Uniform bounds of solutions

Z_δ=β+γ−¹

2 =E^β ×E^γ−12 ⊂C(Ω)×C(Ω). In both of these cases we can solve the problem in any space dimension.

We conclude this section with the following corollary.

Corollary 4.4. Consider the system of equations (4.1)-(4.4). Assume the hypotheses of Theorem 4.3 holds within (4.23)

α=θβ+ (1−θ)(γ−1

2), θ ∈[0,1]. (4.30)

Then,

(i) (4.24) holds in Z_β =E^β×E^β+12 ×E^β+12.

(ii) If 2β+γ > ^3N₄ ,then the solution to the problem (4.1)-(4.4), satises

U ∈C(0,∞, C^θ(Ω))∩C(0,∞, C^2+θ(Ω))∩C¹(0,∞, C^θ(Ω)), for someθ∈(0,1) and is a classical solution.

Proof. To prove (i), it suces to note that if α is as given in (4.30), then β > α if and only if β > γ− ¹₂, and also α > β if and only if γ − ¹₂ > β. Combining the two we nd that γ = β+ ¹₂. We prove (ii) of this corollary in the next section of the paper. As an alternative, using a classical approach, since by (4.11), U₀ ∈L^∞(Ω), the conclusion follows by [5, 38, 51, 76, 88, 92] and Theorem 4.3. The proof of the corollary is complete.

Lemma 4.5. Consider the evolution problem (4.1)-(4.4) in context of the Theorem 4.3.

Assume that the initial data of the system of equationsU0= (u0, v0, w0)^>∈Z_β(γ)∩[L^∞(Ω)]³, and that u ∈L^∞(0, T, L^r(Ω)), for some r > ^N₂ are nitely bounded in norms. Then v, w∈ L^∞(Ω×(0, T)) are also nitely bounded in norm of the given function space.

Proof. It suces only to consider one of either of the last two equations of (4.1)-(4.4) inv or w. We adopt here for simplication to use the function space H¹(Ω). Thus, considering the equation inv and taking the inner product ofL²(Ω)with|v|^r−1v, r >1we get that

1 r+ 1

d dt

Z

Ω

|v|^r+1+ 2√

d2r r+ 1

2Z

Ω

|∇|v|^r+12 |²+λ2

Z

Ω

|v|^r+1 ≤a2| Z

Ω

u|v|^r−1v|

≤ a2

Z

Ω

|v|^N(r+1)N−2

Θ1Z

Ω

|u|^r

Θ2Z

Ω

|v|^r+1 Θ3

≤a2C Z

Ω

(|∇|v|^r+12 |²+ (|v|^r+12 )²

^N_N−2^Θ1 Z

Ω

|u|^r

_N−2^2Θ2 Z

Ω

|v|^r+1 Θ3

,

(4.31)

where in the second inequality above we have used the Nakao-Hölder -Sobolev inequality [3, 77], since there exists ϑ >0 such thatr= ^N₂ +ϑand

Θ₁ = N−2

N+ϑ, Θ₂= 2

N +ϑ, Θ₃ = ϑ N+ϑ,

the third inequality is due to Sobolev space embeddings [2, 12, 25], i.e. (4.10), inα= 1/2.

In what follows we rst note that2r > r+ 1>2, hence after multiplying throughout in (4.31) by r+ 1, and using in the right hand side Young's inequality [12, 77] i.e.

ab≤ηa^s+C_ηb^s0, a, b≥0, η ∈(0,1) since

NΘ₁

N−2 = N

N +ϑ, NΘ₁

N−2 + Θ₃ = 1 we obtain if we let

β0= inf{µ₁,2d2−η,2λ2−η}>0, µ1∈σ(−∆ + 1), Ca2 =a2C

that d dt

Z

Ω

|v|^r+1+β0

Z

Ω

|∇|v|^r+12 |²+ Z

Ω

|v|^r+1

≤(2rCa2)

1

Θ3 sup

(0,T)

Z

Ω

|u|^r

!_ϑ(N−2)⁴ Z

Ω

|v|^r+1 ≤(2rCa2)

1 Θ3

Z

Ω

|v|^r+1,

(4.32)

since the term in brackets from the last inequality right to left is nitely bounded from above, and _Θ¹₃ > _ϑ(N⁴₋₂₎ we have incorporated the bounding from above constant with and/or the givenCa2 ≥0.

Therefore, if ri = 2ⁱ, i∈N, and

Θi= 2(ri+ 1)

N(r_i+ 1)−(N−2)(ri−1+ 1),Θ⁰_i= 1−Θi,

then by the Hölder's inequality as well as from the Sobolev type inclusions [2, 12, 25] and Young's inequality [12, 77], one obtains that

Z

Ω

|v|^ri+1 ≤ Z

Ω

|v|^N(N−2ri+1)

Θ⁰_iZ

Ω

|v|^ri+1 Θi

≤ C Z

Ω

(|v|^ri2+1)²+ Z

Ω

|∇|v|^ri2+1|²

NΘ0 i

N−2 Z

Ω

|v|^ri+1 Θi

. Thus, from (4.32) while still setting C_a2 =C_a2C, it follows that

d dt

Z

Ω

|v|^ri+1+β0(η) Z

Ω

|∇|v|^ri2+1|²+ Z

Ω

|v|^ri+1

≤(2rCa2)

1 Θ3

Z

Ω

(|v|^ri2+1)²+ Z

Ω

|∇|v|^ri+12 |²

NΘ0 i

N−2 Z

Ω

|v|^ri+1 Θi

≤η Z

Ω

(|v|^ri2+1)²+ Z

Ω

|∇|v|^ri2+1|²

+ (2r_iC_a2)

N+2 2Θ3

Z

Ω

|v|^ri−1+1 si

, and because ^N_N−2^Θ0i <1, we have used Young's inequality [12, 77].

Now set si= _r^ri+1

i−1+1 and since ^N+2₂ ≥ ^{N ri−1}_r ^+ri+2

i+2 , we get d

dt Z

Ω

|v|^ri+1+β Z

Ω

|∇|v|^ri2+1|²+ Z

Ω

|v|^ri+1

≤(2r_iC_a2)^σ Z

Ω

|v|^ri−1+1 si

,

whereσ = ^N+2_2Θ

3 , β=β(µ₁, d₂, λ₂,2η)>0.Applying Poincaré inequality and deningy_i(t) = R

Ω|v|^ri+1, we obtain

dy_i

dt +βy_i ≤(r_iC)^σ(yi−1)^si. (4.33) IfM =M(kv₀k_∞)>0is such that yi(0)≤M^{(ri−1+1)si)}. Then, solving (4.33), we get

yi(t)≤(riC)^σ yi(0) + sup

t∈(0,T)

yi−1(t)

!si! . We obtain from(a+b)^p ≤2^p(a^p+b^p), a, b, p≥0 withi=k≥1 that

y_k(t)≤(2C)^{1+2sk+2sk−1sk+...+2s2s3...sk}(2C)^{kσ+(k−1)σsk+...+σs2s3...sk}M^2s1s2...sk+ + (2C)^{1+2sk+2sk−1sk+...+2s2s3...sk}(2C)^{kσ+(k−1)σsk+...+σs2s3...sk} sup

t∈(0,T)

Z

Ω

|v|²

!s1s2...sk

≤(2C)^2Ak(2C)^σBkM^2χk + (2C)^2Ak(2C)^σBk sup

t∈(0,T)

Z

Ω

|v|²

!χk

, whereχ_k=s₁. . . s_k≤ ^rk₂⁺¹,

A_k = 1 +s_k+s_ksk−1+. . .+s_ksk−1. . . s₁ ≤(r_k+ 1)

∞

X

i=1

1 r_i+ 1, Bk = k+ (k−1)sk+ (k−2)sksk−1+. . .+sksk−1. . . s1≤(rk+ 1)

∞

X

i=1

i ri+ 1, and the series in the right hand sides converge since r_i = 2ⁱ. We then let

ω₁ =

∞

X

i=1

1

ri+ 1, ω₂=

∞

X

i=1

i ri+ 1, to conclude that

y_k(t) ≤



(2C)^2ω1(2C)^σω2M+ (2C)^2ω1(2C)^σω2 sup

t∈(0,T)

Z

Ω

|v|²

!¹₂



rk+1

≤ (2C)^2ω1(2C)^σω2M sup

t∈(0,T)

Z

Ω

|v|^{2 1}

2

+ 1

!!rk+1

. This implies

sup

Ω

|v(t, v₀)| ≤ lim

k→0

Z

Ω

|v|^{rk+1 1}

rk+1

≤ (2C)^2ω1(2C)^σω2M sup

t∈(0,T)

Z

Ω

|v|^{2 1}₂

+ 1

! ,

and the proof of the lemma is complete. Note that this proof in general scales of spaces E^α, α∈Rimplies the results since γ ≥β ≥1/2, but this will be more complicated.

Next we observe that due to the linearity of the system of equations in v, wand Lemma 4.5 we have, as a corollary, the following:

Corollary 4.6. Consider the evolution problem (4.1)-(4.4) in the context of Theorem 4.3 and Lemma 4.5. Assume the initial data u₀ ∈ L^∞(Ω), ∇v₀,∇w₀ ∈ [L^∞(Ω)]^N, and that

∇u∈L^∞(0, T, L^r(Ω)), r > ^N₂ are nitely bounded in norms of the given spaces. Then, the gradient solutions∇v,∇w∈L^∞(Ω×(0, T)), andu∈L^∞(Ω×(0, T)are also nitely bounded in norms.

Proof. It suces to note that from one of v, w system equations of (4.1)-(4.4), if one dif- ferentiate these equations with respect to the x variable, and takes as test function say

|∇v|^r−1∇v ∈ H¹(Ω), r > 1. Then, Lemma 4.5 holds due to the linearity of these system equations and weak coupledness.

So we only need to prove that u∈L^∞(Ω×(0, T))is nitely bounded in norm. To this end, consider theu equation of the system (4.1)-(4.4) and take the inner product of L²(Ω) with test function |u|^r−1u∈H¹(Ω), r >1 to nd that

1 (r+ 1)

d dt

Z

Ω

|u|^r+1+ 2√

d₁r r+ 1

2Z

Ω

|∇|u|^r+12 |²

=χ₂r Z

Ω

|u|^r−1u∇u∇v−χ₃r Z

Ω

|u|^r−1u∇u∇w

≤(C1χ2+C2χ3)r Z

Ω

||u|^r∇u|= 2(C1χ2+C2χ3)r r+ 1

Z

Ω

||u|^r+12 ∇|u|^r+12 |.

This yields that d dt

Z

Ω

|u|^r+1+ 2d₁ Z

Ω

|∇|u|^r+12 |²≤2(C₁χ₂+C₂χ₃)r Z

Ω

||u|^r+12 ∇|u|^r+12 |

≤ η Z

Ω

|∇|u|^r+12 |²+ (C₁χ₂+C₂χ₃)r 2η

Z

Ω

|u|^r+1, (4.34)

by Young's inequality

ab≤ηa^s+ (ηs)^−s

0

ss⁰⁻¹b^s0, a, b≥0,1 s+ 1

s⁰ = 1,

and ifβ₀ := 2d₁−η >0or0< η1is adequately chosen, then Poincaré inequality implies that

d dt

Z

Ω

|u|^r+1+β Z

Ω

|u|^r+1≤ (C1χ2+C2χ3)r 2η

Z

Ω

|u|^r+1, (4.35) where for µ₁∈σ(−∆), we denedβ:=µ₁β₀>0.

From this point, one can proceed as in the proof of Lemma 4.5 to conclude that u ∈ L^∞(Ω)is nitely bounded in norm.

Alternatively, we notice that by interpolation, for ϕ∈H¹(Ω), it holds that

kϕ−ϕk²_L2(Ω)≤Ck∇ϕk^2θ_L2(Ω)kϕk^2(1−θ)_L1(Ω), (4.36) whereθ= _N+2^N , and Young's inequality yields

kϕk²_L2(Ω)≤η₀k∇ϕk²_L2(Ω)+C(1 +η⁻

N 2

0 )kϕk²_L1(Ω).

Consequently, settingϕ=|u|^r+12 ,η₀ = _(r+1)^r2Cη with C_η = ^(C1χ2_2η^+C2χ3), we obtain that rC_η

Z

Ω

|u|^r+1≤ r (r+ 1)²

Z

Ω

|∇|u|^r+12 |²+ (2²C_η)^N2 C(1 +r^N) Z

Ω

|u|^r+12 2

, since r >1,2r > r+ 1, and0< η 1suciently small impliesC_η 1.

Therefore,C_η ≥C and, because1 +r^N ≤(1 +r)^N, we obtain from (4.34) the following iterative inequality type of (4.35), with β0−¹₄ >0,

d dt

Z

Ω

|u|^r+1+β Z

Ω

|u|^r+1≤(2C_η)^N(1 +r)^N Z

Ω

|u|^r+12 2

implying

=⇒ Z

Ω

|u|^r+1 ≤ Z

Ω

u^r+1₀ + (2C_η)^N(1 +r)^N sup

(0,T)

Z

Ω

|u|^r+12 2

. Next, dening

K(p) := max (

ku₀k_L∞(Ω),sup

(0,T)

Z

Ω

|u|^{p 1}_p)

(4.37)

leads to that

K(r+ 1)≤[(2Cη)^N(1 +r)^N]^r+11 K(r+ 1

2 ), ∀r≥1, so that, if we letr_i+ 1 = 2ⁱ, i∈N^∗, then we conclude that

K(2ⁱ) ≤ (2C_η)^N2−i(2ⁱ)^N2iK(2ⁱ⁻¹)≤. . .≤(2C_η)^NPik=12−k(2ⁱ)^2−iN. . .(1 + 2)^2−1NK(1)

≤ (2Cη)^N h

2^i2−iN(2⁻ⁱ)^2−iN i

. . . . h

2^2−1N(2⁻¹)^2−1N i

K(1)

≤ (2Cη)^N2^NPik=1k2−k×2^NPik=12−kK(1)≤C2^3NK(1).

Consequently, taking the limit asi→ ∞ yields kuk_L∞(Ω)≤C2^3NK(1)≤C2^3Nmax

ku₀k_L∞(Ω),ku₀k_L1(Ω) <∞, and the proof of the corollary is complete.

The following is a particular converse lemma to Corollary 4.6, since its conclusion holds for allr ∈[2,∞].

Lemma 4.7. Consider the evolution problem (4.1)-(4.4). Assume that the hypotheses of Corollary 4.6 hold, and that ∇u₀ ∈ L^∞(Ω) is nitely bounded in norm. If ∇v,∇w ∈ L^∞(Ω×(0, T)) are nitely bounded in norm, then ∇u ∈ L^∞(0, T, L^r(Ω)),∀r > ^N₂ is also nitely bounded in norm.

Proof. Dierentiate the system equation inuwith respect tox. Then, take the inner product

of L²(Ω)with the test function|∇u|^r∇u∈H¹(Ω), r≥0to nd that 1

r+ 2 d dt

Z

Ω

|∇u|^r+2+4d1(r+ 1) (r+ 2)²

Z

Ω

|∇(|∇u|^r+22 |²

= r+ 2 2 χ2

Z

Ω

|∇u|^r+22 ∇v∆u+u|∇u|^{r+22 −2}∇u∆u∆v

−

−r+ 2 2 χ₃

Z

Ω

|∇u|^r+22 ∇w∆u+u|∇u|^{r+22 −2}∇u∆u∆w

≤ r+ 2 2 χ2

Z

Ω

||∇u|^r+22 ∇v∆u|+ Z

Ω

|u|∇u|^{r+22 −2}∇u∆u∆v|

+ +r+ 2

2 χ3

Z

Ω

||∇u|^r+22 ∇w∆u|+ Z

Ω

|u|∇u|^{r+22 −2}∇u∆u∆w|

≤ r+ 2 2

C_χ

Z

Ω

|∇(|∇u|^r+22 )∇u|+C_u Z

Ω

||∇u|^r2∆u(∆v|+ ∆w|)

,

(4.38)

where Cχ = (χ2C∇v +χ3C∇w) ≥0, with C∇v, C∇w, Cu constants for the upper bounds of the variables in L^∞(Ω). Multiplying throughout by r+ 2, and since ¹₂ +_r+2¹ ≤1, for any r≥0, we get, by Holder's inequality, that

d dt

Z

Ω

|∇u|^r+2+ 2d₁ Z

Ω

|∇(|∇u|^r+22 |²

≤ (r+ 2)² 2

Cχ

Z

Ω

|∇(|∇u|^r+22 )∇u|+Cu

Z

Ω

||∇u|^r2∆u(∆v|+ ∆w|)

≤ (r+ 2)²

2 Cχ

Z

Ω

|∇(|∇u|^r+22 )|^{2 1}

2Z

Ω

|∇u|^{r+2 1}

r+2

+ + 2Cu

r+ 2 Z

Ω

|∇(|∇u|^r+22 )|^{2 1}

2

"

Z

Ω

|∆v|^{2 1}

2

+ Z

Ω

|∆w|^{2 1}

2

#!

.

(4.39)

We recall at this point the Young's inequality ab≤ηa^s+η^−s

0

ss⁰⁻¹b^s0, a, b≥0, η ∈(0,1),1 s+ 1

s⁰ = 1

with which, if we let η1 := _(r+2)^2η22Cχ, where by Nirenberg-Gagliardo's inequality [30, 25]

0< η2 ≤1 and Z

Ω

|∇u|^r+2 ≤η2

Z

Ω

|∇(|∇u|^r+22 )|²+CΩη^−m₂ Z

Ω

|∇u|^r+22 2

,

for m > ^N₂,C_Ω=C(Ω, m),η₃ := ^16η22

(r+2)⁴C_χ²|Ω|^1−r+22 , then we get that (r+ 2)²

2 Cχ

Z

Ω

|∇(|∇u|^r+22 )∇u| ≤ (r+ 2)² 2 Cχ

Z

Ω

|∇(|∇u|^r+22 )|^{2 1}₂ Z

Ω

|∇u|^{2 1}₂

≤ (r+ 2)² 2 Cχ

η1

Z

Ω

|∇(|∇u|^r+22 )|²+ (4η1)⁻¹ Z

Ω

|∇u|²

≤η₂ Z

Ω

|∇(|∇u|^r+22 )|²+(r+ 2)²

2 C_χ|Ω|^1−r+22

8η2

(r+ 2)²C_χ −1Z

Ω

|∇u|^r+2

≤2η2

Z

Ω

|∇(|∇u|^r+22 )|²+(r+ 2)²

2 Cχ|Ω|^1−r+22

8η2

(r+ 2)²Cχ

−1

CΩ(η3)^−m×

× Z

Ω

|∇u|^r+22 2

= 2η2

Z

Ω

|∇(|∇u|^r+22 )|²+ (r+ 2)^4mΓ1CΩ

Z

Ω

|∇u|^r+22 2

,

whereΓ1 =

C_χ²|Ω|^1−r+22 16η²₂

^m+1 .

As for the last expression in (4.39), we need a control from above of the integrals involving

−∆ of v, w. To this end, multiplying either stationary equations in v or w by −∆ of the variable, to obtain that

d2

Z

Ω

|∆v|²+λ2

Z

Ω

|∇v|² =a2

Z

Ω

∇u∇v≤a2

Z

Ω

|∇u|^{2 1}₂Z

Ω

|∇v|^{2 1}₂

≤η₄ Z

Ω

|∇u|²+a₂(4η₄)⁻¹ Z

Ω

|∇v|²≤η₄|Ω|^1−r+22 Z

Ω

|∇u|^r+2+a₂(4η₄)C_∇v² |Ω|

≤η₄|Ω|^1−r+22 µ1

Z

Ω

|∇(|∇u|^r+22 )|²+a₂(4η₄)⁻¹C_∇v² |Ω|.

Note that this remains true even if one had considered the entire equation involving the time derivative, since by Theorem 4.3, the solutions are continuous in time. Letη₄ ≤η₅ and set

η₆ := 1 (r+ 2)C_u

2η₅+ η₄ 2µ₁η₅

1 d₂ + 1

d₃

|Ω|^1−r+22

.

Then we nd that (r+ 2)C_u

Z

Ω

|∇(|∇u|^r+22 )|^{2 1}₂ "

Z

Ω

|∆v|^{2 1}₂

+ Z

Ω

|∆w|^{2 1}₂#

≤(r+ 2)C_u

2η₅ Z

Ω

|∇(|∇u|^r+22 )|²+ 1 4η5

Z

Ω

|∆v|²+ Z

Ω

|∆w|²

≤(r+ 2)Cu

2η5+ η4

2µ1η5

1 d2

+ 1 d3

|Ω|^1−r+22 Z

Ω

|∇(|∇u|^r+22 )|² + +

a2(C∇v+C∇w)|Ω|

16η₅η₄

≤ η₆ Z

Ω

|∇(|∇u|^r+22 )|²+

a₂(r+ 2)(C∇v+C∇w)|Ω|C_u 16η²₄

. Thus, from (4.39), if we let η₇ = 2η₂ +η₆, Γ₂ = ^a2(C∇v+C_16η^∇w2 ^)|Ω|Cu

4 and Γ = max{Γ₁,Γ₂}, then we are led to conclude that

d dt

Z

Ω

|∇u|^r+2+ 2d₁ Z

Ω

|∇(|∇u|^r+22 |²

≤η7

Z

Ω

|∇(|∇u|^r+22 )|²+ (r+ 2)^4mΓ Z

Ω

|∇u|^r+22 2

+ 1

!

implying

=⇒ d dt

Z

Ω

|∇u|^r+2+β Z

Ω

|∇u|^r+2≤(r+ 2)^4mΓ Z

Ω

|∇u|^r+22 2

+ 1

!

(4.40)

implying

=⇒ Z

Ω

|∇u|^r+2 ≤ Z

Ω

|∇u₀|^r+2+ (r+ 2)^4mΓ



 sup

(0,T)

Z

Ω

|∇u|^r+22

!2

+ 1



, following a use of Poincaré inequality and thatβ := 2d1−η7 >0.

Next, proceeding as either in the proof of Lemma 4.5 or as in the proof of Corollary 4.6, yields that∇u∈L^∞(Ω×(0, T)) is nitely bounded in norm.

To complete the ideas, consider (4.37) in gradient functions and take p =r+ 2, to get that

K(r+ 2)≤[Γ(r+ 2)^4m]^r+21 K(r+ 2

2 ),∀r≥0.

Now, if we letr_i+ 2 = 2ⁱ, i∈N, then we obtain that

K(2ⁱ) ≤ Γ²⁻ⁱ(2ⁱ)^4m2−iK(2ⁱ⁻¹)≤. . .≤Γ^Pik=12−k(2ⁱ)^2−i4m. . .(2)^2−14mK(1)

≤ Γ h

2^i2−i4m(2⁻ⁱ)^2−i4m i

. . . . h

2^2−14m(2⁻¹)^2−14m i

K(1)

≤ Γ2^{4mPik=1k2−k}×2^4mPik=12−kK(1)≤C2^12mK(1).

Thus taking the limit asi→ ∞ yields

k∇uk_L∞(Ω)≤C2^12mK(1)≤C2^12mmax

k∇u₀k_L∞(Ω),k∇u₀k_L1(Ω) <∞, and the proof of the lemma is complete.

Now we prove (ii) of Corollary 4.4.

Proof. We use a bootstrap argument. By Theorem 4.3 taking into account that the space inclusions (4.11) imply E^β, E^γ ⊂ C(Ω), we get that u ∈ C(Ω ×(0, T)). Thus, viewing either equations in v or w variables, we get for example that vt ∈ L^∞(Ω), consequently g(t) :=a₂u−v_t ∈L^p(Ω)for all p≥1. Thus,v ∈W^2,p(Ω)for all p ≥1 and ∇v ∈W^1,p(Ω) for all p ≥1 In particular, ∇v∈ W^1,p(Ω) for somep > N, yielding ∇v ∈C^θ(Ω), for some θ >0.

In fact, if p < N, as ∇v ∈ W^1,p(Ω) ⊂ L^q1(Ω), q1 = _N−p^pN if p > ^N₂, then q1 > N and the above statements hold. If not we repeat the process, with W^1,p(Ω) ⊂ L^q2(Ω), q₂ = _N−q^q1N

1 = _N−2p^pN and ifp > ^N₃ we are done asq₂> N. In the otherwise case we repeat the iterative process to ndqm= _N^qm−1_−q ^N

m−1 = _N^pN_−mp and ifp > _m+1^N then we are done. Thus in a nite number of steps it is always possible to getqm > N and the above Hölder smoothness of gradient solutions are obtained.

The next immediate result from Corollary 4.6 is also that u ∈ L^∞(Ω). Now, if ∇u₀ ∈ L^∞(Ω), then Lamma 4.7 implies that ∇u∈L^∞(Ω). Furthermore,

f(t) =−div(u ~d(∇v,∇w))∈W^2,p(Ω), ∀p >1.

Thus viewing the equation in u as an elliptic problem, we get ∇u ∈W^1,p(Ω)⊂C^θ(Ω) for some θ > 0, since, in particular, using a bootstrap iteration argument as in above lines W^1,p(Ω) ⊂ L^qm(Ω), it is possible to get q_m > N, provided that p > _m+1^N . Consequently, u∈C^2+θ(Ω). Getting back to the v equation, we obtain g(t)∈C^θ(Ω)and so v ∈C^2+θ(Ω) for some θ >0. By similarity, of the equations we also havew∈C^2+θ(Ω)for some θ >0.

Combining all of the above, we conclude that the solution to the problem (4.1)-(4.4) veries regularity properties given in Corollary 4.4, and that it is a classical solution.