Ammari H., Benkirane A., Touzani A. (editors) Recent Developments in Nonlinear Analysis

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where k(x, t) is a positive function in L

(Q), and α, β are strictly positive

constants. We recall that, for k > 1 and s in R, the truncation T

is deﬁned

as,

(s) =

(

s if s ≤ k

|s|

if |s| > k.

And we deﬁne the function that will be used later ϕ

(r) =

(s)ds.

3. Main results

Consider the problem

∈ L

(Ω), f ∈ L

(Q)

∂u

∂t

− div(a(x, t, Du)) = f in Q

u = 0 on ∂Ω×]0, T [,

u(x, 0) = u

on Ω.

(12)

Deﬁnition 3.1. Let f ∈ L

(Q) and u

∈ L

(Ω). A real-valued function u

deﬁned on Ω × [0, T ] is a renormalized solution of problem (12) if:

u ∈ C([0, T ]; L

(Ω)); (13)

(u) ∈ L

(0, T ; W

1, p

(Ω, w)) for all (k ≥ 0); (14)

∀ c ≥ 0, T

k+c

(u)−T

(u)→0 strongly in L

(0, T ; W

1, p

(Ω, w)) as k →+∞

(15)

∂S(u)

∂t

− div (S

(u)a(x, t, Du)) + S

(u)a(x, t, Du)Du = fS

(u) in D

(Q)

(16)

for all functions S ∈ C

∞

(R) such that S

has a compact support in R (i.e

∈ D(R)),

u(t = 0) = u

. (17)

Remark 1. Equation (16) is formally obtained by pointwise multiplication

of equation (12) by S

(u). Nevertheless, this process cannot be justiﬁed in

general, so that weak solutions are not always renormalized solutions. In

order terms, equation (16) is nothing but

−

S(u)

∂ϕ

∂t

(u)a(Du)Dϕ +

(u)a(Du)Duϕ =

(u)ϕ,

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for all ϕ ∈ C

∞

(Q) and all S ∈ C

∞

(R) with S

∈ C

∞

(R). This formally

corresponds to using in problem (12) the test function ϕS(u), but once

again, this is only formal.

Indeed, if M is such that suppS

⊂ [−M, M ], the following identiﬁcations

are made in (16):

• S(u) belongs to L

∞

(Q) since S is a bounded function.

• S

(u)a(x, t, Du) identiﬁes with S

(u)a(x, t, DT

(u)) a.e in Q. Since

(u) ≤ M a.e in Q and S

(u) ∈ L

∞

(Q), we obtain from (8) and (14)

that

(u)a(x, t, DT

(u)) ∈

i=1

(Q, w

∗

)

• S

(u)a(x, t, Du)Du identiﬁes with S

(u)a(DT

(u))DT

(u)

and S

(u)a(DT

(u))DT

(u) ∈ L

(Q).

• S

(u)f belongs to L

(Q) .

The above considerations show that equation (16) holds in D

(Q) and

∂S(u)

∂t

belongs to L

(0, T ; W

−1, p

(Ω, w

∗

)) + L

(Q). It follows that u belongs to

(0, T ; L

(Ω)) so that the initial condition (17) makes sense.

Theorem 3.1. Let f ∈ L

(Q) and u

∈ L

(Ω). Assume that (H1) and

(H2), there exists a unique renormalized solution u of problem (12) .

Proof of the existence result

Step 1: The approximate problem

Consider the approximate problem:

∈ L

(0, T ; W

1, p

(Ω, w))

∂u

∂t

− div(a(x, t, Du

)) = f

in D

(Q),

(t = 0) = u

(18)

with (f

) be a sequence of smooth functions such that f

= T

(f) and

→ f in L

(Q), and (u

) be a sequence such that u

= T

) and

→ u

in L

(Ω). f

and u

belong, respectively to L

∞

(Q) and

∞

(Ω).

Let X = L

(0, T ; W

1, p

(Ω, w)), X

∗

= L

(0, T ; W

−1,p

(Ω, w

∗

)), H =

(Ω, σ) and W

(0, T, V, H) = {v ∈ X : v

∈ X

∗

The operator A is bounded, demicontinuous, pseudomonone with respect

to D(L) where D(L) = {v ∈ X; v

∈ X

∗

, v(0) = 0} and strongly coercive.

So all conditions of theorem 5 in

are met. Therefore, there exists a solution

December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent

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∈ D(L) of the evolution equation (18) for any f

∈ X

∗

. By the deﬁnition

of D(L) and see,

we obtain

D(L) ⊆ W

(0, T, V, H) ⊆ C([0, T ], H).

Which implies that u

∈ C([0, T ], H).

Step 2: Some estimations about the truncated sequence of solu-

tions.

Using in (18) the test function T

)χ

(0,τ)

, we get, for every τ ∈ [0, T ]

(Since |ϕ

(r)| ≤ k |r|).

∂u

∂t

)dx ≥

Ω

(τ))dx − k ku

(Ω)

Then we deduce that,

Ω

(τ))dx +

Ω

a(x, t, Du

)DT

)dxdt

≤ k(ku

(Ω)

+ kf

(Q)

) ≤ ck.

Thanks to (10) and by the fact that ϕ

(τ)) ≥ 0, we deduce that,

i=1

(x)



∂T

)

∂x



dxdt

≤ ck, ∀k ≥ 1.

(19)

Then, T

) is bounded in L

(0, T ; W

1, p

(Ω, w)), and T

) * v

(0, T ; W

1, p

(Ω, w)). Using the compact imbedding (3.3) we get,

) → v

strongly in L

(Q, σ) and a.e in Q.

Let k > 0 large enough and B

be a ball of Ω, we have,

k meas({|u

| > k} ∩ B

× [0, T ]) =

{|u

|>k}∩B

)|dxdt

≤

)|dxdt

≤ T c

i=1

(x)



∂T

)

∂x



dxdt

≤ ck

So, we have

lim

k→+∞

(meas({|u

| > k} ∩ B

× [0, T ])) = 0.

Consider now a function non decreasing g

∈ C

(R) such that g

(s) =

s for |s| ≤

and g

(s) = k f or |s| ≥ k

Multiplying the approximate equation by g

), we get

∂g

)

∂t

− div(a(x, t, Du

)) + a(x, t, Du

) = f

)

December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent

333

in the sense of distributions. This implies, thanks to the fact g

has compact

support, that g

) is bounded in X, while it’s time derivative

∂g

)

∂t

bounded in X

∗

+ L

(Q).

Hence see Lemma 3.8 in

allows us to conclude that g

) is compact in

loc

(Q, σ). Thus, for a subsequence, it also converges in measure and almost

every where in Q (since we have, for every λ > 0, )

meas({|u

− u

| > λ} ∩ B

× [0, T ]) ≤ meas({|u

| > k} ∩ B

× [0, T ])

+meas({|u

| > k} ∩ B

× [0, T ]) + meas({|g

) − g

)| > λ})

Let ε > 0, then, there exist k(ε) > 0 such that,

meas({|u

− u

| > λ} ∩ B

× [0, T ]) ≤ ε f or all n, m ≥ n

(k(ε), λ, R).

This proves that (u

) is a Cauchy sequence in measure in B

×[0, T ], thus

converges almost everywhere to some measurable function u. Then for a

subsequence denoted again u

, we can deduce from (19) that,

) * T

(u) weakly in L

(0, T ; W

1, p

(Ω, w)) (20)

and then, the compact imbedding (3.3) gives,

) → T

(u) strongly in L

(Q, σ) and a.e in Q.

Which implies, by using (8), for all k > 0 that there exists a function

∈

i=1

(Q, w

∗

), such that

a(x, t, DT

)) * h

weakly in

i=1

(Q, w

∗

) (21)

We now establish that u belongs to L

∞

(0, T ; L

(Ω)). Using (19), (20) and

passing to the limit-inf as n tends to +∞ and we obtain

Ω

(u)(t)dx ≤ k[kfk

(Ω)

+ ku

(Ω)

]

By the deﬁnition of ϕ

, we deduce from the above inequality that

Ω

|u(x, t)|dx ≤

meas(Ω) + k[kfk

(Ω)

+ ku

(Ω)

]

for almost any t in (0,T), which gives that u belong to L

∞

(0, T ; L

(Ω)) .

Step 3: In this step we identify the weak h

in (21) and we prove the

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weak-L

convergence of the ”truncated” energy a(x, t, DT

))DT

)

as n tends to + ∞. We can prove easily the limits

lim

m→+∞

lim sup

n→+∞

{m≤|u

|≤m+1}

a(Du

)Du

dxdt = 0. (22)

We deﬁne for all µ ≥ 0 and all (x, t) ∈ Q,

= µ

∞

˜v(x, s) exp(µ(s − t))ds where ˜v(x, s) = v(x, s)χ

(0,T )

(s).

See proposition 3.1-3.3 in.

Lemma 3.1. Let k ≥ 0 be ﬁxed. Let (T

(u))

the molliﬁcation of T

(u).

Let S be an increasing C

∞

(R)-function such that S(r) = r for |r| ≤ k

and supp S

is compact. Then,

lim

µ→+∞

lim

n→+∞



∂S(u

)

∂t

, (T

) − (T

(u))

)



dtds ≥ 0, (23)

where h., .i denotes the duality pairing between

(Ω) + W

−1,p

(Ω, w

∗

) and L

∞

(Ω) ∩ W

1, p

(Ω, w).

Proof of Lemma 3.1. Let k ≥ 0 be ﬁxed. Let us ﬁrst recall that since

supp S

is compact, we have

S(u

) ∈ X ∩ L

∞

(Q), S(u) ∈ X ∩ L

∞

(Q) and

∂S(u

)

∂t

∈ L

(Q) + X

∗

Moreover, since S is increasing and S(r) = r for |r| ≤ k,

(S(u

)) = T

) and T

(S(u)) = T

(u) a.e. in Q.

As a consequence, (T

(S(u)))

= (T

(u))

a.e. in Q for any µ > 0. It

follows that under the notation v

= S(u

) and v = S(u), we have



∂S(u

)

∂t

, (T

) − (T

(u))

)



dtds



∂(v

)

∂t

, (T

) − (T

(v))

)



dtds



∂(v

− (T

(v))

)

∂t

, (v

− (T

(v))

)



dtds

−



∂(v

)

∂t

, (v

− T

))



dtds

Ω

∂(T

(v))

∂t

− (T

(v))

)dxdtds,

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335

Ω

|(v

− (T

(v))

dxdt −

Ω

|(v

− (T

(v))

(t = 0)dx

−

Ω

− (T

))|

dxdt +

Ω

|(v

− (T

)))|

(t = 0)dx

Ω

∂(T

(v))

∂t

− (T

(v))

)dxdtds,

(24)

since

(s − T

(s))ds =

|r − T

(r)|

To pass to the limit in (24) as n tends to +∞, we ﬁrst observe that since S

is bounded, the deﬁnition of v

and v together with convergence of u

a.e

in Q imply that

→ v strongly in L

(Q) and in L

∞

weak − ∗.

Moreover, the initial condition for u

gives

(t = 0) = S(u

)(t = 0) = S(u

) a.e in Q,

so that the strong convergence of u

to u

in L

(Ω) implies that

(t = 0) → S(u

) in L

(Ω).

Passing to the limit as n tends to +∞ in (24) is an easy task and leads to

lim

n→+∞



∂S(u

)

∂t

, (T

) − (T

(u))

)



dtds

Ω

|(v − (T

(v))

dxdt −

Ω

|(S(u

) − (T

(v))

(t = 0)dx

−

Ω

|v − (T

(v))|

dxdt +

Ω

|S(u

) − T

(S(u

))|

Ω

∂(T

(v))

∂t

(v − (T

(v))

)dxdtds,

(25)

for any µ > 0.

In order to pass to the limit-inf as µ tends to inﬁnity in (25), we now use

the deﬁnition of (T

(u))

, in terms of T

(v), rewrite as

∂(T

(v))

∂t

+ µ((T

(v))

− T

(v)) = 0 in D

(Q), (26)

(v))

(t = 0) = v

in Ω, (27)

(v))

→ T

(v) in L

(Q) (28)

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336

(v))

(t = 0) → T

(S(u

)) in L

(Q) (29)

as µ tends to +∞, since again T

(S(u

)) = T

) a.e in Ω.

In view of (26), (28), (29) passing to the limit-inf as µ tends to +∞ in (25)

leads to

lim inf

µ→+∞

lim

n→+∞



∂S(u

)

∂t

, (T

) − (T

(u))

)



dtds

= lim inf

µ→+∞

Ω

(v) −(T

(v))

)(v − (T

(v))

)dxdtds.

(30)

The proof of lemma 3.1 then reduces to show that the right-hand side of

(30) is nonnegative. To this end we just remark that

(v) −(T

(v))

)(v − (T

(v))

) ≥ 0 a.e in Q.

We prove the following lemma, which is the key point in the monotonicity

arguments that will be developed in undo lemma .

Lemma 3.2. The subsequence of u

satisﬁes for any k ≥ 0

lim sup

n→+∞

Ω

a(DT

))DT

)dxdtds ≤

Ω

(u)dxdtds,

(31)

where h

is deﬁned in (21).

In the following we adapt the above-mentioned method to problem (12)

and we ﬁrst introduce a sequence of increasing C

∞

(R)-functions S

such

that

(r) = r if |r| ≤ m,

suppS

⊂ [−(m + 1), m + 1],

∞

≤ 1, for any m ≥ 1.

Pointwise multiplication of equation (18) by S

) (which is licit) leads

∂S

)

∂t

− div (S

)a(x, t, Du

))

)a(x, t, Du

)Du

= f

) in D

(Q)

(32)

which is nothing but the renormalized of formulation (16) for u

with

S = S

. Remark that (32) implies that

∂S

(u)

∂t

∈ L

(Q) + X

∗

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337

The use of the test function W

= T

) − (T

(u))

, k ≥ 0 in

equation.(32) we obtain upon integration over (0,t) and then over (0,T):



∂S

)

∂t

, W



dtds +

Ω

)a(Du

)DW

Ω

)a(Du

)Du

dxdtds =

Ω

(33)

We pass to the limit in (33) as n tends to +∞, µ tends to +∞ and then m

tends to +∞, the real number k ≥ 0 being ﬁxed. In order to perform this

task we prove below the following results for ﬁxed k ≥ 0 :

lim inf

µ→+∞

lim

n→+∞



∂S

)

∂t

, W



dtds ≥ 0, for any m ≥ k, (34)

lim

m→+∞

lim sup

µ→+∞

lim sup

n→+∞



Ω

)a(Du

)Du

dxdtds



= 0, m ≥ 1

(35)

lim

µ→+∞

lim

n→+∞

Ω

dxdtds = 0 f or any m ≥ 1 (36)

Proof of (34). The function S

belongs to C

∞

(R) and is increasing. We

have for m ≥ k, S

(r) = r for |r| ≤ k while suppS

is compact.

In view of the deﬁnition of W

, lemma 3.1 applies with S = S

for ﬁxed

m ≥ k. As a consequence (34) holds true.

Proof of (35) In order to avoid repetitions in the proofs of (36), let us

summarize the properties of W

. For ﬁxed µ > 0

* T

(u) −(T

(u))

weakly in L

(0, T ; W

1, p

(Ω, w)), as n → +∞



∞

(Q)

≤ 2k, for any n > 0 and f or any µ > 0

we deduce that for ﬁxed µ > 0

→ T

(u)−(T

(u))

a.e in Q and in L

∞

(Q)weak −∗, as n → +∞

one has suppS

⊂ [−(m + 1), −m] ∪ [m, m + 1] for any m ≥ 1. As a

consequence



Ω

)a(Du

)Du



≤ T kS

∞



∞

{m≤|u

|≤m+1}

a(Du

)Du

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for any m ≥ 1, any µ > 0 and any n ≥ 1

it possible to obtain

lim sup

µ→+∞

lim sup

n→+∞



Ω

)a(Du

)Du

dxdtds



≤ C lim sup

n→+∞

{m≤|u

|≤m+1}

a(Du

)Du

dxdt,

for any m ≥ 1, where C is a constant independent of m. Appealing now to

(22) it possible to pass the limit as m tends to +∞ to establish (35).

Proof of (36) Lebesque’s convergence theorem implies that for any µ > 0

and any m ≥ 1

lim

n→+∞

Ω

dxdtds

Ω

(u)(T

(u) −(T

(u)

))dxdtds

Now, for ﬁxed m ≥ 1, using

makes it possible to pass to the limit as

µ → +∞ in the above equality to obtain (36).

We now turn back to the proof of lemma 3.2. Due to (34), (35) and (36),

we are in a position to pass the limit-sup when n tends to +∞, then to the

limit-sup when µ tends +∞ and then to the limit as m tends to +∞ in

(33). Using the deﬁnition of W

we obtain that for any k ≥ 0

lim

m→+∞

lim sup

µ→+∞

lim sup

n→+∞

Ω

)a(Du

)

(DT

) − D(T

(u))

)dxdtds ≤ 0.

Since for k ≤ n and k ≤ m, the above inequality implies that for k ≤ m

lim sup

n→+∞

Ω

a(Du

)DT

)dxdtds

≤ lim

m→+∞

lim sup

µ→+∞

lim sup

n→+∞

Ω

)a(Du

)D(T

(u))

dxdtds

(37)

The right-hand side of (37) is computed as follows. We have for n ≥ m + 1:

)a(Du

) = S

)a(DT

m+1

)) a.e in Q.

Due to the weak convergence of a(DT

m+1

)) it follows that for ﬁxed

m ≥ 1

)a(Du

) * S

m(u

m+1

weakly in

i=1

(Q, w

∗

)

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when n tends to +∞. The strong convergence of (T

(u))

to T

(u) in

(0, T ; W

1, p

(Ω, w)) as µ tends to +∞, then we have

lim

µ→+∞

lim

n→+∞

Ω

)a(Du

)D(T

(u))

Ω

m+1

(u),

(38)

as soon as k ≤ m, S

m(r) = 1 for |r| ≤ m. Now for k ≤ m we have,

a(DT

m+1

))χ

{|u

|<k}

= a(DT

))χ

{|u

|<k}

a.e in Q,

which implies that, passing to the limit as n → +∞,

m+1

{|u

|<k}

= h

{|u

|<k}

a.e in Q − {|u| = k} for k ≤ m. (39)

As a consequence of (39) we have for k ≤ m,

m+1

(u) = h

(u) a.e in Q. (40)

Using (37), (38), (40) we obtain (31) and the proof of lemma 3.2 is complete.

Let k ≥ 0 be ﬁxed. Using (9) we have

lim

n→+∞

Ω

[a(DT

)) −a(DT

(u))][DT

) −DT

(u)]dxdtds ≥ 0.

(41)

By lemma 3.2, weak convergence of T

) and a(DT

)) we pass to the

limit-sup as n → +∞ in (41) and obtain the result :

lim

n→+∞

Ω

[a(DT

)) −a(DT

(u))][DT

) −DT

(u)]dxdtds = 0.

(42)

Lemma 3.3. For ﬁxed k ≥ 0, we have

= a(DT

(u)) a.e in Q, (43)

a(DT

))DT (u

) * a(DT

(u))DT

(u) weakly in L

(Q). (44)

Proof of Lemma 3.3. The proof is standard, from weak convergence of

)) and a(DT

)), (42) and (9) we deduce the result.

Step 4: Convergence of (u

) in C([0, T ], L

(Ω))

Proposition 3.1. The sequence (u

) is a Cauchy sequence in

C([0, T ], L

(Ω)).