COUPLING REQUIREMENTS FOR WELL POSED AND STABLE MULTI-PHYSICS PROBLEMS

VI International Conference on Computational Methos for Couple Problems in Science an Engineering COUPLED PROBLEMS 15 B. Schrefler, E. Oñate an M. Paparakakis(Es) COUPLING REQUIREMENTS FOR WELL POSED AND STABLE MULTI-PHYSICS PROBLEMS Jan Norström, Fatemeh Ghasemi Department of Mathematics, Computational Mathematics, Linköping University, SE-581 83 Linköping, Sween. e-mail: jan.norstrom@liu.se, web page: http://www.mai.liu.se/ janno11/ Department of Mathematics, Computational Mathematics, Linköping University, SE-581 83 Linköping, Sween e-mail: fatemeh.ghasemi@liu.se - Web page: http://www.mai.liu.se/ fatgh43/ Key wors: multi-physics problem, well pose problems, stability, coupling proceure, high orer finite ifferences, summation-by-parts operatorsm, weak interface conitions Abstract. We iscuss well-poseness an stability of multi-physics problems by stuying a moel problem. By applying the energy metho, bounary an interface conitions are erive such that the continuous an semi-iscrete problem are well-pose an stable. The numerical scheme is implemente using high orer finite ifference operators on summation-by-parts (SBP) form an weakly impose bounary an interface conitions. Numerical experiments involving a spectral analysis corroborate the theoretical finings. 1 INTRODUCTION Roughly speaking, a well pose initial bounary value problem require that a unique solution that can be estimate in terms of the ata, exist. The most common proceure for showing well poseness is the so calle energy-metho where one multiply the governing partial ifferential equations (PDEs) with the solution, integrate by parts an impose bounary conitions [1]. The same general knowlege is not wie-sprea when it comes to the mathematical coupling of multi-physics problems. The reason for that is the more complex an to some extent more unclear nature of coupling conitions compare to imposing bounary conitions. Firstly, accuracy relations must exist such that combinations of variables for one set of PDEs at the interface is equal to combinations of variables for the other set. Seconly, the number of accuracy relations must fit both problems. Too many conitions ruin existence an too few ruin uniqueness. If the number of accuracy relations are too few, aitional conitions requiring external ata must be ae. Thirly, the accuracy relations must be such that no artificial growth or ecay is generate. 1

We will investigate the problems mentione above an generalize the investigation in [, 3] where we erive the coupling conitions by only emaning a well pose problem. Coupling of hyperbolic PDEs of ifferent size at the interface will be our primary focus. Once the coupling conitions are known for the continuous multi-physics problem we will iscretize using high orer finite ifferences on summation-by-parts form an inclue the coupling conitions weakly using the SAT technique [4, 5]. THE MODEL PROBLEM We will consier the following system, an the scalar equation u t + Au x =, 1 x, t >, (1) u(x, ) = f(x), v t + bv x =, x 1, t >, () v(x, ) = g(x). In (1), u = (u 1, u ) T is a vector of unknowns, f(x) = (f 1 (x), f (x)) T is a vector of given ata an for simplicity we choose [ ] a A =, a >. (3) a Two bounary/interface conitions are neee for the system (1) while equation () nees one bounary/interface conition..1 The interface conitions We apply the energy metho to both equations an a them together to get t ( u + α v ) = u T Au x= + αbv x= = w T Ew, (4) where α is positive free weight, w = [u 1, u, v] T an E is a E = a. (5) αb In (4), the bounary terms at the outer bounaries x = ±1 are ignore. The eigenvalues of E are {a, a, αb}. If b <, one of the eigenvalues is positive an we nee one conition at x =, otherwise we nee two conitions.

In orer to couple the problems we nee at least one accuracy conition. Let The relation (6) inserte in equation (4) leas to v = C T u, C = [c 1, c ] T. (6) t ( u + α v ) = u T (, t)du(, t), (7) where D = (αbcc T A). The characteristic polynomial relate to the eigenvalues λ of D is λ αb(c 1 + c )λ + αabc 1 c a. (8) To simplify the following iscussion we let s 1 = αb(c 1 + c ) an s = αabc 1 c a, which yiels the roots λ 1, = s 1 ± s 1 s. (9) First we consier b <. This leas to a positive s 1. If c 1 c a/bα, then s an both roots of the characteristic polynomial are negative, which means that D is negative efinite. This means that if c 1 an c have opposite sign, the couple problems satisfy an energy estimate for all choices of α. But if c 1 an c have the same sign, the energy estimate is not satisfie for any value of α. Consequently the couple problems with the interface conition v = C T u satisfy an energy estimate for b < if an only if c 1 an c have opposite signs. Next, consier b >. This leas to negative s 1 an at least one of the eigenvalues must be positive, which means that we nee an aitional conition. As mentione above, two conitions are neee at x =. One of them is an interface conition an the other one must be such that the right-han sie of (7) is negative semi-efinite. We will refer to this aitional conition as a bounary conition. If c 1 c a/bα, then s an one of the eigenvalues of D is positive (λ + ) an the other one is negative (λ ). Let D = Y ΛY T an rewrite (7) as t ( u + α v ) = u T (, t)(y ΛY T )u(, t), (1) where Λ = iag{λ +, λ } an Y is the matrix of eigenvectors to D. Let Λ = Λ + + Λ, where Λ + = iag{λ +, } an Λ = iag{, λ }. Furthermore we have D = D + + D where D + = Y Λ + Y T an D = Y Λ Y T. Then (1) leas to t ( u + α v ) = (Y T u(, t)) T (Λ + + Λ )(Y T u(, t)), (11) 3

The most general conition base on (11) is (Y T + R r Y T )u(, t) = h(t), x =, (1) where Y + an Y are the eigenvectors relate to the positive an negative eigenvalues, respectively. Letting h(t) = an inserting (1) into (11) leas to t ( u + α v ) = (λ + R rλ + )(Y T u(, t)). (13) If λ + R rλ +, then the right-han sie of (13) is boune an we have a well-pose coupling. Note that with R r = we have the so calle characteristic bounary conitions. Consequently an energy estimate is obtaine if c 1 an c are chosen such that c 1 c a/bα. This means that c 1 an c must be less than an arbitrary positive number that we can choose. In short: all values of c 1 an c lea to a well-pose problem if b >.. The semi-iscrete problem Let A be an M N matrix an B a P R matrix. The Kronecker prouct of these matrices is efine as a 11 B a 1N B A B =. (14) a M1 B a MN B First, we consier b <. The semi-iscrete SBP-SAT formulations of (1) an () are, u t + (D u A)u = (P 1 u E u N Σ)(C T ũ N v )e u N, v t + bd v v = P 1 v σ(v C T ũ N )e v. (15) In (15), the outer bounary conitions are ignore as in the continuous case, D u,v = Pu,v 1 Q u,v are the ifference operators, P u,v are positive efinite matrices an Q u,v satisfy Q u,v + Q T u,v = iag[ 1,, 1]. The iscrete gri functions, relate to the gri vectors x u = (x = 1,, x N = ) an x v = (y =,, y M = 1) are u = (u 1, u, u 1N, u N ), v = (v,, v M ). (16) The vectors e u N = (,,, 1, 1)T an e v = (1,, ) T are N 1 an M 1, respectively. EN u = iag[,, 1] an Ev = iag[1,, ] are N N an M M, respectively. The penalty matrix Σ is given by [ ] σ1 σ Σ =, (17) σ 3 σ 4 4

σ is penalty parameter an ũ N = [u 1N, u N ] T. Next, consier b >. The semi-iscrete SBP-SAT formulations of (1) an () are, u t + (D u A)u = (Pu 1 EN u Σ)(C T ũ N v )e u N + (Pu 1 EN u Ξ)((I N H)u e u N h), v t + bd v v = Pv 1 σ(v C T ũ N )e v. (18) where the penalty matrix Ξ an H are given by [ ] [ χ1 χ Ξ = 1 Rr, H = χ 3 χ 4 ] Y T. (19) The bounary ata h is efine as h = [, h] T. Also, in the following analysis we will use the iscrete norms u P u I = u T (P u I)u, v P v = v T P v v. ()..1 Stability conitions at the interface First we consier b <. The iscrete energy metho is applie to (15) by multiplying the two equations with u T (P u I) an v T P v, respectively. The SBP properties of D u,v yiels t ( u P u I + α v P v ) = ũ T NAũ N + α bv + ũ T NΣH + α σv (v C T ũ N ). (1) In(1), α is a positive weight (not necessarily the same as in the continuous case) an H = [C T ũ N v, C T ũ N v ] T. In orer to mimic the continuous case, we choose σ = c 1 αb/ an σ 4 = c αb/ an σ 1 = σ 3 =. The final penalty matrix in block form becomes Σ = αb/ [ C ]. By inserting that into (1) we get t ( u P u I + α v P v ) =ũ T NDũ N + α v (b + σ) σv C T ũ N (αb + α σ). () If we choose σ = αb/α, for α α the right-han sie of () will be boune ue to the continuous result above. Next, we consier b > an let h(t) =. Multiplying (18) by u T (P u I) an v T P v leas to t ( u P u I + α v P v ) ũ T N(D + Ξ H + (Ξ H) T )ũ N, (3) where we have chosen Σ an σ as for the case b <. By using Y Y T = I, we can rewrite the right-han sie of (3) as ũ T N(D + Ξ H + (Ξ H) T )ũ N =(Y T ũ N ) T (Λ + (Y T Ξ HY ) + (Y T Ξ HY ) T )(Y T ũ N ). (4) 5

Let Ξ = Y T Ξ an choose Ξ such that Ξ = iag( χ 1, χ ). We also use the following split, [ ] (Y (Y T T ũ N ) = + ũ N ) (Y T. (5) ũ N ) Now, we can rewrite (4) as, [ ũ T N(D + Ξ H + (Ξ H) (Y T T )ũ N = + ũ N ) (Y T ũ N ) ] T [ λ + + χ 1 R r χ 1 R r χ 1 λ ] [ (Y T + ũ N ) (Y T ũ N ) ]. (6) By the choice χ 1 = λ +, the right-han sie of (6) can be rewritten as (λ + R rλ + )(Y T ũ N ) λ + ((Y T + ũ N ) R r (Y T ũ N )), (7) which is negative ue to the continuous result. Consequently if we choose Σ = αb/ [ C ] [ ] λ, σ = αb/α, Ξ = Y +, then for α α, (18) is stable... Stability conitions at the left bounary In orer to have a well-pose problem, we nee one conition at x = 1. We consier the homogeneous bounary conition with R l < 1. The SAT term at x = 1 is (Pu 1 [ ] π1 π Π =, Ĥ = (X T + R l X T )u( 1, t) =, (8) π 3 π 4 E u Π)(I N Ĥ)u, where [ 1 Rl ] X T, (9) an X is the matrix of eigenvectors to A. It can be shown that an energy estimate is obtaine if π 1 = a/(4(r l + 1)), π =, π 3 = a/(4(r l 1)), π 4 =. (3)..3 Stability conitions at the right bounary For the case b <, one conition at x = 1 is also neee. We choose the homogeneuos v(1, t) =. The SAT term at x = 1 is Pv 1 where θ satisfies θv N ev N θ b/. (31) 6

3 THE SPECTRUM In this section, we consier the continuous an iscrete spectrum for our problem. 3.1 The spectrum for the continuous problem By applying the Laplace transform to (1) an () we get the following system of orinary ifferential equations sû + Aû x =, 1 x, sˆv + bˆv x =, x 1. (3) We have ignore the initial conitions, since they o not influence the spectra an make the ansatz û = e kx ψ an ˆv = e k3x ψ 3. This leas to [ ] A (si + Bk)Ψ =, B =, (33) b where Ψ = [ψ 1, ψ, ψ 3 ] T. This system of equations have a non-trivial solution only when et(si + Bk) =, which leas to k 1 = s, k a = s an k a 3 = s. b The general solution incluing the eigenvectors is ŵ = α 1 e s a x 1 1 + α e s a x 1 1 + α 3 e s b x 1, (34) where ŵ = [û 1, û, ˆv] T. The unknowns α 1, α an α 3, will be etermine by the bounary an interface conitions. First we consier b <, with the conitions (X T + R l X T )u( 1, t) =, C T u(, t) v(, t) =, (35) v(1, t) =, an R l 1. The interface an bounary conitions are such that the couple problem is well-pose. By applying these conitions to (34), we obtain c 1 + c c 1 c 1 Eα =, E = e s a R l e s a, (36) e s b where α = [α 1, α, α 3 ] T. A non-trivial solution, require et(e) = e s b (e s a (c c 1 ) R l e s a (c1 + c )) =. (37) 7

The zeros of et(e) which form the spectrum of (3) are s = { a ln( R l(c 1 +c ) c c 1 ) + naπi, n Z, if R l(c 1 +c ) c c 1 >, a l(c 1 +c ) ) + naπi + aπi, n Z, if R l(c 1 +c ) c c 1 <. c c 1 (38) The real part of s is negative if R l(c 1 +c ) c c 1 < 1. It is easy to verify that this hols for R l < 1 an c 1, c with opposite sign. This means that if we choose c 1, c an R l such that the couple problem leas to an energy estimate, then the real part of s will be negative. Recall that this require that c 1 an c must have opposite signs. Next, consier b >. In this case we have the conitions (X T + R l X T )u( 1, t) =, C T u(, t) v(, t) =, (39) (Y T + R r Y T )u(, t) =, an λ + Rrλ + an Rl 1. The couple problems with the conitions (39) satisfy an energy estimate. By applying (39) to (34) leas to the following system of equations c 1 + c c 1 c 1 Eα =, E = e s a R l e s a, (4) y 1 R r y 11 + 1 R r y 1 R r y 11 1 + R r where α = [α 1, α, α 3 ] T. The zeros of et(e) in this case are { a s = ln( R l(y 1 R ry 11 +1 R r) y 1 +R ry 11 +1 R r ) + naπi, n Z, if R l(y 1 R ry 11 +1 R r) y 1 +R ry 11 +1 R r >, a ln( R l(y 1 R ry 11 +1 R r) ) + naπi + aπi, n Z, if R l(y 1 R ry 11 +1 R r) y 1 +R ry 11 +1 R r <. y 1 +R ry 11 +1 R r The real part of s is negative if (41) R l(y 1 R r y 11 + 1 R r ) y 1 + R r y 11 + 1 R r < 1. (4) Note that the eterminant of E is inepenent of c 1 an c. This means that if (4) hols, then the real part of s is negative for all c 1 an c. Recall that for all values of c 1 an c, suitable choices of α such that the couple problems satisfy an energy estimate coul be foun. This implies that there is no limitation on c 1 an c in both the energy an spectral analysis. However, recall that this require an aitional bounary conition. 3. The semi-iscrete spectrum Consier b <. The SBP-SAT approximation of (1) an (), incluing (35) is u t + (D u A)u = (Pu 1 E u 1 Π)(I N Ĥ)u + (Pu EN u Σ)(C T ũ N v )e u N, v t + bd v v = P 1 v σ(v C T ũ N )e v + P 1 v θv N e N. (43) 8

In orer to etermine the semi-iscrete spectrum, we follow [6] an rewrite (43) in matrix form as where W = [u 1, u, v] T an [ (Qu A) H i = bq v W t = P 1 (H i + H c )W, (44) ] [ P, P 1 1 u I = Pv 1 ]. (45) The penalty matrix H c which is zero except at the bounaries an interface has the structure ΠĤ... EC H c = T E σc T, (46) σ... θ where E = [σ, σ 4 ] T. Next, consier b >. The SBP-SAT approximation of (1) an (), with conitions (39) is u t + (D u A)u = (Pu 1 E u Π)(I N 1 Ĥ)u + (Pu EN u Σ)(C T ũ N v )e u N + (Pu 1 EN u Ξ)(I N H)u, v t + bd v v = Pv 1 σ(v C T ũ N )e v. (47) The approximation (47) can be written on the the form (44) where in this case ΠĤ... H c = EC T + Ξ H E σc T σ, (48)... while H i is the same as before an given above. The eigenvalues of the matrix P 1 (H i +H c ) form the iscrete spectrum of (43) an (47). 4 NUMERICAL RESULTS In this section, we use the metho of manufacture solution in orer to test the accuracy of the approximations. RK3 is use to iscretize time. We also iscuss the relation between the continuous an semi-iscrete spectrum. 9

SBP 1 SBP 4 SBP 63 SBP 84 N error rate error rate error rate error rate e- - e-3 - e-3-1e-3-4 6e-3 1.877 3e-4 3.6 1e-4 4.35 3e-5 5.376 8 1e-3.46 3e-5 3.4 8e-6 4.4 8e-7 5.39 16 4e-4 1.985 3e-6 3.5 4e-7 4.47 e-8 5.113 3 1e-4.4 4e-7 3.1 e-8 4.375 6e-1 5.91 64 e-5 1.998 6e-8 3.13 1e-9 4.77 e-11 5.47 Table 1: error an rate q u for b <. SBP 1 SBP 4 SBP 63 SBP 84 N error rate error rate error rate error rate e-1-3e- - 3e- - 4e-3-4 4e-.148 3e-3 3.154 1e-3 4.469 e-4 4.88 8 1e-.5 4e-4 3.46 6e-5 4.74 8e-6 4.741 16 e-3.14 5e-5 3.11 e-6 4.668 3e-7 4.916 3 6e-4.5 6e-6 3.3 1e-7 4.474 9e-9 4.798 64 e-4.1 7e-7 3. 4e-9 4.467 3e-1 4.83 Table : error an rate q v for b <. 4.1 Accuracy The analytical solution that we use in the metho of manufacture solution is q u = ln u 1 (x, t) = u (x, t) = cos(π(x t)), v(x, t) = sin(3π(x bt)). (49) The rate of convergence is calculate as ( (u N 1 1, u N 1 ) (u 1, u ) Pu I (u N 1, u N ) (u 1, u ) Pu I ) / ln ( N1 N ), q v = ln ( v N 1 v Pv v N v Pv ) / ln ( N1 N ),(5) where u 1, u an v are the analytical solutions an u N i 1, u N i an v N i are the corresponing numerical solutions with N i gri points. First, we consier b <. The choosen coefficients are α = α = 1, a = 1, b = 1. To have a well-pose problem, we choose R l < 1 an c 1, c such that c 1 c 1/. Let R l =.5 an c 1 = 1, c =. Tables 1 an show the error an convergence rate q u an q v, respectively, for SBP operators with th, 3 th, 4 th an 5 th orer. Next, we consier b >. We again choose α = α = 1, a = 1, b = 1 an take R l =.5, R r =.5, c 1 = 1 an c = 1 in orer to have a well-pose problem. Tables 3 an 4 show the error an convergence rates for q u an q v, respectively. Clearly, the esign orer of accuracy is obtaine. 1

SBP 1 SBP 4 SBP 63 SBP 84 N error rate error rate error rate error rate e- - 5e-3-3e-3 - e-3-4 6e-3 1.885 6e-4.951 e-4 4.186 1e-4 4.39 8 1e-3.6 8e-5.979 8e-6 4.367 3e-6 5.5 16 4e-4 1.983 1e-5.994 4e-7 4.49 8e-8 5. 3 9e-5.5 1e-6.999 e-8 4.39 e-9 5. 64 e-5 1.998 1e-7.999 1e-9 4.31 5e-11 5.193 Table 3: error an rate q u for b >. SBP 1 SBP 4 SBP 63 SBP 84 N error rate error rate error rate error rate 3e- - 8e-3-7e-3-3e-3-4 7e-3. 1e-3 3.83 4e-4 4.87 e-4 4.47 8 1e-3.3 1e-4.954 1e-5 4.53 7e-6 4.54 16 4e-4. e-5.983 7e-7 4.44 3e-7 4.773 3 1e-4. e-6.989 3e-8 4.437 9e-9 4.753 64 3e-5. 3e-7.994 e-9 4.436 3e-1 4.749 Table 4: error an rate q v for b >. 4. The spectrum of the continuous an semi-iscrete operators Figures 1-3 show the iscrete an continuous spectrum for ifferent gris using the SBP4 operator. One can clearly see the convergence of the iscrete spectrum to the continuous one as the gris are refine. This convergence hol both for positive an negative b an show that the solutions of the semi-iscrete scheme converge to the continuous one. 5 SUMMARY AND CONCLUSIONS We have iscusse well-poseness an stability of multi-physics problems by analyzing a moel problem. It was shown that for ceartain wave spees, only interface conitions were require, while in other cases aitional information in the form of bounary conitions must be supplie. By applying the energy metho, we erive bounary an interface conitions such that the continuous an semi-iscrete problem are well-pose an stable. The numerical scheme was implemente using high orer finite ifference operators on SBP form an weakly impose bounary an interface conitions using the SAT technique. It was shown that we obtaine esign orer of accuracy, an that the spectrum of the 11

6 4 N=8 N=16 N=3 N=64 Continuous 8 6 4 N=8 N=16 N=3 N=64 Continuous 4 4 6 6 1 9 8 7 6 5 4 3 1 8 6 5 4 3 1 Figure 1: Global view: the iscrete an continuous spectrum, b < (left) an b > (right). 1 5 5 15 1 Imaginary part 5 N=8 N=16 N=3 N=64 Continuous 5 5 N=8 N=16 N=3 N=64 Continuous 1 1 15 1.3 1.3 1.8 1.6 1.4 1. 1. 1.18 Real part 1.48 1.46 1.44 1.4 1.4 1.38 1.36 1.34 1.3 Figure : Meium view: the iscrete an continuous spectrum, b < (left) an b > (right). 4 4 6 8 1 1 N=8 N=16 N=3 N=64 Continuous 14 1.3 1.3 1.8 1.6 1.4 1. 1. 1.18 8 6 4 4 6 8 N=8 N=16 N=3 N=64 Continuous 1.43 1.4 1.41 1.4 1.39 1.38 1.37 1.36 1.35 1.34 Figure 3: Zoome view: the iscrete an continuous spectrum, b < (left) an b > (right). 1

iscrete operator converge to the spectrum of the continuous operator. The numerical experiments in combination with the theoretical erivations showe what type of analysis that is require to obtain accurate numerical simulations of multi-physics problems. Future work will inclue a generalization of this investigation for hyperbolic problems, an an extension to coupling of incompletely parabolic problems such as the compressible Navier-Stoke s equations. REFERENCES [1] Norström, J. an Svär, M. Well-pose bounary conitions for the Navier-Stokes equations. SIAM J. Numer. Anal. (5) 43:131-155. [] Norström, J. an Eriksson, S. Flui structure interaction problems: the necessity of a well pose, stable an accurate formulation. Commun. comput. phys(cicp). (1) 8:1111-1138. [3] Linström, J. an Norström, J. A stable an high orer accurate conjugate heat transfer problem. J. Comput. Phys. (1) 9:544-5456. [4] Norström, J. an Berg, J. Conjugate heat transfer for the unsteay compressible Navier-Stokes equations using a multi-block coupling. Comput Fluis. (13) 7:- 9. [5] Norström, J., Gong, J., Van er Weie, E. an Svär, M. A stable an conservative high orer multi-block metho for the compressible Navier-Stokes equations. J. Comput. Phys. (9) 8:9-935. [6] Norström, J. an Carpenter, M. H. High-orer finite ifference methos, multiimensional linear problems an curvilinear coorinates. J. Comput. Phys. (1) 173:149-174. 13