Existence, uniqueness, and boundedness of the component u in the

In this section, we prove existence, uniqueness, and a prioriL^∞ estimates on the solutions of (3.50) and (3.52) for all three choices of the test spaceV. First we observe that these weak

formulations for both 2-term and 3-term splittings can be written in the common form Find u∈H_g¹(Ω) such thatb(x, u+w)v∈L¹(Ω) for allv ∈V and

a(u, v) + Z

Ω

b(x, u+w)vdx= Z

Ω

f · ∇vdx, for all v∈V. (3.41) In the case of the 2-term splitting we have

w=G, f =χΩs(m−s)∇G, and g=g−G on∂Ω (3.42) whereas in the case of the 3-term splitting we have

w= 0, f =−χ_Ωmm∇u^H+χ_Ωsm∇G, and g=g on ∂Ω. (3.43) We assume that the function gspecifying the Dirichlet boundary condition on ∂Ω is given as the trace of some H²(Ω) function (see the discussion in Section 3.2.3). Note that w ∈ L^∞(Ωions) and f = (f1, f2, . . . , f_d) ∈ [L^s(Ω)]^d with s > d, since G is smooth in Ωs

and ∇u^H ∈[L^s(Ω_m)]^d, s > d if Γ isC^0,1 (see the discussion in Section 3.2.3).

Uniqueness

First, we prove uniqueness of a solution to (3.41) for all three choices of the test space V. Suppose that u1 andu2 are two solutions of (3.41). Then, we have

a(u1−u2, v) + Z

Ω

(b(x, u1+w)−b(x, u2+w))vdx= 0,∀v∈V. (3.53)

In the case of V =H₀¹(Ω),u1−u2 ∈V and thus we can test (3.53) with v :=u1 −u2 to obtain

a(u₁−u₂, u₁−u₂) + Z

Ω

(b(x, u₁+w)−b(x, u₂+w)) (u₁−u₂)dx= 0.

Since a(·,·) is coercive and b(x,·) is monotone increasing, we obtain u1 −u2 = 0. When V =H₀¹(Ω)∩L^∞(Ω) and V =C₀^∞(Ω), the differenceu₁−u₂ is not necessarily inV and we cannot test with it in (3.53). In this case, V is a dense subspace ofH₀¹(Ω) which contains C₀^∞(Ω). Since a(u,·) defines a bounded linear functional over H₀¹(Ω), from (3.53) it follows that the linear functionalT_cdefined by

hT_c, vi:= (b(x, u₁+w)−b(x, u₂+w), v),∀v∈V

is bounded overV. Moreover,c∈L¹_loc(Ω) andc(u1−u₂)≥0∈L¹(Ω) a.e. x∈Ω. This means that we can apply Theorem 2.18 to the functionalT_cand the test functionv=u₁−u₂∈H₀¹(Ω) (see also Remark 2.19 after Theorem 2.18). Again, using the coercivity of a(·,·) and the

monotonicity of b(x,·) (see (3.39)), we conclude that u1−u2= 0.

3.3. POISSON-BOLTZMANN EQUATION 53 Existence

We continue with the proof of existence of a solution to (3.41) simultaneously for the test spacesV = H₀¹(Ω)∩L^∞(Ω) and V =C₀^∞(Ω). The existence of a solution to (3.41) with V =H₀¹(Ω) will follow from the existence for the first two choices ofV once we prove the a prioriL^∞ bound on their common solution.

It is convenient to split u into u^L and u^N, i.e., u =u^L+u^N, where u^L satisfies the linear nonhomogeneous interface elliptic problem

Find u^L∈H_g¹(Ω) such that a(u^L, v) = Z

Ω

f · ∇vdx, for all v∈V (3.46)

and the componentu^N has to satisfy

Findu^N ∈H₀¹(Ω) such thatb(x, u^N+u^L+w)v∈L¹(Ω) for all v∈V and a(u^N, v) +

Z

Ω

b(x, u^N+u^L+w)vdx= 0, for all v∈V. (3.47) Problem (3.46) is linear and hence all weak formulations for the three choices of V are obviously equivalent. From the Lax-Milgram Theorem it is clear that there exists a unique u^L ∈ H_g¹(Ω) satisfying (3.46). Moreover, from Theorem 2.32 it follows that u^L ∈ L^∞(Ω).

Indeed, letu^L_g ∈H²(Ω) be such that γ₂(u^L_g) =g on∂Ω. Then we consider the homogenized version of (3.46) given by

a(u^L₀, v) = Z

Ω

f · ∇vdx− Z

Ω

∇u^L_g · ∇vdx, for allv∈V, (3.54)

where u^L₀ ∈ H₀¹(Ω) and u^L := u^L_g +u^L₀. Since ∇u^L_g ∈ H¹(Ω) it follows by the Sobolev embedding theorem (Theorem 2.21) that∇u^L_g ∈

L⁶(Ω)d

ford= 2,3. Therefore, by recalling thatf ∈[L^s(Ω)]^d, s > d we can apply Theorem 2.32 to obtain thatu^L₀ ∈L^∞(Ω) and hence u^L∈L^∞(Ω).

To show existence of u^N satisfying (3.47), we introduce the energy functionalJ^N :H₀¹(Ω)→ R∪ {+∞}, defined by

J^N(v) = 1

2a(v, v) + Z

Ω

B(x, v+u^L+w)dx, (3.55)

where it is understood thatJ^N(v) = +∞whenever B(x, v+u^L+w)∈/L¹(Ω), i.e.,

J^N(v) :=









 1

2a(v, v) + Z

Ω

B(x, v+u^L+w)dx, ifB(x, v+u^L+w)∈L¹(Ω), +∞, ifB(x, v+u^L+w)∈/ L¹(Ω).

(3.56)

We consider the variational problem

Find u^N_min∈H₀¹(Ω) such thatJ^N(u^N_min) = min

v∈H₀¹(Ω)

J^N(v). (3.57)

Notice that for d≤2 it holds e^v∈L²(Ω) for allv∈H₀¹(Ω) (e.g., see [110, 179]) and therefore dom(J^N) =

v∈H₀¹(Ω) such that J^N(v)<∞ is a linear space coinciding with H₀¹(Ω).

However, in dimension d = 3, dom(J^N) is only a convex set and not a linear space (see Remark 3.23). In fact, dom(J^N) is not even a closed subspace ofH₀¹(Ω). Indeed, dom(J^N) containsC₀^∞(Ω) which is dense inH₀¹(Ω). If dom(J^N) were closed, it would coincide with H₀¹(Ω) and we know that this is not true in dimensiond≥3 (see the examples in Remark 3.23).

Since dom(J^N) is convex and obviously J^N is convex overdom(J^N) it follows that J^N is convex overH₀¹(Ω) (see Remark 2.31 and [65]). To show existence of a minimizer of J^N over the reflexive Banach spaceH₀¹ one has to verify the following assertions:

(1) J^N is proper, i.e., J^N is not identically equal to +∞and does not take the value−∞;

(2) J^N is sequentially weakly lower semicontinuous (s.w.l.s.c.), i.e., if {v_n}^∞_n=1 ⊂ H₀¹(Ω) and vn* v (weakly inH₀¹(Ω)) then J^N(v)≤lim inf

n→∞ J^N(vn);

(3) J^N is coercive, i.e., lim

n→∞J^N(v_n) = +∞whenever kv_nk_H1(Ω)→ ∞.

Assertion (1) is obvious since R

Ω

B(x, u+u^L+w)dx ≥ 0 and J^N(0) is finite. To see that (2) is fulfilled, notice that J^N is the sum of the functionals A(v) := ¹₂a(v, v) and R

Ω

B(x, v+u^L+w)dx. The former is convex and Gateaux differentiable, and therefore s.w.l.s.c. (for the proof of this implication, see, e.g. Corollary 2.4 in [169]). Indeed, by using the symmetry of a(·,·), for anyz, v∈H₀¹(Ω) and any λ∈[0,1], for the convexity of ¹₂a(v, v) we obtain

A(λz+ (1−λ)v)

=1

2a(λz+ (1−λ)v, λz+ (1−λ)v) = 1

2 λ²a(z, z) + 2λ(1−λ)a(z, v) + (1−λ)²a(v, v)

≤1

2 λ²a(z, z) +λ(1−λ) (a(z, z) +a(v, v)) + (1−λ)²a(v, v)

(3.58)

=λA(z) + (1−λ)A(v),

where we have used the inequality a(z, v)≤p

a(z, z)p

a(v, v)≤ ¹₂(a(z, z) +a(v, v)). To see thatA(·) is Gateaux differentiable, we use the symmetry of a(·,·) together with its linearity

3.3. POISSON-BOLTZMANN EQUATION 55 in each argument. For anyz∈H₀¹(Ω) and any direction v∈H₀¹(Ω) we have

λ→0lim⁺

A(z+λv)−A(z)

λ = lim

λ→0⁺

1 2

a(z+λv, z+λv)−a(z, z) λ

= lim

λ→0⁺

1 2

a(z, z) +λ²a(v, v) + 2λa(z, v)−a(z, z) λ

= lim

λ→0⁺

1

2λa(v, v) +a(z, v)

=a(z, v).

(3.59)

Sincea(z,·) is a bounded linear functional overH₀¹(Ω), by the definition of Gateaux differen-tiability (see Definition 2.27), it follows thatAis Gateaux differntiable at z with a Gateaux differential atz equal to the functionalA⁰(z), defined by hA⁰(z), vi:=a(z, v),∀v∈H₀¹(Ω).

However, ford= 3, the functionalR

Ω

B(x, v+u^L+w)dxis not Gateaux differentiable (see Remark 3.25). Nevertheless, one can still show that the functional R

Ω

B(x, v+u^L+w)dx is s.w.l.s.c. using Fatou’s lemma and the compact embedding of H₀¹(Ω) into L²(Ω) (see Theorem 2.22) as follows. Let {v_n}^∞_n=1 ⊂H₀¹(Ω) be a sequence which converges weakly in H₀¹(Ω) to an element v ∈ H₀¹(Ω), i.e., v_n * v. Since the embedding H₀¹(Ω) ,→ L²(Ω) is compact it follows thatvn→v (strongly) inL²(Ω), and therefore we can extract a pointwise almost everywhere convergent subsequencev_nm(x)→v(x) (see Theorem 4.9 in [96]). Since B(x,·) is a continuous function for any x ∈ Ω and x 7→ B(x, s) is measurable for any s ∈ R it means that B is a Carath´eodory function and as a consequence the function x7→B(x, v_nm(x)) is measurable for allk∈N (see Proposition 3.7 in [55]). By noting that B(x, z(x) +u^L(x) +w(x))≥0 for allz∈H₀¹(Ω) and using the fact thatB(x,·) is a continuous function for anyx∈Ω, from Fatou’s lemma we obtain

lim inf

m→∞

Z

Ω

B(x, vnm(x) +u^L(x) +w(x))dx≥ Z

Ω

lim inf

m→∞ B(x, vnm(x) +u^L(x) +w(x))dx

= Z

Ω

B(x, v(x) +u^L(x) +w(x))dx. (3.60) Now it is clear that if{v_nm}^∞_m=1 is an arbitrary subsequence of{v_n}^∞_n=1, then there exists a further subsequence{v_nms}^∞_s=1 for which (3.60) is satisfied. This means that (3.60) is also satisfied for the whole sequence{v_n}^∞_n=1, and hence R

Ω

B(x, v+u^L+w)dxis s.l.w.s.c. (see Remark 3.22).

The coercivity of J^N follows by observing that J^N(v) = 1

2a(v, v) + Z

Ω

B(x, v+u^L+w)dx≥_mink∇vk²_L2(Ω)≥ _min

1 +C_P²kvk_H1(Ω), where CP is the constant in the inequality kvk_L2(Ω)≤ CPk∇vk_L2(Ω),∀v ∈ H₀¹(Ω). Now, the existence of a minimizer u^N_min of J^N over the reflexive Banach space H₀¹(Ω) follows by

Theorem 2.30. Moreover, since a(v, v) is a strictly convex functional it follows that J^N is also strictly convex, and therefore this minimizer is unique.

Theorem 3.17

There exists a unique u^N_min∈H₀¹(Ω)such that J^N(u^N_min) = min

v∈H₀¹(Ω)

J^N(v).

We will show that the minimizeru^N_min of J^N overH₀¹(Ω) satisfies (3.47) for V =C₀^∞(Ω) and V =H₀¹(Ω)∩L^∞(Ω). Notice that J^N is not Gateaux differentiable at each point in H₀¹(Ω) and thus we cannot conclude straightforwardly that the minimizer u^N_min is a solution to the weak formulation (3.47) (see Remark 3.25).

Now by using the Lebesgue dominated convergence theorem and the fact that at the unique minimizeru^N_min of J^N we haveB(x, u^N_min+u^L+w)∈L¹(Ω), we will show thatu^N_min is also a solution to (3.47). We have that J^N(u^N_min+λv)−J^N(u^N_min) ≥0 for all v ∈H₀¹(Ω) and all λ≥0, i.e.,

1

2a(u^N_min+λv, u^N_min+λv) + Z

Ω

B(x, u^N_min+λv+u^L+w)dx

−1

2a(u^N_min, u^N_min)− Z

Ω

B(x, u^N_min+u^L+w)dx≥0,

which, by using the symmetry of a(·,·), is equivalent to λa(u^N_min, v) +λ

2a(v, v) + Z

Ω

B(x, u^N_min+λv+u^L+w)−B(x, u^N_min+u^L+w)

dx≥0.

(3.61) Divide both sides of (3.61) by λ >0 and letλ→0⁺ to obtain

a(u^N_min, v) + lim

λ→0⁺

Z

Ω

B(x, u^N_min+λv+u^L+w)−B(x, u^N_min+u^L+w)

λ dx≥0. (3.62)

To compute the limit in the second term of (3.62), we will apply the Lebesgue dominated convergence theorem. We have

fλ(x) := B(x, u^N_min(x) +u^L(x) +w(x) +λv(x))−B(x, u^N_min(x) +u^L(x) +w(x)) λ

λ→0⁺

−−−−→b(x, u^N_min(x) +u^L(x) +w(x))v(x) for a.e x∈Ω

(3.63)

By the mean value theorem we have

f_λ(x) =b(x, u^N_min+u^L(x) +w(x) + Θ(x)λv(x))v(x), where Θ(x)∈(0,1),∀x∈Ω

3.3. POISSON-BOLTZMANN EQUATION 57 and hence, ifv∈L^∞(Ω), we can obtain the following bound onfλ:

|f_λ(x)|=

−4πe²₀ k_BTv(x)

Nions

X

j=1

Mj(x)ξje^−ξj(^uNmin(x)+u^L(x)+w(x)+Θ(x)λv(x))

≤max

j |ξ_j|kvk_L^∞_(Ω)4πe²₀ k_BT

Nions

X

j=1

Mj(x)e^−ξj(^uNmin(x)+u^L(x)+w(x))^−ξjΘ(x)λv(x)

≤max

j |ξ_j|max

j e^{|ξj|kvkL∞(Ω)}kvk_L∞(Ω)

4πe²₀ k_BT

Nions

X

j=1

M_j(x)e^−ξj(^uNmin(x)+u^L(x)+w(x))

= max

j |ξ_j|max

j e^{|ξj|kvkL∞(Ω)}kvk_L∞(Ω)B(x, u^N_min(x) +u^L(x) +w(x))∈L¹(Ω),∀λ≤1.

(3.64) From the Lebesgue dominated convergence theorem, by using (3.63) and (3.64), it follows that the limit in (3.62) is equal toR

Ω

b(x, u^N_min+u^L+w)vdx, and therefore we obtain a(u^N_min, v) +

Z

Ω

b(x, u^N_min+u^L+w)vdx≥0,∀v∈H₀¹(Ω)∩L^∞(Ω). (3.65) This means thatu^N =u^N_minis a solution to the weak formulation (3.47) forV =H₀¹(Ω)∩L^∞(Ω) andV =C₀^∞(Ω). The uniqueness of the solutionu^N of (3.47) is done in the same way as the uniqueness of (3.41). Now, it is clear thatu=u^N +u^L∈H_g¹ is the unique solution to (3.41) for the test spacesV =H₀¹(Ω)∩L^∞(Ω) and V =C₀^∞(Ω).

To show thatu^N =u^N_min is also a solution to (3.47) with V =H₀¹(Ω) and thatu=u^N +u^L is a solution to (3.41) withV =H₀¹(Ω) we will prove the a priori L^∞ bound on the solution of (3.47) withV =H₀¹(Ω)∩L^∞(Ω) and V =C₀^∞(Ω).

A priori L^∞ bound on the component u^N

Next, we show that the solution to Problem (3.47) is essentially bounded regardless of which test spaceV we have.

Proposition 3.18

The unique weak solution u^N to Problem (3.47) belongs to L^∞(Ω). Moreover, there is a positive constant e > 0, depending only on d, Ω, _min := min

x∈Ω(x), such that ku^Nk_L∞(Ω)≤ ku^L+wk_L∞(Ωions)+e. If the charge neutrality condition (3.36) holds, then e= 0.

Proof. To prove thatu^N is bounded we apply Theorem 2.18 once again. The first step is to show that (3.47) holds for the test functions

v=G_s(u^N) := sgn(u^N) max{

u^N

−s,0}=







u^N −s forx∈

u^N > s ,

0 forx∈

u^N ∈[−s, s] , u^N +s forx∈

u^N <−s .

(3.66)

with s≥ ku^L+wk_L^∞_(Ωions). We notice that similar test functions Gs(u^N) have been used in [112, Theorem B.2] in the context of linear elliptic problems.

It is easy to see that Gs(0) = 0 and thatGs(·) is Lipschitz continuous for anys. Therefore, by Stampacchia’s theorem (e.g., see [90, 112]) it follows thatGs(u^N)∈H₀¹(Ω) and that the weak partial derivatives of G_s(u^N) are given by

∂G_s(u^N)

∂xi

=







∂u^N

∂xi forx∈

u^N > s , 0 forx∈

u^N ∈[−s, s] ,

∂u^N

∂xi forx∈

u^N <−s .

(3.67)

Next, the functionalTbdefined byhT_b, vi:=R

Ω

b(x, u^N+u^L+w)vdx, ∀v∈C₀^∞(Ω) is bounded and linear over the dense subspaceC₀^∞(Ω) ofH₀¹(Ω) and b(x, u^N +u^L+w)∈L¹_loc(Ω). This fact follows from (3.47) and the fact that the functionala(u^N,·) belongs toH⁻¹(Ω). Then, in view of Theorem 2.18, to show thathT_b, G_s(u^N)i=R

Ω

b(x, u^N +u^L+w)G_s(u^N)dx it suffices to verify that

b(x, u^N+u^L+w)Gs(u^N)≥f a.e. for some f ∈L¹(Ω). (3.68) By choosing s ≥ ku^L+wk_L^∞_(Ωions), using the monotonicity of b(x,·), and recalling that b(x, s) = 0 for allx∈Ω_m∪Ω_IEL, we obtain

• forx∈Ωions∩

u^N > s :

b(x, u^N +u^L+w)Gs(u^N) =b(x, u^N+u^L+w)(u^N −s)≥b(x,0)(u^N −s);

• forx∈Ω_ions∩

u^N ∈[−s, s] : b(x, u^N +u^L+w)G_s(u^N) = 0;

• forx∈Ω_ions∩

u^N <−s :

b(x, u^N +u^L+w)G_s(u^N) =b(x, u^N+u^L+w)(u^N +s)≥b(x,0)(u^N +s).

(3.69)

By taking into account the equality b(x,0) =− 4π kBT

Nions

X

j=1

M_jξ_j =constfor x∈Ω_ions, (3.70)

we see thatb(x,0)∈L^∞(Ω) and henceb(x,0)(u^N −s)∈L¹ Ω_ions∩

u^N > s and b(x,0)(u^N+s)∈L¹ Ωions∩

u^N <−s . Therefore, from (3.69) it follows that (3.68) holds for the summable function f defined by

f(x) =







0 for x∈Ωm∪ΩIEL∪ Ωions∩

u^N ∈[−s, s] , b(x,0)(u^N −s) for x∈Ω_ions∩

u^N > s , b(x,0)(u^N +s) for x∈Ω_ions∩

u^N <−s .

(3.71)

3.3. POISSON-BOLTZMANN EQUATION 59 Now we are ready to prove thatu^N ∈L^∞(Ω). First, we consider the case when the charge neutrality condition (3.36) holds. From (3.68), (3.70), and (3.71), it follows that

Z

Ω

b(x, u^N +u^L+w)G_s(u^N)dx≥ Z

Ω

b(x,0)G_s(u^N)dx= 0. (3.72) Moreover, by using the definition of a(·,·) and the expression (3.67) for the weak partial derivatives ofGs(u^N), we obtain

a(u^N, G_s(u^N)) = Z

Ω

∇u^N · ∇G_s(u^N) = Z

Ω

∇G_s(u^N)· ∇G_s(u^N)dx

≥ _mink∇G_s(u^N)k²_L2(Ω)≥ min

C_P² kG_s(u^N)k²_L2(Ω), (3.73) where C_P is the constant in the inequality kvk_L2(Ω)≤C_Pk∇vk_L2(Ω) that holds for all v ∈ H₀¹(Ω). Finally, using (3.47), (3.72), and (3.73) we obtain

kG_s(u^N)k²_L2(Ω)≤0, for alls≥ ku^L+wk_L∞(Ωions). Consequently

u^N

≤salmost everywhere for all s≥ ku^L+wk_L∞(Ωions). In the case where the charge neutrality condition (3.36) does not hold, we have

Z

Ω

b(x, u^N +u^L+w)G_s(u^N)dx≥ Z

Ω

b(x,0)G_s(u^N)dx. (3.74) We further estimatea(u^N, G_s(u^N)) from below and−R

Ω

b(x,0)G_s(u^N)dx from above using the Sobolev embedding H¹(Ω) ,→ L^q(Ω) where q < ∞ for d= 2, and q = _d−2^2d for d ≥ 3.

Let q^∗ denote the H¨older conjugate to q. Then, q^∗ = _q−1^q > 1 for d = 2, and q^∗ = _d+2^2d ford≥3. In order to treat both cases in which we are interested simultaneously, namely d= 2,3, we can take q= 6 andq^∗ = 6/5. By CE we denote the embedding constant in the inequalitykvk_L6(Ω)≤C_Ekvk_H1(Ω),∀v∈H¹(Ω), which depends only on the domain Ω andd.

Fora(u^N, Gs(u^N)), we have a(u^N, G_s(u^N)) =

Z

Ω

∇G_s(u^N)· ∇G_s(u^N)dx≥ min

1 +C_P²kG_s(u^N)k²_H1(Ω) (3.75) and for−R

Ω

b(x,0)Gs(u^N)dxwe obtain

− Z

Ω

b(x,0)Gs(u^N)dx=− Z

A(s)

b(x,0)Gs(u^N)dx≤ kb(x,0)k_Lq∗

(A(s))kG_s(u^N)k_Lq(Ω)

≤C_Ekb(x,0)k_Lq∗

(A(s))kG_s(u^N)k_H1(Ω), (3.76) whereA(s) :={x∈Ω :

u^N(x)

> s}. Combining (3.47), (3.74), (3.75), and (3.76), we arrive at the estimate

_min

1 +C_P²kG_s(u^N)k_H1(Ω)≤CEkb(x,0)k_Lq∗

(A(s)). (3.77)

The final step before applying Lemma 2.33 is to estimate the left-hand side of (3.77) from below in terms of |A(h)| for h > s≥ ku^L+wk_L∞(Ωions) and the right-hand side of (3.77) from above in terms of |A(s)|. Again, using the Sobolev embedding H¹(Ω),→ L^q(Ω) and H¨older’s inequality yields

kG_s(u^N)k_H1(Ω)≥ 1 C_E



 Z

Ω

Gs(u^N)

qdx





1 q

= 1 C_E





 Z

A(s)

u^N

−s

qdx







1 q

≥ 1 CE





 Z

A(h)

(h−s)^qdx







1 q

= 1 CE

(h−s)|A(h)|^1q (3.78)

and

kb(x,0)k_Lq∗

(A(s))≤ kb(x,0)k_L2(Ω)|A(s)|^2−q

∗

2q∗ . (3.79)

Combining (3.77), (3.78), and (3.79), we obtain the following inequality for the nonnegative and nonincreasing function ϕ(t) :=|A(t)|

|A(h)| ≤ C_E² 1 +C_P² min

kb(x,0)k_L2(Ω)

!q

|A(s)|^q−22

(h−s)^q , for all h > s≥ ku^L+wk_L∞(Ωions). (3.80) Since ^q−2₂ = 2>1, by applying Lemma 2.33 we conclude that there is some e >0 such that

0< e^q= C_E² 1 +C_P² min

kb(x,0)k_L2(Ω)

!q

A ku^L+wk_L∞(Ωions)

q−4

2 2

q(q−2) q−4

≤ C_E² 1 +C_P²

_min kb(x,0)k_L2(Ω)

!q

|Ω|^q−42 2

q(q−2) q−4 =:e^q and

A ku^L+wk_L^∞_(Ωions)+e

= 0. Hence ku^Nk_L^∞_(Ω)≤ ku^L+wk_L^∞_(Ωions)+e.

From Proposition 3.18 it follows that the solution u^N to (3.47) is essentially bounded for all three choices of the test space V. As a result, the nonlinearity evaluated at u^N is also essentially bounded and thus by a standard density argument the weak formulations (3.47) for all three choices ofV are equivalent. We summarize the obtained in this section results in the following theorem.

Theorem 3.19 If we assume that

(1) the functiong specifying the Dirichlet boundary condition in (3.41)is given as the trace of some function inH²(Ω),

3.3. POISSON-BOLTZMANN EQUATION 61 (2) in the case of the 3-term splitting Γ∈C^0,1,

then the unique u^L ∈ H_g¹(Ω) is also in L^∞(Ω). Moreover, the variational problem (3.57) has a unique minimizer u^N_min∈H₀¹(Ω)∩L^∞(Ω) which coincides with the unique solution of problem (3.47) for all three choices of the test space V. As a result, problem (3.41) has a unique solution u=u^N+u^L∈H_g¹(Ω)∩L^∞(Ω)for all three choices of the test space V.

Existence, uniqueness, and boundedness of u without the additional splitting into u^N and u^L

Finally, we note that one can obtain directly the existence of a solution to problem (3.41) by considering the variational problem

Find umin ∈H_g¹(Ω) such that J(umin) = min

v∈H¹_g(Ω)

J(v), (3.81)

where the functionalJ :H_g¹(Ω)→R∪ {+∞}is defined byJ := ¹₂a(v, v) +R

Ω

B(x, v+w)dx− R

Ω

f · ∇vdx. Again, here it is understood that J(v) = +∞ whenever B(x, v+w)∈/ L¹(Ω), i.e.,

J(v) :=









 1

2a(v, v) + Z

Ω

B(x, v+w)dx− Z

Ω

f · ∇vdx, ifB(x, v+w)∈L¹(Ω), +∞, ifB(x, v+w)∈/L¹(Ω).

(3.82)

It is easy to see thatdom(J) is a convex set inH_g¹(Ω) and that J is convex on dom(J) (the argument is similar to the one in Remark 3.23). Since the set H_g¹ is a closed convex (and therefore weakly closed) subset of the reflexive Banach spaceH¹(Ω), J is strictly convex, proper, coercive, and sequentially weakly lower semicontinuous functional, from Theorem 2.30 it follows that problem (3.81) has a unique minimizerumin over H_g¹. That H_g¹(Ω) is norm closed in H¹(Ω) and convex follows easily by the linearity and boundedness of the trace operatorγ2. That J is strictly convex, proper, and s.w.l.s.c. follows by similar arguments to the ones made aboutJ^N. It is left to see that J is coercive overH_g¹(Ω). Let ug ∈H¹(Ω) be such that γ₂(u_g) =g on∂Ω. For any v∈H_g¹(Ω), we have γ₂(v−u_g) = 0. Since Ω is a bounded Lipschitz domain, it follows by Theorem 2.16 thatv−ug ∈H₀¹(Ω). By applying Poincar´e’s inequality we obtain

kvk_H1(Ω)−ku_gk_H1(Ω)

≤ kv−u_gk_H1(Ω)≤ q

1 +C_P²k∇(v−u_g)k_L2(Ω)

≤ q

1 +C_P² k∇vk_L2(Ω)+k∇u_gk_L2(Ω)

.

(3.83)

After squaring both sides of (3.83) and using the inequality 2ab ≤ a² +b²,∀a, b ∈ R we obtain the estimate

kvk²_H1(Ω)−2kvk_H1(Ω)ku_gk_H1(Ω)+ku_gk_H1(Ω)≤2 1 +C_P²

k∇vk²_L2(Ω)+k∇u_gk²_L2(Ω)

. (3.84)

Now, the coercivity of J follows by recalling that B(x, s) ≥ 0,∀x ∈ Ω,∀s ∈ R and using (3.84):

J(v) = 1

2a(v, v) + Z

Ω

B(x, v+w)dx− Z

Ω

f· ∇vdx≥ _min

2 k∇vk²_L2(Ω)−kfk_L2(Ω)k∇vk_L2(Ω)

≥ min

4 1 +C_P²

kvk²_H1(Ω)−2kvk_H1(Ω)ku_gk_H1(Ω)+ku_gk_H1(Ω)

(3.85)

−min

2 k∇u_gk²_H1(Ω)−kfk_L2(Ω)kvk_H1(Ω)→+∞ whenever kvk_H1(Ω)→ ∞.

We can summarize the above considerations in the following theorem.

Theorem 3.20

Problem (3.81) has a unique solution u_min ∈H_g¹(Ω).

By varying the functionalJ with directions v∈H₀¹(Ω)∩L^∞(Ω) one can show in a similar way to the approach for J^N that the unique minimizerumin of J overH_g¹(Ω) is a solution to (3.41) for the test spacesV =H₀¹(Ω)∩L^∞(Ω) and V =C₀^∞(Ω). To show thatu_min is also a solution to (3.41) with V =H₀¹(Ω) one has to show the a priori L^∞ bound on the unique solution of (3.41) with V =H₀¹(Ω)∩L^∞(Ω) and V =C₀^∞(Ω) (see p. 52 for the uniqueness of a solution to (3.41)). This can be done in a similar way to the proof of Proposition 3.18 or by observing that u=u^N +u^L∈L^∞(Ω). The a priori L^∞ estimate also follows from the more general Theorem 3.29 which we prove in Section 3.4.

Once we know that the solutionu of (3.41) withV =H₀¹(Ω)∩L^∞(Ω) orV =C₀^∞(Ω) is in L^∞(Ω), it follows thatb(x, u+w)∈L^∞(Ω) and therefore by a standard density argument we obtain that u is also the unique solution of (3.41) withV =H₀¹(Ω). We can summarize the above considerations in the following theorem.

Theorem 3.21

The variational problem (3.81)has a unique minimizerumin ∈H_g¹(Ω)which coincides with the unique solution of problem (3.41)for the test spaces V =C₀^∞(Ω) and V =H₀¹(Ω)∩L^∞(Ω).

Moreover, if we assume that

(1) the functiong specifying the Dirichlet boundary condition in (3.41)is given as the trace of some function inH²(Ω),

(2) in the case of the 3-term splitting Γ∈C^0,1, thenu_min ∈L^∞(Ω).

Remark 3.22 Denote F(v) := R

Ω

B(x, v+u^L+w)dx. We have a sequence {v_n}^∞_n=1 such that for each subsequence {v_nm}^∞_m=1 one can find a further subsequence {v_nms}^∞_s=1 for which

3.3. POISSON-BOLTZMANN EQUATION 63 F(v)≤lim inf

s→∞ F(vn_ms). We claim that F(v) ≤ lim inf

n→∞ F(vn). To see this, suppose to the contrary that F(v) > lim inf

n→∞ F(vn). Then, there exists a subsequence {v_nm}^∞_m=1 such that lim inf

n→∞ F(v_n) = lim

k→∞F(v_nm) and therefore F(v) > lim

m→∞F(v_nm). But for this subsequence, according to the assumptions on {v_n}^∞_n=1, we can find a further subsequence {v_nms}^∞_s=1 such that

F(v)≤lim inf

s→∞ F(v_nms) = lim

s→∞F(v_nms) = lim

m→∞F(v_nm)< F(v), which is a contradiction with the assumption that F(v)>lim inf

n→∞ F(vn).

Remark 3.23

It is worth noting thatdom(J^N) is a linear subspace of H₀¹(Ω) for d≤2 and not a linear subspace of H₀¹(Ω) if d≥3. In dimension d≤2, from [110, 179] we know thate^v ∈L²(Ω) for anyv∈H₀¹(Ω)and thus e^λv1+µv2 ∈L²(Ω)for any λ, µ∈R and any v₁, v₂∈H₀¹(Ω).

On the other hand, ifd≥3, first observe thatdom(J^N) =

v∈H₀¹(Ω) : B(x, v+w)∈L¹(Ω) , where w := u^L+w ∈ L^∞(Ω_ions). For simplicity we consider the case of the PBE, i.e.,

B(x, v+w) = k²cosh(v+w). Let Ω = B(0,1) and let Ωions = B(0, r) for some r < 1, where B(0, r) denotes the ball in R^d, d≥3 with radius r and a center at 0. We consider the

function v= ln_|x|¹ ∈H₀¹(B(0,1)). Sincee^v = _|x|¹ ∈L¹(Ω_ions) and e^λv= ¹

|x|^λ ∈/ L¹(Ω_ions) for any λ≥d¹, we obtain

Z

Ω

k²cosh(v+w)dx= Z

Ωions

k²_ions e^v+w+ e^−v−w

2 dx

≤1

2k²_ionse^{kwkL∞(Ωions)} Z

Ωions

e^v+ e^−v

dx≤ 1

2k²_ionse^{kwkL∞(Ωions)}



 Z

Ωions

e^vdx+|Ω_ions|



<∞ and

Z

Ω

k²cosh(λv+w)dx≥ 1 2

Z

Ωions

k²_ionse^λv+wdx≥ 1

2k²_ionse^{−kwkL∞(Ωions)} Z

Ωions

e^λvdx=∞ for anyλ > d.

This means that v∈dom(J^N), but λv /∈dom(J^N) for any λ≥d. Thereforedom(J^N) is not a linear space.

However, dom(J^N) ⊂ H₀¹(Ω) is a convex set. To see this, let v₁, v₂ ∈ dom(J^N), i.e., B(x, v1+w), B(x, v2+w)∈L¹(Ω). Since B(x,·) is convex it follows that for almost each x∈Ω and every λ∈[0,1] we have

B(x, λv₁(x) + (1−λ)v₂(x) +w(x))≤λB(x, v₁(x) +w(x)) + (1−λ)B(x, v₂(x) +w(x)).

1For anyd, using spherical coordinates, we have R

B(0,1) 1

|x|^λdx∼

1

R

0 1

ρ^λρ^d−1dρ=

1

R

0 1

ρ^λ−d+1dρ <∞if and only ifλ−d+ 1<1, i.e., if and only ifλ < d.

By integrating the above inequality over Ω we obtain Z

Ω

B(x, λv1+ (1−λ)v2+w)dx≤λ Z

Ω

B(x, v1+w)dx+ (1−λ) Z

Ω

B(x, v2+w)dx <∞ and thus λv₁+ (1−λ)v₂∈dom(J^N) for allλ∈[0,1]. Hencedom(J^N) is convex.

Remark 3.24

One can also see that the functional ¹₂a(v, v) is sequentially w.l.s.c. by noting that it is convex and continuous in H₀¹(Ω). From the continuity inH₀¹(Ω) it follows that it is also sequentially lower semicontinuous. Now, since ¹₂a(v, v) is convex and s.l.s.c. it follows that it is also sequentially w.l.s.c. (see, e.g. Corollary 3.9 in [96] or Corollary 2.2 in [65]). To see that

1

2a(v, v) is continuous in H₀¹(Ω)observe that 1

2(a(v, v)−a(u, u)) = 1

2(a(v, v) +a(v, u)−a(v, u)−a(u, u))

=1

2(a(v, v−u) +a(v−u, u))≤max kvk_H1(Ω)kv−uk_H1(Ω)+kv−uk_H1(Ω)kuk_H1(Ω)

, where _max:=kk_L∞(Ω).

Remark 3.25

In dimension d = 3, the functional R

Ω

B(x, v+w)dx is not Gateaux differentiable at any u∈H₀¹(Ω)∩L^∞(Ω), where we have denotedw:=u^L+w∈L^∞(Ωions). In factR

Ω

B(x, v+w)dx is discontinuous at every u∈H₀¹(Ω)∩L^∞(Ω). This is easy to demonstrate with the paradigm of the following example. LetΩions =B(0,2)⊂Ωbe the ball centered at 0 with radius 2 and for simplicity let B(x, v+w) =k²cosh(v+w). There exists a function z∈H₀¹(Ω) such that

R

Ωions

e^λzdx= +∞ for any λ >0. In particular, we can set z=φ|x|^−1/3, where φis a smooth function equal to 1 inB(0,1) and 0 inR³\B(0,2). Then z∈H₀¹(Ω), bute^λz ∈/ L¹(Ω_ions) for anyλ >0 since e^λz >|x|⁻³ for small enough|x|. In this case, for any u∈H₀¹(Ω)∩L^∞(Ω) and any λ >0 we have

Z

Ω

k²cosh(u+λz+w)dx≥ 1 2

Z

Ωions

k²_ionse^u+λz+wdx≥ k²_ionse^{−ku+wkL∞(Ωions)} 2

Z

Ωions

e^λzdx= +∞.

Remark 3.26

With respect to the general case of (3.41) which includes both the 2-term and the 3-term splittings, for the nonlinearity b(x, u+w), evaluated at the solution u (if it exists), we have that

b(x, u+w)v∈L¹(Ω)for all v∈V, (3.86) whereV is one of the spaces H₀¹(Ω), H₀¹(Ω)∩L^∞(Ω), orC₀^∞(Ω). Moreover, from (3.41) and the fact thata(u,·),R

Ω

f · ∇(·)dx define bounded linear functionals overH₀¹(Ω) it follows that

3.3. POISSON-BOLTZMANN EQUATION 65 the nonlinearity b(x, u+w) defines a bounded linear functional for all elements in the dense subspaceV, i.e.,

Z

Ω

b(x, u+w)v

≤Ckvk_H1(Ω) for all v∈V. (3.87) If we could conclude from (3.86) and (3.87) that b(x, u+w) is in L⁶⁵(Ω) (the inverse of H¨older’s inequality), then a density argument will give us that any solutionu of (3.41) with V = H₀¹(Ω)∩L^∞(Ω) and V = C₀^∞(Ω) is also a solution with V = H₀¹(Ω). However, the following example, showed to the author by C. Remling [106], demonstrates that this might not be necessarily true: take the function z= _1−|x|¹ , where Ωis the unit ball B1 :=B(0,1) in

R³. Then z /∈L¹(Ω), but R

B1

zvdx

.kvk_H1(B1),∀v∈H₀¹(B1). Indeed, we have Z

B1

|zv|dx= Z

B1

1

1− |x||v|dx= Z

B1

1

(1− |x|)^1/4 · |v|

(1− |x|)^3/4dx

≤



 Z

B1

dx (1− |x|)^1/2

Z

B1

v²

(1− |x|)^3/2dx





1/2

=C1



 Z

B1

v²

(1− |x|)^3/2dx





1/2

,

(3.88)

where C₁ = R

B1

dx (1−|x|)^1/2

!1/2

. If v is a smooth function in C₀^∞(B₁), we can obtain the estimate

Z

B1

v²(x)

(1− |x|)^3/2dx≤C2kvk²_H1(B1) (3.89) for someC2 >0. Now, from (3.88) and (3.89) it follows that

Z

B1

|zv|dx≤C₁C₂kvk_H1(B1), ∀v∈C₀^∞(B₁). (3.90) By the density inH₀¹(Ω) of compactly supported smooth functions, (3.90)also holds for all v ∈ H₀¹(B1). To see this, let v ∈ H₀¹(B1) be an arbitrary function. Then there exists a sequence {v_n}^∞_n=1 ⊂ C₀^∞(B₁) such that kv_n−vk_H1(B1)→ 0. Passing to a subsequence we obtain that v_nk(x) → v(x) a.e x ∈ B₁ and also kv_nkk_H1(B1)→ kvk_H1(B1). For each v_nk we have

Z

B1

|zv_nk|dx≤C₁C₂kv_nkk_H1(B1).

We can apply Fatou’s lemma to the measurable and positive functions |zv_nk| to obtain Z

B1

|z(x)v(x)|dx= Z

B1

lim inf

k→∞ |z(x)v_nk(x)|dx≤lim inf

k→∞

Z

B1

|zv_nk|dx

≤lim inf

k→∞ C1C2kv_nkk_H1(B1)=C1C2kvk_H1(B1).

Showing the estimate in (3.89): First observe that for anyx6= 0 it holds

v(x) =v(x)−v(x/|x|) =−

1/|x|

Z

1

∇v(tx)·xdt.

By using the Cauchy-Schwartz inequality and the fact that 0<|x| ≤1, we obtain the estimate

v²(x)≤

1/|x|

Z

1

|x|²dt

1/|x|

Z

0

|∇v(tx)|²dt=|x|(1− |x|)

1/|x|

Z

0

|∇v(tx)|²dt (3.91)

≤ 1

|x|−1

1/|x|

Z

1

|∇v(tx)|²dt. (3.92)

By dividing both sides of (3.91) by (1− |x|)^3/2 and then integrating in the disk A_1/2 :=

B₁\B_1/2 we obtain

Z

A1/2

v²(x)

(1− |x|)^3/2dx≤ Z

A1/2





 1

|x|p 1− |x|

1/|x|

Z

1

|∇v(tx)|²dt





dx≤ Z

A_1/2





 2 p1− |x|

1/|x|

Z

1

|∇v(tx)|²dt





dx.

Now we make a spherical change of variables, i.e., x₁ =rsinθcosϕ x2 =rsinθsinϕ x₃ =rcosθ

θ∈[0, π], ϕ∈[0,2π), r∈[1/2,1]

and obtain Z

A_1/2





 2 p1− |x|

1/|x|

Z

1

|∇v(tx)|²dt





dx

=

1

Z

1/2 π

Z

0 2π

Z

0

2r²sinθ

√1−r

1/r

Z

1

|∇v(trsinθcosϕ, trsinθsinϕ, trcosθ)|²dt

dϕdθdr. (3.93)

We make one more change of variables in the integral in t: tr =l and we can continue (3.93)

3.3. POISSON-BOLTZMANN EQUATION 67 as follows

Z

A_1/2





 2 p1− |x|

1/|x|

Z

1

|∇v(tx)|²dt





dx

=

1

Z

1/2



 2r²

√1−r

π

Z

0 2π

Z

0 1

Z

r

sinθ

r |∇v(lsinθcosϕ, lsinθsinϕ, lcosθ)|²dldϕdθ



dr

≤

1

Z

1/2







√2r 1−r

π

Z

0 2π

Z

0 1

Z

1/2

4l²sinθ|∇v(lsinθcosϕ, lsinθsinϕ, lcosθ)|²dldϕdθ





dr

=

1

Z

1/2

√8r

1−rdr Z

A1/2

|∇v(x)|²dx=C3k∇vk²_L2(A1/2),

(3.94)

where C3 :=

1

R

1/2

√8r

1−rdr. Finally, by using (3.94) we obtain the following estimate:

Z

B1

v²(x)

(1− |x|)^3/2dx= Z

B1/2

v²(x)

(1− |x|)^3/2dx+ Z

A1/2

v²(x) (1− |x|)^3/2dx

≤ Z

B1/2

v²(x)

(1−0.5)^3/2dx+C3k∇vk²_L2(A_1/2)=

√

8kvk²_L2(B_1/2)+C3k∇vk²_L2(A_1/2)

≤√

8kvk²_H1(B1/2)+C3kvk²_H1(A1/2)≤max{√

8, C3}kvk²_H1(B1). Therefore, (3.89) holds withC₂:= max{√

8, C₃}.

3.3.4 Regularity of the component u in the 2-term and 3-term splittings

Im Dokument The Poisson-Boltzmann Equation: Analysis, A Posteriori Error Estimates and Applications / eingereicht von Svetoslav Nakov, MSc. (Seite 61-77)