Numerical Methods, Existence and Uniqueness for the
Thermoelastics Systems with Moving Boundary
B. Santos
1, M. A. Rincon
2,
J. L´ımaco
1 ∗ 1Instituto de Matem´atica, Universidade Federal Fluminense, Rio de janeiro, Brazil
2
Instituto de Matem´atica, Universidade Federal do Rio de Janeiro, Rio de janeiro, Brazil
Abstract
In this work, we are interested in obtaining the existence, uniqueness of the solution and an approximated numerical solution for the model of linear thermoelasticity with moving boundary. We apply the finite element method with a finite difference method to obtain an approximated numerical solution. Some numerical experiments were presented to show the moving boundary’s effects in the problems in linear thermoelasticity.
Keywords: Thermoelasticity system; Moving Boundary; Finite Element Method; Finite Difference Method.
1
Introduction
Let Qt={(x, t) ∈ IR 2
; α(t) < x < β(t), 0 < t < T} be the non-cylindrical domain with boundary Σt=
[
0<t<T
{α(t), β(t)} × {t} and consider the following problem:
(I) ∂2u ∂t2 − ∂2u ∂x2 + η1 ∂θ ∂x = 0, ∀ (x, t) ∈ Qt ∂θ ∂t − k ∂2θ ∂x2 + η2 ∂2u ∂x∂t = 0, ∀ (x, t) ∈ Qt, u = θ = 0, ∀ (x, t) ∈ Σt, u(x, 0) = u0(x), ∂u ∂t(x, 0) = u1(x), θ(x, 0) = θ0(x); α(0) < x < β(0).
Existence and uniqueness of the elasticity linear and nonlinear inside bounded and unbounded cylin-drical domains, have been studied by several authors, among them, [3] and [4].
In this work, we will investigate the existence, uniqueness and approximated solution of the problem (I). We also will show the influence of moving boundary employing numerical examples. For this we consider the following hypotheses:
H1: α, β∈ C2([0, T ); IR),
with 0 < γ0= min
0≤t≤Tγ(t), where γ(t) = β(t)− α(t),
H2: ∃k1∈ IR, such that,
0 < k1< 1− (α0(t) + γ0(t)y) 2
, for 0≤ t ≤ T and 0 ≤ y ≤ 1. H3: k > 0, and η1.η2> 0.
We will now consider a change of variables to transform the domain Qt in a cylindrical domain Q.
Observe that, when (x, t) varies in Qt the point (y, t) of IR2, with y = (x− α(t))/γ(t) varies in the
cylinder Q = (0, 1)× (0, T ). Thus, we have the application, defined by T : Qt→ Q = (0, 1)×]0, T [
(x, t)7→ (y, t) = x − α(t)γ(t) , t,
(1)
belong C2. The inverse
T−1 is also C2. This technique that transforms the equation from moving
boundary in fixing boundary was initially employed by Medeiros at al. in [9] and [10].
Doing the change of variable v(y, t) = u(α(t) + γ(t)y, t) and φ(y, t) = θ(α(t) + γ(t)y, t) and applying in the problema (I), we obtain the following problem equivalent, defined in a cylindrical domain:
(II) ∂2v ∂t2 − ∂ ∂y a1(y, t) ∂v ∂y + a2(t) ∂φ ∂y + a3(y, t) ∂2v ∂y∂t+ a4(y, t) ∂v ∂y = 0, in Q ∂φ ∂t − b1(t) ∂2φ ∂y2 + b2(t) ∂2v ∂y∂t+ b3(y, t) ∂φ ∂y + b4(t) ∂v ∂y + b5(y, t) ∂2v ∂y2 = 0, in Q v = φ = 0; ∀ (y, t) ∈ Σ, v(y, 0) = v0(y), ∂v
∂t(y, 0) = v1(y), φ(y, 0) = φ0(y), for 0 < y < 1. where b1(t) = k/γ(t) 2 , b2(t) = η2/γ(t) , b3(y, t) =−(α0(t) + γ0(t)y)/γ(t) , b4(t) =−γ0(t)/γ(t) 2
, b5(y, t) = b3(y, t)/γ(t) , a1(y, t) = 1/γ(t) 2
−b3(y, t)
2
, a2(t) = η1/γ(t) , a3(y, t) = 2b3(y, t) , a4(y, t) =−(α00(t) + γ00(t)y)/γ(t).
Let (( , )), k · k and ( , ), | · |, be respectively the scalar product and the norms in H1
0(0, 1) and L 2
(0, 1). We denote by a1(t, v, w) and b1(t, v, w) the bilinear forms, continuous, symmetric and coercive, defined
in H1 0(0, 1) by a1(t, v, w) = Z 1 0 a1(y, t) ∂v ∂y ∂w ∂y dy, b1(t, v, w) = Z 1 0 b1(t) ∂v ∂y ∂w ∂y dy. (2)
2
Existence and Uniqueness
We shall first establish the existence and uniqueness of problem (II) as an auxiliary theorem and then prove the original problem (I).
Theorem 1 Under the hypotheses (H1), (H2) and (H3) and given the initial dates {u0, θ0} ∈ H01(Ω0)∩ H2(Ω0), u1∈ H01(Ω0),
there exists functions{u; θ} : Qt→ IR, solution of Problem (I) in Qt, satisfying the following conditions:
1. u∈ L∞(0, T ; H1
0(Ωt)∩ H2(Ωt)), u0 ∈ L∞(0, T ; H01(Ωt)), u00∈ L∞(0, T ; L2(Ωt)),
2. θ∈ L2(0, T ; H1
0(Ωt)∩ H2(Ωt)), θ0∈ L2(0, T ; H01(Ωt)).
Theorem 2 Under the hypotheses (H1), (H2) and (H3) and given the initial dates {v0, φ0} ∈ H01(0, 1)∩ H
2
(0, 1), v1∈ H01(0, 1),
there exists functions{v; φ} : Q → IR, solution of Problem (II) in Q , satisfying the following conditions: 1. v∈ L∞(0, T ; H1 0(0, 1)∩ H 2(0, 1)), v0 ∈ L∞(0, T ; H1 0(0, 1)), v00∈ L∞(0, T ; L 2(0, 1)), 2. φ∈ L2(0, T ; H1 0(0, 1))∩ H 2(0, 1), φ0 ∈ L2(0, T ; H1 0(0, 1)).
Proof of Theorem 2. To prove the theorem, we introduce the approximate solutions. Let T > 0 and denote by Vmthe subspace spanned by{w1, w2, ..., wm}, where {wν, λν; ν = 1,· · · m} are solutions
of the spectral problem ((wi, v)) = µ(wi, v), ∀v ∈ H01(0, 1). If{vm; φm} ∈ Vmthen its can be represented
by vm= m X ν=1 dνm(t)wν(y), φm= m X ν=1 gνm(t)wν(y) (3)
Let us consider{vm; φm} solutions of the system of ordinary differential equations,
(III) (v00 m, w) + a1(t, vm, w) + a2( ∂φm ∂y , w) + (a3 ∂v0 m ∂y , w) + (a4 ∂vm ∂y , w) = 0, (φ0 m, w) + b1(t, φm, w) + b2( ∂v0 m ∂y , w) + (b3 ∂φm ∂y , w) + (2b4 ∂vm ∂y , w) + (b5 ∂vm ∂y , ∂w ∂y) = 0, vm(0) = v0m→ v0, in H01(0, 1)∩ H 2 (0, 1), v0 m(0) = v1m→ v1 in H01(0, 1), φm(0) = φ0m→ φ0 in H01(0, 1)∩ H 2(0, 1),
where w∈ Vm. The system (III) has local solution in the interval (0, Tm). To extend the local solution
to the interval (0, T ) independent of m, the following estimates are necessary: A Priori Estimate
Taking w = v0
mand w = φm in the equation (III)1and (III)2, respectively, we get
1 2 d dt|v 0 m| 2 + a1(t, vm, vm0 ) + a2 ∂φm ∂y , v 0 m +a3 ∂v0 m ∂y , v 0 m +a4 ∂vm ∂y , v 0 m = 0 (4) 1 2 d dt|φm| 2 + b1(t, φm, φm) + b2 ∂v0 m ∂y , φm + b3 ∂φm ∂y , φm + 2b4 ∂vm ∂y , φm + + b5 ∂vm ∂y , ∂φm ∂y = 0 (5)
Note that, we have the following equalities and inequality: a1(t, vm, v0m) = 1 2 d dt a1(t, vm, vm)− 1 2 a0 1 ∂vm ∂y , ∂w ∂y , a2 ∂φm ∂y , v 0 m =−η1 η2 b2 ∂v0 m ∂y φm , a3 ∂v0 m ∂y , v 0 m =−γ 0 γ|v 0 m| 2 , b3 ∂φm ∂y , φm = γ 0 2γ|φm| 2 , (b5 ∂vm ∂y , ∂φm ∂y ) ≤ ckvmk 2 +1 2b1(t, φm, φm) (6)
Multiplying (4) by (η2/η1), adding to (5) and using (6) we have;
η2 2η1 d dt |v0 m| 2 + a1(t, vm, vm) + η1 η2|φ m|2 + b1(t, φm, φm) ≤ C|v0 m| 2+ kv0mk2+|φm|2 (7)
So, integrating (7), using that a1(t, v, w) and b1(t, v, w) are coercive forms and applying the Gronwall’s inequality, we get |v0 m| 2 +kvmk 2 +|φm| 2 + Z t 0 kφ mk 2 ≤ c1 |v1m| 2 +kv0mk 2 +|φ0m| 2 ec2T. (8) A Second Estimate
Taking the derivative with respect the t, of approximate systems (III)1,2, and also w = v00
m, w = φ0m, respectively, we obtain (v000 m, v 00 m) + a1(t, vm0 , vm00) + a2 ∂φ0 m ∂y , v 00 m +a3 ∂v00 m ∂y , v 00 m +(a0 3+ a4) ∂v0 m ∂y , v 00 m + a0 1(t, vm, v00m) + a0 2 ∂φm ∂y , v 00 m +a0 4 ∂vm ∂y , v 00 m = 0 (9) and (φ00 m, φ 0 m) + b1(t, φ0m, φ 0 m) + b2 ∂v00 m ∂y , φ 0 m +b3 ∂φ0 m ∂y , φ 0 m + (2b4+ b02) ∂v0 m ∂y , φ 0 m −b5 ∂v0 m ∂y , ∂φ0 m ∂y + b0 1(t, φm, φ0m)+ b03 ∂φm ∂y , φ 0 m + 2b04 ∂vm ∂y , φ 0 m +b05 ∂vm ∂y , ∂φ0 m ∂y = 0, (10)
We also have the following equality and inequality, a1(t, vm0 , v00m) = 1 2 d dt a 0 1(t, v 0 m, v 0 m)− 1 2a 0 1(t, v 0 m, v 0 m) a3 ∂v00 m ∂y , v 00 m = γ 0 γ|v 00 m| 2 a0 1(t, vm, v00m) = d dt a0 1 ∂vm ∂y , ∂v0 m ∂y −a00 1 ∂vm ∂y , ∂v0 m ∂y −a0 1 ∂v0 m ∂y , ∂v0 m ∂y a2 ∂φ0 m ∂y , v 00 m =−η1.b2 η2 ∂v00 m ∂y , φ 0 m b3 ∂φ0 m ∂y , φ 0 m =1 2 γ0 γ |φ 0 m| 2 a0 1 ∂vm ∂y , ∂v0 m ∂y ≤ Ckvmk2+ η2 4η1 a1(t, v0m, v0m), (11)
Multiplying (9) by (η1/η2), adding to (10) and using (11), we obtain
η2 2η1 d dt n |v00 m| 2 +a1(t, vm0 , v0m) + a0 1 ∂vm ∂y , ∂v0 m ∂y +α β |φ 0 m| o + b1(t, φ0m, φ 0 m) ≤ Ckvmk2+kv0mk 2+ |v00 m| 2+ kφmk2+|φ0m| 2 (12)
But from (III)1,5 we have that|v00 m(0)|
2and
|φ0 m(0)|
2 are bounded. Hence, integrating (12) with respect
a t, and applying the Gronwall’s inequality, we get kv0 mk 2 +|v00 m| 2 +|φ0 m| 2 + Z t 0 kφ 0 mk 2 ≤ C (13)
A Third Estimate Taking w =− ∂2
vm/∂y2and w =− ∂2φm/∂y2, in the approximate systems (III)1,2, we have,
v00 m,− ∂2 vm ∂y2 + a1 t, vm,− ∂2 vm ∂y2 + a2 ∂φm ∂y ,− ∂2 vm ∂y2 + a3 ∂v0 m ∂y ,− ∂2v m ∂y2 +a4 ∂vm ∂y ,− ∂2v m ∂y2 = 0 (14) and φ0 m,− ∂2φ m ∂y2 + b1 t, φm,− ∂2φ m ∂y2 + b2 ∂v0 m ∂y ,− ∂2φ m ∂y2 + 2b4 ∂vm ∂y ,− ∂2 φm ∂y2 +b5 ∂vm ∂y ,− ∂3 φm ∂y3 = 0 (15)
Note that, we have the following equalities: a1 t, vm,− ∂2v m ∂y2 = a1 t,∂vm ∂y , ∂vm ∂y + ∂a1 ∂y ∂vm ∂y , ∂2v m ∂y2 b1 t, φm,− ∂2φ m ∂y2 = b1 t,∂φm ∂y , ∂φm ∂y b5 ∂vm ∂y ,− ∂3 φm ∂y3 =b5 ∂2 vm ∂y2 , ∂2 φm ∂y2 − ∂b∂y5 ∂v∂ym,∂ 2 φm ∂y2 (16)
From (14), (15 and (16) and since that a1(t, v, w) and b1(t, v, w) are coercive forms, we obtain;
∂ 2 vm ∂y2 2 ≤ c6 kφmk2+|v00m| 2 +kv0 mk 2 +kvmk2 (17) ∂ 2 φm ∂y2 2 ≤ c7 |φ0 m| 2 +kφmk2+|v00m| 2 +kv0 mk 2 +kvmk2 (18)
The estimates obtained in (8), (13), (17) and (18), permit to pass the limits in the approximate systems (III)1,2, to the Galerkin method and then we get the solutions {v, φ} in the sense defined in
the Theorem 2.
Uniqueness of Solution
Let{ˆv, ˆφ} and {ev, eφ} be two solutions of Problem (II). Then v = ˆv − ev and φ = ˆφ− eφ are also solutions of Problem (II), with initial condition nulls. Then, multiplying the equation (II)1,2, respectively by
(η1/η2)v and φ, we obtain; |v0 |2 +kvk2 +|φ|2 ≤ c Z t 0 |v0 |2 +kvk2 +|φ|2 (19) So, applying Gronwall Lemma, we have|v0
|2+
kvk2+
|φ|2= 0 and therefore, we conclude that v = φ = 0
for all 0 < t < T . This, finishes the proof of Theorem 2. tu The Original Problem (I)
Proof of Theorem 1. Let{v, φ} be solution of Problem (II), with initial data, given by; v0(y) = u0 α(0) + γ(0)y, φ0(y) = θ0 α(0) + γ(0)y, v1(y) = u1 α(0) + γ(0)y+α0(0) + γ0(0)yu0 0 α(0) + γ(0)y
Considerer the functions u(x, t) = v(y, t) and θ(x, t) = φ(y, t), where x = α(t) + γ(t)y. To verify that u(x, t) and θ(x, t), under the hypotheses of Theorem 1, are a solution of problem (I), it is sufficient to observe that the mapping: (x, t)→ ((x − α)/γ , t) of the domain Qtinto Q = (0, 1)× (0, T ) is of class
C2. Since that 1. ∂u 2 ∂x2 = 1 γ ∂v2 ∂y2 , η1 ∂θ ∂x = a2 ∂θ ∂y, 2. u00= v00 −∂y∂ a1 ∂v ∂y + a3 ∂2v ∂y∂t+ a4 ∂v ∂y + 1 γ2 ∂2v ∂y2 3. k∂ 2θ ∂x2 = b1 ∂2φ ∂y2 , ∂θ ∂t = ∂φ ∂t + b3 ∂φ ∂y, 4. η2 ∂2 u ∂x∂t = b2 ∂2 v ∂y∂t+ b4 ∂v ∂y,
and from problem (II) we also have that{u, θ} satisfy the problem (I).
The regularity of {v(y, t), φ(y, t)} given by Theorem 2 implies that {u(x, t), θ(x, t)} is a solution of problem (I) and the uniqueness of the solution of problem (I) is a direct consequence of the uniqueness of problem (II).tu
3
Approximate Solution
Our goal in this section is the numerical implementation of approximate solutions. To obtain the numerical approximate solutions we will use both finite element method and finite difference method. Besides, some numerical experiments will be presented to analyze the effect of the moving boundary in the systems thermoelastics.
For convenience, our numerical analysis using finite elements methods approximation will be based on the equivalent problem (II) in the rectangular domain, in place the problem (I) where the domain is dependent of the time. We also will consider in the numerical simulations the following change in the boundary functions, α(t) =−K(t) and β(t) = K(t). Of course, that all the results are still valid.
Note that, now we have Qt=
(x, t)∈ IR2
; x = K(t)y, y∈ (−1, 1), t ∈ (0, T ) (20) being the non-cylindrical domain with boundary Σt=
[
0<t<T
{−K(t), K(t)} × {t}, and consequently we have the cylindrical domain Q = (−1, 1) × (0, T ). This form we obtain the following relation between the functions; u(x, t) = v(y, t) = v x K(t), t and θ(x, t) = φ(y, t) = φ x K(t), t . (21)
3.1
Variational Form of the Problem
Let us consider the following variational form, given by (III)1,2
(v00 m, w) + a1(t, vm, w) + a2 ∂φm ∂y , w +a3 ∂v0 m ∂y , w +a4 ∂vm ∂y , w = 0 (22)
and (φ0 m, w) + b1(t, φm, w) + b2 ∂v0 m ∂y , w +b3 ∂φm ∂y , w + 2b4 ∂vm ∂y , w −b5 ∂vm ∂y , ∂w ∂y = 0, ∀w ∈ Vm (23)
where now, using (20), the functions bi and ai are given by,
b1= k/K2(t), b2= η2/K(t), b3=−K0(t)y/K(t), b4=−K0(t)/K 2 (t), b5=−b3/K(t), a1= 1/K2(t)− b23, a2= η1/K(t), a3= 2b3, a4=−K00(t)y/K(t) (24)
Galerkin Methods and Approximation
Consider the functions{vm; φm} ∈ Vm defined in (3). Taking w = ϕj(y) and substituting in (22) and
(23), we obtain the systems of ordinary equations, given by A d00(t) +B(t) + E(t)d(t) + (a 2C)g(t) + D(t) d0(t) = 0 A g0(t) +b 1F + G(t) g(t) + (b2C)d0(t) + 2b4C + R(t) d(t) = 0 (25) where A = Z 1 −1
ϕi(y) ϕj(y) dy, B(t) =
Z 1 −1 a1 ∂ϕi(y) ∂y ∂ϕj(y) ∂y dy, C = Z 1 −1 ∂ϕi(y)
∂y ϕj(y) dy, D(t) = Z 1
−1
a3
∂ϕi(y)
∂y ϕj(y) dy, E(t) =
Z 1 −1
a4
∂ϕi(y)
∂y ϕj(y) dy, F = Z 1 −1 ∂ϕi(y) ∂y ∂ϕj(y) ∂y dy, G(t) = Z 1 −1 b3 ∂ϕi(y)
∂y ϕj(y) dy, R(t) = Z 1 −1−b 5 ∂ϕi(y) ∂y ∂ϕj(y) ∂y dy. (26) d(t) =d1(t),· · · dm(t) t and g(t) =g1(t),· · · gm(t) t .
For numerical convenient, using (24), we shall rewrite the B(t) matrix in a form B = B1+ B2, where
B1(t) = 1 K(t)2 Z 1 −1 ∂ϕi(y) ∂y ∂ϕj(y) ∂y dy, B 2 (t) =− K 0(t) K(t) 2Z 1 −1 (y)2 ∂ϕi(y) ∂y ∂ϕj(y) ∂y dy
3.2
Finite Element Approximation
We now present a semi-discrete formulation for problem (25) using the Galerkin finite element method to discretize the spatial variable. To boundary values not dependent of the time t, if we add (increase) that the boundary values are variables with the time t, then we obtain, like the Problem (II), coefficients aiand biall dependents of the time t and spatial variable y, with more reason, same for regular domains
Ω, is quite difficult to solve explicitly. Then some numerical methods may be used to find the solution approximately; the finite element method is just one such method which we now summarize. We first applied the method to obtain the approximation solution of the exact solution v(y, t) of the Problem (II) and after, using the transformation (21) we have the approximation solution of the u(x, t) for the Problem (I) in the domain Qt.
First, we divide the domain Ω = (0, 1) in local domain Ωi= (xi, xi+1). Then, Ω = int ∪mi=1Ω¯i
and Ωi∩ Ωj=∅, if i 6= j. In finite element methods, the ϕi are piecewise polynomials of some degree in
Ω and vanish on ∂Ω. More specifically, in this work, we have used the hat function, i.e, ϕi(y) = y− yi−1 h , ∀ y ∈ [yi−1, yi] yi+1− y h , ∀ y ∈ [yi, yi+1] 0, ∀ y /∈ [yi−1, yi+1] (27)
where we are consider the uniform mesh, h = hi = yi+1− yi, i = 1, 2, . . . , m in the discretization in
m-parts, with −1 = y1 < y2 < · · · < ym+1 = 1. Note that, if |i − j| > 2, then (ϕi, ϕj) = 0, and
(∂ϕi/∂y, ∂ϕj/∂y) = 0. Hence all the matrix of systems are tridiagonal.
Matrix Calculation
For each Ωi, we have to calculate each integrals defined in (26), using the functions (24), (27) and its
derived. Doing the calculus, we obtain, respectively the following elements, for each tridiagonal matrix A, B1 , B2 , C, D, E, F , G and R: aii= 4 3m , ai, i+1= ai+1, i= 1 3m , b1 ii= m K2 , b 1 i, i+1= b 1 i+1, i=− m 2K2 , b2 ii=− m(K0)2 3K2 3y2 i + 4 m2 , b2 i, i+1= b 2 i+1, i= m (K0)2 6K2 3y2 i + 6yi m + 4 m2 cii= 0, ci, i+1=− 1 2, ci+1, i= 1 2 dii= 4K0 3mK, di, i+1= K0 3K 4 m+ 3yi , di+1, i=− K0 3K 2 m + 3yi , eii = 2K00 3mK, ei, i+1= K00 6K 4 m+ 3yi , ei+1,i=− K00 6K 2 m+ 3yi , fii= m, fi, i+1= fi+1,i=− m 2 gii = 2K0 3m K, gi, i+1= K0 6K 4 m + 3yi , gi+1,i=− K0 6K 2 m + 3yi , ri, i=− mK0y i K2 , ri, i+1= ri+1, i= K0 2K2 1 + myi , (28)
3.3
Finite Difference Method
The equation (25) represent a system of ordinary differential equations of second order and due to matrices characteristics (dependence of the variables y and t) of system, obtaining the solution is not always possible. So, we will apply a numerical method to obtain the approximated solution for the system (25), using the approximate Newmark’s method (see, for instance, Hugles [6], pp 493).
Let dn = d(t
n) and gn = g(tn) be the approximate solution of the exact solution d(t) and g(t)
of (25)1,2, respectively, where we denote the discrete times in the interval [0, T ] by tn = n∆t, n =
To δ≥ 1/4, with δ ∈ IR, consider the following approximation d∗n= δ dn+1+ (1
− 2δ)dn+ δ dn−1
g∗n= δ gn+1+ (1
− 2δ)gn+ δgn−1, (29)
and for the first and second derivative, we take the difference operator in the following form τ dn= dn+1− dn−1 2∆t , τ g n= gn+1− gn−1 2∆t , δ 2 dn= dn+1− 2dn+ dn−1 ∆t2 (30)
which, for this approximate, the discrete error can be showed of order o(∆t2
). Coupled Systems
Let the systems (25)1,2 at the discrete mesh points tn = n∆t, and substituting the approximation (29)
and (30), we obtain the following coupled systems; b Andn+1+ bBn gn+1= bCdn − bDdn−1 − bEgn − bF gn−1 e Andn+1+ eBn gn+1= − eCdn+ eDdn−1 − eEgn+ eF gn−1 (31) where, we are denoting
b An= A + δ∆t2 (B1 )n+∆t 2 D n, Bbn= an 2 δ∆t 2 C b Cn= 2A − ∆t2 (1− 2δ) (B1 )n+ (B2 )n+ En b Dn= A + δ∆t2 (B1 )n −∆t2 Dn, Ebn= an 2 (1− 2δ)∆t 2 C b Fn= an 2 δ∆t 2 C , Aen= b n 2 2 C, Be n= A 2 + b n 1 δ∆t F e Cn= ∆t2bn 4 C + R n, Den= bn2 2 C e En= ∆tbn 1(1− 2δ) F + G n, Fen= A 2 − b n 1 δ∆t F (32)
To determine the solution {dn, gn
}, the coupled system of algebraic equations (31), may be solved by iteration, as follows: To start the iteration, we first taking n = 0 in (31) and rewrite the system as
b A0 d1 + bB0 g1 = bC0 d0 − bD0 d−1 − bE0 g0 − bF0 g−1 e A0 d1 + eB0 g1 =− eC0 d0 − eD0 d−1 − eE0 g0 + eF0 g−1
where the right-hand side is determined by the (starting) values, since the exact solutions{v(y, t); φ(y, t)} are known at time t = 0 and{v0; φ0
} are just the initial values, i.e, d0= v0(.) = v(., 0), g0= φ0(.) =
φ(., 0), where, we have used (3) and (27).
We can determine a close approximation {d−1; g−1
} by the second order Taylor extrapolation of {v(., t); φ(., t)} from t0= 0, viz; d−1= d0 − ∆t d0(0) +∆t 2 2 d 00(0), g−1= g0 − ∆t g0(0) (33)
that by the initial condition we have,
d0(0) = ∂v
and the values, d00(0) = ∂ 2 v ∂t2 (yi, 0), g 0(0) =∂φ ∂t (yi, 0),
are calculated from the field equation (II)1and (II)2, at t0= 0 and the initial values v0(.) = v(., 0) φ0(.) =
φ(., 0).
The system may be solved uniquely for {d1, g1
}, since its coefficient matrix is non singular. Having determined the values{d1, g1
}, then for n = 1, 2, · · · N, we obtain the approximate solution {dn+1, gn+1
} for the coupled system of algebraic equations (31), which can be rewrite in the block matricial form
b An Bbn e An Ben d n+1 gn+1 = Sn Tn (34)
when the time varies discretely over the interval [0, T ], where Sn = bCdn − bDdn−1 − bEgn − bF gn−1 and Tn= − eCdn+ eDdn−1 − eEgn+ eF gn−1.
The system (34) may be solved uniquely, since the matrix is non-singular. In order to solve the systems we can use de Gauss Elimination, LU factorization, as same as [5, 8] or Uzwa method [5].
Note that each square matrix of linear system have (m− 1) ordem, since that all matrix defined by (32) are (m− 1) ordem. So the linear systems have 2(m − 1) × 2(m − 1), with the block matrix and right-hand known by before iteration.
Uncoupled Systems Since{dn, gn
} must be solved jointly at each time step, the preceding numerical scheme is computation-ally coupled. From the numerical standpoint the coupled system is larger and hence harder to solve than an uncoupled system involving only dn+1 or only gn at each time step t
n. In order to get uncoupled
systems, see [1] and [11], we replace the central difference by the backward extrapolation for the first derivate, d0(t n) = 1 2∆t 3dn − 4dn−1+ dn−2 (35)
then substituting in the systems (25)1,2 together with (29) and (30), we obtain, after some simple
calculation, b An dn+1= bBndn − bCndn−1 − bDn gn+1 − bEngn − bFngn−1 e An gn+1= − eBndn+ eCn dn−1 − eDndn−2 − eEn gn+ eFngn−1 (36) where b An= A + δ∆t2 Bn 1 + ∆t 2 D n, Bbn= 2A − ∆t2 (1− 2δ)Bn 1 + B n 2 + E n, b Cn= A + δ∆t2 Bn 1 − ∆t 2 D n, Dbn = δan 2∆t 2 C, b En= (1− 2δ)an 2∆t 2 C, Fbn= δan2∆t 2 C, e An= 1 2 A + δb n 1∆tF, Ben= 1 2 3bn 2 + 4bn4∆t C + ∆tRn, e Cn= 2bn 2 C, Den = 1 4 Ce n, e En= ∆t(1 − 2δ) bn 1F + G n, Fen= 1 2 A− δb n 1∆tF. (37)
To start the iteration, we first taking n = 0 in (36)2 and after n = 0 in (36)1, so we have that e A0g1= − eB0d0+ eC0d−1 − eD0d−2 − eE0g0+ eF0g−1 b A0 d1 = bB0 d0 − bC0 d−1 − bD0 g1 − bE0 g0 − bF0 g−1 (38)
The terms of right-hand side of the (38)1, now involve the values
n d0
, d−1, d−2, g0
, g−1oare known
by (33) a least the term d−2, that can be calculated, taking n = 0 in (35);
d−2=
−3d0
+ 4d−1+ 2∆t d0(0) (39)
So, the value of g1is calculated in (38)
1 and substituting in (38)2 we obtain the value of d1. Having
determined the values d1
, then taking n = 1 in (36)2 we obtain the value g2 and taking n = 1 in
(36)1, we obtain de value of d2. Then the numerical scheme may be moved forward alternately between
(36)2 and (36)1. This numerical system is computationally uncoupled, since at each step tn, we have
two separate systems. Following the preceding numerical scheme, we obtain the values{gn+1, dn+1
} for n = 2, 3· · · , N. These values together with the starting values, constitute the finite element approximate solution to the initial boundary value problem based on Problem (II)
4
Numerical Simulation
A numerical example will be given to illustrate some features of the present model, using the method developed for the uncoupled systems that is faster. In the example, we need to determine the constants η1, η2and k, which give rise to the coupling of the parabolic and hyperbolic equation in the thermoelastic
system (I). The constants are given by the following formulas η1= α(3λ + 2µ)√θ0 p (λ + 2µ)ρ c l2, η2= lη1, k = k c lpρ(λ + 2µ).
where c the is specific heat; α the is coefficient of thermal expansion; k the is thermal conductivity; l = 2K(0) the is the length of the string; ρ the is density of the string; θ0 the is initial temperature; λ
and µ are the coefficients of Lam´e.
For the numerical example, these values will be calculated from the physical properties of aluminum. In this case, we have µ = 26.24× 109and λ = 58.41
× 109. Using the thermal and mechanics properties
from aluminum, we obtain the approximate values η1= 0.164, η2= 0.161 and k = 0.177.
Let us considerer in (29) the weight δ = 0.5, Ω = (−K(t), K(t)) divided in m subintervals, i.e, h = 2/m and ∆t = T /N , for different values of N and T for the discrete time. To calculate the coefficients defined in (24) step by step, the function K(t) that defines the time dependence of the boundary for the non-cylindrical domain Qt in (20) must be given. In this example it is given by
K(t) = 1− 1/ exp (t + 1). Note that, in this case, Qttends to Q rapidly as t increases. This particular
function was taken in order to satisfy the hypothesis H2, i.e, K0(t)
≈ 1. From the physical point of view, we require that the speed of the end points be less than the ”characteristic” speed of the system. When we consider only a wave equation for small vibrations of elastic string or beam equation, both with moving boundaries, are not require the monotonicity of those functions, see [2] and [7].
We consider the initial temperature, the initial position and velocity given by
φ0(y) = 0.033(1− y2), v0(y) = 0.057(y2− 1) and v1(y) = 0. (40)
In all the figures we are considering the change of variables y = (x− α(t))/γ(t)).
To obtain the figures Fig.1 - Fig.4, we have used ∆t = 0.03 and h = 0.02, with N = m = 100 and T = 3. Fig.1 and Fig.2, respectively, shows the temperature θ(x, t) and the displacement u(x, t) in the midpoint x = 0.
20 15 10 5 0 0.04 0.03 0.02 0.01 0 -0.01
Fig.1: Temperature at midpoint θ(0, t)
20 15 10 5 0 0.06 0.04 0.02 0 -0.02 -0.04 -0.06
Fig.2: Displacement at midpoint u(0, t)
Fig.3 and Fig.4, shows the approximate solution θ(x, t∗) and u(x, t∗), in the interval [0, T ] = [0, 3] for
different values of t∗, t∗= 0, 0.25, 0.75, 1.5, 2.25, 3.0. In these figures, we can see that the maximum
temperature and displacement decrease as time increases.
1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 0.04 0.02 0 -0.02 Fig.3: θ(x, t∗) at t∗= 0, 0.25, 0.75, 1.5, 2.25, 3.0
1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 0.06 0.04 0.02 0 -0.02 -0.04 -0.06 Fig.4: u(x, t∗) at t∗= 0, 0.25, 0.75, 1.5, 2.25, 3.0
Note that the interval of the boundary has varied from [−0.63, 0.63] to [−0.98, 0.98].
To obtain Fig.5 and Fig.6, we have used ∆t = 0.125 and h = 0.05. In Fig.5 and Fig.6 the evolution of the displacement function u(x, t) and the evolution of the temperature function θ(x, t) are plotted, showing the profile of the displacement and temperature, where the time varies from 0 to 8.
u(x, t) 0.04 0.03 0.02 0.01 0 -0.01 x 1 0.5 0 -0.5 -1 t 4 5 6 7 8 3 2 1 0
Fig.5: Displacement with 80 steps of time
In Fig.5, we show a continuous profile of the displacement in time, which show clearly the amplitude of the displacement decreases. Similar results are valid also for the temperature profile.
θ(t, x) 0.06 0.03 0 -0.03 -0.06 x 1 0.5 0 -0.5 -1 t 5 6 7 4 3 2 1 0
Fig.6: Temperature with 80 steps of time
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[2] Clark, H. R.; Rincon, M. A., Rodrigues, R. D.: Beam Equation with Weak-Internal Damping in Domain with Moving Boundary, Applied Numerical Mathematics, Vol. 47, Fasc.2, (2003), 139-157. [3] Dafermos, C.M.:On the existence and the asymptotic stability of solution to the equations of linear
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[4] Dassios, G. ; Grillakis, M.: Dissipation rates and partition of energy in thermoelasticity, Arch. Rational Mech. Anal. 87-1, (1984), 49-91.
[5] G.H. Golub ; C.F Van Loan: Matrix Computations, 3 rd ed., Johns Hopkins U. Press, Bltimore, (1996).
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