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Iterative methods for ternary diffusions

Boundary Value Problems20142014:87

https://doi.org/10.1186/1687-2770-2014-87

  • Received: 31 January 2014
  • Accepted: 9 April 2014
  • Published:

Abstract

We apply iterative methods to three-component diffusion equations and study theirconvergence in L 2 and in the Sobolev space W 1 , . The system is parabolic and mass-conservative.Newton’s method converges very fast and its iterations do not leave theset of admissible functions.

MSC: 35K51, 35K57, 65M12, 65M80.

Keywords

  • diffusion
  • iterative method
  • Newton’s method

1 Introduction

Since its discovery and later analysis by Darken [1], the Kirkendall effect [2] has been found in various alloy systems, and studies on lattice defectsand diffusion developed significantly. The Danielewski-Holly method [3] extends the Darken standard theory of interdiffusion and describes theprocess in the bounded mixture showing constant concentration. Under certainregularity assumptions and quantitative condition Danielewski and Holly proved theexistence and uniqueness of solution to PDE describing the interdiffusion phenomena.Further developments have been presented in numerous articles; e.g.[4, 5].

In the paper we apply Newton’s method (see [68]) to three-component diffusion equations and study the convergence in L 2 and Sobolev space W 1 , . The system of equations is strongly coupled,however, the maximum principle presented in Section 1 confirms its parabolictype. Parabolicity is additionally confirmed by our convergence result for iterativemethods. This falsifies the nonparabolicity hypothesis by Danielewski and Holly [3], where they construct an initial concentration whose L 2 norm increases in time, at least on some interval.The Newton method, known as quasilinearization method, is very useful inmodern numerical methods for solving PDE’s; see [9]. We apply this method to strongly coupled parabolic systems describingdiffusing mixtures. This strong parabolicity might have caused weird phenomena, butwe have discovered a kind of maximum principle and some conservation laws in thissystem, hence the iterative methods proposed here behave very well. Our result isvery useful in numerical simulations when one wants to construct reliable and fastconvergent approximations. Since Newton’s method produces linear PDE’ssatisfying maximum principles and a priori estimates of the respectiveGreen functions or Cauchy kernels, one can find errors estimates much better thanthose obtained from the Newton-Kantorovich theorem, cf. [10, 11].

Consider a mixture composed of three different components. Let t 0 , x [ L , L ] , v i : [ 0 , ) × [ L , L ] R denote the velocity field of the i thcomponent and c i : [ 0 , ) × [ L , L ] R its molar density or molar concentration. It is ameasure of the number of particles contained in any volume, c 1 + c 2 + c 3 const . The component diffusion flux is a Fickian flow:
J i d ( t , x ) : = D i grad c i ,
where D i is the intrinsic diffusitivity of the i thcomponent which we assume to satisfy D 1 > D 2 > D 3 > 0 . Denote D i : = D i D 3 for i = 1 , 2 . The overall i th component flux is a sum ofdiffusion and convection fluxes:
J i : = J i d + c i v D ,
where v D stands for a drift velocity. By the mass conservationlaw:
c i t = div J i
and upon denoting u = c 1 , v = c 2 , w = c 3 we arrive at the following system of equations:
u t = D 1 u x x ( u [ D 1 u x + D 2 v x ] ) x , v t = D 2 v x x ( v [ D 1 u x + D 2 v x ] ) x , w t = D 3 w x x ( w [ D 1 u x + D 2 v x ] ) x
(1.1)
with the initial condition
u ( 0 , x ) = u 0 ( x ) , v ( 0 , x ) = v 0 ( x ) , w ( 0 , x ) = w 0 ( x ) = 1 u 0 ( x ) v 0 ( x )
(1.2)
for x [ L , L ] and the Neumann boundary condition
u n = 0 , v n = 0 , w n = 0 for  t 0 , x { L , L } .
(1.3)

Let denote the space consisting of triples of functions ( u , v , w ) satisfying

u , v , w C 1 , 2 ,

u x , u x x , v x , v x x are bounded,

u 0 , v 0 , w 0 for t 0 , x [ L , L ] ,

u + v + w = 1 for t 0 , x [ L , L ] ,

u, v, w obey the Neumann boundarycondition.

Remark 1.1 If ( u , v , w ) X then the third equation of (1.1) is not necessary,since w = 1 u v . However, we keep it for a more convenient analysisof some properties of solutions.

Remark 1.2 We call
v D = D 1 u x + D 2 v x = D 1 u x + D 2 v x + D 3 w x

the drift velocity; it describes the marker position.

Lemma 1.3 (Mass conservation)

If ( u , v , w ) X satisfy (1.1), (1.2) then
L L u d x = const. , L L v d x = const. , L L w d x = const.
Proof The relation
d d t L L u d x = 0

can be shown by means of the Neumann boundary condition. □

Lemma 1.4 (Maximum principle)

Suppose that u ( 0 , ) , v ( 0 , ) , w ( 0 , ) C 2 and
u ( 0 , x ) 0 , v ( 0 , x ) 0 , w ( 0 , x ) 0 , u ( 0 , x ) + v ( 0 , x ) + w ( 0 , x ) = 1

for x [ L , L ] . If ( u , v , w ) satisfy (1.1)-(1.3) then ( u , v , w ) X .

Proof Let u ˜ = u + ε e λ t , v ˜ = v + ε e λ t , w ˜ = w + ε e λ t for ε > 0 . We have
u ˜ t = u t + ε λ e λ t , u ˜ x = u x , u ˜ x x = u x x , v ˜ t = v t + ε λ e λ t , v ˜ x = v x , v ˜ x x = v x x .
There exists λ R (sufficiently large) such that we have strongdifferential inequalities:
u ˜ t > D 1 u ˜ x x u ˜ [ D 1 u ˜ x x + D 2 v ˜ x x ] u ˜ x [ D 1 u ˜ x + D 2 v ˜ x ] , v ˜ t > D 2 v ˜ x x v ˜ [ D 1 u ˜ x x + D 2 v ˜ x x ] v ˜ x [ D 1 u ˜ x + D 2 v ˜ x ] , w ˜ t > D 3 w ˜ x x w ˜ [ D 1 u ˜ x x + D 2 v ˜ x x ] w ˜ x [ D 1 u ˜ x + D 2 v ˜ x ] .
We claim that u ˜ > 0 , v ˜ > 0 , w ˜ > 0 in the whole domain. Suppose that this is not trueand take the smallest t > 0 such that u ˜ ( t , x ) = 0 , or v ˜ ( t , x ) = 0 , or w ˜ ( t , x ) = 0 for some x [ L , L ] . Without loss of generality we assume u ˜ ( t , x ) = 0 . Since u ˜ ( t , x ) = min { t t , x [ L , L ] } u ˜ ( t , x ) we have u ˜ x ( t , x ) = 0 , u ˜ t ( t , x ) 0 and u ˜ x x ( t , x ) 0 . Hence
0 u ˜ t ( t , x ) > D 1 u ˜ x x ( t , x ) u ˜ ( t , x ) [ D 1 u ˜ x x ( t , x ) + D 2 v ˜ x x ( t , x ) ] u ˜ x ( t , x ) [ D 1 u ˜ x ( t , x ) + D 2 v ˜ x ( t , x ) ] 0 ,

which is a contradiction. Thus u ˜ > 0 for t 0 , x [ L , L ] . If ε 0 + then u ˜ u . Hence u > 0 . Similarly, v ( t , x ) 0 and w ( t , x ) 0 for t 0 , x [ L , L ] . □

2 Uniqueness

Let X ¯ be the closure of w.r.t. the L 2 norm. The existence and uniqueness of solutions toproblem (1.1)-(1.3) in w.r.t. the Sobolev norm W 1 , 2 is given in [3]. The following proposition concerns the uniqueness of solutions in L 2 . Since the set of C 2 -functions is dense in L 2 , the proof is carried out in . The uniqueness isobtained for weak solutions.

Proposition 2.1 Assume that ( 7 4 3 ) D 2 D 1 ( 7 + 4 3 ) D 2 and ( u 0 , v 0 , w 0 ) X ¯ . Then a weak solution ( u , v , w ) X ¯ to problem (1.1)-(1.3) is unique in L 2 .

Proof Since every L 2 -function can be approximated by a sequence of -functions, it suffices to show the uniqueness of -solutions w.r.t. the L 2 -norm. Let ( u , v , w ) X and ( u ¯ , v ¯ , w ¯ ) X be solutions to (1.1)-(1.3). Denote
Δ u = u u ¯ , Δ v = v v ¯ , Δ w = w w ¯
and observe that
Δ u t = D 1 Δ u x x ( Δ u [ D 1 u x + D 2 v x ] ) x ( u ¯ [ D 1 Δ u x + D 2 Δ v x ] ) x , Δ v t = D 2 Δ v x x ( Δ v [ D 1 u x + D 2 v x ] ) x ( v ¯ [ D 1 Δ u x + D 2 Δ v x ] ) x , Δ w t = D 3 Δ w x x ( Δ w [ D 1 u x + D 2 v x ] ) x ( w ¯ [ D 1 Δ u x + D 2 Δ v x ] ) x .
We have
L L Δ u Δ u t d x = D 1 L L Δ u Δ u x x d x L L Δ u ( Δ u [ D 1 u x + D 2 v x ] ) x d x L L Δ u ( u ¯ [ D 1 Δ u x + D 2 Δ v x ] ) x d x .
Using integration by parts we obtain
2 D 1 L L Δ u Δ u x x d x = 2 D 1 L L ( Δ u x ) 2 d x , 2 L L Δ u ( Δ u [ D 1 u x + D 2 v x ] ) x d x = L L ( Δ u ) 2 [ D 1 u x x + D 2 v x x ] d x , 2 L L Δ u ( u ¯ [ D 1 Δ u x + D 2 Δ v x ] ) x d x = 2 L L Δ u x u ¯ [ D 1 Δ u x + D 2 Δ v x ] d x .
Hence
d d t L L [ ( Δ u ) 2 + ( Δ v ) 2 + ( Δ w ) 2 ] d x = 2 L L ( Δ u Δ u t + Δ v Δ v t + Δ w Δ w t ) d x = 2 L L ( D 1 ( Δ u x ) 2 + D 2 ( Δ v x ) 2 + D 3 ( Δ w x ) 2 ) d x L L ( ( Δ u ) 2 + ( Δ v ) 2 + ( Δ w ) 2 ) [ D 1 u x x + D 2 v x x ] d x + 2 L L ( Δ u x u ¯ + Δ v x v ¯ + Δ w x w ¯ ) [ D 1 Δ u x + D 2 Δ v x ] d x .
By the fact that Δ w x = Δ u x Δ v x we obtain
2 L L ( Δ u Δ u t + Δ v Δ v t + Δ w Δ w t ) d x = L L ( ( Δ u ) 2 + ( Δ v ) 2 + ( Δ w ) 2 ) [ D 1 u x x + D 2 v x x ] d x 2 L L ( D 1 + D 3 D 1 ( u ¯ w ¯ ) ) ( Δ u x ) 2 d x 2 L L ( D 2 + D 3 D 2 ( v ¯ w ¯ ) ) ( Δ v x ) 2 d x 2 L L ( 2 D 3 D 1 ( v ¯ w ¯ ) D 2 ( u ¯ w ¯ ) ) Δ u x Δ v x d x .
We examine the nonnegative definiteness of the matrix:
A = [ D 1 + D 3 D 1 ( u ¯ w ¯ ) D 3 1 2 D 1 ( v ¯ w ¯ ) 1 2 D 2 ( u ¯ w ¯ ) D 3 1 2 D 1 ( v ¯ w ¯ ) 1 2 D 2 ( u ¯ w ¯ ) D 2 + D 3 D 2 ( v ¯ w ¯ ) ]
i.e.
D 1 + D 3 D 1 ( u ¯ w ¯ ) 0 , D 2 + D 3 D 2 ( v ¯ w ¯ ) 0 , and det ( A ) 0 .
The first two inequalities are true due to the relations:
1 u ¯ w ¯ 1 , 1 v ¯ w ¯ 1 , D 1 > D 2 > D 3 > 0 .
The condition
( 7 4 3 ) D 2 D 1 ( 7 + 4 3 ) D 2

implies det ( A ) 0 for all admissible u ¯ , w ¯ . □

3 Iterative methods

Recall that
D 1 u x + D 2 v x = D 1 u x + D 2 v x + D 3 w x , D 1 u x x + D 2 v x x = D 1 u x x + D 2 v x x + D 3 w x x .
Assume that ( u ( 0 ) , v ( 0 ) , w ( 0 ) ) coincides with ( u 0 , v 0 , w 0 ) at t = 0 and formulate an iterative method for (1.1)-(1.3):
u t ( k + 1 ) = D 1 u x x ( k + 1 ) ( u ( k ) [ D 1 u x ( k + 1 ) + D 2 v x ( k + 1 ) + D 3 w x ( k + 1 ) ] ) x , v t ( k + 1 ) = D 2 v x x ( k + 1 ) ( v ( k ) [ D 1 u x ( k + 1 ) + D 2 v x ( k + 1 ) + D 3 w x ( k + 1 ) ] ) x , w t ( k + 1 ) = D 3 w x x ( k + 1 ) ( w ( k ) [ D 1 u x ( k + 1 ) + D 2 v x ( k + 1 ) + D 3 w x ( k + 1 ) ] ) x ,
(3.1)
with the initial condition
u ( k + 1 ) ( 0 , x ) = u 0 ( x ) , v ( k + 1 ) ( 0 , x ) = v 0 ( x ) , w ( k + 1 ) ( 0 , x ) = w 0 ( x )
(3.2)
for x [ L , L ] and the Neumann boundary condition. Moreover, assumethat
u 0 ( x ) + v 0 ( x ) + w 0 ( x ) = 1
(3.3)
for x [ L , L ] . Denote
Δ u ( k ) = u ( k + 1 ) u ( k ) , Δ v ( k ) = v ( k + 1 ) v ( k ) , Δ w ( k ) = w ( k + 1 ) w ( k ) .

Lemma 3.1 Assume u 0 , v 0 , w 0 X , ( u ( 0 ) , v ( 0 ) , w ( 0 ) ) = ( u 0 , v 0 , w 0 ) at t = 0 and u ( 0 ) + v ( 0 ) + w ( 0 ) = 1 . If ( u ( k ) , v ( k ) , w ( k ) ) fulfills (3.1) with (3.2), the Neumann boundary conditionand (3.3), then u ( k ) + v ( k ) + w ( k ) = 1 .

Proof It suffices to show u ( k ) + v ( k ) + w ( k ) = 1 u ( k + 1 ) + v ( k + 1 ) + w ( k + 1 ) = 1 . We assume the induction hypothesis u ( k ) + v ( k ) + w ( k ) = 1 . Thus
u t ( k + 1 ) + v t ( k + 1 ) + w t ( k + 1 ) = D 1 u x x ( k + 1 ) + D 2 v x x ( k + 1 ) + D 3 w x x ( k + 1 ) ( u x ( k ) + v x ( k ) + w x ( k ) ) [ D 1 u x ( k + 1 ) + D 2 v x ( k + 1 ) + D 3 w x ( k + 1 ) ] ( u ( k ) + v ( k ) + w ( k ) ) [ D 1 u x x ( k + 1 ) + D 2 v x x ( k + 1 ) + D 3 w x x ( k + 1 ) ] 0 .

Hence the statement is proved. □

The following theorem establishes a convergence of the iterative method(3.1)-(3.2).

Theorem 3.2 Suppose ( u 0 , v 0 , w 0 ) X and ( u ( 0 ) , v ( 0 ) , w ( 0 ) ) = ( u 0 , v 0 , w 0 ) at t = 0 . If u x ( k ) , v x ( k ) , w x ( k ) are C 2 and
0 u ( k ) 1 , 0 v ( k ) 1 , 0 w ( k ) 1 for  k = 1 , 2 ,

then the sequence ( u ( k ) , v ( k ) , w ( k ) ) defined by (3.1), (3.2) converges to the solution ( u , v , w ) of (1.1), (1.2) in the Sobolev space W 1 , .

Proof As in the previous section denote the increments Δ u ( k ) = u ( k + 1 ) u ( k ) , Δ v ( k ) = v ( k + 1 ) v ( k ) , Δ w ( k ) = w ( k + 1 ) w ( k ) . From (3.1) we have the following differentialequations:
Δ u t ( k + 1 ) = D 1 Δ u x x ( k + 1 ) ( u ( k + 1 ) [ D 1 Δ u x ( k + 1 ) + D 2 Δ v x ( k + 1 ) ] ) x ( Δ u ( k ) [ D 1 u x ( k + 1 ) + D 2 v x ( k + 1 ) ] ) x , Δ v t ( k + 1 ) = D 2 Δ v x x ( k + 1 ) ( v ( k + 1 ) [ D 1 Δ u x ( k + 1 ) + D 2 Δ v x ( k + 1 ) ] ) x ( Δ v ( k ) [ D 1 u x ( k + 1 ) + D 2 v x ( k + 1 ) + D 3 w x ( k + 1 ) ] ) x .
Using the Green functions G 1 , k , G 2 , k corresponding to the differential operators
[ t D 1 2 x 2 + u ( k + 1 ) D 1 2 x 2 D 2 2 x 2 D 1 2 x 2 t D 2 2 x 2 + v ( k + 1 ) D 2 2 x 2 ]
(3.4)
we have
[ Δ u ( k + 1 ) ( t , x ) Δ v ( k + 1 ) ( t , x ) ] = 0 t L L [ G 1 , k ( t , s , x , y ) P 1 , k ( s , y ) G 2 , k ( t , s , x , y ) P 2 , k ( s , y ) ] d y d s , [ Δ u x ( k + 1 ) ( t , x ) Δ v x ( k + 1 ) ( t , x ) ] = 0 t L L [ G x 1 , k ( t , s , x , y ) P 1 , k ( s , y ) G x 2 , k ( t , s , x , y ) P 2 , k ( s , y ) ] d y d s ,
where P i , k ( s , y ) depend on Δ u ( k ) , Δ v ( k ) , Δ u x ( k ) , Δ v x ( k ) , Δ u x ( k + 1 ) , Δ v x ( k + 1 ) for i = 1 , 2 . The Green functions G i , k depend on u ( k ) , v ( k ) and have the uniform estimates
L L | G i , k ( t , s , x , y ) | d y C , L L | G x i , k ( t , s , x , y ) | d y C t s ,
(3.5)
with some generic constant C not depending on k. By Lemma 3.1there exists M 0 such that
u x ( k + 1 ) ( t , ) L M , v x ( k + 1 ) ( t , ) L M , u x x ( k + 1 ) ( t , ) L M , v x x ( k + 1 ) ( t , ) L M .
(3.6)
Since
P 1 , k ( t , ) L M ( D 1 + D 2 ) Δ u ( k ) ( t , ) L + M ( D 1 + D 2 ) Δ u x ( k ) ( t , ) L + M D 1 Δ u x ( k + 1 ) ( t , ) L + M D 2 Δ v x ( k + 1 ) ( t , ) L
we get
( Δ u ( k + 1 ) , Δ v ( k + 1 ) ) ( t , ) W 1 , : = Δ u ( k + 1 ) ( t , ) W 1 , + Δ v ( k + 1 ) ( t , ) W 1 , 0 t 2 C 1 t s ( Δ u ( k ) , Δ v ( k ) ) ( s , ) W 1 , d s + 0 t 2 C 1 t s ( Δ u ( k + 1 ) , Δ v ( k + 1 ) ) ( s , ) W 1 , d s .
Applying Lemma A.1 we have ( Δ u ( 0 ) , Δ v ( 0 ) ) ( t , ) W 1 , K 0 and by induction: ( Δ u ( k ) , Δ v ( k ) ) ( t , ) W 1 , K k t k 2 for k = 1 , 2 ,  . Hence
K k + 1 = K k C 1 2 C T 1 / 2 0 1 θ k 1 θ d θ .
Notice that
K k + 1 K k = C 1 2 C T 1 / 2 0 1 θ k 1 θ d θ 0 as  k .

By the d’Alembert’s ratio test the convergence radius is+∞. □

We give sufficient conditions for the successive approximations to remain in .

Proposition 3.3 Assume that u 0 , v 0 C 4 , 0 < ε 0 u 0 ( x ) 1 ε 0 < 1 , 0 < ε 0 v 0 ( x ) 1 ε 0 < 1 and the sequence ( u ( k ) , v ( k ) , w ( k ) ) defined by (3.1) with the first element given by
u ( 0 ) ( t , x ) = u 0 ( x ) + t k u ( x ) and v ( 0 ) ( t , x ) = v 0 ( x ) + t k v ( x ) ,
where k u , k v X are of the form
k u ( x ) = D 1 u 0 ′′ ( x ) ( u 0 ( x ) [ D 1 u 0 ( x ) + D 2 v 0 ( x ) ] ) x , k v ( x ) = D 2 v 0 ′′ ( x ) ( v 0 ( x ) [ D 1 u 0 ( x ) + D 2 v 0 ( x ) ] ) x ,
converges to the solution ( u , v , w ) of (1.1), (1.2) in the Sobolev space W 1 , . If
k = 0 K k t k / 2 ε 0 , where  K k = K k 1 C 1 2 C T 1 / 2 0 1 θ k 1 1 θ d θ ,

then 0 u ( k ) ( t , x ) 1 and 0 v ( k ) ( t , x ) 1 , k = 0 , 1 ,  .

Proof We have
u t ( 0 ) ( t , x ) = k u ( x ) , u x ( 0 ) ( t , x ) = u 0 ( x ) + t k u ( x ) , u x x ( 0 ) ( t , x ) = u 0 ′′ ( x ) + t k u ′′ ( x ) .
Hence
k u ( x ) = D 1 u 0 ′′ ( x ) u 0 ( x ) [ D 1 u 0 ( x ) + D 2 v 0 ( x ) ] u 0 ( x ) [ D 1 u 0 ′′ ( x ) + D 2 v 0 ′′ ( x ) ] = D 1 ( u x x ( 0 ) ( t , x ) t k u ′′ ( x ) ) ( u x ( 0 ) ( t , x ) t k u ( x ) ) [ D 1 u x ( 0 ) + D 2 v x ( 0 ) t ( D 1 k u ( x ) + D 2 k v ( x ) ) ] ( u ( 0 ) ( t , x ) t k u ( x ) ) [ D 1 u x x ( 0 ) + D 2 v x x ( 0 ) t ( D 1 k u ′′ ( x ) + D 2 k v ′′ ( x ) ) ] .
Thus we get
Δ u t ( 0 ) ( t , x ) = u t ( 1 ) ( t , x ) u t ( 0 ) ( t , x ) = D 1 u x x ( 1 ) ( t , x ) u x ( 0 ) ( t , x ) [ D 1 u x ( 1 ) ( t , x ) + D 2 v x ( 1 ) ( t , x ) ] u ( 0 ) ( t , x ) [ D 1 u x x ( 1 ) ( t , x ) + D 2 v x x ( 1 ) ( t , x ) ] k u ( x ) = D 1 Δ u x x ( 0 ) ( t , x ) ( u ( 0 ) ( t , x ) [ D 1 Δ u x ( 0 ) ( t , x ) + D 2 Δ v x ( 0 ) ( t , x ) ] ) x + t R 1 ( x ) + t 2 R 2 ( x ) ,
where
R 1 ( x ) : = D 1 k u ′′ ( x ) u x ( 0 ) ( t , x ) ( D 1 k u ( x ) + D 2 k v ( x ) ) k u ( x ) [ D 1 u x ( 0 ) ( t , x ) + D 2 v x ( 0 ) ( t , x ) ] u ( 0 ) ( t , x ) ( D 1 k u ′′ ( x ) + D 2 k v ′′ ( x ) ) k u ( x ) [ D 1 u x x ( 0 ) ( t , x ) + D 2 v x x ( 0 ) ( t , x ) ] , R 2 ( x ) : = k u ( x ) ( D 1 k u ( x ) + D 2 k v ( x ) ) + k u ( x ) ( D 1 k u ′′ ( x ) + D 2 k v ′′ ( x ) ) .
For 0 t T we have
R L = t R 1 + t 2 R 2 L T ( R 1 L + T R 2 L ) = : K 0 .
Thus
( Δ u ( 0 ) , Δ v ( 0 ) ) ( t , ) W 1 , 0 t C t s [ ( Δ u ( 0 ) , Δ v ( 0 ) ) ( s , ) W 1 , + K 0 ] d s .
Applying Lemma A.1 we have
( Δ u ( 0 ) , Δ v ( 0 ) ) ( t , ) W 1 , K 1 t 1 / 2
and by induction ( Δ u ( k ) , Δ v ( k ) ) ( t , ) W 1 , K k + 1 t k + 1 2 for k = 1 , 2 ,  . Hence
K k + 1 = K k C 1 2 C T 1 / 2 0 1 θ k 1 θ d θ .

 □

Remark 3.4 The functions k u , k v X can be slightly perturbed near the lateral boundaryin order to fulfill the Neumann boundary condition.

4 Convergence of the Newton method

As in the previous section denote
Δ u ( k ) = u ( k + 1 ) u ( k ) , Δ v ( k ) = v ( k + 1 ) v ( k ) , Δ w ( k ) = w ( k + 1 ) w ( k ) .
We assume that ( u ( 0 ) , v ( 0 ) , w ( 0 ) ) = ( u 0 , v 0 , w 0 ) at t = 0 and formulate the Newton method for (1.1)-(1.3):
u t ( k + 1 ) = D 1 u x x ( k + 1 ) ( u ( k ) [ D 1 u x ( k ) + D 2 v x ( k ) ] ) x ( Δ u ( k ) [ D 1 u x ( k ) + D 2 v x ( k ) ] ) x ( u ( k ) [ D 1 Δ u x ( k ) + D 2 Δ v x ( k ) ] ) x , v t ( k + 1 ) = D 2 v x x ( k + 1 ) ( v ( k ) [ D 1 u x ( k ) + D 2 v x ( k ) ] ) x ( Δ v ( k ) [ D 1 u x ( k ) + D 2 v x ( k ) ] ) x ( v ( k ) [ D 1 Δ u x ( k ) + D 2 Δ v x ( k ) ] ) x , w t ( k + 1 ) = D 3 w x x ( k + 1 ) ( w ( k ) [ D 1 u x ( k ) + D 2 v x ( k ) ] ) x ( Δ w ( k ) [ D 1 u x ( k ) + D 2 v x ( k ) ] ) x ( w ( k ) [ D 1 Δ u x ( k ) + D 2 Δ v x ( k ) ] ) x ,
(4.1)

with the initial condition (3.2) and the Neumann boundary condition.

Lemma 4.1 Assume u 0 , v 0 , w 0 X , ( u ( 0 ) , v ( 0 ) , w ( 0 ) ) = ( u 0 , v 0 , w 0 ) at t = 0 and u ( 0 ) + v ( 0 ) + w ( 0 ) = 1 . If ( u ( k ) , v ( k ) , w ( k ) ) fulfills (4.1) with (3.2) and the Neumann boundarycondition, then u ( k ) + v ( k ) + w ( k ) = 1 .

Proof We show u ( k ) + v ( k ) + w ( k ) = 1 u ( k + 1 ) + v ( k + 1 ) + w ( k + 1 ) = 1 . The only solution to the differential equation
u t ( k + 1 ) + v t ( k + 1 ) + w t ( k + 1 ) = ( u x ( k + 1 ) + v x ( k + 1 ) + w x ( k + 1 ) ) [ D 1 u x ( k ) + D 2 v x ( k ) + D 3 w x ( k ) ] ( u ( k + 1 ) + v ( k + 1 ) + w ( k + 1 ) 1 ) [ D 1 u x x ( k ) + D 2 v x x ( k ) + D 3 w x x ( k ) ]

is u ( k + 1 ) + v ( k + 1 ) + w ( k + 1 ) 1 . □

The following theorem establishes the convergence of the Newton method.

Theorem 4.2 Suppose ( u 0 , v 0 , w 0 ) X and ( u ( 0 ) , v ( 0 ) , w ( 0 ) ) = ( u 0 , v 0 , w 0 ) at t = 0 . If u x ( k ) , v x ( k ) , w x ( k ) are C 2 and
0 u ( k ) 1 , 0 v ( k ) 1 , 0 w ( k ) 1 for  k = 1 , 2 , ,

then the sequence ( u ( k ) , v ( k ) , w ( k ) ) defined by (4.1), (3.2) converges to the solution ( u , v , w ) of (1.1)-(1.3) with respect to the norms in the Sobolev space W 1 , .

Proof We have the following differential equations:
Δ u t ( k + 1 ) = D 1 Δ u x x ( k + 1 ) ( Δ u ( k ) [ D 1 Δ u x ( k ) + D 2 Δ v x ( k ) ] ) x ( Δ u ( k + 1 ) [ D 1 u x ( k + 1 ) + D 2 v x ( k + 1 ) ] ) x ( u ( k + 1 ) [ D 1 Δ u x ( k + 1 ) + D 2 Δ v x ( k + 1 ) ] ) x , Δ v t ( k + 1 ) = D 2 Δ v x x ( k + 1 ) ( Δ v ( k ) [ D 1 Δ u x ( k ) + D 2 Δ v x ( k ) ] ) x ( Δ v ( k + 1 ) [ D 1 u x ( k + 1 ) + D 2 v x ( k + 1 ) ] ) x ( v ( k + 1 ) [ D 1 Δ u x ( k + 1 ) + D 2 Δ v x ( k + 1 ) ] ) x , Δ w t ( k + 1 ) = D 3 Δ w x x ( k + 1 ) ( Δ w ( k ) [ D 1 Δ u x ( k ) + D 2 Δ v x ( k ) ] ) x ( Δ w ( k + 1 ) [ D 1 u x ( k + 1 ) + D 2 v x ( k + 1 ) ] ) x ( w ( k + 1 ) [ D 1 Δ u x ( k + 1 ) + D 2 Δ v x ( k + 1 ) ] ) x .
By the Green functions G 1 , k , G 2 , k :
Δ u ( k + 1 ) ( t , x ) = 0 t L L G 1 , k ( t , s , x , y ) ( Δ u ( k ) ( s , y ) [ D 1 Δ u y ( k ) ( s , y ) + D 2 Δ v y ( k ) ( s , y ) ] ) y d y d s + 0 t L L G 1 , k ( t , s , x , y ) Δ u y ( k + 1 ) ( s , y ) [ D 1 u y ( k + 1 ) ( s , y ) + D 2 v y ( k + 1 ) ( s , y ) ] d y d s + 0 t L L G 1 , k ( t , s , x , y ) Δ u ( k + 1 ) ( s , y ) [ D 1 u y y ( k + 1 ) ( s , y ) + D 2 v y y ( k + 1 ) ( s , y ) ] d y d s + 0 t L L G 1 , k ( t , s , x , y ) u y ( k + 1 ) ( s , y ) [ D 1 Δ u y ( k + 1 ) ( s , y ) + D 2 Δ v y ( k + 1 ) ( s , y ) ] d y d s .
Using the integration by parts we get
0 t L L G 1 , k ( t , s , x , y ) ( Δ u ( k ) ( s , y ) [ D 1 Δ u y ( k ) ( s , y ) + D 2 Δ v y ( k ) ( s , y ) ] ) y d y d s = 0 t L L G y 1 , k ( t , s , x , y ) Δ u ( k ) ( s , y ) [ D 1 Δ u y ( k ) ( s , y ) + D 2 Δ v y ( k ) ( s , y ) ] d y d s .
From the following property:
L L | G y i , k ( t , s , x , y ) | d y C t s ,
estimates like (3.5), (3.6), and x y 1 2 ( x 2 + y 2 ) we obtain
Δ u ( k + 1 ) ( t , ) L 0 t C 2 2 t s Q 1 , k ( s ) d s ,
where
Q 1 , k ( s ) = Δ u ( k ) ( s , ) L 2 + Δ u y ( k ) ( s , ) L 2 + Δ v y ( k ) ( s , ) L 2 + Δ u y ( k + 1 ) ( s , ) L + Δ u ( k + 1 ) ( s , ) L d s + Δ u y ( k + 1 ) ( s , ) L + Δ v y ( k + 1 ) ( s , ) L .
Similarly
Δ v ( k + 1 ) ( t , ) L 0 t C 2 2 t s Q 2 , k ( s ) d s ,
where
Q 2 , k ( s ) = Δ v ( k ) ( s , ) L 2 + Δ u y ( k ) ( s , ) L 2 + Δ v y ( k ) ( s , ) L 2 + Δ v y ( k + 1 ) ( s , ) L + Δ v ( k + 1 ) ( s , ) L + Δ u y ( k + 1 ) ( s , ) L + Δ v y ( k + 1 ) ( s , ) L .
We have
Δ u x ( k + 1 ) ( t , x ) = 0 t L L G x 1 , k ( t , s , x , y ) ( Δ u ( k ) ( s , y ) [ D 1 Δ u y ( k ) ( s , y ) + D 2 Δ v y ( k ) ( s , y ) ] ) y d y d s + 0 t L L G x 1 , k ( t , s , x , y ) Δ u y ( k + 1 ) ( s , y ) [ D 1 u y ( k + 1 ) ( s , y ) + D 2 v y ( k + 1 ) ( s , y ) ] d y d s + 0 t L L G x 1 , k ( t , s , x , y ) Δ u ( k + 1 ) ( s , y ) [ D 1 u y y ( k + 1 ) ( s , y ) + D 2 v y y ( k + 1 ) ( s , y ) ] d y d s + 0 t L L G x 1 , k ( t , s , x , y ) u y ( k + 1 ) ( s , y ) [ D 1 Δ u y ( k + 1 ) ( s , y ) + D 2 Δ v y ( k + 1 ) ( s , y ) ] d y d s .
By integration by parts:
0 t L L G x 1 , k ( t , s , x , y ) ( Δ u ( k ) ( s , y ) [ D 1 Δ u y ( k ) ( s , y ) + D 2 Δ v y ( k ) ( s , y ) ] ) y d y d s = 0 t L L G x y 1 , k ( t , s , x , y ) Δ u ( k ) ( s , y ) [ D 1 Δ u y ( k ) ( s , y ) + D 2 Δ v y ( k ) ( s , y ) ] d y d s .
Since Δ u ( k ) ( s , y ) , Δ u y ( k ) ( s , y ) , Δ v y ( k ) ( s , y ) satisfy the Lipschitz condition, we have theestimates (see [12])
L L | G x y i , k ( t , s , x , y ) Δ u ( k ) ( s , y ) Δ u y ( k ) ( s , y ) | d y 0 t C 3 ( t s ) 3 / 4 Δ u ( k ) ( s , ) L Δ u y ( k ) ( s , ) L d s
and
L L | G x y i , k ( t , s , x , y ) Δ u ( k ) ( s , y ) Δ v y ( k ) ( s , y ) | d y 0 t C 3 ( t s ) 3 / 4 Δ u ( k ) ( s , ) L Δ v y ( k ) ( s , ) L d s .
(4.2)
Hence
Δ u x ( k + 1 ) ( t , ) L 0 t C 4 ( t s ) 3 / 4 ( Δ u ( k ) ( s , ) L 2 + Δ u y ( k ) ( s , ) L 2 + Δ v y ( k ) ( s , ) L 2 ) d s + 0 t C 5 t s ( Δ u y ( k + 1 ) ( s , ) L + Δ u ( k + 1 ) ( s , ) L + Δ u y ( k + 1 ) ( s , ) L + Δ v y ( k + 1 ) ( s , ) L ) d s .
Similarly
Δ v x ( k + 1 ) ( t , ) L 0 t C 4 ( t s ) 3 / 4 ( Δ v ( k ) ( s , ) L 2 + Δ u y ( k ) ( s , ) L 2 + Δ v y ( k ) ( s , ) L 2 ) d s + 0 t C 5 t s ( Δ v y ( k + 1 ) ( s , ) L + Δ v ( k + 1 ) ( s , ) L + Δ u y ( k + 1 ) ( s , ) L + Δ v y ( k + 1 ) ( s , ) L ) d s .
We have
( Δ u ( k + 1 ) , Δ v ( k + 1 ) ) ( t , ) W 1 , 0 t C 4 ( t s ) 3 / 4 ( Δ u ( k ) , Δ v ( k ) ) ( s , ) W 1 , 2 d s + 0 t C 5 t s ( Δ u ( k + 1 ) , Δ v ( k + 1 ) ) ( s , ) W 1 , d s .
We apply Lemma A.1:
K k + 1 t r k + 1 ( 1 2 C 5 T 1 / 2 ) C 4 K k t 2 r k + 1 + 3 / 4 0 1 θ 2 r k ( 1 θ ) 3 / 4 d θ ,
where
r k = 3 4 ( 2 k 1 ) 2 k and K k A 2 k .

 □

We give sufficient conditions for the successive approximations to remain in .

Proposition 4.3 Assume that u 0 , v 0 C 4 , 0 < ε 0 u 0 ( x ) 1 ε 0 < 1 , 0 < ε 0 v 0 ( x ) 1 ε 0 < 1 and the sequence ( u ( k ) , v ( k ) , w ( k ) ) defined by (4.1) with the first element given by
u ( 0 ) ( t , x ) = u 0 ( x ) + t k u ( x ) and v ( 0 ) ( t , x ) = v 0 ( x ) + t k v ( x ) ,
where k u , k v X are of the form
k u ( x ) = D 1 u 0 ′′ ( x ) ( u 0 ( x ) [ D 1 u 0 ( x ) + D 2 v 0 ( x ) ] ) x , k v ( x ) = D 2 v 0 ′′ ( x ) ( v 0 ( x ) [ D 1 u 0 ( x ) + D 2 v 0 ( x ) ] ) x ,
converges to the solution ( u , v , w ) of (1.1), (1.2) in the Sobolev space W 1 , . If
k = 0 K ˜ k t 3 k 4 ε 0 , K ˜ k : = K ˜ k 1 C ˜ 1 2 C T 1 / 2 0 1 θ k 1 ( 1 θ ) 3 / 4 d θ ,

then 0 u ( k ) ( t , x ) 1 and 0 v ( k ) ( t , x ) 1 , k = 0 , 1 ,  .

Proof We have
Δ u t ( 0 ) ( t , x ) = u t ( 1 ) ( t , x ) u t ( 0 ) ( t , x ) = D 1 u x x ( 1 ) ( u ( 0 ) [ D 1 u x ( 0 ) + D 2 v x ( 0 ) ] ) x ( Δ u ( 0 ) [ D 1 u x ( 0 ) + D 2 v x ( 0 ) ] ) x ( u ( 0 ) [ D 1 Δ u x ( 0 ) + D 2 Δ v x ( 0 ) ] ) x k u ( x ) = D 1 Δ u x x ( 0 ) ( Δ u ( 0 ) [ D 1 u x ( 0 ) + D 2 v x ( 0 ) ] ) x ( u ( 0 ) [ D 1 Δ u x ( 0 ) + D 2 Δ v x ( 0 ) ] ) x + t R 1 ( x ) + t 2 R 2 ( x ) ,
where R 1 ( x ) and R 2 ( x ) are of the same form as in the proof ofProposition 3.3. Since
t R 1 + t 2 R 2 L T ( R 1 L + T R 2 L ) = : K ˜ 0
we have
( Δ u ( 0 ) , Δ v ( 0 ) ) ( t , ) W 1 , 0 t C t s ( Δ u ( 0 ) , Δ v ( 0 ) ) ( s , ) W 1 , d s + 0 t C ˜ ( t s ) 3 / 4 K ˜ 0 d s .
Applying Lemma A.1 we get
( Δ u ( 0 ) , Δ v ( 0 ) ) ( t , ) W 1 , K ˜ 1 t 3 / 4
and by induction ( Δ u ( k ) , Δ v ( k ) ) ( t , ) W 1 , K ˜ k + 1 t 3 ( k + 1 ) 4 for k = 1 , 2 ,  . Hence
K ˜ k + 1 : = K ˜ k C ˜ 1 2 C T 1 / 2 0 1 θ k ( 1 θ ) 3 / 4 d θ .

 □

5 Conclusions

Assume that ( u ( 0 ) , v ( 0 ) , w ( 0 ) ) coincides with ( u 0 , v 0 , w 0 ) at t = 0 and consider the following iterative scheme for(1.1)-(1.3):
u t ( k + 1 ) = D 1 u x x ( k + 1 ) ( u ( k + 1 ) [ D 1 u x ( k ) + D 2 v x ( k ) + D 3 w x ( k ) ] ) x , v t ( k + 1 ) = D 2 v x x ( k + 1 ) ( v ( k + 1 ) [ D 1 u x ( k ) + D 2 v x ( k ) + D 3 w x ( k ) ] ) x , w t ( k + 1 ) = D 3 w x x ( k + 1 ) ( w ( k + 1 ) [ D 1 u x ( k ) + D 2 v x ( k ) + D 3 w x ( k ) ] ) x
with the initial condition
u ( k + 1 ) ( 0 , x ) = u 0 ( x ) , v ( k + 1 ) ( 0 , x ) = v 0 ( x ) , w ( k + 1 ) ( 0 , x ) = w 0 ( x )
for x [ L , L ] and the Neumann boundary condition. Denote
Δ u ( k ) = u ( k + 1 ) u ( k ) , Δ v ( k ) = v ( k + 1 ) v ( k ) , Δ w ( k ) = w ( k + 1 ) w ( k ) .
Convergence problems occur in L 2 and the Sobolev norm W 1 , . Our attempt to obtain the following relation for theincrements Δ u ( k ) , Δ v ( k ) , Δ w ( k ) :
d d t A k + 1 ( t ) C [ A k + 1 ( t ) + A k ( t ) ] , A k ( t ) = L L [ ( Δ u ( k ) ) 2 + ( Δ v ( k ) ) 2 + ( Δ w ( k ) ) 2 ] d x
was unsuccessful as it is difficult to estimate the following component:
L L Δ u ( k + 1 ) ( u ( k + 1 ) [ D 1 Δ u x ( k ) + D 2 Δ v x ( k ) + D 3 Δ w x ( k ) ] ) x d x .

This example of iterations shows that strongly coupled systems cause seriousproblems with their approximation. We think that the ternary system and suitableapproximations to it will be somehow expressed in an abstract way, based on aConti-Opial type theorem, like in [13].

In order to illustrate fast convergence of Newton’s iterations we providenumerical examples with D 1 = 1 , D 2 = 0.5 , D 3 = 0.2 , and u 0 , v 0 being sample piecewise polynomial functions takingvalues in [ 0.2 , 0.8 ] . We check the differences u k + 1 u k and v k + 1 v k for k = 0 , 1 , 2 , 3 , 4 . Our computer programs are performed by implicitfinite difference methods with steps h 0 = h 1 = 0.01 ; see Table 1 (directiterations), Table 2 (Newton’s iterations).
Table 1

Maximal differences between successive approximations u ( k ) by direct iterations ( 3.1 ) with D 1 = 1 , D 2 = 0.5 , D 3 = 0.2 , h = 0.01 , h 0 = 0.01

t

| u ( 1 ) u ( 0 ) |

| u ( 2 ) u ( 1 ) |

| u ( 3 ) u ( 2 ) |

| u ( 4 ) u ( 3 ) |

| u ( 5 ) u ( 4 ) |

0.00

0.000000e+00

0.000000e+00

0.000000e+00

0.000000e+00

0.000000e+00

0.05

1.023328e−01

3.742852e−03

1.383520e−04

7.409068e−06

3.256123e−07

0.10

1.476173e−01

6.990620e−03

2.908263e−04

1.835346e−05

1.004908e−06

0.15

1.783626e−01

9.324930e−03

4.031573e−04

2.746097e−05

1.675459e−06

0.20

2.028066e−01

1.096861e−02

4.788241e−04

3.397007e−05

2.215127e−06

0.25

2.239067e−01

1.206830e−02

5.253484e−04

3.806322e−05

2.601687e−06

0.30

2.425240e−01

1.273201e−02

5.499511e−04

4.011546e−05

2.844953e−06

0.35

2.593736e−01

1.304597e−02

5.585483e−04

4.051817e−05

2.967818e−06

0.40

2.745434e−01

1.307759e−02

5.556250e−04

3.970391e−05

2.985870e−06

Table 2

Maximal differences between successive approximations u ( k ) by Newton’s method ( 4.1 ) with D 1 = 1 , D 2 = 0.5 , D 3 = 0.2 , h = 0.01 , h 0 = 0.01

t

| u ( 1 ) u ( 0 ) |

| u ( 2 ) u ( 1 ) |

| u ( 3 ) u ( 2 ) |

| u ( 4 ) u ( 3 ) |

| u ( 5 ) u ( 4 ) |

0.00

0.000000e+00

0.000000e+00

0.000000e+00

0.000000e+00

0.000000e+00

0.05

9.970638e−02

1.703717e−03

6.949251e−07

1.628697e−13

8.038015e−14

0.10

1.430019e−01

3.927916e−03

4.076982e−06

4.423240e−12

5.245804e−14

0.15

1.723550e−01

5.877059e−03

9.749392e−06

2.192441e−11

3.896883e−14

0.20

1.956575e−01

7.536084e−03

1.659814e−05

1.148864e−10

3.987088e−14

0.25

2.155355e−01

8.989691e−03

2.404146e−05

4.279036e−10

3.957945e−14

0.30

2.332049e−01

1.029947e−02

3.161596e−05

1.167791e−09

3.042011e−14

0.35

2.489078e−01

1.153614e−02

4.250313e−05

2.625514e−09

1.676437e−14

0.40

2.631780e−01

1.269348e−02

5.972126e−05

5.163286e−09

2.378098e−13

Appendix

Lemma A.1 Assume that
z ( t ) 0 t C t s z ( s ) d s + 0 t C ˜ ( t s ) α p ( s ) d s and p ( s ) K s m .

Then z ( t ) K ˜ t m + α for t [ 0 , T ] , where 1 2 α < 1 and 1 2 C T 1 / 2 > 0 .

Proof We have
z ( t ) 0 t C t s z ( s ) d s + 0 t C ˜ ( t s ) α K s m d s 0 t C t s K ˜ t m + α d s + 0 t C ˜ ( t s ) α K s m d s
for 0 t T and 1 2 α < 1 . We claim that
2 K ˜ C t m + α T 1 / 2 + C ˜ K t m + α 0 1 θ m ( 1 θ ) α d θ K ˜ t m + α , K ˜ t m + α ( 1 2 C T 1 / 2 ) C ˜ K t m + α 0 1 θ m ( 1 θ ) α d θ .
It suffices to take
K ˜ : = K C ˜ 1 2 C T 1 / 2 0 1 θ m ( 1 θ ) α d θ .
(A.1)

 □

Declarations

Acknowledgements

This work was supported by the National Science Center (Poland) decision no.DEC-2011/02/A/ST8/00280.

Authors’ Affiliations

(1)
Institute of Mathematics, University of Gdańsk, Wita Stwosza 57, Gdańsk, 80-952, Poland

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Copyright

© Leszczyński and Wrzosek; licensee Springer. 2014

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