An Application of Homotopy Analysis Method for Estimation the Diaphragm Deflection in MEMS Capacitive Microphone

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1 ISSN (print, (online International Journal of Nonlinear Science Vol.17(2014 No.1,pp.3-13 An Application of Homotopy Analysis Metho for Estimation the Diaphragm Deflection in MEMS Capacitive Microphone M. M. Khaer 1,2, N. H. Sweilam 3 1 Department of Mathematics an Statistics, College of Science, Al-Imam Mohamma Ibn Sau Islamic University (IMSIU, Riyah 11566, Saui Arabia 2 Department of Mathematics, Faculty of Science, Benha University, Benha, Egypt 3 Department of Mathematics, Faculty of Science, Cairo University, Giza, Egypt (Receive 15 January 2013, accepte 2 December 2013 Abstract: In this article, homotopy analysis metho (HAM is introuce to analyze the eflection of polysilicon iaphragm of Micro Electro Mechanical Systems (MEMS capacitive microphone. The resiual stresses in the material use to make the iaphragm change the vibrational characteristics of the microphone iaphragm an consequently influence the microphone s first resonant frequency, cutoff frequency an sensitivity. The most successful evices use poly-silicon as a iaphragm material, because of its resiual stress is controllable by high temperature annealing after ion implantation by boron or phosphorous. We prove the existence an the uniqueness of the solution of consiere problem using the theory of semi-group. Series solutions of the problem uner consieration are evelope by means of HAM an the recurrence relations are given explicitly. The numerical examples show the rapi convergence of the series constructe by this metho to the exact solution. Moreover, this technique oes not require any iscretization, linearization or small perturbations. Test problems have been consiere to ensure that HAM is accurate an efficient compare with the variational iteration metho (VIM. Keywors: Analytical solution; Diaphragm eflection; Capacitive microphone; Semi-group theory; Homotopy analysis metho. PACS coe: x, a, i, e. 1 Introuction It is well known that most of the scientific phenomena are moele using orinary or partial ifferential equations. Analytical solutions of these equations may well escribe the various phenomena in science an nature, such as vibrations, solitons an propagation with a finite spee. The homotopy analysis metho is an analytical technique for solving nonlinear ifferential equations evise by Shi-Jun Liao in 1992 [1]. This metho has been successfully applie to solve many types of nonlinear problems in science an engineering by many authors ([2]-[9], an references therein. We aim in this work to effectively employ HAM to establish the estimation of iaphragm eflection in MEMS capacitive microphone ([10]-[12]. By the presente metho, numerical results can be obtaine with using a few iterations [1]. Moreover, HAM contains the auxiliary parameter, which provies us with a simple way to ajust an control the convergence region of solution series [13]. Therefore, HAM hanles linear an nonlinear problems without any assumption an restriction. The paper has been organize as follows. In section 2, the eflection of iaphragm in ifferential form is presente. In section 3, some notations an preliminaries results concerning semigroup are introuce. In section 4, the existence an the uniqueness of the solution of the bounary value problem are prove. In section 5, the basic iea of homotopy analysis metho is escribe. In section 6, applying HAM to analyze the eflection of polysilicon iaphragm of micro-electro mechanical systems capacitive microphone is presente. In section 7, the simulation results is given. Also, in section 8, the convergence of the exact solution is illustrate. Discussion an conclusions are presente in section 9. Corresponing author. aress: mohamemb@yahoo.com Copyright c Worl Acaemic Press, Worl Acaemic Union IJNS /778

2 4 International Journal of Nonlinear Science, Vol.17(2014, No.1, pp Deflection of iaphragm in ifferential form The performance of the microphone epens on the size, stress an eflection of the iaphragm. The iaphragm eflection u can be approximate by the following ifferential equation ([10], [11] α 4 u + β 2 u = 2 u t 2, (1 where α, β are the flexural rigiity, tensile force per unit length, an is the ensity (mass per unit area of the iaphragm, respectively. 2 is the Laplacian operator an 4 inicates to 2 ( 2. For the first funamental moe, we can assume the eflection of the square iaphragm is [10] ( πy u(x, y, t A sin sin e 2πt, (2 where is the iaphragm with. Substitution of Eq.(2 into Eq.(1 yiels the first resonant frequency for the iaphragm [10] ( 1 απ 2 f res = 4 + β 2 2. (3 3 Notations an preliminaries results concerning semigroup One of our aims in this paper is to stuy the existence an the uniqueness for solution of BVP (1. We recall the basic efinitions an the most important properties of semi-groups theory [14]. These recalls are only intene to fix some notations an references an are confine to what will be useful in the sequel. For what concerns the results on semi-groups we refer to [15]. We collect firstly the known results concerning the theory of semi-group in the following efinitions. Definition 1 A family of linear, continuous operators G(t t 0 (B, B, B is a Banach space, is calle a semigroup if G(t M(t, G(0 = 1, G(t 1 + t 2 = G(t 1.G(t 2. It is sai to be strongly continuous if for each v B the function t G(tv is continuous in [0, ]. Definition 2 An operator A from B to B is sai to be generator of semigroup G(t if it in the following form G(tv v Av = lim, t 0 t whose omain D(A is the set of the elements v of B which the right han sie exists. Definition 3 The semigroup G(t is calle contraction semigroup if G(t 1. Definition 4 An operator A in a Hilbert space H is sai to be an accretive (resp. issipative operator if Re(Av, v 0(resp. 0. Proposition 1 If A is a generator of semigroup G(t, then D(A is ense in B; moreover, if v D(A, then G(tv D(A for t 0 an it is a continuously ifferential function. It satisfies (G(tv = A (G(tv = G(tAv, t 0. t This Proposition shows that if v D(A, then G(tv is a solution of the following evolution initial value problem u(t = Au(t, u(0 = v. (4 t In fact, semigroups are tool for the resolution of problem of the form (4, in fact, if G(t is known, G(tv is efine for v B, then G(tv is the limit of solutions of (4 with initial value in D(A. It is natural to efine generalize solution of (4 with u(0 = v B by G(tv. It is important to have necessary an sufficient conitions for an operator A to be the generator of a semigroup G. IJNS for contribution: eitor@nonlinearscience.org.uk

3 M. M. Khaer, N. H. Sweilam: An Application of Homotopy Analysis Metho for Estimation the Diaphragm Deflection 5 Theorem 2 (Hille-Yosia If G(t is a continuous semigroup on B, with then, its generator A satisfies [i] D(A is ense in B, A is close; G(t Me ωt, [ii] The semi-infinite interval λ > ω, (λ real belongs to the resolvent set of A. Reciprocally, if A is an operator in the Banach space B, satisfying [i] an [ii], it is the generator of a continuous semigroup G(t. Theorem 3 (Hille-Yosia If G(t is a continuous semigroup of contractions, its generator A satisfies [i] D(A is ense in B, A is close; [ii] The semi-infinite interval λ > ω, (λ real belongs to the resolvent set of A. Reciprocally, if A is an operator in the Banach space B, satisfying [i] an [ii], it is the generator of contraction semigroup G(t in B. Theorem 4 (Lumer-Philips Let A be a linear operator in Hilbert space H with omain D(A ense in H. Then [i] If A is the generator of a contraction semi-group in H, then A is issipative an the range R(λ A of λ A is the whole space H for all λ > 0; [ii] If A is issipative an there exists λ > 0 such that the range of λ A is the whole space H, then A is the generator of contraction semigroup in H. An important property of semigroup If A is the generator of continuous semigroup G(t in B one may consier the nonhomogeneous initial value problem u t = Au + f, u(0 = v, (5 where f is a continuous function of t with values in B. Then, the unique generalization solution of (5 with v B for t 0 is given by u(t = G(tv + t 0 G(t sf(ss, t 0. (6 This is a continuous function of t with values in B. Moreover, if v D(A an f(t is continuously ifferentiable with values in B an u(t D(A for t 0. Then we obtain the classical solution of (5. 4 Existence an uniqueness of the solution of the bounary value problem (1 The bounary value problem (1 can be rewritten in the following form with the bounary conitions an the following initial conitions 2 u t 2 = u(x, y, 0 = sin Let us introuce the following Sobolev spaces ( α 4 u + ( β 2 u, (7 u = u = 0, u Ω, (8 n ( πy sin, u(x, y, 0 t H = L 2 (Ω; V 0 = H 1 0 (Ω; V 1 = H 2 0 (Ω, = 0. IJNS homepage:

4 6 International Journal of Nonlinear Science, Vol.17(2014, No.1, pp an the bilinear forms a 0 (u, v = ( β scalar prouct We have the following continuous an ense embeing Ω u v x x xy, a 1(u, v = 2 u 2 v xy, a = β Ω a 0 + a 1. (9 We note that all these forms verify the hypotheses of Lax-Milgram theorem [16]. The space V 0 is a Hilbert space with the a 0 (u, v, let V enotes the space V 1 equippe with the scalar prouct a = a 0 + a 1. V V 0 H H V 0 V. Let us efine (as in first representation Riesz theorem [16] the unboune self-ajoint operators ( β A 0 = β 2 ; A 1 = 4 ; A = β 2 + α 4, associate to the forms β a 0, a 1, a, respectively. Proposition 5 The BVP (7 is equivalent to the classical equation u t = A u, moreover the operator A is generator of a continuous semigroup in V H. Proof. First we suppose that u = (u 1, u 2, where u 1 = u, u 2 = u ( 0 I t, A = then (8 can be written in the -A 0 form u = A u. (10 t So, it suffices to prove that the conitions of Lumer-Philips are satisfie, that means A is generator of a contraction semigroup in V H. 1. D(A is ense in V H, on the other han, we can write Re(A u, u = Re[a(u 2, u 1 + ( Au 1, u 2 H ] = 0, then the operator A is issipative. 2. The range of (λ A is the all space V H because for any given w from V H, we can fin an element u from D(A verifie λ u A u = w for all real number λ > 0. (11 Therefore, after some manipulations, we have (A + λ 2 u 1 = w 2 + λw 1. (12 The left han sie is an operator associate to a Hermitian an coercive form in V by the Lax-Milgram theorem, there exists u 1 from V solution of (12, then u 2 exists too. Consequently, we foun u from D(A solution of (11. Problem (7 has an unique generalize solution as given in (6. By application the Proposition 2, we are immeiately in the framework of Eq.(6, then the Proposition is prove. 5 Basic iea of HAM To illustrate the basic iea of HAM ([1]-[8], we consier the following ifferential equation N[u(l, t] = 0, (13 where N is a linear operator for this problem, l an t enote inepenent variables, u(l, t is an unknown function. For simplicity, we ignore all bounary an initial conitions, which can be treate in the similar way. IJNS for contribution: eitor@nonlinearscience.org.uk

5 M. M. Khaer, N. H. Sweilam: An Application of Homotopy Analysis Metho for Estimation the Diaphragm Deflection Zeroth-orer eformation equation Liao [1], constructe the so-calle zeroth-orer eformation equation (1 q [ϕ(l, t; q u 0 (l, t] = q N[ϕ(l, t; q], (14 where is an auxiliary linear operator, u 0 (l, t is an initial guess, 0 is an auxiliary parameter an q [0, 1] is the embeing parameter. Obviously, when q = 0 an q = 1, it hols, respectively ϕ(l, t; 0 = u 0 (l, t, ϕ(l, t; 1 = u(l, t. (15 Thus, as q increasing from 0 to 1, the solution ϕ(l, t; q various from u 0 (l, t to u(l, t. Expaning ϕ(l, t; q in Taylor series with respect to the embeing parameter q, one has where ϕ(l, t; q = u 0 (l, t + u m (l, tq m, (16 m=1 u m (l, t = 1 m ϕ(l, t; q q=0 m! q m. (17 Assume that the auxiliary linear operator, the initial guess an the auxiliary parameter are selecte such that the series (16 is convergent at q = 1, Then at q = 1 an by (15, the series (16 becomes u(l, t = u 0 (l, t The m th orer eformation equation u m (l, t. (18 m=1 Define the vector u n (l, t = [u 0 (l, t, u 1 (l, t,..., u n (l, t]. (19 Differentiating Eq.(14 m times with respect to the embeing parameter q, then setting q = 0 an iviing them by m!, finally using (17, we have the so-calle m th -orer eformation equations [u m (l, t δ m u m 1 (l, t] = R m ( u m 1, (20 where an R m ( u m 1 = δ m = 1 m 1 N[ϕ(l, t; q] q=0 (m 1! q m 1, (21 { 0, m 1; 1, m > 1. (22 Applying 1 on both sie of Eq.(20, we get In this way, it is easily to obtain u m for m 1, at N th orer, we have u m (l, t = δ m u m 1 (l, t + 1 [R m ( u m 1 ]. (23 u(l, t N u m (l, t. (24 m=0 When N, we get an accurate approximation of the original Eq.(1. For the convergence of the propose metho we refer the reaer to Liao ([1]-[8]. If Eq.(1 amits unique solution, then this metho will prouce the unique solution. If Eq.(1 has not possess unique solution, HAM will give a solution among many other (possible solutions. IJNS homepage:

6 8 International Journal of Nonlinear Science, Vol.17(2014, No.1, pp Applications the propose metho We will apply HAM to estimate iaphragm eflection by solving Eq.(1 with three cases to illustrate the strength of the metho an to establish the approximate solutions for these problems. We introuce a comparison with the VIM ([17]- [19]. 6.1 HAM for case stuy 1 The iaphragm compliance an iaphragm eflection epen on its flexural rigiity α, an tension, β. The flexural rigiity of the iaphragm is given by Eτ 3 α = 12(1 δ 2, (25 where E is the Young s moulus of elasticity, τ is the iaphragm thickness an δ is the Poisson s ratio. We want to analyze iaphragm eflection ieally, when E 0 α 0. Eq.(1 when α 0 converts to β 2 u = 2 u t 2. (26 Now, to implement HAM for solving (26, we choose the linear operator [ϕ(x, y, t; q] = 2 ϕ(x, y, t; q t 2, (27 with the property, [a 1 + a 2 t] = 0, where a 1, a 2 are constants. We now efine a linear operator as ( N[ϕ(x, y, t; q] = 2 ϕ(x, y, t; q β t 2 2 ϕ(x, y, t; q. (28 Using above efinition, we construct the zeroth-orer eformation equation For q = 0 an q = 1, we can write Thus, we obtain the m th orer eformation equations where (1 q [ϕ(x, y, t; q u 0 (x, y, t] = q N[ϕ(x, y, t; q]. (29 ϕ(x, y, t; 0 = u 0 (x, y, t, ϕ(x, y, t; 1 = u(x, y, t. (30 [u m (x, y, t δ m u m 1 (x, y, t] = R m ( u m 1, R m ( u m 1 = 2 ϕ(x, y, t; q t 2 Now the solution of the m th orer eformation equations for m 1 becomes ( β 2 ϕ(x, y, t; q. (31 u m (x, y, t = δ m u m 1 (x, y, t + 1 [R m ( u m 1 ]. (32 This in turn gives the first few components of the approximate solution. We start with initial approximation ( πy u 0 (x, y, t = sin sin. (33 Since ũ = u 0 + u 1 + u From the above equations (32, we can obtain u n s as follows ( βπ 2 t 2 ( πy u 1 (x, y, t = 2 sin sin, ( βπ 2 t 2 u 2 (x, y, t = 2 + β 2 π 2 t β2 2 π 4 t 4 ( πy sin sin, ( β π 2 t 2 u 3 (x, y, t = 2 + β 2 π 2 t β2 2 π 4 t π 2 t 2 β( ( (1 + 2 π 2 t 2 β + 2 β 2 π 4 t 4 ( πy sin sin, other components of the approximate solution can be obtaine in the same manner. IJNS for contribution: eitor@nonlinearscience.org.uk

7 M. M. Khaer, N. H. Sweilam: An Application of Homotopy Analysis Metho for Estimation the Diaphragm Deflection HAM for case stuy 2 The tension, β, is etermine by the resiual stress of the iaphragm material, θ an τ, is the iaphragm thickness which satisfy β = θτ. (34 We want to investigate iaphragm eflection ieally, when θ 0 β 0. Eq.(1 when β 0 converts to α 4 u = 2 u t 2. (35 Now, to implement HAM for solving (35, we choose the same linear operator efine in (27 an efine a linear operator as ( N[ϕ(x, y, t; q] = 2 ϕ(x, y, t; q α t ϕ(x, y, t; q. (36 Using above efinition, we construct the zeroth-orer eformation equation For q = 0 an q = 1, we can write Thus, we obtain the m th orer eformation equations where (1 q [ϕ(x, y, t; q u 0 (x, y, t] = q N[ϕ(x, y, t; q]. (37 ϕ(x, y, t; 0 = u 0 (x, y, t, ϕ(x, y, t; 1 = u(x, y, t. (38 [u m (x, y, t δ m u m 1 (x, y, t] = R m ( u m 1, R m ( u m 1 = 2 ϕ(x, y, t; q t 2 + Now the solution of the m th orer eformation equations for m 1 becomes ( α 4 ϕ(x, y, t; q. (39 u m (x, y, t = δ m u m 1 (x, y, t + 1 [R m ( u m 1 ]. (40 This in turn gives the first few components of the approximate solution. We use arbitrary an initial approximation that satisfies the initial conitions as (33. Substituting Eq.(33 into Eq.(40 an summarize it, we have ( 2 απ 4 t 2 u 1 (x, y, t = 4 u 2 (x, y, t = u 3 (x, y, t = sin sin ( πy, ( 2 απ 4 t α 2 π 8 t απ4 t 2 4 sin ( 2 απ 4 t α 2 π 8 t απ4 t 2 4 sin ( πy, + 2α 2 π 4 t 2 ( ( α(1 + 2 π 4 t 2 + 2α 2 π 8 t sin other components of the approximate solution can be obtaine in the same manner. sin ( πy, 6.3 HAM for case stuy 3 In this subsection, we estimate the iaphragm eflection at this practical substrates in the general case (α β 0. Now, to implement HAM for solving (1, we choose the same linear operator (27. We now efine a linear operator as N[ϕ(x, y, t; q] = 2 ϕ(x, y, t; q t 2 + ( α 4 ϕ(x, y, t; q ( β 2 ϕ(x, y, t; q. (41 IJNS homepage:

8 10 International Journal of Nonlinear Science, Vol.17(2014, No.1, pp Using above efinition, we construct the zeroth-orer eformation equation (1 q [ϕ(x, y, t; q u 0 (x, y, t] = q N[ϕ(x, y, t; q]. (42 For q = 0 an q = 1, we can write ϕ(x, y, t; 0 = u 0 (x, y, t, ϕ(x, y, t; 1 = u(x, y, t. (43 Thus, we obtain the m th orer eformation equations [u m (x, y, t δ m u m 1 (x, y, t] = R m ( u m 1, where R m ( u m 1 = 2 ϕ(x, y, t; q t 2 + ( α 4 ϕ(x, y, t; q ( β 2 ϕ(x, y, t; q. (44 Now the solution of the m th orer eformation equations for m 1 becomes u m (x, y, t = δ m u m 1 (x, y, t + 1 [R m ( u m 1 ]. (45 This in turn gives the first few components of the approximate solution. We start with initial approximation (33. Since ũ = u 0 + u From the above equations (45, we can obtain u n s as follows u 2 (x, y, t = u 1 (x, y, t = 4 ( 2απ 4 t 2 + π 2 t 2 β sin sin ( πy, ( 4 (2απ4 t 2 + π 2 t 2 β (r2 t 2 (2D0r 2 + a1 2 T ( απ 4 t π 2 t 2 β sin sin(πy, other components of the approximate solution can be obtaine in the same manner. 7 Simulation results In orer to comparison iaphragm eflection u which obtaine by HAM with three conitions with exact solution in (1, we have numerical examples in 2D. For each three escribe equations we have three examples which mae by y = 0.1, y = 1, y = 10. In orer to comparison exact solution with HAM, magnitue of exponential function in (1 consiere one, e 12πt = 1. In all examples we have some constant parameters = 0.1, = 2, t = 0.5. Simulation 2 imensional (α = 0: Fig. 1 presents the behavior of the approximate solution with = 1.66 an the exact solution at y = 0.1, 1, 10 at the final time t = 0.5. From this figure we can conclue that the solution by using the propose metho an the exact are in excellent agreement. It is note that our approximate solutions converges at = 1.66 an = The explicit, analytic expression given by Eq.(32 contains the auxiliary parameter, which gives the convergence region an rate of approximation for HAM. However, the errors can be further be reuce by calculating higher orer approximations. This proves that HAM is a very useful analytic metho to get accurate analytic solutions to linear an strongly nonlinear problems. Simulation 2 imensional (β = 0: Figure 2 shows comparison between the exact solution an HAM results, analysis of iaphragm eflection over x-axis where α = 1, A = IJNS for contribution: eitor@nonlinearscience.org.uk

9 M. M. Khaer, N. H. Sweilam: An Application of Homotopy Analysis Metho for Estimation the Diaphragm Deflection 11 (a (b Figure 1: Comparison between the exact solution an HAM results when α = 0, β = 100, A = an = 0.1, = 2, t = 0.5. (a (b Figure 2: Comparison between the exact solution an HAM results when β = 0, α = 1, A = an = 0.1, = 2, t = 0.5. Simulation 2 imensional (general case: Figure 3 shows comparison between the exact solution an HAM results, analysis of iaphragm eflection over x-axis where α = 1, β = 100, A = IJNS homepage:

10 12 International Journal of Nonlinear Science, Vol.17(2014, No.1, pp (a (b Figure 3: Comparison between the exact solution an HAM results when α = 1, β = 100, A = an = 0.1, = 2, t = Convergence of the exact solution Liao [1] showe that whatever a solution series converges it will be one of the solutions of consiere problem. Liao ([1]- [8] presente to be controlle by the auxiliary parameter the rate of convergence the approximate solutions obtaine by HAM. HAM an the VIM [11] solutions of two-imensional iaphragm eflection in MEMS capacitive microphone when = 1.66 an h = Conclusions In this Letter, we use HAM for obtaining the iaphragm eflection of MEMS capacitive microphone in first funamental moe an using the PC-base Mathematica package for illustrate examples. By this metho a rapi convergent series is prouce. The results show that this metho provies excellent approximations to the solution of relate equation to iaphragm eflection with high accuracy an impression of influence parameter in iaphragm eflection will be more sensible. Finally, it has been attempte to show the capabilities an facile applications of HAM in comparison with the exact solution. Results show a goo agreement between the results obtaine using HAM an VIM. The numerical results showe that this metho has very accuracy an reuctions of the size of calculations compare with the VIM ([17]-[20] an the homotopy perturbation metho [21]. It may be conclue that this methoology is very powerful an efficient technique in fining exact solutions for wie class of problems. It is also worth noting to point out that the avantage of this methoology shows a fast convergence of the solutions by means of the auxiliary parameter, = 1.66, = HAM is very easy applie to both ifferential equations an linear or nonlinear ifferential systems. The approximate solutions were almost ientical to analytic solutions of the nonlinear evolution equations. Acknowlegments The authors are very grateful to the eitor an the referees for carefully reaing the paper an for their comments an suggestions which have improve the paper. References [1] S. J. Liao, The propose homotopy analysis technique for the solution of nonlinear problems, PhD thesis, Shanghai Jiao Tong University, IJNS for contribution: eitor@nonlinearscience.org.uk

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