Peano Type Theorem for Abstract Parabolic Equations

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www.elsevier.com/locate/anihpc

Peano type theorem for abstract parabolic equations

^✩

Oleg Zubelevich

¹

Department of Differential Equations and Mathematical Physics, Peoples Friendship University of Russia, Ordzhonikidze st., 3, 117198, Moscow, Russia

Received 16 December 2007; received in revised form 30 June 2008; accepted 19 December 2008 Available online 8 January 2009

Abstract

We consider parabolic problems with non-Lipschitz nonlinearity in the different scales of Banach spaces and prove local-in-time existence theorem.

MSC:35K90; 35R05; 35R10

Keywords:Peano theorem; Abstract Cauchy problem; Nonlocal problems; Functional-differential equations; Integro-differential equations;

Quasilinear parabolic equations

1. Introduction

This paper is devoted to quasi-linear parabolic equations with a non-Lipschitz nonlinearity. In the classical setup a quasi-linear initial value parabolic problem has the form

u_t=f

t, u,∇^ku

+Au, u|t=0= ˆu. (1.1)

HereAis a linear elliptic operator of ordernand the term∇^kusymbolizes the derivatives ofuup to orderk. Besides this, Eq. (1.1) must be provided with the boundary conditions.

If the functionuˆbelongs to a suitable space, the mappingf is Lipschitz in a certain sense andk < nthen problem (1.1) has a unique local-in-time solution. This simple observation easily follows from the contracting mapping principle.

We consider the case when the functionf is non-Lipschitz. It is well known that in general situation, in infinitely dimensional Banach space, an initial value problem for differential equation with non-Lipschitz right-hand side does not have solutions [8,18,14]. Nevertheless, as a rule, the initial value problem lives not in a single Banach space but in a scale of Banach spaces and in addition this scale is completely continuous embedded. Such scales for example are the scale of Sobolev spaces, the scale of analytic functions. This observation prompts that to find a solution one should study the problem in the whole scale.

✩ Partially supported by grants RFBR 02-01-00400, ScienSch-691.2008.1.

E-mail address:ozubel@yandex.ru.

1 Current address: 2-nd Krestovskii Pereulok 12-179, 129110, Moscow, Russia.

doi:10.1016/j.anihpc.2008.12.002

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The main mathematical tool we use is a locally convex space version of the Schauder fixed point theorem and theory of scales of Banach spaces. Another approaches to the abstract parabolic problems in the Lipschitz setup contain in [5]. Non-Lipschitz setup has been considered in [7]. There is some extra hypothesis of partial order in the Banach space employed in this article.

2. Main theorem

Consider two scales of Banach spaces{E_s, · ^E_s }s>0and{G_s, · ^G_s }s>0such thatE_s⊆G_sfor alls >0. All the embeddingsE_s₊_δ⊆E_s,δ >0 are completely continuous and

· ^Es · ^Es+δ. (2.1)

The parameters may not necessarily be ran through all the positive real numbers. We do not use the spacesE_s, G_s with bigsand one can assume for example thats∈(0,1). It is just for simplicity’s sake that we considers >0.

Introduce constantsC, T , R >0,φ, α0.

LetS^t:G_s→E_s,t >0 be a strongly continuous linear semigroup in the following sense. For anyu∈E_sone has S^tu−u^E

s →0 ast0 and S^tu^E

s Cu^Es.

Definition 1.The semigroupS^t is said to be parabolic if there exists a constant γ >1 such that for anyδ, t >0, δ^γ< t < T we have

S^tu^E

s+δC

t^φu^G_s . (2.2)

LetBs(r)be an open ball of the spaceEs with radiusr and center at the origin. Suppose a functionf:(0, T] × Bs+δ(R)→Gs to be continuous and such that if(s+δ)^γ < tT andu∈Bs+δ(R)then the following inequality holds

f (t, u)^G

s C

δ^α. (2.3)

Remark 1.A case when f (t, u)^G

s C

t^βδ^α, β >0,

is rather usual but sinceδ^γ < tthis case reduces to (2.3):C/(t^βδ^α)C/δ^βγ⁺^α.

We proceed with two setups of our problem. The first one is a classical setup and we find classical solutions and the second one is a generalized setup to obtain generalized solutions.

In the generalized setup we are looking for solutions to the following integral equation u(t )=

t

0

S^t⁻^ξf ξ, u(ξ )

dξ. (2.4)

In the classical setup we make several additional assumptions. Namely, suppose thatG_s =E_s. Introduce a linear operatorA:E_s₊_δ→E_s and assume that the semigroupS^t is generated by this operator:S^t=e^At such that for any u∈Es+δwe have

hlim→0+

1 h

e^Ah−idEs+δ

u−Au ^E

s

=0. (2.5)

In the classical setup our problem has the form

u_t=f (t, u)+Au, (2.6)

u|t=0=0. (2.7)

The sense of initial condition (2.7) will be clear in the sequel.

Now we give a definition.

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Definition 2.We shall say that problem (2.6) or (2.4) is parabolic if the semigroupS^t is parabolic and χ=φ+α

γ <1.

In case of Remark 1χ=φ+β+α/γ .

Let a spaceE¹(T ),T >0, be given by the formula E¹(T )=

0<s^γ<τ <T

C¹

(τ, T ), E_s

. (2.8)

This space consists of all functionsu that map any numbert ∈(0, T )to the element u(t )∈

0<s^γ<tEs and the restrictionu|(τ,T )belongs to the spaceC¹((τ, T ), Es)for alls∈(0, τ^1/γ).

Theorem 1.

1. Classical setup. Suppose that problem (2.6) is parabolic. Then there exists a constant T_∗>0 such that this problem has a solutionu(t )∈E¹(T_∗),and for any constantc∈(0,1)one has

u(t )^E

ct^1/γ →0 ast0. (2.9)

The functionu(t )also solves Eq.(2.4).

2. Generalized setup. Suppose that problem(2.4)is parabolic. Then there exists a constantT_∗>0 such that this problem has a solution

u(t )∈E(T_∗)=

0<s^γ<τ <T_∗

C

(τ, T_∗), E_s .

In both cases the constantsT_∗depends only onC, α, γ , φ.

The proof of Theorem 1 contains in Sections 3, 4.

Then we apply our theorem to some functional differential equation. To compare our result with the known one we also consider the Navier–Stokes equation.

IfAis the classical Laplace operator and the parabolic equation is considered in a suitable domain thenγ=2 and the inequality from formula (2.8) takes the form 0< s²< τ.

The parameter s symbolizes a spatial variable, so that this inequality specifies the parabolic domain in the plane(τ, s). This endows the term “parabolic equation” with the new sense.

Let us remark that ifG_s =E_s=R^m, · ^E_s = | · |,s >0 andA=0 then Theorem 1 generalizes classical Peano’s theorem to the case when the right side of the equation satisfies (2.3) withs=δ=(t /3)^1/γ,α=0.

3. Preliminaries on functional analysis

In this section we collect several facts from functional analysis. These facts will be useful in Section 4 when we prove Theorem 1.

Consider the spaces C

[τ, T], E_μτ1/γ

, 0< μ <1, 0< τ < T ,

with standard norms. Now we construct the projective limit of these spaces. Define a spaceE(T )as follows

E(T )=

0<μ<1

0<τ <T

C

[τ, T], E_μτ1/γ

.

There is another equivalent definition of the spaceE(T ):

E(T )=

0<s^γ<τ <T

C

[τ, T], E_s .

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Being endowed with a collection of seminorms uτ,μ= max

τξT

u(ξ )^E

μτ^1/γ, u∈E(T ), (3.1)

the spaceE(T )becomes a locally convex topological space.

These seminorms obviously satisfy the following inequalities

uτ,μuτ,μ+δ, δ >0, (3.2)

uτ,rμur^γτ,μ, 0< r1. (3.3)

Indeed, formula (3.2) follows from (2.1) directly. Formula (3.3) is a result of the estimate uτ,rμ= max

τξT

u(ξ )^E

μ(r^γτ )^1/γ max

r^γτξT

u(ξ )^E

μ(r^γτ )^1/γ = ur^γτ,μ.

Formulas (3.2), (3.3) imply that the spaceE(T )is first countable: the topology of this space can be defined by the seminorms (3.1) only withμ, τ∈Q.

Recall the Arzela–Ascoli theorem [17]:

Theorem 2.LetH⊂C([0, T], X)be a set in the space of continuous functions with values in a Banach spaceX.

Assume that the set H is closed, bounded, uniformly continuous and for every t∈ [0, T] the set{u(t )∈X} is a compact set in the spaceX. Then the setHis a compact set in the spaceC([0, T], X).

Now we shall establish an analogue of this result.

Proposition 1.Suppose that a setK⊂E(T )is closed. ThenKis a compact set if the following two conditions are fulfilled.

The setKis bounded.

For anyε >0and for anyτ ∈(0, T ),μ∈(0,1)there exists a constantδ >0such that ift, t∈ [τ, T],|t−t|< δ then

sup

u∈K

u(t)−u(t)^E

μτ^1/γ < ε.

(This means thatKis a uniformly continuous set.) First prove a lemma.

Lemma 1.Let{vj} ⊆Kbe a sequence. Then for anyτ ∈(0, T )the sequence{vj}contains a subsequence that is convergent in all the norms · τ,μ,μ∈(0,1).

Proof. Indeed, take an increasing sequenceμk→1,μ1>0 and fix any value ofτ∈(0, T ). Since the sequence{vj} is bounded and uniformly continuous inC([τ, T], E_μ₂_τ1/γ)then by Theorem 2 it contains a subsequence{v¹_j}that is convergent inC([τ, T], E_μ

1τ^1/γ).

Further since the sequence{v_j¹}is bounded and uniformly continuous inC([τ, T], E_μ

3τ^1/γ)one can pick a subsequence{v²_j} ⊆ {v¹_j}such that the sequence{v_j²}is convergent inC([τ, T], E_μ

2τ^1/γ), etc.

By inequality (3.2) the diagonal sequence{v_j^j}converges in all the norms · τ,μ,μ∈(0,1)with this fixedτ. 2 Proof of Proposition 1. A setP =Q∩(0, T )is countable. So we can number its elements as followsP = {τ_i}i∈N.

We must show that any sequence{u_j} ⊆Kcontains a convergent subsequence{u_j_k}.

By Lemma 1 there is a subsequence{u¹_j} ⊆ {uj}that is convergent in all the norms · τ₁,μμ∈(0,1).By the same argument there is a subsequence{u²_j} ⊆ {u¹_j}that is convergent in all the norms · τ₂,μμ∈(0,1), etc.

The diagonal sequence{u^j_j}is convergent in all the norms · τ_k,μ,k∈N,μ∈(0,1).

By inequality (3.3) the sequence{u^j_j}is convergent in all the norms · τ,μ,τ∈(0, T ),μ∈(0,1).

Proposition 1 is proved. 2

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Lemma 2. Let X, Y be Banach spaces. Suppose thatA_a:X→Y, a> a >0 is a collection of bounded linear operators such that for eachx∈Xwe have

sup

a>a>0

A_axY <∞, A_axY →0 asa→0.

Then for any compact setB⊂Xit follows that sup

x∈BA_axY →0 asa→0.

This result is a direct consequence of the Banach–Steinhaus theorem [17,9].

Let us recall a generalized version of the Schauder fixed point theorem.

Theorem 3.(See [6].) LetW be a closed convex subset of the locally convex spaceE. Then a compact continuous mappingf:W→W has a fixed pointuˆ i.e.f (u)ˆ = ˆu.

4. Proof of Theorem 1 By definition put

W (T_∗)=

u∈E(T_∗)| uτ,νR, 0< τ < T_∗,0< ν <1 . The constantT_∗>0 will be defined.

First we find a fixed point of a mapping F (u)=

t

0

S^t⁻^ξf ξ, u(ξ )

dξ.

This fixed point is the generalized solution announced in the second part of the theorem. Then by using formula (2.5) we show that this fixed point is the desired solution to problem (2.6).

Lemma 3.If the constantT_∗is small enough then the mappingF takes the setW (T_∗)to itself.

Proof. Let constants t, s be taken as follows 0< s < t^1/γ,t T_∗. Supposeu∈W (T_∗) then estimate a function v(t )=F (u):

v(t )^E

s

t

0

S^t⁻^ξfξ, u(ξ )^E

s dξ=X+Y, (4.1)

here we use the notation X=

t−s^γ

0

s dξ, Y =

t

t−s^γ

s dξ.

To estimateXtake constantsεandμsuch that 0< ε < s

t^1/γ < μ <1. (4.2)

The constantεis assumed to be small and the constantμis assumed to be close to 1.

Let the variablesδandδbe given by the formulas δ=s−εξ^1/γ, δ=ξ^1/γ(μ−ε).

Taking into account thatξ ∈(0, t−s^γ]we see that the variablesδ, δare positive and

s−δ >0, s−δ+δ< ξ^1/γ, δ < (t−ξ )^1/γ. (4.3)

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The inequality in the middle implies that

u(ξ )∈B_s₋_δ₊_δ(R) (4.4)

and thus the termXis estimated as follows XC

t−s^γ

0

(t−ξ )⁻^φfξ, u(ξ )^G

s−δdξC²

t−s^γ

0

1 δ^α(t−ξ )^φdξ

C²

(μ−ε)^α

t−s^γ

0

dξ (t−ξ )^φξ^α/γ

ξ=yt

= C²t¹⁻^χ (μ−ε)^α

1−s^γ/t

0

dy (1−y)^φy^α/γ

C²J t¹⁻^χ

(μ−ε)^α, J= 1 0

dy

(1−y)^φy^α/γ. (4.5)

We shall estimate the termY.

Introduce a functionψby the formula ψ (y)=y^1/γ+(1−y)^1/γ−1.

The functionψis positive on the interval(0,1). Define a constantI as follows I=

1 0

dy (1−y)^φ(ψ (y))^α.

Let the constantμbe as above. We redefine the variablesδ, δby the formulas δ=μ(t−ξ )^1/γ, δ=μξ^1/γ+δ−s.

Now the variableξ belongs to the interval[t −s^γ, t]and thus the variables δ, δ are positive and satisfy inequalities (4.3).

It is only not trivial to show that the variableδis positive. Let us prove this. Indeed, δ=μξ^1/γ +μ(t−ξ )^1/γ −s=t^1/γ

μy^1/γ +μ(1−y)^1/γ − s t^1/γ

, (4.6)

recall thaty=ξ /t.From (4.6) it follows that

δ> t^1/γμψ (y). (4.7)

By the same argument as above, inclusion (4.4) is fulfilled with the newδandδ. We are ready to estimate the termY. By (4.7) it follows that

Y C t

t−s^γ

(t−ξ )⁻^φfξ, u(ξ )^G

s−δdξC² t

t−s^γ

dξ (t−ξ )^φδ^α

C²t¹⁻^χ μ^α

1 1−s^γ/t

dy

(1−y)^φ(ψ (y))^α C²I

μ^α t¹⁻^χ. (4.8)

Now the assertion of the lemma follows from formulas (4.1), (4.5) and (4.8). 2

Corollary 1.Formulas(4.5), (4.8)imply that if0< s^γ< tT_∗andv(t )=F (u),u∈W (T_∗)then v(t )^E

s c₂t¹⁻^χ,

herec₂is a positive constant independent onu, t, s.

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Lemma 4.The setF (W (T_∗))is precompact inE(T_∗).

Proof. By Proposition 1 it is sufficient to prove that the setF (W (T_∗))is uniformly continuous.

Take a functionu∈W (T_∗)and letv(t )=F (u). We must show that ift, tτ,τ ∈(0, T_∗)then for anyμ∈(0,1) one has

sup

u∈W (T_∗)

v(t)−v(t)^E

μτ^1/γ →0, as|t−t| →0.

Indeed, for definiteness assume thatt> tthen

v(t)−v(t)=

t

S^t⁻^ξf (ξ, u) dξ+

S^t⁻^t−idE_s

^t

0

S^t⁻^ξf (ξ, u) dξ, s^γ< τ. (4.9) Choose a positive constantδsuch that(s+δ)^γ< τ and using the parabolicity of the semigroupS^t estimate the first term from the right side of this formula

t

S^t⁻^ξf (ξ, u) dξ

E

s

C

t

(t−ξ )⁻^φf (ξ, u)^G

s−δdξ

C²

t

dξ

δ^α(t−ξ )^φ= C²

δ^α(1−φ)(t−t)¹⁻^φ. So that the first term in the right side of (4.9) is vanished uniformly.

Consider a set

U=

τtT_∗

t

0

S^t⁻^ξf (ξ, u) dξu∈W (T_∗)

.

By Lemma 3 the setUis bounded in any spaceE_μ_τ1/γ with 1> μ> μthus it is compact inE_μτ1/γ. By Lemma 2 we get

sup

w∈U

S^t⁻^tw−w^E

μτ^1/γ →0, ast−t→0.

This shows that the second term in the right side of formula (4.9) is vanished uniformly. 2 Corollary 2.The setF (W (T_∗))is uniformly continuous with respect to the variablet.

Lemma 5.The mappingF:W (T_∗)→W (T_∗)is continuous with respect to the topology of the spaceE(T_∗).

Proof. Suppose a sequence{v_l} ⊂W (T_∗)to be convergent to the elementv∈W (T_∗)asl→ ∞.We need to show that for anys^γ < τ < T_∗the sequence

sup

τtT_∗

t

0

S^t⁻^ξf

ξ, vl(ξ ) dξ−

t

0

S^t⁻^ξf ξ, v(ξ )

dξ

E

s

vanishes asl→ ∞.

By Corollary 2 the sequence t

0

S^t⁻^ξf

ξ, v_l(ξ ) dξ

(4.10)

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is uniformly continuous on the interval [τ, T_∗]. The uniform convergence of such a sequence is equivalent to its pointwise convergence [17]. Thus it is sufficient to prove that sequence (4.10) is convergent inE_s for eacht∈ [τ, T_∗].

Fixt∈ [τ, T_∗]and let constantsε, μsatisfy inequality (4.2). Then using the argument of Lemma 3 write

t

0

S^t⁻^ξ f

ξ, vl(ξ )

−f

ξ, v(ξ ) dξ

E

s

C

t−s^γ

0

(t−ξ )⁻^φf

ξ, vl(ξ )

−fξ, v(ξ )^G

εξ^1/γdξ

+C t

t−s^γ

(t−ξ )⁻^φf

ξ, v_l(ξ )

−fξ, v(ξ )^G

s−μ(t−ξ )^1/γdξ. (4.11)

Since the functionf is continuous, for a fixedξ we have:

(t−ξ )⁻^φf

ξ, vl(ξ )

−fξ, v(ξ )^G

εξ^1/γ →0, ξ∈

0, t−s^γ , (t−ξ )⁻^φf

ξ, vl(ξ )

−fξ, v(ξ )^G

s−μ(t−ξ )^1/γ →0, ξ∈

t−s^γ, t , asl→ ∞.

Moreover by formulas (4.5), (4.8) both of these expressions are majorized with theL¹-integrable function:

(t−ξ )⁻^φf

ξ, vl(ξ )

−fξ, v(ξ )^G

εξ^1/γ (t−ξ )⁻^φf

ξ, vl(ξ )^G

εξ^1/γ+fξ, v(ξ )^G

εξ^1/γ

2C²

(μ−ε)^αξ^α/γ(t−ξ )^φ, and

(t−ξ )⁻^φf

ξ, v_l(ξ )

−fξ, v(ξ )^G

s−μ(t−ξ )^1/γ 2C²

t^α/γμ^α(ψ (ξ /t ))^α(t−ξ )^φ.

Therefore by the dominated convergence theorem the integrals in the right side of (4.11) are vanished asl→ ∞. 2 So by Theorem 3 and Lemmas 3–5 we obtain a fixed point of the mappingF, sayu:

F (u)=u∈W (T_∗).

This proves the second part of Theorem 1.

To prove the first one let us show that this fixed point is the solution to problem (2.6). Suppose thatt, t+h > s^γ. First consider the caseh >0. Differentiate the functionu(t )explicitly:

u_t(t )=lim

h→0h⁻¹ t+h

0

e^A(t⁺^h⁻^{ξ )}f ξ, u(ξ )

dξ− t

0

e^A(t⁻^{ξ )}f ξ, u(ξ )

dξ

=lim

h→0h⁻¹

t+h

t

e^A(t⁺^h⁻^{ξ )}f ξ, u(ξ )

dξ+lim

h→0h⁻¹

e^Ah−idEs

^t

0

e^A(t⁻^{ξ )}f ξ, u(ξ )

dξ. (4.12)

Lemma 3 implies that_t

0e^A(t⁻^{ξ )}f (ξ, u(ξ )) dξ∈E_s withs^γ < s^γ < t, t+hhence formula (2.5) gives h⁻¹

e^Ah−idE_s

^t

0

e^A(t⁻^{ξ )}f ξ, u(ξ )

dξ→A t

0

e^A(t⁻^{ξ )}f ξ, u(ξ )

dξ (4.13)

inEsash→0.

Let us prove that h⁻¹

t+h

t

e^A(t⁺^h⁻^{ξ )}f ξ, u(ξ )

dξ→f t, u(t )

(4.14) inE_sash→0.

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Indeed, observe that h⁻¹

t+h

t

e^A(t⁺^h⁻^{ξ )}f ξ, u(ξ )

dξ−f t, u(t )

=h⁻¹ t+h

t

e^A(t⁺^h⁻^{ξ )} f

ξ, u(ξ )

−f t, u(t )

dξ+

t+h

t

e^A(t⁺^h⁻^{ξ )}−idE_s

f t, u(t )

dξ

.

The first integral in the right side of this formula is estimated as follows:

t+h

t

e^A(t⁺^h⁻^{ξ )} f

ξ, u(ξ )

−f t, u(t )

dξ

E

s

Ch max

tξt+h

f ξ, u(ξ )

−ft, u(t )^E

s =o(h).

Since the semigroupe^At is strongly continuous for the second integral we get

t+h

t

e^A(t⁺^h⁻^{ξ )}−idEs

f t, u(t )

dξ

E

s

h max

tξt+h

e^A(t⁺^h⁻^{ξ )}−idEs

ft, u(t )^E

s =o(h).

Ifh <0 then instead of formula (4.12) one must use the following expression u_t(t )=lim

h→0h⁻¹

idEs−e⁻^Ah^t⁺^h

0

e^A(t⁺^h⁻^{ξ )}f ξ, u(ξ )

dξ− t

t+h

e^A(t⁻^{ξ )}f ξ, u(ξ )

dξ

.

In this case only the proof of the formula

hlim→0h⁻¹

idEs−e⁻^Ah^t⁺^h

0

e^A(t⁺^h⁻^{ξ )}f ξ, u(ξ )

dξ=A t

0

e^(t⁻^{ξ )}f ξ, u(ξ )

dξ

differs from the previous argument.

Let us prove this formula. Obviously we have idEs−e⁻^Ah^t⁺^h

0

e^A(t⁺^h⁻^{ξ )}f ξ, u(ξ )

dξ=

idEs −e⁻^Ah u(t )+

idEs−e⁻^Ah

u(t+h)−u(t )

. (4.15) The set

V =

u(t+h)−u(t ) u(t+h)−u(t )^E_s

h∈(h,0)

withh<0 close to zero is bounded inE_s,s^γ < s^γ< t+h. ConsequentlyV is a compact set inEs. By Lemma 2 the set

(A₋h−A)V , A₋h=1 h

idE_s−e⁻^Ah is bounded inE_s and thus the setA₋_hV is also bounded.

Thus taking into account that the functionu(t )is continuous we yield 1

h

idE_s−e⁻^Ah

u(t+h)−u(t )

E s

=u(t+h)−u(t )^E

s · A₋h

u(t+h)−u(t ) u(t+h)−u(t )^E_s

^E

s

=o(1).

For the second term of the right side of (4.15) this implies idEs−e⁻^Ah

u(t+h)−u(t )^E

s =o(h).

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The first term of the right side of formula (4.15) is estimated as follows idEs −e⁻^Ah

u(t )−hAu(t )^E

s =o(h).

Substituting formulas (4.13) and (4.14) to (4.12) we see that the functionuis a solution to Eq. (2.6).

Formula (2.9) follows from Corollary 1.

Theorem 1 is proved.

5. Applications

In the sequel we denote all the inessential positive constants by the same letterc.

5.1. Parabolic equation with gradient nonlinearity In this section we consider a model example.

LetM⊂R^mbe a bounded domain with smooth boundary∂M.

Consider the following equation

u_t=f (∇u)+u, u|t=0= ˆu∈H₀^1,q(M), u(t, ∂M)=0, t >0, (5.1) hereq >1.

The functionf is continuous inR^m and for allz∈R^m we have|f (z)|c(|z|^p+1),qp1. Note that the functionf may not necessarily be a Lipschitz function.

Let us show that if

m(p−1) < q (5.2)

then problem (5.1) has a generalized solution fromC([0, T], H₀^1,q(M)),the constantT >0 depends onu.ˆ

If the functionf is a Lipschitz function then inequality (5.2) is well known: it corresponds to the subcritical case in the sense of Fujita.

After the change of the unknown functionu=e^tuˆ+vour problem takes the form v_t=g(t, x,∇v)+v, v|t=0=0, g(t, x,∇v)=f

∇

e^tuˆ+v

. (5.3)

Introduce the following fractional order spacesX^s,p=(−)⁻^s/2(L^p(M)). Here by(−)⁻¹hwe denote the solution to the problem

u= −h, u(∂M)=0.

The properties of the spacesX^s,pare described in [2–4,15]. See also [5] and references therein. Particularly the spaces X^s,p satisfy almost all embedding theorems which the Sobolev spacesH^s,p(M)do;X^k,p=H^k,p(M)∩H₀^1,p(M), k∈N; and the following semigroup lemma is fulfilled.

Lemma 6.Lete^tustands for the solution to initial value problem v_t=v, v|t=0=u.

Regarding boundary conditions we assume either M⊂R^m is a bounded domain with smooth boundary and then v(t, ∂M)=0, orMis anm-dimensional smooth compact manifold without boundary.

Upon these assumptions one has an estimate e^tuX^l,a(M) c

t^μuX^h,b(M), μ=m(1/b−1/a)+l−h

2 . (5.4)

The real numbersa, b, h, l are such thatlh,1< ba. Inequality(5.4)holds for allt belonging to some interval (0,t˜], constantcdoes not depend onuandt.

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Consider problem (5.3) in the scales

E_s=X¹⁺^s⁰⁺^s,q(M), · s= · _X¹+s0+s,q(M), s∈(0, S), and

G_s=X⁻^λ,q(M), · ^G= · X⁻^λ,q(M),

this means that all the spacesG_scoincide with each other, the constantsS >0, s00 and 0λ < m(1−1/q)to be defined.

Introduce a constant r= qm

m+λq∈(1, q]. It is easy to see that

L^r(M)⊂X⁻^λ,q(M). (5.5)

Indeed, like in the case of Sobolev spaces [1], for the fractional order spaces we have X₀^λ,q(M)⊂L

mq

m−λq(M), 1 q +1

q =1.

For conjugated spaces this gives L^r(M)⊂X⁻^λ,q(M), 1

r +m−λq mq =1.

Then using formula (5.5) estimate the functiong:

g(t, x,∇v)^Gcg(t, x,∇v)

L^r(M)c∇

e^tuˆ+v^p

L^pr(M)+1 ce^tuˆ^p

X^1,pr(M)+ v^p_X1,pr(M)+1

. (5.6)

Choose a constants₀as follows s0=m

1 q − 1

rp

.

Then the conditionX¹⁺^s⁰^,q(M)⊆X^1,pr(M)is satisfied.

Here we assume that the constantλis such that we haveq < rp. By Lemma 6 one has e^tuˆ^p

X^1,pr(M)ct⁻^β ˆu^p_X1,q(M), β=m 2

p q −1

r

.

Ifv∈B_s= {h∈E_s| h^E_s 1}then by all these argument formula (5.6) implies g(t, x,∇v)^G c

t^β. Another inequality we need is

e^tw

sct⁻^φw^G, φ=1+s₀+s+λ

2 ,

this formula also follows from Lemma 6.

Proposition 2.The mapping(t, v)→g(t, x,∇v)is a continuous mapping of(0, T )×B_s toG_s.

Proof. Assume the converse: there exists a sequence(t_k, v_k)such thatt_k →t∈(0, T ),v_k →v inE_s as k→ ∞, v, v_k∈B_s and

g_k(x)−g(x)^Gc >0, (5.7)

here we putg_k(x)=g(t_k, x,∇v_k),g(x)=g(t, x,∇v).

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By the argument above formula (5.7) imply g_k(x)−g(x)

L^r(M)c >0.

Since∇v_k→ ∇vinL^pr(M)then there exists a subsequence{v_k} ⊆ {v_k}such that∇v_k→ ∇valmost every where inM. Thus|g_k(x)−g(x)|^r→0 almost everywhere inM. Consequently|g_k(x)−g(x)|^r→0 in measure.

It remains to show that the sequence |g_k(x)−g(x)|^r is uniformly integrable. If we do this then by the Vitali convergence theorem [11,10] it follows thatg_k(x)−g(x)L^r(M)→0 and this contradiction proves the proposition.

Note that since v_k, v ∈E_s, s >0, we actually have ∇v_k → ∇v, in L^pr⁺^σ(M) with small σ >0. Thus the functions g_k(x)−g(x) belong not only to L^r(M) but also to L^r⁺^ε(M) with small ε >0 and the sequence g_k(x)−g(x)L^r⁺^ε(M) is bounded (these observations follow from the same argument as above). The last observation can be rewritten as follows:

sup

k

M

g_k(x)−g(x)^ræg_k(x)−g(x)dx=sup

k

g_k(x)−g(x)^r⁺^ε

L^r+ε(M)<∞

with æ(y)=y^ε. Since the function æ is monotone and unbounded inR+, this proves the uniform integrability of the sequence|g_k(x)−g(x)|^r. 2

Now we see thatα=0 and to apply Theorem 1 we needχ=φ+β <1. It is easy to show that the last inequality follows from (5.2) if only the constantSis sufficiently small and the constantλis chosen to make the expressionpr to be sufficiently close toq.

Indeed, let us rewrite inequality (5.2) asλ₀=m(p−1)/q <1. Ifp=1 then we setλ=1. Otherwise, ifqp >1, we takeλ∈(0, λ0)such that 0< λ₀−λ1.Then

1 r =1

q + λ m<1

q +λ₀ m =p

q, 0<p q −1

r 1.

This implies 0< pr−q1, 0< s₀1, 0< β1,and χ=φ+β < 1+s₀+S+λ₀

2 +β <1 ifS1.

5.2. The scale of analytic functions

LetT^m=R^m/(2πZ)^mbe them-dimensional torus. All the technique developed below can be transferred almost literally to the case of the problem with zero boundary conditions on them-dimensional cube.

Byx=(x1, . . . , xm)denote an element ofR^m. Let

T^ms =

z=x+iy∈C^mx∈T^m, |y| = _m

j=1

|y_j|² ¹

2

< s

be the complex neighborhood of the torusT^m.

Define a set Es,s >0, as followsEs =C(T^ms )∩O(T^ms). HereO(T^ms )stands for the set of analytic functions inT^ms .

The setEsis a Banach space with respect to the normus=max_z_∈Tm

s |u(z)|. By the Montel theorem the embed- dingsEs+δ⊂Es,δ >0 are completely continuous. By definition putE0=C(T^m)and · 0= · C(T^m).

ByH^k,p(T^ms )we denote a subspace ofO(T^ms ), this subspace consists of functionsu(x, y)=u1(x, y)+iu2(x, y) with finite Sobolev norm.

For example, the spaceH¹(T^ms )consists of functionsuwith finite norm u²_H1(T^ms)=

T^ms

u(x, y)²dx dy+

T^ms

m j=1

∂u(x, y)

∂x_j

²dx dy (5.8)