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A free boundary problem for the Navier–Stokes equations with measure data
Nonlinear Analysis 64 (2006) 1 – 21 www.elsevier.com/locate/na
A free boundary problem for the Navier–Stokes equations with measure data Bui An Ton∗ Department of Mathematics, University of British Columbia, Vancouver, BC, Canada V6 T1 Z2 Received 17 July 2003; accepted 14 May 2005
Abstract The existence of a weak solution of an initial boundary-value problem for the plane nonstationary Navier–Stokes equations with Radon measure data on the free boundary, is established. The problem may be considered as a model of the blood flow around the heart valves. An inverse problem is studied, it allows us to find the boundary forces acting on the valve from the observed values of the velocity of the fluid in a fixed subregion. 䉷 2005 Elsevier Ltd. All rights reserved. MSC: 35K60; 76D05; 76D27; 76Z05 Keywords: Navier–Stokes; Free boundary; Radon measure; Heart valves; Blood flow
1. Introduction The purpose of this paper is to study a free boundary problem for the plane nonstationary Navier–Stokes equations with measure data. The problem may be considered as a model of blood flow around the heart valves. The main characteristic features of the problem are: • The force field. It is zero everywhere except on the free boundary where it is infinite there but its integral over any finite region containing the free boundary is finite (like a two-dimensional impulse on the free boundary). In the blood flow model, the heart valve ∗ Tel.: +1 604 822 2666; fax: +1 604 822 6074.
E-mail address: [email protected]. 0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2005.05.038
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B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
exerts forces on the fluid (i) forces which stop the flow when the valve is closed and (ii) forces which shear the flow to form vortices when the valve is open. • The free boundary. It is given in terms of Lagrangian coordinates with the fluid velocity being only a weak solution of the Navier–Stokes equations. A model for the flow of blood around the heart valves was developed by Peskin in [5,6]. Approximating the Dirac delta measure by L1 -functions and linearizing the problem through successive approximations, Peskin treated the problem numerically in [5,6]. Let be a bounded open subset of the plane with a smooth boundary and consider the initial boundary-value problem u − u + (u.∇)u = ∇p + (u, F) ∇.u = 0
in × (0, T )
in Q = × (0, T )
u(x, t) = 0 on j × (0, T ),
u(x, 0) = u0 in
(1.1)
with d X(, t) = u(X(, t); t), dt and
T
(u, F), =
(u;t)
0
X(, 0) =
(1.2)
˜ f((t))(X((t); u)) d(t) dt
∀ ∈ C0 (Q).
(1.3)
We denote by X(; t), the position of the particle of the heart valve at the time t, which was at , at t = 0. The material parameter for the free boundary (u, t), describing the heart valve is (t), and ˜ ˜ (u, t) = {X((t); u) : X((t); u) = X(, t; u), ∈ , with () = c}, where () = c is the free boundary at the initial time. We consider boundary forces f with F ∈ L∞ (Q), F|(u,t) = f,
|f((t))| 1,
Mb (Q) 1,
supp (u; F) ⊂ (u; t).
The set of Radon measures in Q is denoted by Mb (Q). Elliptic and parabolic partial differential equations with measure data have been the subject of extensive investigations by Betta et al. [1], Boccardo and Gallouet [2], Boccardo et al. [3], Ton [7] and by others. In Section 2, the notations, the main assumptions are given. With field forces being a Radon measure, a solution u of (1.1) may only be in Lr (0, T ; J01,r ()) with r ∈ (1, 4/3). Thus a solution of (1.2), if it exists, would be in C(0, T ; J01,r ()) and the expression u(X(, t), t) may be meaningless. Since X is the material position of a point of the valve, it is in Q and therefore is in L∞ (Q). We are led to consider the set K = {y : y Lr (0,T ;J 1,r ()) C, y L∞ (0,T ;C()) 1, 0
y L1 (0,T ;(J 1,r ()∩W 3,2 ())∗ ) C}. 0
(1.4)
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
3
It is a closed bounded, convex subset of the Hilbert space L2 (0, T ; (J01,2 ()∩W 3,2 ())∗ ) and the projection P of L2 (0, T ; (J01,2 ()∩W 3,2 ())∗ ) onto K is defined. Instead of (1.2), we shall study the initial-value problem d X(; t) = Pv(X(, t); t) in × (0, T ), dt
X(; 0) = in
(1.5)
for a given v ∈ Lr (0, T ; J01,r ()) ∩ L∞ (0, T ; L1 ()) with J01,r () = {u : u ∈ W01,r (); ∇.u = 0}. It is a Banach space with the obvious norm. The existence of a solution of the modified equation (1.5) is carried out in Section 3. In Section 4, we shall consider the following initial boundary-value problem for the plane Navier–Stokes equations w − w + (w.∇)w = ∇p + gn in Q w = 0 on j × (0, T ) ∇.w = 0 in Q, w(x; 0) = u0n (x) in
(1.6)
with gn L1 (Q) 1, u0n L1 () u0 L1 () ,
{gn , u0n }→ {(v), u0 } in D (Q) × D ().
The Radon measure is given by (v; F), = 0
T
(v,t)
˜ f((t))(X((t))) d(t) dt
∀ ∈ C0 (Q)
(1.7)
with F|(u,t) = f. It will be shown that for each n there exists a unique solution {wn , wn } ∈ Lr (0, T ; J01,r ()) × L1 (0, T ; (J01,r ∩ W 3,2 ())∗ ). Let BC be the convex set BC = {v : v Lr (0,T ;J 1,r ()) C; v L1 (0,T ;(J 1,r ()∩W 3,2 ())∗ ) C}. 0
(1.8)
0
It is a sequentially compact subset of Lr (0, T ; L2r/(2−r) ()) and BC ⊂ C(0, T ; (J01,2 ˜ C be the Lr (0, T ; L2r/(2−r) ())-closure of the set BC and let () ∩ W 3,2 ())∗ ). Let B v ∈ B˜ C , then in Section 5 we consider the nonlinear mapping T(v) = wn . It will be shown that T satisfies the hypotheses of the Schauder fixed point theorem and thus Tun = un for some un . The existence of a weak solution of the free boundary problem for the Navier–Stokes equations is obtained by letting n → ∞. The problem of the unicity of the solution as well as the three-dimensional cases is open. Finally in Section 6, we shall study an optimization problem. It allows us to determine the boundary forces from the observed values of u in a fixed subregion.
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B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
2. Notations, assumptions, known results Throughout the paper we shall assume that • {x : |x|1} is in the interior of . • There exists F with F L∞ (Q) 1, such that F|(u,t) = f, is the boundary force. Let BC be as in (1.9). It is a convex subset of Lr (0, T ; L2r/(2−r) ()) and by Aubin’s ˜ C , the clotheorem, the set is sequentially compact in Lr (0, T ; L2r/(2−r) ()). Thus, B r 2r/(2−r) sure of BC , in the L (0, T ; L ())-norm, is a compact convex subset of Lr (0, T ; 2r/(2−r) L ()). ˜ C ⊂ L2 (0, T ; (J 1,2 () ∩ W 3,2 ())∗ ). Moreover, it is easy to check that B 0 The following decomposition of L2 () and of W01,2 () are known L2 () = J0 () ⊕ G(),
W01,2 () = J01,2 () ⊕ (J01,2 ())⊥ .
Thus for v ∈ W01,2 (), we have v = v˜ + vˆ , v˜ ∈ J01,2 (),
vˆ ∈ (J01,2 ())⊥ .
On the other hand, using the unique decomposition of L2 () into J0 () and G(), we deduce that v = v˜ + ∇q,
∇q ∈ W01,2 () ∩ (J01,2 ())⊥ .
We shall now introduce the Lorentz spaces Lr,s (Q). Let Lr,∞ (Q), r ∈ (1, ∞), be the space of Lebesgue measurable functions such that f Lr,∞ (Q) = sup h{meas{(x, t) ∈ Q; |f (x, t)| > h}}1/r < ∞. h>0
For s ∈ (1, ∞), the Lorentz space Ls,1 (Q) is the space of Lebesgue measurable functions f such that |Q| f ∗ ()−1+1/s d. f Ls,1 (Q) = 0
The decreasing rearrangement f ∗ () of f is defined by f ∗ () = inf{ 0 : meas{(x, t) ∈ Q : |f (x, t)| > } < h},
h ∈ [0, |Q|].
It is known that for 1 < s < r < ∞, we have Lr,1 (Q) ⊂ Lr,r (Q) = Lr (Q) ⊂ Lr,∞ (Q) ⊂ Ls,1 (Q). The following generalized Holder inequality holds r,∞ f g dxdt f Lr,∞ (Q) g r/(r−1),1 (Q) × Lr/(r−1),1 (Q) L (Q) ∀{f, g} ∈ L Q
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
5
with r ∈ (1, ∞). Let k (s) = k for s > k,
k (s) = s for |s| k,
k (s) = −k otherwise.
Lemma 2.1. Let w be in C(0, T ; C() ∩ W01,2 ()) and suppose that T |∇k (w(x, t))|2 dx dt C1 k + C2 ∀k. 0
Then w Ls/2,∞ (Q) C(|Q|){C1 + C2 }1/2 ,
2 < s < ∞.
The constants C, C1 , C2 are all independent of k. Proof. It is clear that k (w) is in L2 (0, T ; W01,2 ()). From the Sobolev imbedding theorem, we get k (w(., t)) Ls () ∇k (w(., t)) L2 () for all t ∈ [0, T ]. With k = h, we get {meas[x : x ∈ , |w(x, t)| > h]}
2/s
=
2/s {x:x∈,|w(x,t)|>h} dx
−2
Ch
∇h (w(., t)) 2L2 () .
Since w ∈ C(Q), we have meas{(x, t) : (x, t) ∈ Q, |w(x, t)| > h} =
T
meas{x : x ∈ , |w(x, t)| > h} dt.
0
Moreover, meas{x : x ∈ , |w(x, t)| > h} is a continuous function of t. Hence, there exists t˜ ∈ [0, T ] such that {meas{(x, t) : (x, t) ∈ Q, |w(x, t)| > h}}2/s = T 2/s {meas{x : x ∈ , |w(x, t˜)| > h}}2/s Ch−2 ∇h (w)(., t˜) 2L2 ()) . Since by hypothesis, w is in C(0, T ; W01,2 ()), we get for any r ∈ (1, 2) | ∇w(., tn ) rLr (t ) − ∇w(., tn ) rLr (t ) | ∇w(., tn ) rLr (t /t ) n n ∇w rC(0,T ;L2 ()) {meas(tn /t )}(2−r)/2 C(2−r)/2
with tn = {x : x ∈ , |w(x, tn )| < h},
t =
n
tn
and
tn → t.
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B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
Thus ∇h (w)(., t) Lr () is continuous in t. We obtain ∇h (w)(., t˜) Lr () lim −1 →0 −1
(
t˜+
∇h (w)(., s) Lr () ds
+ 1){C1 h + C2 }1/2
∀r ∈ (1, 2).
Let r → 2 and we get ∇h (w)(., t˜) L2 () ( −1 + 1){C1 h + C2 }1/2 . We have w Ls/2,∞ (Q) = sup h{meas{(x, t) : (x, t) ∈ Q; |w(x, t)| > h}}2/s h>0
sup
h{meas{(x, t) : (x, t) ∈ Q; |w(x, t)| > h}}2/s
0
+ sup h{meas{(x, t) : (x, t) ∈ Q; |w(x, t)| > h}}2/s
h0 h Ch0 |Q|2/s
The lemma is proved.
+ C −1 (C1 + C2 h−1 0 ).
3. The initial-value problem (1.5) The main result of the section is the following theorem. ˜ c , then for any given ∈ , there exists a solution X of (1.5) Theorem 3.1. Let v be in B with X(.; v) C(0,T ;L∞ ()) + X (.; v) L∞ (0,T ;L∞ ()) C{1 + ||}. The constant C is independent of v. Since K, defined by (1.4) is a closed, bounded closed convex subset of the Hilbert space L2 (0, T ; (J01,2 () ∩ W 3,2 ())∗ ), the projection P of L2 (0, T ; (J01,2 () ∩ W 3,2 ())∗ ) is well-defined and is a non-expansive mapping, i.e. Pf − Pg L2 (0,T ;(J 1,2 ()∩W 3,2 ())∗ ) f − g L2 (0,T ;(J 1,2 ()∩W 3,2 ())∗ ) 0
0
for all f, g ∈ L2 (0, T ; (J01,2 () ∩ W 3,2 ())∗ ). Proof of Theorem 3.1. (1) Let v ∈ BC and consider the differential equation d X(t) = Pv(X(t), t) on [0, T ], dt
X(0) = .
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
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Since vˆ (., t) = Pv(., t) is a continuous function on for almost all t, the existence of a solution X of (1.5) on [0, T∗ ] follows from the Peano theorem. Since t ˆv(X, s) L∞ () ds || + T , X(., t) L∞ () || + 0
we deduce that T∗ = T . (2) The theorem on the continuous dependence on a parameter of the solution of an ordinary differential equation, shows that the function X of × [0, T ] → given by X(, t) = X(t) is in C 1 ( × [0, T ]). It is clear that X L∞ (0,T ;L∞ ()) Pv L∞ (0,T ;L∞ ()) 1. ˜ C , then v ∈ L2 (0, T ; (J 1,2 () ∩ W 3,2 ())∗ ) ∩ Lr (0, T ; L2r/(2−r) ()) (3) Let v be in B 0 and there exists {vk } ∈ BC such that vk → v
in Lr (0, T ; L2r/(2−r) ()).
With vk , we have from the previous part Xk = Pvk (Xk , t) on [0, T ],
Xk (, 0) =
and Xk (t) − Xk (s) C() |t − s| Pvk L∞ (Q) |t − s|. It follows that {Xk } is equicontinuous and there exists a subsequence such that Xk → in C(0, T ; C())). With K as stated, then Xk → in (L∞ (Q))weak∗ . On the other hand, we have vˆ k (Xk ) − vˆ () = vˆ k (Xk ) − vˆ (Xk ) + vˆ (Xk ) − vˆ (). • Since vˆ (., t) = Pv(., t) is continuous on and since Xk → uniformly in × (0, T ), we get vˆ (Xk ) → vˆ () a.e. in Q. Thus it follows from the definition of K and from the Lebesgue convergence theorem that vˆ (Xk ) → vˆ () in Lr (0, T ; L2r/(2−r) ()). • With vˆ k ∈ K, it follows from Aubin’s theorem that vˆ k →
in Lr (0, T ; L2r/(2−r) ()) and a.e. in Q.
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B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
On the other hand since P is a non-expansive mapping of L2 (0, T ; (J01,2 ()∩W 3,2 ())∗ ) into K, we get vˆ k → vˆ
in L2 (0, T ; (J01,2 () ∩ W 3,2 ()∗ )).
Hence = vˆ . We have {[, t] : |ˆvk (Xk (), t) − vˆ (Xk (), t)| > ; = Xk ()} ⊂ {[, t] : |ˆvk (, t) − vˆ (, t)| > }. Therefore, vˆ k (Xk , t) − vˆ (Xk , t) → 0
a.e. in Q
With ˆvk (Xk ) − vˆ (Xk ) L∞ (Q) C, the Lebesgue convergence theorem gives vˆ k (Xk ) − vˆ (Xk ) → 0 in Lr (Q). • Combining the two parts we obtain vˆ k (Xk ) − vˆ () → 0 in Lr (Q). It follows that = vˆ (); |t=0 = . Thus, = X(, t; v).
Now we shall consider the dependence of the solutions of (1.5) on v. The results will be needed when we apply the Schauder theorem to the nonlinear mapping T given by (1.7). ˜ C and let X(., t; vk ) be the solutions of the initial-value problem Theorem 3.2. Let vk ∈ B (1.5) with vk instead of v. Suppose that vk → v
in Lr (0, T ; L2r/(2−r) ()),
then there exists a subsequence such that {vk , X(.; vk )} → {v, X(.; v)} in Lr (0, T ; L2r/(2−r) ()) ∩ (Lr (0, T ; J01,r ()))weak∗ × (L∞ (Q))weak∗ . Moreover, X(., t; v) is a solution of the initial-value problem (1.5). Proof. With the estimates of Theorem 3.1, it is clear that there Xk (., t) − Xk (., s) C() |t − s|. Thus there exists a subsequence such that {Xk (., t), Xk (., t)} → {, }
in C(Q) × (L∞ (Q))weak∗ .
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
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with Xk = Xk (., t; vk ). On the other hand, with vˆ k = Pvk we have |ˆvk (Xk (t), t) − vˆ ((t), t)| |ˆvk (Xk (t), t) − vˆ (Xk (t); t)| + |ˆv(Xk (t), t) − vˆ ((t), t)|. As in part (3) of the proof of Theorem 3.1, we get in Lr (Q).
vˆ k (Xk ) − vˆ () → 0
It follows that = vˆ (); |t=0 = i.e. = X(, t; v). Let (v; t) = {X(, t; v) : () = c} with () = c describing the valve at the initial time. Then in view of Theorem 3.2, we have the following lemma. Lemma 3.1. Let {vk , v, X( , t; vk ), X( ., t; v)} and be as in Theorem 3.2. Then (v; t) =
∞
(vk , t).
k k0
We now study the dependence of the source (v) on v and on F. Lemma 3.2. Let Fk be in L∞ (Q) and let vk , v, X(., t; vk ), X(., t; v) be as in Theorem 3.2 with (vk , Fk ) Mb (Q) 1, Suppose that Fk → F in
(v, F) Mb (Q) 1.
(L∞ (Q))
weak ∗ ,
then (vk , Fk ) → (v, F) in D (Q).
Proof. Let be in C0 (Q), then we have T fk ((t))(Xk ( ., t; vk )) d dt (vk , Fk ), = 0
=
0
T
(vk ,t)
(v,t) T
+ 0
fk ((t))(Xk (., ; vk )) d dt
(vk ,t)/(v,t)
fk (){(Xk ( ., vk ))} d dt.
Therefore, |(vk , Fk ) − (v, Fk ), |C (X( .; v)) − (Xk ( .; vk )) C(Q) + Cmeas{(v)/(vk )}. On the other hand, we have
T
(v, Fk ) − (v, F), = 0
(v,t)
{fk − f}(X( ., v)) d dt
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B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
for all ∈ C0 (Q). Since Fk → F in (L∞ (Q))weak∗ , it now follows from Theorem 3.2 that (vk , Fk ) → (v, F) in D (Q). ˜ C and suppose that vk → v in Lr (0, T ; L2r/(2−r) ()). Then Lemma 3.3. Let vk , v be in B 1 there exists {gk } ∈ C (Q) ∩ BC such that |(vk , F) − gk , | + |(v, F) − gk , |3/k
∀ ∈ C0 (Q).
˜ C there exists gk ∈ C 1 (Q) ∩ BC such that Proof. Since vk is in B 1 ∀, C0 (Q) 1. k On the other hand, from Lemma 3.2 we have |gk − (vk , F), |
|(vk , F) − (v, F), |
1 k
∀, C0 (Q) 1.
Hence, 2 k
|gk − (v, F), | The lemma is proved.
∀, C0 (Q) 1.
4. The Navier–Stokes equations Let ˜ C × L1 () × Mb (Q), {v, u0 , (v, F)} ∈ B
∇.u0 = 0,
(v, F) Mb (Q) 1.
Then there exists {g , u0 } in C01 (Q) × C02 () ∩ J01,2 () such that g L1 (Q) 1,
u0 L1 () u0 L1 ()
g − (v, F) Mb (Q) ,
u0 → u0 in D ().
In this section, we shall consider the initial boundary-value problem u − u + (u .∇)u = ∇p + g in Q ∇.u = 0 in Q u = 0 on j × (0, T ), u (x, 0) = u0 (x) in .
(4.1)
The main result of the section is the following theorem. Theorem 4.1. Let {v, u0 , g , u0 } be as above. Then there exists a unique solution u of (4.1) in C(0, T ; J01,2 () ∩ W 2,2 ()) with u ∈ L2 (0, T ; J01,2 ()) and u Lr (0,T ;J 1,r ()) + u L1 (0,T ;(J 1,r ()∩W 3,2 ())∗ ) 0
0
C{1 + u0 L1 () + (v, F) Mb (Q) } for r ∈ (1, 4/3). The constant C is independent of , v.
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
11
The existence of a unique solution u of (4.1) is well-known (cf. [6, pp. 79–81]. We shall now establish the estimates of the theorem. Lemma 4.1. Let u be as in the theorem. Then u L∞ (0,T ;L1 ()) C{1 + || + u0 L1 () }. The constant C is independent of , v. Proof. (1) Since u is in C(Q), it follows that supQ |uj, (x, t)|= j, exists. Let =inf j j, and let if inf{ , 1} < s, inf{1, } = s if |s| inf{ , 1}, − inf{1, } if − inf{ , 1} < s. With u ∈ C 1 (0, T ; C()) ∩ L2 (0, T ; J01,2 ()), we now consider the testing function (u ) = ((u1, ), (u2, )). Set {(x, t) :, (x, t) ∈ Q; |uj, (x, t)| inf{1, }} ⊂ supp(uj, ) D1 = j
j
is a non-empty subset and j × (0, T ) ⊂ jD1 . For each t ∈ [0, T ], we have (u ) ∈ W01,2 () and thus we get the decomposition (u (., t)) = (u ˜ (., t)) + ∇q(., t),
(u ˜ (., t)) ∈ J01,2 (),
∇q ∈ W01,2 ().
It is clear that maps L2 () into J0 () and W01,2 () into J01,2 (). Since (u (., t)) L∞ () 1, we have (u ˜ (., t)) L∞ ()
inf
|((u ˜ (., t)), h)|
inf
|((u (., t)), h)|
h L1 () 1;∇.h=0 h L1 () 1; ∇.h=0
(u (., t)) L∞ () 1. Thus, (u ˜ ) L∞ (0,T ;L∞ ()) 1. We have ˜ (., t)) = (u (., t), (u (., t)) − ∇q) = (u , (u (., t))) (u (., t), (u as (u (., t), ∇q) = −
2 j,k=1
(Dj uk, (., t), Dk Dj q) = (div(u (., t)), q) = 0.
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B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
We have used the fact that ∇q ∈ W01,2 (). Since (v) W 1,2 () (v) ˜ J 1,2 () 0
0
v W 1,2 () , 0
∀v ∈ W01,2 (),
we deduce that ˜ (v) L∞ () 1
∀v ∈ W01,2 ().
With u ∈ L2 (0, T ; J01,2 ()), we get ˜ )) = (u , (u ) + ∇q) = (u , (u )). (u (., t), (u (2) Set
s
(s) =
(r) dr.
0
Multiplying (4.1) by (u ˜ ) and we obtain 2 d (u (x, t)) dx + (uj, )|Dm uj, |2 dx dt j,m=1 2 + uj, Dj um, (u ˜ m, ))(x, t) dx = g (v)(u ˜ ) dx. j,m=1
Since ∇.u = 0, we have 2 2 ˜ m, ) dx = − uj, Dj um, (u um, um, ˜ (um, ) Dj um, dx j,m=1
j,m=1
= −
2
j,m=1
=
2 m=1
uj, Dj
s ˜ (s) ds
dx
0
div(u )
um,
um,
s ˜ (s) ds
dx = 0.
0
It follows that d (u (x, t)) dx g (v)(u ˜ (x, t)) dx. dt Hence
u (., t) L1 () C 1 + || +
(u ) dx ds
C{1 + || + g (v) L1 () + u0 L1 () } C{1 + || + u0 L1 () }. The constant C is independent of v, .
(4.2)
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
13
Lemma 4.2. Suppose all the hypotheses of Theorem 4.1 are satisfied. Let u be as in Lemma 4.1, then u Lr (0,T ;J 1,r ()) C{1 + || + u0 L1 () } for r ∈ (1, 4/3). 0
The constant C is independent of v, . Proof. (1) Let
⎧ 1 ⎪ ⎪ ⎪ ⎨s − k k (s) = 0 ⎪ ⎪ ⎪ ⎩s + k −1
if if if if if
1 + k < s, k s 1 + k, − k s k, − 1 − k s − k, s − 1 − k.
• Let k be a positive integer with k < = inf j
sup |uj, (x, t)| , Q
k K = sup j
sup |uj, (x, t)| . Q
Let Bk =
2
{(x, t) : (x, t) ∈ Q; k |uj, (x, t)| k + 1}
j =1
then Bk ∩ { j supp(uj, )} is non-empty. With k (u ) ∈ W01,2 () we have as in Section 2, the decomposition ˜ k (u ) + ∇q, k (u ) =
{ ˜ k (u ), ∇q} ∈ J01,2 () × (G() ∩ W01,2 ()).
Moreover, ˜ k (u (., t)) L∞ () 1
∀t ∈ [0, T ]
˜ k (u (., t)) L∞ () 1
∀t ∈ [0, T ].
and
As in the Proof of Lemma 4.1, we have ˜ k (u (., t)) = (u (., t) − u (., t), k (u )(., t)). (u (., t) − u , Taking the pairing of (4.1) with ˜ k (u ), and we get 2 d k (u (x, t)) dx + k (uj, )|Dm uj, (x, t)|2 dx dt j,m=1 2 + um, Dm uj, ˜ k (uj, (x, t)) dx C(1 + ||) + (v, F) Mb (Q) ) j,m=1
14
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
with
s
k (s) = 0
k (r) dr.
An integration by parts yields 2 ˜ (um, (x, t)) dx uj, Dj um, k j,m=1
=−
2 j,m=1
=−
2
j,m=1
=
2 m=1
˜ (um, )Dj um, dx uj, um, k uj, Dj
0
div(u )
um, (x,t)
um,
0
˜ (s) ds s k
˜ (s) ds s k
dx
dx = 0.
It follows that 2 k (u (x, t)) dx + k (um, )|Dj um, |2 dx
j,m=1
C{1 + || + (v, F) Mb (Q) + u0 L1 () } and hence Bk
|∇u |2 dx C{1 + || + u0 L1 ()) }.
(4.3)
Let r ∈ (1, 4/3) then an application of the Holder inequality yields k|Bk | |uj, (x, t)| dx dt u L3r/2 (Bk ) |Bk |(3r−2)/3r . Bk
Thus, |Bk | k −3r/2 u L3r/2 (B ) . 3r/2
(4.4)
k
Again an application of the Holder inequality together with the estimate (4.4) gives ∇u rLr (Bk ) |Bk |(2−r)/2 ∇u rL2 (B Ck
−3r(2−r)/4
k) 3r(2−r)/4 u L3r/2 (B ) k
× {1 + || + u0 L1 () + (v, F) Mb (Q) }. • We now consider the case
= inf j sup |uj, (x, t)| < k. Q
(4.5)
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
15
Since = supQ us, , the set supp(us, ) {(x, t) : k |us, (x, t) k + 1} is empty. We have only to study the case when supQ |uj, (x, t)| > k with j = s. An identical argument as before gives again the estimate (4.5). (2) Let k0 be a fixed positive number and let be the truncated function if s > inf{k0 , }, inf{k0 , } (s) = s if |s| inf{k0 , }, − inf{k0 , } if s < inf{k0 , }. Then as in Lemma 4.1, we have |∇u |2 dx dt C{k0 + u0 L1 () + (v, F) Mb (Q) }. Dk0
Now the Holder inequality gives r/2
∇u rLr (Dk ) ||(2−r)/2 ∇u L2 (D
k0 )
0
C{k0 + ||(1 + u0 L1 () + (v, F) Mb (Q) )}. (3) It follows from (4.4) to (4.5) that ∇u rLr (Q) C(k0 + || + u0 L1 () + (v, F) Mb (Q) + C(1 + || + u0 L1 () + (v, F) Mb (Q) ×
K k=k0
u L3r/2 (B ) k −3r(2−r)/4 u L3r/2 (Q) . 3r(2−r)/4
3r/2
k
(4.6)
Applying the Holder inequality to (4.6) with exponents 2/r, 2/(2 − r) and we obtain ∇u rLr (Q) C{1 + || + u0 L1 () + (v, F) Mb (Q) ) ⎛ ⎧ ⎫r/2 ⎞ K ⎨ ⎬ ⎟ ⎜ 3r/2 k −3(2−r)/2 × ⎝1 + u L3r/2 (Q) ⎠ ⎩ ⎭ k=k0
C{1 + || + u0 L1 () + (v, F) Mb (Q) } ⎧ ⎫r/2 ∞ ⎨ ⎬ 3r/2 × k −3(2−r)/2 u L3r/2 (Q) . ⎩ ⎭ k=k0
(4.7) The series converges since r ∈ (1, 4/3). (4) An application of the Holder inequality gives 1/3
2/3
u L3r/2 () u L1 () u L2r/(2−r) () .
(4.8)
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B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
Taking the estimate of Lemma 4.1 into account, we get 3r/2
u L3r/2 (Q) C{1 + || + u0 L1 ()
+ (v, F) Mb (Q) } × u rLr (0,T ;L2r/(2−r) ()) .
(4.9)
(5) The Sobolev imbedding theorem yields u rLr (0,T ;L2/(2−r) ()) C u r r
L (0,T ;W01,r ())
.
It follows from (4.6) to (4.9) that u rLr (0,T ;L2/(2−r) ()) C{1 + || + u0 L1 () + (v, F) Mb (Q) } ⎛ ⎧ ⎫r/2 ⎞ ⎨ ⎬ ⎟ ⎜ 3r/2 k −3(2−r)/2 × ⎝1 + u L3r/2 (Q) ⎠ ⎩ ⎭ k=k0
C(1 + || + u0 L1 () + (v, F) Mb (Q) ) ⎛ ⎧ ⎫r/2 ⎞ ⎨ ⎬ ⎟ ⎜ k −3(2−r)/2 × ⎝1 + u rLr (0,T ;L2/(2−r) ()) ⎠. ⎩ ⎭ k=k0
(4.10) Since the series converges for r ∈ (1, 4/3), there exists k0 such that u rLr (0,T ;L2/(2−r) ()) C{1 + || + u0 L1 () + (v, F) Mb (Q) }.
(4.11)
Using the estimate (4.11) in (4.9) and we have 3r/2
u L3r/2 (Q) C{1 + || + u0 L1 () + (v, F) Mb (Q) }.
(4.12)
With (4.12) in (4.7) and we obtain u Lr (0,T ;J 1,r ()) C{1 + || + u0 L1 () + (v, F) Mb (Q) }. 0
The lemma is proved.
Lemma 4.3. Suppose all the hypotheses of Theorem 4.1 are satisfied. Then u L1 (0,T ;(J 1,r ()∩W 3,2 ())∗ ) C{1 + || + u0 L1 () + (v, F) Mb (Q) } 0
with r ∈ (1, 4/3). The constant C is independent of , v, F. Proof. Let w be an element of J01,r () ∩ W 3,2 () and consider the equation (u (., t), w) + (∇u (., t), ∇w) = −((u .∇)u (., t), w) + (g , w). With the estimate of Lemma 4.2, we have from Lemma 2.1 u Ls/2,∞ (Q) C{1 + || + u0 L1 () + (v, F) Mb (Q) },
2 < s < ∞.
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
17
We have |uj, Dj uk, wk (x)| dx dt wk L∞ () uj, Ls/2,∞ (Q) Dj uk Ls/(s−2),1 (Q) Q
C w W 3,2 () {1 + || + u0 L1 () + (v, F) Mb (Q) )} × Dj uk, Ls/(s−2),1 (Q) . With s > 2r/(r − 1) > 8, we have Lr,1 (Q) ⊂ Lr (Q) ⊂ Ls/(s−2),1 (Q). Therefore |uj, Dj uk, wk (x)| dx dt C{1 + || + u0 L1 () + (v, F) Mb (Q) } Q
× u Lr (0,T ;J 1,r ()) w W 3,2 () 0
C{1 + u0 L1 () + (v, F)} w W 3,2 () . With (4.13), it is now trivial to establish the estimate of the lemma.
(4.13)
Proof of Theorem 4.1. The theorem is now an immediate consequence of Lemmas 4.1– 4.4. 5. A free boundary problem The main result of the paper is the following theorem. ˜ C be as in (1.8) Theorem 5.1. Let {u0 , F} be in L1 () × L∞ (Q) with ∇.u0 = 0 in , let B and let P be the projection of L2 (0, T ; (J01,2 () ∩ W 3,2 ())∗ ) onto K defined by (1.4 ). Then there exists ˜ C × C(0, T ; W 1,r ()) × Lr (0, T ; W 1,r ()) {u, X( ., ; u); X ( .; u)} ∈ B with (u.∇)w integrable in Q for all w ∈ S and such that T T T (u, w ) dt + (∇u, ∇w) dt = (u.∇w, u) dt + (u0 , w( ., 0)) − 0
0
0
+ (u, F), w. Moreover, X(, t; u) = +
T
Pu(X(, s; u)) ds.
0
The set S is given by S = {w : {w, w } ∈ C(0, T ; J01,r () ∩ W 3,2 ()) × Lr/(r−1) (Q); w( ., T ) = 0}. The force field (u, F) is in Mb (Q) with support on (u, t) = {X(, t; u) : ∈ , () = c}. The initial position of the free boundary is the given curve () = c.
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B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
The proof of the main result of the paper is carried out in this section. Let {v, u0 } ∈ ˜ C × L1 (), and let {g , u } be in C 1 (Q) × C 1 () with B 0 0 0 g L1 (Q) (v, F) Mb (Q) 1,
u0 L1 () u0 L1 ()
and g − (v, F) Mb (Q) + u0 − u0 L1 () . ˜ C × L1 () and {g , u } as From Theorem 4.1, we know that for any given {v, u0 } ∈ B 0 ˜ C . We define the nonlinear mapabove, there exists a unique solution u of (4.1) with u ∈ B ˜ C , considered as a subset of Lr (0, T ; L2r/(2−r) ()) into Lr (0, T ; L2r/(2−r) ()) ping A of B by A(v) = u .
(5.1)
˜ C of Lr (0, T ; L2r/(2−r) ()) into It is clear that A maps the compact convex subset B itself. To apply the Schauder fixed point theorem, it suffices to show that A is continuous. Lemma 5.1. Let A be as in (5.1), then it is a continuous mapping of Lr (0, T ; L2r/(2−r) ()) into Lr (0, T ; L2r/(2−r) ()). ˜ C , vn → v in Lr (0, T ; L2r/(2−r) ()). Then there exists, as done Proof. (1) Let {vn } ∈ B in Lemma 3.4, {gn , g } ∈ C 1 (Q) such that gn − g L2 (Q) 1/n,
g − (v, F) Mb (Q) .
An application of Lemma 3.3 gives g − (vn ) Mb (Q) 2/n + . Let A(vn ) = un where un is the unique solution of (4.1) with g , v replaced by gn , vn . From the estimate of Theorem 4.1, we have un Lr (0,T ;J 1,r ()) + (un ) L1 (0,T ;(J 1,r ()∩W 3,2 ())∗ ) C{1 + || + u0 L1 () }. 0
0
It follows from the above estimate and from Aubin’s theorem that there exists a subsequence such that un → u
in (Lr (0, T ; J01,r ())weak ∩ Lr (0, T ; L2r/(2−r) ()) and a.e. in Q.
(2) With {gn , un } ∈ C01 (Q) × C 1 (), we have the standard estimate un L∞ (0,T ;L2 ()) + un L2 (0,T ;J 1,2 ()) C{1 + gn L2 (Q) + u0 L2 ()) } 0
C{1 + g L2 (Q) + u0 L2 () }.
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
19
The constant C is independent of n, . Let {w, w } be in C(0, T ; J01,r () ∩ W 3,2 ()) × 1,r/(r−1) ()) with w( ., T ) = 0, then we have Lr/(r−1) (0, T ; J0 T T T − (un , w ) dt + (∇un , ∇w) dt = (u0 , w( ., 0)) + ((un .∇)w, un ) 0 0 0 T + (gn , w) dt. 0
Let n → ∞ and we get T − (u , w ) dt + 0
T 0
T (∇u , ∇w) dt = (u0 , w( ., 0)) + ((u .∇)w, u ) dt 0 T + (g , w) dt. 0
It follows that u − u + (u .∇)u = ∇p + g in D (Q) ∇.u = 0, u = 0 on j × (0, T ) u ( ., 0) = u0 in . It is known that there exists a unique solution of the problem. Hence A(v) = u and the operator A is continuous from Lr (0, T ; L2r/(2−r) ()) into itself. ˜ C such that A(u ) = u . Lemma 5.2. There exists u ∈ B Proof. The lemma is now an immediate consequence of the Schauder fixed point theorem. Proof of Theorem 5.1. Let u be as in Lemma 5.2. Then we have u − u + (u .∇)u = ∇p + g (u ) in Q ∇.u = 0 in Q, u = 0 on j × (0, T ) u ( ., 0) = u0 in
(5.2)
with d X(, t; u ) = Pu (X(, t; u ); t) on (0, T ), dt
X(, 0; u ) = .
(5.3)
˜ C is a compact subset of Lr (0, T ; L2r/(2−r) ()), it follows from Theorem (1) Since B 4.1 that there exists a subsequence such that u → u
in Lr (0, T ; J01,r ())weak ∩ Lr (0, T ; L2r/(2−r) ()) and a.e. in Q.
Moreover, u → u
in D (0, T ; (J01,r () ∩ W 3,2 ())∗weak ).
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B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
(2) Let → 0 in (5.2) and it follows from Theorem 3.1 that there exists X ∈ C(0, T ; J01,r ()) with X ∈ Lr (0, T ; J01,r ()) such that d X(, t; u) = Pu(X(, t; u); t) on (0, T ), dt
X(, 0; u) = .
(5.4)
(3) Let w be in C 1 (0, T ; J01,r () ∩ W 3,2 ()) and consider the term T T ((u .∇)u , w) dt = − ((u .∇)w, u ) dt. 0
0
Since u → u a.e. in Q, the set ⎧ ⎫ 2 ⎨ ⎬ S = (x, t) : (x, t) ∈ Q; |{uj, (x, t)uk, (x, t) − uj (x, t)uk (x, t)}| 1 ⎩ ⎭ j,k=1
are increasing as → 0 and meas(S ) → |Q|. Thus the Lebesgue convergence theorem gives
T
lim
→0
2
j,k=1
0
{uj, uk, − uj uk }Dj wk (x, t)} dx dt = 0
for all w ∈ C(0, T ; J01,r () ∩ W 3,2 ()). On the other hand, we have T
(u , w) dt = − 0
T
(u , w ) dt + (u0 , w(., 0)) ∀w ∈ S.
0
Thus,
T
0
(u , w) dt → −
T
(u, w ) dt + (u0 , w(., 0))
∀w ∈ S.
0
From (5.2), we deduce that
T
lim
0
=−
2
j,k=1 T
uj, uk, Dj wk (x, t) dx dt
T
(u, w ) dt +
0
(∇u, ∇w) dt + (u, F), w + (u0 , w(., 0))
0
for all w ∈ S. It follows that T (u.∇)w.u dx dt = − 0
T
T
(u, w ) dt +
0
(∇u, ∇w) dt
0
+ (u, F), w + (u0 , w(., 0))
for all w ∈ S. The theorem is proved .
(5.5)
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
21
6. An inverse problem Let G be a bounded subset of the plane with G ⊂ and let be a L1 (0, T ; L1 (G))function, representing the observed values of u in G × (0, T ). We consider the function T J(u, u0 , F) = |u − | dx dt, 0
G
where {u, X( , t; u)} is a solution of the free boundary problem of Theorem 5.1. Let V (u0 ) = inf{J(u, u0 , F) : ∀u solution of (1.1), (1.3).(1.5), ∀F, F L∞ (Q) 1}. We now have the following result which allows us to find the boundary force from the observed values of u in a fixed subregion. Theorem 6.1. Let u0 ∈ L1 () with ∇.u0 = 0. Then there exists ˜ C × C(0, T ; W 1,r ()) ∩ L∞ (Q) × Lr (0, T ; W 1,r ()) × L∞ (Q) ˜ ∈B ˜ X ˜ , F} ˜ X, {u, ˜ ˜ u0 , F). such that V (u0 ) = J(u, Proof. (1) Let {un , Fn } be a minimizing sequence of the above optimization problem, with V (u0 )J(un , u0 , Fn ) V (u0 ) +
1 , n
˜ C , Fn L∞ (Q) 1. un ∈ B
˜ C is a compact subset of Lr (0, T ; L2r/(2−r) ()), there exists a subsequence such Since B that un → u˜ in Lr (0, T ; L2r/(2−r) ()). With Fn → F˜ in (L∞ (Q))weak∗ , we deduce from Lemma 3.2 that ˜ ˜ F) in D (Q). (un , Fn ) → (u, ˜ in (2) From Theorem 3.2, we get X(., t; un ) → X(., t; u) (Lr (0, T ; J01,r ()))weak ∩ (L∞ (Q))weak∗ ∩ Lr (0, T ; L2r/(2−r) ()). Now as in part (3) of the Proof of Theorem 5.1, we deduce that u˜ is a weak solution of ˜ ˜ u0 , F). (1.1) and thus V (u0 ) = J(u, References [1] M.F. Betta, A. Mercaldo, F. Murat, M.M. Porszio, Existence of renormalized solutions to nonlinear elliptic equations with lower order terms and right hand side measure, J. Math. Pures Appl. 81 (2002) 533–566. [2] L. Boccardo, T. Gallouet, Nonlinear elliptic and parabolic equations involving measure data, J. Funct. Anal. 87 (1989) 149–169. [3] L. Boccardo, T. Gallouet, L. Orsina, Existence and uniqueness of entropy solutions for nonlinear elliptic equations with measure data, Ann. Inst. Henri Poincare Anal. Non Lineaire 13 (1996) 539–551. [5] C. Peskin, Flow patterns around heart valves, J. Comput. Phys. 10 (2) (1972) 252–271. [6] C. Peskin, Mathematical Aspects of Heart Physiology, Courant Institute of Mathematical Sciences, New York University, 1973–1974. [7] B.A. Ton, Time dependent Stokes equations with measure data, Abstr. Appl. Anal. 17 (2003) 953–973.
A free boundary problem for the Navier–Stokes equations with measure data Bui An Ton∗ Department of Mathematics, University of British Columbia, Vancouver, BC, Canada V6 T1 Z2 Received 17 July 2003; accepted 14 May 2005
Abstract The existence of a weak solution of an initial boundary-value problem for the plane nonstationary Navier–Stokes equations with Radon measure data on the free boundary, is established. The problem may be considered as a model of the blood flow around the heart valves. An inverse problem is studied, it allows us to find the boundary forces acting on the valve from the observed values of the velocity of the fluid in a fixed subregion. 䉷 2005 Elsevier Ltd. All rights reserved. MSC: 35K60; 76D05; 76D27; 76Z05 Keywords: Navier–Stokes; Free boundary; Radon measure; Heart valves; Blood flow
1. Introduction The purpose of this paper is to study a free boundary problem for the plane nonstationary Navier–Stokes equations with measure data. The problem may be considered as a model of blood flow around the heart valves. The main characteristic features of the problem are: • The force field. It is zero everywhere except on the free boundary where it is infinite there but its integral over any finite region containing the free boundary is finite (like a two-dimensional impulse on the free boundary). In the blood flow model, the heart valve ∗ Tel.: +1 604 822 2666; fax: +1 604 822 6074.
E-mail address: [email protected]. 0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2005.05.038
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B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
exerts forces on the fluid (i) forces which stop the flow when the valve is closed and (ii) forces which shear the flow to form vortices when the valve is open. • The free boundary. It is given in terms of Lagrangian coordinates with the fluid velocity being only a weak solution of the Navier–Stokes equations. A model for the flow of blood around the heart valves was developed by Peskin in [5,6]. Approximating the Dirac delta measure by L1 -functions and linearizing the problem through successive approximations, Peskin treated the problem numerically in [5,6]. Let be a bounded open subset of the plane with a smooth boundary and consider the initial boundary-value problem u − u + (u.∇)u = ∇p + (u, F) ∇.u = 0
in × (0, T )
in Q = × (0, T )
u(x, t) = 0 on j × (0, T ),
u(x, 0) = u0 in
(1.1)
with d X(, t) = u(X(, t); t), dt and
T
(u, F), =
(u;t)
0
X(, 0) =
(1.2)
˜ f((t))(X((t); u)) d(t) dt
∀ ∈ C0 (Q).
(1.3)
We denote by X(; t), the position of the particle of the heart valve at the time t, which was at , at t = 0. The material parameter for the free boundary (u, t), describing the heart valve is (t), and ˜ ˜ (u, t) = {X((t); u) : X((t); u) = X(, t; u), ∈ , with () = c}, where () = c is the free boundary at the initial time. We consider boundary forces f with F ∈ L∞ (Q), F|(u,t) = f,
|f((t))| 1,
Mb (Q) 1,
supp (u; F) ⊂ (u; t).
The set of Radon measures in Q is denoted by Mb (Q). Elliptic and parabolic partial differential equations with measure data have been the subject of extensive investigations by Betta et al. [1], Boccardo and Gallouet [2], Boccardo et al. [3], Ton [7] and by others. In Section 2, the notations, the main assumptions are given. With field forces being a Radon measure, a solution u of (1.1) may only be in Lr (0, T ; J01,r ()) with r ∈ (1, 4/3). Thus a solution of (1.2), if it exists, would be in C(0, T ; J01,r ()) and the expression u(X(, t), t) may be meaningless. Since X is the material position of a point of the valve, it is in Q and therefore is in L∞ (Q). We are led to consider the set K = {y : y Lr (0,T ;J 1,r ()) C, y L∞ (0,T ;C()) 1, 0
y L1 (0,T ;(J 1,r ()∩W 3,2 ())∗ ) C}. 0
(1.4)
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
3
It is a closed bounded, convex subset of the Hilbert space L2 (0, T ; (J01,2 ()∩W 3,2 ())∗ ) and the projection P of L2 (0, T ; (J01,2 ()∩W 3,2 ())∗ ) onto K is defined. Instead of (1.2), we shall study the initial-value problem d X(; t) = Pv(X(, t); t) in × (0, T ), dt
X(; 0) = in
(1.5)
for a given v ∈ Lr (0, T ; J01,r ()) ∩ L∞ (0, T ; L1 ()) with J01,r () = {u : u ∈ W01,r (); ∇.u = 0}. It is a Banach space with the obvious norm. The existence of a solution of the modified equation (1.5) is carried out in Section 3. In Section 4, we shall consider the following initial boundary-value problem for the plane Navier–Stokes equations w − w + (w.∇)w = ∇p + gn in Q w = 0 on j × (0, T ) ∇.w = 0 in Q, w(x; 0) = u0n (x) in
(1.6)
with gn L1 (Q) 1, u0n L1 () u0 L1 () ,
{gn , u0n }→ {(v), u0 } in D (Q) × D ().
The Radon measure is given by (v; F), = 0
T
(v,t)
˜ f((t))(X((t))) d(t) dt
∀ ∈ C0 (Q)
(1.7)
with F|(u,t) = f. It will be shown that for each n there exists a unique solution {wn , wn } ∈ Lr (0, T ; J01,r ()) × L1 (0, T ; (J01,r ∩ W 3,2 ())∗ ). Let BC be the convex set BC = {v : v Lr (0,T ;J 1,r ()) C; v L1 (0,T ;(J 1,r ()∩W 3,2 ())∗ ) C}. 0
(1.8)
0
It is a sequentially compact subset of Lr (0, T ; L2r/(2−r) ()) and BC ⊂ C(0, T ; (J01,2 ˜ C be the Lr (0, T ; L2r/(2−r) ())-closure of the set BC and let () ∩ W 3,2 ())∗ ). Let B v ∈ B˜ C , then in Section 5 we consider the nonlinear mapping T(v) = wn . It will be shown that T satisfies the hypotheses of the Schauder fixed point theorem and thus Tun = un for some un . The existence of a weak solution of the free boundary problem for the Navier–Stokes equations is obtained by letting n → ∞. The problem of the unicity of the solution as well as the three-dimensional cases is open. Finally in Section 6, we shall study an optimization problem. It allows us to determine the boundary forces from the observed values of u in a fixed subregion.
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B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
2. Notations, assumptions, known results Throughout the paper we shall assume that • {x : |x|1} is in the interior of . • There exists F with F L∞ (Q) 1, such that F|(u,t) = f, is the boundary force. Let BC be as in (1.9). It is a convex subset of Lr (0, T ; L2r/(2−r) ()) and by Aubin’s ˜ C , the clotheorem, the set is sequentially compact in Lr (0, T ; L2r/(2−r) ()). Thus, B r 2r/(2−r) sure of BC , in the L (0, T ; L ())-norm, is a compact convex subset of Lr (0, T ; 2r/(2−r) L ()). ˜ C ⊂ L2 (0, T ; (J 1,2 () ∩ W 3,2 ())∗ ). Moreover, it is easy to check that B 0 The following decomposition of L2 () and of W01,2 () are known L2 () = J0 () ⊕ G(),
W01,2 () = J01,2 () ⊕ (J01,2 ())⊥ .
Thus for v ∈ W01,2 (), we have v = v˜ + vˆ , v˜ ∈ J01,2 (),
vˆ ∈ (J01,2 ())⊥ .
On the other hand, using the unique decomposition of L2 () into J0 () and G(), we deduce that v = v˜ + ∇q,
∇q ∈ W01,2 () ∩ (J01,2 ())⊥ .
We shall now introduce the Lorentz spaces Lr,s (Q). Let Lr,∞ (Q), r ∈ (1, ∞), be the space of Lebesgue measurable functions such that f Lr,∞ (Q) = sup h{meas{(x, t) ∈ Q; |f (x, t)| > h}}1/r < ∞. h>0
For s ∈ (1, ∞), the Lorentz space Ls,1 (Q) is the space of Lebesgue measurable functions f such that |Q| f ∗ ()−1+1/s d. f Ls,1 (Q) = 0
The decreasing rearrangement f ∗ () of f is defined by f ∗ () = inf{ 0 : meas{(x, t) ∈ Q : |f (x, t)| > } < h},
h ∈ [0, |Q|].
It is known that for 1 < s < r < ∞, we have Lr,1 (Q) ⊂ Lr,r (Q) = Lr (Q) ⊂ Lr,∞ (Q) ⊂ Ls,1 (Q). The following generalized Holder inequality holds r,∞ f g dxdt f Lr,∞ (Q) g r/(r−1),1 (Q) × Lr/(r−1),1 (Q) L (Q) ∀{f, g} ∈ L Q
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
5
with r ∈ (1, ∞). Let k (s) = k for s > k,
k (s) = s for |s| k,
k (s) = −k otherwise.
Lemma 2.1. Let w be in C(0, T ; C() ∩ W01,2 ()) and suppose that T |∇k (w(x, t))|2 dx dt C1 k + C2 ∀k. 0
Then w Ls/2,∞ (Q) C(|Q|){C1 + C2 }1/2 ,
2 < s < ∞.
The constants C, C1 , C2 are all independent of k. Proof. It is clear that k (w) is in L2 (0, T ; W01,2 ()). From the Sobolev imbedding theorem, we get k (w(., t)) Ls () ∇k (w(., t)) L2 () for all t ∈ [0, T ]. With k = h, we get {meas[x : x ∈ , |w(x, t)| > h]}
2/s
=
2/s {x:x∈,|w(x,t)|>h} dx
−2
Ch
∇h (w(., t)) 2L2 () .
Since w ∈ C(Q), we have meas{(x, t) : (x, t) ∈ Q, |w(x, t)| > h} =
T
meas{x : x ∈ , |w(x, t)| > h} dt.
0
Moreover, meas{x : x ∈ , |w(x, t)| > h} is a continuous function of t. Hence, there exists t˜ ∈ [0, T ] such that {meas{(x, t) : (x, t) ∈ Q, |w(x, t)| > h}}2/s = T 2/s {meas{x : x ∈ , |w(x, t˜)| > h}}2/s Ch−2 ∇h (w)(., t˜) 2L2 ()) . Since by hypothesis, w is in C(0, T ; W01,2 ()), we get for any r ∈ (1, 2) | ∇w(., tn ) rLr (t ) − ∇w(., tn ) rLr (t ) | ∇w(., tn ) rLr (t /t ) n n ∇w rC(0,T ;L2 ()) {meas(tn /t )}(2−r)/2 C(2−r)/2
with tn = {x : x ∈ , |w(x, tn )| < h},
t =
n
tn
and
tn → t.
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B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
Thus ∇h (w)(., t) Lr () is continuous in t. We obtain ∇h (w)(., t˜) Lr () lim −1 →0 −1
(
t˜+
∇h (w)(., s) Lr () ds
+ 1){C1 h + C2 }1/2
∀r ∈ (1, 2).
Let r → 2 and we get ∇h (w)(., t˜) L2 () ( −1 + 1){C1 h + C2 }1/2 . We have w Ls/2,∞ (Q) = sup h{meas{(x, t) : (x, t) ∈ Q; |w(x, t)| > h}}2/s h>0
sup
h{meas{(x, t) : (x, t) ∈ Q; |w(x, t)| > h}}2/s
0
+ sup h{meas{(x, t) : (x, t) ∈ Q; |w(x, t)| > h}}2/s
h0 h Ch0 |Q|2/s
The lemma is proved.
+ C −1 (C1 + C2 h−1 0 ).
3. The initial-value problem (1.5) The main result of the section is the following theorem. ˜ c , then for any given ∈ , there exists a solution X of (1.5) Theorem 3.1. Let v be in B with X(.; v) C(0,T ;L∞ ()) + X (.; v) L∞ (0,T ;L∞ ()) C{1 + ||}. The constant C is independent of v. Since K, defined by (1.4) is a closed, bounded closed convex subset of the Hilbert space L2 (0, T ; (J01,2 () ∩ W 3,2 ())∗ ), the projection P of L2 (0, T ; (J01,2 () ∩ W 3,2 ())∗ ) is well-defined and is a non-expansive mapping, i.e. Pf − Pg L2 (0,T ;(J 1,2 ()∩W 3,2 ())∗ ) f − g L2 (0,T ;(J 1,2 ()∩W 3,2 ())∗ ) 0
0
for all f, g ∈ L2 (0, T ; (J01,2 () ∩ W 3,2 ())∗ ). Proof of Theorem 3.1. (1) Let v ∈ BC and consider the differential equation d X(t) = Pv(X(t), t) on [0, T ], dt
X(0) = .
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
7
Since vˆ (., t) = Pv(., t) is a continuous function on for almost all t, the existence of a solution X of (1.5) on [0, T∗ ] follows from the Peano theorem. Since t ˆv(X, s) L∞ () ds || + T , X(., t) L∞ () || + 0
we deduce that T∗ = T . (2) The theorem on the continuous dependence on a parameter of the solution of an ordinary differential equation, shows that the function X of × [0, T ] → given by X(, t) = X(t) is in C 1 ( × [0, T ]). It is clear that X L∞ (0,T ;L∞ ()) Pv L∞ (0,T ;L∞ ()) 1. ˜ C , then v ∈ L2 (0, T ; (J 1,2 () ∩ W 3,2 ())∗ ) ∩ Lr (0, T ; L2r/(2−r) ()) (3) Let v be in B 0 and there exists {vk } ∈ BC such that vk → v
in Lr (0, T ; L2r/(2−r) ()).
With vk , we have from the previous part Xk = Pvk (Xk , t) on [0, T ],
Xk (, 0) =
and Xk (t) − Xk (s) C() |t − s| Pvk L∞ (Q) |t − s|. It follows that {Xk } is equicontinuous and there exists a subsequence such that Xk → in C(0, T ; C())). With K as stated, then Xk → in (L∞ (Q))weak∗ . On the other hand, we have vˆ k (Xk ) − vˆ () = vˆ k (Xk ) − vˆ (Xk ) + vˆ (Xk ) − vˆ (). • Since vˆ (., t) = Pv(., t) is continuous on and since Xk → uniformly in × (0, T ), we get vˆ (Xk ) → vˆ () a.e. in Q. Thus it follows from the definition of K and from the Lebesgue convergence theorem that vˆ (Xk ) → vˆ () in Lr (0, T ; L2r/(2−r) ()). • With vˆ k ∈ K, it follows from Aubin’s theorem that vˆ k →
in Lr (0, T ; L2r/(2−r) ()) and a.e. in Q.
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B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
On the other hand since P is a non-expansive mapping of L2 (0, T ; (J01,2 ()∩W 3,2 ())∗ ) into K, we get vˆ k → vˆ
in L2 (0, T ; (J01,2 () ∩ W 3,2 ()∗ )).
Hence = vˆ . We have {[, t] : |ˆvk (Xk (), t) − vˆ (Xk (), t)| > ; = Xk ()} ⊂ {[, t] : |ˆvk (, t) − vˆ (, t)| > }. Therefore, vˆ k (Xk , t) − vˆ (Xk , t) → 0
a.e. in Q
With ˆvk (Xk ) − vˆ (Xk ) L∞ (Q) C, the Lebesgue convergence theorem gives vˆ k (Xk ) − vˆ (Xk ) → 0 in Lr (Q). • Combining the two parts we obtain vˆ k (Xk ) − vˆ () → 0 in Lr (Q). It follows that = vˆ (); |t=0 = . Thus, = X(, t; v).
Now we shall consider the dependence of the solutions of (1.5) on v. The results will be needed when we apply the Schauder theorem to the nonlinear mapping T given by (1.7). ˜ C and let X(., t; vk ) be the solutions of the initial-value problem Theorem 3.2. Let vk ∈ B (1.5) with vk instead of v. Suppose that vk → v
in Lr (0, T ; L2r/(2−r) ()),
then there exists a subsequence such that {vk , X(.; vk )} → {v, X(.; v)} in Lr (0, T ; L2r/(2−r) ()) ∩ (Lr (0, T ; J01,r ()))weak∗ × (L∞ (Q))weak∗ . Moreover, X(., t; v) is a solution of the initial-value problem (1.5). Proof. With the estimates of Theorem 3.1, it is clear that there Xk (., t) − Xk (., s) C() |t − s|. Thus there exists a subsequence such that {Xk (., t), Xk (., t)} → {, }
in C(Q) × (L∞ (Q))weak∗ .
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
9
with Xk = Xk (., t; vk ). On the other hand, with vˆ k = Pvk we have |ˆvk (Xk (t), t) − vˆ ((t), t)| |ˆvk (Xk (t), t) − vˆ (Xk (t); t)| + |ˆv(Xk (t), t) − vˆ ((t), t)|. As in part (3) of the proof of Theorem 3.1, we get in Lr (Q).
vˆ k (Xk ) − vˆ () → 0
It follows that = vˆ (); |t=0 = i.e. = X(, t; v). Let (v; t) = {X(, t; v) : () = c} with () = c describing the valve at the initial time. Then in view of Theorem 3.2, we have the following lemma. Lemma 3.1. Let {vk , v, X( , t; vk ), X( ., t; v)} and be as in Theorem 3.2. Then (v; t) =
∞
(vk , t).
k k0
We now study the dependence of the source (v) on v and on F. Lemma 3.2. Let Fk be in L∞ (Q) and let vk , v, X(., t; vk ), X(., t; v) be as in Theorem 3.2 with (vk , Fk ) Mb (Q) 1, Suppose that Fk → F in
(v, F) Mb (Q) 1.
(L∞ (Q))
weak ∗ ,
then (vk , Fk ) → (v, F) in D (Q).
Proof. Let be in C0 (Q), then we have T fk ((t))(Xk ( ., t; vk )) d dt (vk , Fk ), = 0
=
0
T
(vk ,t)
(v,t) T
+ 0
fk ((t))(Xk (., ; vk )) d dt
(vk ,t)/(v,t)
fk (){(Xk ( ., vk ))} d dt.
Therefore, |(vk , Fk ) − (v, Fk ), |C (X( .; v)) − (Xk ( .; vk )) C(Q) + Cmeas{(v)/(vk )}. On the other hand, we have
T
(v, Fk ) − (v, F), = 0
(v,t)
{fk − f}(X( ., v)) d dt
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B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
for all ∈ C0 (Q). Since Fk → F in (L∞ (Q))weak∗ , it now follows from Theorem 3.2 that (vk , Fk ) → (v, F) in D (Q). ˜ C and suppose that vk → v in Lr (0, T ; L2r/(2−r) ()). Then Lemma 3.3. Let vk , v be in B 1 there exists {gk } ∈ C (Q) ∩ BC such that |(vk , F) − gk , | + |(v, F) − gk , |3/k
∀ ∈ C0 (Q).
˜ C there exists gk ∈ C 1 (Q) ∩ BC such that Proof. Since vk is in B 1 ∀, C0 (Q) 1. k On the other hand, from Lemma 3.2 we have |gk − (vk , F), |
|(vk , F) − (v, F), |
1 k
∀, C0 (Q) 1.
Hence, 2 k
|gk − (v, F), | The lemma is proved.
∀, C0 (Q) 1.
4. The Navier–Stokes equations Let ˜ C × L1 () × Mb (Q), {v, u0 , (v, F)} ∈ B
∇.u0 = 0,
(v, F) Mb (Q) 1.
Then there exists {g , u0 } in C01 (Q) × C02 () ∩ J01,2 () such that g L1 (Q) 1,
u0 L1 () u0 L1 ()
g − (v, F) Mb (Q) ,
u0 → u0 in D ().
In this section, we shall consider the initial boundary-value problem u − u + (u .∇)u = ∇p + g in Q ∇.u = 0 in Q u = 0 on j × (0, T ), u (x, 0) = u0 (x) in .
(4.1)
The main result of the section is the following theorem. Theorem 4.1. Let {v, u0 , g , u0 } be as above. Then there exists a unique solution u of (4.1) in C(0, T ; J01,2 () ∩ W 2,2 ()) with u ∈ L2 (0, T ; J01,2 ()) and u Lr (0,T ;J 1,r ()) + u L1 (0,T ;(J 1,r ()∩W 3,2 ())∗ ) 0
0
C{1 + u0 L1 () + (v, F) Mb (Q) } for r ∈ (1, 4/3). The constant C is independent of , v.
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
11
The existence of a unique solution u of (4.1) is well-known (cf. [6, pp. 79–81]. We shall now establish the estimates of the theorem. Lemma 4.1. Let u be as in the theorem. Then u L∞ (0,T ;L1 ()) C{1 + || + u0 L1 () }. The constant C is independent of , v. Proof. (1) Since u is in C(Q), it follows that supQ |uj, (x, t)|= j, exists. Let =inf j j, and let if inf{ , 1} < s, inf{1, } = s if |s| inf{ , 1}, − inf{1, } if − inf{ , 1} < s. With u ∈ C 1 (0, T ; C()) ∩ L2 (0, T ; J01,2 ()), we now consider the testing function (u ) = ((u1, ), (u2, )). Set {(x, t) :, (x, t) ∈ Q; |uj, (x, t)| inf{1, }} ⊂ supp(uj, ) D1 = j
j
is a non-empty subset and j × (0, T ) ⊂ jD1 . For each t ∈ [0, T ], we have (u ) ∈ W01,2 () and thus we get the decomposition (u (., t)) = (u ˜ (., t)) + ∇q(., t),
(u ˜ (., t)) ∈ J01,2 (),
∇q ∈ W01,2 ().
It is clear that maps L2 () into J0 () and W01,2 () into J01,2 (). Since (u (., t)) L∞ () 1, we have (u ˜ (., t)) L∞ ()
inf
|((u ˜ (., t)), h)|
inf
|((u (., t)), h)|
h L1 () 1;∇.h=0 h L1 () 1; ∇.h=0
(u (., t)) L∞ () 1. Thus, (u ˜ ) L∞ (0,T ;L∞ ()) 1. We have ˜ (., t)) = (u (., t), (u (., t)) − ∇q) = (u , (u (., t))) (u (., t), (u as (u (., t), ∇q) = −
2 j,k=1
(Dj uk, (., t), Dk Dj q) = (div(u (., t)), q) = 0.
12
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
We have used the fact that ∇q ∈ W01,2 (). Since (v) W 1,2 () (v) ˜ J 1,2 () 0
0
v W 1,2 () , 0
∀v ∈ W01,2 (),
we deduce that ˜ (v) L∞ () 1
∀v ∈ W01,2 ().
With u ∈ L2 (0, T ; J01,2 ()), we get ˜ )) = (u , (u ) + ∇q) = (u , (u )). (u (., t), (u (2) Set
s
(s) =
(r) dr.
0
Multiplying (4.1) by (u ˜ ) and we obtain 2 d (u (x, t)) dx + (uj, )|Dm uj, |2 dx dt j,m=1 2 + uj, Dj um, (u ˜ m, ))(x, t) dx = g (v)(u ˜ ) dx. j,m=1
Since ∇.u = 0, we have 2 2 ˜ m, ) dx = − uj, Dj um, (u um, um, ˜ (um, ) Dj um, dx j,m=1
j,m=1
= −
2
j,m=1
=
2 m=1
uj, Dj
s ˜ (s) ds
dx
0
div(u )
um,
um,
s ˜ (s) ds
dx = 0.
0
It follows that d (u (x, t)) dx g (v)(u ˜ (x, t)) dx. dt Hence
u (., t) L1 () C 1 + || +
(u ) dx ds
C{1 + || + g (v) L1 () + u0 L1 () } C{1 + || + u0 L1 () }. The constant C is independent of v, .
(4.2)
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
13
Lemma 4.2. Suppose all the hypotheses of Theorem 4.1 are satisfied. Let u be as in Lemma 4.1, then u Lr (0,T ;J 1,r ()) C{1 + || + u0 L1 () } for r ∈ (1, 4/3). 0
The constant C is independent of v, . Proof. (1) Let
⎧ 1 ⎪ ⎪ ⎪ ⎨s − k k (s) = 0 ⎪ ⎪ ⎪ ⎩s + k −1
if if if if if
1 + k < s, k s 1 + k, − k s k, − 1 − k s − k, s − 1 − k.
• Let k be a positive integer with k < = inf j
sup |uj, (x, t)| , Q
k K = sup j
sup |uj, (x, t)| . Q
Let Bk =
2
{(x, t) : (x, t) ∈ Q; k |uj, (x, t)| k + 1}
j =1
then Bk ∩ { j supp(uj, )} is non-empty. With k (u ) ∈ W01,2 () we have as in Section 2, the decomposition ˜ k (u ) + ∇q, k (u ) =
{ ˜ k (u ), ∇q} ∈ J01,2 () × (G() ∩ W01,2 ()).
Moreover, ˜ k (u (., t)) L∞ () 1
∀t ∈ [0, T ]
˜ k (u (., t)) L∞ () 1
∀t ∈ [0, T ].
and
As in the Proof of Lemma 4.1, we have ˜ k (u (., t)) = (u (., t) − u (., t), k (u )(., t)). (u (., t) − u , Taking the pairing of (4.1) with ˜ k (u ), and we get 2 d k (u (x, t)) dx + k (uj, )|Dm uj, (x, t)|2 dx dt j,m=1 2 + um, Dm uj, ˜ k (uj, (x, t)) dx C(1 + ||) + (v, F) Mb (Q) ) j,m=1
14
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
with
s
k (s) = 0
k (r) dr.
An integration by parts yields 2 ˜ (um, (x, t)) dx uj, Dj um, k j,m=1
=−
2 j,m=1
=−
2
j,m=1
=
2 m=1
˜ (um, )Dj um, dx uj, um, k uj, Dj
0
div(u )
um, (x,t)
um,
0
˜ (s) ds s k
˜ (s) ds s k
dx
dx = 0.
It follows that 2 k (u (x, t)) dx + k (um, )|Dj um, |2 dx
j,m=1
C{1 + || + (v, F) Mb (Q) + u0 L1 () } and hence Bk
|∇u |2 dx C{1 + || + u0 L1 ()) }.
(4.3)
Let r ∈ (1, 4/3) then an application of the Holder inequality yields k|Bk | |uj, (x, t)| dx dt u L3r/2 (Bk ) |Bk |(3r−2)/3r . Bk
Thus, |Bk | k −3r/2 u L3r/2 (B ) . 3r/2
(4.4)
k
Again an application of the Holder inequality together with the estimate (4.4) gives ∇u rLr (Bk ) |Bk |(2−r)/2 ∇u rL2 (B Ck
−3r(2−r)/4
k) 3r(2−r)/4 u L3r/2 (B ) k
× {1 + || + u0 L1 () + (v, F) Mb (Q) }. • We now consider the case
= inf j sup |uj, (x, t)| < k. Q
(4.5)
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
15
Since = supQ us, , the set supp(us, ) {(x, t) : k |us, (x, t) k + 1} is empty. We have only to study the case when supQ |uj, (x, t)| > k with j = s. An identical argument as before gives again the estimate (4.5). (2) Let k0 be a fixed positive number and let be the truncated function if s > inf{k0 , }, inf{k0 , } (s) = s if |s| inf{k0 , }, − inf{k0 , } if s < inf{k0 , }. Then as in Lemma 4.1, we have |∇u |2 dx dt C{k0 + u0 L1 () + (v, F) Mb (Q) }. Dk0
Now the Holder inequality gives r/2
∇u rLr (Dk ) ||(2−r)/2 ∇u L2 (D
k0 )
0
C{k0 + ||(1 + u0 L1 () + (v, F) Mb (Q) )}. (3) It follows from (4.4) to (4.5) that ∇u rLr (Q) C(k0 + || + u0 L1 () + (v, F) Mb (Q) + C(1 + || + u0 L1 () + (v, F) Mb (Q) ×
K k=k0
u L3r/2 (B ) k −3r(2−r)/4 u L3r/2 (Q) . 3r(2−r)/4
3r/2
k
(4.6)
Applying the Holder inequality to (4.6) with exponents 2/r, 2/(2 − r) and we obtain ∇u rLr (Q) C{1 + || + u0 L1 () + (v, F) Mb (Q) ) ⎛ ⎧ ⎫r/2 ⎞ K ⎨ ⎬ ⎟ ⎜ 3r/2 k −3(2−r)/2 × ⎝1 + u L3r/2 (Q) ⎠ ⎩ ⎭ k=k0
C{1 + || + u0 L1 () + (v, F) Mb (Q) } ⎧ ⎫r/2 ∞ ⎨ ⎬ 3r/2 × k −3(2−r)/2 u L3r/2 (Q) . ⎩ ⎭ k=k0
(4.7) The series converges since r ∈ (1, 4/3). (4) An application of the Holder inequality gives 1/3
2/3
u L3r/2 () u L1 () u L2r/(2−r) () .
(4.8)
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B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
Taking the estimate of Lemma 4.1 into account, we get 3r/2
u L3r/2 (Q) C{1 + || + u0 L1 ()
+ (v, F) Mb (Q) } × u rLr (0,T ;L2r/(2−r) ()) .
(4.9)
(5) The Sobolev imbedding theorem yields u rLr (0,T ;L2/(2−r) ()) C u r r
L (0,T ;W01,r ())
.
It follows from (4.6) to (4.9) that u rLr (0,T ;L2/(2−r) ()) C{1 + || + u0 L1 () + (v, F) Mb (Q) } ⎛ ⎧ ⎫r/2 ⎞ ⎨ ⎬ ⎟ ⎜ 3r/2 k −3(2−r)/2 × ⎝1 + u L3r/2 (Q) ⎠ ⎩ ⎭ k=k0
C(1 + || + u0 L1 () + (v, F) Mb (Q) ) ⎛ ⎧ ⎫r/2 ⎞ ⎨ ⎬ ⎟ ⎜ k −3(2−r)/2 × ⎝1 + u rLr (0,T ;L2/(2−r) ()) ⎠. ⎩ ⎭ k=k0
(4.10) Since the series converges for r ∈ (1, 4/3), there exists k0 such that u rLr (0,T ;L2/(2−r) ()) C{1 + || + u0 L1 () + (v, F) Mb (Q) }.
(4.11)
Using the estimate (4.11) in (4.9) and we have 3r/2
u L3r/2 (Q) C{1 + || + u0 L1 () + (v, F) Mb (Q) }.
(4.12)
With (4.12) in (4.7) and we obtain u Lr (0,T ;J 1,r ()) C{1 + || + u0 L1 () + (v, F) Mb (Q) }. 0
The lemma is proved.
Lemma 4.3. Suppose all the hypotheses of Theorem 4.1 are satisfied. Then u L1 (0,T ;(J 1,r ()∩W 3,2 ())∗ ) C{1 + || + u0 L1 () + (v, F) Mb (Q) } 0
with r ∈ (1, 4/3). The constant C is independent of , v, F. Proof. Let w be an element of J01,r () ∩ W 3,2 () and consider the equation (u (., t), w) + (∇u (., t), ∇w) = −((u .∇)u (., t), w) + (g , w). With the estimate of Lemma 4.2, we have from Lemma 2.1 u Ls/2,∞ (Q) C{1 + || + u0 L1 () + (v, F) Mb (Q) },
2 < s < ∞.
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
17
We have |uj, Dj uk, wk (x)| dx dt wk L∞ () uj, Ls/2,∞ (Q) Dj uk Ls/(s−2),1 (Q) Q
C w W 3,2 () {1 + || + u0 L1 () + (v, F) Mb (Q) )} × Dj uk, Ls/(s−2),1 (Q) . With s > 2r/(r − 1) > 8, we have Lr,1 (Q) ⊂ Lr (Q) ⊂ Ls/(s−2),1 (Q). Therefore |uj, Dj uk, wk (x)| dx dt C{1 + || + u0 L1 () + (v, F) Mb (Q) } Q
× u Lr (0,T ;J 1,r ()) w W 3,2 () 0
C{1 + u0 L1 () + (v, F)} w W 3,2 () . With (4.13), it is now trivial to establish the estimate of the lemma.
(4.13)
Proof of Theorem 4.1. The theorem is now an immediate consequence of Lemmas 4.1– 4.4. 5. A free boundary problem The main result of the paper is the following theorem. ˜ C be as in (1.8) Theorem 5.1. Let {u0 , F} be in L1 () × L∞ (Q) with ∇.u0 = 0 in , let B and let P be the projection of L2 (0, T ; (J01,2 () ∩ W 3,2 ())∗ ) onto K defined by (1.4 ). Then there exists ˜ C × C(0, T ; W 1,r ()) × Lr (0, T ; W 1,r ()) {u, X( ., ; u); X ( .; u)} ∈ B with (u.∇)w integrable in Q for all w ∈ S and such that T T T (u, w ) dt + (∇u, ∇w) dt = (u.∇w, u) dt + (u0 , w( ., 0)) − 0
0
0
+ (u, F), w. Moreover, X(, t; u) = +
T
Pu(X(, s; u)) ds.
0
The set S is given by S = {w : {w, w } ∈ C(0, T ; J01,r () ∩ W 3,2 ()) × Lr/(r−1) (Q); w( ., T ) = 0}. The force field (u, F) is in Mb (Q) with support on (u, t) = {X(, t; u) : ∈ , () = c}. The initial position of the free boundary is the given curve () = c.
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B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
The proof of the main result of the paper is carried out in this section. Let {v, u0 } ∈ ˜ C × L1 (), and let {g , u } be in C 1 (Q) × C 1 () with B 0 0 0 g L1 (Q) (v, F) Mb (Q) 1,
u0 L1 () u0 L1 ()
and g − (v, F) Mb (Q) + u0 − u0 L1 () . ˜ C × L1 () and {g , u } as From Theorem 4.1, we know that for any given {v, u0 } ∈ B 0 ˜ C . We define the nonlinear mapabove, there exists a unique solution u of (4.1) with u ∈ B ˜ C , considered as a subset of Lr (0, T ; L2r/(2−r) ()) into Lr (0, T ; L2r/(2−r) ()) ping A of B by A(v) = u .
(5.1)
˜ C of Lr (0, T ; L2r/(2−r) ()) into It is clear that A maps the compact convex subset B itself. To apply the Schauder fixed point theorem, it suffices to show that A is continuous. Lemma 5.1. Let A be as in (5.1), then it is a continuous mapping of Lr (0, T ; L2r/(2−r) ()) into Lr (0, T ; L2r/(2−r) ()). ˜ C , vn → v in Lr (0, T ; L2r/(2−r) ()). Then there exists, as done Proof. (1) Let {vn } ∈ B in Lemma 3.4, {gn , g } ∈ C 1 (Q) such that gn − g L2 (Q) 1/n,
g − (v, F) Mb (Q) .
An application of Lemma 3.3 gives g − (vn ) Mb (Q) 2/n + . Let A(vn ) = un where un is the unique solution of (4.1) with g , v replaced by gn , vn . From the estimate of Theorem 4.1, we have un Lr (0,T ;J 1,r ()) + (un ) L1 (0,T ;(J 1,r ()∩W 3,2 ())∗ ) C{1 + || + u0 L1 () }. 0
0
It follows from the above estimate and from Aubin’s theorem that there exists a subsequence such that un → u
in (Lr (0, T ; J01,r ())weak ∩ Lr (0, T ; L2r/(2−r) ()) and a.e. in Q.
(2) With {gn , un } ∈ C01 (Q) × C 1 (), we have the standard estimate un L∞ (0,T ;L2 ()) + un L2 (0,T ;J 1,2 ()) C{1 + gn L2 (Q) + u0 L2 ()) } 0
C{1 + g L2 (Q) + u0 L2 () }.
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
19
The constant C is independent of n, . Let {w, w } be in C(0, T ; J01,r () ∩ W 3,2 ()) × 1,r/(r−1) ()) with w( ., T ) = 0, then we have Lr/(r−1) (0, T ; J0 T T T − (un , w ) dt + (∇un , ∇w) dt = (u0 , w( ., 0)) + ((un .∇)w, un ) 0 0 0 T + (gn , w) dt. 0
Let n → ∞ and we get T − (u , w ) dt + 0
T 0
T (∇u , ∇w) dt = (u0 , w( ., 0)) + ((u .∇)w, u ) dt 0 T + (g , w) dt. 0
It follows that u − u + (u .∇)u = ∇p + g in D (Q) ∇.u = 0, u = 0 on j × (0, T ) u ( ., 0) = u0 in . It is known that there exists a unique solution of the problem. Hence A(v) = u and the operator A is continuous from Lr (0, T ; L2r/(2−r) ()) into itself. ˜ C such that A(u ) = u . Lemma 5.2. There exists u ∈ B Proof. The lemma is now an immediate consequence of the Schauder fixed point theorem. Proof of Theorem 5.1. Let u be as in Lemma 5.2. Then we have u − u + (u .∇)u = ∇p + g (u ) in Q ∇.u = 0 in Q, u = 0 on j × (0, T ) u ( ., 0) = u0 in
(5.2)
with d X(, t; u ) = Pu (X(, t; u ); t) on (0, T ), dt
X(, 0; u ) = .
(5.3)
˜ C is a compact subset of Lr (0, T ; L2r/(2−r) ()), it follows from Theorem (1) Since B 4.1 that there exists a subsequence such that u → u
in Lr (0, T ; J01,r ())weak ∩ Lr (0, T ; L2r/(2−r) ()) and a.e. in Q.
Moreover, u → u
in D (0, T ; (J01,r () ∩ W 3,2 ())∗weak ).
20
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
(2) Let → 0 in (5.2) and it follows from Theorem 3.1 that there exists X ∈ C(0, T ; J01,r ()) with X ∈ Lr (0, T ; J01,r ()) such that d X(, t; u) = Pu(X(, t; u); t) on (0, T ), dt
X(, 0; u) = .
(5.4)
(3) Let w be in C 1 (0, T ; J01,r () ∩ W 3,2 ()) and consider the term T T ((u .∇)u , w) dt = − ((u .∇)w, u ) dt. 0
0
Since u → u a.e. in Q, the set ⎧ ⎫ 2 ⎨ ⎬ S = (x, t) : (x, t) ∈ Q; |{uj, (x, t)uk, (x, t) − uj (x, t)uk (x, t)}| 1 ⎩ ⎭ j,k=1
are increasing as → 0 and meas(S ) → |Q|. Thus the Lebesgue convergence theorem gives
T
lim
→0
2
j,k=1
0
{uj, uk, − uj uk }Dj wk (x, t)} dx dt = 0
for all w ∈ C(0, T ; J01,r () ∩ W 3,2 ()). On the other hand, we have T
(u , w) dt = − 0
T
(u , w ) dt + (u0 , w(., 0)) ∀w ∈ S.
0
Thus,
T
0
(u , w) dt → −
T
(u, w ) dt + (u0 , w(., 0))
∀w ∈ S.
0
From (5.2), we deduce that
T
lim
0
=−
2
j,k=1 T
uj, uk, Dj wk (x, t) dx dt
T
(u, w ) dt +
0
(∇u, ∇w) dt + (u, F), w + (u0 , w(., 0))
0
for all w ∈ S. It follows that T (u.∇)w.u dx dt = − 0
T
T
(u, w ) dt +
0
(∇u, ∇w) dt
0
+ (u, F), w + (u0 , w(., 0))
for all w ∈ S. The theorem is proved .
(5.5)
B.A. Ton / Nonlinear Analysis 64 (2006) 1 – 21
21
6. An inverse problem Let G be a bounded subset of the plane with G ⊂ and let be a L1 (0, T ; L1 (G))function, representing the observed values of u in G × (0, T ). We consider the function T J(u, u0 , F) = |u − | dx dt, 0
G
where {u, X( , t; u)} is a solution of the free boundary problem of Theorem 5.1. Let V (u0 ) = inf{J(u, u0 , F) : ∀u solution of (1.1), (1.3).(1.5), ∀F, F L∞ (Q) 1}. We now have the following result which allows us to find the boundary force from the observed values of u in a fixed subregion. Theorem 6.1. Let u0 ∈ L1 () with ∇.u0 = 0. Then there exists ˜ C × C(0, T ; W 1,r ()) ∩ L∞ (Q) × Lr (0, T ; W 1,r ()) × L∞ (Q) ˜ ∈B ˜ X ˜ , F} ˜ X, {u, ˜ ˜ u0 , F). such that V (u0 ) = J(u, Proof. (1) Let {un , Fn } be a minimizing sequence of the above optimization problem, with V (u0 )J(un , u0 , Fn ) V (u0 ) +
1 , n
˜ C , Fn L∞ (Q) 1. un ∈ B
˜ C is a compact subset of Lr (0, T ; L2r/(2−r) ()), there exists a subsequence such Since B that un → u˜ in Lr (0, T ; L2r/(2−r) ()). With Fn → F˜ in (L∞ (Q))weak∗ , we deduce from Lemma 3.2 that ˜ ˜ F) in D (Q). (un , Fn ) → (u, ˜ in (2) From Theorem 3.2, we get X(., t; un ) → X(., t; u) (Lr (0, T ; J01,r ()))weak ∩ (L∞ (Q))weak∗ ∩ Lr (0, T ; L2r/(2−r) ()). Now as in part (3) of the Proof of Theorem 5.1, we deduce that u˜ is a weak solution of ˜ ˜ u0 , F). (1.1) and thus V (u0 ) = J(u, References [1] M.F. Betta, A. Mercaldo, F. Murat, M.M. Porszio, Existence of renormalized solutions to nonlinear elliptic equations with lower order terms and right hand side measure, J. Math. Pures Appl. 81 (2002) 533–566. [2] L. Boccardo, T. Gallouet, Nonlinear elliptic and parabolic equations involving measure data, J. Funct. Anal. 87 (1989) 149–169. [3] L. Boccardo, T. Gallouet, L. Orsina, Existence and uniqueness of entropy solutions for nonlinear elliptic equations with measure data, Ann. Inst. Henri Poincare Anal. Non Lineaire 13 (1996) 539–551. [5] C. Peskin, Flow patterns around heart valves, J. Comput. Phys. 10 (2) (1972) 252–271. [6] C. Peskin, Mathematical Aspects of Heart Physiology, Courant Institute of Mathematical Sciences, New York University, 1973–1974. [7] B.A. Ton, Time dependent Stokes equations with measure data, Abstr. Appl. Anal. 17 (2003) 953–973.