Regular expansion and consistency

not be altered. Therefore we get also the asymptotic similarity of the derivatives, simply by multiplying the equation

∂_t(∂_x^kd_ǫ) +ǫ⁻¹S∂_x(∂_x^kd_ǫ) =ǫ⁻²J ∂^k_xd_ǫ−ǫ^α∂_x^kr_ǫ, 1≤k≤m−1

with B∂^k_xd_ǫ and performing exactly the same steps as in the proof of the previous theorem.

Corollary 3.1. Suppose the assumptions of theorem 3.1 are satisfied. If addition-ally fǫ,ˆfǫ ∈ Cper^m (X^T,F) then the two functions are for ǫ↓ 0 asymptotically similar inC T,H^m−1(X,F)

with a residual of orderα, i.e. there is a constantK >0such that

sup

t∈Tk∂_x^kfǫ(t,·)−∂_x^kˆfǫ(t,·)k_L²(X;F)< Kǫ^α 0≤k≤m−1.

Sobolev’s embedding theorem⁹ tells us, that any weakly differentiable function on an interval of the real axis is continuous (strictly speaking its equivalence class with respect to almost everywhere equality contains a continuous representative).

Moreover its supremum-norm can be estimated by theL²-norm of the function and its weak derivative. Especially this entails in our case form≥2

sup

t∈Tk∂_x^kf_ǫ(t,·)−∂_x^kˆf_ǫ(t,·)k_L²(X;F) = O(ǫ^α), 0≤k≤m−1

=⇒ sup

t∈T

sup

x∈Xk∂_x^kf_ǫ(t, x)−∂_x^kˆf_ǫ(t, x)kF = O(ǫ^α), 0≤k≤m−2 and allows us to formulate a second corollary.

Corollary 3.2. Under the hypothesis of corollary 3.1, f_ǫ andˆf_ǫ are asymptotically similar in Cper^m−2(XT,F), i.e. with respect to the supremum-norm overXT =T × X.

3.2 Regular expansion and consistency

This section deals with the explicit construction of a comparison (prediction) func-tion in the form of a truncated regular expansion for some arbitrary but fixed n∈N₀:

ˆf_ǫ^[n]:=f⁽⁰⁾+ǫf⁽¹⁾+. . .+ǫⁿf⁽ⁿ⁾ . (3.12) In general it is not possible to choose theasymptotic order functionsf^(k), 0≤k≤n, in such a way that the comparison function becomes an exact solution of the lattice-Boltzmann equation (3.2). Hence we try to determine thef^(k)’s so that theresidual¹⁰

∂_tˆf_ǫ^[n] − ǫ⁻¹S∂_xˆf_ǫ^[n] − ǫ⁻²J₀ˆf_ǫ^[n] + chˆf_ǫ^[n],1iw − gw =: ˜r^[n]_ǫ

is asymptotically minimized for ǫ ↓ 0. ˜r^[n]ǫ inherits the structure of a finite power series starting however with ǫ⁻². In order to make the residual vanish as fast as possible for ǫ→0, we equate terms of equal power in ǫwith 0. The f^(k)’s must be

9[4] page 129

10This expression is obtained by plugging the comparison function ˆfǫ^[n]into the lattice-Boltzmann equation and putting all terms on the left hand side.

determined in such a manner that the obtained equations are satisfied. A priori, the residual comprises n+ 3 different orders fromǫ⁻² toǫⁿ which yieldsn+ 3 different equations. As we have to determine n+ 1 asymptotic order functions, we expect that all equations but two can be fulfilled by an appropriate choice of thef^(k)’s. We pick the equations referring to ǫ⁻² up to ǫⁿ⁻² which ensures the residual to be of magnitude O(ǫⁿ⁻¹). So the subsequent equations result

ǫ⁻² : 0 =J₀f⁽⁰⁾ ǫ⁻¹ : S∂_xf⁽⁰⁾ =J₀f⁽¹⁾

ǫ^k : ∂_tf^(k)+S∂_xf^(k+1) =J₀f^(k+2)−chf^(k),1iw+gδ_k0w





 (3.13) where the index k runs through the set {0, ..., n−2}. Apart from the two first equations, the obtained equations share the same algebraic structure.

If we achieve to satisfy the above equations by a special selection of f⁽⁰⁾, . . . ,f⁽ⁿ⁾, then only terms of the order n−1 andnwill survive.

ǫⁿ⁻¹∂tf⁽ⁿ⁻¹⁾+ǫⁿ⁻¹S∂xf⁽ⁿ⁾+ǫⁿ∂tf⁽ⁿ⁾=−ǫⁿ⁻¹chf⁽ⁿ⁻¹⁾,1iw−ǫⁿchf⁽ⁿ⁾,1iw+ ˜r^[n]_ǫ These remaining terms have to be compensated by the actual residual, if equality shall hold. Resolving thus for ˜r^[n]ǫ yields

˜r^[n]_ǫ =ǫⁿ⁻¹n

∂_tf⁽ⁿ⁻¹⁾+S∂_xf⁽ⁿ⁾+ǫ∂_tf⁽ⁿ⁾+chf⁽ⁿ⁻¹⁾,1iw+ǫchf⁽ⁿ⁾,1iw

| {z }

residue function r^[n]_ǫ

o. (3.14)

Iff⁽ⁿ⁻¹⁾ andf⁽ⁿ⁾ are sufficiently smooth, the residue function is uniformly bounded for ǫ ∈ (0,1]. Hence we find the ansatz (3.12) consistent of order n−1 provided that we manage to solve system (3.13).

For ease of notation we define two differential operators D_t:=−τ ∂_t, D_x :=−τ S∂_x

mapping Cper^m (XT;F) to Cper^m−1(XT;F). Let us collect those terms in (3.13) on the left hand side, which contain the highest occurring asymptotic order function. By introducing the linear map

M :F → F, M :=I− h·,1iw=−τ J₀ (3.15) the equations (3.13) are recast in the form:

Mf⁽⁰⁾ = 0 (3.16)

Mf⁽¹⁾ = D_xf⁽⁰⁾ (3.17)

Mf^(k+2) = D_tf^(k)+D_xf^(k+1)−τ chf^(k),1iw+τ gδ_k0w (3.18) Before we continue, a quick study of the properties ofM seems to be recommended.

Lemma 3.2. The linear operator M, as defined above, is a projector. For given y ∈ F the equation Mx = y is only solvable if hy,1i = 0. Its general solution is given by x= kerM +y=Rw+y.

3.2. Regular expansion and consistency 135

Proof: Since hw,1i= 1 we get:

M²x = Mx− hMx,1iw = x− hx,1iw−

x− hx,1iw,1 w

= x−2hx,1iw+hx,1ihw,1iw = x− hx,1iw

= Mx

So, we see that M is a projector, indeed. Its nullspace is obviously given by kerM = span{w}. Hence M is not invertible and the equation admits only a solution ify∈rangeM. ByFredholm’s alternative for Euclidean vector spaces (see the proposition below) we have rangeM = (kerM^⊤)_⊥. Let us therefore compute the adjoint operator M^⊤:

hMx,zi =

x− hx,1iw, z

= hx,zi − hx,1ihw,zi = hx,zi − hx,hz,wi1i

=

x, z− hz,wi1

= hx, M^⊤zi ⇒ M^⊤=I− h·,wi1

Evidently, we have kerM^⊤ = span{1}, which yields the necessary solvability con-dition hy,1i = 0 ⇔ y∈ (kerM^⊤)_⊥. Furthermore being a projector, M acts like the identity¹¹ on its range. So y ∈ rangeM is a specific solution. As the general solution is given as a sum of a particular solution plus an arbitrary solution of the homogeneous equation being an element of kerM = span{w}, we find the asserted

formula verified.

Proposition 3.2. LetA:V →V be an endomorphism on an Euclidean vector spaceV. Then the range (image) of A is the orthogonal complement to the kernel (nullspace) of the adjoint operator A^⊤, i.e.:

rangeA= (kerA^⊤)⊥.

Proof: Letw∈kerA^⊤andu∈rangeA, i.e. there exists av∈V withu=Av. Then we haveu⊥v, since hu, wi = hAv, wi = hv, A^⊤wi = hv,0i = 0.

This shows rangeA ⊂ (kerA^⊤)_⊥. In order to prove the equality we must take recourse to the finite dimensionality ofV. Therefore it is possible to consider a matrix representation ofAandA^⊤. We note that a matrix and its transpose have the same rank, since row rankandcolumn rank are equal for any matrix. Thus we conclude: dim(rangeA) = dim(rangeA^⊤). Moreover the following relations hold true:

i) V = kerA^⊤⊕(kerA^⊤)⊥ ⇒ dim`

(kerA^⊤)⊥

´ = dimV −dim(kerA^⊤) ii) dimV = dim(kerA^⊤) + dim(rangeA^⊤) ⇔ dim(rangeA^⊤) = dimV −dim(kerA^⊤) Hence we get: dim(rangeA) = dim`

(kerA^⊤)_⊥´

. So rangeAis a subspace of (kerA^⊤)_⊥having the same dimension. This enforces, however, the equality of the two spaces.

By means of the preceding lemma, equations (3.16),(3.17) and (3.18) are easily solved. As f⁽⁰⁾ is in the kernel of M according to equation (3.16), we conclude

f⁽⁰⁾ =u⁽⁰⁾w, (3.19)

with someu⁽⁰⁾:XT →Rto be specified more precisely later. If we take this expres-sion for f⁽⁰⁾ and remember hSw,1i=0, we notice immediately, that the right hand

11Let y ∈ rangeM with y = Mx. The defining property of projectors, M² = M, implies y−My=Mx−M²x=Mx−Mx= 0. ThereforeM acts like the identity on rangeM.

side of equation (3.17) fulfills the solvability conditionhD_xf⁽⁰⁾,1i= 0 independently of u⁽⁰⁾. Therefore we obtain

f⁽¹⁾ =u⁽¹⁾w+D_xf⁽⁰⁾ (3.20) with another unknown functionu⁽¹⁾ remaining also undetermined so far. Similarly, we proceed with the higher asymptotic orders. This yields for k∈ {0, ..., l−2}the following recursion relation

f^(k+2) =u^(k+2)w+Dtf^(k)+Dxf^(k+1)−τ chf^(k),1iw+τ gδ_k0w, (3.21) provided the solvability condition

D_tf^(k)+D_xf^(k+1)−τ chf^(k),1iw+τ gδ_k0w, 1

= 0 (3.22)

is fulfilled. Obviously, for any order a new scalar function comes into play. To all appearances, the undetermined F-valued asymptotic order f^(k) is replaced by an-other unknown but scalar-valued function u^(k), that seems to be quite arbitrary at first glance. Therefore the impression imposes, that we are not winning much, since the complete meaning of these scalar-valued functions remains a little bit vague.

So far it is only known that u^(k) = hf^(k),1i due to the solvability condition; thus u^(k) is the mass moment of f^(k). However, we already anticipate, that the solvabil-ity condition inflicts certain constraints on the u^(k)’s, which make them not at all arbitrary. In order to demonstrate this, it is necessary, that we substitute in equa-tion (3.21) the asymptotic orders recursively by their corresponding mass moments.

To do so, it is convenient to define the following family of differential operators by recurrence, which spares the cumbersome writing of lengthy terms.

P_k:= 0 fork <0, P₀ :=I, P_k:=D_xP_k−1+D_tP_k−2 fork≥1 (3.23) We consider P_k as a differential operator from Cper^m+k(X^T;F) to Cper^m (X^T;F) for m∈N₀. In the sequel of this section we tacitly assume enough smoothness, so that all occurring differential operators can be applied in a meaningful sense.

Remark 3.1. (Parity) An essential property of the P_k’s is, that they exhibit a well defined parity, if we interpret them as polynomials with respect to the operator S. By a simple induction it is verified for j ∈N₀ that even powers of S including S⁰ = I occur only in P_2j, whereas P_2j+1 contains odd ones. Hence the relations S^2j+1=S and hSw,1i = 0 entail

hP_2j+1fw,1i= 0 (3.24)

for any functionf ∈ Cper^2j+1(XT;R).

Remark 3.2. (Homogeneity)Another characteristic feature should not stay un-mentioned, for it will become of much use further down. From a formal point of view, we can consider the differential operator P_k as a polynomial in D_t and D_x. Let us therefore write P_k = P_k(D_t, D_x). Moreover, let us replace the differential

3.2. Regular expansion and consistency 137

operatorsD_t, D_x by the formal variablesχ, ξ, so that we can define the polynomial p_k(χ, ξ) :=P_k(χ², ξ). Evidently, these polynomials obey the recursion relation

p₀(χ, ξ) = 1, p₁(χ, ξ) =ξ, p_k(χ, ξ) :=ξ p_k−1(χ, ξ) +χ²p_k−2(χ, ξ). By an induction argument one can check, that these polynomials are homoge-neous and their degree of homogeneity is given by their index. Thusp_k(λχ, λξ) = λ^kp_k(χ, ξ). Hence, the following consequence ensues: If we assign to each monomial ofP_k(D_t, D_x) a degree given bytwice the order of the time derivative plus the order of the spatial derivative, it turns out, that we obtain for every monomial of P_k the same degree. Introducing the function spaces

Cⁿ

per(XT;F) :=

f ∈ Cper(XT;F) : ∂_tⁱ∂_x^jf ∈ Cper(XT;F) if 2i+j≤n , it is now straightforward to see, that P_k maps C_per^n+k(XT;F) to C_perⁿ (XT;F).

The objective of the next lemma is to resolve the recursion relation (3.21) for the asymptotic order functions and to express them in terms of their mass moments.

Lemma 3.3. f^(k+2),0≤k≤n−2, admits the following explicit representation f^(k+2)=

k+2X

j=0

P_k+2−ju^(j)w − τ c Xk j=0

P_k−ju^(j)w + τ P_kgw. (3.25) Proof: The proof is done by induction.

Start: It is easily checked that the formula holds even true forf⁽⁰⁾ and f⁽¹⁾, i.e. for k∈ {−2,−1}, if we setu^(j)= 0 forj <0. Usingδ₀₀= 1 we get fork= 0:

f⁽²⁾ = u⁽²⁾w+D_tf⁽⁰⁾+D_xf⁽¹⁾−τ chf⁽⁰⁾,1iw+τ gw

= u⁽²⁾w+D_tu⁽⁰⁾w+D_xu⁽¹⁾w+D²_xu⁽⁰⁾w−τ cu⁽⁰⁾hw,1iw+τ gw

= P0u⁽²⁾w+P1u⁽¹⁾w+P2u⁽⁰⁾w−τ cP0u⁽⁰⁾w+τ P0gw Thus we obtain the claimed formula also in this case.

Step: Assuming formula (3.25) to be true fork≥0, we show that it is also retrieved fork+1>0. Note that the Kroncker delta vanishes then. Setting in equation (3.21) kto k+ 1 yields:

f^{(k+3) !}= u^(k+3)w+Dxf^(k+2)+Dtf^(k+1)−τ cu^(k+1)w.

Now the induction hypothesis is used to substitute f^(k+1) and f^(k+2) by analogous formulas:

f^(k+3) = u^(k+3)w+D_{x k+2}

P

j=0

P_k+2−ju^(j)w−τ cP^k

j=0

P_k+1−ju^(j)w+τ P_kgw

+D_{t k+1}

P

j=0

P_k+2−ju^(j)w−τ c^kP⁻¹

j=0

P_k+1−ju^(j)w+τ P_k−1gw

−τ cu^(k+1)w.

Next, corresponding sums inside each of the square brackets are rewritten such that the index j runs everywhere over the same range. So the sums can be condensed:

f^(k+3)= P₀u^(k+3)w+P₁u^(k+2)w+^k+1P

Finally the recursion relation (3.23) is applied f^(k+3)= ^k+3P

j=0

P_k+3−ju^(j)w−τ c^k+1P

j=0

P_k+1−ju^(j)w+τ P_k+1gw,

to recover formula (3.25) mutatis mutandiswhere kis replaced by k+ 1.

At last, we are prepared to analyze the solvability condition. Equation (3.25) is equivalent to Taking all addends containing u^(k+1) and u^(k) out of the sums, we obtain

hP₁u^(k+1)w,1i + hP₂u^(k)w,1i − τ chP₀u^(k)w,1i =

we arrive at a diffusion-reaction equation for u^(k) with a sourceR_k:

∂_tu^(k)−^τ_θ∂_x²u^(k)+cu^(k) = ωDkP−1

Let kbe even. Due to equation (3.24) only those terms survive the scalar product, where k+ 2−j is even too. Thus we obtain for k = 2n at the right hand side

3.2. Regular expansion and consistency 139

an expression containing only u⁽⁰⁾, u⁽²⁾, ..., u⁽²ⁿ⁻²⁾ and g as well as certain (mostly mixed) partial derivatives of these quantities. For oddkthe situation is completely analogous except that g does not occur. This results in a decoupling of even and odd orders, that emerges as the responsible feature for the second order convergence mentioned in section 3. Setting either k= 2l ork= 2l+ 1 we obtain the following evolution equations from the above one:

∂_tu^(2l)−^τ_θ∂_x²u^(2l)+cu^(2l)=R_2n u⁽⁰⁾, u⁽²⁾, ..., u^(2l−2), g

, (3.26)

∂_tu^(2l+1)−^τ_θ∂_x²u^(2l+1)+cu^(2l+1)=R_2l+1 u⁽¹⁾, u⁽³⁾, ..., u^(2l−1)

. (3.27)

In the case of the two lowest orders we get in particular:

∂_tu⁽⁰⁾−^τ_θ∂_x²u⁽⁰⁾+cu⁽⁰⁾ = g (3.28)

∂_tu⁽¹⁾−^τ_θ∂_x²u⁽¹⁾+cu⁽¹⁾ = 0 (3.29) These evolution equations respresent the desired conditions, we have to impose on the mass moments, in order to make the ansatz (3.12) “as consistent as possible”.

Together with appropriate initial conditions, this hierarchy of diffusion-reaction equations determines the mass moments u⁽⁰⁾, ..., u⁽ⁿ⁻²⁾ uniquely. In contrast, no constraints arise from the solvability condition for the mass moments u⁽ⁿ⁻¹⁾, u⁽ⁿ⁾. Besides sufficient regularity requirements, they can be freely chosen. That is why we are allowed to set them to zero.

Remark 3.3. (Vanishing of odd orders) The homogeneity of equation (3.29) has an important consequence, asu⁽¹⁾remains zero once and for all, if it is initialized so. By equation (3.27) higher odd orders depend entirely on lower ones. Therefore we conclude, that all subsequent odd orders vanish too, provided they are initialized by zero. This fact emanates from the circumstance that the inhomogeneitygwof the lattice-Boltzmann equation (3.2) appears only in the zeroth order as O(1)-term. Of course, things would change, if some O(ǫ)-term were admitted as additional source.

Remark 3.4. (Construction of the comparison function) How can we con-struct a truncated regular expansion, that reaches on the one hand a prescribed order of consistency but that is on the other hand as simple as possible? The pre-vious considerations give the following suggestion, that will serve as guideline for the next subsection. In order to distinguish easily between even and odd order functions we considerf^[2n+3] (instead of f^[n]) from now on.

a) Find sufficiently smooth functionsu⁽⁰⁾, u⁽²⁾, ..., u⁽²ⁿ⁾satisfying (3.28) and (3.27) respectively.

b) Set u⁽¹⁾, u⁽³⁾, ..., u⁽²ⁿ⁺¹⁾, u⁽²ⁿ⁺³⁾ andu⁽²ⁿ⁺²⁾ equal to 0 and compute by equa-tion (3.25) the 2m+ 4 asymptotic orders f⁽⁰⁾,f⁽¹⁾, ...,f⁽²ⁿ⁺³⁾.

c) Set ˆf^[2n+3]:=P_2n+3

k=0 ǫ^kf^(k), which is then consistent of order 2n+ 2.

d) Asu⁽²ⁿ⁺²⁾, u⁽²ⁿ⁺³⁾ = 0 the mass moments off⁽²ⁿ⁺²⁾,f⁽²ⁿ⁺³⁾vanish. According to equation (3.14) the residual is therefore simplified to

˜r_ǫ^[2n+3]=ǫ²ⁿ⁺²n

∂_tf⁽²ⁿ⁺²⁾+S∂_xf⁽²ⁿ⁺³⁾+ǫ∂_tf⁽²ⁿ⁺³⁾o

. (3.30)

The proposed construction of the comparison function has a nice impact on the internal structure of the asymptotic orders. It ensures that even asymptotic orders are exclusively composed of terms containing an even power of s (i.e. 1 ≡ s⁰,s²).

In contrast, the addends pertaining to odd asymptotic order functions are always proportional to s ≡ s¹, which is the only odd power. We prove this property by exploiting the orthogonality ofs∈ F with respect to the subspace span{1,s²} ⊂ F. Lemma 3.4. Suppose that the asymptotic orders are constructed as described above, such that the mass moments of the odd orders vanish all. Then the following rela-tions hold true for j∈ {0,1, ...,2n+ 3}.

hf^(2j),si= 0, hf^(2j+1),s²i= 0, hf^(2j+1),s²−¹_θi= 0 (3.31) Proof: The proof is done by induction.

Start: Usingu⁽¹⁾ = 0 we get from (3.19) and (3.20)

f⁽⁰⁾=u⁽⁰⁾w, f⁽¹⁾ =D_xf⁽⁰⁾ =−τ ∂_xu⁽⁰⁾sw.

Recalling the orthogonality of w,s∈ F, (3.31) becomes directly clear forj= 0.

Step: Let us now suppose, that the relations are true for the asymptotic ordersf^(2j) andf^(2j+1) for somej∈N₀. Applying the recursive representation (3.21) fork= 2j we find

hf^(2j+2),si = u^(2j+2)hw,si+hD_tf^(2j),si+hD_xf^(2j+1),si −τ chf^(2j),1i hw,si

= 0 − τ ∂thf^(2j),si − τ ∂xhsf^(2j+1),si − 0 = 0

Using hsf^(2j+1),si = hf^(2j+1),s²i we see, that both scalar products vanish due to the induction hypothesis. Analogously we consider equation (3.21) for k= 2j+ 1 using the assumption of the lemma (u^(2j+3) = 0) and the induction hypothesis hf^(2j+1),1i=u^(2j+1) = 0.

hf^(2j+3),s²i = hD_tf^(2j+1),s²i + hD_xf^(2j+2),s²i

= −τ ∂_thf^(2j+1),s²i − τ ∂_xhsf^(2j+2),s²i = 0

The first scalar product vanishes by the induction hypothesis. Shifting s from the left to the right hand side of the second scalar product and usings³ =s, we obtain hf^(2j+2),si. But this vanishes by what we have just proved above.

The third asserted relation is an immediate consequence of the second one and the assumption that the odd mass moments are zero.

hf^(2j+1),s²−¹_θi=hf^(2j+1),s²i −¹_θhf^(2j+1),1i= 0−0 = 0 These relations are exploited in the next section, when we show that the moments with respect to sand s²− ¹_θ exhibit also a privileged convergence behavior besides the mass moment.

Im Dokument Analysis of Lattice-Boltzmann Methods : Asymptotic and Numeric Investigation of a Singularly Perturbed System (Seite 145-152)