Numercal Algorthms 14 (1997) 69 93 69 Immersed nterface methods for movng nterface problems Zhln L Department of Mathematcs, Unversty of Calforna at Los Angeles, Los Angeles, CA 90095, USA E-mal: zhln@math.ucla.edu Receved May 1995; revsed December 1995 Communcated by T. F. Chan A second order dfference method s developed for the nonlnear movng nterface problem of the form u t + λuu x =(βu x) x f(x, t), x [0,α) (α, 1], dα = w(t, α; u, ux), dt where α(t) s the movng nterface. The coeffcent β(x, t) and the source term f(x, t) can be dscontnuous across α(t) and moreover, f(x, t) may have a delta or/and delta-prme functon sngularty there. As a result, although the equaton s parabolc, the soluton u and ts dervatves may be dscontnuous across α(t). Two typcal nterface condtons are consdered. One condton occurs n Stefan-lke problems n whch the soluton s known on the nterface. A new stable nterpolaton strategy s proposed. The other type occurs n a one-dmensonal model of Peskn s mmersed boundary method n whch only jump condtons are gven across the nterface. The Crank Ncolson dfference scheme wth modfcatons near the nterface s used to solve for the soluton u(x, t) and the nterface α(t) smultaneously. Several numercal examples, ncludng models of ce-meltng and glacaton, are presented. Second order accuracy on unform grds s confrmed both for the soluton and the poston of the nterface. Keywords: mmersed nterface method, Stefan problem, movng nterface, dscontnuous coeffcents, sngular source term, mmersed boundary method, Cartesan grd, heat conducton. AMS subject classfcaton: 65N06, 76T05, 80A, 86A40. Ths work was supported by URI grant #N0001409-J-1890 from ARPA, NSF Grant DMS-9303404, and DOE Grant DE-FG06-93ER5181. Present address: Department of Mathematcs and Statstcs, Msssspp State Unversty, Msssspp State, MS 3976, USA. J.C. Baltzer AG, Scence Publshers
70 Z. L / Movng nterface problems 1. Introducton In ths paper we study the mmersed nterface method for the one-dmensonal movng nterface problem u t + λuu x =(βu x ) x f(x, t), x [0,α) (α, 1], (1) dα dt = w( t, α; u,u +,u ) x,u+ x, t > 0, () where w s a known functon and u, u +, u x,andu + x are the lmtng values of u(x, t) and u x (x, t) from the left and rght hand sde of a movng nterface α(t). The coeffcent β(x, t) > 0 and the source term f(x, t) may be dscontnuous; furthermore, f(x, t) may have a delta or/and delta-prme functon sngularty at the nterface α(t). Ths s a parabolc problem and the soluton s pecewse smooth. The dscontnutes can only occur on the nterface α(t). Intal and boundary condtons, whch are not our man nterests here, may also be mposed. We assume these can be handled usng conventonal technques. The nterface α(t) dvdes the soluton doman nto two parts: 0 x<α(t) and α(t) <x 1. The soluton n each doman [0,α(t)) and (α(t), 1] s smooth, but coupled wth the soluton on the other sde by nterface condtons (or nternal boundary condtons) whch usually take one of the followng forms: Case 1. The soluton on the nterface s gven. One example s a mathematcal model for soldfcaton problems. When trackng an nterface of meltng ce, for example, the temperature at the meltng/freezng nterface s gven by the meltng temperature of the ce. In ths case u(α, t) =u 0 s the meltng temperature. When λ 0, ths s the classcal Stefan problem. Varous approaches have been used to solve Stefan or other lnear free or movng nterface/boundary problems numercally [1, 4, 6, 8, 10, 11, 13, 0, 4, 7], and recent work by Chen et al. [5] usng the level set approach. Compared to case dscussed below, the Stefan problem s easer to solve because the value of the soluton on the nterface s known. However, few numercal methods are second order accurate n the nfnty norm for both the soluton and the nterface. Most methods nvolve some transformatons ether for the dfferental equatons or the coordnate system, whch complcates the problem n some way. Some of the methods would exhbt some knd of nstablty near the nterface as explaned n secton 4. The method proposed n sectons 6 s smpler, more stable, and second order accurate both for the soluton u and the nterface α(t) smultaneously for more general equatons. Furthermore, the soluton s obtaned usng a fxed Cartesan grd. Case. Jump condtons of the form [u] def = u(α +,t) u(α,t)=q(t), (3) [βu x ] def = β(α +,t)u x (α +,t) β(α,t)u x (α,t)=v(t) (4)
Z. L / Movng nterface problems 71 are gven. Our nterest n ths case was motvated by the desre to develop a second order accurate algorthm for solvng the ncompressble Naver Stokes equatons obtaned from the formulaton of the mmersed boundary method. Ths method was orgnally developed by Peskn and hs co-workers for studyng blood flow n a beatng heart [, 3], but has been used n a wde varety of other problems [, 7, 9, 5]. In Peskn s method the physcal doman s mmersed n a rectangular regon. The boundary condton s treated as a forcng term whch s only supported on the boundary, therefore the forcng term s sngular. Case s a one-dmensonal model for the mmersed boundary method formulaton wth a more general equaton for the moton. A lnear model n one dmenson would be u t =(βu x ) x C(t)δ ( x α(t) ) Ĉ(t)δ ( x α(t) ) f(x, t). From ths equaton, assumng f(x, t) s contnuous, we can derve the followng jump condtons [u] = Ĉ(t) β +β +, [βu x]=c(t), where β and β + are the lmtng values of β(x, t) from left and rght of the nterface α(t). In many other cases the jump condton can be obtaned from physcal reasonng. For nstance, f u stands for temperature, then [u] =0; and [βu x ]=v(t)means that the net heat flux across the nterface s equal to the source strength v(t). One of the advantages of usng the jump condtons s that we do not need to dscretze the delta functon, an approach descrbed brefly n the next paragraph. Peskn and many other people use a dscrete delta functon approach to spread the sngular force term to the nearby grd ponts and use the dscrete delta functon agan to nterpolate the velocty feld to move the boundary. Whle ths approach seems to work, t s usually only frst order accurate and dffcult to analyze especally n two or hgher dmensons. Varous attempts have been made to analyze and mprove the dscrete delta functon approach or to try to solve the resultng dfferental equatons dfferently. Beyer and LeVeque [3] studed varous one-dmensonal movng nterface problems for the heat equaton assumng a pror knowledge of the nterface. A dscrete delta functon s carefully selected and some correcton terms are added f necessary n ther approach to get second order accuracy. Wegmann and Bube [6] recently appled the mmersed nterface method for certan one-dmensonal nonlnear problem wth a fxed nterface. However, for the nterface problems dscussed here the nterface s unknown and movng, and the dscrete dfference scheme s a nonlnear system of equatons nvolvng both the soluton and the nterface. In two dmensons, the dscrete delta functon approach generally s only frst order accurate. And t seems mpossble to fnd a generc dscrete delta functon whch makes the algorthm second order accurate. A new approach, the mmersed nterface method [15, 17], s ntended to solve general PDEs wth dscontnuous coeffcents and/or sngular sources wth second order accuracy at all grd ponts ncludng those
7 Z. L / Movng nterface problems whch are close to or on the nterface. The man dea s to ncorporate the known jumps n the soluton or ts dervatves nto the fnte dfference scheme, obtanng a modfed scheme whose soluton s second order accurate at all grd ponts on the unform grd even for a qute arbtrary nterface. Ths method has been mplemented for several dfferent applcatons n one, two and three dmensons [16, 18, 19]. By mplementng ths method for the one dmensonal model here, we not only effectvely solved some general one-dmensonal movng nterface problems, but also hope to get some nsghts nto the method as well as the problems. We are currently workng on second order accurate mmersed nterface methods for the full Naver Stokes equaton wth a movng boundary, where we have to deal wth the nonlnear term u u.that s one of the reasons why we have the nonlnear term λuu x n our model equaton (1). Case (wth λ = 0) s also a model of heat conducton wth an nterface between two dfferent materals. Now u s the temperature and hence s contnuous, meanng q(t) 0 n (3). The net heat flux across the nterface s v(t) n (4), the source strength on the nterface. Agan, n ths case we do not know the value of the soluton on the nterface but only the jump condtons. For many classcal Stefan problems, the moton of the nterface s proportonal to the flux across the nterface dα dt = σ(t)[βu x], u(α, t) =u 0, (5) where u 0 s the known temperature at the nterface. Ths knd of problem fts both cases 1 and and wll be dscussed n secton 4.3.. Computatonal frame We use a unform grd x = h, = 0, 1,...,N, x 0 =0, x N = 1, where h s the step sze n space. We use k as the step sze n tme and assume the rato k/h s a constant so that we can wrte O(k) as O(h) or vce versa. Usng the Crank Ncolson scheme, the sem-dscrete dfference scheme for (1) can be wrtten n the followng general form, u n+1 = 1 u n k Q n+1/ + λ ( u n u n x, + u n+1 u n+1 ) x, ( (βux ) n x, +(βu x ) n+1 x, ) 1 (6) ( f n + f n+1 ), where u n x, and (βu x) n x, are dscrete analogues of u x and (βu x ) x at (x,t n ), and Q n+1/ s a correcton term needed when α crosses the grd lne x = x at some tme between t n and t n+1, as dscussed n the next secton. Numercal schemes wll be
Z. L / Movng nterface problems 73 (a) (b) α( t) α() t t n+1 j t n α n τ j τ Fgure 1. Interface crossng the grd. (a) α(t) ncreases wth tme. (b) α(t) decreases wth tme. dsplayed n a box, as llustrated by equaton (6). For smplcty, we wll drop the superscrpt n durng the dscusson of the space dscretzaton f there s no confuson. At a grd pont x, whch s away from the nterface (.e., α [x 1,x +1 ]), the classc central dfferences wll be used u x, = u +1 u 1, (7) h (βu x ) x, = β 1/u 1 (β 1/ + β +1/ )u + β +1/ u +1 h, (8) where β +1/ = β(x +h/, :). We wll dscuss how to dscretze u x and (βu x ) x when α [x j 1,x j+1 ] for cases 1 and n secton 4. The nterface locaton s also determned by the trapezodal method appled to () α n+1 α n k = 1 ( w n + w n+1), (9) where w l = w ( t l,α l ; u,l,u +,l,u,l x,u +,l ) x,andα l,u ±,l, u ±,l x are the approxmatons of α(t l ), u(α ±,t l ),andu x (α ±,t l ) respectvely. These quanttes are only defned on the nterface. We use the followng notaton to express the jump n a functon g(x, t) across the nterface [g] def = g(α +,t) g(α,t); [g] ;t def = g(α, t + ) g(α, t ). It s easy to see that [g] =±[g] ;t and the sgn depends on the moton of the nterface. For example we have mnus sgn for the case n fgure 1(a), and plus sgn for the one n fgure 1(b). The kernel of the algorthm at a tme level t n conssts of the followng: Determne Q n+1/ j t n+1. f the nterface crosses the grd lne x = x j from tme t n to
74 Z. L / Movng nterface problems Derve the dfference formula for u x and (βu x ) x at the two grd ponts closest to the nterface. Compute the quanttes u ±, u ± x, [u t], etc. whch are only defned on the nterface. Solve the nonlnear system of equatons for u n+1 } and the locaton of the nteface α n+1. Each of these steps wll be descrbed n detal n the remander of the paper. Away from the nterface, the local truncaton errors for the dfference scheme are O(h ).But at a few grd ponts near the nterface, we allow the local truncaton errors to be O(h) based on the fact that the local truncaton error of a dfference scheme on a boundary can be one order lower than those of nteror ponts wthout affectng global second order accuracy. 3. Grd crossng If there s no grd crossng at a grd pont x from tme t n to tme t n+1, meanng that (x,t n )and (x,t n+1 )are on the same sde of the nterface α(t),.e., x (α n,α n+1 ), then we can take Q n+1/ = 0 and we have u(x,t n+1 ) u(x,t n ) = 1 ( ut x,t k [ n+1) ( +u t x,t n)] + O ( k ). (10) However, f the nterface crosses the grd lne x = x j at some tme 1 τ, t n <τ< t n+1, such that x j = α(τ), see fgure 1, then the tme-dervatve of u s not smooth. In ths case even though we can approxmate the x-dervatves well at each tme level (see secton 4), the standard Crank Ncolson scheme needs to be corrected to guarantee second order accuracy. Ths s done by choosng a correcton term Q n+1/ j based on the followng theorem: Theorem 3.1. Suppose the equaton α(t) =x j has a unque soluton τ n the nterval t n <t<t n+1. If we choose Q n+1/ j = [u] ;τ k + 1 ( t n+1/ τ ) [u t ] k ;τ (11) then u(x j,t n+1 ) u(x j,t n ) Q n+1/ j = 1 ( ut xj,t k ( n+1) ( +u t xj,t n)) + O(k). (1) Proof. We expand u(x j,t n ) and u(x j,t n+1 ) n Taylor seres about tme τ from each sde of the nterface to get u ( x j,t n) =u(x j,τ )+ ( t n τ ) u t (x j,τ )+O ( k ), 1 The crossng tme τ really depends on the grd ndex j as well as the tme ndex n, see fgure 1. To smplfy the notaton, τ wll be used to ndcate the crossng tme, wthout explctly showng ts dependence on j and n.
Z. L / Movng nterface problems 75 u ( x j,t n+1) =u(x j,τ + )+ ( t n+1 τ ) u t (x j,τ + )+O ( k ) =u(x j,τ )+[u] ;τ + ( t n+1 τ ) u t (x j,τ )+ ( t n+1 τ ) [u t ] ;τ +O ( k ). Combnng the two expressons above gves u ( x j,t n+1) u ( x j,t n) =ku t (x j,τ )+[u t ] ;τ + ( t n+1 τ ) [u t ] ;τ +O ( k ). (13) On the other hand, we also have ( u t xj,t n) =u t (x j,τ )+O(k), ( u t xj,t n+1) =u t (x j,τ )+[u t ] ;τ +O(k), whch mples ku t (x j,τ )= k ( ut xj,t ( n) ( +u t xj,t n+1)) k [u t] ;τ + O ( k ). (14) Substtutng (14) nto (13) gves u(x j,t n+1 ) u(x j,t n ) k Ths s equvalent to (1). = 1 ( ( ut xj,t n) ( +u t xj,t n+1)) + [u] ;τ k + tn+1/ τ [u t ] ;τ + O(k). k (15) We know [u] ;τ from the jump condtons. However, to compute Q n+1/ j,wealso need to fnd the locaton τ and the jump [u t ] ;τ. Let us frst dscuss how to fnd τ f t exsts. Usng the Crank Ncolson formula twce we get: α τ α n τ t n = 1 ( w n + w τ ), α n+1 α τ t n+1 τ = 1 ( w τ + w n+1). Elmnatng the w τ term we get x j α n τ t n αn+1 x j t n+1 τ = 1 ( w n w n+1). (16) Ths s the equaton for the crossng tme τ and t s coupled wth the equatons (6), (9) and (11). The estmaton of [u t ] ;τ n (11) depends on nterface condtons and wll be dscussed n secton 5. Note that the dscusson above s stll vald even f the nterface crosses several grd ponts durng one tme step. However, t would be better to control the tme step so that the nterface only crosses one grd pont durng one tme step. Ths wll gve a smaller error constant.
76 Z. L / Movng nterface problems 4. Dscretzaton of u x and (βu x ) x near the nterface As we mentoned earler, at grd ponts whch are away from the nterface ( x α >h) we use the central dfference scheme. So only the closest grd ponts from the left and the rght of the nterface need specal treatment at each tme level t n or t n+1. The dscretzaton apparently depends on nterface condtons and wll be dscussed separately n ths secton. 4.1. Case 1: the soluton on the nterface s known Let the soluton on the nterface be u ( α, t ) = r(t). (17) We begn our dscusson wth the very general equaton of moton (). For those Stefan problems n whch the velocty of the nterface s propotonal to the flux, the dscusson s gven n secton 4.3. Snce we know the value of the soluton on the nterface, we could dscretze u x and u xx usng one sded nterpolaton as usual. For example, f x j α n <x j+1 (note that the subscrpt j now has a dfferent meanng from that used n secton 3), then u n x,j = u n x(j 1,j,α n,x j ), where u x (j 1,j,α,x)= x j +α x (x j 1 α)h u j 1+ x j 1+α x (α x j )h u j + x j 1+x j x r(t). (18) (x j α)(α x j 1 ) Ths s a second order approxmaton to u x (x j,t). However, notce that α(t) changes wth tme, so does x j α. If x j α gets too small, the magntudes of the coeffcents n the nterpolaton (18) become very large, sometmes even blow up. Such nstablty s caused from the formulaton of the dervatves, whch s very senstve to the locaton of the nterface. An ntutve fx would be u n x,j = un x (j,j 1,αn,x j ), when x j α s small. A more robust way whch we have been usng successfully s the lnear combnaton of those two above: u n x,j = αn x j u n x h (j 1,j,αn,x j ) + x j+1 α n u n h x(j,j 1,α n,x j ). (19)
Z. L / Movng nterface problems 77 There are several advantages of ths robust approach. Frst of all, the nterpolaton s stll second order accurate. Secondly, f we rewrte (19) as u n x,j = j j=j γ n j u n j + γ n αr ( t n), then the magntudes of the coeffcents γj n and γα n wll always be order O(1/h). Furthermore the truncaton error n such an nterpolaton wll vary smoothly as tme ncreases. Ths s very sgnfcant n two or hgher dmensonal movng nterface problems where we want to avod non-physcal oscllatons. In the same manner, we use the followng nterpolaton to dscretze u n x,j+1 : u n x,j+1 = αn x j u n h x(j +,j+3,α n,x j+1 ) + x j+1 α n u n x h (j +1,j+,αn,x j+1 ). Smlarly, usng the followng second order dscretzaton for u xx, u xx (j 1,j,α,x)= (α x j 1 )h u j 1+ (α x j )h u j (0) + r(t), (1) (x j 1 α)(x j α) we can compute u n xx,j and un xx,j+1 as follows, u n xx,j = αn x j u n h xx(j 1,j,α n,x j ) + x j+1 α n u n xx h (j,j 1,αn,x j ), () u n xx,j+1 = αn x j u n xx h (j +,j+3,αn,x j+1 ) + x j+1 α n u n h xx(j + 1,j+,α n,x j+1 ). At the tme level l = n + 1, we can stll use the same trck when x j s too close to α n+1. But the resultng lnear system (f we freeze the nonlnear term uu x ) s no longer trdagonal snce an addtonal pont j s nvolved at grd pont x j. However, n ths case we can smply set (3) u n+1 j = r ( t n+1), f x j α n+1 h (4) wthout affectng second order accuracy. By dong so we wll stll have a trdagonal system for the lnearzed equatons.
78 Z. L / Movng nterface problems 4.. Case : The jump condtons are known Suppose we know the jump condtons (3), (4) across the nterface, [u] =q(t)and [βu x ]=v(t).we also should have knowledge of [β(t)], [f(t)] across the nterface. As n [15, 17], the mmersed nterface method for the space dscretzaton nvolves the followng steps: Use the jump condtons and the dfferental equaton to get the nterface relatons between the quanttes on each sde of the nterface. Use the nterface relatons to derve a modfed dfference scheme. Derve the correcton term based on the dfference scheme and the nterface relatons. Wth ths process, we have the followng theorem: Theorem 4.1. If x j α(t) <x j+1,then u(α,t) u x (α,t) u xx (α,t) =S u(x j 1,t) u(x j,t) u(x j+1,t) C j,1 + O(h3 ) O(h ), (5) O(h) C j, C j,3 where for k = 1,, 3, C j,k = s k3 q + v β + (x j+1 α) + (x j+1 α) ( β + q +[f] v ( w λu + β + + β x + ) )}, (6) q and v are defned as n (3) and (4), and S = s kj } = A 1, the nverse of the followng matrx 1 1 1 A T = a kj } T x = j 1 α x j α a 3 (x j 1 α) (x j α) ρ(x j+1 α) wth a 3 = ρ(x j+1 α)+ (x j+1 α) ( w(1 ρ)+λ(u + β + ρ u )+βx ρβ x + ), where ρ = β /β +. The proof can be found n the appendx. Usng ths theorem we get a dscretzed form of (βu x ) x λuu x at the grd pont x j, x j α<x j+1, (βu x ) x,j λu j u x,j =(βu x ) x λu u x + O(h) ( β s 31 + s 1 (βx λu ) ) u j 1 + ( β s 3 + s (βx λu ) ) u j + ( β s 33 + s 3 (βx λu ) ) u j+1 ( β C j,3 + C j, (βx λu ) ). (7)
Z. L / Movng nterface problems 79 The attractve aspect of ths approach s that we can stll use a three-pont stencl and the dscretzaton s vald for any locaton α. The theorem above also gves an nterpolaton formula whch can be used to compute u and u x values whch are needed for the computaton of w and the frozen term λuu x (see secton 5). For the grd pont x j+1, x j α<x j+1, there s a smlar formula whch we state as follows: Theorem 4.. Let u(x, t) be the soluton of (1) and () wth jump condtons (3) and (4). If x j α(t) <x j+1,then u(α+,t) u x (α +,t) = S u(x j,t) u(x j+1,t) C j+1,1 C j+1, + O(h3 ) O(h ), (8) u xx (α +,t) u(x j+,t) O(h) where for k = 1,, 3, C j+1,k = s k1 [u] [βu x] β (x j α) C j+1,3 (x j α) ( β q +[f] v ( w λu β + βx ) )}, (9) and S = s kj } = Ã 1, the nverse of the followng matrx wth 1 1 1 Ã T = ã kj } T ã = 1 x j+1 α x j+ α (x j α) (x j+1 α) (x j+ α), ã 1 = (x j α) ρ 4.3. Stefan problems ρ (x j α) ( ( β w 1 1 ) ( ) ) u + λ ρ ρ u+ β x ρ + β+ x. For a number of Stefan problems, the governng equatons have the form u t =(βu x ) x, dα dt = σ(t)[βu x], u(α, t)=u 0. (30) So we know the soluton on the nterface as well as the jump relatons [u] =0and [βu x ]= α/σ.
80 Z. L / Movng nterface problems It s easy to see that equaton (30) can be dscretzed usng ether of the approaches descrbed n sectons 4.1 or 4.. Numercal experments show that the approach usng the jump relatons descrbed n secton 4. s slghtly better. However, we need one more equaton to make the dscrete system closed because now w = σ[βu x ] s unknown. Ths equaton s the restrcton of the soluton on the nterface 1 0 u(x, t)δ(x α) dx = u 0. A second order accurate dscretzaton can be easly derved from theorem 4.1: s 11 u j 1 + s 1 u j + s 13 u j+1 C j,1 = u 0. 5. Computaton of the quanttes defned on the nterface As we mentoned n secton, we need to know some quanttes defned only on the nterface such as u ±,u ± x,[u t ] τ n order to compute (9), (11), (16) and (7). Agan we dstngush the two dfferent cases. 5.1. Case 1: the soluton on the nterface u(α, t) =r(t)s known In ths case, the soluton s contnuous, whch means u = u + = r(t). Wth the knowledge of the computed soluton u n, an estmaton of un+1,andthe soluton on the nterface r(t n ), we use the one sded dfference (19) exchangng the poston between x j and α n, x j α n <x j+1, to compute u,n x, and (0) exchangng the poston between x j+1 and α n to compute u +,n x. The same approach s used for the next tme level t n+1. If the nterface crosses a grd lne x = x at some tme τ, we need to compute [u t ] ;τ n order to get the correcton term Q n+1/. In ths case we smply use [u t ] ;τ = un+1 r(τ) t n+1 τ 5.. Case : the jump condtons are known r(τ) un τ t n. (31) In ths case, u ± and u ± x are computed usng (5) (6) and (8) (9). In order to compute the jump [u t ] τ, we dfferentate the frst jump condton wth respect to t to get.e., u(α +,t) u(α,t)=q(t) ( ux (α +,t) u x (α,t) ) dα dt +u t(α +,t) u t (α,t)=q (t), [u t ]=q (t) [u x ]w. (3)
Z. L / Movng nterface problems 81 We need to express (3) n terms of the quanttes at tme level ether t n or t n+1. If α n α n+1 (see fgure 1(b)), we have, at tme t = τ, Otherwsewehave u = u n j + O(h), u + = u n+1 j u x = un x,j + O(h), u+ x = un+1 x,j +O(h), + O(h). u =u n+1 j +O(h), u + = u n j +O(h), u x = u n+1 x,j + O(h), u + x = u n x,j + O(h), for α n <α n+1. Thus we use the followng scheme to compute [u t ] τ : [u t ] ;τ = q (τ) w ( τ,u n j,un+1 j,u n x,j,un+1 x,j q (τ) w ( τ,u n+1 j,u n j,un+1 x,j,un x,j )( u n+1 x,j u n x,j), f α n α n+1, )( u n+1 x,j u n x,j), f α n <α n+1. (33) 6. Solvng the resultng nonlnear system of equatons From the dscusson above we know that n order to get the soluton u(x, t) at tme t n+1, generally we need to solve the followng nonlnear system: u n+1 u n k Q n+1/ + λ ( u n u n x, + u n+1 u n+1 ) x, ( (βux ) n x, +(βu x ) n+1 x, = 1 α n+1 α n = 1 w k ( n + w n+1), ) 1 ( f n + f n+1 ), where the quanttes u l x, and (βu x) l x, for l = n or n + 1, can be expressed as some lnear combnatons of u l. The coeffcents of such combnaton near the nterface, and the correcton term Q n+1/ to (u n+1 u n j )/k, depend on the nterface poston and the nterface condtons. We have shown how to get these quanttes n prevous sectons for dfferent nterface condtons. Snce we use a fully mplct dscretzaton, the numercal scheme s stable. The local truncaton errors are O(h ) at most grd ponts, but O(h) at two grd ponts whch are closest to the nterface from the left and the rght, and at those grd ponts where the nterface crosses. So the global error n the soluton s second order accurate at all grd ponts. So we have a qute complcated nonlnear system to solve. The dffculty s that some quanttes such as Q n+1/ j, C n+1 j,k are not known untl we know the soluton for
8 Z. L / Movng nterface problems and α n+1. Ths leads to an mplct system, whch must be solved teratvely. One concern about teratve methods s whether there s a tme step restrcton. In our approach we use an mplct dscretzaton for the dffuson term (βu x ) x and a predct-correct approach for the moton of α(t) n whch the CFL restrcton s k h. Therefore we do not need to worry about the tme step restrcton unless the nterface changes rapdly. In that case, the stffness of the moton wll requre that n order to acheve the desred accuracy, we must take a small tme step anyway. An adaptve tme step s chosen for equaton () based on the classc stablty theory u n+1 } k mn h, 1 w/ α. The constrant k h s mposed to mantan second order accuracy both n space and tme. From equaton () we can get Ths mples or n dscrete form α w w t = w,.e., w α = w t /w. } k mn h, w w/ t, k new = mn h, w n } k old w n+1 w n. (34) Below we gve an outlne of our teratve process. Suppose we have obtaned all necessary quanttes at the tme t n, and the current tme step s k (.e., t n+1 = t n + k). We want to get all correspondng quanttes at the tme level t n+1. Unfortunately we have to ntroduce another subscrpt m for each teraton. Determne j 0 such that x j0 α n <x j0 +1. Compute u n x, (βu x) n x at x j 0 accordng to the scheme dscussed n secton 4. Set α n+1 1 = α n + kw ( t n,α n ;u,n,u +,n,u,n x,u +,n ) x. Determne an ntal guess of the soluton u n+1,1 at the tme level t n+1. For m = 1,,..., (**) Determne j m such that x jm α n+1 1 <x jm+1. Determne the coeffcents for un+1 n and the correcton terms for u n+1 x and u n+1 xx at x jm. Substtute u n+1,m the nonlnear term u n+1 u n+1 x,.
Z. L / Movng nterface problems 83 If j 0 j m, then for l = j 0 + 1,...,j m, when j 0 <j m,or for l = j m, j m + 1,..., j 0, when j 0 >j m,frst get τ m usng (16), then determne the correcton Q n+1/ l,m to (u n+1 l,m un l )/k usng the technque descrbed n secton 3. Solve the trdagonal system for u n+1,m+1. Interpolate u n+1,m+1 } to get u±,n+1 m+1, u±,n+1 x,m+1, f necessary. Determne α n+1 m+1 = αn + k ( w n + w n+1 m+1), where w n+1 m+1 = w( t n+1,α n+1 m ;u,n+1 m+1,u+,n+1 m+1,u,n+1 x,m+1 x,m+1),u+,n+1. If αm n+1 α n+1 m+1 >ε,a gven tolerance, then m = m + 1. Go to (**). If αm n+1 α n+1 m+1 <ε,then set all quanttes }n+1 m to } n+1, n other words we drop the m} notaton and accept these values for the tme level t n+1. Determne the next tme step sze k, k = mn h, w n } k w n+1 w n. (35) Go to the next tme step. 7. Numercal examples Here we present three dfferent examples. The frst two are from real applcatons. The thrd one s a constructed example for the nonlnear movng nterface problem. Example 1. Two phase Stefan problem. Ths s a classcal example n trackng a freezng front of ce n water. The descrpton of the problem s excerpted from [11], where Furzeland used ths example to compare dfferent methods. The thermal propertes are k = conductvty, C = c ρ wth c = specfc heat and ρ = densty (assume the same n each phase), σ = Lρ wth L = latent heat. Subscrpt = 1 denotes phase 1 (ce), 0 <x<α(t);= denotes phase (water), α(t) <x<1. The equatons are u C t = k u x, u 1 =u <0, u 1 =u =0 σ α t = k u 1 1 x k u x = 1,,t 0<t, x = 0,t>0, on x = α(t), t 0<t,
84 Z. L / Movng nterface problems Table 1 Grd refnement analyss for example 1 at t = 1.0. N E N rato E α rato 0 4.3067 10 3 1.0941 10 4 40 9.7147 10 4 4.4333.4947 10 5 4.3857 80.3713 10 4 4.0967 5.798 10 6 4.3539 160 5.8160 10 5 4.077 1.388 10 6 4.1434 30 1.413 10 5 4.090 3.371 10 7 4.1009 where the soluton u represents the temperature. Ths problem has an exact soluton α(t)=φ κ 1 t, u 1 (x, t)=u 1 erf(x/ } κ 1 t), erfφ u (x, t)=u 0 1 erfc(x/ } κ t) erfc(φ, κ 1 /κ ) where κ = k /C, erf(x) s the error functon, u and u 0 are two constants, and φ s the root of the transcendental equaton e φ erfφ + k κ1 u 0 e κ1φ/κ k 1 κ u erfc(φ κ 1 /κ ) + φλ π C 1 u = 0, whch can be easly computed, say usng the bsecton method. The exact soluton s used as the ntal condton at tme t 0 = 0.5, as well as the boundary condton at both ends, x = 0, and x = 1. The followng thermal propertes are used k 1 =., k = 0.556, C 1 = 1.76, C = 4.6, σ = 338, wth u = 0 and u 0 = 10 whch gves φ = 0.05469... Table 1 shows the results of a grd refnement analyss, where E N s defned as the nfnty norm of the error at the fxed tme t,.e., E N = max u(x,t) u N where N s the number of grd ponts as defned n secton. We defne u N as the computed soluton at the unform grd ponts x, = 1,,..., at the fnal tme t. E α s the error between α(t) and the computed nterface at the fnal tme t. We see that doublng the number of grd ponts gves a reducton n both errors by a factor of roughly 4, ndcatng second order accuracy. Fgure (a) shows the true soluton and },
Z. L / Movng nterface problems 85 10 The computed temperature, n = 40, t = 1.0 10 x10-4 Error plot: True - Approx, n = 40, t=1.0 5 8 0 6 u -5 4-10 : True soln. o : Approx. soln. -15 0-0 0 0.1 0. 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x (a) - 0 0.1 0. 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x (b) Fgure. The comparson of the exact and computed soluton at t = 1.0 for example 1 wth N = 40. (a) The sold lne s the exact soluton and the dots are the computed results at grd ponts. (b) The correspondng error plot. the computed soluton for N = 40. Fgure (b) shows the correspondng error plot. We see that the error n the soluton u s relatvely large around the nterface compared to other grd ponts but not sgnfcantly so. Globally we obtan second order accurate results at all grd ponts (see table 1). We also tested a smlar example descrbed by Keller [14] for modelng a meltng and freezng problem at a constant speed. The results agan agreed wth our analyss. Example. Ths smulaton shows the temperature profle of an ce sheet durng the process of a glacaton. Hene and McTgue [1] proposed a one-dmensonal thermal model to study the temperature change of a glacer. The ce sheet gradually grows to a thckness of 3000 m over approxmately 10,000 years. At the same tme, a heat source, n the form of a geothermal heat flux, s warmng the glacer from a depth of 4000 m n the rock. The pont of nterest n ths problem s the nterface between ce and rock. If the temperature approaches the meltng pont, the glacer may begn to slde wth catastrophc effect. A smlar problem can be found n [13]. The mathematcal model Hene and McTgue used s smlar to example 1, see fgure 3. T ρc p t = ( k T ), 0 x h(t), t>0, x x where T s the temperature. However, both the heat capacty ρc p, and the thermal conductvty k are functons of T (x, t) now. So ths s a nonlnear model for the temperature. The estmated heght of the ce sheet s h(t) =h 0 +δ h +(h h 0 δ h ) ( 1 e t/tr),
86 Z. L / Movng nterface problems x= h( t) heght of glacer ce x= h 0 rock rock [ T ] = 0 ce T s ht () Tx= q 0 [ kt x ] = 0 heat δ h x = 0 h 0 x x = 0 geothermal heat q 0 Fgure 3. Geometrcal frame of Hene and McTgue s glacer model. where h 0 s the thckness of rock, δ h s an (arbtrarly) small ntal ce layer thckness, h s the fnal total thckness (rock + ce), and t r s the rse tme for the ce sheet growth. The boundary condtons consst of constant heat flux from the nner layer of the earth and constant temperature at the top of the ce sheet: The thermal propertes are T x (0,t)= q 0, T ( h(t),t ) =T s. k ce = C 1 e C (T +73.15), k rock = constant, (ρc p ) ce = r 1 + r T, (ρc p ) rock = constant. The ntal condtons are T h0 + q 0 (h 0 x), f 0 x h 0, k T = rock 1 log ( e C T s C (h 0 + δ h x)γ ), f h 0 <x<h, C where Γ = q 0 C 1 e 73.15C, T h0 = 1 C log ( e C T s C δ h ).
Z. L / Movng nterface problems 87 5 0-5 TEMP (deg. C) -10-15 -0-5 0 0.5 1 1.5.5 3 3.5 TIME (seconds) x10 1 Fgure 4. Temperature on the nterface between the rock and ce. The followng thermal propertes are used C 1 = 9.88, C = 0.0057, r 1 = 1.936 10 6, r = 6.600 10 3, k rock =.50, (ρc p ) rock =.30 10 6, q 0 = 0.05 w/m, T s = 5.0 C. The geometrcal parameters are h 0 = 4000 m, δ h = 1m, h = 7000 m, t r = 1.9 10 1 s. In ths example, we have a fxed nterface x = h 0 wth the natural jump condtons [T ]=0and[k T/ x] =0 snce temperature s contnuous and there s no heat source at the nterface. Ths corresponds to case n our prevous dscusson wth w 0. On the other hand, at the movng boundary h(t) we know the surface temperature T s. So t s also a one-phase nonlnear movng boundary problem. Fgure 4 shows the temperature hstory of the nterface between rock and ce. The geophyscal mplcaton of the result can be found n [1]. To check the correctness of our algorthm, we constructed a problem for whch we have an exact soluton wthn the same geometrcal frame. The numercal results confrmed that our method converges to the exact soluton wth second order accuracy. The detals are omtted here due to space lmtatons.
t 88 Z. L / Movng nterface problems 0.8 0.7 0.6 0.5 0.4 0.3 (,1) (1.8,1) 0. 0.1 0 (5,1) (3,1) (1,1) (1,5) -0.1 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 x Fgure 5. Movng nterface α(t), 0 t 1, from left to rght, (β,β + )=(5,1),(3,1),(,1),(1.8,1), (1,1)and (1, 5). Example 3. Nonlnear movng nterface example for case. In ths example, we take λ = 1 n (1) and construct the followng exact soluton: sn(ω u(x, t) = 1 x)e β ω1 t, sn ( ω ω x ) e β+ ω t, f x α(t), f x>α(t), for some choce of ω 1, ω, β, and β +. The source term f(x, t) s dscontnuous 1 f(x, t) = ω 1sn(ω 1 x)e β ω1 t, f x α(t), 1 ω sn(ω ω x)e β+ ω t, f x>α(t). We assume that the soluton u(x, t) s contnuous across α(t). So the nterface α(t) can be determned from the scalar equaton (36) (37) sn(ω 1 α)e β ω 1 t = sn ( ω ω α ) e β+ ω t. (38) Ths equaton has a unque soluton f we take, for example, π<ω 1 <πand also π<ω <π. Fgure 5 gves the plot of α(t) as the parameters (β,β + ) change on a unform grd. We can see how the nterface crosses the grd. Ths example s adapted from [3].
Z. L / Movng nterface problems 89 The profle of u(x,t), 0 <= t <= 1, N = 160, k=h/ 0.05 u(x,0.1) 1 t=0 0.04 : True soln. 0.8 0.03 o : Approx. soln. u 0.6 0.0 0.4 0.01 0. t=0.1 0 0 0.1 0. 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x (a) 0-0.01 0 0.1 0. 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 (b) x Fgure 6. (a) The profle of the computed soluton u(x, t) from t = 0tot=0.1 wth β = β + = 1, and N = 160. (b) The comparson of the exact and computed soluton at t = 1.0 wth β = 5.0, β + = 1.0, and N = 80. The sold lne s the exact soluton and the dots are the computed results at the grd ponts. Table Grd refnement analyss for example 3 at t = 0.1. N E N rato E α rato 40 6.5460 10 3.605 10 80 6.0108 10 3 1.0890 8.6674 10 3.6081 160 1.1557 10 3 5.009.7449 10 3 3.1575 30.9347 10 4 3.9381 6.9958 10 4 3.938 640 7.3175 10 5 4.0106 1.7508 10 4 3.9957 180 1.837 10 5 4.014 4.375 10 5 4.004 560 4.546 10 6 4.0115 1.0913 10 5 4.0065 510 1.1370 10 6 3.998.7311 10 6 3.9960 The nonlnear ordnary dfferental equaton for the moton s dα dt = (ω 1 ω )u(α, t) u x (α,t) u x (α +,t). (39) The ntal and boundary condtons are u(0,t)=0, u(1,t)=0, (40) sn(ω1 x), f x α(0), u(x, 0)= (41) sn(ω ω x), f x α(0). The jump condtons used are [u]=0, [ βux (α, t) ] = β + ω cos(ω ω α)e β+ ω t β ω 1 cos(ω 1 α)e β ω 1 t.
90 Z. L / Movng nterface problems Fgure 6 shows a typcal computed profle of u(x, t) as tme changes. We can clearly see how the nterface moves and crosses the grd wth tme. Ths example s more challengng than tradtonal examples because the jumps n the coeffcent and n the dervatve of the soluton are sgnfcantly larger than n the prevous examples. Table shows the grd refnement analyss both for the computed soluton and the nterface. In ths case, the nterface crosses several grd ponts durng the frst few tme steps (see fgure 5). However, second order accuracy s stll acheved. 8. Summary In summary we have developed a second order accurate mmersed nterface method for a class of one dmensonal nonlnear movng nterface problems wth two typcal nterface condtons: () Stefan-lke problems for whch we know the soluton on the nterface. () Problems n whch we only know the jump condtons n the soluton and the flux across the nterface. Applcatons nclude the mmersed boundary method and heat conducton n dfferent materals. Numercal experments have confrmed second order convergence of the methods proposed n ths paper. Currently we are workng on smlar numercal methods for two dmensonal problems. It s well known that the nterface between the two phases n the two dmensonal Stefan problem may develop sngulartes or complcated shapes even f t s ntally smooth. Thus we are gong to explore the possbltes of usng the level set technque [1] to track the nterface. Acknowledgements I am ndebted to Prof. R. J. LeVeque for the orgnal motvaton and many useful dscussons. I would also lke to thank hm and D. Calhoun for readng, correctng and commentng on ths paper. I am grateful to Prof. D. McTgue for useful dscussons concernng hs glacer model. It s very benefcal talkng to B. Merrman, S. Osher, J. Zou and other people about the problems n ths paper. Appendx. Proof of theorem 4.1 Here we present a bref proof of theorem 4.1. We frst derve one more nterface relaton usng the known nformaton and then gve the proof of the theorem. Lemma A1. Let u(x, t) be the soluton of (1) () wth the jump condtons (3) and (4), then q +[f] v ) u + xx = ρu xx + 1 β + where ρ = β /β +. + u x ( w λu + β + + β x + (w(1 ρ)+λ(u + ρ u )+β x β β +β+ x )}, (4)
Z. L / Movng nterface problems 91 Proof. From equaton (1) we know β + u + xx + β + x u + x λu + u + x f + u + t = β u xx + β x u x λu u x f u t. (43) Pluggng (3) and the second jump condton (4) nto (43) and arrangng terms, we get (4). Proof of theorem 4.1. Expandng u(x j 1,t), u(x j,t) from the left hand sde, and u(x j+1,t) from the rght hand sde of α, we have So j+1 k=j 1 u(x j 1,α)=u +(x j 1 α)u x + 1 (x j 1 α) u xx + O ( h 3), (44) u(x j,α)=u +(x j α)u x + 1 (x j α) u xx + O( h 3), (45) u(x j+1,α)=u + +(x j+1 α)u + x + 1 (x j+1 α) u + xx + O ( h 3). (46) ( s,k j+ u(x k,t)=s 1 u +(x j 1 α)u x + 1 (x j 1 α) u ) xx ( + s u +(x j α)u x + 1 (x j α) u ) xx ( + s 3 u + +(x j+1 α)u + x + 1 (x j+1 α) u + ) xx + O ( h 4 ), (47) for = 1, and 3. Notce that we have used the fact that s j s of order O(h 1 ). From nterface relatons (3), (4) and (4) we can solve for u +, u + x,andu+ xx n terms of u, u x,andu xx and substtute them nto the expresson above and collect terms to obtan j+1 k=j 1 s,k j+ u(x k,t) = ( ) s 1 +s +s 3 u + ( s 1 (x j 1 α ) +s (x j α)+s 3 a 3 )u x ( (xj 1 α) + s 1 + (x j α) s + (x j+1 α) ρs 3 )u xx + s 3 q + (x j+1 α)v β + + (x j+1 α) ( β + q +[f] v ( w λu + β + + β x + ) )} + O ( h 4 )
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