WITH MEASURE DATA

WITH MEASURE DATA

BUI AN TON

Received 22 August 2003

We establish the existence of a unique solution of an initial boundary value prob- lem for the nonstationary Stokes equations in a bounded fixed cylindrical do- main with measure data. Feedback laws yield the source and its intensity from the partial measurements of the solution in a subdomain.

1. Introduction

LetΩbe a bounded open subset ofR³ with a smooth boundary and consider the initial boundary value problem

u−∆u+ (w· ∇)u= ∇p+g(t)µ inΩ×(0, T),

∇ ·u=0 inQ=Ω×(0, T), u=0 on∂Ω×(0, T), u(x,0)=u0(x) inΩ,

(1.1)

wherewis a given solenoidal smooth vector function,µ=(µ1, µ2, µ3) is a Radon measure of bounded variation onΩ, andu0is inL¹(Ω).

Letχbe a given function inL¹(0, T;L¹(G)) whereGis an interior subset ofΩ.

We associate with (1.1) the cost function Jg;µ;u0;τ=

_T

τ

G|u−χ|dx dt. (1.2)

The purpose of this paper is twofold:

(1) to establish the existence of a unique weak solution of (1.1),

(2) to determine the source µ and the intensity of the source g(t) from the partial measurementsχof the solutionuin a fixed subdomainG× (0, T).

Copyright©2003 Hindawi Publishing Corporation Abstract and Applied Analysis 2003:17 (2003) 953–973 2000 Mathematics Subject Classification: 35K20, 49J20, 76B75 URL:http://dx.doi.org/10.1155/S1085337503308012

Nonlinear elliptic boundary value problems with Radon measure data have been the subject of extensive investigations by Betta et al. [1], Boccardo and Gallou¨et [2], Boccardo et al. [3], and others. Parabolic initial boundary value problems in cylindrical domains with Radon measure data were studied by Boc- cardo and Gallou¨et [2], Br´ezis and Friedman [4]. In all the cited works, scalar- valued functions are involved. For vector-valued functions with a constraint im- posed on the solution, the generic truncated function used in Boccardo- Gallou¨et treatment in [2] does not seem applicable. As is well known, for par- tial differential equations with measure data, the lower-order terms give rise to difficulties. In this paper, we consider the nonstationary Stokes equations with measure data, the case of the full-time-dependent Navier-Stokes equations is open. Feedback laws for distributed systems of parabolic initial boundary value problems, with data inL²(Q), were obtained by Popa in [5,6] and for interacting controls by the author in [7].

The existence of a unique weak solution of (1.1) is established inSection 2.

The value function associated with (1.1) and (1.2) is studied inSection 3. Feed- back laws are established inSection 4. The results of this paper seem new.

2. Existence theorem

We denote byM_b(Ω) the set of all Radon measures of bounded variation onΩ, byW^1,p(Ω) the usual Sobolev spaces, and byJ₀^1,p(Ω) the Banach space

J₀^1,p(Ω)=

u:u=

u1, u2, u3

∈W₀^1,p(Ω),∇ ·u=0 inΩ, 1< p <∞, (2.1) with the obvious norm.

Throughout the paper, we assumewto be inC(0, T;C(Ω))∩L²(0, T;J0^1,2(Ω)).

LetᏳbe a compact convex subset of L²(0, T) and letᐁbe a convex subset of Mb(Ω), compact in the vague topology, that is,µ_nMb(Ω)≤C, then there exists a subsequence such thatµ_nk→µin the distribution sense inΩandµ_nMb(Ω)≤C.

Set

Ᏹg;µ;u0

=u0

L¹(Ω)+gL²(0,T)µMb(Ω). (2.2) Let

g,u0,µ,w∈L²(0, T)×L¹(Ω)×Mb(Ω)×C0, T;C(Ω)∩L²0, T;J0^1,2(Ω), (2.3) then there exists{gn,uⁿ0,fⁿ,wⁿ} ∈C0^∞(0, T)×C0^∞(Ω)×C¹(0, T;C0¹(Ω)) with∇ · wⁿ=0 such that

gn,uⁿ₀,fⁿ,wⁿ−→

g,u0,µ,w (2.4)

inL²(0, T)×L¹(Ω)×Ᏸ(Ω)×C(0, T;C(Ω))∩L²(0, T;J₀^1,2(Ω)) and uⁿ0

L¹(Ω)≤u0

L¹(Ω), fⁿ_L1(Ω)≤ µMb(Ω), g_nL2(0,T)≤ gL²(0,T). (2.5) Consider the initial boundary value problem

u_n−∆un+wⁿ· ∇

un= ∇p+gn(t)fⁿ inQ,

∇ ·un=0 inQ=Ω×(0, T), un(x, t)=0 on∂Ω×(0, T),

un(x,0)=uⁿ₀(x) inΩ.

(2.6)

With the approximating system (2.6), we have the following known result.

Lemma2.1. Let

gn,fn,uⁿ₀,wⁿ∈C₀^∞(0, T)×

C^∞₀(Ω)²×C¹0, T;C¹₀(Ω), ∇ ·wⁿ=0.

(2.7) Then there exists a unique solutionunof (2.2) inC¹(0, T;C(Ω))withun∈L²(0, T;

J0^1,2(Ω)∩W^3,2(Ω))andu_n∈L²(0, T;L²(Ω)).

Sinceun∈C(Q), sup_(x,t)∈Q|uj,n(x, t)| =αj,nexists. Denote α_n= inf

1≤j≤3α_j,n. (2.8)

The following decompositions ofL²(Ω) andW0^1,2(Ω) are known:

L²(Ω)=J0(Ω)⊕G(Ω), W0^1,2(Ω)=J0^1,2(Ω)⊕

J0^1,2(Ω)^⊥. (2.9) Thus, forv∈W₀^1,2(Ω), we get

v=v˜+ ˆv, v˜∈J₀^1,2(Ω),vˆ∈

J₀^1,2(Ω)^⊥. (2.10) We deduce from the unique decomposition ofL²(Ω) intoJ0(Ω)mandG(Ω) that

v=v˜+∇q, v˜∈J0^1,2(Ω), vˆ= ∇q∈W0^1,2(Ω). (2.11) Lemma2.2. Let{gn,fn,uⁿ₀,un}be as inLemma 2.1. Then

un

L^∞(0,T;L¹(Ω))+∇un²

L²(D_n¹)≤C1 +|Ω|+Ᏹg;µ;u0

(2.12)

with

D¹_n=

3

j=1

(x, t) : (x, t)∈Q;u_{n, j}(x, t)≤inf1, α_n. (2.13)

The constantCis independent ofn.

Proof. (1) Let

ϕ=











infα_n,1, if infα_n,1< s,

s, if−infαn,1≤s≤infαn,1,

−infαn,1, ifs <−infαn,1.

(2.14)

Withun∈C¹(0, T;C(Ω)) and with∇ ·un=0 in Q, we now consider the test- ing vector functionϕ(un)=(ϕ(un,1), . . ., ϕ(un,3)). It is clear that

D¹_n⊂

3

j=1

suppun, j

(2.15)

is nonempty and ∂Ω×(0, T)⊂∂D_n¹. For eacht∈[0, T], we have ϕun(·, t)∈ W0^1,2(Ω) and using the decomposition (2.11), we write

ϕu_n(·, t)=ϕ˜u_n(·, t)+∇q(·, t) (2.16) with

ϕ˜u(·, t)∈J₀^1,2(Ω), ∇q∈W₀^1,2(Ω), ∀t∈[0, T]. (2.17) It is clear thatϕmapsL²(Ω) intoJ0(Ω) andW₀^1,2(Ω) intoJ₀^1,2(Ω). Since

ϕun(·, t)_L∞(Ω)≤1, (2.18) we obtain, by using the decomposition (2.11),

ϕ˜un(·, t)_L∞(Ω)≤ inf

h_{L1 (Ω)}≤1;∇·h=0

ϕ˜un(·, t),h

≤ inf

h_{L1 (Ω)≤}1;_∇·h₌0

ϕun(·, t),h

≤ϕun(·, t)_L∞(Ω), ∀t∈[0, T].

(2.19)

Hence,

ϕ˜un

L^∞(0,T;L^∞(Ω))≤1. (2.20)

We have

∆un(·, t),ϕ˜un(·, t)=

∆un(·, t),ϕun(·, t)− ∇q

=

∆un(·, t),ϕun(·, t)+ 3 j,k=1

D_ju_n,k, D_jD_kq

=

∆un(·, t),ϕun(·, t)− 3 j,k=1

Dkun,k,∆q

=

∆un(·, t),ϕun(·, t)

(2.21)

as

un(·, t),∇q∈J0^1,2(Ω)∩W^3,2(Ω)×W0^1,2(Ω),

∇ ·

∆un(·, t)=0 fort∈[0, T]. (2.22) Since

Dj

ϕ(v)˜ _L²(Ω)≤Dj

ϕ(v)_L²(Ω)

≤ϕ(v)_L∞(Ω)D_jv_L2(Ω)

≤Djv_L2(Ω) ∀v∈W0^1,2(Ω),

(2.23)

we deduce that

ϕ˜(v)_L∞(Ω)≤1 ∀v∈W0^1,2(Ω). (2.24) Withu_n∈C¹(0, T;C(Ω)),∇ ·u_n=0, and thusu_n∈L²(0, T;J0(Ω)), we get

u_n,ϕ˜un(·, t)=

u_n,ϕun(·, t)− u_n,∇q

=

u_n,ϕun(·, t). (2.25) (2) Set

φ(s)= _s

0ϕ(r)dr. (2.26)

Multiplying (2.6) by ˜ϕ(un), we obtain d

dt

Ωφun(x, t)dx+ 3 j,m=1

Ωϕun, jDmun, j(x, t)²dx +

3 j,k=1

Ωwⁿ_jDjun,kϕ˜un,k

(x, t)dx

=

Ωgn(t)fn,kϕ˜un,k dx.

(2.27)

Sincewⁿ∈J₀^1,2(Ω), we get by integration by parts that 3

j,k=1

Ωwⁿ_jDjun,kϕ˜un,k dx

= − 3 j,k=1

Ωwⁿ_jun,kϕ˜un,k

Djun,k(x, t)dx

= − 3 j,k=1

Ωw_j(x, t)D_j

_un,k(x,t)

0 sϕ˜(s)ds

dx

= 3 k=1

Ωdivwⁿ

_un,k(x,t)

0 sϕ˜(s)ds

dx=0.

(2.28)

It follows that d dt

Ωφun(x, t)dx≤

Ωgnfnϕ˜un(x, t)dx. (2.29) Hence,

Ω

un(x, t)dx≤C

1 +|Ω|+

Ωφun(x, t)dx

≤C

|Ω|+ _t

0

Ωφu0(x)dx

+Ᏹg;µ;u0

≤C1 +|Ω|+Ᏹg;µ;u0 .

(2.30)

The constantC is independent of nand of the data (g;µ;u0). The second

assertion of the lemma is now obvious.

Lemma2.3. Suppose all the hypotheses ofLemma 2.2are satisfied. Then un

L^p(0,T;W0^1,r(Ω))≤C1 +|Ω|+Ᏹg;f;u0

(2.31) withr <1<5/4. The constantCis independent ofnand of the data(g;µ;u0).

Proof. (1) Letψkbe the function

ψ_k(s)=





















1, if 1 +k < s, s−k, ifk≤s≤1 +k, 0, if −k≤s≤k, s+k, if −1−k≤s≤ −k,

−1, ifs≤ −1−k.

(2.32)

(1) Letkbe a positive integer with αn=inf

j

sup

Q

un, j(x, t)

> k, k≤K_n=sup

j

sup

Q

u_{n, j}

.

(2.33)

Let

Bk=

3

j=1

(x, t) : (x, t)∈Q;k≤un, j(x, t)≤1 +k, (2.34)

then

Bk∩



 j

suppun, j



= ∅. (2.35)

Withψ_k(u_n(·, t))∈W₀^1,2(Ω) fort∈[0, T], we use the decomposition (2.11), and as in the first part of the proof ofLemma 2.2, we have

ψ_kun(·, t)=ψ˜_kun(·, t)+∇q (2.36) with

∇q∈ J0^1,2

_⊥

, ψ˜_kun(·, t)∈J0^1,2(Ω) ∀t∈[0, T]. (2.37) Moreover,

ψ˜_kun_{L∞(0,T;L∞(Ω))}≤k, ψ˜_kun_{L∞(0,T;L∞(Ω))}≤1. (2.38) As inLemma 2.2, we have

u_n,ψ_kun

=

u_n,ψ˜_kun

, ∆un,ψ_kun

=

∆un,ψ˜_kun

. (2.39) Taking the pairing of (2.6) with ˜ψ_k(un(x, t)), we obtain

d dt

ΩΦk

un(x, t)dx+ 3 j,m=1

Ωψ_ku_{n, j}(x, t)D_mu_{n, j}(x, t)²dx +

3 j,m=1

Ωwⁿ_j·D_ju_n,mψ˜_ku_n,m(x, t)dx

≤C1 +Ᏹg;µ;u0

,

(2.40)

where

Φk(s)= _s

0ψ_k(σ)dσ. (2.41)

Sincewⁿ∈J0^1,2(Ω), we get by integration by parts that 3

j,m=1

Ωwⁿ_jDjun,mψ˜k

un,m(x, t)dx

= − 3 j,m=1

Ωwⁿ_jun,m ψ˜k

un,m

Djun,mdx

= − 3 j,m=1

Ωwⁿ_jDj

_un,m(x,t)

0 sψ˜k

(s)ds

dx

= 3 m=1

Ωdivwⁿ

_un,m(x,t)

0 sψ˜k

(s)ds

dx=0.

(2.42)

It follows that

ΩΦk

un(x, t)dx +

3 j,m=1

Ω

ψ_ku_n,m(x, t)D_ju_n,m(x, t)²dx≤C1 +|Ω|+Ᏹg;µ;u0

. (2.43) Therefore,

Bk

∇un(x, t)²dx dt≤C1 +|Ω|+Ᏹg;µ;u0

. (2.44)

Letr∈(1,5/4), then an application of the H¨older inequality gives kBk≤

Bk

un, j(x, t)dx dt

≤un

L^4r/3(Bk)Bk^(4r−3)/4r.

(2.45)

Therefore,

B_k≤k^−4r/3u_n^4r/3_L4r/3(Bk). (2.46) An application of the H¨older inequality, together with (2.45) and (2.46), yields

∇u_n^r_Lr(Bk)≤B_k^(2−r)/2∇u_n^r_L2(Bk)

≤Ck^{−2r(2−r)/3}un^2r(2−r)/3

L^4r/3(Bk)

1 +|Ω|+Ᏹg;µ;u0

. (2.47)

(2) We now consider the case α_n=inf

j

sup

Q

u_{n, j}(x, t)

< k. (2.48)

Sinceαn=sup_Q|un,s|< k, suppu_n,s∩

(x, t) :k≤u_n,s(x, t)≤k+ 1= ∅. (2.49) We have two subcases:

(i) infj=s{sup_Q|un, j(x, t)|}> kand we treat exactly as before, (ii) infj=s{sup_Q|u_{n, j}|} =sup_Q|u_n,m|< k.

As above, we only have to consider the case sup

Q

un,r> k, r=m, s, (2.50)

and again, as in the first part, the same argument carries over.

Repeating the argument done before, we obtain

Bk

∇u_n(x, t)²dx dt≤C1 +|Ω|+Ᏹg;µ;u0

. (2.51)

As before, let 1< r <5/4, and an identical argument as in the first part yields ∇un^r

L^r(Bk)≤∇un^r

L²(Bk)B_k^(2−r)/2

≤Ck^{−2r(2−r)/3}un^2r(2−r)/3

L^4r/3(Bk)

1 +|Ω|+Ᏹg;µ;u0

. (2.52)

Combining the two cases, we get from (2.47) and (2.52), ∇un^r

L^r(Bk)≤Ck^{−2r(2−r)/3}un^2r(2−r)/3

L^4r/3(Bk)

|Ω|+Ᏹg;µ;u0

. (2.53)

The constantCis independent ofn,kand ofg,µ,u0.

(2) Letk0be a fixed positive number and letϕnbe the truncated function

ϕ_n(s)₌











infk0, α_n, ifs >infk0, α_n, s, if−infk0, αn

≤s≤infk0, αn

,

−infk0, αn

, ifs <−infk0, αn .

(2.54)

Then as inLemma 2.2, we have

D^k0n

∇un²dx dt≤Ck0+|Ω|+Ᏹg;µ;u0

. (2.55)

Now a proof as in that ofLemma 2.2gives

D^k0n

∇un^rdx dt≤ |Ω|^(2−r)/2∇un^r/2

L²(D^k0n)|Ω|^(2−r)/2

≤Ck0+|Ω|+Ᏹg;µ;u0

,

(2.56)

whereDn^k0is as inLemma 2.2.

(3) Combining (2.53) and (2.56), we obtain ∇un^r

L^r(0,T;L^r(Ω))≤Ck0+ 1 +|Ω|+Ᏹg;µ;u0 +C1 +|Ω|+Ᏹg;µ;u0

×

Kn

k=k0

un^2r(2−r)/3

L^4r/3(Bk) k^{−(2−r)2r/3}.

(2.57)

Applying the H¨older inequality to (2.57), we get ∇un^r

L^r(0,T;L^r(Ω))≤C1 +|Ω|+Ᏹg;µ;u0

+C1 +|Ω|+Ᏹg;µ;u0

×un^4r/3

L^4r/3(0,T;L^4r/3(Ω))





Kn

k=k0

k^{−4(2−r)/3}





r/2

≤C1 +|Ω|+Ᏹg;µ;u0un^4r/3

L^4r/3(0,T;L^4r/3(Ω))

×



 ∞ k=k0

k^{−4(2−r)/3}





r/2

.

(2.58)

The series converges since 1< r <5/4.

(4) Again, the H¨older inequality gives un

L^4r/3(Ω)≤un^1/4

L¹(Ω)un^3/4

L^3r/(3−r)(Ω). (2.59)

With the estimate ofLemma 2.2, we obtain un^2r(2−r)/3

L^4r/3(0,T;L^4r/3(Ω))≤C1 +|Ω|+Ᏹg;µ;u0

+C1 +|Ω|+Ᏹg;µ;u0

×un^r(2−r)/2

L^r(0,T;L^3r/(3−r)(Ω)).

(2.60)

(5) The Sobolev imbedding theorem yields un^r

L^r(0,T;L^3r/(3−r)(Ω))≤Cun^r

L^r(0,T;W0^1,r(Ω)). (2.61)

It follows from (2.57), (2.59), and (2.60) that un^r

L^r(0,T;L^3r/(3−r)(Ω))≤C1 +|Ω|+Ᏹg;µ;u0 +C1 +|Ω|+Ᏹg;µ;u0

×un^4r/3

L^4r/3(0,T;L^4r/3(Ω))





k=k0

k^{−4(2−r)/3}





r/2

≤C1 +|Ω|+Ᏹg;µ;u0

+Cun^r

L^r(0,T;L^3r/(3−r)(Ω))





k=k0

k^{−4(2−r)/3}





r/2

.

(2.62)

Since 1< r <5/4, the series converges and there existsk0such that un^r

L^r(0,T;L^3r/(3−r)(Ω))≤C1 +|Ω|+Ᏹg;µ;u0

. (2.63)

Using the estimate (2.63) in (2.60), we obtain un^2r(2−r)/3

L^4r/3(0,T;L^4r/3(Ω))≤C1 +|Ω|+Ᏹg;µ;u0

_r(4−r)/2

. (2.64)

Taking the estimate (2.64) into (2.57), we have un^r

L^r(0,T;J^1,r0 (Ω))≤C1 +|Ω|+Ᏹg;µ;u0

. (2.65)

The lemma is proved.

Lemma2.4. Suppose all the hypotheses of Lemmas2.1and2.2are satisfied. Then u_nLr(0,T;(J0^1,r(Ω))^∗)≤C1 +|Ω|+Ᏹg;µ;u0

. (2.66)

The constantCis independent ofn,g,u0, andµ.

Proof. Sincer∈(1,5/4), we haveW^1,r/(r−1)(Ω)⊂L^∞(Ω) asΩis a bounded open subset ofR³. Letvbe inL^r/(r−1)(0, T;J0^1,r(Ω)), then we get from (2.2)

u_n,v+ν∇un,∇v+wⁿ· ∇

un,v=

gfn,v. (2.67) Thus,

_T

0

u_n,vdt≤u_nLr(0,T;J0^1,r(Ω))vL^r/(r−1)(0,T;J^1,r/(r−1)(Ω))

+wL^∞(0,T;L^∞(Ω))u_nLr(0,T;J0^1,r(Ω))v_Lr/(r−1)(0,T;J0^1,r/(r−1)(Ω))

+CᏱg;µ;u0

vL^r/(r−1)(0,T;L^∞(Ω)).

(2.68)

It follows from the estimate ofLemma 2.2that _T

0

u_n,vdt≤C1 +|Ω|+Ᏹg;µ;u0

v_Lr/(r−1)(0,T;J0^1,r/(r−1)(Ω)). (2.69) Hence,

u_nLr(0,T;(J0^1,r/(r−1)(Ω))^∗)≤C1 +|Ω|+Ᏹg;µ;u0

. (2.70)

The lemma is proved.

The main result of this section is the following theorem.

Theorem2.5. Let{g,µ,u0}be inL²(0, T)×Mb(Ω)×L¹(Ω)and letwbe in L^∞0, T;L^∞(Ω)∩L²0, T;J₀^1,2(Ω). (2.71) Then there exists a unique solutionuof (1.1). Moreover,

uL^∞(0,T;L¹(Ω))+u_Lr(0,T;J0^1,r(Ω))+u_Lr(0,T;(J0^1,r(r−1)(Ω))^∗)≤CᏱg;µ;u0 (2.72) forr∈(1,5/4). The constantCis independent ofg,µ, andu0.

Proof. Letu_nbe as in Lemmas2.1,2.2, and2.3. Then from the estimates of those lemmas, we get, by taking subsequences,

un,u_n−→

u,u (2.73)

in

L^∞0, T;L¹(Ω)_weak^∗∩L^r0, T;L^r(Ω)∩

L^r0, T;J₀^1,r(Ω)_weak

×

L^r0, T;J0^1,r/(r−1)(Ω)^∗weak. (2.74)

It is now clear thatuis a solution of (1.1) with

u,u∈L^∞0, T;L¹(Ω)∩L^r0, T;J₀^1,r(Ω)×L^r0, T;J₀^1,r/(r−1)(Ω)^∗. (2.75) It is trivial to check that the solution is unique. Since the solution is unique, an application of the closed-graph theorem yields the stated estimate. The constant

Cdepends onΩ.

3. Value function

Letχ be an element ofL¹(0, T;L¹(Ω)) representing the observed values of the solutionuof (1.1) in an interior subdomain G. Let

Ᏻ=

g:gW^1,2(0,T)≤1, ᐁ=

µ:µMb(Ω)≤1. (3.1)