Control of homogeneous shear flow of multimode Maxwell fluids

J. Non-Newtonian Fluid Mech. 165 (2010) 136–142

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Control of homogeneous shear flow of multimode Maxwell fluids Evgeny Savelev, Michael Renardy ∗ Department of Mathematics, Virginia Tech, Blacksburg, VA 24061-0123, USA

a r t i c l e

i n f o

Article history: Received 5 November 2008 Received in revised form 28 August 2009 Accepted 5 October 2009

Keywords: Parallel shear flow Controllability

a b s t r a c t We consider a homogeneous parallel shear flow of a multimode Maxwell fluid. This problem results in a set of ordinary differential equations for the stresses. In this system, we view the shear rate as a control and consider the problem of steering the system to a given state of stress. The objective is to steer the system from given initial stresses to a final state of stress, allowing the shear rate to vary in an arbitrary fashion. We show that this problem is related to a calculus of variations problem. For the case of two modes, we obtain a characterization of the set of achievable streses. © 2009 Elsevier B.V. All rights reserved.

1. Parallel shear flow as a control system One of the questions posed in the field of control theory is whether a system can be steered from a given initial state to a desirable final state with a given class of control inputs. Such issues arise, for instance, in chosing injection rates to fill a mold, or in the use of boundary deformations or injection of fluid to influence turbulence. In general, the question which outputs can be achieved with a given class of inputs poses interesting and challenging mathematical problems. An extensive mathematical literature on controllability has been developed for classical systems of continuum mechanics, such as the Navier–Stokes equations or elasticity. A widely studied question of controllability is whether a body force prescribed in a part of the domain or boundary conditions on a part of the boundary are sufficient to control the state of the system (i.e. the deformation and velocity for elasticity and the velocity field for Newtonian fluids) at a given future time. A number of papers have analyzed the control of viscoelastic media in the same setting, i.e. controlling deformation and velocity for a solid or velocity for a fluid [3–7]. However, in the context of viscoelastic flows, in contrast to Newtonian fluids, the state of the system is not characterized by the velocity field alone. For fluids with constitutive relations of Maxwell type, we can characterize a state by velocities and stresses, and controllability problems are naturally posed by requiring that stresses be controlled in addition to the velocities [1] (controlling stresses can actually be more

∗ Corresponding author at: Department of Mathematics, Virginia Tech, 460 McBryde Hall, Blacksburg, VA 24061-0123, USA. Tel.: +1 540 231 8001; fax: +1 540 231 5960. E-mail address: [email protected] (M. Renardy). 0377-0257/$ – see front matter © 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.jnnfm.2009.10.006

important than controlling velocities; for instance in a manufacturing process stresses can survive solidification and cause subsequent deformation or damage). The governing equations consist of balance laws and constitutive relations, and it is natural to stipulate that control inputs should be restricted to act on the momentum equation or on the boundary conditions, since controlling the constitutive law would require an alteration of the material. In that context, linear viscoelastic flows can be shown to be only partially controllable [9]. This raises the question what happens in the nonlinear case. In general, the question to what extent viscoelastic stresses are accessible to attempts to control them and to what extent they are uncontrollable is likely to be very difficult. One of the simplest situations that can be considered is parallel shear flow. In this flow, there are only shear stresses if we restrict ourselves to linear viscoelasticity. Nonlinear effects, however, lead to normal stresses, and the question of controlling these becomes of interest; in particular, we can infer nothing here from the study of the linearized problem. In [10], a number of models are analyzed in homogeneous shear flows. Here, if we control the motion of the boundaries, we control the shear rate, and thus the shear rate has the role of the control. The question then is, to what stresses can we steer the fluid if we allow arbitrary variations of shear rate. For a variety of constitutive models, this question is answered in [10]. In particular, for the upper convected Maxwell (UCM) model, the achievable stresses are characterized by a positive definiteness condition. On the other hand, the situation for inhomogeneous shear flows is far more complex and not fully understood at this point [11]. If multimode Maxwell fluids are considered, there is obviously a positive definiteness condition for each stress component. However, it is easy to see that this necessary condition is not sufficient to characterize the set of achievable stresses. Below, we shall show how the characterization of achievable stresses is linked to

E. Savelev, M. Renardy / J. Non-Newtonian Fluid Mech. 165 (2010) 136–142

a problem in calculus of variations. The analysis of this calculus of variations problem will lead us to a complete characterization of this set in the case of a two mode Maxwell fluid. We shall also discuss extensions of the analysis and open issues for the case of more than two modes. 2. Governing equations and preliminaries

It is obvious from the above that each Si (T ) is non-negative. To determine the possible range of the Si , an essential step will be to find the minimum value of linear combinations of S1 and S2 . This leads to a calculus of variations problem. We can put this problem in the context of second order ordinary differential equations, but in order to do so, we need to make a substitution to introduce a new control variable in place of the shear rate ux . We note that the combination

We shall consider a two mode Maxwell fluid, i.e. the stress tensor has the form

q = 2 1 − 1 2

T = T1 + T2 ,

satisfies the equation

(1)

T˙ i + (v · ∇ )Ti − (∇ v)Ti − Ti (∇ v)T + i Ti = i (∇ v + (∇ v)T ).

(2)

We focus on the special case of parallel shear flow, where v = (0, u(x, t), 0),

Ti =

0 i 0

i i 0

0 0 0



.

(3)

The constitutive law above then specializes to ˙ i − 2i ux + i i = 0, ˙ i + i i = i ux .

(4)

Here i and i are the relaxation rate and modulus corresponding to the two modes. They are positive, and 1 = / 2 . We shall be concerned with the case where the shear flow is homogeneous, i.e. the shear rate ux and the stress components i and i are independent of x. In the analysis which follows, we view the shear rate as a control variable, and we pose the question which values of the stresses can be reached. More precisely: For given initial values 1 (0), 2 (0), 1 (0) and 2 (0), and with an arbitrary time dependence ux (t) of the shear rate, which values of the stresses 1 (T ), 2 (T ), 1 (T ) and 2 (T ) can be reached at a later time T ? As it turns out, the shear stresses i can take arbitrary values, and the main question is to characterize the achievable values of the normal stresses. These values depend on the initial values, and on the final values of the shear stresses, but we can partially remove this dependence by introducing “shifted” normal stresses, which below will be denoted by Si . The values Si are restricted by a positivity condition; for instance, it is well known that no matter what we do with the shear rate history in shear flow, the first normal stress difference will never be negative. The objective of the following analysis is basically the derivation of precise bounds which specify “how positive” the normal stress has to be. The special structure of the system (4), which is unique for the UCM model, gives the shear stress as a solution of a linear equation, and then the normal stress as a result of solving a second linear equation. The resulting closed form solution is



i (T ) = i (0)e−i T + e−i T

(7)

q˙ = 1 2 2 − 2 1 1 .

and Ti satisfies the constitutive equation



137

T

i ei t ux (t) dt, 0



i i (T )−i2 (T )=e−i T (i i (0)−i2 (0))+i e−i T

If 1 and 2 are square integrable, then not only q, but also its derivative are square integrable; that is, q is in the Sobolev space H 1 (0, T ). For the analysis, we shall formulate the problem in terms of q as the unknown variable, rather than ux . We can rewrite the shear stress components as 1 =

2 q + q˙ , (2 − 1 )2

2 =

1 q + q˙ , (2 − 1 )1

(9)

and the shear rate is given by ux =

q˙ + (1 + 2 )q˙ + 1 2 q . (2 − 1 )1 2

(10)

Since we want to steer the system from given initial shear stresses i (0) to given final values i (T ), we have, according to (7) and (8), prescribed values q(0) = Q0 ,

q(T ) = QT ,

˙ q(0) = P0 ,

˙ ) = PT . q(T

(11)

In terms of the q variable, the Si become



T

S1 (T ) = 1 e−1 T

e1 t 0

 S2 (T ) = 2

e−2 T

T

e

2 t

0

  q + q˙ 2 2 (2 − 1 )2

  q + q˙ 2 1 (2 − 1 )1

dt, (12) dt.

Our problem is now to determine which value of Si (T ) can be achieved with a q that satisfies given boundary conditions (11). We note that any given q can be changed in such a way that the boundary conditions for q˙ assume arbitary values while the H 1 norm of q (and consequently q(0), q(T ), S1 (T ) and S2 (T )) changes by an arbitrarily small amount. Consequently we shall henceforth ignore the boundary conditions for q˙ and consider the following problem: For given values q(0) = Q0 , q(T ) = QT , and q otherwise arbitary in H 1 (0, T ), characterize the set of possible values (S1 (T ), S2 (T )) given by (12). Let A be the set of admissible functions q, i.e.

(5)

T

A = {q ∈ H 1 (0, T ) | q(0) = Q0 , q(T ) = QT },

ei t i2 (t) dt. 0

This form of the solution makes it natural to introduce the new variable Si (t) = i i (t) − i (t)2 − e−i t (i i (0) − i (0)2 ).

(8)

(6)

For Si (T ) to be finite, it is obviously necessary and sufficient that i is square integrable, i.e. i ∈ L2 (0, T ), and hence our analysis will be based on shear stresses in that space. Note that Si (T ) is essentially i (T ), up to a shift that depends on i (T ) and the initial conditions. Our results are more conveniently stated in terms of Si rather than i .

(13)

and let S be the set of reachable values of (S1 , S2 ), S = {(S1 (T ), S2 (T )) | q ∈ A}.

(14)

It is instructive to consider the simple case Q0 = QT = 0. In that case, A is the Hilbert space H01 (0, T ), i.e. the set of all functions with square integrable derivative which vanish at the end points. The norm of a function in H 1 (0, T ) is defined as



1/2

T 2

q =

2

˙ dt |q| + |q| 0

,

(15)

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E. Savelev, M. Renardy / J. Non-Newtonian Fluid Mech. 165 (2010) 136–142

but on H01 we can define an equivalent norm by



1/2

T

˙ 2 dt |q|

q =

,

(16)

0

see e.g. Theorem 7.32 in [8]. Each of the Si is bounded, i.e. there is a constant C1 such that Si (T ) ≤ C1 q2 ,

(17)

and coercive, i.e. there is a constant C2 such that 2

C2 q ≤ Si (T ).

(19)

For a proof of these facts, see Lemma 3.1 below. The best possible constants for C3 and C4 can be found by minimizing, or, respectively, maximizing S2 for a given value of S1 . By the Lagrange multiplier method, this turns out to be equivalent to minimizing (maximizing) S2 + ˛S1 , and the Euler–Lagrange equations for this minimization problem lead to an eigenvalue problem for a second order ODE (with ˛ as the eigenvalue). The reachable set is a sector in the first quadrant of the (S1 , S2 ) plane bounded by the lines S2 /S1 = C3 and S2 /S1 = C4 . Whether the boundaries are included in S depends on whether the extremal values of S2 /S1 are actually achieved. The case of inhomogeneous boundary conditions is not quite as simple. To characterize S , we shall, in the following section, first consider the problem of finding extrema for linear combinations of S1 and S2 . Finding such extrema can be linked to linear ordinary differential equations. Whenever a linear combination of S1 and S2 has a minimum or maximum, this defines a half plane which contains S . The intersection of all such half planes is a convex set. Obviously, S is a subset of this convex set. We shall prove in Section 4 that, except possibly for boundary points, S conincides with this convex set. In Section 5, we shall give some numerical examples. In the final section of the paper, we discuss extensions to more than two relaxation modes. We can partly generalize the analysis, and consider the minimization of linear combinations of the Si to obtain a convex set which must contain S . A proof that S actually exhausts this set, however, remains elusive at this point. 3. Minimizing linear combinations of S1 and S2 As a preparation for the following analysis, we shall need the following lemma. Let



T

f (u, v) =

f (t)(u˙ + 2 u)(v˙ + 2 v) dt, 0

 g (u, v) =

T

vn (t) =

exp(1 ( − t))wn () d,

(1) We note that this in particular proves that I1 (v) is coercive, setting 1 = 0 we obtain I1 (v) ≥ Kv˙ 2L2 ≥ Mv2 1 . 0

(2) We want to consider I1 on the set A defined in (13) instead of H01 . For this, simply pick any fixed q0 ∈ A and set v = q − q0 , so that v ∈ H01 . We then have I1 (q) = I1 (q0 ) + 2f (q0 , v) + I1 (v).

It follows immediately from the lemma above that S1 (T ) and S2 (T ) are strictly convex and coercive on the set A and hence have a unique minimum. Let us now consider linear combinations of the form V˛ = S2 (T ) + ˛S1 (T ).

P(t)q¨ + Q (t)q˙ + R(t)q = 0, q(0) = Q0 , q(T ) = QT . 2 e2 (t−T ) 21 22 e2 (t−T ) 21

with given positive continuous functions f (t) and g(t) and arbitrary real numbers 1 and 2 . Then there is a constant C such that I2 (v) ≥ CI1 (v) for every v ∈ H01 (0, T ). Note that S1 (T ) and S2 (T ) are precisely of same form as I1 and I2 considered in the lemma. The expressions f and g are the blinear forms naturally associated with the quadratic forms I1 and I2 .

+˛

1 e1 (t−T )

+˛

22

21

,

21 e1 (t−T ) 22

e2 (t−T ) 1 2 (2 − 1 )

T

g(t)(v˙ + 1 v) dt

(23)

where

R(t) =

0

(22)

If ˛ is positive, this linear combination is convex, and by the lemma above it still remains convex when ˛ is slightly negative. On the other hand, if ˛ is negative with a large magnitude, then V˛ is concave and hence has a unique maximum. In between there is a range (˛1 , ˛2 ) in which V˛ is neither convex nor concave and hence assumes all real values. We know that ˛1 and ˛2 are both negative, and we can get bounds on them by looking at the Euler–Lagrange equations (see e.g. [2]). The Euler–Lagrange equations for an extremum of V˛ on the set A are

2

I2 (v) = g (v, v) =

(21)

Hence convexity of I1 on A is equivalent to convexity on H01 .

f (t)(v˙ + 2 v) dt, 0

(20)

H

2



t

which implies that the L2 norm of vn can be bounded in terms of the L2 -norm of wn . Remarks:

Q (t) =

T

I1 (v) = f (v, v) =

2

0

g(t)(u˙ + 1 u)(v˙ + 1 v) dt,



2

2f (t)[(v˙ n + 1 vn ) + (2 − 1 ) v2n ] dt,

it follows that the L2 -norm of vn tends to infinity. On the other hand, let wn = v˙ n + 1 vn . By assumption, the L2 -norm of wn is bounded. But we have

P(t) =

0

T

0

It follows that there are constants C3 and C4 such that

Lemma 3.1.



I1 (vn ) ≤



(18)

C3 S1 (T ) ≤ S2 (T ) ≤ C4 S1 (T ).

For the proof of the lemma, assume vn is such that I2 (vn ) = 1, but I1 (vn ) → ∞. Also, we shall assume 2 = / 1 , otherwise there is nothing to prove. Since

(24)

,

+˛

e1 (t−T ) 1 2 (1 − 2 ) 22

.

It is well known that a necessary condition for V˛ to be convex (concave) is that P(t) is always nonnegative (nonpositive); this is known as Legendre’s necessary condition, see e.g. [2]. As a consequence we deduce that ˛1 ≤ ˇ1 = −

2 22 1 21

,

˛2 ≥ ˇ2 = −

here we have assumed 2 > 1 .

2 22 1 21

e(1 −2 )T ;

(25)

E. Savelev, M. Renardy / J. Non-Newtonian Fluid Mech. 165 (2010) 136–142

It is natural to ask if these bounds for ˛1 and ˛2 are sharp. We shall now show that in general they are not. The functional that we investigate for convexity is



T

V˛ (q) = S2 (T ) + ˛S1 (T ) =

˙ dt, f (t, q, q)

(26)

0

139

is indeed negative for this combination of values. This function does not satisify v(0) = 0. Indeed, v(0) = 1 for any coefficient  we use. We can, however, circumvent this issue by making a small correction. Let



v˜ (t) =

v(t2 )w(t), if t < t2 , v(t), if t ≥ t2 .

(36)

˙ is where f (t, q, q) ˙ = f (t, q, q)

˙ 2 e2 (t−T ) (1 q + q) 2

(2 − 1 ) 21

2

+˛

˙ 1 e1 (t−T ) (2 q + q)

Here w(t) is continuous on [0, t2 ] and has the following properties:

2

2

(2 − 1 ) 22

.

(27)

Convexity means that



w(t) = 0, ı , t w(t) = 1,

˙ w(t) =

t ∈ [0, t1 ], t ∈ [t1 , t2 ],

(37)

t = t2 .

T

2 e

2 (t−T )

(v˙ + 1 v)

2

22

+ ˛1 e

1 (t−T )

(v˙ + 2 v)

2

21

dt ≥ 0

(28)

0

for v ∈ H01 . The value for ˇ2 found from the Euler–Lagrange equations above is ˇ2 =



(29)

T

˙ w(t) dt = ı (ln(t2 ) − ln(t1 )) = 1.



t w˙ 2 (t) dt = ı2 (ln(t2 ) − ln(t1 )) = ı.

(30)

Therefore



T

(31)

0

This integral is, in general, not easy to analyze, but we can study a limiting case. If 2 and 1 are close (1 ≈  ≈ 2 ), we can say that approximately: 1 2 − 1



T

(e2 t − e1 t )v˙ 2 − (2 − 1 )(1 e2 t + 2 e1 t )v2 dt 0

T

t e t v˙ 2 − 2e t v2 dt.

≈

(32)

0

For simplicity, we put T = 1. To see if any function makes this functional negative, we minimize it over the set of functions that satisfy

1

e t v2 dt = 1. Therefore, the problem converts into the minimization of the following integral: 0

T

t e t v˙ 2 + e t v2 dt

(33)

0

where  is a Lagrange multiplier. For this minimization problem, the Euler–Lagrange equations can be solved in closed form, yielding:

v(t) = e− t (C1 U(−n, 1,  t) + C2 Ln ( t)) n =

− . 

(34)

Here U is a confluent hypergeometric function and Ln is a Laguerre function. Since we are looking for a fuunction with zero boundary condition, we drop the U function, because it is singular at zero. We start with v(t) = e− t Ln ( t). It can be verified that for  = 2, there is a root of Ln () = 0 for  approximately equal to 2.65896, and that



1

te2t (Ln (2t)) − 4e2t (Ln (t))2 dt 2

0

1

2



1

t e t v˙ 2 − 4e t v2 dt.

(35)

(40)

t2

Here M = maxt ∈ [0,1] |Ln (2t)|.

After an integration by parts, this inequality transforms to (e2 t − e1 t )v˙ 2 − (2 − 1 )(1 e2 t + 2 e1 t )v2 dt < 0.

(39)

0

0

+ (21 e2 t − 22 e1 t )v(t)2 dt < 0.



t2

t e2t v˜˙ − 2e2t v˜ 2 dt ≤ ıe2t2 M 2 +

0



(38)

Now we have

(e2 t − e1 t )v˙ (t)2 + 2(1 e2 t − 2 e1 t )v(t)v˙ (t)



1

0

2 22 −e−T (2 −1 ) . 1 21

(provided 2 > 1 ). We want to show that the functional can already be nonconvex at this value. That is, we need to exhibit a v for which



We choose t2 and ı in advance so that ı and t2 are small. Then we choose t1 so that

(41)

If we choose ı and t2 sufficiently small, this integral is indeed negative. 4. Characterization of the reachable set Let q˛ denote the unique minimizer of V˛ if ˛ > ˛2 , or, respectively the unique maximizer if ˛ < ˛1 . We can include the limit ˛ = ∞ by letting q∞ denote the minimizer of S1 . If ˛ is outside the interval [˛1 , ˛2 ], the Euler–Lagrange equations have a unique solution and since the solutions of ordinary differential equations depend analytically on the coefficients, q˛ is an analytic function of ˛. The corresponding values (S1 (T, ˛), S2 (T, ˛)) also depend analytically on ˛ and they trace a curve in the (S1 , S2 ) plane which is analytic except for points where dS1 (T, ˛)/d˛ = dS2 (T, ˛)/d˛ = 0. It is clear that the slope of this curve is −˛. Points where both derivatives are zero can occur only for isolated values of ˛ unless both derivatives are identically zero. This occurs only if S1 and S2 have a common minimizer, as is the case when Q0 = QT = 0. In that case, the entire curve collapses into a single point. We want to know how this curve behaves as ˛ → ˛2 + (an analogous discussion applies for ˛ → ˛1 −). To avoid double indices in our notations, we shall write for ˛2 . The functional V is still convex, but not necessarily strictly convex. It may or may not have a minimum. There are three possible cases which may arise: (1) V does not have a minimum. In that case, the curve goes to infinity. (2) V has a minimum, but the minimum is not unique. In that case, the curve can be extended by a straight line segment that goes to infinity. (3) V has a unique minimum (this case is possible only if ˛2 = ˇ2 ). In that case the curve can also be extended by a straight line, the only difference to the previous case is that the line does not belong to S , but points arbitrarily close to it do. We now fill in the specifics of the argument.

140

E. Savelev, M. Renardy / J. Non-Newtonian Fluid Mech. 165 (2010) 136–142 Table 1 Parameter values used for the numerical examples. Variable Value

T 1

q(0) 2

q(T ) 1

1 1

2 2

1 1

2 2

and (U2 , W ) in S is in S . To verify this claim, consider the set {q ∈ A | S1 (T ) = U}.

(42) H 1 (0, T )

This set is the intersection of the closed subspace A of with a level set of S1 , which is topologically equivalent to a sphere. This intersection is connected; hence its projection onto the (S1 , S2 ) plane is connected. In summary, we have characterized the set of reachable values of S1 and S2 as a convex set bounded by the curve obtained by finding minima of linear combinations ˛1 S1 + ˛2 S2 and possibly straight line segments extending the ends of this curve to infinity. Fig. 1. Schematic drawing of the reachable set.

Consider a sequence ˛n → + and let qn be the minimizer of V˛n . Then qn is a minimizing sequence for V . If qn is bounded in H 1 , then it contains a weakly convergent subsequence, and the limit is a minimizer for V . If V does not have a minimum, qn must therefore be unbounded and consequently S1 (T, ˛n ) and S2 (T, ˛n ) diverge to infinity. In this case, therefore, the curve goes to infinity. Now consider the case where V has a minimum. If V is not strictly convex, then this minimum is not unique; rather it is assumed along an entire line in A. In the (S1 , S2 ) plane, this line corresponds to a half line which extends the curve to infinity. If the minimum is unique, then V is strictly convex but not coercive. Hence there are points q ∈ A with arbitrarily large S1 and S2 for which V (q) is arbitrarily close to its minimum value. In this case we can also extend by a half line extending to infinity; the only difference is that the points on the half line are not in S , but points arbitarily close to them are in S . The curve and, if applicable, the half lines extending it, bound a convex set in the (S1 , S2 ) plane which looks qualitatively like Fig. 1. At this point, we do not know if the interior of this convex set is contained in S or if there may be holes. To rule out the possibility of holes, assume that there are two points (U, W1 ) and (U, W2 ) which are both in S . We claim the line connecting these two points is also in S . Likewise, the line connecting any two points (U1 , W )

5. Numerical results In this section, we present some specific examples for the set S and show trends for its variation with parameters. For the computations, we used a combination of two methods: Solution of the Euler–Lagrange equations using a shooting method, and direct discretization of the minimization problem using a Ritz-Galerkin method. The former method provides the more efficient calculation, but the latter method ensures that our solution is actually a minimizer. For interpreting these results, recall that S1 and S2 are essentially the two components of the normal stress. Normal stresses in shear flow of the UCM fluid are restricted by a positive definiteness constraint, for instance, no matter what we do with the shear rate history, the first normal stress difference will never be negative. The Si are a precise measure of exactly “how positive” the two components of the normal stress have to be. The actual normal stress component i (T ) is of the form Si (T )/i + ki , where the shift ki depends on T , i (0), i (0) and i (T ). Hence a plot of the reachable set in the (1 (T ), 2 (T )) plane would simply be an appropriate shift of the set S . The parameters determining the set S are the boundary conditions for q and the fluid parameters i and i . In our numerical experiments, we start from the “base values” given in Table 1 and then vary the parameters one at a time. Fig. 2 shows the effect of varying Q0 or QT . If we increase Q0 from 0, we first see an expansion of the set S , and then eventually

Fig. 2. The sets S for various Q0 , QT . (a) Q0 variations; (b) QT variations.

E. Savelev, M. Renardy / J. Non-Newtonian Fluid Mech. 165 (2010) 136–142

141

Fig. 3. The sets S for various 1 , 1 . (a) 1 variations; (b) 1 variations.

a leftward shift. If, on the other hand, we increase QT from zero, we see a rightward shift and a rounding of the tip. As discussed above, the boundary of S always approaches a constant slope at infinity, and this slope is independent of Q0 and QT (since convexity of linear combinations of S1 and S2 does not depend on the boundary conditions). Fig. 3 shows the effect of varying 1 and 1 . We see that the set S shrinks considerably as 1 approaches 2 . Indeed, for this graph the values of the boundary conditions were changed to Q0 = 0.2 and QT = 0.1; if we had used the values from the table, the curves for the higher values of 1 would have moved out of the picture. We note that the difference 2 − 1 appears in the denominator of the expressions relating the i to q; this results in large values of S1 (T ) and S2 (T ) when this difference is small. We also note that in the limit 2 = 1 , we have S2 /S1 = 22 /21 , and hence we expect S to collapse to a line with slope 4. The graph indeed shows this trend. The original values from the table were used for the second part of the figure. The effect of varying 1 is primarily a rotation of S .

The formulation of the calculus of variations problem becomes a little more complicated if there are more than two relaxation modes, and we shall end this paper with a discussion of this aspect. Instead of the function q above, we now have the family of functions qi,j = j i − i j .

(44)

It is easily checked that k qi,j + i qj,k + j qk,i = 0,

therefore it suffices to consider the functions q1,i for i > 1, the remaining functions are linear combinations of these. Moreover, the q1,i are related to each other by differential relations, because 1 =

q˙ 1,i + i q1,i . (i − 1 )i

(46)

The quantities Si are defined as follows:





T

S1 (T ) = 1 e−1 T

e1 t 0



6. More than two relaxation modes We expect a similar picture when there are more than two relaxation modes. However, at this point, we do not have a complete analysis. For concreteness, we shall focus the following discussion on three relaxation modes; it is clear how it would be extended to the general case. Let Si for each relaxation mode be defined as in (6). As above, we can consider extreme values of linear combinations ˛1 S1 + ˛2 S2 + ˛3 S3 on the “admissible” set. Each time such a linear combination has a maximum or minimum, the reachable set S is confined in a half space, and, as above, the intersection of these half spaces is a convex set. Unfortunately, our argument above showing that the interior of this convex set does indeed belong to S does not extend to more than two relaxation modes, and we do not know how to prove this or if it is always true. We note that the values of more than two positive definite quadratic functionals do not necessarily form a convex set, as elementary examples show. For, instance, in three dimensional space, consider the three functions S2 = S3 =

2x12

+ x22 + x32 , (x1 − 1)2 + x22 + 3x32 .

S2 (T ) = 2

(43)

We note that (1, 2, 0) and (1, 2, 4) are possible combinations of values (assumed for x1 = ±1, x2 = x3 = 0). However, (1, 2, 2) is not possible, and neither is any point close to it. (Note that S1 = 1, S2 = 2 implies x1 = ±1, x2 = x3 = 0, and hence S3 is either 0 or 4.)



T

e−2 T

e

2 t

0





T

S3 (T ) = 3 e−3 T

e3 t 0

2 q1,2 + q˙ 1,2 (2 − 1 )2 1 q1,2 + q˙ 1,2 (2 − 1 )1 1 q1,3 + q˙ 1,3 (3 − 1 )1

2 dt,

2 dt,

(47)

2 dt.

The functions q1,2 and q1,3 are dependent on 1 through the following differential equations: q˙ 1,2 = −2 q1,2 + 2 (2 − 1 )1 , q˙ 1,3 = −3 q1,3 + 3 (3 − 1 )1 .

(48)

They act as a constraint on our minimization problem, therefore we include them into the functional to be minimized with appropriate Lagrange multipliers (p1,2 (t), p1,3 (t)). The Euler–Lagrange equations for the minimization problem will have a more elegant and compact form if we first replace q˙ 1,j in (47) by the right hand side of (48). The resulting equations now look like:



S1 = x12 + x22 + x32 ,

(45)

S1 (T ) = 1

T

e−1 T

e1 t (1 )2 dt, 0

 S2 (T ) = 2

T

e−2 T

e2 t 0

 S3 (T ) = 3 e−3 T

T

e3 t 0

   − q 2 2 1 1,2 1

   − q 2 3 1 1,3 1

dt,

dt.

(49)

142

E. Savelev, M. Renardy / J. Non-Newtonian Fluid Mech. 165 (2010) 136–142



We combine the original linear combination ˛i Si with the constraints multiplied by the appropriate Lagrange multipliers into a single functional J(q). This leads to the following minimization problem: Minimize J(q) =

T 0

f (t, q1,2 , q1,3 , q˙ 1,2 , q˙ 1,3 , 1 ) dt where

f (t, . . . , 1 ) = ˛1 F1 + ˛2 F2 + ˛3 F3 + p1,2 (t)E1 + p1,3 (t)E2 , F1 = 1 e1 (t−T ) (1 )2 , F2 = 2 e2 (t−T ) F3 = 3 e

   − q 2 2 1 1,2

   −1 q 2 3 1 1,3 3 (t−T ) 1

, (50) ,

E1 = −q˙ 1,2 − 2 q1,2 + 2 (2 − 1 )1 , E2 = −q˙ 1,3 − 3 q1,3 + 3 (3 − 1 )1 . Subject to the following boundary conditions: 0 , q1,i (0) = Q1,i

i = 2, 3,

The minimizer for this problem will have to satisfy the Euler–Lagrange equations, that briefly can be expressed as d D f = Dq12 f, dt q˙ 12 d D f = Dq13 f, dt q˙ 13

(52)

Finally, we couple the equations above with the equations from (48):

 p˙ 1,3 = 3

We have posed the problem of characterizing the set of stresses which can be achieved in a homogeneous shear flow of a multimode UCM fluid, with given initial stresses, and with the shear rate arbitrary and at our disposal. There are no restrictions on the shear stresses, but the normal stresses must satisfy a certain positivity condition. The precise nature of this restriction is not trivial. For the case of two modes, we have formulated an equivalent calculus of variations problem. A comprehensive analysis of this problem allows a complete characerization of the set of reachable stresses. We have also discussed the analogous method for more than two modes, partial results which can be obtained by it, and the open question which remains.

This research was supported by the National Science Foundation under Grant DMS-0707727. References

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0 = D1 f



7. Conclusions

Acknowledgment

T . q1,i (T ) = Q1,i

p˙ 1,2 = 2

This is a differential system of total order four, which must then be supplemented with the four boundary conditions for the q1.i .

2˛2 e2 (t−T ) (−q1,2 + 2 1 ) 21 2˛3 e3 (t−T ) (−q1,3 + 3 1 ) 21



+ 2 p1,2 ,

 + 3 p1,3 ,

0 = −21 (˛1 1 21 e1 (t−T ) + ˛2 2 22 e2 (t−T ) +˛3 3 23 e3 (t−T ) )+2(˛2 2 2 e2 (t−T ) q1,2 + ˛3 3 3 e3 (t−T ) q1,3 ) + (1 − 2 )21 2 p1,2 + (1 − 3 )21 3 p1,3 , q˙ 1,2 = −2 q1,2 + 2 (2 − 1 )1 , q˙ 1,3 = −3 q1,3 + 3 (3 − 1 )1 .

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