Immersed interface methods for moving interface
problems
∗
Zhilin Li∗∗
Department of Mathematics, University of California at Los Angeles, Los Angeles, CA 90095, USA
E-mail: [email protected]
Received May 1995; revised December 1995 Communicated by T. F. Chan
A second order difference method is developed for the nonlinear moving interface problem of the form
ut+λuux= (βux)x−f(x, t), x∈[0, α)∪(α,1], dα
dt =w(t, α; u, ux),
whereα(t)is the moving interface. The coefficientβ(x, t)and the source termf(x, t)can be discontinuous acrossα(t) and moreover, f(x, t) may have a delta or/and delta-prime function singularity there. As a result, although the equation is parabolic, the solution u
and its derivatives may be discontinuous acrossα(t). Two typical interface conditions are considered. One condition occurs in Stefan-like problems in which the solution is known on the interface. A new stable interpolation strategy is proposed. The other type occurs in a one-dimensional model of Peskin’s immersed boundary method in which only jump conditions are given across the interface. The Crank–Nicolson difference scheme with modifications near the interface is used to solve for the solution u(x, t)and the interface
α(t) simultaneously. Several numerical examples, including models of ice-melting and glaciation, are presented. Second order accuracy on uniform grids is confirmed both for the solution and the position of the interface.
Keywords: immersed interface method, Stefan problem, moving interface, discontinuous coefficients, singular source term, immersed boundary method, Cartesian grid, heat conduc-tion.
AMS subject classification: 65N06, 76T05, 80A22, 86A40.
∗This work was supported by URI grant #N00014092-J-1890 from ARPA, NSF Grant DMS-9303404,
and DOE Grant DE-FG06-93ER25181.
∗∗Present address: Department of Mathematics and Statistics, Mississippi State University, Mississippi
State, MS 39762, USA.
1. Introduction
In this paper we study the immersed interface method for the one-dimensional moving interface problem ut+λuux = (βux)x−f(x, t), x∈[0, α)∪(α,1], (1) dα dt =w t, α;u −, u+, u− x, u+x , t >0, (2)
where w is a known function and u−, u+, u−x, and u+x are the limiting values of
u(x, t)andux(x, t) from the left and right hand side of a moving interfaceα(t). The
coefficientβ(x, t)>0 and the source termf(x, t)may be discontinuous; furthermore,
f(x, t) may have a delta or/and delta-prime function singularity at the interfaceα(t). This is a parabolic problem and the solution is piecewise smooth. The discontinuities can only occur on the interface α(t). Initial and boundary conditions, which are not our main interests here, may also be imposed. We assume these can be handled using conventional techniques.
The interface α(t) divides the solution domain into two parts: 0 6 x < α(t)
and α(t) < x 6 1. The solution in each domain [0, α(t)) and (α(t),1] is smooth,
but coupled with the solution on the other side by interface conditions (or internal boundary conditions) which usually take one of the following forms:
Case 1. The solution on the interface is given. One example is a mathematical model for solidification problems. When tracking an interface of melting ice, for example, the temperature at the melting/freezing interface is given by the melting temperature of the ice. In this caseu(α, t) =u0 is the melting temperature. When λ≡0, this is
the classical Stefan problem.
Various approaches have been used to solve Stefan or other linear free or moving interface/boundary problems numerically [1, 4, 6, 8, 10, 11, 13, 20, 24, 27], and recent work by Chen et al. [5] using the level set approach. Compared to case 2 discussed below, the Stefan problem is easier to solve because the value of the solution on the interface is known. However, few numerical methods are second order accurate in the infinity norm for both the solution and the interface. Most methods involve some transformations either for the differential equations or the coordinate system, which complicates the problem in some way. Some of the methods would exhibit some kind of instability near the interface as explained in section 4. The method proposed in sections 2–6 is simpler, more stable, and second order accurate both for the solution
uand the interface α(t) simultaneously for more general equations. Furthermore, the solution is obtained using a fixed Cartesian grid.
Case 2. Jump conditions of the form
[u]def= u(α+, t)−u(α−, t) =q(t), (3) [βux]
def
=β(α+, t)u
are given. Our interest in this case was motivated by the desire to develop a sec-ond order accurate algorithm for solving the incompressible Navier–Stokes equations obtained from the formulation of the immersed boundary method. This method was originally developed by Peskin and his co-workers for studying blood flow in a beat-ing heart [22, 23], but has been used in a wide variety of other problems [2, 7, 9, 25]. In Peskin’s method the physical domain is immersed in a rectangular region. The boundary condition is treated as a forcing term which is only supported on the boundary, therefore the forcing term is singular. Case 2 is a one-dimensional model for the immersed boundary method formulation with a more general equation for the motion. A linear model in one dimension would be
ut= (βux)x−C(t)δ x−α(t)
−Cb(t)δ0 x−α(t)−f(x, t).
From this equation, assumingf(x, t)is continuous, we can derive the following jump conditions
[u] = 2Cb(t)
β−+β+, [βux] =C(t),
whereβ−andβ+are the limiting values ofβ(x, t) from left and right of the interface
α(t). In many other cases the jump condition can be obtained from physical reasoning. For instance, ifu stands for temperature, then[u] =0; and [βux] =v(t) means that
the net heat flux across the interface is equal to the source strength v(t). One of the advantages of using the jump conditions is that we do not need to discretize the delta function, an approach described briefly in the next paragraph.
Peskin and many other people use a discrete delta function approach to spread the singular force term to the nearby grid points and use the discrete delta function again to interpolate the velocity field to move the boundary. While this approach seems to work, it is usually only first order accurate and difficult to analyze especially in two or higher dimensions.
Various attempts have been made to analyze and improve the discrete delta function approach or to try to solve the resulting differential equations differently. Beyer and LeVeque [3] studied various one-dimensional moving interface problems for the heat equation assuminga priori knowledge of the interface. A discrete delta
function is carefully selected and some correction terms are added if necessary in their approach to get second order accuracy. Wiegmann and Bube [26] recently applied the immersed interface method for certain one-dimensional nonlinear problem with a fixed interface. However, for the interface problems discussed here the interface is unknown and moving, and the discrete difference scheme is a nonlinear system of equations involving both the solution and the interface.
In two dimensions, the discrete delta function approach generally is only first order accurate. And it seems impossible to find a generic discrete delta function which makes the algorithm second order accurate. A new approach, the immersed interface
method [15, 17], is intended to solve general PDEs with discontinuous coefficients
which are close to or on the interface. The main idea is to incorporate the known jumps in the solution or its derivatives into the finite difference scheme, obtaining a modified scheme whose solution is second order accurate at all grid points on the uniform grid even for a quite arbitrary interface. This method has been implemented for several different applications in one, two and three dimensions [16, 18, 19]. By implementing this method for the one dimensional model here, we not only effectively solved some general one-dimensional moving interface problems, but also hope to get some insights into the method as well as the problems. We are currently working on second order accurate immersed interface methods for the full Navier–Stokes equation with a moving boundary, where we have to deal with the nonlinear termu· ∇u. That is one of the reasons why we have the nonlinear termλuux in our model equation (1).
Case 2 (withλ=0) is also a model of heat conduction with an interface between two different materials. Now u is the temperature and hence is continuous, meaning
q(t)≡0 in (3). The net heat flux across the interface isv(t)in (4), the source strength on the interface. Again, in this case we do not know the value of the solution on the interface but only the jump conditions.
For many classical Stefan problems, the motion of the interface is proportional to the flux across the interface
dα
dt =σ(t)[βux], u(α, t) =u0, (5)
where u0 is the known temperature at the interface. This kind of problem fits both
cases 1 and 2 and will be discussed in section 4.3.
2. Computational frame We use a uniform grid
xi =ih, i=0,1, . . . , N, x0=0, xN =1,
where h is the step size in space. We use k as the step size in time and assume the ratio k/h is a constant so that we can write O(k) as O(h) or vice versa. Using the Crank–Nicolson scheme, the semi-discrete difference scheme for (1) can be written in the following general form,
uni+1−un i k −Q n+1/2 i + λ 2 u n iunx,i+uni+1u n+1 x,i = 1 2 (βux) n
x,i+ (βux)nx,i+1 − 1 2 f n i +fin+1 , (6)
where unx,i and (βux)nx,i are discrete analogues of ux and (βux)x at (xi, tn), and
Qni+1/2is a correction term needed whenαcrosses the grid linex=xi at some time
αn α( )t tn tn+1 τ j α( )t τ j (a) (b)
Figure 1. Interface crossing the grid. (a)α(t)increases with time. (b)α(t)decreases with time.
displayed in a box, as illustrated by equation (6). For simplicity, we will drop the superscriptnduring the discussion of the space discretization if there is no confusion. At a grid pointxi,which is away from the interface (i.e.,α6∈[xi−1, xi+1]), the classic
central differences will be used
ux,i= ui+1−ui−1 2h , (7) (βux)x,i= βi−1/2ui−1−(βi−1/2+βi+1/2)ui+βi+1/2ui+1 h2 , (8)
whereβi+1/2=β(xi+h/2,:). We will discuss how to discretizeux and(βux)xwhen
α∈[xj−1, xj+1]for cases 1 and 2 in section 4.
The interface location is also determined by the trapezoidal method applied to (2)
αn+1−αn k = 1 2 w n+wn+1, (9) wherewl =w tl, αl; u−,l, u+,l, u−,l x , u+x,l , and αl,u±,l,u±,l
x are the approximations
ofα(tl), u(α±, tl), andux(α±, tl) respectively. These quantities are only defined on
the interface. We use the following notation to express the jump in a functiong(x, t) across the interface
[g]def= g(α+, t)−g(α−, t); [g];t
def
= g(α, t+)−g(α, t−).
It is easy to see that[g] =±[g];t and the sign depends on the motion of the interface.
For example we have minus sign for the case in figure 1(a), and plus sign for the one in figure 1(b).
The kernel of the algorithm at a time leveltn consists of the following:
• Determine Qnj+1/2 if the interface crosses the grid line x=xj from time tn to
• Derive the difference formula for ux and(βux)x at the two grid points closest
to the interface.
• Compute the quantitiesu±,u±x,[ut], etc. which are only defined on the interface. • Solve the nonlinear system of equations for {uni+1} and the location of the
intefaceαn+1.
Each of these steps will be described in detail in the remainder of the paper. Away from the interface, the local truncation errors for the difference scheme are O(h2). But at a few grid points near the interface, we allow the local truncation errors to be O(h) based on the fact that the local truncation error of a difference scheme on a boundary can be one order lower than those of interior points without affecting global second order accuracy.
3. Grid crossing
If there is no grid crossing at a grid pointxi from timetn to time tn+1, meaning that
(xi, tn)and(xi, tn+1)are on the same side of the interfaceα(t),i.e.,xi6∈(αn, αn+1),
then we can takeQni+1/2=0 and we have
u(xi, tn+1)−u(xi, tn) k = 1 2 ut xi, tn+1 +ut xi, tn +O k2. (10)
However, if the interface crosses the grid linex=xj at some time1 τ,tn< τ <
tn+1, such thatx
j =α(τ), see figure 1, then the time-derivative of u is not smooth.
In this case even though we can approximate the x-derivatives well at each time level (see section 4), the standard Crank–Nicolson scheme needs to be corrected to guarantee second order accuracy. This is done by choosing a correction termQnj+1/2
based on the following theorem:
Theorem 3.1. Suppose the equationα(t) =xj has a unique solutionτ in the interval
tn< t < tn+1. If we choose Qnj+1/2= [u];τ k + 1 k t n+1/2−τ[u t];τ (11) then u(xj, tn+1)−u(xj, tn) k −Q n+1/2 j = 1 2 ut xj, t n+1+u t xj, tn +O(k). (12)
Proof. We expandu(xj, tn) and u(xj, tn+1) in Taylor series about time τ from each
side of the interface to get
u xj, tn =u(xj, τ−) + tn−τ ut(xj, τ−) +O k2 , 1
The crossing time τ really depends on the grid index j as well as the time index n, see figure 1. To simplify the notation, τ will be used to indicate the crossing time, without explicitly showing its dependence onjandn.
u xj, tn+1 =u(xj, τ+) + tn+1−τ ut(xj, τ+) +O k2 =u(xj, τ−) + [u];τ + tn+1−τ ut(xj, τ−) + tn+1−τ [ut];τ +O k2 .
Combining the two expressions above gives
u xj, tn+1 −u xj, tn =kut(xj, τ−) + [ut];τ+ tn+1−τ [ut];τ +O k2 . (13)
On the other hand, we also have
ut xj, tn =ut(xj, τ−) +O(k), ut xj, tn+1 =ut(xj, τ−) + [ut];τ+O(k), which implies kut(xj, τ−) = k 2 ut xj, t n+u t xj, tn+1 −k 2[ut];τ+O k 2. (14)
Substituting (14) into (13) gives
u(xj, tn+1)−u(xj, tn) k = 1 2 ut xj, t n+u t xj, tn+1 +[u];τ k +t n+1/2−τ k [ut];τ +O(k). (15) This is equivalent to (12).
We know[u];τ from the jump conditions. However, to computeQnj+1/2, we also
need to find the locationτ and the jump[ut];τ. Let us first discuss how to findτ if it
exists. Using the Crank–Nicolson formula twice we get:
ατ−αn τ −tn = 1 2 w n+wτ, αn+1−ατ tn+1−τ = 1 2 w τ+wn+1.
Eliminating thewτ term we get
xj−αn τ −tn − αn+1−xj tn+1−τ = 1 2 w n−wn+1. (16)
This is the equation for the crossing timeτ and it is coupled with the equations (6), (9) and (11).
The estimation of [ut];τ in (11) depends on interface conditions and will be
discussed in section 5.
Note that the discussion above is still valid even if the interface crosses several grid points during one time step. However, it would be better to control the time step so that the interface only crosses one grid point during one time step. This will give a smaller error constant.
4. Discretization ofux and(βux)x near the interface
As we mentioned earlier, at grid points which are away from the interface (|xi−α|> h)
we use the central difference scheme. So only the closest grid points from the left and the right of the interface need special treatment at each time level tn or tn+1.
The discretization apparently depends on interface conditions and will be discussed separately in this section.
4.1. Case 1: the solution on the interface is known
Let the solution on the interface be
u α, t=r(t). (17)
We begin our discussion with the very general equation of motion (2). For those Stefan problems in which the velocity of the interface is propotional to the flux, the discussion is given in section 4.3. Since we know the value of the solution on the interface, we could discretizeux and uxx using one sided interpolation as usual. For
example, if xj 6 αn < xj+1 (note that the subscriptj now has a different meaning
from that used in section 3), then
unx,j =unx(j−1, j, αn, xj), where ux(j−1, j, α, x) = xj+α−2x (xj−1−α)h uj−1+ xj−1+α−2x (α−xj)h uj + xj−1+xj−2x (xj−α)(α−xj−1) r(t). (18)
This is a second order approximation toux(xj, t). However, notice thatα(t) changes
with time, so doesxj−α. If|xj−α|gets too small, the magnitudes of the coefficients
in the interpolation (18) become very large, sometimes even blow up. Such instability is caused from the formulation of the derivatives, which is very sensitive to the location of the interface. An intuitive fix would be
unx,j =unx(j−2, j−1, αn, xj),
when|xj −α|is small. A more robust way which we have been using successfully
is the linear combination of those two above:
unx,j = α n−x j h u n x(j−1, j, αn, xj) +xj+1−α n h u n x(j−2, j−1, αn, xj). (19)
There are several advantages of this robust approach. First of all, the interpolation is still second order accurate. Secondly, if we rewrite (19) as
unx,j=
j X j=j−2
γjnunj +γαnr tn,
then the magnitudes of the coefficients γjn and γαn will always be order O(1/h). Furthermore the truncation error in such an interpolation will vary smoothly as time increases. This is very significant in two or higher dimensional moving interface problems where we want to avoid non-physical oscillations.
In the same manner, we use the following interpolation to discretizeun x,j+1: unx,j+1= α n−x j h u n x(j+2, j+3, αn, xj+1) +xj+1−α n h u n x(j+1, j+2, αn, xj+1). (20)
Similarly, using the following second order discretization foruxx,
uxx(j−1, j, α, x) = 2 (α−xj−1)h uj−1+ 2 (α−xj)h uj + 2 (xj−1−α)(xj −α) r(t), (21)
we can computeunxx,j andunxx,j+1 as follows,
unxx,j = α n−x j h u n xx(j−1, j, αn, xj) +xj+1−α n h u n xx(j−2, j−1, αn, xj), (22) un xx,j+1= αn−xj h u n xx(j+2, j+3, αn, xj+1) +xj+1−α n h u n xx(j+1, j+2, αn, xj+1). (23)
At the time level l = n+1, we can still use the same trick when xj is too
close to αn+1. But the resulting linear system (if we freeze the nonlinear term uux)
is no longer tridiagonal since an additional point j−2 is involved at grid point xj.
However, in this case we can simply set
unj+1 =r tn+1, ifx
j−αn+16h2 (24)
without affecting second order accuracy. By doing so we will still have a tridiagonal system for the linearized equations.
4.2. Case 2: The jump conditions are known
Suppose we know the jump conditions (3), (4) across the interface, [u] = q(t) and [βux] = v(t). We also should have knowledge of [β(t)], [f(t)] across the interface.
As in [15, 17], the immersed interface method for the space discretization involves the following steps:
• Use the jump conditions and the differential equation to get the interface relations between the quantities on each side of the interface.
• Use the interface relations to derive a modified difference scheme.
• Derive the correction term based on the difference scheme and the interface relations.
With this process, we have the following theorem: Theorem 4.1. Ifxj 6α(t)< xj+1,then u(α −, t) ux(α−, t) uxx(α−, t) =S uu(x(xj−j, t1, t)) u(xj+1, t) − CCj,j,12 Cj,3 + O(h 3) O(h2) O(h) , (25) where fork=1,2,3, Cj,k=sk3 q + v β+(xj+1−α) +(xj+1−α) 2 2β+ q0+ [f]− v β+ w−λu ++β+ x , (26)
q and v are defined as in (3) and (4), and S = {skj} = A−1, the inverse of the
following matrix AT ={akj}T = 1 1 1 xj−1−α xj−α a23 (xj−1−α)2 2 (xj−α)2 2 ρ(xj+1−α)2 2 with a23=ρ(xj+1−α) + (xj+1−α)2 2β+ w(1−ρ) +λ(u +ρ−u−) +β− x −ρβx+ ,
whereρ=β−/β+. The proof can be found in the appendix.
Using this theorem we get a discretized form of(βux)x−λuux at the grid point
xj,xj 6α < xj+1, (βux)x,j−λujux,j = (βux)−x −λu−u−x +O(h) ≈ β−s31+s21(βx−−λu−) uj−1+ β−s32+s22(βx−−λu−) uj + β−s33+s23(βx−−λu−) uj+1− β−Cj,3+Cj,2(βx−−λu−) . (27)
The attractive aspect of this approach is that we can still use a three-point stencil and the discretization is valid for any locationα. The theorem above also gives an interpolation formula which can be used to compute u− and u−x values which are needed for the computation ofwand the frozen term λuux (see section 5).
For the grid point xj+1, xj 6 α < xj+1, there is a similar formula which we
state as follows:
Theorem 4.2. Letu(x, t) be the solution of (1) and (2) with jump conditions (3) and (4). If xj 6α(t)< xj+1,then u(α +, t) ux(α+, t) uxx(α+, t) =Se uu(x(xj+j, t1, t)) u(xj+2, t) − CCjj++11,,12 Cj+1,3 + O(h 3) O(h2) O(h) , (28) where fork=1,2,3, Cj+1,k =esk1 −[u]−[βux] β− (xj−α) −(xj−α)2 2β− q0+ [f]− v β− w−λu −+β− x , (29)
andSe={eskj}=Ae−1, the inverse of the following matrix e AT ={eakj}T = 1 1 1 ea21 xj+1−α xj+2−α (xj−α)2 2ρ (xj+1−α)2 2 (xj+2−α)2 2 , with ea21 = (xj −α) ρ − (xj−α)2 2β− w 1− 1 ρ +λ u− ρ −u + −βx− ρ +β + x . 4.3. Stefan problems
For a number of Stefan problems, the governing equations have the form
ut= (βux)x,
dα
dt =σ(t)[βux],
u(α, t) =u0.
(30)
So we know the solution on the interface as well as the jump relations [u] = 0 and [βux] = ˙α/σ.
It is easy to see that equation (30) can be discretized using either of the approaches described in sections 4.1 or 4.2. Numerical experiments show that the approach using the jump relations described in section 4.2 is slightly better. However, we need one more equation to make the discrete system closed because now w = σ[βux] is
unknown. This equation is the restriction of the solution on the interface Z 1
0
u(x, t)δ(x−α)dx=u0.
A second order accurate discretization can be easily derived from theorem 4.1:
s11uj−1+s12uj+s13uj+1−Cj,1=u0.
5. Computation of the quantities defined on the interface
As we mentioned in section 2, we need to know some quantities defined only on the interface such as u±, u±x,[ut]τ in order to compute (9), (11), (16) and (27). Again
we distinguish the two different cases.
5.1. Case 1: the solution on the interfaceu(α, t) =r(t) is known
In this case, the solution is continuous, which meansu−=u+=r(t).
With the knowledge of the computed solutionuni, an estimation ofuni+1, and the solution on the interface r(tn), we use the one sided difference (19) exchanging the position betweenxj andαn,xj 6αn< xj+1, to computeu−x,n, and (20) exchanging
the position between xj+1 and αn to compute u+x,n. The same approach is used for
the next time leveltn+1.
If the interface crosses a grid linex =xi at some time τ, we need to compute
[ut];τ in order to get the correction term Qni+1/2. In this case we simply use
[ut];τ =
uni+1−r(τ)
tn+1−τ −
r(τ)−uni
τ −tn . (31)
5.2. Case 2: the jump conditions are known
In this case, u± and u±x are computed using (25)–(26) and (28)–(29). In order to compute the jump[ut]τ, we differentiate the first jump condition
u(α+, t)−u(α−, t) =q(t)
with respect to tto get
ux(α+, t)−ux(α−, t) dα dt +ut(α +, t)−u t(α−, t) =q0(t), i.e., [ut] =q0(t)−[ux]w. (32)
We need to express (32) in terms of the quantities at time level either tn ortn+1. If
αn>αn+1 (see figure 1(b)), we have, at time t=τ,
u−=unj +O(h), u+=unj+1+O(h),
u−x =unx,j+O(h), u+x =unx,j+1+O(h).
Otherwise we have
u−=unj+1+O(h), u+ =unj +O(h),
u−x =unx,j+1+O(h), u+x =unx,j+O(h),
forαn< αn+1. Thus we use the following scheme to compute[ut]τ:
[ut];τ = ( q0(τ)−w τ, unj, unj+1, unx,j, ux,jn+1 unx,j+1−ux,jn , ifαn>αn+1, q0(τ)−w τ, unj+1, un j, unx,j+1, unx,j unx,j+1−un x,j , ifαn< αn+1. (33)
6. Solving the resulting nonlinear system of equations
From the discussion above we know that in order to get the solution u(x, t) at time
tn+1, generally we need to solve the following nonlinear system:
uni+1−uni k −Q n+1/2 i + λ 2 u n iunx,i+uni+1unx,i+1 = 1 2 (βux) n
x,i+ (βux)nx,i+1 −1 2 f n i +fin+1 , αn+1−αn k = 1 2 w n+wn+1,
where the quantitiesulx,i and (βux)lx,i for l=nor n+1, can be expressed as some
linear combinations of uli. The coefficients of such combination near the interface, and the correction term Qni+1/2 to (uin+1−unj)/k, depend on the interface position and the interface conditions. We have shown how to get these quantities in previous sections for different interface conditions.
Since we use a fully implicit discretization, the numerical scheme is stable. The local truncation errors are O(h2)at most grid points, but O(h)at two grid points which are closest to the interface from the left and the right, and at those grid points where the interface crosses. So the global error in the solution is second order accurate at all grid points.
So we have a quite complicated nonlinear system to solve. The difficulty is that some quantities such asQnj+1/2,Cj,kn+1 are not known until we know the solution for
uni+1 and αn+1. This leads to an implicit system, which must be solved iteratively. One concern about iterative methods is whether there is a time step restriction. In our approach we use an implicit discretization for the diffusion term (βux)x and a
predict-correct approach for the motion ofα(t)in which the CFL restriction isk∼h. Therefore we do not need to worry about the time step restriction unless the interface changes rapidly. In that case, the stiffness of the motion will require that in order to achieve the desired accuracy, we must take a small time step anyway. An adaptive time step is chosen for equation (2) based on the classic stability theory
k6min h, 1 ∂w/∂α .
The constraintk6h is imposed to maintain second order accuracy both in space and time. From equation (2) we can get
∂α ∂w ∂w ∂t =w, i.e., ∂w ∂α = ∂w ∂t /w. This implies k6min h, w ∂w/∂t , or in discrete form knew =min h, w nk old wn+1−wn . (34)
Below we give an outline of our iterative process.
Suppose we have obtained all necessary quantities at the timetn, and the current time step is k (i.e., tn+1 = tn+k). We want to get all corresponding quantities at
the time leveltn+1. Unfortunately we have to introduce another subscriptmfor each iteration.
• Determinej0 such thatxj0 6α
n< x
j0+1. Computeu n
x,(βux)nx atxj0 according to the scheme discussed in section 4.
• Set
αn1+1=αn+kw tn, αn;u−,n, u+,n, u−x,n, u+x,n.
• Determine an initial guess of the solutionuni,+11 at the time leveltn+1.
Form=1,2, . . . ,
• (**) Determinejm such thatxjm 6α n+1
1 < xjm+1. Determine the coefficients
and the correction terms forunx+1 anduxxn+1 atxjm. Substituteu n+1
i,m foruni+1 in
• If j0 6= jm, then for l = j0+1, . . . , jm, when j0 < jm, or for l = jm, jm+
1, . . . , j0,whenj0> jm,first getτm using (16), then determine the correction
Ql,mn+1/2 to(ul,mn+1−unl)/k using the technique described in section 3.
• Solve the tridiagonal system foruni,m+1+1.
• Interpolate{uni,m+1+1}to get u±m,n+1+1, u±x,m,n++11,if necessary.
• Determine αnm++11=αn+k 2 w n+wn+1 m+1 , where wmn++11=w tn+1, αmn+1;u−m,n++11, u+m,n++11, ux,m−,n++11, u+x,m,n++11. • If|αn+1
m −αnm++11|> ε, a given tolerance, thenm=m+1.Go to (**).
• If |αnm+1 −αnm++11| < ε, then set all quantities { }nm+1 to { }n+1, in other
words we drop the{m}notation and accept these values for the time leveltn+1.
Determine the next time step sizek,
k=min h, w nk wn+1−wn . (35)
Go to the next time step.
7. Numerical examples
Here we present three different examples. The first two are from real applications. The third one is a constructed example for the nonlinear moving interface problem. Example 1. Two phase Stefan problem. This is a classical example in tracking a freezing front of ice in water. The description of the problem is excerpted from [11], where Furzeland used this example to compare different methods. The thermal properties are ki = conductivity, Ci = ciρ with ci = specific heat and ρ = density
(assume the same in each phase), σ = Lρ with L = latent heat. Subscript i = 1 denotes phase 1 (ice), 0< x < α(t); i=2 denotes phase 2 (water), α(t) < x <1. The equations are
Ci ∂ui ∂t =ki ∂2u i ∂x2 , i=1,2, t0< t, u1=u∗<0, x=0, t >0, u1=u2=0 σ∂α ∂t =k1 ∂u1 ∂x −k2 ∂u2 ∂x onx=α(t), t0< t,
Table 1
Grid refinement analysis for example 1 att=1.0.
N kEN k∞ ratio |Eα| ratio 20 4.3067×10−3 1.0941×10−4 40 9.7147×10−4 4.4333 2.4947×10−5 4.3857 80 2.3713×10−4 4.0967 5.7298×10−6 4.3539 160 5.8160×10−5 4.0772 1.3828×10−6 4.1434 320 1.4213×10−5 4.0920 3.3721×10−7 4.1009
where the solutionu represents the temperature. This problem has an exact solution
α(t) =2φ√κ1t, u1(x, t) =u∗ 1−erf(x/2 √ κ1t) erfφ , u2(x, t) =u0 1− erfc(x/2 √ κ2t) erfc(φpκ1/κ2) ,
whereκi=ki/Ci,erf(x) is the error function, u∗ and u0 are two constants, andφis
the root of the transcendental equation e−φ2 erfφ + k2 k1 r κ1 κ2 u0e−κ1φ 2/κ 2 u∗erfc(φpκ1/κ2) +φλ √ π C1u∗ =0,
which can be easily computed, say using the bisection method. The exact solution is used as the initial condition at time t0 = 0.5, as well as the boundary condition at
both ends,x=0,andx=1.The following thermal properties are used
k1=2.22, k2 =0.556, C1=1.762, C2=4.226, σ=338,
withu∗=−20 and u0=10 which givesφ=0.2054269. . ..
Table 1 shows the results of a grid refinement analysis, wherekENk∞is defined
as the infinity norm of the error at the fixed time t,i.e.,
kENk∞=max
i u(xi, t)−u N i ,
where N is the number of grid points as defined in section 2. We define uNi as the computed solution at the uniform grid points xi, i=1,2, . . . , at the final time t. Eα
is the error betweenα(t) and the computed interface at the final time t.We see that doubling the number of grid points gives a reduction in both errors by a factor of roughly 4, indicating second order accuracy. Figure 2(a) shows the true solution and
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -20 -15 -10 -5 0 5 10 x u
The computed temperature, n = 40, t = 1.0
__ : True soln. o : Approx. soln. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -2 0 2 4 6 8 10x 10 -4 x
Error plot: True - Approx, n = 40, t = 1.0
(a) (b)
Figure 2. The comparison of the exact and computed solution att=1.0 for example 1 withN =40. (a) The solid line is the exact solution and the dots are the computed results at grid points. (b) The corresponding error plot.
the computed solution for N =40. Figure 2(b) shows the corresponding error plot. We see that the error in the solutionuis relatively large around the interface compared to other grid points but not significantly so. Globally we obtain second order accurate results at all grid points (see table 1). We also tested a similar example described by Keller [14] for modeling a melting and freezing problem at a constant speed. The results again agreed with our analysis.
Example 2. This simulation shows the temperature profile of an ice sheet during the process of a glaciation. Heine and McTigue [12] proposed a one-dimensional thermal model to study the temperature change of a glacier. The ice sheet gradually grows to a thickness of 3000 m over approximately 10,000 years. At the same time, a heat source, in the form of a geothermal heat flux, is warming the glacier from a depth of 4000 m in the rock. The point of interest in this problem is the interface between ice and rock. If the temperature approaches the melting point, the glacier may begin to slide with catastrophic effect. A similar problem can be found in [13].
The mathematical model Heine and McTigue used is similar to example 1, see figure 3. ρcp ∂T ∂t = ∂ ∂x k∂T ∂x , 06x6h(t), t >0,
where T is the temperature. However, both the heat capacity ρcp, and the thermal
conductivity k are functions of T(x, t) now. So this is a nonlinear model for the temperature. The estimated height of the ice sheet is
h(t) =h0+δh+ (h∞−h0−δh) 1−e−t/tr
x=0 geothermal heat q0 x=h0 ice x=h t( ) height of glacier h t( ) Ts x h0 δh ice rock heat Tx= −q0 [kTx] 0= [ ] 0T = rock x=0
Figure 3. Geometrical frame of Heine and McTigue’s glacier model.
whereh0is the thickness of rock,δhis an (arbitrarily) small initial ice layer thickness,
h∞ is the final total thickness (rock + ice), and tr is the rise time for the ice sheet
growth. The boundary conditions consist of constant heat flux from the inner layer of the earth and constant temperature at the top of the ice sheet:
Tx(0, t) =−q0, T h(t), t
=Ts.
The thermal properties are
kice=C1e−C2(T+273.15), krock=constant,
(ρcp)ice=r1+r2T, (ρcp)rock=constant.
The initial conditions are
T = Th0+ q0 krock (h0−x), if 06x6h0, − 1 C2 log e−C2Ts−C 2(h0+δh−x)Γ , if h0 < x < h∞, where Γ= q0 C1 e−273.15C2, T h0 =− 1 C2 log e−C2Ts−C 2δh .
0 0.5 1 1.5 2 2.5 3 3.5 x 1012 -25 -20 -15 -10 -5 0 5 TIME (seconds) TEMP (deg. C)
Figure 4. Temperature on the interface between the rock and ice.
The following thermal properties are used
C1=9.828, C2 =0.0057, r1=1.936×106, r2=6.600×103,
krock =2.50, (ρcp)rock =2.30×106, q0=0.05 w/m2, Ts=−25.0◦C.
The geometrical parameters are
h0=4000 m, δh =1 m, h∞=7000 m, tr=1.9×1012 s.
In this example, we have a fixed interfacex=h0with the natural jump conditions
[T] =0 and[k∂T /∂x] =0 since temperature is continuous and there is no heat source at the interface. This corresponds to case 2 in our previous discussion with w ≡ 0.
On the other hand, at the moving boundaryh(t) we know the surface temperatureTs.
So it is also a one-phase nonlinear moving boundary problem.
Figure 4 shows the temperature history of the interface between rock and ice. The geophysical implication of the result can be found in [12].
To check the correctness of our algorithm, we constructed a problem for which we have an exact solution within the same geometrical frame. The numerical results confirmed that our method converges to the exact solution with second order accuracy. The details are omitted here due to space limitations.
0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 x t (5,1) (3,1) (2,1) (1.8,1) (1,1) (1,5)
Figure 5. Moving interfaceα(t), 06t61, from left to right,(β−, β+) = (5,1),(3,1),(2,1),(1.8,1),
(1,1)and(1,5).
Example 3. Nonlinear moving interface example for case 2. In this example, we take
λ=1 in (1) and construct the following exact solution:
u(x, t) = ( sin(ω1x)e−β −ω2 1t, ifx6α(t), sin ω2−ω2x e−β+ω22t, ifx > α(t), (36)
for some choice ofω1, ω2,β−,andβ+.The source termf(x, t)is discontinuous
f(x, t) = ( −1 2ω1sin(2ω1x)e− 2β−ω2 1t, ifx6α(t), 1 2ω2sin(2ω2−2ω2x)e−2 β+ω2 2t, ifx > α(t). (37)
We assume that the solution u(x, t) is continuous across α(t). So the interface α(t) can be determined from the scalar equation
sin(ω1α)e−β
−ω2
1t=sin ω2−ω2αe−β
+ω2
2t. (38)
This equation has a unique solution if we take, for example,π < ω1 <2π and also
π < ω2<2π. Figure 5 gives the plot ofα(t)as the parameters (β−, β+)change on a
uniform grid. We can see how the interface crosses the grid. This example is adapted from [3].
0 0.2 0.4 0.6 0.8 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 t=0 t=0.1
The profile of u(x,t), 0 <= t <= 1, N = 160, k=h/2
x u 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -0.01 0 0.01 0.02 0.03 0.04 0.05 x u(x,0.1) __ : True soln. o : Approx. soln. (a) (b)
Figure 6. (a) The profile of the computed solutionu(x, t)fromt=0 tot=0.1 withβ−=β+=1, and
N=160. (b) The comparison of the exact and computed solution att=1.0 withβ−=5.0,β+=
1.0, andN =80. The solid line is the exact solution and the dots are the computed results at the grid points.
Table 2
Grid refinement analysis for example 3 att=0.1.
N kENk∞ ratio |Eα| ratio 40 6.5460×10−3 2.2605×10−2 80 6.0108×10−3 1.0890 8.6674×10−3 2.6081 160 1.1557×10−3 5.2009 2.7449×10−3 3.1575 320 2.9347×10−4 3.9381 6.9958×10−4 3.9238 640 7.3175×10−5 4.0106 1.7508×10−4 3.9957 1280 1.8237×10−5 4.0124 4.3725×10−5 4.0042 2560 4.5462×10−6 4.0115 1.0913×10−5 4.0065 5120 1.1370×10−6 3.9982 2.7311×10−6 3.9960
The nonlinear ordinary differential equation for the motion is dα dt = (ω2 1−ω22)u(α, t) ux(α−, t)−ux(α+, t) . (39)
The initial and boundary conditions are
u(0, t) =0, u(1, t) =0, (40)
u(x,0) =
sin(ω1x), ifx6α(0),
sin(ω2−ω2x), ifx>α(0). (41)
The jump conditions used are [u] =0, βux(α, t) =−β+ω2cos(ω2−ω2α)e−β +ω2 2t−β−ω1cos(ω1α)e−β−ω21t.
Figure 6 shows a typical computed profile ofu(x, t) as time changes. We can clearly see how the interface moves and crosses the grid with time. This example is more challenging than traditional examples because the jumps in the coefficient and in the derivative of the solution are significantly larger than in the previous examples.
Table 2 shows the grid refinement analysis both for the computed solution and the interface. In this case, the interface crosses several grid points during the first few time steps (see figure 5). However, second order accuracy is still achieved.
8. Summary
In summary we have developed a second order accurate immersed interface method for a class of one dimensional nonlinear moving interface problems with two typical interface conditions: (i) Stefan-like problems for which we know the solution on the interface. (ii) Problems in which we only know the jump conditions in the solution and the flux across the interface. Applications include the immersed boundary method and heat conduction in different materials. Numerical experiments have confirmed second order convergence of the methods proposed in this paper. Currently we are working on similar numerical methods for two dimensional problems. It is well known that the interface between the two phases in the two dimensional Stefan problem may develop singularities or complicated shapes even if it is initially smooth. Thus we are going to explore the possibilities of using the level set technique [21] to track the interface.
Acknowledgements
I am indebted to Prof. R. J. LeVeque for the original motivation and many useful discussions. I would also like to thank him and D. Calhoun for reading, correcting and commenting on this paper. I am grateful to Prof. D. McTigue for useful discussions concerning his glacier model. It is very beneficial talking to B. Merriman, S. Osher, J. Zou and other people about the problems in this paper.
Appendix. Proof of theorem 4.1
Here we present a brief proof of theorem 4.1. We first derive one more interface relation using the known information and then give the proof of the theorem.
Lemma A1. Letu(x, t) be the solution of (1)–(2) with the jump conditions (3) and (4), then u+xx =ρu−xx+ 1 β+ q0+ [f]− v β+ w−λu ++β+ x + u−x w(1−ρ) +λ(u+ρ−u−) +βx−−β − β+β + x , (42) whereρ=β−/β+.
Proof. From equation (1) we know
β+u+xx+β+xu+x −λu+u+x −f+−u+t
=β−u−xx+βx−u−x −λu−u−x −f−−u−t . (43) Plugging (32) and the second jump condition (4) into (43) and arranging terms, we
get (42).
Proof of theorem 4.1. Expanding u(xj−1, t), u(xj, t) from the left hand side, and
u(xj+1, t) from the right hand side ofα, we have
u(xj−1, α) =u−+ (xj−1−α)u−x +12(xj−1−α) 2u− xx+O h3 , (44) u(xj, α) =u−+ (xj−α)u−x +12(xj −α) 2u− xx+O h3 , (45) u(xj+1, α) =u++ (xj+1−α)u+x +12(xj+1−α) 2u+ xx+O h3 . (46) So j+1 X k=j−1 si,k−j+2u(xk, t) =si1 u−+ (xj−1−α)u−x + 12(xj−1−α) 2u− xx +si2 u−+ (xj−α)u−x +12(xj−α) 2u− xx +si3 u++ (xj+1−α)u+x +12(xj+1−α) 2u+ xx +O h4−i, (47)
for i = 1,2 and 3. Notice that we have used the fact that sij is of order O(h1−i).
From interface relations (3), (4) and (42) we can solve foru+, u+x, andu+xx in terms of u−, u−x, and u−xx and substitute them into the expression above and collect terms to obtain j+1 X k=j−1 si,k−j+2u(xk, t) = si1+si2+si3 u−+ si1(xj−1−α +si2(xj−α) +si3a23)u−x + (xj−1−α)2 2 si1+ (xj−α)2 2 si2+ (xj+1−α)2 2 ρsi3 u−xx +si3 q+(xj+1−α)v β+ + (xj+1−α)2 2β+ q0+ [f]− v β+ w−λu ++β+ x +O h4−i
= 3 X k=1 sikak1, 3 X k=1 sikak2, 3 X k=1 sikak3 ! · u − u−x u−xx +Cj,i+O h4−i =eTi · u − u−x u−xx +Cj,i+O h4−i ,
whereei is the unit vector inR3, for example,e2T = [0,1,0]. This completes the proof
the theorem.
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