DISTRIBUTED AND BOUNDARY CONTROL OF THE VISCOUS ...

DISTRIBUTED AND BOUNDARY CONTROL OF THE VISCOUS BURGERS' EQUATION Hung V. Ly

Department of Mathematics, University of California Irvine, CA 92697-3875, USA. Email: [email protected]. Kenneth D. Mease

Department of Mechanical & Aerospace Engineering, University of California Irvine, CA 92697, USA. Email: [email protected]. Edriss S. Titi

Department of Mathematics and Department of Mechanical & Aerospace Engineering, University of California Irvine, CA 92697-3875, USA. Email: [email protected].

ABSTRACT Earlier results for distributed and boundary controls of the viscous Burgers' equation were established by Burns et al. and Byrnes et al.. In their results there are technical restrictions on the sizes of the initial data. In this paper we relax these restrictions, as well as treat the Burgers' equation with other nonlinear boundary conditions. AMS Subject Classi cation: 93D15, 93D20, 93D25, 35K05, 35K55 & 35K60

1 INTRODUCTION AND NOTATION Simple and yet containing many features of uid dynamics Burgers', equation has received much attention and special interest from the control community (see e.g. [1] [3]-[11] [13] [16] [24] [28] [29]). With the greater ambition of controlling nonlinear evolutionary equations, like the Navier Stokes equation, many people from both elds of mathematics and control ([12] [20]-[26]) have used the Burgers' equation as a model for their analytical and numerical studies. In [5] [6] [29], Burns and Kang considered the viscous Burgers equation with the Dirichlet boundary conditions on the interval [0; l]. They showed that if the solution starts out \small enough", it will go to zero exponentially. However, the rate at which the solution goes to zero depends on the viscosity, which could be very small. Thus they were interested in driving solutions to zero at a desired faster exponential rate. They succeeded by using distributed and boundary feedback control (via linear quadratic regulator). However, it was required that solution be \suciently small" initially. The rst part of this paper is devoted to removing the restriction on the \smallness" of initial data for the viscous Burgers equation with the Dirichlet boundary conditions in [5] [6] [29]. Byrnes et al. in [7] [8] [9] [11] considered the viscous Burgers equation with the Neumann boundary conditions. They showed by using boundary control that if the solution starts out \small enough", it will go to zero exponentially. Our second goal in this paper is to relax the restriction of \small initial data" in Byrnes et al. [7] [8] [9] [11] and to prove their result for a larger set of initial data. This paper is structured in the following fashion. In Sections 2 we consider the viscous Burgers equation with the Dirichlet boundary conditions on the interval [0; l]. We show analytically how to remove the restriction of the \smallness" of the initial data in [5] [6] [29] for the open-loop problem and prove the global existence and uniqueness for the close-loop problem for any initial data in L2. We establish in Section 3 the existence, uniqueness and regularity of solutions to the Burgers' equation with Neumann homogeneous boundary conditions (the open-loop problem) for any initial data in L2. In particular, we show that these solutions satisfy some kind of a maximumprinciple which guarantees the global existence. In fact, we show that the solutions are classical. We also discuss in Section 3 the boundary-controlled viscous Burgers equation (the closed-loop problem). In Section 4 we will discuss the well-posedness of the Burgers' equation subject to some nonlinear boundary conditions suggested by J.A. Burns [4]. The results of a preliminary version of this paper have been reported in [30]. We end this section with some notations and well-known inequalities which 2

will be used throughout the paper. Let Lp(0; l) and H m (0; l) denote the usual Lp-Lebesque space of integrable functions and H m -Sobolev spaces respectively for 1  p < 1 and m 2 IR. The inner product in L2(0; l) will be denoted by (
;
). We also denote by H01(0; l) the closure of C01 in the H 1(0; l)-topology, and by H ?1(0; l) the dual space of H01(0; l). (see [2] and [39] for details). For any Banach space X and 1  p < 1 we de ne Lp((a; b); X ) = fw(t) 2 X a.e. t 2 (a; b) j ba kw(t)kpX dt < 1g: Here Lp((a; b); X ) again is a Banach space with the norm kwkLp((a;b);X ) = ( ba kw(t)kpX dt)1=p: Let X  be the dual space of X , we denote h; uiX  the action of the functional  2 X  on u 2 X . Because the following inequalities are used frequently in this paper, we will give complete formulations of them here and only refer to their names whenever necessary. Young's inequality Suppose A and B are nonnegative real numbers, then for every  > 0, we have p q AB  1p Ap + q Bq ; where 1 < p < 1 and 1p + q1 = 1: (1.1) R

R

Embedding Theorems (see for instance [2] [31] and [36])

1. For every u 2 H m (0; l) where m < 21 , there are constants c0 and c1 depending on l and m such that

kukLq (0;l)  c0kukH m(0;l)  c1kukmH (0;l)kuk1L?(0m;l) 1

2

(1.2)

where 1q = 12 ? m. 2. If m1 < m2, then H m (0; l) is compactly embedded in H m (0; l). 2

1

3. If m > 21 + j , then H m(0; l) is compactly embedded in CBj [0; l] for j  0: Agmon's inequality for IR1 (see for example [39]) For every u 2 H 1(0; l), there is a constant c2 depending only on l such that

kukL1(0;l)  c2kuk1L=2(0;l)kuk1H=2(0;l):

(1.3)

@u k2 1 k u k L (0;l)  k l @x L (0;l) for any u 2 H0 (0; l);

(1.4)

2

Poincare inequalities 1 2 2

2

1

2

1 ku ? uk2  k @u k2 1 L (0;l) l2 @x L (0;l) for any u 2 H (0; l); where u denotes the average of u over [0; l]. 2

2

3

(1.5)

2 BURGERS' EQUATION WITH DIRICHLET BOUNDARY CONDITIONS

2.1 Stability of the Uncontrolled System

In this subsection we consider the dynamical system comprised of the viscous Burgers' equation with Dirichlet boundary conditions on a nite interval [0; l] and t > 0. @w (x; t) ?  @ 2w (x; t) + w(x; t) @w (x; t) = 0; 0 < x < l; t > 0 (2.1) @t @x2 @x w(x; 0) = w0(x) (2.2) w(0; t) = w(l; t) = 0; for t > 0 (2.3) where  > 0 is the viscosity. It was shown in [5] [6] that the solution of the system (2:1)- (2:3) with small initial data in H01(0; l) will go to zero exponentially. Here we will rst establish the well-posedness of the system (2:1)-(2:3) for any initial data in L2(0; l), then we show that for any initial data in H 1(0; l), its solution will go to zero asymptotically regardless of the size of the H 1(0; l)-norm of the initial data, an improvement over the results in [5] [6]. We set A = ? dxd subject to the Dirichlet boundary conditions. A is an unbounded linear self-adjoint positive operator from D(A) = f 2 H 2(0; l) j (0) = (l) = 0g = H 2(0; l) \ H01(0; l) onto L2(0; l). In addition the operator A can be extended in a unique fashion as a map A : H01(0; l) ?! H ?1 (0; l) such that l @u @v @v ): (y) @x (y)dy = ( @u ; hAu; viH ? = 0 @x @x @x One can further show that A?1 : L2(0; l) ?! L2(0; l) exists and is a self-adjoint positive compact operator. Thus there exists a complete orthonormal set n of eigenfunctions of A such that An = n n and 0 < 1  2  : : :, where n = ( nl )2. In fact, we have 2

2

Z

1

s

n (x) = 2l sin( n l x);

0 0 be the desired exponential rate at which we want the solutions to the heat equation to go to zero. Let w(x; t) be the solution to the Heat equation

@w ?  @ 2w = 0 @t @x2 8

with We set

w(x; 0) = w0(x) ; w(0; t) = w(l; t) = 0: z(x; t) = w(x; t)et;

(2.24)

then z(x; t) will satisfy @z (x; t) =  @ 2z (x; t) + z(x; t) 0 < x < l; t > 0 (2.25) @t @x2 z(x; 0) = z0(x) (2.26) z(0; t) = z(l; t) = 0: (2.27) Notice that if we just control z(x; t) to remain bounded uniformly, then w(x; t) will approach zero exponentially at the desired rate of at least . We also note that the operator A; := (@xx + I ), with Dirichlet boundary conditions, induces a set of eigenvalues and eigenfunctions 2 2 n ;n =  ? l2 (2.28) (2.29) ;n = 2l sin( nl x); respectively. In fact f;ng's form an orthonormal basis in L2(0; l). If ;n  0 for all n = 1; 2; : : :, there is nothing to prove. Hence we assume that n is the largest positive integer for which we have ;n > 0. Thus there exists an integer n such that for each n = 1; 2; : : : ; n, n =  ( nl )2 <  and the corresponding f;1; ;2; : : : ; ;n g are the unstable modes that we want to control. Burns and Kang [5] [6] modi ed (2:25), (2:26) and (2:27) to the following control driven system @z (x; t) =  @ 2z (x; t) + z(x; t) + B (u) 0 < x < l; t > 0 (2.30) @t @x2 y(t) = Cz(t) (2.31) where B (u) is the control input into the system and y(t) = Cz(t) is the output observation; both of which are de ned as follows. u(t) is the feedback control that we are after to drive solution of (2:30) to zero asymptotically to zero and B (u(t)) = b(x)u(t); (2.32) is the bounded control input linear operator from U := IR to L2(0; l) and b(x) 2 L2(0; l). @z ( ); : : :; @z ( )) (2.33) y(t) = Cz(t) = (z(x1); : : : ; z(xk ); @x 1 @x m s

g

e

e

9

g

is the output observation operation mapping from H01 to Y := IRk+m , and xi 2 (0; l) for 1  i  k, j 2 (0; l) for 1  j  m. Also xi + z(xi) = 21 z(s)ds; (2.34) xi ? @z ( ) = 1 j + @z (s)ds = z(j + ) ? z(j ? ) (2.35) @x j 2 j ? @x 2 and  is chosen so that (xi ? ; xi + )  (0; l) for 1  i  k and (j ? ; j + )  (0; l) for 1  j  m. With respect to the basis f;ng's, we can have a better observation at the changing values of the sensors via the following: xi + z(xi ) = 21 x ? z(s)ds i 1 1 xi + ;n(s)ds ( z;  = ;n ) 2 Z

e

Z

g

Z

e

Z

X

xi ?

n=1 p 1 2l X

nxi ) sin ( n )(z;  ) sin( ;n l l n=1 2

=

(2.36)

@z ( ) = 1 j + @z (s)ds @x j 2 j ? @x   = z ( j +  ) ? z (  j ?  ) 2 1 (z; ;n) 21 (;n(j + ) ? ;n(j ? )) = n=1 p 1 2  p cos ( nl j ) sin ( n = (2.37) l )(z; ;n): n=1  l Operating within the Linear Quadratic Regulator (LQR) framework (see, for instance, Banks and Kunish [3], Gibson [22], and Pritchard and Salamon [35]), we seek u(
) 2 L2(0; 1 : U ) that minimizes the cost functional Z

g

X

X

J (u) =

Z

1

0

fky(t)k2Y + ku(t)k2U gdt < 1

subject to the control system (2:30)-(2:31). We remark here that the stabilizability condition bn = (b(x); ;n) 6= 0 for all n = 1; 2; : : : ; n; (2.38) is to ensure that we can aect every unstable mode. Thus by assuming the stabilizability condition (2:38), we are guaranteed that no linear unstable mode 10

will not be controlled by the feedback control. Also if we examine (2:36) and (2:37), we will see that the detectability condition For each n = 1; 2; : : : ; n; there exists at least one xi or j such that xi 62 f kln : k = 1; 2; :::; n ? 1g or (2.39) l : k = 0; 1; 2; :::; n ? 1g:  j 62 f (2k2+1) n is equivalent to making sure that we can observe the unstable modes from the output. Otherwise the coecient of sines and cosines in (2:36) and (2:37) are zero, and hence y(t) = 0. If we assume both the stabilizability (2:38) and detectability (2:39) conditions, then this LQR problem has unique optimal solution u(
) 2 L2(0; 1 : U ) (see Banks and Kunish [3] and Pritchard and Salamon [35]) given by

u(t) = ?B ; z(t) 8t  0

(2.40)

where ; 2 L(L2(0; l)) is the unique nonnegative self-adjoint operator satisfying the algebraic Riccati equation (@xx + I ); z + ; (@xx + I )z ? ; BB ; z + C Cz = 0: (2.41) Moreover, for every  2 spec (A; ? ; ),   ? for some  > 0. In other words, under the in uence of the LQR control, the solution z(x; t) of (2:25)(2:27) will tend to zero as t tends to in nity. Note that the rate at which z(x; t) tends to zero is now  which again depends on  ; however the rate that w(x; t) of the heat equation tends to zero will be  +   , as desired. We next write the LQR control in terms of w(x; t) of the heat equation and then apply it to the nonlinear Burgers' equation. Using (2:24) and (2:32), one can write (2:30) as

@w (x; t) =  @ 2w (x; t) + b(x)e?tu(t): @t @x2

(2.42)

De ning the feedback operator

K; = ?B ;

(2.43)

and applying

u(t) = e?tu(t); (2.44) where u(t) is the rate-controlled feedback for the heat equation, we have from (2:40), (2:43) and (2:44) that u = e?tu = ?B ; ze?t = ?B ; w = K; w 11

(2.45)

Note that K; is a linear functional on L2(0; l), therefore by the Riesz representation theorem there exists a unique feedback gain function k; (
) 2 L2(0; l) such that l (2.46) K; w = k; (s)w(s; t)ds; Z

and so by (2:45) u(t) =

lk

0

s)w(s; t)ds: Now

0 ; (

R

Z l ? t ? t B (u)e = b(x)u(t)e = b(x)u = b(x) k

; (s)w(s; t)ds;

0

(2.47)

therefore from (2:42) and (2:47) the rate-controlled heat equation will then be @w (x; t) =  @ 2w (x; t) + b(x) l k (s)w(s; t)ds ; @t @x2 0 w(x; 0) = w0(x) ; w(0; t) = w(l; t) = 0: Note that R := BK; is a bounded linear operator L2(0; l) into L2(0; l) where Z

Z

l

Rv := BK; v = b(x) k; (s)v(sds 0

for every v 2 L2(0; l). Moreover, with this feedback control and the stabilizability and detectability conditions we have for any  2 spec (@xx + BK; ) that   ?( + ). Applying this control to the fully nonlinear viscous Burgers' equation we obtain @w (x; t) =  @ 2w (x; t) ? w(x; t) @w (x; t) + b(x) l k (s)w(s; t)ds (2.48) ; @t @x2 @x 0 w(x; 0) = w0(x) (2.49) w(0; t) = w(l; t) = 0: (2.50) We will rst show the existence and uniqueness of the solution to the system (2:48)-(2:50) for any initial data w0(x) 2 L2(0; l), then we will show that its solution will go to zero at a desired exponential rate . We can then write (2:48)-(2:50) as the functional evolution equation dw + Aw + N (w; w) + Rw = 0 (2.51) dt w(x; 0) = w0(x); where equation (2:51) holds weakly in H ?1(0; l) for almost every t > 0. Namely, h dw dt ; viH ? +  (wx; vx) + (N (w; w); v) l l ? ( k; (s)w(s; t)ds)( b(s)v(s)ds) = 0 (2.52) Z

1

Z

Z

0

0

12

for almost every t > 0, for all v 2 H01(0; l), and

w(x; 0) = w0(x) 2 L2(0; l):

(2.53)

De nition 2.2 For any T > 0, a weak solution of equations (2:48)-(2:50) is

a function

2 ?1 w 2 C ([0; T ]; L2(0; l)) \ L2([0; T ]; H01(0; l)); @w @t 2 L ([0; T ]; H (0; l)) (2.54)

satisfying (2:52) and (2:53). Furthermore, if in addition

2 (0; l); 2 L w 2 L1([0; T ]; H01(0; l)) \ L2([0; T ]; D(A)); @w @t for almost every t > 0, and dw + Aw + N (w; w) + Rw = 0 in L2(0; l) for a.e. t > 0; dt then w is called a strong solution of (2:1)-(2:3).

(2.55)

Because of the additional distributed control term in the controlled equation, we need the following estimate.

kRwkL (0;l)  kbkL (0;l) kk; kL (0;l) kwkL (0;l) 2

2

2

2

8w 2 L2(0; l);

(2.56)

Then we can use Theorem III.3.1 in [39] again to get the following theorem

Theorem 3 For any w0 2 L2(0; l) the problem (2:48)-(2:50) has a unique weak solution w for any T > 0. Furthermore, if w0 2 H01(0; l), then w is actually the strong solution and w 2 C ([0; T ]; H01(0; l)) \ L2 ([0; T ]; D(A)):

3 BURGERS' EQUATION WITH NEUMANN BOUNDARY CONDITIONS 3.1 Stability of the Uncontrolled System

We consider the viscous Burgers' equation with Neumann boundary conditions. @w (x; t) ?  @ 2w (x; t) + w(x; t) @w (x; t) = 0 0 < x < l; t > 0 (3.1) @t @x2 @x w(x; 0) = w0(x) (3.2) 13

@w (0; t) = @w (l; t) = 0; t > 0: (3.3) @x @x We are going to establish the existence of a solution to (3.1)-(3.3) via the Galerkin procedure. The approach here is similar to those in [14], [37] or [38], where it was done for the Navier-Stokes equations. First let us de ne @ (l) = 0g (3.4) A := (?@xx + I ) ; D(A) = f 2 H 2(0; l) j @ (0) = @x @x @ (l) = 0g V := f 2 C 1(0; l) j @ (0) = @x @x V := Closure of V in the H 1 ? norm  H 1(0; l) H := Closure of V in the L2 ? norm  L2(0; l); and we denote by k
kV := k
kH (0;l) and k
kH := k
kL (0;l): A is a linear unbounded self-adjoint positive operator, with A?1 : L2(0; l) ?! L2(0; l) is compact. Thus there exists a complete orthonormal basis for the space H , f j g1j=0, consisting of eigenfunctions of A with corresponding the eigenvalues 0 < 1  2  : : :, i.e. A j = j j . In fact, we have j = ( jl )2 + 1, jx 1 j = l cos( l ) for j = 0; 1; 2; : : :. Note that f j g1 j =0 form an orthonormal basis for H and an orthogonal basis for V and D(A). Moreover, note that 1

2

q

kA1=2wk2H = kwk2V ;

for every w 2 V:

(3.5)

Proposition 3.1 The operator A can be extended in a unique fashion as a map A : V ?! V  such that hAw; viV  = (wx ; vx) + (w; v) for all w; v 2 V . Furthermore, A1=2 can be extended in a unique fashion as a map A1=2 : H ?! V  such that for any w 2 H (3.6) kA1=2wk2V  = kwk2H : Proof. The proof is omitted. Thus it follows from equations (3:5) and (3:6) that

kAwk2V  = kA1=2wk2H = kwk2V ; for every w 2 V:

(3.7)

Moreover, equation (3:1) can be written in the form of dw + Aw ? w + N (w; w) = 0 (3.8) dt w(x; 0) = w0(x) (3.9) where A is de ned in (3:4) and N (w; w) = w @w @x is a bilinear form on V  V , with values in H . 14

De nition 3.2 For any T > 0, a weak solution of (3:1)-(3:3) is a function w(x; t) 2 CW ([0; T ]; H ) \L2([0; T ]; V ) \L2loc ((0; T ]; D(A)) such that @w @t 2 2  L (0; T ; V ) and  (3.10) h dw dt ; viV +  (A w; A v) ?  (w; v) + (N (w; w); v) = 0 for all v 2 V and a.e. t 2 [0; T ] and w(0) = w0 2 L2(0; l): (3.11) 1 2

1 2

We remark here that this de nition of weak solutions is equivalent to the weak solution in the sense of distribution, as it was justi ed in the case of the Navier-Stokes equation in Chapter III of [38]. Therefore, this de nition of weak solutions is equivalent to the one de ned in [9]. Moreover, the boundary conditions (3:3) are satis ed by our formulation of the weak solutions because w 2 D(A)  H 2(0; l) for almost every t > 0 and therefore the trace of the derivative makes sense in this case. In the next theorem, we claim that unique weak solution exists for some short time for any initial data in H .

Theorem 4 Let w0 2 H , then there exists T = T (w0;  ) > 0, such that the system (3:1) - (3:3) has a unique weak solution w in the interval [0; T ]. Moreover, w 2 C ([0; T ]; H ) \ L2 ([0; T ]; V ) \ Cloc ((0; T ]; V ): Proof. Let m be a positive integer and Hm = spanf 1; : : :; mg where as before f 1; : : : ; mg are the rst m eigenfunctions of A. We de ne Pm : H ! Hm to be the L2 orthogonal projection. Let wm 2 Hm be the solution to the

approximate Galerkin system dwm + Aw ? w + P N (w ; w ) = 0 m m m m m dt with initial value wm (x; 0) = wm0 (x) = Pm w0: Note that since wm 2 Hm , we have wm = mj=1 j (t) j (x) and

(3.12) (3.13)

P

kwm0 k2H =

m

X

j =1

j2(0) 

1

X

j =1

j2(0) = kw0k2H :

(3.14)

Moreover, fj (t)gmj=1 satisfy

dj +  ( ? 1) + m (N ( ; ); )  = 0 j j k l j k l dt l;k=1

(3.15)

j (0) = (w0; j ):

(3.16)

X

15

Notice that the system (3:15)-(3:16) is a nite system of initial-valued ordinary dierential equations with a Lipschitz nonlinearity (quadratic in this case) whose solution j (t), for 1  j  m, exists in some interval around t = 0. Now that we have shown wm (t) exists for some interval about t = 0, we would like to nd the lower bound on the maximal interval of existence of positive time. It is well-known from the theory of O.D.E. that in this case the solution exists as long as it is nite. Thus, let us nd estimates on the size of wm(t). We take the L2 inner product of (3:12) with wm to get 1 d kw k2 +  kw k2 + (P N (w ; w ); w ) =  kw k2 (3.17) m V m m m m m H 2 dt m H Note that l m (Pm N (wm; wm); wm) = (N (wm; wm); Pm wm) = wm2 @w @x dx; 0 and l 2 @wm j wm @x jdx  kwmk2L (0;l)k @w @x kH by Cauchy-Schwarz inequality 0  kwmk3H=2kwmk3V=2 by an interpolation inequality  13 kwmk6H + 2 kwmk2V by Young's inequality: Z

Z

4

Therefore (3:17) implies that d kw k2 +  kw k2  2 kw k2 + 2 kw k6 m V m H dt m H 3 m H If we let k0 = max(2; 2 ) and ym(t) = 1 + kwm(t)k2H , then we have dym = d 1 + kw k2  k kw k2 (1 + kw k2 )2  k y3 : m H 0 m H m H 0 m dt dt Integrating the above inequality we get 2 1 : ym2 (t)  1 ? 2yym2(0) for 0  t < 2 2ym(0)k0 m (0)k0 t Since ym(0)  1 + kw0k2H , we have 1 + kw0k2H kwm(t)k2H  ym(t)  1 ? 2(1 + kw0k2H )2k0t

(3.18)

3





(3.19)

q

for 0  t < 2(1+kw1 kH ) k : In particular, 0 2

2

0

kwm(t)k2H  R0 for every m  0 and for all t 2 [0; T ]; 16

(3.20)

p

where R0 := 2(1 + kw0k2H ) and T (; kw0 kH ) = 4(1+kw1 kH ) k > 0. Thus 0 2

2

0

wm 2 L1([0; T ]; H ) for all m  0:

(3.21)

By integrating (3:18) over the interval [0; T ] and using (3:20), we get

kwm(T )kH +  2

T

Z

0

kwm( )k2V d  kw0k2H + T k0(R0 + R30) := R1: (3.22)

Hence wm is bounded uniformly, i.e. independent of m, in the L2([0; T ]; V )norm. Consequently, Awm is bounded uniformly in the L2([0; T ]; V )-norm due to (3:7). Next we will show that PmN (wm ; wm) is bounded uniformly in the L2([0; T ]; V )-norm. Note that for any v 2 V , one can use Holder and Sobolev inequalities to obtain

j(Pm N (wm; wm); v)j  kwmkH kwmkV kvkL1  k1kwmkH kwmkV kvkV : where k1 depends only on l. Hence kPm N (wm; wm)k2V   k1kwmk2H kwmk2V and so by (3:20) and (3:22) we get Z

0

T

kPm N (wm; wm)kV  dt  k1R0 2

Z

T

kwm(t)k2V dt  k1R0R1:

0

(3.23)

Because of the above, equation (3:12) allows us to deduce that dwdtm is bounded uniformly in L2([0; T ]; V ). Also since L2([0; T ]; V ) is a Hilbert space and wm is bounded uniformly in 2 L ([0; T ]; V ), by the weak compactness theorem, there exist w 2 L2([0; T ]; V ) and a subsequence fwm0 g of fwmg such that fwm0 g converges weakly to w in L2([0; T ]; V ). Also by Aubin's compactness Theorem (see, e.g., [14] and [38]) fwm0 g converges strongly to w in L2([0; T ]; H ). Let us from now on denote all the subsequences of fwmg by itself by extracting and relabeling. Thus fwmg converges strongly to fwg in H almost everywhere. Similarly, since Awm and dwm are bounded uniformly in the L2 ([0; T ]; V  )-norm, we have Aw and dwm m dt dt dw 2  converge weakly to Aw and dt in L ([0; T ]; V ), respectively. Speci cally, for almost every t; s 2 [0; T ], t > s and for every v 2 V Z

t

1=2

1=2

(A wm( ); A v)d = s Z

and

Z

t

hAwm( ); viV  d

sZ

t

t

?! s hAw( ); viV  d = s (A1=2w( ); A1=2v)d;

(3.24)

dwm ( ) ; vi  d ?! th dw( ) ; v6 h V d d s s

(3.25)

Z

t

Z

17

V  d;

as m ! 1. Let v 2 V be arbitrary. We let equation (3:12) act as an element in V  on v and integrate over the interval [s; t]  [0; T ] to get t dwm t 1=2 1=2 h ; v i V  d +  (A wm ; A v )d s dt s t t (3.26) ?  (wm; v)d + (PmN (wm ; wm); v)d = 0 Z

Z

Z

Z

s

s

Using equations (3:24) and (3:25) along with the fact that fwmg converges strongly to fwg in H almost everywhere. Therefore, we can pass to the limits for all except the last terms on the left-hand side of equation (3:26). Next we will show that Z

t

(Pm N (wm; wm); v)d ?! s

Z

s

t

(N (w; w); v)d

as m ! 1 for almost every t; s 2 [0; T ]. Notice that Z

t s

@wm (P v ? v)dxd w m @x m s 0 t l m + s 0 (wm ? w) @w (3.27) @x vdxd t l @wm @w w( @x ? @x )vdxd : +

(PmN (wm ; wm); v) ? (N (w; w); v)d 

t l

Z Z



Z Z



Z Z



s 0

We will show that each term on the right hand side of (3:28) will go to zero as m ! 1. First, t t l @wm k kP v ? vk d m k w ( P wm @w m kL1 (0;l)k m v ? v )dxd  H @x @x H m s s 0 t  kPm v ? vkH s k1kwmk2V d  k1R1kPm v ? vkH by (3:22): Z Z



Z

Z

Since kPm v ? vkH ?! 0, we have t l

Z Z

s 0

m wm @w @x (Pm v ? v)dxd ?! 0 as m ! 1:

Similarly, for the second term of (3:28) t @wm k d m (wm ? w) @w vdxd  k v k L1(0;l) kwm ? wkH k @x @x H 0 s  k1kvkV kwm ? wkL ([0;T ];H )kwmkL (0;T ;V )  k1 R1kvkV kwm ? wkL ([0;T ];H ) by (3:22):

t l

Z Z

s

Z



2

2

q

2

18

Since kwm ? wkL ([0;T ];H ) ! 0, it follows that t l m (wm ? w) @w (3.28) @x vdxd ?! 0 as m ! 1: s 0 We treat the last integral in (3:28) dierently. Since w 2 L2([0; T ]; H ) and v 2 V , we have wv 2 L2([0; T ]; H ) as 2

Z Z



Z t t l 2 2 w v dxdt  kvk2L1(0;l) kwk2H d s s 0

Z Z

 R0T k12kvk2V ;

by Sobolev Embedding Theorem and by (3.20). Recall that wm converges 2 weakly to w in L2([0; T ]; V ), thus @w@xm converges weakly to @w @x in L ([0; T ]; H ). 2 As shown above wv 2 L ([0; T ]; H ), so that t l @wm @w ( ? @x )wv dxdt ?! 0 as m ! 1: s 0 @x Consequently, all three terms on the right hand side of (3:28) converge to zero and we have for every v 2 V and a.e. t; s 2 [0; T ] Z Z

Z

s

t



(Pm N (wm; wm); v)d ?!

t

Z

s

(N (w; w); v)d

as m ! 1:

(3.29)

Now we use (3:24) (3:25) and (3:29) along with the fact that wm converges strongly to w in L2([0; T ]; H ) to pass to the limits for (3:26) to obtain t dw h dt ; viV  +  (A1=2wm; A1=2v) ?  (w; v) + (N (w; w); v) d = 0; (3.30) s for every v 2 V and a.e. t; s 2 [0; T ] and so (3:10) is satis ed. It also follows from Chapter III Lemma 1.1 of [37] that dtd (w(t); v) = h @w @t ; v iV  , hence equation (3:30) becomes Z

o

n

Z

t

Z

t

(w(t); v) +  s (A1=2wm; A1=2v)d Z

t

?  s (w; v)d + s (N (w; w); v)d = (w(s); v): (3.31) Since V is dense in H , equation (3:31) implies that w 2 CW ([0; T ]; H ). Moreover, since wm converges to w strongly in L2([0; T ]; H ), we have from (3:20) that for a.e. t 2 [0; T ] 2 kw(t)k2H = mlim (3.32) !1 kwm (t)kH  R0 : Integrating (3:18) over any interval [s; t]  [0; T ] and using (3:32) we have Z

t

kwm(t)kH +  s kwm( )k2V d  kwm(s)k2H + k0(R0 + R30 )(t ? s): (3.33) 2

19

Note that wm converges to w weakly in L2([0; T ]; V ) implies that t

Z

s

Z

t

kw(t)k2V d  lim inf kwm( )k2V d for every [s; t]  [0; T ]: (3.34) m!1 s

Due to (3:32), (3:34) and the following property lim sup(an + bn )  lim sup(an) + lim inf(bn); n!1 n!1 n!1

for any sequences an, bn  0, inequality (3:33) becomes Z

t

kw(t)kH +  s kw( )k2V d  kw(s)k2H + k0 (R0 + R30 )(t ? s); (3.35) for a.e. s; t 2 [0; T ] such that t > s. Now one can rearrange (3:35) to get 2





kw(t)k2H ? kw(s)k2H  

Z

s

t



kw( )k2V d + k0(R0 + R30)(t ? s);

(3.36)

for almost every s; t 2 [0; T ]; t > s. Using (3:36) together with the fact that w also belongs to CW ([0; T ]; H ), we get w 2 C ([0; T ]; H ). Going back to equation (3:35), and using (3:22), we attain Z

0

T

kw( )k2V d  R1; and so w 2 L2([0; T ]; V ):

(3.37)

Furthermore, since each wm belongs to D(A), we can take the L2-inner product of (3:12) with Awm and obtain 1 d kw k2 +  kAw k2 =  (w ; Aw ) ? (N (w ; w ); Aw ) (3.38) m H m m m m m 2 dt m V We use Cauchy-Schwarz, Young's, and Agmon's inequalities and the Sobolev embedding theorem to estimate the right hand side as follows,  (wm; Awm)   kwmk2H + 4 kAwmk2H ; and 2 j(N (wm ; wm); Awm)j  k1 kwmk4V + 4 kAwmk2H : Therefore (3:38) becomes d kw k2 +  kAw k2   kw k2 + k12 kw k4 : (3.39) m H m H dt m V  mV By dropping the term  kAwmk2H and letting k2 := maxf k ;  g and ym(t) := 1 + kwm (t)k2V , we have dym  k y2 : (3.40) 2 m dt 2 1

20

Equation (3:40) can be written as R

t

ym(t)  ym(s)ek s ym ( )d ; 0  s < t  T : (3.41) Thus, (3:37) and (3:41) imply that for all s; t 2 (0; T ) ym(t)  ym(s)ek fT +R g: (3.42) Now by integrating (3:42) with respect to s on the interval [0; t] and using (3:37) again, we get 2

2

Z

t

1

tym(t)  ( ym(s)ds)ek fT +R g  (T + R1)ek fT +R g: 2

1

2

1

0

Thus, we have for all t 2 (0; T ] 0 kwm(t)k2V  T +t R1 ek fT +R g  Rt1 independent of m: (3.43) Since wm converges weakly to w in V , by extracting a subsequence and relabeling, we have wm converges weakly to w in V pointwise almost everywhere in [0; T ], i.e. on a set E  [0; T ] such that j[0; T ]nE j = 0. Now since wm converges weakly to w in V on this set E , for every s > 0, t  s and t 2 E , we have from (3:43) that 0 0 2  R1  R1 ; kw(t)k2V  lim inf k w ( t ) k V m?!1 m t s and hence w 2 L1([s; T ]; V ) for any s > 0, i.e. w 2 L1loc(0; T ]; V ). Integrating (3:39) over (s; t)  (0; T ) and using (3:43) and (3:20), we see that 0 2 t kwm (t)k2V ? kwm(s)k2V +  s kAwm( )k2H d  (t ? s)[R0 + (Rst1) ]; (3.44) uniformly in m. Moreover, by rearranging (3:44) and using (3:43) we have 0 2 t k Awm( )k2H d  1 kwm(s)k2V + (t ? s)[R0 + (Rst1) ] s 0 0 2  Rs1 + (t ? s)[R0 + (Rs12) ]; (3.45) 2

1

Z

Z

for every s > 0, t  s and t 2 E . Now it follows from equation (3:45) that for every s > 0 there exists a subsequence wm0 , which depends on s, of wm such that wm0 converges weakly to w in L2([s; T ]; D(A)). Hence for every s > 0, we have established 0 )2 0 t t ( R R 1 2 2 0 k Aw( )kH d  lim inf kAwm ( )kH d  s + (t ? s)[R0 + s12 ]; 0 m !1 s s Z

Z

21

and so we have w 2 L2loc((0; T ]; D(A)). In fact, it also follows from (3:31) and 2 (3:44) that w 2 C ((s; T ]; V ) and from (3:1) that @w @t (x; t) 2 L (s; T ; H ) for any s 2 (0; T ). Consequently, we have established the existence of a weak solution and its regularity for short time T (; kw0kH ) > 0, where w(x; t) 2 L2([0; T ]; V ) \ C ([0; T ]; H ) \ Cloc((0; T ]; V ) \ L2loc((0; T ]; D(A)): We will next prove the uniqueness of weak solutions. Let w1 and w2 be any two weak solutions of (3:1) (3:2) and (3:3). Thus dw1 + Aw ? w + N (w ; w ) = 0 (3.46) 1 1 1 1 dt dw2 + Aw ? w + N (w ; w ) = 0: (3.47) 2 2 2 2 dt Let w = w1 ? w2 then dw + Aw ? w + w @w ? w @w2 = 0: (3.48) 1 dt @x @x Note that because of the regularity of weak solution, equation (3:48) holds almost everywhere in V . Hence, we can let it act on w(t) 2 V to yield l @w l 2 @w2 2 2 h @w ; w i V  +  kwkV ?  kwkH + w1 wdx + w @t @x @x dx = 0; (3.49) 0 0 for a.e. t 2 [0; T ]. Again by the regularity of weak solution, h @w @t ; w@wiV  makes sense and it follows from Chapter III Lemma 1.2 of [37] that h @t ; wiV  = 1 d kwk2 : By H older and Young inequalities and the Sobolev embedding theH 2 dt orem we have l @w2 k   kwk2 + k12 kwk2 kw k2 ; 2 1 j 0 w2 @w k w k k dx j  k w k H L @x @x H 4 V  H 2 V and l @w k   kwk2 + k12 kwk2 kw k2 : j 0 w1 @w wdx j  k w 1 kL1 kwkH k @x @x H 4 V  H 1 V By using the above, equation (3.49) becomes d kwk2  d kwk2 +  kwk2  2[ + 1 (kw k2 + kw k2 )] kwk2 ; 2 V V H dt H dt H  1V for a.e. t 2 [0; T ]. By Gronwall's lemma and equation (3.37), we have for a.e. t 2 [0; T ] Z

Z

Z

Z

RT 0 2 2 0 [ + 1 (kw1 ( )k2V +kw2 ( )k2V )]d

kw(t)kH  kw kH e 2

22

 kw0k2H e2(T + R ): (3.50) 2 1

Since kw0kH = kw1(0) ? w2(0)kH = 0, the above inequality implies that kw(t)kH = 0 for a.e. t 2 [0; T ]. As a consequence, equation (3.50) implies the continuous dependence of the solution on the initial data, in the H -topology, for any nite time interval [0; T ]. In short, we have

Corollary 3.3 The problem (3:1)-(3:3) is well-posed in the space H . De nition 3.4 For any T > 0, a weak solution of (3:1)-(3:3) in [0; T ] is called a strong solution if w 2 L1 ([0; T ]; V ) \ L2 ([0; T ]; D(A)) such that @w 2 L2 ([0; T ]; H ) and @t dw + Aw ? w + N (w; w) = 0 in H for a.e. 0 < t < T; dt with w(x; 0) = w0 (x) 2 H:

Note that if w0 2 V , then equation (3:40) implies that there exists some T (kw0kV ) > 0 such that

kwm(t)kV  2kw0kV for all t 2 [0; T ]: In particular, by passing to the limit of a subsequence one concludes as before that kw(t)kV  2kw0kV for all t 2 [0; T ] \ E; where E is de ned earlier. If one goes back to equation (3:42) and replaces s with zero, one can show similarly to Theorem 4 that the weak solution w also 2  belongs to C ([0; T ]; V ) \ L2([0; T ]; D(A)) and that @w @t (x; t) 2 L ([0; T ]; H ). But then from De nition (3:4), this weak solution is strong. In particular, we have proved

Theorem 5 Let w0 2 V , then there exists T  = T (w0;  ) > 0, such that (3:1)-(3:3) has a unique strong solution w on [0; T ]. Moreover, w 2 C ([0; T ]; V ). Let us remark that the results mentioned above are still valid in the presence of a forcing function, f (x; t), on the right hand side of the Burgers' equation. This is true provided that f 2 L2([0; T ]; V ) for every T > 0, in the case of weak solution, and in the case of strong solution, f 2 L2([0; T ]; H ) for every T > 0. (This is the same as in the case for the Navier- Stokes equations, see, for example, [14], [37] or [38] for details.) To achieve the global existence of solutions, we will need to show that the solutions remain bounded in the appropriate norms for all times. To establish this fact we will show that the solutions satisfy some version of the Maximum 23

Principle. But to implement the classical Maximum Principle we would need to insure that the solutions possess certain level of regularity. Next we will follow the work of Foias and Temam in [17] for the Navier- Stokes equations and show that under certain assumptions on the forcing function, the solutions are analytic in time. To be speci c, we will prove: Theorem 6 Let w0 2 V and w be the strong solution of (3:1)-(3:3) on [0; T ], then w is analytic in time with values in D(A) in certain neighborhood of (0; T ] in the complex plane Cl . Proof. Let HCl := fw + iv j u 2 H; v 2 H g be the complexi cation of the space H . Similarly VCl, VCl and D(A)Cl are the complexi cation spaces of V , V  and D(A), respectively. We also de ne the inner product for elements in HCl by: (w1 + iw2; v1 + iv2)Cl = (w1; v1) + (w2; v2) + i[(w2; v1) ? (w1; v2)]: Let us consider the Galerkin system of the complexi ed Burgers' equation dwm( ) + Aw ( ) ? w ( ) + P N (w ( ); w ( )) = 0 (3.51) m m m m m d wm(x; 0) = wm0 (x) (3.52) where  2 Cl and wm( ) = mj=1 j ( ) j (x). (3:51) and (3:52) is an analytic system of O.D.E. which admits a unique analytic solution wm( ) in the neighborhood of 0 of the complex plane. Since w0 is real, the solutions wm( ) with  restricted to the neighborhood of (0; T ] in the real line will coincide with the usual Galerkin solutions. We are going to show that the apriori norm estimates are uniformly independent of m. Then we will pass to the limit to obtain the time-complexi ed solution. Let  = sei where  2 (? 2 ; 2 ) and s  0. We take the scalar inner product in HCl of (3:51) with Awm( ), then multiply by ei and take the real part to get 1 d kw k2 +  cos kAw k2  j(P N (w ; w ); Aw ) j +  j(w ; Aw ) j: m H m m m m Cl m m Cl 2 ds m V (3.53) Recall that the nonlinear term can be estimated similarly in the proof of Theorem 4. That is,  kAw k2 + k12 kw k4 : j(Pm N (wm; wm); Awm)Clj   cos m H 4  cos  m V Also by Cauchy-Schwarz and Young's inequalities we have  kAw k2 + 1 kw k2 ;  j(wm; Awm)Clj   cos m H 4  cos  m H P

C l

C l

C l

C l

24

C l

C l

so that (3:53) gives d kw k2 +  cos kAw k2  k3 (1 + kw k2 )2; where k = 1 + k2: m H 3 m V 1 ds m V  cos  By letting ym ( ) = 1+ kwm ( )k2V and neglecting the term  cos kAwmk2H  k y 2 ( ); and therefore: 0, it follows that dsd ym( )   cos  m ym(0) cos  provided 0  s <  cos  : ym( )   cos  ? k3ym (0)s k3ym(0) Since ym(0)  1 + kw0k2V , kwm( )k2V  2(1 + kw0k2V ) := R2; for 0  s  2k (1 +coskw 0k2 ) : (3.54) 3 V Hence the domain of analyticity of wm ( ) is D = f = sei j jj < 2 & 0   cos  . Let us denote by j
j the Euclidean norm of s   g and 0 <  = 2k (1+ Cl kw kV ) the complex vector. Since wm( ) is analytic in D and for any x 2 [0; l] jwm(x;  )j2Cl  sup fjRewm (x;  )j2 + jImwm(x;  )j2g  kwm ( )k2V  R2; C l

C l

C l

C l

C l

3

C l

0 2

3

x2[0;l]

C l

it follows from the vector-valued version of the Vitali's Convergence Theorem (see, for example, [32] page 450) that wm ( ) converges uniformly to w( ) on every compact subset of D and w( ) is analytic in D . Note that w( ) satis es the same estimates as wm( ) in D . In particular, kw( )k2H  kw( )k2V  R2: Moreover, w( ) satis es the equation @w( ) + Aw( ) ? w( ) + N (w( ); w( )) = 0: (3.55) @ Let ? be the boundary of the closed disk of radius r in D and centered at  . Then by the Cauchy integral formula w dz: dk w ( ) = k! (3.56) k d 2i ? (z ?  )k+1 Thus k (3.57) k ddwk ( )kV  2k! ? kwkV k+1 jdzj  rkk! R2 : jz ?  jCl Let S be a compact subset of D and rS = distance (S ; @D ). Particularly for k = 1 and r smaller than rS , we have from (3:57) that ( )kV  1 R2 for all  2 S : (3.58) k dw d r C l

C l

Z

Z

C l

C l

C l

1 2

25

1 2

Then from (3:55) we have kAw( )k2H  1 [k dwd( ) k2H + kN (w( ); w( ))k2H ] + kw( )k2H (3.59) for all  2 S . Moreover, one can use Agmon's inequality to show that C l

C l

C l

C l

kN (w( ); w( ))k2H  k12kw( )k4V  k12R22:

(3.60)

C l

Combining (3:58) and (3:60) in (3:59) we get kAw( )k2H  1 [ r12 R2 + k12R22] + R2  R23 for all  2 S : (3.61) Therefore it follows from the Cauchy integral formula (3:56) and (3:61) that 1 kAw( )kH jdzj  R3 for all  2 S : (3.62) kA dw (  ) k H  d 2 ? jz ?  j2Cl r Note that the domain D depends only on  and kw0kV , we can repeat this argument with  instead of zero to extend the domain of analyticity all the way to T . C l

Z

C l

C l

C l

Proposition 3.5 (Regularity) Let w0 2 V and w(x; t) be the strong solution of (3:1)-(3:3) on [0; l]  [0; T ], then the solution is classical on [0; l]  (0; T ]. Moreover, w(x; t) 2 CB2+ ([0; l]  [s; T ]) for every s > 0 and  2 (0; 1). Proof. Let s 2 (0; T ] be arbitrary. It follows from Theorem 6 that wt 2 L2([s; T ]; D(A)). Thus we have wtx 2 L2([s; T ]; V ) and wtxx 2 L2([0; l][s; T ]). Let us go back and take the spatial weak derivative of the Burgers equation (3:1) with respect to x once and twice to get

wxxx = wtx ? wwxx ? wx2 in D0(0; l) (3.63) 0 wxxxx = wtxx ? wwxxx ? 3wxwxx in D (0; l); (3.64) respectively. Since wtx 2 L2([s; T ]; V ) and w 2 L2([0; T ]; D(A)) \ C ([0; T ]; V ), it follows from (3:63) that wxxx 2 L2([0; l]  [s; T ]). Similarly, because w 2 L2([0; T ]; D(A)) \ C ([0; T ]; V ), wtxx 2 L2([0; l]  [s; T ]) and wxxx 2 L2([0; l]  [s; T ]), equation (3:64) implies that wxxxx 2 L2([0; l]  [s; T ]), i.e. w belongs to L2([0; T ]; H 4(0; l)). So far we have shown that wxxxx; wxxx; wtxx; wtx 2 L2([0; l]  [s; T ]): Now because w is analytic in time with values in H 4(0; l), we also have wxxxx; wtxxx; wttxx; wtttx; wtttt 2 L2([0; l]  [s; T ]); 26

i.e., w 2 H 4([0; l]  [s; T ]). Therefore, by Sobolev Embedding Theorem we have w 2 CB2+([0; l]  [s; T ]) for every  2 (0; 1). Remark: By the same argument one can show that w 2 H m ([0; l]  [s; T ]) for any m > 0 and hence by the Sobolev Embedding Theorem w 2 C 1([0; l]  [s; T ]) Proposition 3.6 (Maximum Principle) Let w0 2 V and w(x; t) be the strong solution of (3:1)-(3:3) on [0; l]  [0; T ], then for all t 2 (0; T ): (3.65) kw(
; t)kL1  kw(
; 0)kL1 < 1 Proof. Recall from Proposition 3.5 that the solution of (3:1)-(3:3) is classical instantly. Let t0 2 (0; T ) be xed. Let us consider the following cases: Suppose that w(x; t0) attains its positive maximum at some x0 2 (0; l). Then since w(x0; t0) is a local maximum, we have @w (x ; t) = 0 and @ 2w (x ; t)  0: @x 0 t=t @x2 0 t=t Thus (3:1) implies that @w (x ; t) =  @ 2w (x ; t) ? w(x ; t) @w (x ; t)  0; 0 @t 0 t=t @x2 0 t=t @x 0 t=t and so w(x0; t0) is non-increasing. Suppose otherwise that w(x; t0) attains its positive maximum at one of the end points. Let us assume without loss of generality that w(x; t0) attains its positive maximum at x = 0. Applying the Taylor's expansion formula we have, @ 2w (c; t) x2; quadfor some 0 < c < x: w(x; t0) ? w(0; t0) = @w (0 ; t ) x + t=t t=t @x @x2 But because at the boundary, @w @x (0; t) = 0 for all t > 0 and also w(0; t0) is a positive maximum, it results that @@xw (c; t) t=t  0: As x ! 0+ , we have c ! 0+ and therefore we get @@xw (0; t) t=t  0: Consequently, equation (3:1) implies that @w @t (0; t) t=t  0, and hence w(0; t0 ) is non-increasing. Similarly, one can claim that if w(x0; t0) attains its negative minimum at some x0 2 [0; l], then w(x0; t0) is non-decreasing. We have proved that for every t  t0 > 0, kw(
; t)kL1  kw(
; t0)kL1 . Since w 2 C ([0; T ]; V ), we have w 2 C ([0; T ]; L1) and by letting t0 & 0+, we conclude (3.65). Recall from Theorem 5 that we have short time existence of the strong solution for any initial data w0 2 V . Next we show that the strong solution exists globally.





0



0





0



0



2

2



2



2

0

27

0

0

0

0

0

Let us assume that [0; Tmax) is the maximal interval of existence. Then ? kw(t)kV = either Tmax = 1, and we are done, or Tmax < 1 and lim supt!Tmax 2 1. We suppose the latter. Let us take the L -inner product of (3.1) with wxx and use Young's inequality. We have for any t 2 [0; Tmax) d kw(t)k2 +  kw (t)k2  1 kw(t)k2 kw(t)k2  1 kw0k2 kw(t)k2 ; xx V H  L1 (0;l) V L1 (0;l) V dt  where the last inequality is attained using the Maximum Principle of Proposition 3.6. It follows from Gronwall's Lemma that

kw(t)k2V  kw0k2V e  kw kL1 1

0 2

;l t

(0 )

for any t 2 [0; Tmax) and hence, lim sup kw(t)kV  kw0k2V e  kw kL1 ? 1

0 2

t!Tmax

;l Tmax

(0 )

0 1. The system (3:1)-(3:3) has a unique strong solution w, where

w 2 C ([0; T ]; V ) \ L2([0; T ]; D(A)): 2. (Analyticity in time) w is analytic in time with values in D(A) in a neighborhood of (0; T ] of the complex plane Cl . 3. (Regularity) The solution is classical instantly and w(x; t) 2 C 1 ([0; l]  (0; T ]). 4. (Maximum Principle) kw(
; t)kL1  kw(
; 0)kL1 < 1 for all t 2 [0; T ].

Let us go back to the weak solution. Suppose w0 2 H , then we have from Theorem 4 that there exists a unique weak solution for some short time T > 0 and, in particular, the solution is in L2([0; T ]; V ). Thus w(t) 2 V for almost every t 2 (0; T ). Hence there exists a t0 2 (0; T ) arbitrarily close to zero such that w(t0) 2 V . Also since w 2 C ([0; T ]; H ), w(t0) is uniquely determined as an element of V . Now using Theorem 7 and w(t0) 2 V as an initial data, we have unique strong solution exists thereafter. Consequently the weak solution exists globally and becomes classical instantaneously. We summarize with the following theorem: 28

Theorem 8 Let w0 2 H , then for any T > 0 1. The system (3:1)-(3:3) has a unique weak solution w, where

w 2 L2([0; T ]; V ) \ C ([0; T ]; H ) \ Cloc((0; T ]; V ) \ L2loc((0; T ]; D(A)): 2. (Instantly strong) The weak solution w of (3:1)-(3:3) is strong instantaneously. In particular, for any t0 2 (0; T )

w 2 C ([t0; T ]; V ) \ L2([t0; T ]; D(A)): 3. (Analyticity in time) For any t0 2 (0; T ), w is analytic in time with values in D(A) in a neighborhood of [t0; T ] of the complex plane Cl . 4. (Regularity) The solution is classical instantly and w(x; t) 2 C 1 ([0; l]  (0; T ]). 5. (Maximum Principle) For any t0 2 (0; T ) we have

kw(
; t)kL1  kw(
; t0)kL1 < 1;

for all t 2 [t0; T ]:

Remark: We cannot conclude part 5 of Theorem 8 for t0 = 0 simply because the initial data is not in L1 to start with. However, for initial data in L1 it is not true that the weak solution belongs to C ([0; T ]; L1). Speci cally, since the weak solution is classical for all t > 0, by the uniform continuity the initial data must be continuous if the weak solution were in C ([0; T ]; L1). Therefore, there is a need for an alternative proof of the Maximum Principle with initial data in L1 and we will show it in the next Proposition. The technique we employ in the proof was inspired by the work of Foias-Manley-Temam [18]. Proposition 3.7 Let w0 2 L1 (0; l) and w be the weak solution of (3:1) (3:2) and (3:3) in [0; T ] with   w0(x)  , a.e. in [0; l]: Then for every (x; t) 2 [0; l]  (0; T ] (3.66)   w(x; t)   and

for every t 2 [0; T ].

kw(t)kL1(0;l)  kw0kL1(0;l)

Proof. Let

M (x; t)  [w(x; t)?]+  maxfw(x; t)?; 0g = 29

8 < :

w(x; t) ?  0

(3.67)

w(x; t) >  otherwise,

then M (x; t)  0 for any (x; t) 2 [0; l]  [0; T ] and M (x; 0) = M0(x)  0. Notice also that for every t > 0,

@M (x; t) = @x@ (w(x; t) ? ) = @w @x (x; t) @x 0 8
 otherwise.

:

Since w is a weak solution, as we have seen already for every t > 0, w 2 D(A) dw and dw dt 2 H (actually dt 2 D (A) as well). In particular, (w ? ) 2 V for every t > 0. It is also well-known that M (t) = [w(t) ? ]+ 2 V (see [23] p. 153). Moreover, one can rewrite equation (3:1) as: @ (w(x; t) ? ) ?  @ 2 (w(x; t) ? ) + w(x; t) @w (x; t) = 0; (3.68) @t @x2 @x where the above equation holds in H for every t 2 (0; T ]. Multiply equation (3:68) by M (x; t) and integrate over [0; l] we get 1 d kM k2 ?  ([w ? ] ; M ) = l wM @w dx: (3.69) xx H 2 dt @x 0 Now by using the Generalized Stokes Formula (see e.g. [14] and [38]) and the boundary conditions (3:3), l @M k2 =  k @M k2 : ? ([w ? ]xx; M ) = ?M @M +  k (3.70) @x 0 @x H @x H Moreover, l l @M dx = l(w(t) ? )M @M dx +  l M @M dx wM dx = wM @w @x @x @x @x 0 0 0 0 l 2 @M l @M = M @x dx +  M @x dx 0 0 @M k  kM kL1 kM kH k @M k H + jj kM kH k @x @x H 2 2  1 kM k2L1 kM k2H +  kM k2H + 2 k @M @x kH 2  1 (c0kM k2V + 2)kM k2H + 2 k @M @x kH : Thus equation (3.69) becomes d kM k2  1 (c kM k2 + 2)kM k2 ; H dt H  0 V Z



Z

Z

Z

Z

Z

Z

R

T

and by Gronwall's Lemma kM (t)kH  kM0kH e  (c kM kV + )d : Since T 2 2 0 (c0 kM kV +  )d is bounded from the de nition of weak solution and since 1

R

30

0

0

2

2

kM0kH  0, it follows that kM (t)kH  0 and thus M (x; t)  0 and w(x; t)   for every (x; t) 2 [0; l]  (0; T ]. Similarly, by letting m(x; t)  [w(x; t) ? ]?  minfw(x; t) ? ; 0g one can repeat the above argument to show that w(x; t)   for every (x; t) 2 [0; l]  (0; T ].

We remark here that the zero solution of the uncontrolled Burgers' equation (3:1) (3:2) with Neumann boundary conditions (3:3) is not asymptotically stable. In fact from the proof of the Proposition 3:6, we can see that if w0(x) = , then w(x; t)   is the steady state and solution will not go to zero regardless how small  is. Moreover this is true even for the Heat equation; that is the zero solution of the heat equation with Neumann boundary condition is not asymptotically stable. Thus there is a need for stabilizing the zero solution for both the linear and nonlinear equations.

3.2 Stabilization of Zero Solution by Boundary Control

One way of stabilizing the zero solution for the linearized Burgers' equation is to use the radiation boundary condition at each end. Following [33], we consider the system @w (x; t) ?  @ 2w (x; t) = 0 0 < x < l; t > 0 (3.71) @t @x2 w(x; 0) = w0(x) (3.72) @w (0; t) = k w(0; t); @w (l; t) = ?k w(l; t); t > 0 (3.73) 1 2 @x @x where k1; k2  0 are the boundary control parameters. One can interpret this kind of stabilization as follows. We use the two ends of the rod as locations of the sensors and actuators; we heat up the rod if it is cold and cool it down if it is hot. Byrnes et al. [7]-[9] applied the same control strategy to the fully nonlinear equation. They considered the initial-valued boundary-controlled viscous Burgers' equation @w (x; t) ?  @ 2w (x; t) + w(x; t) @w (x; t) = 0 0 < x < l; t > 0 (3.74) @t @x2 @x w(x; 0) = w0(x) (3.75) @w (0; t) = k w(0; t); @w (l; t) = ?k w(l; t); t > 0 (3.76) 1 2 @x @x where k1; k2  0 are the gain parameters or control parameters. The existence and uniqueness of the above uncontrolled problem (3:74)-(3:76) were established in Byrnes et al. [11]; however it was required that the initial data in L2(0; l) to be \suciently small." In this section, we will prove the existence 31

and uniqueness in general without the restricting the sizes of initial data. In addition, for this system (3:74)-(3:76), Byrnes et al. [9] proved that for positive gains and suciently small initial data in H 1(0; l), the H 1(0; l)-norm of the solution will go to zero exponentially. More recently, Byrnes et al. [11] improved their result by replacing \suciently small data in H 1(0; l)" with \suciently small initial data in L2(0; l)". Here we will prove the latter result of Byrnes et al. and relax the technical restriction \suciently small initial data in H 1(0; l)". Let us denote 2 jjw(
; t)jj2 := k1w2(0; t) + k2w2(l; t) + k @w (3.77) @x (
; t)kH Lemma 3.8 If k1; k2 > 0, then jj
jj and k
kV are equivalent norms. Proof. Assume that ' 2 C 1[0; l]. From the Fundamental Theorem of Calculus we have '(0) = '(x) ? 0x @' @s (s)ds; so by Young's and Cauchy-Schwarz inequalities, we have l 2g  2(j'(x)j2 + lk @' k2 ): j dx ) j'(0)j2  2fj'(x)j2 + ( 0 j @' @x @x H Now we integrate both sides on the interval [0; l] to get 2 )  2 maxf1; l2gk'k2 : j'(0)j2  2l (k'k2H + l2k @' k V @x H l Similarly, j'(l)j2  2l maxf1; l2gk'k2V ; and so jj'jj2  C1k'k2V , where C1 = 1+(k1 + k2) 2l maxf1; l2g: On the other hand, one can obtain the following again 2 from the Fundamental Theorem of Calculus k'k2H  2l(j'(0)j2 + lk @' @x kH ); and so 2  2l[j'(0)j2 + (l + 1 )k @' k2 ] k'k2V = k'k2H + k @' k @x H 2l @x H 2 )  1 jj'jj2  C1 (k1'2(0) + k @' k @x H C R

Z

2

where

1

C2

2

= 2l maxf k1 ; l + 21l g: Consequently, 1

C2k'k2V  jj'jj2  C1k'k2V

(3.78)

Because C 1[0; l] is dense in V , equation (3:78) holds for all ' 2 V . Let us recall the uncontrolled system (3:1)-(3:3) from the previous subsection. The distinction between the uncontrolled and the controlled ones are the boundary conditions. Namely, the boundary conditions wx(0; t) = wx(l; t) = 0 are replaced by wx(0; t) = k1w(0; t) and wx(l; t) = ?k2w(l; t). 32

For the controlled system, we rede ne the operator A as

A := (?@xx) along with the boundary conditions in (3:76)

(3.79)

with

@w (l; t) = ?k w(l; t)g D(A) = f 2 H 2(0; l) j @w (0 ; t ) = k 1 w(0; t); 2 @x @x

@w (l; t) = ?k w(l; t)g (0 ; t ) = k V := f 2 C 1(0; l) j @w 1 w(0; t); 2 @x @x V := Closure of V in the H 1 ? norm  H 1(0; l) H := Closure of V in the L2 ? norm  L2(0; l) Notice that a(
;
) : V  V ! IR is a bilinear form de ned by a(w; v) = hAw; viV  = k1w(0; t)v(0; t) + k2w(l; t)v(l; t) + (wx; vx): Moreover, a(
;
) is continuous and coercive. In fact, by using Cauchy-Schwarz inequality, Sobolev Embedding Theorem and equation (3.78), we have for any w; v 2 V

ja(w; v)j  (k1 + k2)kwkL1 kvkL1 + kwxkH kvxkH  f(k1 + k2)a20 +1gkwkV kvkV and

ja(w; w)j = jjwjj2  C2kwk2V ;

where a0 depends only on l. The operator A is self-adjoint, positive, with compact A?1 : H ! H . Consequently, there exist an orthonormal basis f j g1j=1 of H and a discrete set of positive eigenvalues fj g1j=1 such that A j = j j , and 0 < 1  2  : : :, j % 1. Since f j g1j=1 are also elements of V and D(A), they form an orthogonal basis for V and D(A) as well. Note that for any w 2 V

kA1=2wk2H = hAw; wiV  = k1w2(0) + k2 w2(l) + (wx; wx) = jjwjj2: As before we denote N (w; w) = w @w @x as a bilinear form on V  V with values

in H . We de ne weak solutions as follows.

De nition 3.9 A weak solution of (3:74)-(3:76) is a function w 2 CW ([0; T ]; H ) \ L2([0; T ]; V ) \L2loc ((0; T ]; D(A)) such that @w 2 L2 ([0; T ]; V  ) and @t  h dw dt ; viV +  (A w; A v) + (N (w; w); v) = 0 1 2

1 2

33

a.e. t 2 [0; T ]

and for all v 2 V with w(0) = w0 2 H: Furthermore, if in addition

w 2 L1 ([0; T ]; V ) \ L2([0; T ]; D(A)); dw dt 2 H

for a.e. t > 0;

and

dw + Aw + N (w; w) = 0 in H for a.e. t > 0; dt with w(x; 0) = w0(x) 2 V; then w is called a strong solution of (3:74)-(3:76). Let us remark again that our de nition of weak solution allows us to justify the boundary condition (3:76) because w 2 D(A))  H 2 for almost every t > 0 and hence the trace of the derivative for such functions makes sense. With A; D(A); V; V ; H; N (w; w); fj g's and f j g's de ned above, one can repeat step by step the Galerkin procedure for the uncontrolled problem (3:1)(3:3) in the previous section to establish the following theorems. We omit the proofs.

Theorem 9 Let w0 2 V , then for any T > 0 1. The system (3:74) (3:75) and (3:76) has a unique strong solution w, where w 2 C ([0; T ]; V ) \ L2([0; T ]; D(A)): 2. (Analyticity in time) w is analytic in time with values in D(A) in a neighborhood of (0; T ] of the complex plane Cl . 3. (Regularity) The solution is classical instantly and w(x; t) 2 C 1 ((0; T ]  [0; l]). 4. (Maximum Principle) kw(
; t)kL1  kw(
; 0)kL1 < 1 for all t 2 (0; T ):

Theorem 10 If w0 2 H , then for any T > 0 1. The system (3:74)-(3:76) has a unique weak solution w, where

w 2 L2([0; T ]; V ) \ C ([0; T ]; H ) \ Cloc((0; T ]; V ) \ L2loc((0; T ]; D(A)): 2. (Instantly strong) The weak solution w(x; t) of (3:1)-(3:3) is strong instantaneously. In particular, for any t0 2 (0; T ) we have

w 2 C ([t0; T ]; V ) \ L2([t0; T ]; D(A)): 3. (Analyticity in time) For any t0 2 (0; T ), w is analytic in time with values in D(A) in a neighborhood of [t0; T ] of the complex plane Cl .

34

4. (Regularity) The solution is classical instantly and w(x; t) 2 C 1 ([0; l]  (0; T ]). 5. (Maximum Principle) For any t0 2 (0; T )

kw(
; t)kL1  kw(
; t0)kL1 < 1

for all t 2 [t0; T ):

Remark: As mentioned earlier in the Remark after the statemnent of The-

orem 8, we similarly cannot conclude part 5 of the Theorem 10 for t0 = 0. However, the next proposition will enable us to attain such result.

Proposition 3.10 Let w0 2 L1 (0; l) and w be the weak solution of (3:74)(3:76) on [0; T ], such that   w0(x)  , for almost every x 2 [0; l]: Then for every (x; t) 2 [0; l]  (0; T ]   minf; 0g  w(x; t)  maxf; 0g   (3.80) and for every t 2 (0; T ].

kw(t)kL1(0;l)  kw0kL1(0;l)

(3.81)

Proof. Here the proof is identical to the proof in Proposition 3.7. We will let M (x; t)  [w(x; t) ? ]+ instead. Keep in mind that since we have dierent l boundary conditions, the term ?M (x; t) @M may no longer be zero. ( x; t ) @x 0 However, we will show that this is not possible. Moreover, the proof will l be complete if we can show that the term ?M (x; t) @M @x (x; t) 0 in equation (3.70) equals to zero. Notice rst that for any t > 0, if M (l; t) = 0, then M (l; t) @M @x (l; t) = 0. So suppose otherwise that w(l; t) > , then



@M (l; t) = @w (l; t) = ?k w(l; t)  ?k   0; 2 2 @x @x and hence M (l) @M 2 M (l)  0. By the similar @x (l) = ?k2 w(l)M (l)  ?k@M argument, for w(0; t) >  we have M (0; t) @x (0; t) = k1w(0; t)M (0; t)  k1M (0; t)  0. Thus l ?M (x; t) @M ( x; t ) =  fk2w(l; t)M (l; t) + k1w(0; t)M (0; t)g  0 0 @x and equation (3.70) can be written as l @M k2   k @M k2 : +  k ? ([w ? ]xx; M ) = ?M @M 0 @x @x H @x H



35

By repeating the argument in the proof of Proposition 3.7, one concludes that M (x; t)  0 for every t > 0. Let us also recall from Proposition 3.7 that for any t > 0, M (t) 2 V which can be embedded compactly in CB [0; l] for any  2 (0; 21 ). Thus for any t > 0, M (x; t)  0 for all x 2 [0; l], i.e. w(l; t)   for all x 2 [0; l] which contradicts the assumption that w(l; t) >  or w(0; t) > .

Theorem 11 (Byrnes and Gilliam [9]) Consider equation (3:74) along with (3:75) and (3:76). If kw0kH is suciently small, then kw(x; t)kH & 0 as t ! 1: Proof. From Theorem 10 we have w(x; t) 2 C ([0; T ]; H ) \ L2([0; T ]; V ) for

any T > 0. Let us rst take the L2-inner product of (3:74) and w(x; t) and using the boundary conditions (3:76) to get 1 d kw(t)k2 +  jjw(t)jj2 + l w2 @w dx = 0: (3.82) H 2 dt @x 0 If we estimate the last term on the left hand side of equation (3:82) as follows l 2 @w 2 k j w @x jdx  kwkL1 kwkH k @w H  jjwjj kwkH ; @x 0 it will result that 1 d kw(t)k2 +  jjw(t)jj2  jjw(t)jj2kw(t)k : H H 2 dt Let us choose kw0kH suciently small, say kw0kH  4 . Then by the continuity of the solution for some short interval of time [0; T], kw(t)kH  2 for all t 2 [0; T] and so we have 1 d kw(t)k2 +  jjw(t)jj2  0: H 2 2 dt Since C1kw(t)k2H  jjw(t)jj2, it follows that d kw(t)k2 + C kw(t)k2  0; 1 H H dt and by Gronwall's lemma, kw(t)kH  e ? C tkw0kH  4 for all t 2 [0; T]: (3.83) Thus (3:83) implies, by repeating the above argument, that kw(t)kH will remain small for all time and consequently, kw(t)kH & 0 as t ! 1: Z

Z

2

1

36

Remark: Even though the weak solution with initial data w0 2 H becomes strong, the fact that kw(t)kH & 0 does not imply that k @w @x kH remain bounded uniformly in time. Therefore at this stage we do not know that kw(t)kV & 0 or kw(t)kL1 & 0 when w0 is only an element of H , but not more regular, with kw0kH small. However, as a result of the Maximum Principle (Theorem 10) we have the following corollary: Corollary 3.11 Let w0 2 H and under the same assumptions of Theorem 11, we have kw(t)kLp(0;l) & 0 as t ! 1 for any 2  p < 1: Proof. For any t > 0, let t0 2 (0; t) be xed and independent of t, then from the Maximum Principle of Theorem 10, we have

kw(t)kLp(0;l)  kw(t)kHp kw(t)kL1?1p  kw(t)kHp kw(t0)kL1?1p : 2

2

2

2

As t ! 1, the right-hand side goes to zero which implies our corollary. Proposition 3.12 Let w0 2 W 1;1(0; l)  V and w be the strong solution of (3:74)-(3:76) on [0; T ] with   @w@x(x)  ; on [0; l]. Then for every (x; t) 2 [0; l]  (0; T ]   @w (3.84) @x (x; t)   and @w0 k 1 ; k @w ( t ) k for every t 2 [0; T ]: (3.85) L1 (0;l)  k @x @x L (0;l) Proof. Let M (x; t)  [ @w @x (x; t) ? ]+ @w (x; t) >  @w (x; t) ?  @w @x @x  maxf @x (x; t) ? ; 0g = 0 otherwise, then M (x; t)  0 for any (x; t) 2 [0; l]  [0; T ] and M (x; 0) = M0(x)  0. Notice also that for every t > 0, M (t) 2 V and @w (x; t) >  @w @M (x; t) = @x@ ( @w @x (x; t) ? ) = @x (x; t) @x @x 0 otherwise. Since w is a strong solution, it follows from Theorem 9 that w is classical for any t > 0. In addition, for any xed x0 2 [0; l] one can show that M (x0; t) is continuous and M 2(x0; t) is dierentiable. In fact, one can verify that d M 2 (x ; t) = 2M (x ; t) d @w (x ; t): (3.86) 0 0 dt dt @x 0 0

8 < :

8
0, then we have d jj w(t)jj 2 + jj w(t)jj2  0: 1 dt Thanks to Gronwall's lemma, jj w(t)jj 2  jj w0jj 2e? t. And because jj
jj is equivalent to k
kV ; we have kw(t)k2V  C5kw0k2V e? t: Consequently, kw(t)kV & 0 as t ! 1: Therefore it does not matter how large kw0kL1 or kw0kH are, Theorem 12 allows us to drive solutions to zero asymptotically in the stronger norm H 1 as long as k @w@x kL1 is suciently small. Namely, we do not have any restriction 0

1

1

0

40

on kw0kH , but only on the gradient, k @w@x kL1 . In particular, our result applies to any spatially homogeneous solutions and solutions that initially have large amplitude but do not have sharp gradients. However this condition on initial data can be relaxed even more as we will prove a result that allows us to control, in H 1-norm, initial data which is small in the L1 norm, but has sharp gradients, such as the very ne saw-tooth ones, or even the ones with jumps, i.e. initial data that do not belong to V . Theorem 13 Consider the problem (3:74)-(3:76), and let w0 2 L1 (0; l) with kw0kL1(0;l) suciently small. Then kw(x; t)kV & 0 and kw(
; t)kL1(0;l) & 0 as t ! 1. Proof. For such initial data we have unique weak solution which instantly becomes a classic solution (Theorem 10). We estimate the last term of (3:82) l 2 @w 2 1 j w @x jdx  kwkL1 kwkH k @w @x kH  kwkL jjw(t)jj ; 0 then (3:82) becomes 1 d kw(t)k2 +  jjw(t)jj2  kwk 1 jjw(t)jj2; for t > 0: (3.98) L H 2 dt Now we add (3:96) and (3:98) to get d jj w(t)jj 2 +  (k @ 2w k2 + jjw(t)jj2)  1 kwk2 k @w k2 + kwk 1 jjw(t)jj2; L dt @x2 H  L1 @x H where jj w(t)jj2 := jjw(t)jj2 + 21 kw(t)k2H as before. By using the Maximum Principle as in Proposition 3.10, that is kw(t)k2L1  kw0k2L1 , and the equivalent norms of jj w(t)jj 2, we have d jj w(t)jj 2 +  kw(t)k2  ( 1 kw0k2 + kw0k 1 )jj w(t)jj2: L H L1 dt  Note that kw(t)k2V  ( 2l )2kw(t)k2H , thus by the equivalent norms of jj w(t)jj2 again we have d jj w(t)jj 2 + C ( 2 )2jj w(t)jj 2  ( 1 kw0k2 + kw0k 1 )jj w(t)jj2: 6 L L1 dt l  Let 2 = C6( 2l )2 ? ( 1 kw0k2L1 + kw0kL1 ) and assume that kw0kL1 (0;l) suciently small so that 2 > 0. Then d jj w(t)jj 2 + jj w(t)jj2  0: 2 dt and we have jj w(t)jj 2  jj w0jj 2e? t. Because jj
jj is equivalent k
kV ; we have kw(t)kV & 0 as t ! 1: Finally since kw(t)kL1 (0;l)  C7kw(t)kV , we also get kw(t)kL1(0;l) & 0 as t ! 1: 0

Z

2

2

2

41

4 BURGERS' EQUATION WITH NONLINEAR BOUNDARY CONDITIONS

We end this paper with the discussion of the Burgers equation subject to the nonlinear boundary conditions proposed by J.A. Burns [4]. Namely, we consider the following problem: @w (x; t) ?  @ 2w (x; t) + w(x; t) @w (x; t) = 0; 0 < x < l; t > 0 (4.1) @t @x2 @x w(x; 0) = w0(x) (4.2) @w (l; t) = 0: (0 ; t ) = w ( l; t ) (4.3) w(0; t) @w @x @x This kind of boundary conditions is interesting simply because it possesses the uncertainty of the zero value of the function itself and its derivative at the end points, as the system evolves in time. However, we will show that this initial-valued problem is ill-posed and has multiple solutions for the same initial value. For motivation we will give an example which shows that solutions to the Heat equation subject to the nonlinear boundary conditions (4.3) are not unique. Later, we will prove the same result for the Burgers' equation. Let us consider the Heat equation @w (x; t) ?  @ 2w (x; t) = 0 0 < x < l; t > 0 @t @x2 with the initial data w0(x) = 1 ? cos( 2l x): If subject to the Dirichlet boundary conditions of w(0; t) = w(l; t) = 0; (4.4) this Heat equation possesses a unique solution w1(x; t); and if subject to the Neumann boundary conditions @w (0; t) = @w (l; t) = 0; (4.5) @x @x the Heat equation has a unique solution w2(x; t) = 1 ? e?( l ) t cos( 2l x). Since w2(x; t) does not satisfy the boundary conditions (4.4) for any t > 0, we have w2(x; t) 6 w1(x; t). But both w1(x; t) and w2(x; t) satisfy the Heat equation subject to the nonlinear boundary conditions (4.3) and so solutions to this problem are not unique. Proposition 4.1 Let w0 2 H02 (0; l) (i.e. w0 2 H 2(0; l) and w0 (0) = w0 (l) = @w (0) = @w (l) = 0). Let w and w be solutions of the Burgers' equation 1 2 @x @x (4.1)-(4.2) subject to the Dirichlet (4.4) and the Neumann (4.5) boundary conditions, respectively. Suppose w1(x; t)  w2(x; t) for all t > 0 and x 2 [0; l], then w1 (x; t)  w2(x; t)  0 and w0  0: 2

0

0

42

2

Proof. We rst extend w0 to the whole real line to be odd and periodic of period 2l. Let us denote by w0 the extension of w0 and by w1 the solution to (4.1) with initial data w0 subject to periodic boundary conditions. Notice also that the space of periodic odd functions is invariant under the solution operator for the Burgers' equation on IR with periodic boundary condition, i.e. if the initial value is an odd periodic function, then the solution remains so for all t > 0. Thus w1 is the odd periodic solution and hence it also satis es the Dirichlet boundary condition (4.4). Moreover, since w0 2 H 1(0; l), then for any t > 0 the solution w1 belongs to a Gevrey class of regularity (see [15] and [19]), hence w1 is spatial analytic. Now for any t > 0 we can write the Taylor expansion of w1 about x = 0 and we obtain: 1 n w1(x; t) = an(t)xn where an(t) = n1! @@xwn1 (0; t): n=0 X

Keep in mind that w1 also satis es (4.1)-(4.2) and (4.4). Therefore by uniqueness of solution (Theorem 1), w1(x; t) = w1(x; t) [0;l]. Now for any t > 0 we have a0 = w1(0; t) = w1(0; t) = 0: Also since we assume w1(x; t)  w2(x; t), @w1 (0; t) = @w2 (0; t) = 0: 1 a1 = @w (0 ; t ) = @x @x @x Because w1(0; t)  0 for all time, we have @w@t (0; t) = 0 and hence from the equation (4.1)

1

@ 2w1 (0; t) = 1 @w1 (0; t)+ 1 w (0; t) @w1 (0; t) = 0 and a = 1 @ 2w1 (0; t) = 0: 2 @x2  @t  1 @x 2 @x2 w (0; t) = 0. In addition, @ w (0; t) = 0 and Note that @w@x (0; t)  0 implies @@t@x @x so 2w 3 2 1 1 2 + w (0; t) @ w 1 (0; t)] = 0: a3 = 16 @@xw31 (0; t) = 61 [ @@t@x (0; t) + ( @w (0 ; t )) 1 @x @ 2x Proceeding in this fashion, we obtain by induction that an(t) = 0 for all n  0 and consequently w1  0 in the neighborhood of x = 0. But since w1 is analytic, we get w1  0 on all IR. In particular, w1(x; t)  w2(x; t)  0 in [0; l]. Now since w0 2 H02(0; l), it follows from Theorem 1 that for all T > 0, w1 2 C ([0; T ]; H01(0; l)). In particular, kw1(t) ? w0k2H (0;l) & 0 as t ! 0+, therefore w0 must be zero as well. 1

2

2

1

1 2

1

Corollary 4.2 Suppose w0 2 H02(0; l) and w0 6 0. Let w1 and w2 be de ned as in Proposition 4.1, then w1(x; t) 6 w2(x; t). 43

Corollary 4.3 Let w0 2 H02(0; l) and w0 6 0, then solutions to the Burgers equation subject to the nonlinear boundary conditions (4.1)-(4.3) are not unique.

Proof. Let w1 and w2 be de ned as in Proposition 4.1. If w1(x; t)  w2(x; t), then by Proposition 4.1 w0  0, which contradicts the assumption. As a result, w1(x; t) 6 w2(x; t). But both w1(x; t) and w2(x; t) are solutions of (4.1)-(4.3). So we have multiple solutions.

ACKNOWLEDGEMENTS

We would like to thank Panagiota Daskalopoulos for the interesting discussions. This work was partially supported by the NSF Grant No. DMS 9308774 and by the Joint University California-Los Alamos National Laboratory INCOR program. H.V. Ly acknowledges the support of the University of California-Irvine Regents Dissertation Fellowship. Also, E.S. Titi acknowledges the partial support of the AFOSR Contract No. F49620-95-C-0027 through BEAM Engineering and Applied Research.inh-thuong

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