Analysis of a Mixed-Order BVP

We consider the spatial operator A(∂x). In order to match the assumptions in [34] which are necessary later we rewriteA(∂x) to an equivalent operatorA(D) with respect to (u^d,· · ·, u¹, u⁰) in the following way constant coefficients and the entries A_jk(D) are linear differential operators defined on Ω of order not exceeding s_j +m_k. For ξ = (ξ₁,· · ·, ξ_d) ∈ R^d let A(ξ) denote the symbol, ˚A(ξ) the principal symbol ofA(D) respectively( ˚A_jk(ξ) consists of the terms in A(ξ) which are of the ordersj+m_k), then

We compute det ˚A(ξ) and infer

det ˚A(ξ) = 0⇐⇒ξ = 0.

According to [5]A(D) is an elliptic differential matrix operator of mixed-order(a general system of Agmon-Douglis-Nirenberg type).

Next we shall be concerned with the following boundary value problem with a parameter η in some closed sector in the complex plane with vertex at the origin with respect to the unknown functions

u(x) := (u^d, u^d−1,· · ·, u¹, u⁰) (A(D)u(x)−ηu(x) =f(x) in Ω,

B(D)u(x) =g(x) on ∂Ω,

(3.2.32) wheref(x) = (f1(x),· · · , fd+1(x))^T are (d+ 1)×1 matrix functions defined in Ω, g(x) = (g₁(x),· · ·, g_d+1(x))^T defined on ∂Ω. The (d+ 1)×(d+ 1) matrix operator B(D) describes the Dirichlet boundary conditions and is defined as follows

B(D) :=I_d+1,

here letI_d+1denote the (d+ 1)×(d+ 1) identity matrix. As forB(D) we define another sequence of integers{r_j}^d+1₁ such that

r₁ =r₂ =· · ·=r_d= 0, r_d+1=−1, (3.2.33) then it is easy to see that the entries of B(D) are linear differential operators defined on∂Ω of order not exceeding rj+mk withrj < m:=sj+mj = 2 and defined to be zero ifrj+m_k<0.

Before starting our main result we study the pullback of a linear differential operator. We first consider the linear differential operator whose domain of definition and range are scalar-valued functions.

Definition 3.2.1. For two nonempty bounded open sets Ω, G⊂Rⁿ letΦ : Ω→ Gbe a bijection. Given a function u:G→C then the function Φ^∗u:=u◦Φ : Ω→Cis called the pullback of u. Assume Φ : Ω→G is a C^m-diffeomorphism 1 < p < ∞ and let A(x, D) be a linear differential operator on G of order m with domain W_p^m(G), then the operator B(x, D) := Φ^∗A := A(· ◦Φ⁻¹)◦Φ defined onΩ with domain W_p^m(Ω)is called the pullback of A(x, D).

Theorem 3.2.4. In definition 3.2.1 assume the Jacobimatrix Φ⁰ of Φ is con-stant in x∈Ω, i.e., ∂jΦ⁰ ≡0(j= 1,· · ·, n).Then

B(x, D) =A(Φ(x),[Φ⁰(x)]^−TD); (3.2.34) B˚(x, D) = ˚A(Φ(x),[Φ⁰(x)]^−TD), (3.2.35) where[·]^−T := ([·]⁻¹)^T,(·)^T denotes the transpose, B(x, D),˚ A(x, D)˚ denote the principal parts of B(x, D) and A(x, D) respectively.

Proof. Without loss of generality assume (A(y, D)u)(y) =a(y)(D_i1· · ·D_isu)(y) with a functiona: G→ C and i1,· · · , is ∈ {1,· · ·, n}. Let v ∈W_p^m(Ω), then

v◦Φ⁻¹ ∈W_p^m(G)(it follows from theorem 3.14 in [1]) and

∂i1(v◦Φ⁻¹)(y) =

n

X

j1=1

(∂j1v)(Φ⁻¹(y))Sj1i1, (3.2.36) where (Sjk)j,k=1,···,n:= (Φ⁻¹)⁰. By product rule:

∂_i2∂_i1(v◦Φ⁻¹)(y) =

n

X

j2=1 n

X

j1=1

(∂_j2∂_j1v)(Φ⁻¹(y))S_j2i2S_j1i1. (3.2.37) Then it follows from iteration:

∂i1· · ·∂im(v◦Φ⁻¹)(y) =

n

X

jm=1

· · ·

n

X

j1=1

(∂j1· · ·∂jmv)(Φ⁻¹(y))

m

Y

l=1

Sjlil, (3.2.38) which implies

(A(y, D)(v◦Φ⁻¹))(y)

=a(y)

n

X

jm=1

· · ·

n

X

j1=1

(D_j1· · ·D_jmv)(Φ⁻¹(y))

m

Y

l=1

S_jlil

=a(Φ(x)) _m

Y

l=1

((S_jk)j,k=1,···,n)^T(D₁ D₂· · ·D_n)^T

il

! v

! (x).

Thus we obtain (3.2.34), and (3.2.35) follows as a direct consequence.

Compared to the scalar case thepullback of a vector field is defined by a different way. We consider only the special case that the transform Φ : Ω → G is a rotation and therefore Φ is a rotation matrix. Let u : G → Cⁿ be a vector field, then the pullback of u is Φ^∗u:= Φ⁻¹◦u◦Φ : Ω→Cⁿ. Now we give two important examples of the pullback of relevant operators on a vector field (or the range is a vector field).

Example 3.2.1. Let Φ : Ω → G be a rotation, u : G → C a scarlar field, then we obtain the pullback of u u◦Φ : Ω→C from definition 3.2.1. Let A be the gradient of u, then the corresponding differential operator on u◦Φ is also gradient, i.e.,

∇(u◦Φ)(x) = Φ⁻¹∇u(y) with y= Φ(x).

Example 3.2.2. Suppose u :G → Cⁿ is a vecor field, A is the divergence of u, then the divergence ofΦ⁻¹◦u◦Φ : Ω→Cⁿ equals to A(u), i.e.,

div(Φ⁻¹◦u◦Φ)(x) =divu(y) with y= Φ(x).

Remark 3.2.1. From the aspects of theory of fields the gradient of a scalar field and the divergence of a vector field describe the feature of the scalar field and the vector field respectively and they are independent of the choice of the coordinate system (via rotation).

After these preparations mentioned above we have the following

Theorem 3.2.5. The boundary problem (3.2.32) is elliptic with parameter in any sectorL in the complex plane with vertex at the origin and

L ⊂S₁∪S₂ where

S1:=

η∈C, −π+arctg 2ν0

<argη <−arctg 2ν0

, S2:=

η∈C, arctg 2ν0

<argη < π−arctg 2ν0

. Assume

2ν0

≤√ 3,

then the boundary problem (3.2.32) is elliptic with parameter in any sector L where

L ⊂

z∈C, argz6= 0,±arctg 2ν0

in the sense of Definition 2.2.2.

Proof. Ford= 1,

det( ˚A(ξ)−ηI) = 0 =⇒η=ν₀|ξ|²±i 2|ξ|²; ford >1,

det( ˚A(ξ)−ηI) = 0 =⇒η =ν0|ξ|² or η=ν0|ξ|²±i 2|ξ|². Then the first condition of defintion 2.2.2 follows immediately.

Fix now x₀ ∈ ∂Ω, let Φ_x0 denote the coordinate transformation from the original coordinate system into a local coordinate system atx0 (here Φx0(0) = x₀ and ν 7→ e_n, where ν is the interior normal to ∂Ω at x₀ and (e₁,· · ·, e_n) denotes the standard basis in Rⁿ). We achieve this coordinate transformation via translation and rotation. Notice that the considered operator reads





−ν₀4 −div

−T₀∇+ ²

4∇4 −ν₀4+τ⁻¹



, (3.2.39)

then from the examples 3.2.1, 3.2.2 and4=∇ · ∇we can rewrite this operator in the local coordinate system associated with x₀ in the same form as (3.2.39).

Thus the principal parts of the pullback ofA(D) reads the same as A(D).

Checking Lopatinskii-Shapiro condition (the second condition of Definition 2.2.2)will be carried out with the help of [5](45-51) since we consider the fol-lowing problem











A(ξ˚ ⁰, D_d)v(t)−ηv(t) = 0 for t=x_d>0, v(t) = 0 at t= 0,

|v(t)| −→0 as t−→ ∞

(3.2.40)

which is a system of linear ordinary differential equations with constant (com-plex) coefficients and

A(ξ˚ ⁰, D_d)

:=























ν₀|ξ⁰|²+ν₀D_d² 0 · · · 0 A˚_d 0 ν0|ξ⁰|²+ν0D_d² · · · ... A˚d−1

... ... . .. 0 ...

0 0 · · · ν₀|ξ⁰|²+ν₀D²_d A˚₁

−iD_d −iξ_d−1 · · · −iξ₁ ν0|ξ⁰|²+ν0D²_d























with

A˚_d: =−i²

4|ξ⁰|²D_d−i² 4D³_d, A˚_j : =−i²

4ξ_j|ξ⁰|²−i²

4ξ_jD²_d, j= 1,· · ·, d−1.

We apply the results of S. Agmon, A. Douglis, and L. Nirenberg in [5] pp. 45-51 and record the corresponding theorem here for easy reference.

Theorem(see[5])Let L(D_d),L^jk(D_d)denote det

A(ξ˚ ⁰, D_d)−ηI_d+1

,and the adjoint to the matrix A(ξ˚ ⁰, D_d)−ηI_d+1respectively. Assume the polynomialL(r) has exactly p⁺ complex roots with Imr >0 and L(r) =L⁺(r)L⁻(r) where the roots of L⁺(r) are those of L(r) for which Imr >0. Letξ⁰ ∈R^d−1,η∈ L and

|ξ⁰|+|η| 6= 0. Then the following statements are all equivalent to one another: 1. Complementing Condition 2 in [5](48) holds, i.e., The lines of the p⁺×

(d+ 1)matrix

Bσj(r)L^jk(r)

are linear independent modulo the polynomial L⁺(r);i.e., the equations X

σ

CσBσj(r)L^jk(r)≡0 ( modL⁺(r))

imply the constants C_σ are all zero,where B_σj(r) is a submatrix of B(r) with σ= 1,· · · , p⁺,j= 1,· · ·, d+ 1.

2. The problem











A(ξ˚ ⁰, D_d)v(t)−ηv(t) = 0 for t=x_d>0,

Bσj(D)vj(t) =aσ at t= 0(σ = 1,· · ·, p⁺),

|v(t)| −→0 as t−→ ∞(exponentially)

(3.2.41)

has a solution for arbitrary a_σ ∈C. 3. The problem











A(ξ˚ ⁰, D_d)v(t)−ηv(t) = 0 for t=x_d>0,

Bσj(D)vj(t) = 0 at t= 0(σ = 1,· · ·, p⁺),

|v(t)| −→0 as t−→ ∞(exponentially)

(3.2.42)

has a unique solution: v_j = 0 is the only exponentially decaying solution.

We will apply this theorem to complete the proof.

After some calculations we obtain L(r) = (ν0(|ξ⁰|²+r²)−η)^d−1

(ν0(|ξ⁰|²+r²)−η)²+²

4(|ξ⁰|²+r²)²

and

L(r) = 0⇐⇒r² = η ν0

− |ξ⁰|² orr² = 4 4ν₀²+²

ν0−i

2

η− |ξ⁰|² orr² = 4

4ν₀²+²

ν₀+i 2

η− |ξ⁰|². Since according to the assumptions the arguments of η don’t equal to zero and ±arctg

2ν₀ L(r) has exactly d+ 1 complex roots with positive imaginary part. Definey:= r²+|ξ⁰|², let q₁, q₂, q₃ denote the roots of L(r) with positive imaginary part where q²₁ = η

ν₀ − |ξ⁰|², q₂² = 4 4ν₀²+²

ν₀−i

2

η− |ξ⁰|², q₃² = 4

4ν₀²+²

ν0+i 2

η− |ξ⁰|², then

L⁺(r) = (r−q1)^d−1(r−q2)(r−q3).

Now we compute the (d+ 1)×(d+ 1) matrixBσj(r)L^jk(r) ford= 2 andd= 3 separately.

d= 2:

B_σj(r)L^jk(r)

=











(ν0y−η)²+²

4|ξ⁰|²y −²

4rξ1y i(ν0y−η)² 4ry

−²

4ξ₁ry (ν₀y−η)²+²

4r²y i(ν₀y−η)² 4ξ₁y ir(ν₀y−η) iξ₁(ν₀y−η) (ν₀y−η)²











d= 3:

B_σj(r)L^jk(r) = (B_σj(r)L^jk(r))_ij(i, j= 1,· · · ,4) here the entries (B_σj(r)L^jk(r))_ij read as follows:

(B_σj(r)L^jk(r))₁₁= (ν₀y−η)

(ν₀y−η)²+² 4|ξ⁰|²y

, (B_σj(r)L^jk(r))₁₂=−(ν₀y−η)²

4ξ₂ry, (B_σj(r)L^jk(r))₁₃=−(ν₀y−η)² 4ξ₁ry, (Bσj(r)L^jk(r))14=i(ν0y−η)²²

4ry, (Bσj(r)L^jk(r))21=−(ν₀y−η)² 4ξ2ry, (B_σj(r)L^jk(r))₂₂= (ν₀y−η)

(ν₀y−η)²+²

4y(r²+ξ₁²)

, (B_σj(r)L^jk(r))₂₃=−(ν₀y−η)²

4ξ₁ξ₂y, (B_σj(r)L^jk(r))₂₄=i(ν₀y−η)²² 4ξ₂y, (Bσj(r)L^jk(r))31=−(ν₀y−η)²

4ξ1ry, (Bσj(r)L^jk(r))32=−(ν₀y−η)² 4ξ1ξ2y, (B_σj(r)L^jk(r))₃₃= (ν₀y−η)

(ν₀y−η)²+²

4(r²+ξ₂²)y

, (B_σj(r)L^jk(r))₃₄=i(ν₀y−η)²²

4ξ₁y, (B_σj(r)L^jk(r))₄₁=ir(ν₀y−η)², (B_σj(r)L^jk(r))₄₂=iξ₂(ν₀y−η)², (B_σj(r)L^jk(r))₄₃=iξ₁(ν₀y−η)², (B_σj(r)L^jk(r))₄₄= (ν₀y−η)³.

The following cases need to be considered.

1. Case 1: η= 0.

According to the assumption |ξ⁰|+|η| 6= 0 we have |ξ⁰| 6= 0. From the calculations above we obtain q1 =q2 =q3 =i|ξ⁰|and

L⁺(r) = (r−i|ξ⁰|)^d+1.

We consider two cases:

(a) d= 2:

The equations X

σ

a_σB_σj(r)L^jk(r)≡0 mod L⁺(r) (a_σ ∈C, σ= 1,2,3) imply that i|ξ⁰| is a root of multiplicity 3 of the linear combination of the first column ofB_σj(r)L^jk(r):

l1(r) : =a1ν₀²y²+a1

²

4|ξ⁰|²y−a2

²

4ξ1ry+ia3rν0y,

=y

a₁ν₀²y+a₁²

4|ξ⁰|²−a₂²

4ξ₁r+ia₃rν₀

,

= (r−i|ξ⁰|)(r+i|ξ⁰|)×

×

a₁ν₀²y+a₁²

4|ξ⁰|²−a₂²

4ξ₁r+ia₃rν₀

. Since a1ν₀²y+a1

²

4|ξ⁰|²−a2

²

4ξ1r+ia3rν0 is a polynomial of order 2 with respect tor and a₁ν₀²y+a₁²

4|ξ⁰|² −a₂²

4ξ₁r+ia₃rν₀ has a factor (r−i|ξ⁰|)² it follows

a₁ν₀²y+a₁²

4|ξ⁰|²−a₂²

4ξ₁r+ia₃rν₀ =a₁ν₀²(r−i|ξ⁰|)², which is equivalent to

a₁ν₀²r²−i2a₁ν₀²|ξ⁰|r−a₁ν₀²|ξ⁰|²

=a1ν₀²r²+

ia3ν0−a2

² 4ξ1

r+a1|ξ⁰|²

ν₀²+² 4

. By coefficient comparison we obtain

−a₁ν₀²|ξ⁰|²=a1|ξ⁰|²

ν₀²+² 4

, (3.2.43)

which implies

a₁ = 0. (3.2.44)

Then the linear combination of the second column of B_σj(r)L^jk(r) is reduced as:

l₂(r) : =a₂ν₀²y²+a₂²

4r²y+ia₃ξ₁ν₀y,

= (r−i|ξ⁰|)(r+i|ξ⁰|)

a₂ν₀²+a₂² 4

r²+a₂ν₀²|ξ⁰|²+ia₃ξ₁ν₀

.

By a same reasoning

a₂ν₀²+a₂² 4

r²+a₂ν₀²|ξ⁰|²+ia₃ξ₁ν₀ =

a₂ν₀²+a₂² 4

(r−i|ξ⁰|)², from which it follows

a₂ = 0. (3.2.45)

Then a3 = 0 follows immediately. Combining (3.2.44) and (3.2.45) the lines ofB_σj(r)L^jk(r) are linear independent modulo the polyno-mialL⁺(r).

(b) d= 3 :

On the basis of the same procedure as case of d= 2 the equations X

σ

aσBσj(r)L^jk(r)≡0 mod L⁺(r) (aσ ∈C, σ= 1,2,3,4) imply that i|ξ⁰|is a root of multiplicity 4 of the linear combination of the first column of Bσj(r)L^jk(r):

L1(r) : =a1ν0y

ν₀²y²+² 4|ξ⁰|²y

−a2ν0

²

4ξ2ry²−a3ν0

²

4ξ1ry²+ +ia₄ν₀²ry²

= a1ν₀³r²+

ia4ν₀²−a2ν0

²

4ξ2−a3ν0

² 4ξ1

r+a1ν₀³|ξ⁰|²+

+a1ν0

² 4|ξ⁰|²

!

(r−i|ξ⁰|)²(r+i|ξ⁰|)².

(3.2.46) The first factor of L1(r) is a polynomial of order 2 and has a factor (r−i|ξ⁰|)² thus

a1ν₀³r²+

ia4ν₀²−a2ν0

²

4ξ2−a3ν0

² 4ξ1

r+a1ν₀³|ξ⁰|²+a1ν0

² 4|ξ⁰|²

=a₁ν₀³(r−i|ξ⁰|)²

=a1ν₀³r²−i2a1ν₀³|ξ⁰|r−a1ν₀³|ξ⁰|². Coefficient comparison yields

a₁ = 0. (3.2.47)

Then the linear combination of the second column of Bσj(r)L^jk(r) is reduced as:

L2(r) : =a2ν0y

ν₀²y²+²

4y(r²+ξ₁²)

−a3ν0

²

4ξ1ξ2y²+ia4ξ2ν₀²y²

= (r−i|ξ⁰|)²(r+i|ξ⁰|)²×

×

a2ν₀³y+a2ν0

²

4(r²+ξ₁²)−a3ν0

²

4ξ1ξ2+ia4ξ2ν₀²

= (r−i|ξ⁰|)²(r+i|ξ⁰|)²R, where

R:=

a₂ν₀³+a₂ν₀² 4

r²+a₂ν₀³|ξ⁰|²+a₂ν₀²

4ξ₁²−a₃ν₀²

4ξ₁ξ₂+ia₄ξ₂ν₀². Ris a polynomial inr of order 2 and it has a factor (r−i|ξ⁰|)² thus

R=

a₂ν₀³+a₂ν₀² 4

(r−i|ξ⁰|)². Coefficient comparison yields

a2 = 0. (3.2.48)

Then the linear combination of the third column ofB_σj(r)L^jk(r) is reduced as:

L₃(r) : =a₃ν₀y

ν₀²y²+²

4(r²+ξ₂²)y

+ia₄ξ₁ν₀²y²

= (r−i|ξ⁰|)²(r+i|ξ⁰|)²×

×

a₃ν₀³+a₃ν₀² 4

r²+a₃ν₀³|ξ⁰|²+a₃ν₀²

4ξ₂²+ia₄ξ₁ν₀²

! . A same reasoning provides

a3ν₀³+a3ν0

² 4

r²+a3ν₀³|ξ⁰|²+a3ν0

²

4ξ₂²+ia4ξ1ν₀²

=

a3ν₀³+a3ν0

² 4

(r−i|ξ⁰|)². Coefficient comparison yields

a₃ = 0. (3.2.49)

Then the linear combination of the fourth column ofBσj(r)L^jk(r) is L₄(r) : =a₄ν₀³(r−i|ξ⁰|)³(r+i|ξ⁰|)³.

That L4(r) has a factor (r−i|ξ⁰|)⁴ implies

a₄ = 0. (3.2.50)

Thus (3.2.47)-(3.2.50) yield the lines of Bσj(r)L^jk(r) are linear in-dependent modulo the polynomial L⁺(r).

2. Case 2: η6= 0.

In this caseq1, q2, q3 are pairwise unequal and we still consider two cases.

(a) d= 2:

The equations X

σ

aσBσj(r)L^jk(r)≡0 modL⁺(r) (aσ ∈C, σ= 1,2,3) are equivalent to that l_i(q_j) = 0(i, j = 1,2,3) where letl_i(r) denote the corresponding linear combination of theith column. Sinceq₁is a root of the linear combination of the second column ofBσj(r)L^jk(r)

l2(r) :=−a₁²

4rξ1y+a2

(ν0y−η)²+² 4r²y

+a₃iξ₁(ν₀y−η).

(3.2.51)

Then l₂(q₁) = 0 is equivalent to

a1ξ1−a2q1 = 0. (3.2.52) By the same reasoning q₂ q₃ are roots of the linear combination of the third column ofB_σj(r)L^jk(r)

l3(r) :=a1i(ν0y−η)²

4ry+a2i(ν0y−η)²

4ξ1y+a3(ν0y−η)². l3(q2) = 0 is equivalent to

a₁

2q₂+a₂

2ξ₁−a₃= 0; (3.2.53) l3(q3) = 0 is equivalent to

a₁

2q₃+a₂

2ξ₁+a₃= 0. (3.2.54)

Now we computel1(q1) = 0, l1(q2) = 0, l1(q3) = 0, l2(q2) = 0, l2(q3) = 0, l₃(q₁) = 0 then we obtain

l1(q1) = 0⇐⇒ξ1(a1ξ1−a2q1) = 0;

l1(q2) = 0⇐⇒a1

2q2+a2

2ξ1−a3= 0;

l1(q3) = 0⇐⇒a1

2q3+a2

2ξ1+a3= 0;

l2(q2) = 0⇐⇒

a1

2q2+a2

2ξ1−a3

ξ1 = 0;

l₂(q₃) = 0⇐⇒

a₁

2q₃+a₂

2ξ₁+a₃

ξ₁ = 0;

l₃(q₁) = 0⇐⇒(a₁, a₂, a₃)^T ∈R³ are arbitrary.

Finally it is easy to verify thatli(qj) = 0 (i, j= 1,2,3) are equivalent to (3.2.52), (3.2.53) and (3.2.54).

The system of equations (3.2.52), (3.2.53) and (3.2.54) with respect to (a1, a2, a3)^T has only the trivial solution iff.

det















ξ₁ −q₁ 0

2q₂

2ξ₁ −1

2q3

2ξ1 1













 6= 0,

which is equivalent to

ξ₁²+1

2q1(q2+q3)6= 0.

Thus we obtain that the following two statements are equivalent i. X

σ

a_σB_σj(r)L^jk(r)≡0 mod L⁺(r) (a_σ ∈C, σ= 1,2,3) imply that the constantsaσ are all zero,

ii. ξ₁²+1

2q1(q2+q3)6= 0.

(b) d= 3:

Since each entry ofB_σj(r)L^jk(r) has a factorν₀y−η =ν₀(r+q₁)(r− q1) the equationsX

σ

aσBσj(r)L^jk(r)≡0 mod L⁺(r) (aσ ∈C, σ= 1,2,3,4) are equivalent to that Li(qj) = 0(i= 1,2,3,4 j = 1,2,3) whereL_i(r) := l_i(r)

ν0y−η and let l_i(r) denote the corresponding linear combination of the ith column of Bσj(r)L^jk(r). Some calculations

yield

Thus we obtain that

L_i(q_j) = 0(i= 1,2,3,4 j= 1,2,3) are equivalent to the equations

a₁|ξ⁰|²−a₂ξ₂q₁−a₃ξ₁q₁= 0,

We consider the following linear equations:

B(a₁, a₂, a₃, a₄)^T =0,

here

Recall the case ofd= 2 we obtain the conclusion that for both cases(d= 2,3) and η6= 0 the lines of thep⁺×(d+ 1) matrix

B_σj(r)L^jk(r)

are linear independent modulo the polynomialL⁺(r); i.e., the equations X

σ

CσBσj(r)L^jk(r)≡0 ( modL⁺(r)) imply the constantsC_σ are all zero iff.

|ξ⁰|²+1

2q₁(q₂+q₃)6= 0. (3.2.56) In case ofξ⁰ =0 (3.2.56) holds for allη ∈ Lsinceq₁6= 0, q₂+q₃ 6= 0 thus we consider only the case ofξ⁰6=0. We have the reasoning

|ξ⁰|²+ 1

2q1(q2+q3) = 0

=⇒(2|ξ⁰|²+q₁(q₂+q₃))(2|ξ⁰|²−q₁(q₂+q₃))×

×(2|ξ⁰|²+q₁(q₂−q₃))(2|ξ⁰|²−q₁(q₂−q₃)) = 0,

(3.2.57)

and then expand the right equation of (3.2.57) with respect to η to find (2|ξ⁰|²+q1(q2+q3))(2|ξ⁰|²−q1(q2+q3))(2|ξ⁰|²+q1(q2−q3))×

×(2|ξ⁰|²−q₁(q₂−q₃)) = 0, which implies

²c²

ν₀² η³−2²c²|ξ⁰|² ν0

η²+|ξ⁰|⁴(²c²+ 16c)η−16|ξ⁰|⁶

ν₀c+ 1 ν0

= 0, which is equivalent to

P(η) :=η³−2ν₀|ξ⁰|²η²+ν₀²|ξ⁰|⁴

5 +16ν₀^{2 2}

η

−16|ξ⁰|⁶ν₀² c²

2ν₀+ ² 4ν0

= 0

(3.2.58)

wherec:= 4 4ν₀²+².

P(η) is a polynomial in η of order 3 and we check that P(0) =−16|ξ⁰|⁶ν₀²

c²

2ν₀+ ² 4ν₀

<0

thenP(η) has a positive real root. The discriminant ∆ ofP(η) reads

∆ = p³ 27 +q²

4, with

p: =ν₀²|ξ⁰|⁴ 11

3 +16ν₀^{2 2}

q : =ν₀²|ξ⁰|⁶

−16

27ν₀+2ν₀ 3

5 +16ν₀^{2 2}

− 16 c²

2ν₀+ ² 4ν₀

. Then ∆>0 and from algebraic laws P(η) = 0 has one real and a pair of complex conjugate roots. FactorP(η) as

P(η) = (η−α)(η²+βη+γ)

where let αdenote the positive real root ofP(η) = 0 andβ, γ denote the well-founded coefficients. Coefficient comparison yields

β =α−2ν₀|ξ⁰|².

Substitute 2ν₀|ξ⁰|² intoP(η) we obtainP(2ν₀|ξ⁰|²)<0; sinceα >0 is the only real root of P(η) = 0 we conclude 2ν0|ξ⁰|² < αthusβ >0.Then the

two complex conjugate roots of P(η) = 0 are situated in the second and the third quadrants in the complex plane respectively.

Notice that (3.2.56) holds for any η ∈S₁∪S₂ where

S1: =

x∈C| −π+ arctg 2ν0

<argx <−arctg 2ν0

, S2: =

x∈C|arctg 2ν0

<argx < π−arctg 2ν0

since 1

2q1(q2+q3) always has a negative imaginary part for η∈

x∈C| −π+ arctg

2ν₀ <argx <−arctg 2ν₀

and a positive imaginary part for η∈

x∈C|arctg

2ν₀ <argx < π−arctg 2ν₀

. Furthermore fromP(η) = 0 we see

η ²c²

ν₀² η²+|ξ⁰|⁴(²c²+ 16c)

= 2|ξ⁰|^{2 2}c²

ν₀² η²+ 8|ξ⁰|⁴

ν₀c+ 1 ν₀

. A necessary condition of P(η) = 0 is the argument of the complex root ofP(η) = 0 with positive imaginary part is greater equal than 0 and less than 2

3π otherwise the lie of η

²c²

ν₀² η²+|ξ⁰|⁴(²c²+ 16c)

is in the first or the second quadrant, and 2|ξ⁰|²

²c²

ν₀² η²+ 8|ξ⁰|⁴

ν0c+ 1 ν₀

lies in the third or the fourth quadrant, i.e. will be situated in different quadrants.

Assume 2ν0

≤√

3 then the complex conjugate roots ofP(η) = 0 are in S1∪S2,

consequently (3.2.56) holds for all η ∈ C exclusive the nonnegative real axis.

Theorem 3.2.6. Let 1< p <∞, there exists a η0 >0 such that for allη ∈ L with|η| ≥η₀ where Lis defined in theorem 3.2.5 the boundary problem (3.2.32) has a unique solution u(x) = (u^d, u^d−1,· · · , u¹, u⁰)∈Q_d+1

j=1Wp^mj+m(Ω)for any f = (f1,· · · , f_d+1) ∈ Qd+1

j=1Wp^mj(Ω) and g = (g1,· · · , g_d+1)Qd+1

j=1Wp^3/2−rj(Ω) where the sequences of integers {m_j}^d+1₁ and {r_j}^d+1₁ are defined in (3.2.30) and (3.2.33)respectively. Further the a priori estimate

d

X

j=0

|||u_j|||_{md+1−j+2,p,Ω}

≤c





d+1

X

j=1

|||f_d+2−j|||_{md+2−j,p,Ω}+

d+1

X

j=1

|||g_d+2−j|||_2−r

d+2−j−_p¹,p,∂Ω





(3.2.59)

holds where the constant c does not depend upon f, g, η and we use the norms depending on a parameter η ∈ C/{0}, namely let v ∈ W_p^s(Ω), g ∈ W^s−

1

p p(∂Ω) where s is an integer satisfying 1≤s≤m,

|||v|||_s,p,Ω :=kvk_Ws

p(Ω)+|η|^mskvk_Lp(Ω), (3.2.60)

|||g|||_s−1

p,p,∂Ω :=kgk

W^s−

1p p (∂Ω)

+|η|

s−1 p

m kgk_Lp(∂Ω). (3.2.61) Proof. Since the boundary problem (3.2.32) is elliptic with parameter in L from Theorem 3.2.5, the existence and the a priori estimate follow directly from Theorem2.2.1.

Im Dokument Solutions to the Multi-Dimensional Viscous Quantum Hydrodynamic Equations for Semiconductors (Seite 38-54)