The Kačanov method for some nonlinear problems

MATHEMATICS

ELSEVIER

Applied Numerical Mathematics 24 (1997) 57-79

The Ka anov method for some nonlinear problems Weimin H a n

a,*, S©ren Jensenb,1, Igor Shimanskyb,2

a Department of Mathematics, University oflowa, Iowa City, IA 52242, USA b Department of Mathematics, University of Maryland, Baltimore County, Baltimore, MD 21228, USA

Abstract The Ka~anov method is an iteration method for solving some nonlinear partial differential equation problems. In each iteration, a linear problem is solved. In this paper, we discuss the use of the Karanov method in the context of two model problems. We show the convergence of the Ka~anov iteration sequences, and derive a posteriori error estimates for the Karanov iterates. Numerical examples are given showing the convergence of the method and the effectiveness of the a posteriori error estimates. ¢ 1997 Elsevier Science B.V.

1. Introduction The Ka~anov method is an iteration method for solving nonlinear problems, via linearization. An early reference on the method is [25], where the method is applied to compute a stationary magnetic field in nonlinear media. Convergence of the method is proved in the context of the particular application there, though the technique of the proof is rather general. A numerical example in [25] shows that the method is quite effective for certain problems. See also [14]. Later, the method is applied to a nonlinear elasticity problem in [27]. For applications of the method in solving variational inequalities for transonic flows in gas dynamics, see [9,15]. A general convergence result of the method is presented in [28] (see also [33]) for solving a nonlinear variational inequality of the first kind (i.e., it is an inequality because the problem is posed over a non-empty closed convex subset rather than over a whole space). In [21], convergence of the method is proved for solving an even more general nonlinear variational inequality of the mixed kind (i.e., it is an inequality both because the problem is posed over a non-empty closed convex subset and because the problem contains non-differentiable terms), and the result is used in solving a nonlinear variational inequality of the second kind arising in elastoplasticity. We will return to give five more concrete examples where the method is applicable later in this introduction. * Corresponding author. E-mail: [email protected]. Partially supported by Research Grant Council of Hong Kong. J Partially supported by ONR grant N00014-90-J-1238. 2E-mail: [email protected]. 0168-9274/97/$17.00 © 1997 Elsevier Science B.V. All rights reserved. PI1 S0168-9274(97)00009-3

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W. Han et al. / Applied Numerical Mathematics 24 (1997) 57-79

In this paper, we apply the KaSanov method (at times called the secant modules method) to solve some nonlinear problems. Convergence of the method follows from the abstract result proved in [28,33]. The main purpose of the paper is to derive a posteriori error estimates for the Ka6anov iterates--for the class in which they have been shown to converge, including the above examples and two more explicit ones below--and exhibit the efficiency of estimates in the concrete examples. We now know of only one other work where this has been attempted. Recently, after the completion of the present work we find that in [31] is given a different a posteriori error estimate--also via duality theory, but in a different manner. That estimator is seemingly more involved to compute as it requires the computation of another linear problem roughly of the same size as the one required to produce the most recent term Uk+ 1 in the Ka6anov sequence. There, no numerical results are given. As an introductory example, let us consider the problem - d i v ( a ( [ V u [ ) Vu) = f u = 9 c~([Vul2)(0u/On) = h

in ~ , on/"1, on/"2,

(1.1)

where, ~ C ]RN is a nonempty, open bounded domain with a Lipschitz boundary ~O = Tl tO T2, F1 fq/"2 = 0. We will impose conditions on the data in Section 2 for the unique solvability of the problem (1.1). The Kaeanov method for solving (1.1) is the following. Given an initial guess u0 with u0 = g on F1, we define a sequence of iterates {uk} by

-div(a(lVukle) vuk+l) = f Uk+l = 9 O~(l~uk[2)(OUk+l/~n)

= h

in J'2, on F1,

(1.2)

on /"2,

for k = 0, 1, . . . . Notice that in (1.2), we solve a sequence of linear problems instead of the nonlinear problem (1.1). Under suitable assumptions on the given data, the Ka6anov method is convergent, see next section. We now mention a convergence result when the Ka6anov method is applied to solve a general variational inequality problem. Let V be a Hilbert space, K C V a nonempty closed convex subset. Let E : K --+ IR be Gateauxdifferentiable, l E V* a continuous linear functional on V. Then let us consider the constrained minimization problem: find u E K,

such that

E ( u ) - l ( u ) = inf { E ( v ) - l ( v ) } . vCK

(1.3)

Since E is G~teaux-differentiable, a solution of (1.3) satisfies the variational inequality

u • K,

( E ' ( u ) , v - u} > / l ( v - u),

Vv • K.

(1.4)

When E(v) is quadratic in v, the differential operator associated with the left side of (1.4) is linear. When E(v) is not quadratic in v, the differential operator associated with the left side of (1.4) is nonlinear, which makes the variational inequality problem more difficult to be solved. Assume for E'(u), we can find a functional B : K × V × V -+ R, such that

(E'(u),v-w} = B ( u ; u , v - w ) ,

Vu,

• K,

(1.5)

W. Han et al. / Applied Numerical Mathematics 24 (1997) 57-79

59

and for fixed u E K, (v,w) ~-~ B ( u ; v , w ) is a bilinear form on V. Then, (1.4) can be rewritten as

u E K,

B(u; u, v - u) >~ l(v - u),

V v E K,

and the Ka~anov method for (1.4) reads: for k = 0, l, . . . .

uk+l E K,

Vv E K,

B(uk;uk+l,V -- Uk+l) >~ l ( v - - uk+l),

(1.6)

where u0 is an initial guess chosen from K. A proof of the following convergence theorem is found in [33].

Theorem 1.1. (a) Assume for each u E K, the bilinear form (v, w) ~ B(u; v, w) is symmetric from V x V to N. Assume there are constants (~1 > 0 and 6o > O, such that

I (u;v,w)l <

~lllvll Ilwll,

v u E K, v,w ~ Is,

(1.7)

and B(u;v-

w,v-

w) >6oil v - w l [ 2,

Vu, v, w E K.

(1.8)

Then the problem (1.6) has a unique solution uk+l E K, which is also the unique solution of the quadratic minimization problem 1

B ( u k ; v , v ) -- l(v) --+ inf,

v E K.

(1.9)

(b) Further assume E p : K -+ V* is continuous and strongly monotone, i.e., for a constant Po > O,

(E'(u)- E'(v),u-v)

>>.poilu-vii 2,

Vu, v E K.

(1.10)

Also assume the following key inequality holds: E(v)-

E(u) <~ l ( B ( u ; v , v ) -

B(u;u,u)),

Vu, v E K.

(1.11)

Then (1.3) has a unique solution u E K, which is also the unique solution of the variational inequality problem (1.4). The Kaganov method converges, i.e., uk~uinY,

ask-+ec.

We quickly preview the kind of fairly restrictive---conditions on the function a appearing in the problem (1.1) known to guarantee the hypotheses in this theorem (see Theorem 2.1): The function a : R+ --+ I~+ is C 1, and there are positive constants a, c and d, such that

a <. a(s) .< c,

Vs

>0.

(n2)

We note that (H2) is a standard assumption in the proof of Theorem 1.1, cf. [28,33]. Remarkably, there exist several applications in which (H2) is reasonable. Needless (or so one should think) to say, on the other hand, not all a in nonlinear, elliptic equations in the divergence form given in (1.1) satisfy (H2). For example, when a is increasing with some growth conditions, it is appropriate to use different techniques, monotone operators, e.g., to obtain a priori estimates for convergence; cf. [5,10-12,17,23] and the references there. Some work on a posteriori estimates for that case can be found in [20,24]. We are satisfied here to solve those problems with ct that do satisfy (H2), after next giving five additional examples found in the literature where (H2) is satisfied. (After all, no nonlinear resolution method--be

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W. Han et al. / Applied Numerical Mathematics 24 (1997) 57-79

it Newton, FEM, nonlinear Galerkin, or what have you--is expected to work globally and well for all problems.) Hopefully, these examples will convince the reader that, despite the restrictions, the study in hand is worthwhile, at least for some specific classes of interesting problems. We note that in some applications, an upper bound for the gradient of the solution IWl is known or can be computed. Then the inequalities in the assumption (H2) are required to hold only in the actual range of the quantity s = [Vu[. Outside this range, in case needed, we can easily modify the function a in such a way that (H2) is satisfied. Then we apply the Kaeanov method to solve the problem with the modified coefficient function a, which admits the same solution as the original problem. Example 1.2. In [26], the minimal surface equation is treated numerically, where (with s = [Vu[ 2) 1 Ct(.S) -- 1,1""-1-8"1/2'} "

8 > O.

They there explicitly assume that V u c L cc with a uniform bound of the the form: ]Vu] ~< ~ < 1, in which case all the assumptions in (H2) are fulfilled. In fact, a = (1 + ~2)-1/2, c = 1, and d = (1 +

Example 1.3. In [18,25], a nonlinear magnetostatic field problem is treated numerically, u is then the magnetic scalar potential associated with the magnetic vector potential A = (0, 0, u). Two of Maxwell's equations yield the magnetic induction B = curl A and the relation u B = H , where u = 1/# is the reciprocal of the magnetic permeability # and H is the magnetic field. Eq. (1.1) comes about by another of Maxwell's equations curl H -- j = (0, 0, f ) , where j is the magnetic current. The reciprocal of the magnetic permeability u--which in this paper would be denoted a - - i s taken (and empirically shown) to be a function of In[ 2 = IVu[ 2. In [18, Fig. 5], the function u is depicted as dependent upon [XTu[2. This function is later fitted with a monotonically increasing function. However, in the range all computations of [18] lie (if the units are consistent from one section to another in [18]), it turns out that IVul 1 and, in that region, u is actually decreasing. If one fits the function in this section, the resulting function clearly satisfies (H2), compare [18, Fig. 5] with [33, Fig. 25.10]. The authors of [18] briefly consider using the Karanov method for their function, which does not satisfy (H2), and they observe that they indeed had trouble obtaining convergence. It is noteworthy that with the EGSN (Extrapolated Gauss-Seidel-Newton) method used in [18], they count the number of iterations until convergence in hundreds. Example 1.4. In [30,32] a nonlinear heat equation (Stikker's) is derived for the problem of heat conduction in steel coils during the batch annealing process. Here--somewhat unusually for heat equations--the heat conductivity is a function of the temperature gradient XYu. Consider as in [32] a one-dimensional example with =

+ _dye'

> O,

where they explicitly assume that v/S = ux lies in an interval on the nonnegative half of the real axis that does not contain any singularities of a. Quoting from [30], "Both an infinite flux and a simultaneous occurrence of a zero flux and a nonzero temperature gradient are nonphysical situations",

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W. Han et al. / Applied Numerical Mathematics 24 (1997) 57-79

gives the underlying reason. Physical assumptions on _a, _b, c and d given in inequalities (14) in [32] are ad

a

b-~>l

and

~>0,

which then implies that b and c are of the same sign. If a_b < 0, then (H2) is satisfied if the gradient ]Vu I is bounded from above (which is a reasonable assumption made in [32] considering the physical meaning of the problem). If a b > 0, then no such bound is needed for (H2) to be satisfied. Example 1.5. In [29] (1.1) arises from Forchheimer flow in porous media [1-3,6] and Ergun's law for incompressible fluid flow [4,8]. In the case of Forchheimer's law (again with s = IVul2),

a(s) =

2 _.C-~ 4_C 2 -}- d81/2'

s / > 0,

in the absence of gravity and where _c_= # / k (> is the viscosity of the fluid, k is the permeability of the medium) and _d = 4bp (b is a dynamic viscosity and p is the fluid d e n s i t y ) f a l l taken to be positive constants here for simplicity. Under the constraint, as posed in [29], that s is uniformly bounded from above, (H2) is once more satisfied. Example 1.6. (H2) is also satisfied in the elastoplasticity problem treated in [28]. The main purpose of the paper is then to provide a posteriori error analysis for a convergent Ka~anov sequence. We will derive error estimates of the form--recall {uk} defined in (1.6)--

Ilu - ukll

f(uk,uk-l),

where the upper bound f ( u k , uk-l ) of the error depends only on two iterates uk and u a - l , which are available after we solve the two corresponding linearized variational inequalities (1.6). We will use the duality theory in convex analysis (cf. [7]). To do this, we need to introduce an auxiliary space Q, and a continuous linear operator A : V -+ Q, such that the original variational problem (1.3) can be rewritten as

u E V,

J(u, Au) = inf J(v, Av) vCV

(1.12)

with a functional J : V × Q --+ R, the extended real line. Define the conjugate function of J by

J*(v*,q*)=

sup

[(v,v*)+(q,q*)-J(v,q)].

vCV, qeQ

The following result is proved in [20]; there the duality theory technique is used in deriving a posteriori error estimates for linearization of nonlinear problems. The technique has also been used in a posteriori error analysis for regularization methods for solving variational inequalities of the second kind, cf. [19,22]. Theorem 1.7. Assume."

(1) V is a reflexive Banach space, Q a normed space. (2) J " V × Q --+ R is a proper, lower semi-continuous, strictly convex function. (3) 3 uo E V such that J(uo, Auo) < e~ and q ~-+ J(uo, q) is continuous at Auo.

W. Han et al. /Applied Numerical Mathematics 24 (1997) 57-79

62

(4) J(v, Av) --+ +co, as Ilvll ~ ~ , v e v. Then problem (1.12) has a unique solution u E V. Assume further (5) J is G~teaux-differentiable at u, and define

D ( u , v ) = J(v, Av) - J(u, Au) - (J'(u, A u ) , ( v - u, Av - Au))

(1.13)

for any v C V with J(v, Av) < oc. Then, D ( u , v ) ~ J(v, Av) + J*(A*q*,-q*),

Vq* E Q*.

(1.14)

A general procedure for a posteriori error analysis of the Ka6anov method then consists of two steps. First we find a suitable lower bound for D(u, uk), which measures the error u - uk. In the second step, we take an appropriate q* so that the estimate (1.14) is accurate. Usually q* can be chosen in terms of the approximate solutions uk and Uk-1 by checking the relation satisfied by uk and uk-1. In the next two sections, we will apply this strategy for a posteriori error analysis of the Ka~anov method for two important nonlinear problems arising in practice: a stationary conservation law, and a subsonic flow problem. In Section 4, we apply the Ka~anov method for the finite element discretizations of the stationary conservation law. Again, we have the convergence of the method and an a posteriori error estimate for the iterates. The discussion on applying the Ka~anov method for solving discrete analogues of the subsonic flow problem is similar, and is omitted. We present some numerical examples in the last section. The examples confirm the convergence of the Ka~anov method and show the effectiveness of the a posteriori error estimates.

2. The Ka~anov method for a stationary conservation law In this section, we consider the stationary conservation law (1.1) and the Ka~anov method (1.2). Let us define V = H'(S~),

K={vEV:

v=gonF1}, 82

1 f c~(t) dr, o

= f 9(IV, l) dx, =/

(IVul 2) V v V w d x ,

a,2

s?

1"2

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W. Han et al. / Applied Numerical Mathematics 24 (1997) 57-79

Then, the problems (1.1) and (1.2) can be rewritten as

E(u) - l(u) = min!, !2 B(uk; uk+l Uk+l) - l(uk+l) ----min!,

uEK; uk+l E K.

(2.1) (2.2)

We will need the following assumptions. C R N bounded and Lipschitz continuous, F1 ~ 0, f E H 1(g2)*, 9 E n I (~2), and h E H1/2(/"2)*.

(H1)

The function a : ~ + -+ ll~+ is C 1, and there are positive constants a, e and d, such that

a <. a(s) <~c,

fl"(s) = 2 a ' ( s 2) s 2 + oz(82) ~ d,

a'(s) <. 0,

Vs >>.O.

(H2)

Here in (H1), we assume that the given Dirichlet datum g is the trace of a g E H I ( ~ ) on/"1. We note that (H2) is a standard assumption in applications such as nonlinear magnetostatic field problem [25] or elastoplasticity problem [28]. It is proved in [33] that T h e o r e m 2.1. Assume (H1) and (H2). Then, (2.1) has a unique solution u. Let uo E K be given. Then for k = 0, 1, . . . . (2.2) has a unique solution Uk+l, and uk --+ u in H 1($2),

as k --+ cx~.

For an a posteriori error analysis, we introduce Q = (L2(~'2)) u × u

× L2(/"2),

Av = (Vv, vie2) , and identify Q* with Q. For q E Q, we will use the notation q = (ql, q2) with ql E (L2(j'2)) y x g and q2 E L2(/"2). A similar notation is used for q* --- (q~, q~) E Q*. N o w we define a functional from V×Qto~:

d(v.~q)

= ~ f~? j~ql[2½ a ( ~ ) d ~ d x [

fr2 h q 2 d s - fs~fvdx,

+c~,

if v E K , otherwise.

Then, from the definition of a conjugate function, we can find

J*(a*q* - q * ) = '

fs~(q~Vg+ f g + l q l •l b ( I q •l l ) - f°b(Iqtl) 2 ~1a ( ~ ) d ~ ) d x if divq~ = f in ~ , -q~n = q~ = h on/"2, +co,

otherwise,

where, t = b(iq~'[) is the unique solution of the equation Iq;I = °L(t2) t" N o w let us consider the difference D(u, v). For a lower bound, we have

q- frz ghds,

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W. Han et al. / Applied Numerical Mathematics 24 (1997) 57-79

D(u, v) = J (v, Av) - J (u, Au) - ( J' (u, Au), (v - u, Av - Au) ) 1

1

SSI o

1"2

d [a(iVu+tTV(v_u)12)(Vu+tTV(v_u))]V(v_u)dTdxdt

o

1

>1

o s2 For an upper bound, we have, for q~ = h and any

q~ (L2(j'~))NxN,such that div q~ = f E

in [2 and

-q~n = h on/"2, D(u, uk ) <~J(uk, Auk) - J*(A*q*,-q*) IVukl2

o

~

r2

b(Iq~l)2

+S (q~Vg+fg+lq~[b([q~[)- i o

$2

½a(~)d~)dx+ f g h d s 5

IV~kl2 Y2 b(]qT])2

t2

52

1"2

IV-kl 2

: S

i

10l(~)d~dx -~-S

b(IqTI)2

=f[

f 7 kl2

(q~-~-Iq~]b(Iq~I))dx-

S O~([~Tuk-.12)~ " k ~ 7 ( U k - g ) d x

a

+ Ow -,l )w )vg+lq;tb(Iq;O

b(IqTI)2

- ~(Iw~_,l ~) Iw~l ~] dx.

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W. Han et al. /Applied Numerical Mathematics 24 (1997) 57-79

Hence, I ? k12

ldllV(u~-u)ll~2(m <.f [ n

½~(~)d~+ (qr + ~(IW'k-,I 2) Vuk) Vg -+-Iq~lb(Iq~l) b([qTI)2 - a(IVUk_l[ 2) IVukl 2 dx,

Vq~ E (L2(y2)) Nxg,

divq~ = f in ~2,

-q~n = h on/"a.

(2.3)

In particular, we can take q~ =

-a(IVuk-ll 2) Vuk,

then we have the estimate, !2 a(~)

-

n

d~

b(c~(IVuk_ ~I2) IVukl)2

q + a(lVuk-ll 2) IVukl (b(o~([VUk-l[ 2) IVukl) -IVukl)] dx. (2.4) 1

When a is a positive constant, it is easy to verify that the right side of the error estimate (2.4) vanishes, i.e., the error estimate is exact if the differential operator associated with the original problem is linear. By a continuity argument, the right hand side of (2.4) is guaranteed to be small if the coefficient function a is close to a positive constant (and satisfies (H2)). We remark that the Ka6anov method can be applied to solve some nonlinear elasticity problems as well, cf. [27]. Besides the convergence of the Ka~anov method from Theorem 1.1, we can derive similar a posteriori error estimates for the Ka~anov sequences as we do for the problem (2.1) in this section.

3. The Ka[anov method for a subsonic flow problem

Consider the flow of a perfect compressible irrotational fluid in a smooth domain f2 C N N. Denote by u the potential. Then the velocity of the flow is Vu. Assume the flow density is

IV,/Z[2

p(lVul 2) =

1-[(7

+ 1 ) - ~ - - 1)]c 2

) 1/("/--1) '

(3.1)

where 7 is the ratio of specific heats, 7 > 1, c, is the critical velocity, and for convenience, we have assumed the density of the fluid in the motionless state to be 1. The flow is said to be subsonic if IVul < c, in S2. Let 0I? = / " 1 N/"e. We will specify the boundary conditions 0u u = 91 on/"1, P([Vu] 2 ) ~ = 92 o n / " 2 ,

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W. Han et al. / Applied Numerical Mathematics 24 (1997) 57-79

where gl E H1/2(FI), g2 E L2(F2). Denote V ---- H I ( Q ) ,

Vg, -- {v C V: v = gl on/~l},

g~o = {v c Vm"

IVvl~
c0
In practice, cO is taken as close to c, as feasible, naturally keeping a certain positive distance (the nature of the problem changes as IVu I ~ c,, after all). Then, the subsonic flow problem we consider in this section is

~ K~o, f p(IVul2) w v ( v - u)dx > f g2(v - u)ds,

g v C K~ o.

(3.2)

V2 Assume gl is such that

K~o ¢ 0,

(3.3)

which is guaranteed if 91 is small. It is proved in [16] that the variational inequality problem (3.2) has a unique solution u E K~ 0, and it is the unique solution of the minimization problem

u E K~ o,

E(u) - l ( u ) =

inf

{E(v) -/(v)},

(3.4)

vCKc o

where

E(v)_ 7+1c2, f (1 27

[Vvl2 )~/(,-1)

[(7 + 1 ) / ( 7 -- 1)] c 2

dx,

(3.5)

¢2

l(v) = -- / g2v ds.

(3.6)

The Ka~anov method for solving the problem (3.2) is: with an initial guess u0 E K~ o, we compute a sequence of iterates {uk} through

Uk+IE

K~o,

/p(l~7Ukl 2)~7Uk+l ~7(V-

'/Lk+1) dx ~ /ff2(v-/Zk+l)d8 ,

s~

V v E Kco. (3.7)

F2

To verify the convergence of the sequence {uk}, we notice that (1.5) is satisfied with

B(u;v,w) =

/ p(IVul2)V v V w d x .

(3.8)

$2 With the chosen density function p and the set K~ 0' it is easy to see that the inequalities (1.7), (1.8) and (1.10) are satisfied. So in order to apply Theorem 1.1 for the convergence of the Ka~anov sequence, it remains to prove the key inequality (1.11). L e m m a 3.1. Let U C (0, 1), 7 > 1. Then the function

f ( V ) = (1 - V) 7/(7-1) -+-

7 7-1

(1 - U)I/('r-UV

defined for V E [0, 1], has its minimum at V = U.

W. Han et al. / Applied Numerical Mathematics 24 (1997) 57-79

67

Proof. From

f'(Y) - 721

"7

(1 - V)l/(3"-')(-1) +

(1

-- U ) 1/(3"-1) ~--- 0

we find V -- U. Since

f " ( V ) - (7 7- 1 ) 2 ( 1 - V)(2-3")/(7-1) > 0 ,

f o r V E (0, 1),

we conclude that f ( V ) has its minimum at V = U.

[]

From Lemma 3.1, we have f ( V ) ~ f(U), i.e., (1

- V ) 3"/(3"-1) --

(1

-

U) ~/(~-1) ~ -

-

(1 - U)I/(3"-I)(V- U).

7

7

1

Taking

IVul 2 IVvl2 U = [(7 + 1)/(7 - 1)] c2. [(7 + 1)/(7 - 1)] d ' in the above inequality and integrating over Y2, we then obtain the key inequality (1.11). Now we can apply Theorem 1.1 to obtain V =

T h e o r e m 3.2. The Ka6anov method for the subsonic problem (3.2) converges, i.e., uk--+u

ask--+oo.

To perform an a posteriori error analysis for the Ka~anov sequence, define

Q = (L2(~2)) N×N x L2(F2),

Av = (Vv, vlr2) ,

and a functional from V x Q to R,

J(v,q)=

{ 7+1 f( ~

~

1-

'ql' 2 [(7+1)/(7-1)]c2.

) "//(3'-1)

f

dx - ]g2q2 ds,

if

v

E

/Cc~,

* ]

~2

F2 otherwise.

q-OO,

From the definition of the conjugate function, we find that

vE K~°

£2

+ / 7 +~1

F2

(t(Iq~l) 2

t(iq~l)2

+ c2")(1 - [(7 + 1)~(~-- 1)] c2,)

1/(3"-l)

dx,

Y2

if q~ = g2 on F2.

(3.9)

In the formula above, t = t(Iq~ I) is the unique solution of the algebraic equation t2

1-

[(7 + 1 ) / ( 7 -

1/(3"--l)

1)]c 2

t = ]q~l.

(3.10)

W. Han et al. / Applied Numerical Mathematics 24 (1997) 57-79

68

Now let us consider the quantity D(u, v) for u, v E Kco. For a lower bound, we first analyze the function t2 ) "T/('T--1) f(t) -- 7- -+C 1 2 1 - [ ( ' 7 + 1)/('y- 1)]c2, t~>O. 27 W e obtain t2 (2-"/)/(7-1) f"(t) = Thus,

D(u, v) >~dollV(u-

2 v)llL2( ),

(3.11)

where do=

(

c02~ m i n { 1 , ( l _ 1-c2,,]

c2 ~(2-7)/(~-1)} [(?+ 1)/("/- 1)]c2,] "

(3.12)

For an upper bound of D(u, uk), we notice that if we choose (3.13)

q~=--P(IVuk-,12) Vuk, then from (3.7), we get the inequality

/ V v q ~ d x + / v 9 2 d s <<.- /P(lVuk-l12) ]Vuk12dz+ f 92ukds, n

F2

n

Vv E/4co.

F2

Then, n(u, ~k) ~<2(~k, A~k) + J*(A*q*,-q*) =

~

c,

[(-y + 1)/(-r- 1)]d

D

iVUk_l[2 -

1 -

[(3' + 1 ) / ( ~ / -

) 1/(7-1)]Vuk12 1)]c, 2

~r+l(t(iq~l)2+c2,)(1 t(iqtl)2 ) 1/(~-1)] + ~ [('y + 1)/("/- 1)]c2, dx. Combining the above inequality with (3.11), we then have Theorem 3.3. The following a posteriori error estimate for the Ka~anov method for the subsonic flow problem holds."

69

W. Han et al. / Applied Numerical Mathematics 24 (1997) 57-79

2

dollY( u - uk)llL2(n) ~<

f[ $2

"y÷ 1 (

'~Uk' 2

2-~ e2 1 -- [('r + 1)/('y-- 1)]e 2

iVuk_ll2 --

where, q~ is chosen in (3.13),

) 1/(')'-1)

1 - [('r + 1)/('r - 1)] e 2

~+1 + --~

(t(Iq~[) 2+c2,)

) "7/(~'-1)

IVukl2

( 1 - [ ( , y + l ) /t(Iq~l)2 (T_l)]e2

) 1/("l- 1)]

dx

t(IqTI)is the solution of (3.10), and do is defined in (3.12).

To see the effectiveness of the estimate, we observe that when the density function approaches a constant (i.e., 3' --+ oc), the fight side of the error estimate goes to O. The error estimate is exact for problems where the associated differential operators are linear.

4. T h e K a ~ a n o v m e t h o d f o r discrete p r o b l e m s

In actual computations, a nonlinear continuous problem is first discretized, say by the finite element method, and then the discrete nonlinear problem is solved. Here we apply the Ka~anov method to solve discrete nonlinear problems. We will only consider the stationary conservation law discussed in Section 2, since the discussion on the subsonic flow problem in Section 3 is similar. Let Vh be a finite element space approximating Hi(f2), Sh the finite element subspace of Vh consisting of all the functions in Vh which are zero on Fl. We have Sh C H I 1(f2), where H~l(f2) = {v E HI(Y?): v =-0 o n / ' 1 } . For simplicity of exposition, assume the boundary condition function g can be represented exactly by a function from Vh. Then, a finite element solution Uh E Vh for the stationary conservation law is uhlr, = g,

B(

h;Uh, Vh) = l(vh),

'qV h E Sh.

(4.1)

Here, we use the notations for the forms B and l introduced in Section 2. For the convergence of the finite element solution, we have the following Cea's lemrna. L e m m a 4.1. Under the assumptions (H1) and (H2) in Section 2, there holds the inequality

IIv( -

-< c Vh--UhCSh inf I[V( u

-

h)llL2(n)"

Proof. Notice that the solution u of the continuous problem satisfies B(u;u,v)=l(v),

VveH~,(n).

From the above relation and (4.1), we find a useful relation

[ (O~(IVu[2) VU -- Ot(IVUhl2) V U h ) 0

VV h

:0,

V v h • S h.

(4.2)

W. Han et al. / Applied Numerical Mathematics 24 (1997) 57-79

70

Applying the assumption (H2), we have

S (c<(IV~l2) w - ,~(IWhl2) wh) v(~- ~h)a~ 1

iS-"

dt ( I W ~ + t v ( ~ - ~)12) ( W h + t V(~ - ~ ) ) dt V ( u

~7

-

Uh)dX

0

1

S S {,o,(i.o. ~7

(...

0

× (v~,, + t v ( . - ~
I (<.(lwl2) vu - ~(IVuhl2) vuh) v(~ - uh)d.,

$7

= f (<~(lWl2) vu-~(lWhl 2) vuh) v(~-v~)dx $7 1

X7

0 1

0

0

× (v~h + t v ( ~ - ~h)) + ~(IVUh + t V(~ -- ~h)l 2) V(~ -- ~h)) at V(~ - ~h)d~: 1

o

÷ <~(Iv~,, + t v ( ~ - ~h)l ~) ) dt Iv(~ - ~)1 Iv(~ - ",,)1 dx

W. Han et al. / Applied Numerical Mathematics 24 (1997) 57-79

71

1

-< / / ( ~ ( 1 ~

+~ ~(~ - u~)i ~) - ~)~ I~(~

~)l I~(~ - v~)l ~x

S2 0

~<

c IIv(u- uh)ll.(~)llv(u- vh)ll.(~).

Combining the two inequalities, we get c

live,,- u,,)ll.(,~) ~< ~ Ilv(u- v,,)[I.(.),

vv,. ~ v,.,

v,,l~, = 9.

[]

We can apply the KaSanov method to solve the discrete nonlinear problem (4.1). Choosing an initial guess u (°) E Vh such that u(hO)lr, = g, we define a sequence of iterates {U(he) } C Vh, for k = 0, 1, .... through U(k-I-1)

h

F 1 =

g,

B(u(hk); u h(k+l) , Vh) : l(vh),

VB h E Sh.

(4.3)

We can then apply the results and arguments in Section 2 for the discrete problems (4.1) and (4.3) to obtain T h e o r e m 4.2. Assume (H1) and (H2). Then, the problems (4.1) and (4.3) have unique solutions Uh and u. h(k+l) a n d U(hk) --+ Uh in H 1(~2),

as k --+ c~.

Furthermore, we have the a posteriori error estimate

IVu(hk)l2

,'-<~IIv(4~)- u,,)I1~.,(,,)~S[ ¢x

f

,'--(~)< +-(Iw~-')l') Iv4~)l

b(o,(iW~k-,)12 ) iVu(hk)l):

× (b(<~(IVu£k-')12)Iw£k)l)- IW2~)I)] dx, where the function b is the same as that defined in Section 2.

5. Numerical examples In this section, we present some numerical results to show the performance of the Ka~anov method and the effectiveness of the a posteriori error estimates derived in the previous sections. Our model problem is -(a(lu'12)u') : f u(O) = qo,

for x C (0, 1),

~ ( 1 ) = ql.

(5,1)

W. Han et al. / Applied Numerical Mathematics 24 (1997) 57-79

72

Then the Ka6anov method on the continuous level is: given a uo E cl(0, 1) satisfying uo(0) = qo and uo(1) = ql, find {Uk}k~=l recursively defined by

{ -(a(lu~12)U~+l)' = f Uk+l(O) = qo,

for x E (0, 1), (5.2)

U k + l ( 1 ) = ql.

We find that Utk+ 1 :

C k + l - fO f ( s ) d s

a(lu l=)

(5,3)

Since x

Uk+I(x)

= qo + / u~+l(x) dx, 0

the integration constant Ck+l can be computed from the boundary condition at x : 1:

/1

Uk+l(1) = qo + Ck+l

1

a(l~Z~12) dx - j

0

f l fo f(s!ds

a(lu~12) dx = ql.

(5.4)

0

In our numerical examples below, we use the composite trapezoidal role with 401 quadrature points to compute the integrals in (5.4). The a posteriori error estimate to be used is given by (2.4). We will also apply the Ka6anov method to compute finite element approximations of the problem (5.1). The discrete nonlinear system is derived from (4.1), and an a posteriori error estimate is given in Theorem 4.2. In our examples below, finite element approximations are constructed using piecewise linear elements on a uniform mesh with 21 nodes on (0, 1).

Example 5.1.

Let us consider the case when a ( s ) = 1 + A / ( 1 + s ) , 0 ~< A < 8. The assumptions (H1) and (H2) are satisfied, with d = 1 - A/8. Tables 1 and 2 show the convergence of the Ka~anov method and the effectiveness of our a posteriori error estimates for the Ka~anov iterates. In these

II(u ))' - uhllL2(O,1), el is the a posteriori error estimate for el from Theorem 4.2, and 01 : el~el is the effectiveness index. The corresponding values for the Ka~anov method on the tables el =

Table 1 a(s)= l+l/(l+s),

d=7/8,

f=

!

1, q = ( 0 , 0 ) , u 0 : 0

#

el

C1

O1

e2

e2

02

I

2.992e--03

4.372e-03

1.459

3.05 le--03

4.41 l e - 0 3

1.445

2

1.412e--04

2.063e-04

1.454

1.471e--04

2.109e--04

1.433

3

7.221e--06

1.048e--05

1.452

7.673e--06

1.092e--05

1.424

4

3.788e-07

5.494e--07

1.451

4.148e-07

5.873e-07

1.415

5

2.014e--08

2.857e-08

1.415

2.292e--08

3.226e-08

1.407

73

W. Han et al. /Applied Numerical Mathematics 24 (1997) 57-79 Table 2 ~(s)=l+6/(l+s),

a = 1/4, f = l ,

q = (0,0), ~ = 0

#

el

el

01

e2

e2

02

t

1.177e-00

6.200e-04

5.265

1.198e-04

6.254e-04

5.218

2

8.109e - 07

4.267e - 06

5.263

8.413 e - 07

4.367e - 06

5.190

3

5.965e-09

3.297e-08

5.525

6.35 l e - 0 9

3.283e-08

5.169

Table 3 ! c~(s) = 1 + 1/(1 + s 2 ) , d = 7/16, f = 1, q = (0,0), Uo = 0 #

el

el

01

e2

e2

02

1

6.757e-05

1.439e-04

2.130

1.499e-04

3.150e-04

2.101

2

3.828e-07

8.154e-07

2.130

8.996e-07

1.871e-06

2.079

3

2.33%-09

4.979e-09

2.128

5.965e-09

1.229e-08

2.060

Table 4 = 0.20088 Iteration #

e

e

0

3,

1

1.0105e-01

1.0743e-01

1.0632e+00

7.5506e-02

2

1.8553e-02

1.8516e-02

9.9800e-01

1.8360e-01

3

3.2897e-03

3.2772e-03

9.9617e-01

1.7731e-01

4

5.7874e-04

5.7658e-04

9.9628e-01

1.7593e-01

5

1.0167e-04

1.0129e-00

9.9631e-01

1.7567e-01

6

1.7855e-05

1.7789e-05

9.9632e-01

1.7562e-01

7

3.1356e-06

3.1240e-06

9.9632e-01

1.7561e-01

8

5.5065e-07

5.4866e-07

9.9639e-01

1.7561e-01

9

9.6702e-08

9.6645e-08

9.9941e-01

1.7561e-01

10

1.6982e-08

1.7542e-08

1.0330e+00

1.7561e-01

c o n t i n u o u s l e v e l a r e d e n o t e d b y e2, e2 a n d 02. W e set q = (q0, q l ) . T h e first c o l u m n s o f the t a b l e s a r e iteration numbers. W e o b s e r v e that t h e K a ~ a n o v m e t h o d c o n v e r g e s f a s t b o t h f o r s o l v i n g the c o n t i n u o u s p r o b l e m a n d f o r s o l v i n g t h e finite e l e m e n t d i s c r e t i z a t i o n s . F o r the c a s e w h e n )~ = 1, the e f f e c t i v e n e s s i n d i c e s lie b e t w e e n 1.4 a n d 1.5, s h o w i n g the e f f i c i e n c y o f o u r a posteriori e r r o r e s t i m a t e s . W h e n A = 6, the c o n v e r g e n c e o f the K a f i a n o v m e t h o d is still fast, w h i l e the e f f e c t i v e n e s s i n d i c e s a r e n o t c l o s e to 1. T h i s is d u e to the f a c t that in o u r a p o s t e r i o r i e r r o r e s t i m a t e s , the left s i d e s c o n t a i n a f a c t o r o f d

W. Han et al. / Applied Numerical Mathematics 24 (1997) 57-79

74 Table 5 ~=0.1 Iteration #

e

e

8

7

1

4.7263e-02

4.9302e-02

1.0431e+00

3.5316e-02

2

4.1123e-03

4.1002e-03

9.9706e-01

8.7009e-02

3

3.5299e-04

3.5182e-04

9.9672e-01

8.5837e-02

4

3.0260e-05

3.0161e-05

9.9672e-01

8.5725e-02

5

2.5937e-06

2.5852e-06

9.9673e-01

8.5715e-02

6

2.2232e-07

2.2170e-07

9.9723e-01

8.5714e-02

7

1.9056e-08

1.8991e-08

9.9658e-01

8.5714e-02

Iteration #

e

e

0

7

1

4.5793e-03

4.711 l e - 0 3

1.0288e+00

3.4217e-03

2

3.9294e-05

3.9174e-05

9.9694e-01

8.5807e-03

3

3,3681e-07

3.3584e-07

9.97lle-01

8.5716e-03

Iteration #

e

e

0

7

1

4.5659e-04

4.6914e-04

1.0275e+00

3.4117e-04

2

3.9141e-07

3.9026e-07

9.9706e-01

8.5725e-04

Iteration #

e

e

0

7

1

4.5645e-05

4.6893e-05

1.0274e+00

3.4107e-05

2

3.9125e-09

0

0

8.5716e-05

Table 6 = 0.01

Table 7 t¢ = 0.001

Table 8 t¢ = 0.0001

(cf. (2.4) a n d T h e o r e m 4.2). O u r a posteriori error e s t i m a t e s a l w a y s b o u n d the true error, a n d h e n c e , w e e x p e c t the error e s t i m a t e s will b e c o m e less efficient w h e n c e r t a i n p a r a m e t e r i n a p r o b l e m t e n d s to a critical v a l u e (A --~ 8 - i n the c u r r e n t e x a m p l e ) . W e h a v e s i m i l a r n u m e r i c a l results w h e n

a ( s ) = l + A / ( l + s2),

0~
16

I n this case, d -- 1 - 9 A / 1 6 . T a b l e 3 c o n t a i n s the results for the p r o b l e m with A = 1.

75

W. Han et al. / Applied Numerical Mathematics 24 (1997) 5 7 - 7 9

I/

2.5

U j il II

\S

// U'2

x~ ~'t

I/ ii

iI It

~

~'t

~

U'1

~x ~

it

/

¢

1.5

...rtl '

¢1 f

........... ......

0.5

;

~ / ....

.,.....

,,t

"\

.....

.........

........

k ......... • . . . . ' ' " .... !~ ,

........'"

/t

t Ij,

~ ~ ~, x k \

,"l l

\x

/t I , i, , Ii '',

,~,

i

Uo

t tm

iI,

I Jl

"

/

t

0

-0.50

I

I

I

I

I

I

I

I

I

I

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0,8

0.9

1

Fig. 1. Convergence of u~ for problem (5.6).

Example 5.2. In this example, we apply the Ka~anov method to solve the problem

{ -(~(lu'12W)' = 0 u ( o ) = 0,

for x E [0, 1],

(5.5)

~ ( 1 ) = 1.

This problem has an analytic solution u(x) = x. Let us choose a family of a(s) by following: let c(t) -- ~t 3 - 3t,

and positive parameters h and ~, 1+~ forx< l-h, a(x)= l + t ~ c ( ~ A) f o r x e [ 1 - h , l + h ] , 1-t~ forx> l+h. Then fors 2 - 1 c [ - h , h ] , c ( s ) > . - 1 , ct(s)~> 3

{

~"(s)=l+~{c(~--~)s

ands 2~< l + h s o t h a t

2+~c2'(s2h - 1 ) } ~ > 1 - ~ ( 4 + 3 ) .

Fors 2 - 1 ~>hwehave fl"(s) = o~(s2) = 1 - ~ >~ 1 - t~(4 + 3 ) .

76

W. Han et al. / Applied Numerical Mathematics 24 (1997) 57-79

Table 9 Convergence of u~ for problem (5.6) Iteration #

e

e

0

"7

1

3.5940e-01

2.0738e+00

5.7701e+00

1.7767e-01

2

7.3890e-02

3.4719e-01

4.6988e+00

2.0559e-01

3

2.2771 e - 0 2

9.7791 e - 0 2

4.2946e+00

3.0817e-01

4

9.5582e-03

3.9727e-02

4.1563e+00

4.1976e-01

5

4.4861e-03

1.8118e-02

4.0388e+00

4.6934e-01

6

2.2250e-03

8.7958e-03

3.9532e+00

4.9598e-01

7

1.1404e - 03

4.4382e - 0 3

3.8917e+00

5.1255e - 01

8

5.9725e-04

2.2973e-03

3.8465e+00

5.2370e-01

9

3.175 l e - 0 4

1.2103e-03

3.8121e+00

5.3161e-01

10

1.7063e-04

6.4587e-04

3.7852e+00

5.3741e-01

11

9.2454e-05

3.4798e-04

3.7638e+00

5.4184e-01

12

5.0414e-05

1.8887e-04

3.7464e+00

5.4529e-01

13

2.7630e-05

1.0312e-04

3.7323e+00

5.4805e-01

14

1.5204e-05

5.6573e-05

3.7209e+00

5.5029e-01

15

8.3949e-06

3.1160e-05

3.7118e+00

5.5214e-01

16

4.6482e-06

1.7220e-05

3.7047e+00

5.5370e-01

17

2.5799e-06

9.5442e-06

3.6994e+00

5.5503e-01

18

1.4349e-06

5.3030e-06

3.6958e+00

5.5617e-01

19

7.9945e- 07

2.9528e - 06

3.6935e+00

5.5717e - 01

20

4.4612e-07

1.6475e-06

3.6930e+00

5.5803e-01

21

2.4929e-07

9.2128e-07

3.6957e+00

5.5879e-01

A s ~ --4 0, (x t e n d s to 1 u n i f o r m l y a n d w e m i g h t e x p e c t 0 to t e n d to 1. L e t "7 --- e / c / e / c - l , k = 1 , 2 , . . . , b e t h e e r r o r r e d u c t i o n factors. A g a i n , w e o b s e r v e the fast c o n v e r g e n c e o f the K a ~ a n o v m e t h o d , a n d the e f f i c i e n c y o f the a p o s t e r i o r i e r r o r e s t i m a t e s . T a b l e s 4 - 8 s h o w us r e s u l t s f o r h ----7 / 2 a n d d i f f e r e n t ~. N o t e that t h e o r e t i c a l l y , o u r a p o s t e r i o r i e r r o r e s t i m a t e s a l w a y s p r o d u c e u p p e r b o u n d s o n the e x a c t error. S o m e o f the e f f e c t i v e n e s s i n d e x v a l u e s in the t a b l e s a r e s l i g h t l y s m a l l e r t h a n 1, this reflects the i n f l u e n c e o f n u m e r i c a l errors r e s u l t i n g f r o m u s i n g q u a d r a t u r e s in c o m p u t i n g b o t h the e r r o r e s t i m a t e s a n d the true error. A s e x p e c t e d , f o r ~ = 0 w e g e t the s o l u t i o n in o n e iteration.

W. Han et al. /Applied Numerical Mathematics 24 (1997) 57-79

77

Table 10 does not satisfy (H2) Iteration #

e

e

0

q'

1

2.1877e-01

3.283 l e - 0 1

1.5007e+00

1.6347e-01

2

8.4789e-02

1.2465e-01

1.4701e+00

3.8757e-01

3

3.6710e-02

5.3625e-02

1.4608e+00

4.3295e-01

4

1.5401e-02

2.2455e-02

1.4581e+00

4.1953e-01

5

6.5647e-03

9.5728e-03

1.4582e+00

4.2626e-01

6

2.7807e-03

4.0541 e - 0 3

1.4580e+00

4.2359e-01

7

1.1811e-03

1.7221e-03

1.4580e+00

4.2476e-01

8

5.011 l e - 0 4

7.3062e-04

1.4580e+00

4.2426e-01

9

2.1271e-04

3.1013e-04

1.4580e+00

4.2448e-01

10

9.0271e-05

1.3162e-04

1.4580e+00

4.2438e-01

11

3.8313e-05

5.5861e-05

1.4580e+00

4.2442e-01

12

1.6260e-05

2.3708e-05

1.4580e+00

4.2441e-01

13

6.9012e-06

1.0062e-05

1.4580e+00

4.2442e-01

14

2.9290e - 06

4.2704e - 06

1.4580e+00

4.2441 e - 01

15

1.243 l e - 0 6

1.8124e-06

1.4580e+00

4.2442e-01

16

5.2758e-07

7.6911e-07

1.4578e+00

4.2441e-01

17

2.2391e-07

3.2617e-07

1.4567e+00

4.2442e-01

18

9.5032e-08

1.3810e-07

1.4532e+00

4.2441e-01

19

4.0332e - 08

5.7921 e - 08

1.4361 e+00

4.244 1e - 0 1

20

1.7118e-08

2.4025e-08

1.4035e+00

4.2442e-01

E x a m p l e 5.3 A s a f u r t h e r n u m e r i c a l e x a m p l e , w e u s e the K a r a n o v m e t h o d to s o l v e the p r o b l e m

-(a(lu'[2)u')

= 10cos(10x)

u(o) = o ,

u(1) = 1.

for x e [0, 11, (5.6)

Here, a ( s ) - - 27

d

--

7 2

~ a r c t a n ( s ) . W e c a n v e r i f y that the a s s u m p t i o n s (H1) a n d (H2) are satisfied, a n d

]-tSo(1 + 7r/4)

W e take u0 = (e x - 1 ) / ( e - 1). Fig. 1 a n d T a b l e 9 s h o w the c o n v e r g e n c e o f the d e r i v a t i v e u ik for the p r o b l e m (5.6).

78

W. Han et al. / Applied Numerical Mathematics 24 (1997) 57-79

. ..""

1.8

..U;

1.61"

,.'".

1.41 -~ 1.2

~~~

..-"'"

"'"/

U~,

l b[--= , , _ , L , ~ = . , , ; : - . - - - - ; ~ = ~ = , - = . . - - , . ~ ' - ~ , ~ = = = . 2 = = = = = = = = = =

o.eF

...... . . . . . . . . . . . .

I 0.6

O0

u~

.-'""

0.4 02

u,

../'"

..'"'" 0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Fig. 2. ct does not satisfy (H2). E x a m p l e 5.4. The last example is also interesting. Consider the problem { -(~(lu'J2)u') ' : 0

for x ~ [0, 1],

(5.7) u(0) = 0,

u(1) = a.

Here, (~(s) = arctan(s) + 7r/2. We take u0 = x 2. It is easy to see the condition (~'(s) ~< 0 in (H2) is not satisfied. The other conditions in (H2) and (H1) hold, and we can take d = 7r/2. Despite the fact that (~ does not satisfy conditions (H2), we still have the convergence of the Ka~anov iteration method, and the a posteriori error estimate is efficient (see Table 10 and Fig. 2).

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