the operator Tv (;1) m D m t (t D m t v)+ad m t (t;1 D m;1 t v)+pt ;m v = g(t): (.1) Denition. We say that v is a generalized solution of (.1), if the following equality holds (Su v) =(u g) (.) for every u D(S): Let g = t g 1. Denition. of a generalized solutions denes an operator T : L! L (see Denition.18). We can express formula (.) in the form (Su v) = (u g). Since D(S) =D(S) is dense in L,we obtain that in L : T = S Theorem.1 Under the assumptions of Theorem.17, for every g L ; a generalized solution of equation (.1) exists and is unique. Moreover, T ;1 : L! L is compact. Proof. Solvability of the equation Su = f 1 for any right-hand side implies uniqueness of the solution of (.1), while existence of the bounded operator S ;1 (Proposition.19) implies solvability of(.1)for any g L ; (see, for example, [1]). Because of (S ) ;1 =(S ;1 ), compactness of the operator S ;1 implies compactness of the operator T ;1. Theorem.1 is proved. 1

11 Remark. For > 1, and for every generalized solution v of the equation (.1), we have t ;1 (v (m;1) (t)) j t= =: (.3) In fact, replacing g by Tv in equality (.), integrating by parts the second term and using equality (.16) we obtain (.3). Note that for equation (.15) the left-hand side of (.3) for a(;1) m > is only bounded. 3 Operator Equation In this section we consider the operator equation (1.1): L u (;1) m D m t (t D m t u)+ad m;1 t (t ;1 D m t u)+t ;m Pu= f(t) f H: Recall that if a system f' k g 1 P is a Riesz basis in H, every element x Hcan be uniquely 1 represented in the form x = x k' k, and the inequality c 1 X holds, where jj jj stands for the norm in H. jx k j jjxjj c 1 1 X jx k j (3.1) By assumption, the operators A and P appearing in (1.1) have a complete common system of eigenfunctions f' k g 1 that forms Riesz basis in H. So we have u(t) = 1X u k (t) ' k f(t) = 1X f k (t) ' k A' k = a k ' k P' k = p k ' k k N: (3.) Hence, the operator equation (1.1) can be decomposed into an innite chain of ordinary dierential equations L k u k (;1) m D m t (t D m t u k )+a k D m;1 t (t ;1 D m t u k )+t ;m p k u k = f k (t) k N: (3.3) The condition f H implies that f k L ; for k N. For the one{dimensional equation (3.3) we can dene the generalized solutions u k (t) k N (compare with the Denitions.8,.16 and.). Denition 3.1 A function u W m (( b) H) L (( b) H) admitting the representation u(t) = 1X u k (t)' k where u k (t) are the generalized solutions of the one{dimensional equation (3.3) is called a generalized solution of the operator equation (1.1). 11

12 The following theorem is a consequense of the general results of Dezin [1]. Theorem 3. The operator equation (1.1) is uniquely solvable if and only if the one{dimensional equation (3.3) is uniquely solvable and the inequality ju k L jcjf k L ; j = cjg k L j (3.4) is fullled, uniformly with respect to k N, where f k = t g k. Theorems.9,.17 and.1 shows us that a sucient condition for relations (3.4) are either for all k N, with a k =,or p k + d(m ) >"> k N (3.5) k = d(m )+a k =( ; 1)(;1) m d(m ; 1 ; ) + p k >"> (3.6) for all k N, such thata k 6=. Therefore, we can state the following result. Theorem 3.3 Let (3.5) and (3.6) be fullled and let 6= 1 3 ::: m ; 1. Then the operator equation (1.1) has a unique generalized solution for every f H. Proof. Observe that if u is generalized solution of (1.1), then according to (3.1) and (3.4) we have ju L (( b) H)j = t ku(t)k dt X 1 c 1 t ju k (t)j dt c 1 X jf k L ; j cjf Hj : (3.7) Similarly to (.13) we set f = t g. It is clear that g L (( b) H) and jf Hj = jg L (( b) H)j. Inequality (3.7) can be written in the form ju L (( b) H)j c jg L (( b) H)j: (3.8) Analogously to the one{dimensional case the generalized solution of the operator equation (1.1) generates the operator :W m (( b) H) L (( b) H)! L (( b) H): Inequality (3.8) implies that ;1 bounded operator. Hence, we have (), where () is the resolvent set of the operator. : L (( b) H)! W m (( b) H) L (( b) H) is a 1

13 References [1] A. A. Dezin: General questions of the theory of boundary value problems, Nauka, Moscow, 198 [in Russian] Partial Dierential Equations, Springer Verlag, Berlin, [] A. A. Dezin: Degenerate operator equations, Mat. Sb., 115(157), 3(7). (1981), 33{336 [in Russian] English transl. in Math. USSR Sbornik, 43, 3.(198),87{98. [3] P. A. Djarov: Compactness of embeddings in some spaces with power weight, Izvestya VUZ-ov, Matematika, vol. 8. (1988), 8{85. [4] R. Duduchava, D. Elliott and W. L. Wendland: The spline collocation method for Mellin convolution equations, Universitat Stutgart, Sonderforschungsbereich 44, Mehrfeldprobleme in der Kontinuumsmechanik, Bericht 96/4. [5] N. Dunford and J. T. Schwartz: Linear operators, vol. I, Interscience, [6] M. V. Keldis: On certain cases of degenaration of equations of elliptic type on the boundary of a domain, Dokl. Akad. Nauk SSSR 77. (1951), 181{183 [7] L. D. Kudryavtzev: On a variational method of determination of generalized solutions of dierential equations in the functional spaces with power weight, Dierential'nye Uravneniya, 19, 1. (1983), 173{174. [8] L. D. Kudryavtzev: On equivalent norms in the weight spaces, Trudy Mat. Inst. AN SSSR, vol. 17. (1984), 161{19. [9] B.-W. Schulze: Pseudo{dierential operators on manifolds with singularities. North{ Holand, Amsterdam, [1] L. Tepoyan: Degenerate fourth{order dierential{operator equations [in Russian], Differential'nye Urawneniya, 3, 8. (1987), 1366{1376 English transl. in Amer. Math. Society, 3, 8.(1988), 93{939. [11] L. P. Tepoyan: On a dierential{operator equation of higher order [in Russian], Proc. of a Conference, Yerevan, (1993), 1{17. [1] L. P. Tepoyan: On a degenerate dierential{operator equation of higher order, Izvestya Natsionalnoi Akademii Nauk Armenii, Matematika, 34, 5. (1999), 48{56. [13] M. I. Visik: Boundary{value problems for elliptic equations degenerate on the boundary of a region, Mat. Sb., 35(77). (1954), 513{568 English transl. in Amer. Math. Soc. Transl. 35,. (1964). 13

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