On initial-boundary value problems in a strip for generalized two-dimensional Zakharov-Kuznetsov equation

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by  Andrei Faminskii

ON INITIAL-BOUNDARY VALUE PROBLEMS IN A STRIP FOR
GENERALIZED TWO-DIMENSIONAL ZAKHAROV–KUZNETSOV
EQUATION
3
1
0 ANDREIV.FAMINSKIIANDEVGENIYAS.BAYKOVA
2

n
a Peoples’FriendshipUniversityofRussia,Moscow,Russia
J

5
Abstract. Initial-boundary value problems in a strip with different types
1
of boundary conditions for two-dimensional generalized Zakharov–Kuznetsov

equation are considered. Results on global existence and uniqueness of weak
]
P solutionsincertainweightedspacesareestablished.
A
.
h 1. Introduction. Description of main results
t
a
Zakharov–Kuznetsovequation (ZK) on the plane is written as follows:
m
[ ut+uxxx+uxyy+uux =0.

2 Ithasbeenderivedin[17]fordescriptionofion-acousticwaveprocessesinplasma
v
put in the magnetic field. Further, this equation has been considered as a model
6
equation for non-linear waves propagating in dispersive media in the preassigned
9
8 direction (x) and deformed in the transverse direction (y). Zakharov–Kuznetsov
5 equationisoneofthevariationsofmulti-dimensionalgeneralizationsforKorteweg–
. de Vries equation (KdV) u +u +uu =0.
2 t xxx x
1 Boundaryvalueproblemsforthisequation(anditscertaingeneralizations)were
2 usually studied before in domains, which were products of an interval (bounded
1 or non-bounded) on the variable x and the whole line on the variable y. In
:
v particular, there were the initial value problem ([14, 2, 3, 12]), as well as initial-
i boundary value problems in R ×R,R ×R, and I ×R, where I is a bounded
X + −
interval, ([4, 5, 6, 15, 7] and others). On the other hand, it seems more natural
r
a from the physical point of view to consider domains, where the variable y varies
in a bounded interval.
In the present paper we consider initial-boundary value problems in a layer
Π =(0,T)×Σ, where Σ=R×(0,L)={(x,y):x∈R,0<y <L} is a strip of
T
a given width L, for generalized Zakharov–Kuznetsovequation
u +u +u +(g(u)) =f(t,x,y) (1.1)
t xxx xyy x
with an initial condition
u(0,x,y)=u (x,y), (x,y)∈Σ, (1.2)
0
2010 Mathematics Subject Classification. 35Q53,35D30.
Keywordsandphrases. Zakharov–Kuznetsovequation;initial-boundaryvalueproblems;weak
solutions.
1
and boundary conditions of one of the following four types: for (t,x)∈(0,T)×R
whether a) u(t,x,0)=u(t,x,L)=0,
or b) u (t,x,0)=u (t,x,L)=0,
y y
(1.3)
or c) u(t,x,0)=u (t,x,L)=0,
y
or d) u is an L-periodic function with respect to y.
We use the notation ”problem (1.1)–(1.3)” for each of these four cases.
Themainresultsofthispaperaretheoremsonexistenceanduniquenessofglobal
weak solutions (T > 0 – arbitrary). We introduce certain growth restrictions on
the function g as |u|→∞ which are true for ZK equation itself.
The research of global well-posedness is based, firstly, on conservation laws for
equation (1.1) (with f ≡0):
u2dxdy =const, u2 +u2−g∗(u) dxdy =const, (1.4)
ZZ ZZ x y
Σ Σ(cid:0) (cid:1)
wherehereandfurther g∗(u)≡ ug(θ)dθ istheprimitivefor g,whicharetruefor
0
eachofthefourcasesofthebounRdaryconditions,andsecondly,ontheeffectoflocal
smoothing,meaning that solutions havederivatives with respect to space variables
of orders greater by one, than initial functions. For example, if u ∈L (Σ), then
0 2
T r L
(u2 +u2)dydxdt<∞
Z Z Z x y
0 −r 0
for any r >0. Such an effect of local smoothing for the initial value problem was
firstly discoveredin [10, 8] for KdV.
Theresult,basedonconservationlaws(1.4),onexistenceofglobalsolutiontothe
initial value problem, where u ∈ H1(R2), was obtained in [14] (where equations
0
of more general type than (1.1) were considered). Existence of global solution to
theinitialvalueproblemfor u ∈L (R2),basedonthe firstconservationlaw (1.4)
0 2
and on the effect of local smoothing, was obtained in [2] (were equations of more
generaltype were considered too). However,in the study of uniqueness there were
the growth restrictions on the function g, which excluded ZK equation itself.
Classes of global well-posedness for the initial-value problem for ZK equation,
where the initial function u ∈ Hk(R2) for each natural k, were constructed in
0
[3] on the base of the ideas on more accurate study of properties of the linear part
of the equation elaborated earlier in [9] for KdV. Similar results for modified ZK
equation (g(u)=u3/3) were obtained in [12].
The attempt of studying properties of the linear part of ZK equation for the
initial-boundary value problem in the strip Σ with periodic boundary conditions
was made in [13], but the established estimates allowed to prove only local well-
posednesss in the spaces Hs(Σ) for s>3/2.
The recentpaper [11], whereaninitial-boundaryvalue problemfor ZKequation
inahalf-strip R ×(0,L) withzeroDirichletboundaryconditionswasconsidered,
+
should be also mentioned. The usage of exponential weights as x → +∞ allowed
to provethere global well-posedness for such a problem in smooth function spaces.
Introduce now the following notationto describe the main results of the present
paper. For an integer k≥0 let
1/2
|Dkϕ|= (∂k1∂k2ϕ)2 , |Dϕ|=|D1ϕ|.
x y
(cid:16) X (cid:17)
k1+k2=k
2
Let L = L (Σ), Hk = Hk(Σ), x = max(x,0), x = max(−x,0), R =
p p + − +
(0,+∞), R =(−∞,0), Σ =R ×(0,L), Π± =(0,T)×Σ .
− ± ± T ±
For any α≥0 define function spaces
Lα =Lα(Σ)={ϕ∈L :(1+x )αϕ∈L },
2 2 2 + 2
Hk,α =Hk.α(Σ)={ϕ∈Hk :|Djϕ|∈Lα, j =0,...,k}
2
with natural norms (here H0,α =Lα).
2
We shall construct solutions to the considered problems in spaces Xk,α(Π ),
T
k =0 or 1, consisting of functions u(t,x,y) such that
T x0+1 L
u∈C ([0,T];Hk,α), sup |Dk+1u|2dydxdt<∞
w x0∈RZ0 Zx0 Z0
(symbol C denotes the spaceof weakly continuousmappings) andif α>0 then
w
additionally
(1+x)α−1/2|Dk+1u|∈L (Π+)
2 T
(let Xα(Π )=X0,α(Π )).
T T
Theorem 1.1. Let g ∈C1(R) and for certain constants c≥0 and b∈[1,2)
|g′(u)|≤c(1+|u|b) ∀u∈R. (1.5)
Let u ∈ Lα, f ∈ L (0,T;Lα) for certain α ≥ 0 and T > 0. Then there exists
0 2 1 2
a weak solution to each of problems (1.1)–(1.3) in the space Xα(Π ).
T
Theorem 1.2. Let g ∈C2(R) and for certain constants c≥0 and b∈[1,2)
|g′′(u)|≤c(1+|u|b−1) ∀u∈R. (1.6)
Let u ∈H1,α, f ∈L (0,T;H1,α) for certain α≥0 and T >0,
0 1
u | =u | =0, f| =f| =0 in the case a,
0 y=0 0 y=L y=0 y=L
u | =0,f| =0 in the case c,
0 y=0 y=0
u | =u | , f| =f| in the case d.
0 y=0 0 y=L y=0 y=L
Then there exists a weak solution to each of problems (1.1)–(1.3) in the space
X1,α(Π ). If α≥1/2 then the solution is unique in this space.
T
Similar results for the initial value problem for generalized KdV equation were
earlier obtained in [1].
Sincetheconstructedsolutionsareonlyweaktherenaturallyarisesaproblemon
solubilityoftheconsideredproblemsinspacesofsmootherfunctions. Thisproblem
remains open, and one of the obstacles to resolve it is the absence of conservation
laws other than (1.4) in smoother spaces in comparison, for example, with KdV
equation.
Further we use the following auxiliary functions. Let η(x) denotes a cut-off
function, namely, η is an infinitely smooth non-decreasing on R function such
that η(x)=0 when x≤0, η(x)=1 when x≥1, η(x)+η(1−x)≡1.
We say that ρ(x) is an admissible weight function if ρ is an infinitely smooth
positive on R function such that |ρ(j)(x)| ≤ c(j)ρ(x) for each natural j and all
x∈R.
For each α ≥ 0 and β > 0 we introduce an infinitely smooth increasing on R
function ρ (x) as follows: ρ (x)=eβx when x≤−1, ρ (x)=(1+x)α for
α,β α,β α,β
α>0 and ρ (x)=2−(1+x)−1/2 when x≥0, ρ′ (x)>0 when x∈(−1,0).
0,β α,β
3
Note that both ρ and ρ′ are admissible weight functions, and ρ′ (x) ≤
α,β α,β α,β
c(α,β)ρ (x) for all x∈R.
α,β
Note also that if u ∈ Xk,α(Π ) for α ≥ 1/2, then |Dk+1u|ρ (x) ∈
T α−1/2,β
L (Π ) for any β >0.
2 T
Further we need the following interpolating inequality.
Lemma 1.1. Let ρ (x), ρ (x) be two admissible weight functions such that
1 2
ρ (x) ≤ c ρ (x) for some constant c > 0. Let k = 1 or 2, m ∈ [0,k) –
1 0 2 0
integer, q ∈ [2,+∞] if k = 2,m = 0 and q ∈ [2,+∞) in other cases. For
ρ (x ) ρ (x )
2 1 2 2
the case q = +∞ assume also that ≤c if |x −x | ≤ 1. Then
0 1 2
ρ (x ) ρ (x )
1 1 1 2
there exists a constant c>0 such that for every function ψ(x,y), which satisfies
assumptions |Dkψ|ρ1/2(x)∈L , ψρ1/2(x)∈L , the following inequality holds
1 2 2 2
|Dmψ|ρs(x)ρ1/2−s(x) ≤c |Dkψ|ρ1/2(x) 2s ψρ1/2(x) 1−2s+ ψρ1/2(x) ,
(cid:13) 1 2 (cid:13)Lq (cid:13) 1 (cid:13)L2(cid:13) 2 (cid:13)L2 (cid:13) 2 (cid:13)L2
(cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13)(1.7)
m+1 1
where s= − .
2k kq
Proof. If one considers the whole plane R2 instead of the strip Σ the given in-
equality is a special case for a more general interpolating inequality, estimated in
[2] for an arbitrary number of variables and arbitrary values of k. The proof in
this case is similar. (cid:3)
As a rule further we omit limits of integration in integrals over the whole strip
Σ.
The paper is organized as follows. An auxiliary linear problem is considered
in Section 2. Section 3 is dedicated to problems on existence of solutions to the
originalproblem. Results on continuous dependence of solutions on u and f are
0
proved in Section 4. In particular, they imply uniqueness of the solution.
2. An auxiliary linear equation
Consider a linear equation
u +u +u −δ(u +u )=f(t,x,y) (2.1)
t xxx xyy xx yy
for a certain constant δ ∈[0,1].
Introduce certain additional function spaces. Let S(Σ) be a space of infinitely
smooth in Σ functions ϕ(x,y) such that (1+|x|)n|∂k∂lϕ(x,y)|≤c(n,k,l) for
x y
anyintegernon-negative n,k,l andall (x,y)∈Σ. Let S (Σ ) denoteaspaceof
exp ±
infinitely smooth in Σ functions ϕ(x,y) such that en|x||∂k∂lϕ(x,y)|≤c(n,k,l)
± x y
for any integer non-negative n,k,l and all (x,y)∈Σ .
±
Lemma 2.1. Let u ∈ S(Σ)∩S (Σ ), f ∈ C∞ [0,T];S(Σ)∩S (Σ ) and
0 exp + exp +
for any integer j ≥0 (cid:0) (cid:1)
∂2ju | =∂2ju | =0, ∂2jf| =∂2jf| =0 in the case a,
y 0 y=0 y 0 y=L y y=0 y y=L
∂2j+1u | =∂2j+1u | =0, ∂2j+1f| =∂2j+1f| =0 in the case b,
y 0 y=0 y 0 y=L y y=0 y y=L
∂2ju | =∂2j+1u | =0, ∂2jf| =∂2j+1f| =0 in the case c,
y 0 y=0 y 0 y=L y y=0 y y=L
∂ju | =∂ju | , ∂jf| =∂jf| in the case d.
y 0 y=0 y 0 y=L y y=0 y y=L
4
Then there exists a unique solution to each of problems (2.1), (1.2), (1.3) u ∈
C∞ [0,T];S(Σ)∩S (Σ ) .
exp +
(cid:0) (cid:1)
Proof. Let ψ(y), l = 1,2... , be the orthonormal in L (0,L) system of the
l 2
eigenfunctions for the operator (−ψ′′) on the segment [0,L] with corresponding
boundary conditions ψ(0) = ψ(L) = 0 in the case a, ψ′(0) = ψ′(L) = 0 in the
case b, ψ(0) = ψ′(L) = 0 in the case c, ψ(0) = ψ(L),ψ′(0) = ψ′(L) in the case
d, λ be the corresponding eigenvalues. Such systems are well-known and can be
l
written in trigonometric functions. Then with the use of Fourier transform for the
variable x and Fourier series for the variable y a solution to problem (2.1), (1.2),
(1.3) can be written as follows:
+∞
1
u(t,x,y)= eiξxψ(y)u(t,ξ,l)dξ,
2π Z l
X
R l=1
b
where
t
u(t,ξ,l)=u (ξ,l)e(ξ3+ξλl−δ(ξ2+λl))t+ f(τ,ξ,l)e(ξ3+ξλl−δ(ξ2+λl))(t−τ)dτ,
0 Z
0
b c b
u (ξ,l)≡ e−iξxψ (y)u (x,y)dxdy,
0 ZZ l 0
c
f(t,ξ,l)≡ e−iξxψ (y)f(t,x,y)dxdy.
ZZ l
According to the prboperties of the functions u and f this solution u ∈
0
C∞([0,T],S(Σ)).
Next, let v ≡ ∂k∂lu for some k,l ≥ 0. Then the function v satisfies an
x y
equation of (2.1) type, where f is replaced by ∂k∂lf . Let m ≥ 3. Multiplying
x y
this equation by 2xmv and integrating over Σ , we derive an inequality
+
d
xmv2dxdy ≤m(m−1)(m−2) xm−3v2dxdy
dtZZ ZZ
Σ+ Σ+
+δm(m−1) xm−2v2dxdy+2 ∂k∂lfvdxdy.
ZZ ZZ x y
Σ+ Σ+
Let α > 0, n ≥ 3. For any m ∈ [3,n] multiplying the corresponding inequality
by αm/(m!) and summing by m we obtain that for
n (αx)m
z (t)≡ v2(t,x,y)dxdy
n ZZ m!
Σ+mX=0
inequalities
z′(t)≤cz (t)+c, z (0)≤c,
n n n
are valid uniformly with respect to n, whence it follows that
sup eαxv2dxdy <∞.
ZZ
t∈[0,T] Σ+
Thus, u∈C∞([0,T],S (Σ )). (cid:3)
exp +
We now turn to generalized solutions. Let u ∈ S(Σ) ∩ S (Σ ) ′, f ∈
0 exp −
C∞([0,T];S(Σ)∩S (Σ )) ′. (cid:0) (cid:1)
exp −
(cid:0) (cid:1) 5
Definition 2.1. A function u ∈ C∞([0,T];S(Σ)∩S (Σ )) ′ is called a gen-
exp −
eralized solution to corresponding(cid:0)problem (2.1), (1.2), (1.3), i(cid:1)f for any function
ϕ ∈ C∞ [0,T];S(Σ)∩S (Σ ) such that ϕ| = 0 and ϕ| = ϕ| = 0
exp − t=T y=0 y=L
for the ca(cid:0)se a, ϕ | =ϕ | (cid:1)=0 for the case b, ϕ| =ϕ | =0 for the
y y=0 y y=L y=0 y y=L
case c, ϕ| =ϕ| , ϕ | =ϕ | forthecase d,thereholdsthefollowing
y=0 y=L y y=0 y y=L
equality:
hu,ϕ +ϕ +ϕ +δϕ +δϕ i+hf,ϕi+hu ,ϕ| i=0. (2.2)
t xxx xyy xx yy 0 t=0
Lemma 2.2. A generalized solution to each problem (2.1), (1.2), (1.3) is unique.
Proof. The proof is carried out by standard Ho¨lmgren’s argument on the basis of
Lemma 2.1. (cid:3)
Nowwepresentanumberofauxiliarylemmasonsolubilityofthelinearproblems
in non-smooth case.
Introduce a space Xk,α(Π ), that is different from Xk,α(Π ) in that the
T T
condition u ∈ C ([0,T];Hk,α) is substituted by u ∈ C([0,T];Hk,α), and let
w e
Xα(Π )=X0,α(Π ).
T T
Leemma 2.3e. Let u ∈ Lα for some α ≥ 0, f ≡ f + δ1/2f , where f ∈
0 2 0 1x 0
L (0,T;Lα), f ∈L (0,T;Lα). Then there exists a (unique) generalized solution
1 2 1 2 2
to each problem (2.1), (1.2), (1.3) u(t,x,y) from the space Xα(Π ), and δ|Du|∈
T
L (0,T;Lα). Moreover, for any t∈(0,T] uniformly with respect to δ
2 2 e
kukXeα(Πt)+δ1/2(cid:13)|Du|(cid:13)L2(0,t;Lα2)
(cid:13) (cid:13)≤c(T) ku0kLα2 +kf0kL1(0,t;Lα2)+kf1kL2(0,t;Lα2) , (2.3)
(cid:2) (cid:3)
t
u2(t,x,y)ρ(x)dxdy+ |Du|2·(ρ′+δρ)dxdydτ
ZZ Z ZZ
0
t t
≤ u2ρdxdy+c u2ρdxdydτ +2 f uρdxdydτ
ZZ 0 Z ZZ Z ZZ 0
0 0
t
−2δ1/2 f (uρ) dxdydτ, (2.4)
Z ZZ 1 x
0
where ρ is an admissible weight function such that ρ(x)≤const·(1+x )2α and
+
the constant c depends on the properties of the function ρ.
Proof. It is sufficient to consider smooth solutions that were constructed, for ex-
ample, in Lemma 2.1 because of linearity of the problem.
Then multiplying equation (2.1) by 2u(t,x,y)ρ(x) and integrating over Σ we
obtain an equality
d
u2ρdxdy+ (3u2 +u2)ρ′dxdy+2δ (u2 +u2)ρdxdy
dtZZ ZZ x y ZZ x y
− (u2ρ′′′+δu2ρ′′)dxdy =2 f uρdxdy−2δ1/2 f (uρ) dxdy, (2.5)
ZZ ZZ 0 ZZ 1 x
which implies inequality (2.4) by the properties of admissible weight functions.
Next, let ρ(x) ≡ 1+ρ (x−x ) for any x ∈ R and then ρ(x) ≡ ρ (x) if
0,1 0 0 2α,1
α>0. Thus we obtainfrom(2.4) estimate (2.3), which, in particular,allowsus to
prove Lemma in the case of non-smooth solutions. (cid:3)
6
Lemma 2.4. Let δ = 0, u ρ (x) ∈ L for some α > 0 and β > 0, f ≡
0 α,β 2
f +f , where f ρ (x) ∈ L (0,T;L ), f ρ (x) ∈ L (Π ). Then there
0 1x 0 α,β 1 2 1 α+1/2,β 2 T
exists a (unique) generalized solution to each of problems (2.1), (1.2), (1.3) such
that uρ (x) ∈ C([0,T];L ), |Du|ρ′ (x) ∈ L (Π ) and inequality (2.4), where
α,β 2 α,β 2 T
thereis no multiplier δ1/2 in the last term in its right part, holds for any t∈(0,T]
and for ρ≡ρ .
2α,2β
Proof. First of all, note that if uρ ∈ C([0,T];L ) then u belongs to the class
α,β 2
C∞([0,T];S(Σ)∩S (Σ )) ′.
exp −
(cid:0) Next, the functions u ,f ,(cid:1)f canbe regularizedsuch that u ∈L , f +f ∈
0 0 1 0 2 0 1x
L (0,T;Lα) (for example, by substituting u with the function u η(x+1/h)),
1 2 0 0
so one can consider the solution u ∈ Xα(Π ). Equality (2.4) is valid for such
T
solutions, where f should be substituted by f +f , and the term with δ1/2f
0 e 0 1x 1
should be omitted. Also, it follows from (2.5) that |Du|2 can be substituted by
(3u2 +u2) in the second summand of the left part in (2.4).
x y
Note that
ρ2 (x)
2α,2β
≤c(α,β)ρ (x), (2.6)
ρ′ (x) 2α+1,β
2α,2β
and, therefore,
2 f uρ dxdy =− f (uρ ) dxdy ≤ u2ρ′ dxdy
ZZ 1x 2α,2β ZZ 1 2α,2β x ZZ x 2α,2β
ρ2
+c f2 2α,2β dxdy+c (u2+f2)ρ dxdy ≤ u2ρ′ dxdy
ZZ 1ρ′ ZZ 1 2α,2β ZZ x 2α,2β
2α,2β
+c f2ρ dxdy+c (u2+f2)ρ dxdy.
1ZZ 1 2α+1,2β ZZ 1 2α,2β
(cid:3)
Lemma 2.5. Let u ∈ H1,α for some α ≥ 0, f ≡ f + δ1/2f , where
0 0 1
f ∈ L (0,T;H1,α), f ∈ L (0,T;Lα). Assume that u ,f satisfy the same
0 1 1 2 2 0 0
assumptions as the ones for u ,f when y = 0,y = L in Theorem 1.2. Then
0
there exists a (unique) generalized solution to each of problems (2.1), (1.2), (1.3)
u(t,x,y) in the space X1,α(Π ), and δ|D2u|∈ L (0,T;Lα) . Moreover, for any
T 2 2
t∈(0,T] uniformly with respect to δ
e
kukXe1,α(Πt)+δ1/2(cid:13)|D2u|(cid:13)L2(0,t;Lα2)
(cid:13) ≤c(T(cid:13)) ku0kH1,α +kf0kL1(0,t;H1,α)+kf1kL2(0,t;Lα2) , (2.7)
(cid:2) (cid:3)
t
|Du(t,x,y)|2ρ(x)dxdy+ |D2u|2·(ρ′+δρ)dxdydτ
ZZ Z ZZ
0
t
≤ |Du |2ρdxdy+c |Du|2ρdxdydτ
ZZ 0 Z ZZ
0
t t
+2 (f u +f u )ρdxdydτ −2δ1/2 f [(u ρ) +u ρ]dxdydτ,
Z ZZ 0x x 0y y Z ZZ 1 x x yy
0 0
(2.8)
where ρ(x) is the same function as in inequality (2.4).
7
Proof. In the smooth case multiplying (2.1) by −2 u (t,x,y)ρ(x) −
x x
2u (t,x,y)ρ(x) and integrating over Σ we obtain an equality(cid:0) (cid:1)
yy
d
(u2 +u2)ρdxdy+ (3u2 +4u2 +u2 )ρdxdy
dtZZ x y ZZ xx xy yy
+2δ (u2 +2u2 +u2 )ρdxdy− (u2 +u2)(ρ′′′+δρ′′)dxdy
ZZ xx xy yy ZZ x y
=2 (f u +f u )ρdxdy−2δ1/2 f ((u ρ) +u ρ)dxdy.
ZZ 0x x 0y y ZZ 1 x x yy
The rest part of the proof is similar to the end of the proof of Lemma 2.3. (cid:3)
Lemma 2.6. Let the hypothesis of Lemma 2.5 be satisfied for some δ > 0. Con-
sider thesolution u∈X1,α(Π ) constructedtheresuch that |D2u|∈L (0,T;Lα).
T 2 2
Let ρ(x) be the same function as in inequalities (2.4) and (2.8), and, more-
e
over, ρ(x) ≥ 1. Let the function g(u) ∈ C2(R), g(0) = 0, be such that
|g′(u)|,|g′′(u)| ≤ const ∀u ∈ R. Then for any t ∈ (0,T] the following equal-
ity holds:
t
−2 g∗ u(t,x,y) ρ(x)dxdy+2 g′(u)u ·(u +u )ρdxdydτ
ZZ Z ZZ x xx yy
(cid:0) (cid:1) 0
t t
+2 g(u)·(u +u )ρ′dxdydτ −2δ g′(u)·(u2 +u2)ρdxdydτ
Z ZZ xx yy Z ZZ x y
0 0
t t
−2δ g(u)u ρ′dxdydτ =−2 g∗(u )ρdxdy−2 fg(u)ρdxdydτ
Z ZZ x ZZ 0 Z ZZ
0 0
(2.9)
(recall that g∗(u)≡ ug(θ)dθ).
0
R
Proof. . In the smooth case multiplying (2.1) by −2g (u(t,x,y) ρ(x) and inte-
grating one instantly obtains equality (2.9). (cid:0) (cid:1)
In order to obtain this equality in general case we use the passage to the limit.
In this case, the presence of the terms of order higher than quadratic requires
appropriatejustificationforthisprocedure. Notethat |g(u)|≤c|u|, |g∗(u)|≤cu2,
|g∗(u)−g∗(v)|≤c(|u|+|v|)|u−v|, |g(u)−g(v)|≤c|u−v|, |g′(u)−g′(v)|≤c|u−v|.
Then, for example, if we denote by u the corresponding smooth solution
h
t
|g′(u)−g′(u )|·|u |·(|u |+|u |)ρdxdydτ
Z ZZ h x xx yy
0
t
≤c |u−u |·|u |·(|u |+|u |)ρ3/2dxdydτ
Z ZZ h x xx yy
0
1/4 t 1/2 1/2
≤c sup |u−u |4ρ2dxdy u4ρ2dxdy dτ
1τ∈[0,t](cid:16)ZZ h (cid:17) (cid:16)Z0 (cid:16)ZZ x (cid:17) (cid:17)
t 1/2
× |D2u|2ρdxdydτ .
(cid:16)Z0 ZZ (cid:17)
Since uρ1/2,|Du|ρ1/2 ∈ C([0,T];L ), then uρ1/2 ∈ C([0,T];L ) according to
2 4
inequality (1.7) (for q =4, ρ =ρ =ρ) . Moreover, |D2u|ρ1/2 ∈L (Π ) andso
1 2 2 T
u ρ1/2 ∈L (0,T;L ). Thus, the passage to the limit is justified in this case. The
x 2 4
other terms can be considered in a similar way. (cid:3)
8
Lemma 2.7. Let δ =0, u ρ (x)∈H1 for some α>0 and β >0, f ≡f +
0 α,β 0
f , where f ρ (x) ∈ L (0,T;H1), f ρ (x) ∈ L (Π ). Assume that u
1 0 α,β 1 1 α+1/2,β 2 T 0
and f satisfy the same assumptions as the ones for u , f when y =0,y =L in
0 0
Theorem 1.2. Then there exists a (unique) generalized solution to each of problems
(2.1), (1.2), (1.3) u(t,x,y) such that uρ (x) ∈ C([0,T];H1), |D2u|ρ′ (x) ∈
α,β α,β
L (Π ) and for any t ∈ (0,T] inequality (2.8) holds for ρ ≡ ρ , where there
2 T 2α,2β
isnomultiplier δ1/2 inthelasttermintherightpart, andthereshouldbeapositive
coefficient less than 1 before the second term in the left part.
Proof. Regularizethefunctions u , f , f suchthat u , f satisfythehypothesis
0 0 1 0
ofLemma2.5,where f issubstitutedby f +f ,and f ≡0 (wecanalsoassume
0 0 1 1
that when y = 0,y = L the assumptions on the function f are also true for the
0
function f ). Let us consider the corresponding solution u ∈ X1,α(Π ) and
1 T
correspondinginequality (2.8), where f is substituted by f +f andthere is no
0 0 1 e
term δ1/2f .
1
Since
t
(f u +f u )ρ dxdydτ
(cid:12)Z ZZ 1x x 1y y 2α,2β (cid:12)
(cid:12) 0 (cid:12)
(cid:12) t (cid:12)
= f (u ρ ) +u ρ dxdydτ
(cid:12)Z ZZ 1 x 2α,2β x yy 2α,2β (cid:12)
(cid:12) 0 (cid:2) (cid:3) (cid:12)
t (cid:12) 1/2 t ρ2(cid:12) 1/2
≤ (u2 +u2 +u2)ρ′ dxdydτ f2 2α,2β dxdydτ ,
(cid:16)Z0 ZZ xx yy x 2α,2β (cid:17) (cid:16)Z0 ZZ 1ρ′2α,2β (cid:17)
then according to inequality (2.6) one can finish the proof by the passage to the
limit. (cid:3)
3. Existence of weak solutions
Consider an equation of more general than (1.1) type:
u +u +u −δ(u +u )+(g(u)) =f(t,x,y). (3.1)
t xxx xyy xx yy x
Definition 3.1. Afunction u∈L (0,T;L ) iscalledaweaksolutiontoproblem
∞ 2
(3.1), (1.2), (1.3) if the function g(u(t,x,y))∈L ((0,T)×(−r,r)×(0,L)) for any
1
r>0 andforanyfunction ϕ∈C∞(Π ) suchthat ϕ| =0, ϕ(t,x,y)=0 when
T t=T
|x|≥r forsome r >0 and ϕ| =ϕ| =0 inthecasea, ϕ | =ϕ | =0
y=0 y=L y y=0 y y=L
in the case b, ϕ| =ϕ | =0 in the case c, ϕ| =ϕ| ,ϕ | =ϕ |
y=0 y y=L y=0 y=L y y=0 y y=L
in the case d, the following equality holds:
u(ϕ +ϕ +ϕ +δϕ +δϕ )+g(u)ϕ +fϕ dxdydτ
ZZZ t xxx xyy xx yy x
ΠT(cid:2) (cid:3)
+ u ϕ| dxdy =0. (3.2)
ZZ 0 t=0
Σ
Remark 3.1. It is easy to see that if g ≡0 and a function u is a weak solution to
problem (3.1), (1.2), (1.3), then it is a generalized solution to problem (2.1), (1.2),
(1.3) in the sense of Definition 2.1, though the class of test functions ϕ in that
definition is broader, than the one in Definition 3.1.
Remark 3.2. Ifaweaksolutiontoproblem(3.1),(1.2),(1.3) u∈L (0,T;H1) and
∞
the function g has the rate of growth not higher than polynomial when |u|→∞
9
(for example, it satisfies inequality (1.5) for some b > 0), then according to (1.7)
g(u(t,x,y))∈L (0,T;L ) andthusequality(3.2)alsoholdsforthetestfunctions
∞ 2
ϕ from Definition 2.1.
First of all, we prove a lemma on solubility of problem (3.1), (1.2), (1.3) for
spaces Lα in the ”regularized” case.
2
Lemma 3.1. Let δ > 0, g ∈ C1(R) and |g′(u)| ≤ c ∀u ∈ R. Assume that
u ∈ Lα for some α ≥ 0, f ∈ L (0,T;Lα). Then each of problems (3.1), (1.2),
0 2 1 2
(1.3) has a unique solution u∈C([0,T];Lα)∩L (0,T;H1,α).
2 2
Proof. We apply the contractionprinciple. For t ∈(0,T] define a mapping Λ on
0
a set Yα(Π )=C([0,t ];Lα)∩L (0,t ;H1,α) as follows: u=Λv ∈Yα(Π ) is a
t0 0 2 2 0 t0
solution to a linear problem
u +u +u −δu −δu =f −(g(v)) (3.3)
t xxx xyy xx yy x
in Π with boundary conditions (1.2), (1.3).
t0
Note that |g(v)|≤c|v| and, therefore,
kg(v)kL2(0,t0;Lα2) ≤c||v||L2(0,t0;Lα2) <∞.
Thus, according to Lemma 2.3 (where f ≡ δ−1/2g(v)) the mapping Λ exists.
1
Moreover,for functions v,v ∈Yα(Π )
t0
e 1/2
kg(v)−g(v)kL2(0,t0;Lα2) ≤ckv−vkL2(0,t0;Lα2) ≤ct0 kv−v˜kC([0,t0];Lα2).
As a result, accoerding to inequality (2.e3)
kΛv−Λvk ≤c(T,δ)t1/2kv−vk .
Yα(Πt0) 0 Yα(Πt0)
e e (cid:3)
Now we pass to the proof of Theorem 1.1.
Proof of Theorem 1.1. For h∈(0,1] considerasetofinitial-boundaryvalueprob-
lems in Π
T
u +u +u −hu −hu +(g (u)) =f(t,x,y) (3.4)
t xxx xyy xx yy h x
with boundary conditions (1.2), (1.3), where
u 2signθ
g (u)≡ g′(θ)η(2−h|θ|)+g′ η(h|θ|−1) dθ. (3.5)
h Z0 h (cid:0) h (cid:1) i
Note that g (u)≡g(u) if |u|≤1/h and |g′(u)|≤c(h−1) ∀u∈R.
h h
According to Lemma 3.1 there exists a unique solution to this problem u ∈
h
C([0,T];Lα) ∩ L (0,T;H1,α). Moreover, |g′(u)| ≤ c(1 + |u|b) uniformly with
2 2 h
respect to h.
Next, establish appropriate estimates for functions u uniform with respect to
h
h.
10