mur s absorbing boundary condition

... problems, conditions for positiveness of Green s functions, and solutions with various BCs, for example, NBCs The structure of the paper is as follows In Section 2, we review the properties of functional ... determinants and linear functionals We construct a special basis of the solutions in Section and introduce some functions that are independent of this basis The expression of the solution to the second-order ... discrete initial conditions We apply these results to BVPs with NBCs: first, we construct the Green s function for classical BCs, then we can construct Green s function for a problem with NBCs...

Ngày tải lên: 21/06/2014, 11:20

... uλ s ds λ ≤ −bt 2.18 G t, s h s ds λN Letting t gives uλ ≤ −b sh s ds λN 2.19 Since uλ t is positive, we have −b λN sh s ds ≥ 2.20 or, equivalently, λ≥ b N sh s ds 2.21 Hence, the theorem is ... b/M sh s ds, then Tλ K ⊂ K In fact, if z ∈ K, then it is easy to show that Tλ z 0 and Tλ z t ≤ for t ∈ 0, Also if λ > b/M Tλ z −b sh s ds, we have λ sh s f z s ds ≥ −b 2.13 λM sh s ds > 0 Thus ... holds then 2.7 has no positive solution for all λ < b/N and h t are defined as in 2.8 1 sh s ds, where b Proof Suppose that uλ is a positive solution of 2.7 Thus, uλ t −bt T λ uλ t G t, s h s...

Ngày tải lên: 21/06/2014, 16:20

... g ij (25) (26) where γ > is the constant with dimension of the squared mass, α = is the dimensionless constant which specifies the rigidity of the string world sheet is the Laplace-Beltrami operator ... terms Acknowledgements The author would like to thank Profs B.M.Barbashov,V.V.Nesterenko for useful discussions This work was partly supported by Vietnam National Research Programme in National Science ... of Science, Mathematics - Physics 24 (2008) 111-118 118 Conclusion The equations of motion, of the boundary conditions and of the energy - momentum for the classical rigid string are reconserved...

Ngày tải lên: 05/03/2014, 14:20

... Paris: Association Technique de l'Industrie Papetiere (ATIP) 1999, S 153-162 BIENER R., MEIERS R., KUSSI J. -S and H LEIMKÜHLER: Systematische Reduktion des Wasserverbrauchs und des Abwasseranfalls, ... particles and inorganic particles (fillers, lime particles, etc.) These particles are usually referred to as (total) suspended solids (TSS or SS) This is the first type of treatment to be applied ... KNELISSEN: Biologische Kreislaufwasserbehandlung in einer Papierfabrik mit geschlossenem Wasserkreislauf Das Papier 51 (1997) Nr 6A, V153 – V159 DIETZ W and J KAPPEN: Using process simulation as...

Ngày tải lên: 01/04/2014, 00:20

... (P ) is coercive And in Section 4, we consider the case where the energy functional possesses the Mountain Pass geometry Preliminaries In order to discuss problem (P ), we need some theories on ... and s large enough, E(sw) ≤ d1 s p+ −d3 s µ −d4 s µ , thus by (H3 ), We conclude that E(sw) → −∞ as s → +∞ So by the Mountain Pass lemma this theorem is proved By the symmetric Mountain Pass lemma, ... → R is continuous Notice that the function a satisfies (a1 ) may be singular at t = And f , g satisfying (f1 ) f : Ω × R → R satisfies the Caratheodory condition and there exist two constants C1...

Ngày tải lên: 20/06/2014, 21:20

... [?] Putting these expressions into (19), we have 2π γ1 p1 s s γ2 s s sin sin − cos − sp1 sin cos p2 δ1 2p1 2p2 δ2 2p2 2p1 2p2 s 2 p2 s s − cos sin +O , δ2 2p1 2p2 s 0= and using g1δ2p1 = ... − s 1 ⎜ p1 ⎜− sin s − p2 δ1 ⎝ s 2p1 sp1 q(τ ) sin × sin ⎛ + γ2 ⎜ ⎜− cos s − δ2 ⎝ 2p1 p2 × cos ⎛ s − 2p2 p2 ⎡ q(τ ) cos π q(τ ) cos π /2 π γ2 p2 ⎢ ⎢− cos s − δ2 s ⎣ 2p1 p2 × sin s − 2p2 sp2 ... are situated close to entire numbers We shall show that, for large n, only one root (20) lies near to each p1 +p function φ (s) = sin s 2p1 p2 + O(1) φ (s) = sin s p1 +p2 + s p1 +p2 cos s p1...

Ngày tải lên: 20/06/2014, 22:20

... − s) α−1 f (s, u (s) , (φu) (s) , (ψu) (s) ) ds (α) t2 − (t2 − s) α−1 f (s, u (s) , (φu) (s) , (ψu) (s) ) ds (α) η α−1 +ν1 t1 − ⎛ ≤ M⎝ α−1 t2 (α) (η − s) 2α s f (s, u (s) , (φu) (s) , (ψu) (s) )ds ... − s) 2α s |f (s, x (s) , (φx) (s) , (ψx) (s) ) − f (s, 0, 0, 0) (2α − 1) α−1 + |f (s, 0, 0, 0) | ds ⎛ t (t − s) α−1 ≤ sup ⎝ (L1 (s) |x (s) | + L2 (s) | (φx) (s) | + L3 (s) | (ψx) (s) | + M) ds (α) ... ≤ sup ⎝ t∈[0,T] ⎞ (η − s) 2α−2 (L1 (s) |x (s) | + L2 (s) | (φx) (s) | + L3 (s) | (ψx) (s) | + M) ds⎠ (2α − 1) (t − s) α−1 (L1 (s) |x (s) | + γ0 L2 (s) |x (s) | (α) + δ0 L3 (s) |x (s) | + M) ds...

Ngày tải lên: 20/06/2014, 22:20

... solution u Î H We complete the proof ■ Abbreviations (FESDBC): fourth-order elliptic system with Dirichlet boundary condition Acknowledgements This work(Tacksun Jung) was supported by Basic Science ... Hilbert space, which is topological direct sum of the subspaces X1 and X2 Let F Î C1(X, R) Moreover assume: (a) dim X1 0, R >0 and e Î X1, e ≠ such that r < R and sup F < S ... sup F R (e,X2 ) Bρ (X1 ) Let H+ be the subspace of H spanned by the eigenfunctions corresponding to the eigenvalues lk(lk - c) >0 and H- the subspace of H spanned by the eigenfunctions corresponding...

Ngày tải lên: 20/06/2014, 22:20

... Cite this article as: Tahamtani and Peyravi: Spatial estimates for a class of hyperbolic equations with nonlinear dissipative boundary conditions Boundary Value Problems 2011 2011:19 Submit your ... Knowles [1] The interested reader is referred to the papers [2-9] and the reviews by Horgan and Knowles [1,10,11] The energy method is widely used to study such results Spatial growth or decay estimates ... and Schaefer [14] proved such results for some classes of heat conduction problems In [15], Quintanilla investigate the spatial behavior of several nonlinear parabolic equations with nonlinear boundary...

Ngày tải lên: 21/06/2014, 00:20

... both studies, they obtained the necessary and sufficient conditions to the global existence of solutions The fast-slow diffusion equations (there exists i(i = 1, , n) such that ki > 1) is studied ... some sufficient conditions to the global existence and the blow-up of solutions The aim of this article is to study the long-time behavior of solutions to systems (1.1) and provide a simple criterion ... Moreover, this article provides the necessary and sufficient conditions to the global existence and the blow-up of solutions in the general domain Ω Therefore, this article improves the results of [6]...

Ngày tải lên: 21/06/2014, 02:20

... authors express their sincere thanks to the anonymous reviewers for their valuable comments and suggestions for the improvement of the article This study was supported by research grants from ... Ag-water It was found that the Abbreviations List of symbols: c: Constant; Cf: Skin friction coefficient; Cp: Specific heat at constant pressure; f: Dimensionless stream function; hf: Heat transfer coefficient; ... Dynamic viscosity; : Kinematic viscosity; : Fluid density; : Nanoparticle volume fraction; : Stream function; w: Wall shear stress; Subscripts; f: Fluid; nf: Nanofluid; s: Solid Acknowledgements The...

Ngày tải lên: 21/06/2014, 04:20

... paper is as follows We will introduce some lemmas and notations in the rest of this section In Section 2, we present the expression and properties of Green s function associated with boundary ... s y s ds dt 2.12 Substituting 2.12 into 2.10 , we obtain G1 t, s y s ds x t G1 t, s y s ds tα−1 α−1 h t x t dt tα−1 α−11− 1 h t tα−1 dt/ α − 1 ht G1 t, s y s ds dt 2.13 G1 t, s y s ds G2 t, s ... there exists τ > such that G t, s ≥ τ, ∀t, s ∈ θ, − θ , 2.24 where θ ∈ 0, 1/2 Preliminaries In this section, we give some preliminaries for discussing the existence of positive solutions of boundary...

Ngày tải lên: 21/06/2014, 05:20

... subsequence, one says that J satisfies the P S c condition If J satisfies P S c condition for all c ∈ R, one says that J satisfies the P S condition Lemma 2.2 see Assume that u and u are, respectively, ... Mountain pass theorem in half-order intervals, sup-solutions case see v0 < v1 is Suppose that J satisfies (J1 )–(J3 ) v1 < v2 is a pair of strict supersolution of ∇J a subsolution of ∇J Suppose that ... satisfies deformation property when J satisfies P S condition By a similar discussion as in Theorem 1.1, we only need to discuss the critical points of J Now, we construct the sub-sup solutions...

Ngày tải lên: 21/06/2014, 07:20

... These conditions are only necessary for an extremum The question of sufficient conditions for the existence of an extremum is considered in the next section Advances in Difference Equations Special ... definitions used in this paper In Section 3, the necessary optimality conditions are proved for problems 1.1 and 1.2 by giving some special cases which prove the generalization of our problems Sufficient ... b 3.18 Sufficient Conditions In this section, we prove the sufficient conditions that ensure the existence of a minimum maximum Some conditions of convexity concavity are in order Advances in Difference...

Ngày tải lên: 21/06/2014, 11:20

... 2.2 Suppose that u x, t is a weak solution of problem 1.1 – 1.4 and g s satisfies: s sg s ≥ KG s , where K > 2, G s S1 1−p /4 4/ p−1 E0 e where C1 u0 mS1 S1 u1 S2 −2 S1 b/a u0 g ρ dρ, G s ≥ β |s| p ... S can be found in 32 and the norm of L2 S is denoted by • S Boundary Value Problems Blowup of the Solutions In this paper, we always assume that the initial data u0 ∈ H s 1/2 S1 , u1 ∈ H s S1 ... S1 , s > 1, and g ∈ C and that the problem 1.1 – 1.4 possesses a unique local weak solution 2, 3, that is, u is in the class u ∈ L∞ 0, T ; H s ut ∈ L∞ 0, T, H s S1 , Ω , utt ∈ L∞ 0, T ; L2 S1 ,...

Ngày tải lên: 21/06/2014, 11:20

... uk → as k → ∞ possesses a convergent subsequence in E, we say that ϕ satisfies the Palais-Smale condition Let E be a real Banach space Define the set Σ {A | A ⊂ E \ {θ} as symmetric closed set} ... possesses at least n distinct pairs of nontrivial critical points Then, problem 3.19 has at least n distinct pairs of nontrivial classical solutions, that is, problem 1.1 has at least n distinct ... possesses at least n distinct pairs of nontrivial critical points Then, problem 3.35 has at least n distinct pairs of nontrivial classical solutions, that is, problem 1.1 has at least n distinct...

Ngày tải lên: 21/06/2014, 11:20

... bk − ak h s ds h s ds · bk ak bk ak bk ak Gk s, τ σ τ dτ s − ak h s ds bk ak h s bk ak bk ak h s z s ds ds Gk s, τ σ τ dτ ds h s z s ds, 2.19 therefore, bk ak h s z s ds 1− bk ak h s ds r3 bk ak ... ak h s ds h s ds 1 − σ2k ak h s bk r4 − r3 bk − ak bk ak bk ak Gk s, τ σ τ dτ ds bk − s h s dsr3 bk − ak ak bk s − ak h s ds bk ak ak Gk t, s σ s ds s − ak h s dsr4 bk − ak Kk t, s σ s ds bk ... as follows: B1 x t G t, s x s ds where G t, s is given by 2.27 1−t 1− 1 − s h s ds G s, τ h τ x s dτ ds, 4.3 20 Boundary Value Problems By analogous methods, we have the following results Assume...

Ngày tải lên: 21/06/2014, 11:20

... s s, G1 t, s G2 t, s ≤ s ≤ t ≤ 1, ⎩ − s t2 , ≤ t ≤ s ≤ 1, ⎧ ⎨ − t s, ≤ s ≤ t ≤ 1, 2.3 ⎩ − s t, ≤ t ≤ s ≤ Proof Let u be a solution of the BVP 2.1 Then, we may suppose that At2 G1 t, s h s ds ... any t, s ∈ 0, × 0, , t2 1 − s s ≤ G1 t, s ≤ − s s 2 2.7 Boundary Value Problems Lemma 2.3 see 26 For any t, s ∈ 0, × 0, , ≤ G2 t, s ≤ − s s 2.8 In the remainder of this paper, we always assume ... G2 t, s ≤ M2 1 s s M2 t 1−μ G2 τ, s g τ dτ f s, xk s , xk s ds 2.18 1 1−μ G2 τ, s g τ dτ ds M3 , t ∈ 0, , which shows that {yk }∞ is also uniformly bounded This indicates that {yk }∞ is k k equicontinuous...

Ngày tải lên: 21/06/2014, 11:20

... , to get 1 g s s g s y s ds g r y r dr G s, r σ r dr ds sg s 0 g s y s ds g s 3.8 G s, r σ r dr ds Thus, g s y s ds g s 1− G s, r σ r dr ds sg s ds 3.9 Boundary Value Problems Substituting in ... t − s σ s ds, y − y 1 3.4 − s σ s ds Hence 1 y t tg s y s ds t − s σ s ds − t − s σ s ds, 3.5 1 tg s y s ds y t t G t, s σ s ds, 3.6 where G t, s ⎧ s − t if ≤ s ≤ t ≤ 1, ⎩t − s if ≤ t ≤ s ≤ ... G t, s σ s ds 1 g s 1− G s, r σ r dr ds sg s ds 3.10 Therefore y t H t, s σ s ds 3.11 Set g ∗ |1 − g∗ | Note that |G t, s | ≤ for t, s ∈ 0, × 0, 3.12 Our first result reads Theorem 3.3 Assume...

Ngày tải lên: 21/06/2014, 11:20

... therein The study of positive solutions of BVPs that involve Stieltjes integrals has been done, in the case of positive measures, in 21–24 Signed measures were used in 12, 25 ; here, as in 21, ... u cΦ Φ s g s φr s ≥ c T u 2.29 Hence we have T u ∈ K Moreover, the map T is compact since it is a sum of two compact maps: the compactness of k t, s g s f s, u s ds is well known, and since γ1 ... 1, s s s2 : s 2.11 Boundary Value Problems Now, one can see that the derivative of the function k t, s /Φ s with respect to s is non-positive for all s ∈ 0, , that is, the function k t, s /Φ...

Ngày tải lên: 21/06/2014, 11:20