research article on symplectic analysis for the plane elasticity...
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Research ArticleOn Symplectic Analysis for the Plane Elasticity Problem ofQuasicrystals with Point Group 12 mm
Hua Wang, Jianrui Chen, and Xiaoyu Zhang
College of Sciences, Inner Mongolia University of Technology, Hohhot 010051, China
Correspondence should be addressed to Hua Wang; [email protected]
Received 13 March 2014; Accepted 29 May 2014; Published 23 June 2014
Academic Editor: Patricia J. Y. Wong
Copyright © 2014 Hua Wang et al. This is an open access article distributed under the Creative Commons Attribution License,which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
The symplectic approach, the separation of variables based onHamiltonian systems, for the plane elasticity problem of quasicrystalswith point group 12mm is developed. By introducing appropriate transformations, the basic equations of the problem are convertedto two independent Hamiltonian dual equations, and the associated Hamiltonian operator matrices are obtained. The study of theoperator matrices shows the feasibility of the method. Without any assumptions, the general solution is presented for the problemwith mixed boundary conditions.
1. Introduction
Quasicrystals (QCs), a new material and structure, werefirst discovered by the authors in [1] in 1984. QCs thatexhibit excellent physical and mechanical properties, such aslow friction, high hardness, and high wear resistance, havepromising potential applications (cf. [2]). It is well knownthat the general solution of quasicrystal elasticity is veryimportant, but it is difficult to be obtained because of thecomplexity of the basic governing equations. So far, manymethods and techniques have been developed to seek for thegeneral solution (see, e.g., [3–8]). However, some problemsof quasicrystal elasticity have not been solved well due to thecomplicated assumptions of the solution, and the symplecticapproach, developed by Zhong [9], may be helpful in thoseproblems.
The symplectic approach has advantages of avoiding thedifficulty of solving high order differential equations andhaving no any further assumptions and has been applied intovarious research fields such as elasticity [10–12], viscoelastic-ity [13], fluid mechanics [14], piezoelectric material [15], andfunctionally graded effects [16]. In this method, one needsto transform the considered problem into Hamiltonian dualequations and then obtains the desired Hamiltonian operatormatrix. Based on the eigenvalue analysis and eigenfunctionexpansion, the analytical solution of the problem can be
explicitly presented. It should be noted that the feasibility ofthis method depends on the completeness of eigenfunctionsystems of the correspondingHamiltonian operatormatrices.
To the best of the author’s knowledge, there are no reportsof the method on the analysis of QCs. The objective ofthis paper is to propose the symplectic approach for theplane elasticity problem of quasicrystals with point group12mm. After derivation of two independent Hamiltoniandual equations of the problem, we prove the completenessof eigenfunction systems for the corresponding Hamiltonianoperator matrices. Finally, we obtain the analytical solutionwith the use of the eigenfunction expansion.
2. Basic Equations and Their HamiltonianDual Equations
According to the quasicrystal elasticity theory, we havethe deformation geometry equations of the plane elasticityproblem of quasicrystals with point group 12mm
𝜀𝑖𝑗=
1
2
(
𝜕𝑢𝑖
𝜕𝑥𝑗
+
𝜕𝑢𝑗
𝜕𝑥𝑖
) , 𝑤𝑖𝑗=
𝜕𝑤𝑖
𝜕𝑥𝑗
, 𝑖, 𝑗 = 1, 2, (1)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014, Article ID 367018, 7 pageshttp://dx.doi.org/10.1155/2014/367018
2 Abstract and Applied Analysis
the equilibrium equations𝜕𝜎𝑖𝑗
𝜕𝑥𝑗
+ 𝑓𝑖= 0,
𝜕𝐻𝑖𝑗
𝜕𝑥𝑗
+ 𝑔𝑖= 0, (2)
and the generalized Hooke’s law
𝜎𝑥𝑥
= 𝐶12(𝜀𝑥𝑥
+ 𝜀𝑦𝑦) + 2𝐶
66𝜀𝑥𝑥,
𝜎𝑦𝑦
= 𝐶12(𝜀𝑥𝑥
+ 𝜀𝑦𝑦) + 2𝐶
66𝜀𝑦𝑦,
𝜎𝑥𝑦
= 𝜎𝑦𝑥
= 2𝐶66𝜀𝑥𝑦,
𝐻𝑥𝑥
= 𝐾1𝑤𝑥𝑥
+ 𝐾2𝑤𝑦𝑦,
𝐻𝑦𝑦
= 𝐾1𝑤𝑦𝑦
+ 𝐾2𝑤𝑥𝑥,
𝐻𝑥𝑦
= (𝐾1+ 𝐾2+ 𝐾3) 𝑤𝑥𝑦
+ 𝐾3𝑤𝑦𝑥,
𝐻𝑦𝑥
= (𝐾1+ 𝐾2+ 𝐾3) 𝑤𝑦𝑥
+ 𝐾3𝑤𝑥𝑦.
(3)
Here 𝑢𝑖and𝑤
𝑖are the phonon and phason displacements, 𝜎
𝑖𝑗
and 𝜀𝑖𝑗are the phonon stresses and strains,𝐻
𝑖𝑗and𝑤
𝑖𝑗are the
phason stresses and strains, 𝐶12, 𝐶66, 𝐾1, 𝐾2, and 𝐾
3are the
elastic constants, and 𝑓𝑖and 𝑔
𝑖are the body and generalized
body forces, respectively.Substituting (1) and (3) into (2), we get the displacement
equilibrium equations
𝐶66∇2𝑢𝑥+ (𝐶12
+ 𝐶66)
𝜕
𝜕𝑥
(
𝜕𝑢𝑥
𝜕𝑥
+
𝜕𝑢𝑦
𝜕𝑦
) + 𝑓1= 0,
𝐶66∇2𝑢𝑦+ (𝐶12
+ 𝐶66)
𝜕
𝜕𝑦
(
𝜕𝑢𝑥
𝜕𝑥
+
𝜕𝑢𝑦
𝜕𝑦
) + 𝑓2= 0,
𝐾1∇2𝑤𝑥+ (𝐾2+ 𝐾3)
𝜕
𝜕𝑦
(
𝜕𝑤𝑥
𝜕𝑦
+
𝜕𝑤𝑦
𝜕𝑥
) + 𝑔1= 0,
𝐾1∇2𝑤𝑦+ (𝐾2+ 𝐾3)
𝜕
𝜕𝑥
(
𝜕𝑤𝑥
𝜕𝑦
+
𝜕𝑤𝑦
𝜕𝑥
) + 𝑔2= 0,
(4)
where ∇2 = (𝜕2/𝜕𝑥2) + (𝜕2/𝜕𝑦2). Let
𝑞1=
𝐶66
𝐶12
+ 2𝐶66
(
𝜕𝑢𝑥
𝜕𝑦
−
𝜕𝑢𝑦
𝜕𝑥
) ,
𝑞2=
𝜕𝑢𝑥
𝜕𝑥
+
𝜕𝑢𝑦
𝜕𝑦
,
𝑞3=
𝐾1+ 𝐾2+ 𝐾3
𝐾1
(
𝜕𝑤𝑥
𝜕𝑦
+
𝜕𝑤𝑦
𝜕𝑥
) ,
𝑞4= −
𝜕𝑤𝑥
𝜕𝑥
+
𝜕𝑤𝑦
𝜕𝑦
.
(5)
Then (4) can be expressed in the following matrix forms:
𝜕
𝜕𝑦
𝑍1= 𝐻1𝑍1+ 𝐹1, (6)
𝜕
𝜕𝑦
𝑍2= 𝐻2𝑍2+ 𝐹2, (7)
where 𝑍1
= (𝑢𝑥, 𝑢𝑦, 𝑞1, 𝑞2)𝑇, 𝑍2
= (𝑤𝑥, 𝑤𝑦, 𝑞3, 𝑞4)𝑇, 𝐹1
=
−(1/(𝐶12
+ 2𝐶66))(0, 0, 𝑓
1, 𝑓2)𝑇, 𝐹2= (−1/𝐾
1)(0, 0, 𝑔
1, 𝑔2)𝑇,
𝐻1=
(
(
(
(
0
𝜕
𝜕𝑥
𝐶12
+ 2𝐶66
𝐶66
0
−
𝜕
𝜕𝑥
0 0 1
0 0 0 −
𝜕
𝜕𝑥
0 0
𝜕
𝜕𝑥
0
)
)
)
)
,
𝐻2=
(
(
(
(
0 −
𝜕
𝜕𝑥
𝐾1
𝐾1+ 𝐾2+ 𝐾3
0
𝜕
𝜕𝑥
0 0 1
0 0 0
𝜕
𝜕𝑥
0 0 −
𝜕
𝜕𝑥
0
)
)
)
)
.
(8)
Obviously,𝐻𝑇1= 𝐽𝐻1𝐽 and𝐻
𝑇
2= 𝐽𝐻2𝐽, in which 𝐽 = (
0 𝐼2
−𝐼20)
is the symplecticmatrix with 𝐼2being the 2×2 identitymatrix.
Thus,𝐻1and𝐻
2are bothHamiltonian operatormatrices and
(6) and (7) are exactly the Hamiltonian dual equations forthe plane elasticity problem of quasicrystals with point group12mm.
We consider the problem satisfying the mixed boundaryconditions
𝑢𝑥= 0, 𝜎
𝑥𝑦= 0, for 𝑥 = 0, 𝑥 = ℎ,
𝑤𝑥= 0, 𝐻
𝑥𝑦= 0, for 𝑥 = 0, 𝑥 = ℎ.
(9)
From (9) and (3), we have
𝑞1= 𝑞3= 0, for 𝑥 = 0, 𝑥 = ℎ. (10)
3. Theoretical Analysis
In the following, we only discuss (6), and the analysis for (7)is similar.
First, considering the homogeneous equation of (6),
𝜕
𝜕𝑦
𝑍1= 𝐻1𝑍1. (11)
Applying the method of separation of variables to the aboveequation, we write the solution as
𝑍1= 𝑋 (𝑥) 𝑌 (𝑦) , (12)
in which𝑌(𝑦) = 𝑒𝜇𝑦, and 𝜇 and𝑋(𝑥) are the eigenvalue of the
Hamiltonian operator matrix𝐻1and its associated eigenvec-
tor, respectively. They are determined by the equation
𝐻1𝑋(𝑥) = 𝜇𝑋 (𝑥) . (13)
Solving (13) with the boundary conditions (9) and (10) at𝑥 = 0, ℎ, we obtain the eigenvalues of𝐻
1:
𝜇0= 0, 𝜇
𝑛=
𝑛𝜋
ℎ
, 𝜇−𝑛
= −
𝑛𝜋
ℎ
,
𝑛 = 1, 2, . . . ,
(14)
Abstract and Applied Analysis 3
and the associated eigenvectors of 𝜇0, 𝜇𝑛, and 𝜇
−𝑛are
𝑋0
0= (
0
1
0
0
) , 𝑋0
𝑛= (
sin (𝜇𝑛𝑥)
− cos (𝜇𝑛𝑥)
0
0
) ,
𝑋0
−𝑛= (
− sin (𝜇𝑛𝑥)
− cos (𝜇𝑛𝑥)
0
0
) ,
(15)
respectively. From
𝐻1𝑋1
𝑛(𝑥) = 𝜇
𝑛𝑋1
𝑛(𝑥) + 𝑋
0
𝑛(𝑥) , (16)
and the imposed boundary conditions, the first-order Jordanform eigenvectors of 𝜇
0, 𝜇𝑛, and 𝜇
−𝑛can be solved as
𝑋1
0= (
0
0
0
1
) ,
𝑋1
𝑛=
(
(
(
(
(
sin (𝜇𝑛𝑥)
(
𝐶12
+ 3𝐶66
(𝐶12
+ 𝐶66) 𝜇𝑛
− 1) cos (𝜇𝑛𝑥)
2𝐶66
𝐶12
+ 𝐶66
sin (𝜇𝑛𝑥)
2𝐶66
𝐶12
+ 𝐶66
cos (𝜇𝑛𝑥)
)
)
)
)
)
,
𝑋1
−𝑛=
(
(
(
(
(
− sin (𝜇𝑛𝑥)
−(
𝐶12
+ 3𝐶66
(𝐶12
+ 𝐶66) 𝜇𝑛
+ 1) cos (𝜇𝑛𝑥)
−
2𝐶66
𝐶12
+ 𝐶66
sin (𝜇𝑛𝑥)
2𝐶66
𝐶12
+ 𝐶66
cos (𝜇𝑛𝑥)
)
)
)
)
)
,
(17)
respectively. Besides, we can verify that there are no otherhigh-order Jordan form eigenvectors in every chain.
It is easy to prove that the above eigenvectors andJordan form eigenvectors satisfy the symplectic conjugacyand orthogonality; that is,
∫
ℎ
0
𝑋0
0
𝑇
𝐽𝑋0
0𝑑𝑥 = ∫
ℎ
0
𝑋0
0
𝑇
𝐽𝑋1
𝑛𝑑𝑥 = ∫
ℎ
0
𝑋1
0
𝑇
𝐽𝑋0
𝑛𝑑𝑥
= ∫
ℎ
0
𝑋1
0
𝑇
𝐽𝑋1
𝑛𝑑𝑥 = 0,
∫
ℎ
0
𝑋0
𝑛
𝑇
𝐽𝑋0
𝑛𝑑𝑥 = ∫
ℎ
0
𝑋0
𝑛
𝑇
𝐽𝑋1
𝑛𝑑𝑥 = ∫
ℎ
0
𝑋0
𝑛
𝑇
𝐽𝑋0
−𝑛𝑑𝑥
= ∫
ℎ
0
𝑋1
𝑛
𝑇
𝐽𝑋1
−𝑛𝑑𝑥 = 0,
∫
ℎ
0
𝑋0
0
𝑇
𝐽𝑋1
0𝑑𝑥 = ℎ, ∫
ℎ
0
𝑋0
𝑛
𝑇
𝐽𝑋1
−𝑛𝑑𝑥 = −
2𝐶66
𝐶12
+ 𝐶66
ℎ,
𝑛 = ±1, ±2, . . . .
(18)
Next, we will prove the symplectic orthogonal expansiontheorem, that is, the completeness theorem of the generalizedeigenvector system (i.e., the collection of all the eigenvectorsand Jordan form eigenvectors), which shows that the sym-plectic method can be adopted to solve the title problem.
Theorem 1. The generalized eigenvector system
{𝑋0
0, 𝑋1
0} ∪ {𝑋
0
𝑛, 𝑋1
𝑛| 𝑛 = ±1, ±2, . . .} (19)
of the Hamiltonian operator matrix 𝐻1is complete in the
Hilbert space 𝑋; that is, there exist constant sequences {𝑐00, 𝑐1
0},
{𝑐𝑖
𝑛}∞
𝑛=1, and {𝑐
𝑖
−𝑛}∞
𝑛=1(𝑖 = 0, 1) such that
Φ = 𝑐0
0𝑋0
0+ 𝑐1
0𝑋1
0+
+∞
∑
𝑛=1
(𝑐0
𝑛𝑋0
𝑛+ 𝑐1
𝑛𝑋1
𝑛+ 𝑐0
−𝑛𝑋0
−𝑛+ 𝑐1
−𝑛𝑋1
−𝑛)
(20)
for each Φ = (𝜙1(𝑥), 𝜙
2(𝑥), 𝜙
3(𝑥), 𝜙
4(𝑥))𝑇
∈ 𝑋, where 𝑋 =
𝐿2[0, ℎ] × 𝐿
2[0, ℎ] × 𝐿
2[0, ℎ] × 𝐿
2[0, ℎ].
Proof. For anyΦ ∈ 𝑉, in order to prove equality (20), we set
𝑐0
0=
∫
ℎ
0Φ𝑇𝐽𝑋1
0𝑑𝑥
∫
ℎ
0𝑋0
0
𝑇𝐽𝑋1
0𝑑𝑥
=
∫
ℎ
0𝜙2𝑑𝑥
ℎ
,
𝑐1
0=
∫
ℎ
0Φ𝑇𝐽𝑋0
0𝑑𝑥
∫
ℎ
0𝑋1
0
𝑇𝐽𝑋0
0𝑑𝑥
=
∫
ℎ
0𝜙4𝑑𝑥
ℎ
,
𝑐0
𝑛=
∫
ℎ
0Φ𝑇𝐽𝑋1
−𝑛𝑑𝑥
∫
ℎ
0𝑋0
𝑛
𝑇𝐽𝑋1
−𝑛𝑑𝑥
=
∫
ℎ
0𝜙1sin (𝜇
𝑛𝑥) 𝑑𝑥 − ∫
ℎ
0𝜙2cos (𝜇
𝑛𝑥) 𝑑𝑥
ℎ
− (∫
ℎ
0
𝜙3sin (𝜇
𝑛𝑥) 𝑑𝑦+ (
(𝐶12
+ 3𝐶66)
(𝐶12
+ 𝐶66) 𝜇𝑛
+ 1)
×∫
ℎ
0
𝜙4cos (𝜇
𝑛𝑥) 𝑑𝑥) × (
2𝐶66
𝐶12
+ 𝐶66
ℎ)
−1
,
4 Abstract and Applied Analysis
𝑐1
𝑛=
∫
ℎ
0Φ𝑇𝐽𝑋0
−𝑛𝑑𝑥
∫
ℎ
0𝑋1
𝑛
𝑇𝐽𝑋0
−𝑛𝑑𝑥
=
∫
ℎ
0𝜙3sin (𝜇
𝑛𝑥) 𝑑𝑥 + ∫
ℎ
0𝜙4cos (𝜇
𝑛𝑥) 𝑑𝑥
(2𝐶66/ (𝐶12
+ 𝐶66)) ℎ
(21)by the symplectic orthogonal relationship (18). Then,
𝑐0
0𝑋0
0+ 𝑐1
0𝑋1
0+
+∞
∑
𝑛=1
(𝑐0
𝑛𝑋0
𝑛+ 𝑐1
𝑛𝑋1
𝑛+ 𝑐0
−𝑛𝑋0
−𝑛+ 𝑐1
−𝑛𝑋1
−𝑛)
=
(
(
(
(
(
(
(
(
(
+∞
∑
𝑛=1
2
ℎ
∫
ℎ
0
(𝜙1sin 𝑛𝜋𝑥
ℎ
𝑑𝑥) sin 𝑛𝜋𝑥
ℎ
1
ℎ
∫
ℎ
0
𝜙2𝑑𝑥 +
+∞
∑
𝑛=1
2
ℎ
(∫
ℎ
0
𝜙2cos 𝑛𝜋𝑥
ℎ
𝑑𝑥) cos 𝑛𝜋𝑥ℎ
+∞
∑
𝑛=1
2
ℎ
∫
1
0
(𝜙3sin 𝑛𝜋𝑥
ℎ
𝑑𝑥) sin 𝑛𝜋𝑥
ℎ
1
ℎ
∫
ℎ
0
𝜙4𝑑𝑥 +
+∞
∑
𝑛=1
2
ℎ
(∫
ℎ
0
𝜙4cos 𝑛𝜋𝑥
ℎ
𝑑𝑥) cos 𝑛𝜋𝑥ℎ
)
)
)
)
)
)
)
)
)
,
(22)in which the four components of the above expression arethe corresponding Fourier series of 𝜙
1, 𝜙2, 𝜙3, 𝜙4associated
with the orthogonal function system {sin(𝑛𝜋𝑥/ℎ)}+∞𝑛=1
or{cos(𝑛𝜋𝑥/ℎ)}+∞
𝑛=0in 𝐿2[0, ℎ]. Therefore, equality (20) is valid,
which means that the generalized eigenvector system of𝐻1is
complete in the Hilbert space𝑋.
Similarly, for the Hamiltonian operator matrix 𝐻2, we
also have the following completeness theory.
Theorem 2. The generalized eigenvector system
{𝑋
0
0, 𝑋
1
0} ∪ {𝑋
0
𝑛, 𝑋
1
𝑛| 𝑛 = ±1, ±2, . . .} (23)
of the Hamiltonian operator matrix 𝐻2is complete in the
Hilbert space 𝑋, where
𝑋
0
0= (
0
1
0
0
) , 𝑋
0
𝑛= (
sin (𝜆𝑛𝑥)
cos (𝜆𝑛𝑥)
0
0
) ,
𝑋
1
0= (
0
0
0
1
) ,
𝑋
1
𝑛=
(
(
(
(
sin (𝜆𝑛𝑥)
(1 +
2𝐾1+ 𝐾2+ 𝐾3
𝜆𝑛(𝐾2+ 𝐾3)
) cos (𝜆𝑛𝑥)
−(
2𝐾1
𝐾2+ 𝐾3
+ 2) sin (𝜆𝑛𝑥)
(
2𝐾1
𝐾2+ 𝐾3
+ 2) cos (𝜆𝑛𝑥)
)
)
)
)
(24)
are the associated eigenvectors and the first-order Jordan formeigenvectors of the eigenvalue 𝜆
0= 0 and 𝜆
𝑛= 𝑛𝜋/ℎ of𝐻
2.
4. General Solution
By completeness Theorem 1, the general solution of theinhomogeneous equation (6) is represented in the form
𝑍1(𝑥, 𝑦) = 𝑋
0
0𝑌0
0(𝑦) + 𝑋
1
0𝑌1
0(𝑦)
+
+∞
∑
𝑛=1
(𝑋0
𝑛𝑌0
𝑛(𝑦) + 𝑋
1
𝑛𝑌1
𝑛(𝑦) + 𝑋
0
−𝑛𝑌0
−𝑛(𝑦)
+𝑋1
−𝑛𝑌1
−𝑛(𝑦)) .
(25)
The vector 𝐹1can also be expanded as
𝐹1= 𝑋0
0𝐹0
0(𝑦) + 𝑋
1
0𝐹1
0(𝑦)
+
+∞
∑
𝑛=1
(𝑋0
𝑛𝐹0
𝑛(𝑦) + 𝑋
1
𝑛𝐹1
𝑛(𝑦) + 𝑋
0
−𝑛𝐹0
−𝑛(𝑦)
+𝑋1
−𝑛𝐹1
−𝑛(𝑦)) .
(26)
Multiplying both sides of (26) by𝑋10
𝑇
𝐽,𝑋00
𝑇
𝐽,𝑋1−𝑛
𝑇
𝐽,𝑋0−𝑛
𝑇
𝐽,𝑋1
𝑛
𝑇
𝐽, and 𝑋0
𝑛
𝑇
𝐽 and then integrating by 𝑥 from 0 to ℎ,respectively, we have
𝐹0
0(𝑦) =
∫
ℎ
0𝑋1
0
𝑇
𝐽𝐹1𝑑𝑥
∫
ℎ
0𝑋1
0
𝑇𝐽𝑋0
0𝑑𝑥
= 0,
𝐹1
0(𝑦) =
∫
ℎ
0𝑋0
0
𝑇
𝐽𝐹1𝑑𝑥
∫
ℎ
0𝑋0
0
𝑇𝐽𝑋1
0𝑑𝑥
= −
∫
ℎ
0𝑓2𝑑𝑥
(𝐶12
+ 2𝐶66) ℎ
,
𝐹0
𝑛(𝑦) =
∫
ℎ
0𝑋1
−𝑛
𝑇
𝐽𝐹1𝑑𝑥
∫
ℎ
0𝑋1
−𝑛
𝑇𝐽𝑋0
𝑛𝑑𝑥
= (∫
ℎ
0
𝑓1sin (𝜇
𝑛𝑥) 𝑑𝑥 + (
𝐶12
+ 3𝐶66
(𝐶12
+ 𝐶66) 𝜇𝑛
+ 1)
×∫
ℎ
0
𝑓2cos (𝜇
𝑛𝑥) 𝑑𝑥)
× (
2𝐶66(𝐶12
+ 2𝐶66)
𝐶12
+ 𝐶66
ℎ)
−1
,
𝐹1
𝑛(𝑦) =
∫
ℎ
0𝑋0
−𝑛
𝑇
𝐽𝐹1𝑑𝑥
∫
ℎ
0𝑋0
−𝑛
𝑇𝐽𝑋1
𝑛𝑑𝑥
=
−∫
ℎ
0𝑓1sin (𝜇
𝑛𝑥) 𝑑𝑥 − ∫
ℎ
0𝑓2cos (𝜇
𝑛𝑥) 𝑑𝑥
(2𝐶66(𝐶12
+ 2𝐶66) / (𝐶12
+ 𝐶66)) ℎ
𝑛 = ±1, ±2, . . . .
(27)
Abstract and Applied Analysis 5
Table 1: The computed results.
(𝑥, 𝑦) (0.1464, 0.8536) (1.4, 0.55) (2.1, 0.34) (3.42, 0.89) (4.7, 0.27)𝑢𝑥
0.204 −0.08131 −0.237 0.4822 −0.2427𝑢𝑦
−0.06272 −0.01618 −0.02746 −0.0399 −0.04801𝑤𝑥
0.2965 −0.243 −0.583 0.6355 −0.4975𝑤𝑦
−0.02621 −0.006761 −0.01148 −0.01667 −0.02006
Then, substituting (25) and (26) into (6) yields
𝑑𝑌1
0(𝑦)
𝑑𝑦
= 𝐹1
0(𝑦) ,
𝑑𝑌0
0(𝑦)
𝑑𝑦
= 𝑌1
0(𝑦) ,
𝑑𝑌1
𝑛(𝑦)
𝑑𝑦
= 𝜇𝑛𝑌1
𝑛(𝑦) + 𝐹
1
𝑛(𝑦) ,
𝑑𝑌0
𝑛(𝑦)
𝑑𝑦
= 𝜇𝑛𝑌0
𝑛(𝑦) + 𝑌
1
𝑛(𝑦) + 𝐹
0
𝑛(𝑦) .
(28)
Thus, we obtain
𝑌1
0(𝑦) = 𝑐
1
0+ ∫
𝑦
0
𝐹1
0(𝜉) 𝑑𝜉,
𝑌0
0(𝑦) = 𝑐
0
0+ 𝑐1
0𝑦 + ∫
𝑦
0
∫
𝜏
0
𝐹1
0(𝜉) 𝑑𝜉 𝑑𝜏,
𝑌1
𝑛(𝑦) = 𝑐
1
𝑛𝑒𝜇𝑛𝑦+ ∫
𝑦
0
𝐹1
𝑛(𝜉) 𝑒𝜇𝑛(𝑦−𝜉)
𝑑𝜉,
𝑌0
𝑛(𝑦) = (𝑐
0
𝑛+ 𝑐1
𝑛𝑦) 𝑒𝜇𝑛𝑦+ ∫
𝑦
0
𝐹0
𝑛(𝜉) 𝑒𝜇𝑛(𝑦−𝜉)
𝑑𝜉
+ ∫
𝑦
0
∫
𝜏
0
𝐹1
𝑛(𝜉) 𝑒𝜇𝑛(𝑦−𝜉)
𝑑𝜉 𝑑𝜏,
(29)
where 𝑐1
0, 𝑐00, 𝑐1𝑛, and 𝑐
0
𝑛are unknown constants to be deter-
mined by imposing the remaining boundary conditions at 𝑦.Substituting (29) into (25), we have the analytical solutions 𝑢
𝑥
and 𝑢𝑦of (4) given by
𝑢𝑥
=
+∞
∑
𝑛=1
[ (𝑐0
𝑛+ 𝑐1
𝑛+ 𝑐1
𝑛𝑦) 𝑒𝜇𝑛𝑦
− (𝑐0
−𝑛+ 𝑐1
−𝑛+ 𝑐1
−𝑛𝑦) 𝑒−𝜇𝑛𝑦
+ ∫
𝑦
0
((𝐹0
𝑛(𝜉) + 𝐹
1
𝑛(𝜉)) 𝑒𝜇𝑛(𝑦−𝜉)
− (𝐹0
−𝑛(𝜉) + 𝐹
1
−𝑛(𝜉)) 𝑒−𝜇𝑛(𝑦−𝜉)
) 𝑑𝜉
+ ∫
𝑦
0
∫
𝜏
0
(𝐹1
𝑛(𝜉) 𝑒𝜇𝑛(𝑦−𝜉)
−𝐹1
−𝑛(𝜉) 𝑒−𝜇𝑛(𝑦−𝜉)
) 𝑑𝜉 𝑑𝜏] sin 𝜇𝑛𝑥,
𝑢𝑦
= 𝑐0
0+ 𝑐1
0𝑦 + ∫
𝑦
0
∫
𝜏
0
𝐹1
0(𝜉) 𝑑𝜉 𝑑𝜏
−
+∞
∑
𝑛=1
[(𝑐0
𝑛− (
𝐶12
+ 3𝐶66
(𝐶12
+ 𝐶66) 𝜇𝑛
− 1) 𝑐1
𝑛+ 𝑐1
𝑛𝑦) 𝑒𝜇𝑛𝑦
+ (𝑐0
−𝑛+ (
𝐶12
+ 3𝐶66
(𝐶12
+ 𝐶66) 𝜇𝑛
+ 1) 𝑐1
−𝑛+ 𝑐1
−𝑛𝑦) 𝑒−𝜇𝑛𝑦
+ ∫
𝑦
0
((𝐹0
𝑛(𝜉) − (
𝐶12
+ 3𝐶66
(𝐶12
+ 𝐶66) 𝜇𝑛
− 1)
×𝐹1
𝑛(𝜉) ) 𝑒
𝜇𝑛(𝑦−𝜉)
+ (𝐹0
−𝑛(𝜉) + (
𝐶12
+ 3𝐶66
(𝐶12
+ 𝐶66) 𝜇𝑛
+ 1)
×𝐹1
−𝑛(𝜉) ) 𝑒
−𝜇𝑛(𝑦−𝜉)
)𝑑𝜉
+∫
𝑦
0
∫
𝜏
0
(𝐹1
𝑛(𝜉) 𝑒𝜇𝑛(𝑦−𝜉)
+ 𝐹1
−𝑛(𝜉) 𝑒−𝜇𝑛(𝑦−𝜉)
) 𝑑𝜉 𝑑𝜏]
× cos 𝜇𝑛𝑥.
(30)
According to the above procedure for (7), the analyticalsolutions 𝑤
𝑥and 𝑤
𝑦of (4) can be obtained:
𝑤𝑥=
+∞
∑
𝑛=1
[ (𝑑0
𝑛+ 𝑑1
𝑛+ 𝑑1
𝑛𝑦) 𝑒𝜇𝑛𝑦
− (𝑑0
−𝑛+ 𝑑1
−𝑛+ 𝑑1
−𝑛𝑦) 𝑒−𝜆𝑛𝑦
+ ∫
𝑦
0
((𝐹
0
𝑛(𝜉) + 𝐹
1
𝑛(𝜉)) 𝑒𝜆𝑛(𝑦−𝜉)
− (𝐹
0
−𝑛(𝜉) + 𝐹
1
−𝑛(𝜉)) 𝑒−𝜆𝑛(𝑦−𝜉)
) 𝑑𝜉
+ ∫
𝑦
0
∫
𝜏
0
(𝐹
1
𝑛(𝜉) 𝑒𝜆𝑛(𝑦−𝜉)
−𝐹
1
−𝑛(𝜉) 𝑒−𝜆𝑛(𝑦−𝜉)
) 𝑑𝜉 𝑑𝜏] sin 𝜆𝑛𝑥,
6 Abstract and Applied Analysis
𝑤𝑦
= 𝑑0
0+ 𝑑1
0𝑦 + ∫
𝑦
0
∫
𝜏
0
𝐹
1
0(𝜉) 𝑑𝜉 𝑑𝜏
+
+∞
∑
𝑛=1
[(𝑑0
𝑛+ (1 +
2𝐾1+ 𝐾2+ 𝐾3
𝜆𝑛(𝐾2+ 𝐾3)
) 𝑑1
𝑛+ 𝑑1
𝑛𝑦) 𝑒𝜆𝑛𝑦
+ (𝑑0
−𝑛+ (1 −
2𝐾1+ 𝐾2+ 𝐾3
𝜆𝑛(𝐾2+ 𝐾3)
) 𝑑1
−𝑛
+𝑑1
−𝑛𝑦) 𝑒−𝜆𝑛𝑦
+ ∫
𝑦
0
((𝐹
0
𝑛(𝜉) + (1 +
2𝐾1+ 𝐾2+ 𝐾3
𝜆𝑛(𝐾2+ 𝐾3)
)𝐹
1
𝑛(𝜉))
× 𝑒𝜆𝑛(𝑦−𝜉)
+ (𝐹
0
−𝑛(𝜉) +(1 −
2𝐾1+ 𝐾2+ 𝐾3
𝜆𝑛(𝐾2+ 𝐾3)
)𝐹
1
−𝑛(𝜉))
×𝑒−𝜆𝑛(𝑦−𝜉)
) 𝑑𝜉
+ ∫
𝑦
0
∫
𝜏
0
(𝐹
1
𝑛(𝜉) 𝑒𝜆𝑛(𝑦−𝜉)
+ 𝐹
1
−𝑛(𝜉) 𝑒−𝜆𝑛(𝑦−𝜉)
) 𝑑𝜉 𝑑𝜏] cos 𝜆𝑛𝑥,
(31)
where 𝑑1
0, 𝑑0
0, 𝑑1
𝑛, and 𝑑
0
𝑛are unknown constants to be
determined by imposing the remaining boundary conditionsat 𝑦 and
𝐹
1
0=
∫
ℎ
0𝑔2𝑑𝑥
𝐾1ℎ
,
𝐹
1
𝑛=
−∫
ℎ
0𝑔1sin 𝜆𝑛𝑥 𝑑𝑥 + ∫
ℎ
0𝑔2cos 𝜆𝑛𝑥 𝑑𝑥
((2𝐾1+ 𝐾2+ 𝐾3) / (𝐾2+ 𝐾3)) ℎ
,
𝐹
0
𝑛= (∫
ℎ
0
𝑔1sin 𝜆𝑛𝑥𝑑𝑦
−(1 −
2𝐾1+ 𝐾2+ 𝐾3
𝜆𝑛(𝐾2+ 𝐾3)
)∫
ℎ
0
𝑔2cos 𝜆𝑛𝑥 𝑑𝑥)
× (
2𝐾1+ 𝐾2+ 𝐾3
𝐾2+ 𝐾3
ℎ)
−1
.
(32)
5. Numerical Calculations
Compared with [17], the present paper is devoted to thesymplectic analysis of the plane elasticity problem of qua-sicrystals. To guarantee the feasibility of our method, wealso prove the completeness for the eigenfunction systemof the associated Hamiltonian operator matrices. Note thatthe completeness does not always hold for the Hamiltonianoperator matrices.
In order to determine the unknown constants 𝑐𝑘𝑛and 𝑑
𝑘
𝑛
of the analytical solution in (30) and (31), we consider theboundary conditions at 𝑦 = 0, 𝑦 = 𝑙 given by
𝑢𝑥= sin 4𝜋
ℎ
𝑥, 𝑢𝑦= 0, for 𝑦 = 0, 𝑦 = 𝑙,
𝑤𝑥= sin 4𝜋
ℎ
𝑥, 𝑤𝑦= 0, for 𝑦 = 0, 𝑦 = 𝑙.
(33)
In the following, let ℎ = 5, 𝑙 = 1, andwe take the constants𝐶12
= 0.5714, 𝐶66
= 0.88445,𝐾1= 1.22,𝐾
2= 0.24, and𝐾
3=
0.6. The computed results are listed in Table 1 for illustratingpreviousmain results, and the data is the same as that of usingthe treatment in [17].
6. Conclusions
The symplectic approach is established for the plane elasticityproblem of quasicrystals with point group 12mm satisfyingthemixed boundary conditions.The correspondingHamilto-nian operator matrix plays an important role in this method,whose eigenvalues and eigenfunctions need to be obtained.Through calculations, the eigenfunction system is symplecticorthogonal. Based on this, we further verify the feasibilityof this approach. Then the exact analytical solution is givenwith the use of the symplectic eigenfunction method. Wecan know that the method is totally rational and gives us asystematic way to solve physical problems. In addition, thisapproach is expected to apply to other quasicrystal problems.
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper.
Acknowledgments
Theauthors are grateful to the referees for valuable commentson improving this paper. This work is supported by theNational Natural Science Foundation of China (nos. 11261034and 11326239).
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