1.1 Regular and Singular Perturbations . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Evolution Equations in Hilbert Spaces [619783]
Habilitation Thesis
…
Cosma (Barbu) Elena Luminit ¸a
Facultatea de Matematic ˘a s ¸i Informatic ˘a, Universitatea Ovidius din Constant ¸a
Contents
Chapter 1 1
1 Preliminaries 1
1.1 Regular and Singular Perturbations . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Evolution Equations in Hilbert Spaces . . . . . . . . . . . . . . . . . . . . 8
Chapter 2 16
2 On Some Singularly Perturbed Problems 17
2.1 Presentation of the Problems . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.2 Hyperbolic Systems with Algebraic Boundary Conditions . . . . . . . . . . 19
Chapter 3 19
3 Elliptic-like Regularizations of Nonlinear Evolution Equations 20
3.1 Presentation of the Problems . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.2 Asymptotic analysis of problem (P:1)e. . . . . . . . . . . . . . . . . . . . 22
3.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.3.1 Regularization of the semilinear telegraph system . . . . . . . . . . 28
3.3.2 Regularization of the semilinear wave equation . . . . . . . . . . . 31
3.4 Asymptotic analysis of problem (P:2)e. . . . . . . . . . . . . . . . . . . . 34
3.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.5.1 The nonlinear heat equation . . . . . . . . . . . . . . . . . . . . . 41
3.5.2 The nonlinear telegraph system . . . . . . . . . . . . . . . . . . . 43
3.5.3 The nonlinear wave equation . . . . . . . . . . . . . . . . . . . . . 45
3.6 Elliptic regularization of the semilinear telegraph system with nonlinear
boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
3.7 Elliptic regularization of the semilinear heat equation with nonlinear bound-
ary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
Bibliography 66
ii
Chapter 1
Preliminaries
1.1 Regular and Singular Perturbations
In this Section we recall and discuss some general concepts of singular perturbation theory
which will be needed later. Our presentation is mainly concerned with singular perturbation
problems of the boundary layer type, which are particularly relevant for applications.
In order to start our discussion, we are going to set up an adequate framework. Let
DRnbe a nonempty open bounded set with a smooth boundary S. Consider the following
equation, denoted Ee,
Leu=f(x;e);x2D;
where eis a small parameter, 0 <e 1,Leis a differential operator, and fis a given
real-valued smooth function. If we associate with Eesome condition(s) for the unknown u
on the boundary S, we obtain a boundary value problem Pe. We assume that, for each e,Pe
has a unique smooth solution u=ue(x). Our goal is to construct approximations of uefor
small values of e. The usual norm we are going to use for approximations is the sup norm
(or max norm), i.e.,
kgkC(D)=supfjg(x)j;x2Dg;
for every continuous function g:D !R. We will also use the weaker Lp-norm
kgkLp(D)=Z
Djgjpdx1=p
;
where 1p<¥.
In many applications, operator Leis of the form
Le=L0+eL1;
where L0andL1are differential operators which do not depend on e. IfL0does not in-
clude some of the highest order derivatives of Le, then we should associate with L0fewer
1
boundary conditions. So, Pebecomes
L0u+eL1u=f(x;e);x2D;
with the corresponding boundary conditions. Let us also consider the equation, denoted E0,
L0u=f0;x2D;
where f0(x):=f(x;0), with some boundary conditions, which usually come from the origi-
nal problem Pe. Let us denote this problem by P0. Some of the original boundary conditions
are no longer necessary for P0. Problem Peis said to be a perturbed problem (perturbed
model ), while problem P0is called unperturbed (orreduced model ).
Definition 1. Problem Peis called regularly perturbed with respect to some norm kk if
there exists a solution u0of problem P0such that
kue u0k ! 0 ase!0:
Otherwise, Peis said to be singularly perturbed with respect to the same norm.
In a more general setting, we may consider time tas an additional independent variable
for problem Peas well as initial conditions at t=0 (sometimes tis the only independent
variable). Moreover, we may consider systems of differential equations instead of a single
equation. Note also that the small parameter may also occur in the conditions associated
with the corresponding system of differential equations. For example, we will discuss later
some coupled problems in which the small parameter is also present in transmission condi-
tions.Basically, the definition above also applies to these more general cases.
In order to illustrate this definition we are going to consider some examples.
Example 1. Consider the following simple Cauchy problem Pe:
du
dt+eu=f0(t);0<t<T;u(0) =q;
where T2(0;+¥),q2R, and f0:R !Ris a given smooth function. The solution of Pe
is given by
ue(t) =e et
q+Zt
0eesf0(s)ds
;0tT:
Obviously, ueconverges uniformly on [0;T], asetends to 0, to the function
u0(t) =q+Zt
0f0(s)ds;
2
which is the solution of the reduced problem
du
dt=f0(t);0<x<T;u(0) =q:
Therefore, Peis regularly perturbed with respect to the sup norm.
Example 2. LetPebe the boundary value problem
ed2u
dx2+du
dx=2x;0<x<1;u(0) =0=u(1):
Its solution is
ue(x) =x(x 2e)+2e 1
1 e 1=e
1 e x=e
:
Note that
ue(x) = ( x2 1)+e x=e+re(x);
where re(x)converges uniformly to the null function, as etends to 0. Therefore, uecon-
verges uniformly to the function u0(x) =x2 1 on every interval [d;1], 0<d<1, but not
on the whole interval [0;1]. Obviously, u0(x) =x2 1 satisfies the reduced problem
du
dx=2x;0<x<1;u(1) =0;
butkue u0kC[0;1]does not approach 0. Therefore, Peis singularly perturbed with respect
to the sup norm. For a small d,u0is an approximation of uein[d;1], but it fails to be an
approximation of uein[0;d]. This small interval [0;d]is called a boundary layer . Here we
notice a fast change of uefrom its value ue(0) =0 to values close to u0. This behavior of ue
is called a boundary layer phenomenon and Peis said to be a singular perturbation problem
of the boundary layer type . In this simple example, we can see that a uniform approxima-
tion for ue(x)is given by u0(x) +e x=e. The function e x=eis a so-called boundary layer
function (correction ). It fills the gap between ueandu0in the boundary layer [0;d].
Let us remark that Peis a regular perturbation problem with respect to the Lp-norm for
all 1p<¥, sincekue u0kLp(0;1)tends to zero. The boundary layer which we have just
identified is not visible in this weaker norm .
Example 3. LetPebe the following Cauchy problem
edu
dt+ru=f0(t);0<t<T;u(0) =q;
where ris a positive constant, q2Rand f0:[0;T]!Ris a given Lipschitzian function.
3
The solution of this problem is given by
ue(t) =qe rt=e+1
eZt
0f0(s)e r(t s)=eds;0tT;
which can be written as
ue(t) =1
rf0(t)+
q 1
rf0(0)
e rt=e+re(t);0tT;
where
re(t) = 1
rZt
0f0
0(s)e r(t s)=eds:
We have
jre(t)jL
rZt
0e r(t s)=edsL
r2e;
where Lis the Lipschitz constant of f0. Therefore, reconverges uniformly to zero on [0;T]
asetends to 0. Thus ueconverges uniformly to u0(t) = ( 1=r)f0(t)on every interval [d;T],
0<d<T, but not on the whole interval [0;T]iff0(0)6=rq. Note also that u0is the solution
of the (algebraic) equation
ru=f0(t);0<t<T;
which represents our reduced problem. Therefore, if f0(0)6=rq, this Peis a singular pertur-
bation problem of the boundary layer type with respect to the sup norm. The boundary layer
is a small right vicinity of the point t=0. A uniform approximation of ue(t)on[0;T]is
the sum u0(t) +
q 1
rf0(0)
e rt=e. The function
q 1
rf0(0)
e rt=eis a boundary layer
function, which corrects the discrepancy between ueandu0within the boundary layer.
Example 4. InDT=f(x;t); 0<x<1;0<t<Tgwe consider the telegraph system
(
eut+vx+ru=f1(x;t);
vt+ux+gv=f2(x;t);(S)e
with initial conditions
u(x;0) =u0(x);v(x;0) =v0(x);0<x<1; (IC)e
and boundary conditions of the form
(
r0u(0;t)+v(0;t) =0;
u(1;t)+f0(v(1;t)) = 0;0<t<T;(BC)e
where f1;f2:DT!R;f0:R!R,u0;v0:[0;1]!Rare given smooth functions, and r0;r;g
are constants, r0>0;r>0;g0. If in the model formulated above and denoted by Pewe
4
takee=0, we obtain the following reduced problem P0:
(
u=r 1(f1 vx);
vt r 1vxx+gv=f2 r 1f1xinDT;(S)0
v(x;0) =v0(x);0<x<1; (IC)0
(
rv(0;t) r0vx(0;t)+r0f1(0;t) =0;
r f0(v(1;t))+vx(1;t) f1(1;t) =0;0<t<T:(BC)0
In this case, the reduced system (S)0consists of an algebraic equation and a differential
equation of the parabolic type, whereas system (S)eis of the hyperbolic type. The initial
condition for uis no longer necessary. We will derive P0later in a justified manner.
Let us remark that if the solution of Pe, say Ue(x;t) = ( ue(x;t);ve(x;t)), would con-
verge uniformly in DTto the solution of P0, then necessarily
v0
0(x)+ru0(x) =f1(x;0);8×2[0;1]:
If this condition is not satisfied then that uniform convergence is not true and, as we will
show later, Uehas a boundary layer behavior in a neighborhood of the segment f(x;0); 0x1g.
Therefore, this Peis a singular perturbation problem of the boundary layer type with respect
to the sup normkkC(DT)2. However, using the form of the boundary layer functions which
we are going to determine later, we will see that the boundary layer is not visible in weaker
norms, like for instance kkC([0;1];Lp(0;T))2, 1p<¥, and Peis regularly perturbed in such
norms.
Definition 2. Letuebe the solution of some perturbed problem Pedefined in a domain D.
Consider a function U(x;e);x2D1, where D1is a subdomain of D. The function U(x;e)
is called an asymptotic approximation in D1of the solution ue(x)with respect to the sup
norm if
sup
x2D1kue(x) U(x;e)k! 0 as e!0:
Moreover, if
sup
x2D1kue(x) U(x;e)k=O(ek);
then we say that U(x;e)is an asymptotic approximation of ue(x)inD1with an accuracy of
the order ek:We have similar definitions with respect to other norms. In the above definition
we have assumed that Uanduetake values in Rn, andkkdenotes one of the norms of this
space.
For a real-valued function E(e), the notation E(e) =O(ek)means thatjE(e)jMek
for some positive constant Mand for all esmall enough.
5
In the following we are going to discuss the celebrated Vishik-Lyusternik method
[69] for the construction of asymptotic approximations for the solutions of singular per-
turbation problems of the boundary layer type. To explain this method we consider the
problem used in Example 4 above. This problem admits a boundary layer near the side
f(x;0); 0x1gof the rectangle DTwith respect to the sup norm. The existence of
this boundary layer is suggested by the analogous ODE problem considered in Example 3
above. Indeed, eis present only in the first equation of system (S)e. For a given xthis has the
same form as the equation discussed in Example 3, with f0(t):=f1(x;t) vx(x;t). So we
expect to have a singular behavior of the solution near the value t=0 for all x. We will see
that this is indeed the case. We will restrict ourselves to seeking an asymptotic expansion
of the order zero for the solution of Pe, i.e.,
Ue(x;t) =
ue(x;t);ve(x;t)
=
X(x;t);Y(x;t)
+
c0(x;t);d0(x;t)
+
R1e(x;t);R2e(x;t)
;
where
X(x;t);Y(x;t)
is the regular term, t=t=eis the rapid variable for this problem,
c0(x;t);d0(x;t)
is the correction (of order zero), and Re:=
R1e;R2e
is the remainder
(of order zero). The form of the rapid variable tis also suggested by the analogous problem
in Example 3 above. In fact, according to this analogy, we expect a singular behavior near
t=0 only for ue, i.e., d0=0. This will be indeed the case, but we have started with the
above expansion, since our present problem is more complex and ue,veare connected to
each other. We substitute the above expansion into Peand identify the coefficients of the
like powers of e. Of course, we distinguish between the coefficients depending on (x;t)
and those depending on (x;t). We should keep in mind that the remainder components are
small as compared to the other terms. This idea comes from the case when we have a series
expansion of the solution of Pe. So, after substituting the above expansion into (S)e, we
see that the only coefficient of e 1in the second equation is d0t(x;t) = (¶d0=¶t)(x;t).
So, this should be zero, thus d0is a function depending on xonly. Taking also into account
the fact that a boundary layer function should converge to zero as t!¥, we infer that d0
is identically zero, as expected. From the first equation of (S)e, we derive the following
boundary layer equation by identifying the coefficients of e0:
c0t(x;t)+rc0(x;t) =0:
If we integrate this equation and use the usual condition c0(x;t)!0 ast!¥, we get
c0(x;t) =a(x)e rt;
where a(x)will be determined later. Applying the identification procedure to regular terms,
one can see that (X;Y)satisfies the reduced problem P0we have already indicated before.
6
We have in mind that the remainder components should tend to 0 as e!0. From the initial
condition for ue, which reads
u0(x) =X(x;0)+a(x)+R1e(x;0);
we get
u0(x) =X(x;0)+a(x);
which shows how c0(x;t)compensate for the discrepancy in the corresponding initial con-
dition. In fact, at this moment c0(x;t)is completely determined, since the last equation
gives
a(x) =u0(x) 1
r
f1(x;0) v0
0(x)
:
Let us now discuss the boundary conditions. From the equation
r0
X(0;t)+c0(0;t)+R1e(0;t)
+Y(0;t)+R2e(0;t) =0
we derive
r0X(0;t)+Y(0;t) =0;
plus the condition c0(0;t) =0, i.e., a(0) =0. Now, let us discuss the nonlinear boundary
condition, which reads
X(1;t)+c0(1;t)+R1e(1;t)
+f0
Y(1;t)+R2e(1;t)
=0:
We obtain c0(1;t) =0, i.e., a(1) =0. For the regular part we get
X(1;t)+f0(Y(1;t)) = 0:
The same result is obtained if we consider a series (or Taylor) expansion for the solution
ofPe, expand f0
ve(1;t)
around e=0, and then equate the coefficients of e0. Of course,
we need more regularity for f0to apply this method. Therefore it is indeed possible to
show formally that (u;v) = ( X;Y)satisfies P0. We have already determined the corrections
of the order zero, and it is an easy matter to find the problem satisfied by the remainder
components R1e;R2e. So, at this moment, we have a formal asymptotic expansion of the
order zero. The next step would be to show that the expansion is well defined, in particular
to show that, under some specific assumptions on the data, both PeandP0have unique
solutions in some function spaces. Finally, to show that the expansion is a real asymptotic
expansion, it should be proved that the remainder tends to zero with respect to a given norm.
In our applications (including the above hyperbolic Pe) we are going to do even more, to
establish error estimates for the remainder components (in most of the cases with respect to
the sup norm).
7
Let us note that in the above formal derivation procedure we obtained two conditions
for the correction c0(x;t), namely a(0) =0 and a(1) =0. These two conditions assure
thatc0does not create discrepancies at (x;t) = ( 0;0)and(x;t) = ( 1;0), which are corner
boundary points of DT. These conditions can be expressed in terms of our data:
v0
0(0)+ru0(0) =f1(0;0);v0
0(1)+ru0(1) =f1(1;0):
In fact, we will see that these conditions are also necessary compatibility conditions for the
existence of smooth solutions for problems PeandP0.
By the same technique, terms of the higher order approximations can be constructed
as well.
For background material concerning the topics discussed in this chapter we refer the
reader to [31], [45], [48], [54], [53], [65], [66], [67], [69].
1.2 Evolution Equations in Hilbert Spaces
In this chapter we are going to remind the reader of some basic concepts and results in the
theory of evolution equations associated with monotone operators which will be used in the
next chapters. The proofs of the theorems will be omitted, but appropriate references will
be indicated.
Monotone operators
LetHbe a real Hilbert space with its scalar product and associated Hilbertian norm
denoted byh;iandkk, respectively.
By a multivalued operator A:D(A)H!Hwe mean a mapping that assigns to each
x2D(A)a nonempty set AxH. In fact, Ais a mapping from D(A)into 2H, but we prefer
this notation. The situation in which some (or even all) of the sets Axare singletons is not
excluded.
The graph of Ais defined as the following subset of HH
G(A):=f(x;y)2HH;x2D(A);y2Axg:
Obviously, for every MHHthere exists a unique multivalued operator Asuch that
G(A) =M:More precisely,
D(A) =fx2H; there exists a y2Hsuch that (x;y)2Mg;
Ax=fy2H;(x;y)2Mg 8x2D(A):
8
So, every multivalued operator Acan be identified with G(A)and we will simply write
(x;y)2Ainstead of: x2D(A)andy2Ax:
The range R(A)of a multivalued operator A:D(A)H!His defined as the union
of all Ax,x2D(A). For every multivalued operator A, there exists A 1which is defined as
A 1=f(y;x);(x;y)2Ag:
Obviously, D(A 1) =R(A)andR(A 1) =D(A).
Definition 3. A multivalued operator A:D(A)H!His said to be monotone if
hx1 x2;y1 y2i08(x1;y1);(x2;y2)2A: (1.2.1)
Sometimes, having in mind that Acan be identified with its graph, we will say that
Ais a monotone subset of HH. IfAxis a singleton then we will often identify Axwith
its unique element. So, if Ais single-valued (i.e., Axis a singleton for all x2D(A)), then
(1.2.1) can be written as
hx1 x2;Ax1 Ax2i08×1;x22D(A):
A very important concept is the following:
Definition 4. A monotone operator A:D(A)H!His called maximal monotone if A
has no proper monotone extension (in other words, A;viewed as a subset of HH, cannot
be extended to any A0HH,A06=A;such that the corresponding multivalued operator
A0is monotone).
We continue with the celebrated Minty’s characterization for maximal monotonicity:
Theorem 1. (G. Minty) Let A :D(A)H!H be a monotone operator. It is maximal
monotone if and only if R (I+lA) =H for some l>0or, equivalently, for all l>0.
Here Idenotes the identity operator on H. Recall also that for two multivalued operators A,
Bandr2R,A+B:=f(x;y+z);(x;y)2Aand(x;z)2Bg;rA:=f(x;ry);(x;y)2Ag.
Theorem 2. Let A :D(A)H!H be a maximal monotone operator. Assume in addition
that A is coercive with respect to some x 02H, i. e.,
lim
kxk!¥
(x;y)2Ahx x0;yi
kxk= +¥:
Then R (A) =H, i.e., A is surjective.
9
In particular, if Ais strongly monotone, i. e., there exists a positive constant asuch that
hx1 x2;y1 y2iakx1 x2k28(x1;y1);(x2;y2)2A;
then Ais coercive with respect to any x02D(A). Therefore, if Ais maximal monotone and
also strongly monotone, then R(A) =H.
Theorem 3. (R. T. Rockafellar) If A:D(A)H!H and B :D(B)H!H are two maxi-
mal monotone operators such that (Int D(A))\D(B)6=/ 0;then A +B is maximal monotone,
too.
Now, recall that a single-valued operator A:D(A) =H!His said to be hemicontin-
uous if for every x;y;z2H
lim
t!0hA(x+ty);zi=hAx;zi:
Theorem 4. (G. Minty) If A:D(A) =H!H is single-valued, monotone and hemicontin-
uous, then A is maximal monotone.
Let us continue with a brief presentation of the class of subdifferentials. First, let us
recall that a function y:H!( ¥;¥]is said to be proper ifD(y)6=/ 0;where D(y):=
fx2H;y(x)<¥g:D(y)is called the effective domain of y. Function yis said to be
convex if
y(ax+(1 a)y)ay(x)+(1 a)y(y)8a2(0;1);x;y2H;
where usual conventions are used concerning operations which involve ¥.
A function y:H!( ¥;¥]is said to be lower semicontinuous atx02Hify(x0)
liminf x!x0y(x):
Lety:H!( ¥;¥]be a proper convex function. Its subdifferential atxis defined by
¶y(x) =fy2H;y(x) y(v)hy;x vi;8v2Hg:
The operator ¶yHHis called the subdifferential of y. Clearly, its domain D(¶y)is
included in D(y).
Theorem 5. Ify:H!( ¥;¥]is a proper convex lower semicontinuos (on H) function,
then¶yis a maximal monotone operator and, furthermore,
D(¶y) =D(y);Int D(¶y) =Int D(y):
Ify:H!( ¥;¥]is proper and convex, then operator A=¶yiscyclically monotone ,
i. e., for every n2Nwe have
hx0 x1;x
0i+hx1 x2;x
1i++hxn 1 xn;x
n 1i+hxn x0;x
ni0;
10
for all (xi;x
i)2A;i=0;1;;n:
An operator A:D(A)H!His called maximal cyclically monotone ifAcannot be
extended properly to any cyclically monotone operator. Obviously, if y:H!( ¥;¥]is
a proper convex lower semicontinuous function, then A=¶yis maximal cyclically mono-
tone. The converse implication is also true:
Theorem 6. If A:D(A)H!H is a maximal cyclically monotone operator, then there ex-
ists a proper convex lower semicontinuous function y:H!( ¥;¥], uniquely determined
up to an additive constant, such that A =¶y:
The reader may find the detailed proofs of Theorems 1-6 in [17], [22] and [51].
Evolution equations
LetHbe a real Hilbert space. Its scalar product and norm are again denoted by h;i
andkk;respectively. Consider in Hthe following Cauchy problem:
(
u0(t)+Au(t) =f(t);0<t<T;
u(0) =u0;(1.2.2)
where A:D(A)H!His a nonlinear single-valued operator and f2L1(0;T;H):In fact,
the most known existence results are valid for multivalued A’s, but we will consider only
single-valued A’s. This is enough for our considerations.
Definition 5. A function u2C([0;T];H)is said to be a strong solution of equation (1.2.2) 1
if:
(i)uis absolutely continuous on every compact subinterval of (0;T);
(ii)u(t)2D(A)for a. a. t2(0;T);
(iii)usatisfies (1.2.2) 1for a. a. t2(0;T):
If in addition u(0) =u0, then uis called a strong solution of the Cauchy problem (1.2.2).
Definition 6. A function u2C([0;T];H)is called a weak solution of (1.2.2) 1if there exist
two sequencesfungW1;¥(0;T;H)andffngL1(0;T;H)such that :
(k)u0
n(t)+Aun(t) =fn(t)for a. a. t2(0;T);n2N;
(kk)un!uinC([0;T];H)asn!¥;
(kkk)fn!finL1(0;T;H)asn!¥:
Again, if in addition u(0) =u0, then uis called a weak solution of the Cauchy problem
(1.2.2).
11
Theorem 7. (see, e.g., [51], p. 48) If A:D(A)H!H is a maximal monotone operator,
u02D(A)and f2W1;1(0;T;H);then the Cauchy problem (1.2.2) has a unique strong
solution u2W1;¥(0;T;H):Moreover u (t)2D(A)for all t2[0;T];u is differentiable on
the right at every t 2[0;T);and
d+u
dt(t)+Au(t) =f(t)8t2[0;T);
d+u
dt(t)
kf(0) Au0k+Zt
0kf0(s)kds8t2[0;T): (1.2.3)
If u and u are the strong solutions corresponding to (u0;f);(u0;f)2D(A)W1;1(0;T;H);
then
ku(t) u(t)kk u0 u0k+Zt
0kf(s) f(s)kds;0tT: (1.2.4)
Theorem 8. (see, e.g., [51], p. 55) If A:D(A)H!H is a maximal monotone operator,
u02D(A)and f2L1(0;T;H);then the Cauchy problem (1.2.2) has a unique weak solution
u2C([0;T];H):If u and u are the weak solutions corresponding to (u0;f);(u0;f)2D(A)
L1(0;T;H);then u and u still satisfy (1.2.4) .
Remark 1. Both the above theorems still hold in the case of Lipschitz perturbations, i.e.,
in the case in which Ais replaced by A+B, where Ais maximal monotone as before and
B:D(B) =H!His a Lipschitz operator (see [22], p. 105). The only modifications appear
in estimates (1.2.3) and (1.2.4) :
d+u
dt(t)
ewt
kf(0) Au0k+Zt
0e wskf0(s)kds
;0t<T; (1.2.5)
ku(t) u(t)kewt
ku0 u0k+Zt
0e wskf(s) f(s)kds
;0tT; (1.2.6)
where wis the Lipschitz constant of B.
Theorem 9. (H. Br ´ezis, [22]) If A is the subdifferential of a proper convex lower semi-
continuous function j:H!( ¥;+¥];u02D(A)and f2L2(0;T;H);then the Cauchy
problem (1.2.2) has a unique strong solution u ;such that t1=2u02L2(0;T;H);t!j(u(t))
is integrable on [0;T]and absolutely continuous on [d;T];8d2(0;T):If, in addition,
u02D(j);then u02L2(0;T;H);t!j(u(t))is absolutely continuous on [0;T];and
j(u(t))j(u0)+1
2ZT
0kf(t)k2dt8t2[0;T]:
Before stating other existence results let us recall the following definition:
12
Definition 7. LetCbe a nonempty closed subset of H. Acontinuous semigroup of con-
tractions onCis a family of operators fS(t):C!C;t0gsatisfying the following con-
ditions:
(j)S(t+s)x=S(t)S(s)x8x2C;t;s0;
(j j)S(0)x=x8x2C;
(j j j)for every x2C;the mapping t!S(t)xis continuous on [0;¥);
(jv)kS(t)x S(t)ykk x yk 8t0;8x;y2C:
Theinfinitesimal generator of a semigroupfS(t):C!C;t0g;sayG;is given by:
G(x) =lim
h!0
h>0S(h)x x
h;x2D(G);
where D(G)consists of all x2Cfor which the above limit exists.
LetA:D(A)H!Hbe a (single-valued) maximal monotone operator. From Theo-
rem 7 we see that if f0, then for every u02D(A)there exists a unique strong solution
u(t);t0;of the Cauchy problem (1.2.2). We set S(t)u0:=u(t);t0:Then it is easily
seen that S(t)is a contraction (i. e., a non expansive operator) on D(A)(see (1.2.4)) and so
S(t)can be extended as a contraction on D(A);for each t0. Moreover, it is obvious that
the familyfS(t):D(A)!D(A);t0gis a continuous semigroup of contractions and its
infinitesimal generator is G= A. We will say that the semigroup is generated by A. IfA
is linear maximal monotone, then D(A) =H, and the semigroup fS(t):H!H;t0ggen-
erated by Ais aC0-semigroup of contractions. Recall that a family of linear continuous
operatorsfS(t):H!H;t0gis called a C0-semigroup if conditions (j) (j j j)above
are satisfied with C=H. In fact, in this case, continuity at t=0 of the function t!S(t)x,
8x2H, combined with (j) (j j)is enough to derive (j j j). For details on C0-semigroups,
we refer to [2], [37], [56].
Definition 8. LetA:D(A)H!Hbe the infinitesimal generator of a C0-semigroup
fS(t):H!H;t0g, and let f2L1(0;T;H);u02H. The function u2C([0;T];H)given
by
u(t) =S(t)u0+Zt
0S(t s)f(s)ds8t2[0;T]; (1.2.7)
is called a mild solution of the Cauchy problem (1.2.2). Since D(A) =H, the above formula
(which is usually called the variation of constants formula ) makes sense for any u02H.
Obviously, the mild solution has the uniqueness property.
Theorem 10. (see [36], [56], p. 109) Let A :D(A)H!H be the infinitesimal generator
of a C 0-semigroupfS(t):H!H;t0g, and let u 02D(A), f2W1;1(0;T;H):Then, the
mild solution u of (1.2.2) is also its unique strong solution. Moreover, u 2C1([0;T];H),
13
and
f(t) Au(t) =u0(t) =S(t)
f(0) Au0
+Zt
0S(t s)f0(s)ds;0tT: (1.2.8)
We will apply Theorem 10 to the special case of a linear monotone (positive) operator
Afor which A+wIis maximal monotone for some w>0 (Iis the identity of H).
Note that the mild solution of problem (1.2.2) given by the above definition is also a
weak solution of the same problem. Indeed, let (un
0;fn)2D(A)W1;1(0;T;H)approximate
(u0;f)inHL1(0;T;H). Denote by unthe strong solution of (1.2.2) with u0:=un
0and
f:=fn. Then,
un(t) =S(t)un
0+Zt
0S(t s)fn(s)ds;t2[0;T];
which clearly implies our assertion.
The next two results deal with Cauchy problems associated with time-dependent Lipschitz
perturbations of maximal monotone operators.
Theorem 11. Let A :D(A)H!H be a linear maximal monotone operator, u 02H and
let B :[0;T]H!H satisfy the following two conditions:
t!B(t;z)2L1(0;T;H)8z2H; (1.2.9)
and there exists some constant w>0such that
kB(t;x1) B(t;x2)kwkx1 x2k 8t2[0;T];x1;x22H: (1.2.10)
Then there exists a unique function u 2C([0;T];H)which satisfies the integral equation
u(t) =S(t)u0 Zt
0S(t s)B(s;u(s))ds;t2[0;T]:
The solution of this integral equation is called a mild solution of the Cauchy problem
(
u0(t)+Au(t)+B
t;u(t)
=0;0<t<T;
u(0) =u0:(1.2.11)
Remark 2. In the statement of the above theorem assumption (1.2.9) can be replaced by
the following condition:
the function t!B(t;z)is strongly measurable for all z2H;andt!B(t;0)2L1(0;T;H);
which is equivalent with (1.2.9) due to (1.2.10).
The proof of this result is based on Banach’s fixed point principle.
The next theorem is an extension of Theorem 9 to the case of time-dependent Lipschitz
perturbations.
14
Theorem 12. (H. Br ´ezis, [22], pp. 106-107) If A is the subdifferential of a proper convex
lower semicontinuous function j:H!( ¥;+¥];u02D(A) =D(j)and B :[0;T]
D(j)!H satisfies the conditions:
(i)9w0such thatkB(t;x1) B(t;x2)kwkx1 x2k8t2[0;T];8×1;x22D(A);
(ii)8x2D(j);the function t!B(t;x)belongs to L2(0;T;H);
then,8u02D(j);there exists a unique strong solution of the Cauchy problem
(
u0(t)+Au(t)+B(t;u(t)) = 0;0<t<T;
u(0) =u0;
such that t1=2u0(t)2L2(0;T;H):
Now, we are going to recall some existence results concerning fully nonlinear, non-autonomous
(time dependent) evolution equations.
Theorem 13. (T. Kato, [43]) Let A(t):DH!H be a family of single-valued maximal
monotone operators (with D (A(t)) = D independent of t) satisfying the following condition
kA(t)x A(s)xkLjt sj(1+kxk+kA(s)xk)
for all x2D;s;t2[0;T], where L is a positive constant. Then, for every u 02D, there exists
a unique function u 2W1;1(0;T;H)such that u (0) =u0;u(t)2D8t2[0;T];and
u0(t)+A(t)u(t) =0for a. a. t2(0;T):
In fact, Kato’s result is valid in a general Banach space.
Theorem 14. (H. Attouch and A. Damlamian, [4]) Let A (t) =¶f(t;);0tT , where
f(t;):H!( ¥;+¥]are all proper, convex, and lower semicontinuous. Assume further
that there exist some positive constants C 1;C2and a nondecreasing function g:[0;T]!R
such that
f(t;x)f(s;x)+[g(t) g(s)][f(s;x)+C1kxk2+C2]; (1.2.12)
for all x2H;0stT:Then, for every u 02D(f(0;))and f2L2(0;T;H);there
exists a unique function u 2W1;2(0;T;H)such that u (0) =u0and
u0(t)+A(t)u(t) =f(t)for a. a. t2(0;T): (1.2.13)
Moreover, there exists a function h 2L1(0;T)such that
f(t;u(t))f(s;u(s))+Zt
sh(r)dr for all 0stT:
15
Assumption (1.2.12) implies that D(f(t;))increases with t. In all our forthcoming appli-
cations of the above theorem D(f(t;)) = D, a constant set.
The remainder of this chapter is meant for recalling some existence and regularity
results for second order abstract differential equations.
Consider in Hthe following boundary value problem
(
u00(t) =Au(t)+f(t);0<t<T;
u(0) =a;u(T) =b:(1.2.14)
As far as the problem (1.2.14) is concerned, we have the following result of V . Barbu
(see [17], p. 310):
Theorem 15. Let A :D(A)H!H be a maximal monotone operator, a ;b2D(A)and
f2L2(0;T;H):Then problem (1.2.14) has a unique strong solution u 2W2;2(0;T;H):
The above result was extended by R. E. Bruck in [24] to the case a;b2D(A):
Theorem 16. Let A :D(A)H!H be a maximal monotone operator, a ;b2D(A)and f2
L2(0;T;H):Then problem (1.2.14) has a unique strong solution u 2C([0;T];H)TW2;2
loc(0;T;H)
such that
t1=2(T t)1=2u02L2(0;T;H);t3=2(T t)3=2u002L2(0;T;H):
Another extension has been established by A. R. Aftabizadeh and N. H. Pavel [7].
Specifically, let us consider in Hthe problem
(
p(t)u00(t)+r(t)u0(t) =Au(t)+f(t);0<t<T;
u0(0)2b1(u(0) a); u0(T)2b2(u(T) b):(1.2.15)
Theorem 17. If A;b1;b2are maximal monotone in H; 0;a;b2D(A);f2L2(0;T;H);
hAlx Aly;vi08l>08v;x;y2H;x y2D(b1);v2b1(x y);
hAlx Aly;vi08l>08v;x;y2H;x y2D(b2);v2b2(x y);
where A ldenotes the Yosida approximation of A; either D (b1)or D(b2)is bounded; p ;r2
W1;¥(0;T)and p (t)c>08t2[0;T];then problem (1.2.15) has at least one solution
u2W2;2(0;T;H):
If, in addition, at least one of the operators A ;b1;b2is injective, then the solution is
unique.
Note that the last three theorems hold true for multivalued A, and inclusion relation
instead of (1.2.14) 1.
16
Chapter 2
On Some Singularly Perturbed Problems
2.1 Presentation of the Problems
In the first section of this chapter we are interested in some initial-boundary value
problems associated with a partial differential system, known as the telegraph system. This
system describes propagation phenomena in electrical circuits. In an attempt to get as close
as possible to physical reality, we will associate with the telegraph system some linear or
nonlinear boundary conditions which describe specific physical situations. We hope that
the models we are going to present will be of interest to engineers and physicists. Although
our theoretical investigations will be focused on specific models, this will not affect the
generality of our methods. In fact, we are going to address some essential issues which
could easily be adapted to further situations.
Note that the nonlinear character of the problems we are going to discuss creates many
difficulties. Nonlinearities may occur either in the telegraph system or in the boundary
conditions. Our aim here is to develop methods which work for such nonlinear problems.
LetDT:=
(x;t)2R2; 0<x<1;0<t<T
, where T>0 is a given time instant.
Consider the telegraph system in DT:
(
eut+vx+ru=f1(x;t);
vt+ux+gv=f2(x;t);(LS)
with the initial conditions:
u(x;0) =u0(x);v(x;0) =v0(x);0x1; (IC)
and nonlinear algebraic-differential boundary conditions of the form:
(
r0u(0;t)+v(0;t) =0;
u(1;t) cvt(1;t) =f0(v(1;t))+e(t);0tT:(BC:1)
17
Problem (LS);(IC);(BC:1)is a model for transmission (propagation) in electrical circuits
(see [28]).
We will also investigate a more general problem in which the latter equation of the
telegraph system is nonlinear. More precisely, we consider in DTthe following nonlinear
hyperbolic system:(
eut+vx+ru=f1(x;t);
vt+ux+g(x;v) =f2(x;t);(NS)
with initial conditions (IC)and boundary conditions (BC:1). As we will see, the treatment
of this more general problem is more difficult and the results are weaker. The investigation
of this nonlinear problem will show how to deal with the case where the partial differential
system is also nonlinear.
Another problem which will be investigated consists of system (NS), initial conditions
(IC), as well as two integro-differential boundary conditions of the form:
(
u(0;t)+c0vt(0;t) = r0(v(0;t)) l0Rt
0v(0;s)ds+e0(t);
u(1;t) c1vt(1;t) =r1(v(1;t))+l1Rt
0v(1;s)ds e1(t);0tT:(BC:2)
Here, both r0andr1are nonlinear functions. Such boundary (cf. [28]).
Also, we will consider (LS)associated with boundary conditions at the two end points,
which are both algebraic and nonlinear:
(
u(0;t)+r0(v(0;t)) = 0;
u(1;t) =f0(v(1;t));0tT:(BC:3)
The physical relevance of this problem is obvious. In fact, this will be the first model we
are going to address. For each model we need specific methods of investigation.
Of course, there are further boundary conditions which may occur. However, the mod-
els discussed in this section basically cover all possible physical situations and the methods
we use could easily be adapted to further problems, including more complex models. For
example, in the case of integrated circuits, there are ntelegraph systems with 2 nunknowns
which are all connected by means of different boundary conditions. Such models could
be investigated by similar methods and techniques. In some cases, there are negligible pa-
rameters (for instance, inductances) and so we are led to the idea of replacing the original
perturbed models by reduced (unperturbed) models. In order to make sure that the reduced
models still describe well the corresponding phenomena we have to develop an asymptotic
analysis of the singular perturbation problems associated with such transmission processes
in integrated circuits. This analysis can be done by using the same technique as for the case
of a single telegraph system for which we will perform a complete investigation.
Mention should be made of the fact that the models presented above, or similar more
18
complex models, also describe further physical problems, in particular problems which
occur in hydraulics (see V . Barbu [18], V . Hara [38], V . Iftimie [40], I. Straskraba and V .
Lovicar [60], V . L. Streeter and E. B. Wylie [61]).
In the second section of this chapter we are particularly interested in coupled problems
in which a small parameter is present. More precisely, let us consider in the rectangle QT=
(a;c)(0;T); ¥<a<c<¥;0<T<¥, the following system of parabolic equations
(
ut+( eux+a1(x)u)x+b1(x)u=f(x;t)inQ1T;
vt+( m(x)vx+a2(x)v)x+b2(x)v=g(x;t)inQ2T;(S)
with which we associate initial conditions
u(x;0) =u0(x);axb;v(x;0) =v0(x);bxc; (IC)
transmissions conditions at x=b:
(
u(b;t) =v(b;t);
eux(b;t)+a1(b)u(b;t) = m(b)vx(b;t)+a2(b)v(b;t);0tT;(TC)
as well as one of the following types of boundary conditions:
u(a;t) =v(c;t) =0;0tT; (BC:1)
ux(a;t) =v(c;t) =0;0tT; (BC:2)
u(a;t) =0; vx(c;t) =g(v(c;t));0tT; (BC:3)
where Q1T= (a;b)(0;T);Q2T= (b;c)(0;T);b2R;a<b<c,gis a given nonlinear
function and eis a small parameter, 0 <e 1:
Denote by (P:k)ethe problem which consists of (S),(IC),(TC),(BC:k), fork=1;2;3:
It is well-known that transport (transfer) of energy or mass is fundamental to many bio-
logical, chemical, environmental and industrial processes. The basic transport mechanisms
of such processes are diffusion (or dispersion) and bulk flow. Therefore, the corresponding
flux has two components: a diffusive one and a convective one. Here we pay attention to the
case in which the spatial domain is one-dimensional. It is represented by the interval [a;c].
For specific problems describing heat or mass transfer, we refer to A. K. Datta. [29]
(see also the references therein).
2.2 Hyperbolic Systems with Algebraic Boundary Condi-
tions
19
Chapter 3
Elliptic-like Regularizations of
Nonlinear Evolution Equations
3.1 Presentation of the Problems
While in Chapter 2 of the thesis we discussed the possibility to replace singular per-
turbation problems with the corresponding reduced models, in what follows we aim at re-
versing the process in the sense that we replace given problems with singularly perturbed,
higher order (with respect to t) problems, admitting solutions which are more regular and
approximate the solutions of the original problems. More precisely, let us consider the clas-
sical heat equation
ut Du=f(x;t);x2W;0<t<T; (E)
together with the Dirichlet boundary condition
u(x;t) =0 for x2¶W;0<t<T; (BC)
and the initial condition
u(x;0) =u0(x);x2W; (IC)
where Wis a nonempty open bounded subset of Rnwith smooth boundary ¶W, andDis
the Laplace operator with respect to x= (x1;;xn), i.e.,Du=ån
i=1uxixi. Denote by P0this
initial-boundary value problem. If we add eutt, 0<e 1, to equation (E), we obtain an
elliptic equation
eutt+Du=ut f(x;t);x2W;0<t<T: (Ee)
Now, if we associate with the resulting equation the original conditions (BC)and(IC),
then we obtain a new problem. Obviously, this problem is incomplete, since (EE)is of a
higher order with respect to tthan the original heat equation. Therefore, we need to add one
20
additional condition to get a complete problem. We prefer to add a condition at t=Tfor
equation (Ee), either for uor for ut. So, we obtain an elliptic regularization of the original
problem, which is expected to have a more regular solution that approximates in some sense
the solution of problem P0.
In fact, what we have said so far can be expressed in an abstract form. Let Hbe a
Hilbert space with scalar product denoted by h;iand the induced norm kk. Consider the
following evolution problem in H;denoted again P0:
(
u0(t)+Au(t)+Bu(t) =f(t);t2[0;T];(E)
u(0) =u0; (IC)(P0)
where T>0 is a given time instant, u0is a given initial state, f:[0;T]!H;andA:D(A)
H!His a maximal monotone operator, B:H!His a nonlinear operator (see the next
sections for more precise assumptions).
Now, let us consider the following second order equation, again denoted (Ee), Consider
the second order equation
eu00(t)+u0(t)+Au(t)+Bu(t) =f(t);t2[0;T]; (Ee)
with which we associate either
u(0) =u0;u(T) =uT; (BC:1)
or
u(0) =u0;u0(T) =uT: (BC:2)
We denote by (P:k)ethe problem (Ee),(BC:k),k=1;2. Both these problems are ”elliptic
regularizations” of the original problem P0.
Problems (P:k)e,k=1;2, will be investigated in the next sections. Note that we will
require different assumptions for each of the two cases that we will study.
Such kind of approximate problems (regularizations) have been suggested by J. L.
Lions in his monograph on singular perturbations [48] (see p. 407 where the case B=0 is
examined). Lions’ motivation was that such regularizations provide more regular solutions
intapproximating the solution of the original problem (P0)ase!0. Due to the additional
term eu00this method introduced by J. L. Lions is usually called the method of artificial
viscosity. Obviously, from our abstract setting we can derive results for the heat equation
we started with.
21
3.2 Asymptotic analysis of problem (P:1)e
As promised in Section 3.1, we are going to study the first elliptic regularizations of the
following evolution problem in H, denoted (P0):
(
u0(t)+Au(t)+Bu(t) =f(t);t2[0;T];(E)
u(0) =u0; (IC)(P0)
According to our discussion in the previous Section, we examine in what follows the elliptic
regularization of problem (P0), denoted (P:1)e:
(
eu00(t)+u0(t)+Au(t)+Bu(t) =f(t);t2[0;T];
u(0) =u0;u(T) =uT;(P:1)e
where T>0 is a given time instant, u0is a given initial state, uTis a given convenient
vector, f:[0;T]!H;and
(h1)A:D(A)H!His a linear, maximal monotone operator;
B:H!His a nonlinear operator satisfying one of the following two conditions:
(h2)Bis Lipschitzian on H;i. e., there exists L>0 such that
8x;y2H;kBx BykLkx yk;
(h3)Bis monotone and Lipschitzian on every bounded subset of H:
In the first part of this section we present some results on the existence, uniqueness and
regularity of the solutions of the two problems introduced above.
If operator Asatisfies (h1)andBsatisfies either (h2)or(h3);we can derive from
Theorem 7 and Remark 1, Chapter 1 the following result regarding problem (P0):
Lemma 1. If(h1)holds and B satisfies either (h2)or(h3), then, for every f2W1;1(0;T;H);
and u 02D(A), there exists a unique strong solution u 2W1;¥(0;T;H)of problem (P0),
satisfying u (t)2D(A)for all t2[0;T]:
As far as the solution ueof problem (P:1)eis concerned, the next result is also known
(see [7] and [5], Theorem 3.1, including its proof).
Lemma 2. Assume that (h1)and(h3)hold. If f2L2(0;T;H)and u 0;uT2D(A), then for
every e>0problem (P:1)ehas a unique solution u e2W2;2(0;T;H):
Remark 3. Obviously, under the assumptions of Lemma 2 above, operator A +B is maximal
monotone (since B is maximal monotone and D (B) =H). In fact Lemma 2 holds true if (h3)
is replaced by the condition that A +B:D(A)H!H is maximal monotone.
Remark 4. If A is selfadjoint then better existence and regularity results for (P0)and(P:1)e
can be obtained (see [5]). We don’t need such results here since the applications discussed
later involve non-selfadjoint A’s.
22
Remark 5. Under additional conditions on the initial data, higher regularity of u and u e
can be obtained (see [5]). This information might be useful if one looks for asymptotic
expansions of higher orders for u e.
In [5] an asymptotic expansion of order zero for the solution ueof problem (Pe)was
derived. It was done heuristically in a first stage by using classic arguments in perturbation
theory (see [67], [11]). Let us recall this expansion
ue(t) =u(t)+c0(t)+re(t);t2[0;T]; (3.2.1)
where t= (T t)=eis the stretched (fast) variable, uis the regular term (the solution of
problem (P0)),c0is a correction function (or boundary layer function) of order zero, while
redenotes the remainder of order zero. In fact c0has an explicit expression
c0(t) = ( uT u(T))e t8t0; (3.2.2)
while reshould satisfy the problem
(
e(re+u)00+r0
e+Are+Bue=Bu Ac0in[0;T];
re(0) = c0(T=e);re(T) =0:(3.2.3)
Now let us state the main result of this section which is a significant improvement of Theo-
rem 4.1 in [5].
Theorem 18. Assume that (h1)and either (h2)or(h3)are satisfied and, in addition,
f2W1;1(0;T;H);u0;uT2D(A): (3.2.4)
Then, for every e>0(small enough if (h2)is assumed to be satisfied), problem (P:1)e
has a unique solution u e2W2;2(0;T;H)that admits an asymptotic expansion of the form
(3.2.1) , where u is the solution of problem (P0);c0is defined by (3.2.2) , resatisfies problem
(3.2.3) , and the following estimates hold
krekC([0;T];H)=O(e1=4);kue ukL2(0;T;H)=O(e1=2): (3.2.5)
Proof. Throughout this proof we will denote by C1;C2;… some positive constants which
depend on the data, but are independent of e:
The following change of function
bre(t) =re(t)+re(t);re(t) =c0(T=e)(T t)=T;t2[0;T] (3.2.6)
homogenizes the boundary condition (3.2.3) 2:
The proof will be devided into two cases corresponding to (h2)and(h3)as follows
23
Case 1: (h1)and(h2)hold. We cannot use Lemma 2 in this case, so we need the following
result
Lemma 3. Assume that (h1)and(h2)are satisfied. If f2L2(0;T;H);and u 0;uT2D(A),
then there exits an e0>0(which depends on the Lipschits constant L of B) such that for all
e2(0;e0)problem (P:1)ehas a unique solution u e2W2;2(0;T;H).
Proof. Multiply (Ee)bye Mt, with M>0 a constant which will be chosen later. Denoting
v(t) =e Mtu(t), we obtain
(
ee Mt(eMtv)00+v0+Mv+Av+e MtB(eMtv) e Mtf(t) =0;t2[0;T];
v(0) =u0;v(T) =e MTuT=:vT:(3.2.7)
Obviously, vT2D(A). Now, define Qe:D(Qe)L2(0;T;H)!L2(0;T;H)by
D(Qe):=fv2W2;2(0;T;H);v(0) =u0;v(T) =vT;Av2L2(0;T;H)g;
Qe(v):= ee Mt(eMtv)00+v0+M
2v+Av:
Note first that D(Qe)is a nonempty set: for example, the function v(t) =T t
Tu0+t
TvT
belongs to D(Qe).
We also have for all v1;v22D(Qe)
hQev1 Qev2;v1 v2iL2 eZT
0h(eMtv1)00 (eMtv2)00;e Mt(v1 v2)idt+M
2kv1 v2k2
L2
M
2 eM2
kv1 v2k2
L2+ekv0
1 v0
2k2
L2:
We have denoted by h;iL2;kkL2the usual inner product of L2(0;T;H)and the corre-
sponding norm. Thus, if 0 <e<1=(2M);operator Qeis strongly monotone in L2(0;T;H)
with constant M
1
2 eM
.
Moreover, Qeis maximal monotone, i.e., for all g2L2(0;T;H)there exists v2D(Qe)
such thatM
2v+Qev=g(cf. Theorem 1). Indeed, using the substitution u=eMtvwe are led
to the original problem in u,(P:1)e, where B0 and f(t) =eMtg(t), and we know that this
problem has a (unique) solution in W2;2(0;T;H)(cf. Lemma 2).
Next, we define the operator ˜B:L2(0;T;H)!L2(0;T;H);˜Bv:=e MtB(eMtv) +M
2v
e Mtf(t):Obviously, ˜Bis Lipschitzian on L2(0;T;H). Since for all v1;v22H
M
2 L
kv1 v2k2
L2h˜Bv1 ˜Bv2;v1 v2iL2;
it follows that for M:=2LoperatoreBis monotone, hence maximal monotone (since it is
also Lipschitzian).
Note that (P:1)ecan be expressed as (˜B+Qe)(v) =0, where ˜B+Qeis maximal
24
monotone and strongly monotone on D(Qe)L2(0;T;H)with constant L(1 4eL)for
all 0 <e<e0:=1=(4L). Obviously, the equation (˜B+Qe)v=0 has a unique solution
v=ve2D(Qe), i.e., for all e2(0;e0), problem (Pe)has a unique solution u=eMtve2
W2;2(0;T;H). The proof of Lemma 9 is complete.
From Lemmas 1, 2, and 9, (3.2.1), (3.2.2), and (3.2.6) we infer that, for all e>0
(sufficiently small in Case 1),
bre+u
=ue c0+re2W2;2(0;T;H);bre2W1;¥(0;T;H):
An easy computation shows that bresatisfies the problem
(
e(bre+u)00+br0
e+Abre+B(ue) B(u) =he;t2[0;T];
bre(0) =bre(T) =0;(3.2.8)
where
he:= c0(T=e)T 1+Are(t) Ac0(t): (3.2.9)
In order to obtain the estimates (3.2.5) we are going to prove
Lemma 4. Assume that the assumptions in Lemma 9 are fulfilled. If f 2W1;1(0;T;H);
then, for e>0small enough, the following estimates hold
krekL2(0;T;H)=O(e1=2);kr0
ekL2(0;T;H)=O(1): (3.2.10)
Proof. Let us start with (3.2.8). We multiply this equation by e Mt;where the constant
M>0 will be chosen later. Denote ere:=e Mtbre:Taking into account the obvious equation
e Mtbr0
e=Mere+er0
e, we infer
(
ee Mt(eMtere+u)00+er0
e+Mere+Aere+(B(ue) B(u))e Mt=ehe;t2[0;T];
ere(0) =ere(T) =0;
(3.2.11)
whereehe:=e Mthe;t2[0;T]:
Taking the scalar product in Hof equation ( ??)1andereand then integrating over [0;T], we
can derive by using the monotonicity of A
eZT
0h(eMtere+u)00;e Mtereidt+Mkerek2
L2+ZT
0e MthB(ue) B(u);ereidtZT
0hehe;erLeidt:
(3.2.12)
On the other hand, it is easily seen that
krekL2=O(ej);kA(re)kL2=O(ej);8j1;
kc0kL2=O(e1=2);kA(c0)kL2=O(e1=2):(3.2.13)
25
Integrating by parts (3.2.12) yields
eZT
0h(eMtere+u)0;e Mter0
e Me Mtereidt+Mkerek2
L2+ZT
0e MthB(ue) B(u);ereidt
kehekL2kerekL2;
which in view of estimates (3.2.13) implies
M(1 eM)kerek2
L2+eker0
ek2
L2+ZT
0e MthB(ue) B(u);ereidt
C1ekerekL2+C2eker0
ekL2+C3e1=2kerekL2:(3.2.14)
Further, using (h2)and estimates (3.2.13), we infer that
ZT
0e MthB(u) B(ue);ereidtLZT
0e Mtkc0 re+brekkerekdt
LZT
0kc0 rekkerekdt+Lkerek2
L2C4e1=2kerekL2+Lkerek2
L2:
(3.2.15)
Now, choose M=2L:Thus, (3.2.14) and (3.2.15) lead us to
L(1 4eL)kerek2
L2+eker0
ek2
L2C1ekerekL2+C2eker0
ekL2+C5e1=2kerekL2:(3.2.16)
If, for example, 0 <e<minf1;1=(8L)g;then we can deduce from (3.2.16)
L
2kerek2
L2+eker0
ek2
L2C2eker0
ekL2+C6e1=2kerekL2
C7e+L
4kerek2
L2+e
2ker0
ek2
L2;
where we have used the inequality ab(a2+b2)=2. This shows that for all 0 <e 1 we
havekerekL2(0;T;H)=O(e1=2);ker0
ekL2(0;T;H)=O(1):Since re=e2Ltere re;using (3.2.13),
we obtain (3.2.10).
Case 2: (h1)and(h3)hold. We have
Lemma 5. The conclusions of Lemma 4 remain valid if (h2)is replaced by (h3).
Proof. Similar to the proof of Lemma 4. The only difference concerns the evaluation of
26
RT
0e MthB(u) B(ue);ereidt:Using the monotonicity of Bwe can infer
ZT
0e MthB(u) B(ue);ereidt=
ZT
0e 2MthB(u+c0+eMtere re) B(u+c0 re);eMtereidt
+ZT
0e 2Mth B(u+c0 re)+B(u);eMtereidt
ZT
0e 2Mth B(u+c0 re)+B(u);eMtereidt:(3.2.17)
Now, using (3.2.13) and the Lipschitz condition for Bon every bounded subset of H, we
derive
ZT
0e 2Mth B(u+c0 re)+B(u);eMtereidtC8kc0 rekL2kerekL2=O(e1=2)kerekL2
(3.2.18)
(in (3.2.18) the constant C8does not depend on e;since the arguments of Bbelong to a ball
inHcentered at the origin whose radius depends only on kukC([0;T];H);kuTk).
Thus, from (3.2.17) and (3.2.14) we obtain
M(1 eM)kerek2
L2+eker0
ek2
L2+ZT
0e MthB(u+c0 re) B(u);ereidt
C1ekerekL2+C2eker0
ekL2+C3e1=2kerekL2:(3.2.19)
Taking for example M=1 in (3.2.18) and (3.2.19), we derive an estimate similar to (3.2.16).
The conclusion of the lemma follows by arguments similar to those used in the proof
of Lemma 4.
Summarizing all the information above, we see that in both our cases all the terms
of expansion (3.2.1) are well defined and satisfy the stated regularity conditions, while re
satisfies (3.2.10). Consequently, from the obious equality
kre(t)k2= 2ZT
thre(s);r0
e(s)ids8t2[0;T];
we infer thatkrekC([0;T];H)=O(e1=4):
Finally, (3.2.5) 2is an obvious consequence of estimations (3.2.13) and (3.2.10) 1. The
proof of the theorem is complete.
Remark 6. From (3.2.10) we see that r0
e!0ase!0, weakly in L2(0;T;H), i.e.,d
dt
ue
c0
!u0, weakly in L2(0;T;H):
Remark 7. The estimate (3.2.5) 2shows that problem (P:1)eis regularly perturbed of the
zero-th order with respect to the norm of L2(0;T;H), meaning that the boundary layer is
not visible through this weaker norm.
27
3.3 Applications
In this section we illustrate our theoretical results with some significant examples of hyper-
bolic problems.
3.3.1 Regularization of the semilinear telegraph system
Consider in DT:=f(x;t):x2[0;1]andt2[0;T]gthe telegraph system
(
ut+vx+r(x;u) =f1(x;t);
vt+ux+g(x;v) =f2(x;t);(3.3.20)
with initial conditions
u(x;0) =u0(x);v(x;0) =v0(x);0x1; (3.3.21)
and the following type of boundary conditions (Ohm’s law at both the ends of the line)
u(0;t) =r0v(0;t);u(1;t) =f0v(1;t);0tT; (3.3.22)
where r0andf0are given non-negative constants, and r;g:[0;1]R!Rare given func-
tions such that r(;x)andg(;x)belong to L2(0;1)for all x2Rand one of the conditions
(h4),(h5)below holds.
(h4)there exists a positive constant L>0 such that:
jr(x;x) r(x;h)jLjx hj;jg(x;x) g(x;h)jLjx hj;8x;h2R;a. a.x2(0;1);
(h5)x!r(x;x);x!g(x;x)are nondecreasing for a. a. x2(0;1);and for all K>0 there
exists a constant LK>0 such thatjr(x;x) r(x;h)jLKjx hj;jg(x;x) g(x;h)j
LKjx hj;a. a. x2(0;1);8jxj;jhjK.
We consider the Hilbert space H1:=L2(0;1)2;with the usual scalar product defined
by
h(p1;q1);(p2;q2)i=Z1
0(p1p2+q1q2)dx:
Define the operators A1:D(A1)H1!H1by
D(A1) =n
(p;q)2H1(0;1)2; p(0) =r0q(0);p(1) =f0q(1)o
;
A1(p;q) =
q0;p0
;
andB1:H1!H1;B1(p;q) = ( r(;p);g(;q)):
So, in terms of the notations used in the previous sections, problem (3.3.20) (3.3.22)
28
can be expressed as a Cauchy problem in H1:
(
(u0;v0)+A1(u;v)+B1(u;v) = ( f1(;t);f2(;t));in[0;T];
u(0) =u0;v(0) =v0:(P1
0)
If(h4)is satisfied then B1is Lipschitz on H1, soB1+wIis maximal monotone in H1
forw>0 sufficiently large, where Istands for the identity operator on H1. It follows that
A1+B1+wIis maximal monotone (see [39], Proposition 5.1.1, p. 111). Consequently, we
have the following result (cf. [39], Theorem 5.1.1, p. 115) which also works in the case in
which r,gsatisfy assumption (h5)instead of (h4).
Lemma 6. Assume that either (h4)or(h5)is satisfied. Let T >0be fixed. Then, for every
f1;f22W1;1(0;T;L2(0;1));u0;v02H1(0;1);such that
u0(0) =r0v0(0);u0(1) =f0v0(1);
there exists a unique strong solution (u;v)of problem (P1
0)such that
(u;v)2W1;¥(0;T;L2(0;1)2);ux;vx2L¥(0;T;L2(0;1));u;v2L¥(DT);
and u ;v satisfy (3.3.22) for all t2[0;T]:
The elliptic-like regularization of problem (P1
0)is given by the system
(
eutt+ut+vx+r(x;u) =f1(x;t);
evtt+vt+ux+g(x;v) =f2(x;t);inDT;(3.3.23)
subject to
u(x;0) =u0(x);v(x;0) =v0(x);
u(x;T) =uT(x);v(x;T) =vT(x);0x1;(3.3.24)
and the boundary conditions
u(0;t) =r0v(0;t);u(1;t) =f0v(1;t);0tT: (3.3.25)
An abstract form of this problem is the following
(
e(u00;v00)+(u0;v0)+A1(u;v)+B1(u;v) = ( f1;f2);t2[0;T];
u(0) =u0;v(0) =u0;u(T) =uT;v(T) =vT:(P1
e)
Under the assumptions of Lemma 6, if in addition (uT;vT)2D(A1), then for every
e>0 (small enough if (h4)is assumed to be satisfied) problem (P1
e)has a unique solution
(ue;ve)2W2;2(0;T;L2(0;1)2)(see Lemmas 2 and 9).
29
The solution of problem (P1
e)can be written as follows (see (3.2.1)):
ue(x;t) =u(x;t)+c0(x;t)+re(x;t);
ve(x;t) =v(x;t)+d0(x;t)+se(x;t);(3.3.26)
for all (x;t)2DT, where t= (T t)=e,(u;v)is the solution of the problem (3.3.20)
(3.3.22) ;and (cf. (3.2.2))
c0(x;t) = ( uT(x) u(x;T))e t;
d0(x;t) = ( vT(x) v(x;T))e t;8×2[0;1];t0;(3.3.27)
while (cf. (3.2.3)) the remainder terms satisfy the problem
(
e(Re+(u;v))00+R0
e+A1Re+B1(ue;ve) =B(u;v) A1(c0;d0);
Re(0) = (c0(;T=e);d0(;T=e));Re(T) = ( 0;0);(3.3.28)
where Re= (re;se):
Summarizing, Theorem 18 can be formulated in this specific case as follows
Theorem 19. Assume that either (h4)or(h5)is satisfied and, in addition,
(f1;f2)2W1;1(0;T;L2(0;1)2);u0;v0;uT;vT2H1(0;1);
u0(0) =r0v0(0);u0(1) =f0v0(1);
uT(0) =r0vT(0);uT(1) =f0vT(1):(3.3.29)
Then, for every e>0(small enough if (h4)is assumed to be satisfied), problem (P1
e)has a
unique solution (ue;ve)2W2;2(0;T;L2(0;1)2)that admits an asymptotic expansion of the
form (3.3.26) , where (u;v)is the solution of problem (P1
0);c0;d0are defined by (3.3.27) ,
(re;se)satisfies problem (3.3.28) , and the following estimates hold true:
kue u c0kC([0;T];L2(0;1))=O(e1=4);kue ukL2(DT)=O(e1=2);
kve v d0kC([0;T];L2(0;1))=O(e1=4);kve vkL2(DT)=O(e1=2):(3.3.30)
Theorem 19 is a direct consequence of Theorem 18 if (h4)holds, since in this case
operator B1is Lipschitzian on H1. If(h5)holds, then B1is monotone on H1, but it is not
necessarily Lipschitz on bounded subsets of H1:This property has been used to obtain
the estimate (3.2.18). However (3.2.18) can be derived here by using the local Lipschitz
30
condition on r,g. Indeed we have
kB1((u;v)+(c0;d0) (r1;r2)) B1(u;v)k2
H1=
k
r(;u+c0 r1) r(;u)
;
g(;v+d0 r2) g(;v)
k2
H1
=Z1
0
r(;u+c0 r1) r(;u)2
dx+Z1
0
g(;v+d0 r2) g(;v)2
dx;
where r1:=c0(T=e)T t
T;r2:=d0(T=e)T t
T:From Lemma 6 we see that the solution
(u;v)2L¥(DT)2;and since by assumption u0;v0;uT;vT2H1(0;1);we obtain (by using the
local Lipschitz condition on r,g) that there exists an L1>0 depending on f1;f2;u0;v0;uT;vT,
but independent of e, such that
kB1((u;v)+(c0;d0) (r1;r2)) B1((u;v))k2
H1
L1Z1
0(c0 r1)2dx+Z1
0(d0 r2)2dx
=O(e):
This implies estimate (3.2.18), as claimed.
Remark 8. Similar results can be established in the case of linear algebraic-differential
boundary conditions of the form
u(0;t) =r0v(0;t);vt(1;t) u(1;t)+f0v(1;t) =e(t);0tT;
where r 0;f0are given nonnegative constants, and e 2L2(0;T):Such conditions also occur
in specific practical problems.
It is also worth pointing out that in a similar manner further types of linear boundary
conditions can be investigated, such as differential-differential, bilocal or periodic condi-
tions (for more information, see, e.g., [39], Chapter 5).
Even more, our technique is applicable to systems consisting of n couples of telegraph
systems with unknowns u i;vi;i=1;2;:::;n, connected by various linear conditions at x =0
and x =1, as in the case of integrated circuits.
3.3.2 Regularization of the semilinear wave equation
In what follows we are concerned with the wave equation
utt Du+b(ut) =f(x;t);(x;t)2W(0;¥); (3.3.31)
with initial conditions
u(x;0) =u0(x);ut(x;0) =v0(x);x2W; (3.3.32)
31
and the homogeneous Dirichlet boundary condition
u(x;t) =0 on G[0;¥); (3.3.33)
where WRnis a bounded domain with sufficiently smooth boundary G;
(h6)b:R!Ris Lipschitzian on R.
In this case, we consider as usual the Hilbert space (phase space) H2=H1
0(W)L2(W)
with the scalar product defined by
h(p1;q1);(p2;q2)i=Z
WhÑp1;Ñp2idx+Z
Wq1q2dx;
and define the operators A2:D(A2)H2!H2by
D(A2):=f(p;q)2H1
0(W)2;4p2L2(W)g;A2(p;q) = ( q; 4p);
andB2:H2!H2;B2(p;q) = ( 0;b(q)):In fact, D(A2) =
H1
0(W)\H2(W)
H1
0(W).
Obviously, problem (3.3.31) (3.3.33) can be expressed as a Cauchy problem in H2as
follows (
U0(t)+A2U(t)+B2(U(t)) = F(t);t2[0;T];
U(0) =U0;(P2
0)
where U(t):= (u(;t);ut(;t));F(t):= (0;f(;t));U0:= (u0;v0):
According to [51] (Theorem 2.1, p. 209; see also Theorem 2.1, p. 48 and Remark 2.1,
p. 53), we have
Lemma 7. Assume (h6). Let T >0be fixed. Then, for every
f2W1;1(0;T;L2(W));u02H1
0(W)\H2(W);v02H1
0(W);
there exists a unique strong solution u of problem (P2
0)such that
u2W1;¥(0;T;H1
0(W))\W2;¥(0;T;L2(W));
u(;t)2H1
0(W)\H2(W);for all t2[0;T]:
The elliptic-like regularization of problem (P2
0)is
(
eU00+U0+A2U+B2(U) =F(t);t2[0;T];
U(0) =U0;U(T) =UT;(P2
e)
32
where UT:= (uT;vT);or, equivalently,
8
>>>><
>>>>: eutt+ut v=0;
evtt+vt 4u+b(v) =finW[0;T];
u=v=0 onG[0;¥);
u(;0) =u0;v(;0) =v0;u(;T) =uT;v(;T) =vTinW:
According to our results in Section …, if the assumptions of Lemma 7 are fulfilled and in
addition (uT;vT)2D(A2)then, for each e>0 sufficiently small, problem (P2
e)has a unique
solution
Ue:= (ue;ve)2W2;2(0;T;H1
0(W)L2(W)):
In fact, ueis more regular. Indeed, from the equation
e(ue)tt+(ue)t ve=0;
it follows that ue2W4;2(0;T;L2(W)). Even more, from the equation
e(ve)tt+(ve)t 4ue+b(ve) =f;
we derive4ue2L2(0;T;L2(W))and hence
ue2W4;2(0;T;L2(W))\W2;2(0;T;H1
0(W))\L2(0;T;H2(W));
showing that the effect of the regularization is significant.
We also have the asymptotic expansion
ue(x;t) =u(x;t)+l0(x;t)+re(x;t);
ve(x;t) =ut(x;t)+j0(x;t)+se(x;t);x2W;t2[0;T];(3.3.34)
where uis the solution of problem (3.3.31) (3.3.33) ;
l0(x;t) = ( uT(x) u(x;T))e t;
j0(x;t) = ( vT(x) v(x;T))e t;8x2W;t0;(3.3.35)
t= (T t)=e;and the remainder (re;se)satisfies the problem derived from (3.2.3), where
AandBare replaced by A2andB2, respectively.
In view of (h6),B2is a Lipschitz operator on H2. Indeed, for all (u;v);(u1;v1)2H2;
we have
kB2(u;v) B2(u1;v1)kH2=kb(v) b(v1)kL2(W)Lbkv v1kL2(W)Lbk(u;v) (u1;v1)kH2;
33
where Lbis the Lipschitz constant of b.
Taking into account Theorem 18 and the above comments, we can state
Theorem 20. Assume that (h6)is satisfied and, in addition,
f2W1;1(0;T;L2(W));u0;uT2H1
0(W)\H2(W);v0;vT2H1
0(W): (3.3.36)
Then, for every e>0small enough, problem (P2
e)has a unique solution
(ue;ve)2W4;2(0;T;L2(W))\W2;2(0;T;H1
0(W))\L2(0;T;H2(W))W2;2(0;T;L2(W));
which admits an asymptotic expansion of the form (3.3.34) , where (u;v)is the solution of
problem (P2
0);l0;j0are defined by (3.3.35) , and the following estimates hold
kue u l0kC([0;T];L2(W))=O(e1=4);kue ukL2(0;T;H1
0(W))=O(e1=2);
kve ut j0kC([0;T];L2(W))=O(e1=4);kve utkL2(0;T;L2(W))=O(e1=2):(3.3.37)
Remark 9. It is easily seen that a result similar to Theorem 20 above is valid if bis a Lip-
schitz continuous function depending on u. Also, other types of linear boundary conditions
can be analyzed in a similar manner. The case of nonlinear boundary conditions is more
delicate since operator A is non longer linear. We plan to tackle such nonlinear problems
in a forthcoming paper.
3.4 Asymptotic analysis of problem (P:2)e
In what follows we will study the second elliptic regularizations of the same evolution
problem in H, (P0), denoted by (P:2)e, more exactly:
(
eu00(t)+u0(t)+Au(t)+Bu(t)3f(t);t2[0;T]; (Ee)
u(0) =u0;u0(T) =0; (BC)0(P:2)e
where T>0 is a given time instant, u0is a given initial state, f:[0;T]!H;and
(h1)0A:D(A)H!His a (possibly set-valued) maximal monotone operator;
(h2)0B:H!His a Lipschitz continuous operator, i. e., there exists L>0 such that
8x;y2H;kBx BykLkx yk:
Obviously, if (h1)0and(h2)0are satisfied we derive from Lemma 1, that, for every
f2W1;1(0;T;H)andu02D(A), there exists a unique strong solution u2W1;¥(0;T;H)
of problem (P0), satisfying u(t)2D(A)for all t2[0;T].
We also have
Lemma 8. Assume that A is the subdifferential of a proper, convex, lower semicontinuous
34
function j:H!( ¥;+¥]and(h2)0holds. Then, for every f 2L2(0;T;H)and u 02H
withj(u0)<+¥, there exists a unique strong solution u 2W1;2(0;T;H)of problem (P0).
Proof. IfB=0 then the conclusion follows, e.g., from Theorem 9. So we only need to
derive the conclusion when the perturbation Bis a Lipschitz continuous operator, B6=0 .
In this case, if f2L2(0;T;H), one can derive the existence and uniqueness of a solution
u2W1;2(0;T;H)by using the Banach Fixed Point Theorem in the space X=C([0;T];H)
equipped with a convenient Bielecki norm. Indeed, the operator P:X!Xthat associates
with each v2Xthe solution uof the Cauchy problem u0+Au3f Bv;u(0) =u0, has a
fixed point (denoted again u) inXwhich is a strong solution of problem (P0). Its uniqueness
follows by a classic argument (based on the monotonicity of Aand the Lipschitz continuity
ofB).
Next, we are concerned with problem (P:2)e. We start with the following
Proposition 1. Assume that (h1)0holds. If f2L2(0;T;H)and u 02D(A), then for every
e>0problem (P:2)ewith B0has a unique solution u =ue2W2;2(0;T;H):
Proof. Uniqueness can be derived by a classic argument as follows. Let u1;u22W2;2(0;T;H)
be two solutions of problem (P:2)ewith B0. We multiply the obvious equation (inclu-
sion)
e(u1 u2)00 (u1 u2)02Au1 Au2for a. a. t2(0;T)
byu1 u2to obtain (using the monotonicity of A)
eh(u1 u2)00;u1 u2i 1
2d
dtku1 u2k20 for a. a. t2(0;T):
Integrating this inequality over [0;T]yieldsRT
0k(u1 u2)0k2dt0, so u1 u2is a constant
function. In fact u1u2since u1(0) =u2(0) =u0.
For the existence part, we can assume without any loss of generality that 0 2D(A),
02A0, and u0=0. Indeed, if this is not the case, then for some y02Au0, we can replace A
by˜Adefined by
˜A:D(˜A)H!H;D(˜A):=fx;x+u02D(A)g;˜Ax=A(x+u0) fy0g 8x2D(˜A):
Now, using the change w(t):=u(t) u0;problem (Pe)with B0 becomes
(
ew00+w0+˜Aw3f(t) y0;t2[0;T];
w(0) =0;w0(T) =0:(3.4.38)
So in what follows we assume that 0 2D(A), 02A(0), and u0=0. For l>0 consider the
problem (P:2)ewith u0=0,B0, and A:=Al, where Aldenotes the Yosida approxima-
tion of A. Recall that for all l>0Alis monotone and Lipschitz continuous (with Lipschitz
35
constant 1 =l). In our present case, we also have Al0=0 for all l>0. According to [6,
Theorem 3.1 with A0, p. 2929], for all l>0 there exists a unique ul2W2;2(0;T;H)
satisfying the problem
(
eu00
l+u0
l+Alul=f(t);t2[0;T];
ul(0) =0;u0
l(T) =0;(3.4.39)
Multiplying by u00
l, we obtain
eku00
lk2+1
2d
dtku0
lk2+hAlul;u00
li=hf;u00
li;
which gives by integration over [0;T]
eZT
0ku00
lk2dt+1
2ku0
l(0)k2 ZT
0hf;u00
lidt;
sinceh(Alul)0;u0
li0 for a. a. t2(0;T). Therefore, the set fu00
l;l>0gis bounded in
L2(0;T;H)andfu0
l(0);l>0gis bounded in H. It is then easily seen that the set fu0
l;l>
0gis bounded in C([0;T];H)and sofAlul;l>0gis bounded in L2(0;T;H)(see equation
(2:3)1). Now, multiplying the obvious equation
e(ul un)00+(ul un)0+Alul Anun=0;l;n>0;
byul unand then integrating over [0;T]we obtain
eZT
0ku0
l u0
nk2dt+1
2kul(T) un(T)k2+ZT
0hAlul Anun;Jlul Jnunidt= ZT
0hAlul Anun;lAlul nAnunidt;
where Jl= (I+lA) 1. Since the last term in the left-hand side of the above equation is
nonnegative, we have
eZT
0ku0
l u0
nk2dtConst :(l+n);
which shows that u0
lis convergent in L2(0;T;H)asl!0+. Therefore, since ul(t) =
Rt
0u0
l(s)ds, it follows that ulconverges to some uinC([0;T];H)asl!0+. As u00
lis
bounded in L2(0;T;H), we have u2W2;2(0;T;H)and
u0
l!u0inC([0;T];H);u00
l!u00weakly in L2(0;T;H)asl!0+:
Note also that ul Jlul=lAlulconverges to zero in L2(0;T;H), hence Jlulconverges
touinL2(0;T;H)asl!0+. To conclude we can use the demiclosedness of the canonical
extension of AtoL2(0;T;H), which is a maximal monotone operator. More precisely, from
Jlul!ustrongly in L2(0;T;H)andAJlul3Alul!f+eu00 u0weakly in L2(0;T;H)
36
it follows that
eu00+u0+Au3fa. e. in (0;T):
We also have u(0) =0 and u0(T) =0.
Lemma 9. Assume that (h1)0and(h2)0are satisfied. Then for all f 2L2(0;T;H)and u 02
D(A)there exits an e0>0(which depends on the Lipschitz constant L of B) such that
problem (P:2)ehas a unique solution u e2W2;2(0;T;H)for all 0<e<e0.
Proof. For the sake of simplicity, we assume that Ais single-valued (the proof in the set-
valued case is similar, it only requires some technicalities that can be easily manipulated).
We multiply equation (Ee)bye Mtwith M>0 a constant that will be chosen later. So,
denoting v(t) =e Mtu(t), we have
(
ee Mt
eMtv00+v0+Mv+e MtA
eMtv
+e MtB
eMtv
=e Mtf(t)in[0;T];
v(0) =u0;v0(T) = Mv(T):
(3.4.40)
Define the operators Qe:D(Qe)L2(0;T;H)!L2(0;T;H),˜B:L2(0;T;H)!L2(0;T;H);
D(Qe):=n
v2W2;2(0;T;H);v(0) =u0;v0(T) = Mv(T);A(eMtv)2L2(0;T;H)o
;
Qe(v):= ee Mt
eMtv00+v0+M
2v+e MtA(eMtv);˜Bv:=e MtB
eMtv
+M
2v:
From Proposition 1 we deduce that D(Qe)6=/ 0. Indeed, v=e Mtu2D(Qe), where u=ue
is the solution of problem (Pe)with B0:Now for all v1;v22D(Qe)we have
hQev1 Qev2;v1 v2iL2 eZT
0h(eMtv1)00 (eMtv1)00;e Mt(v1 v2)idt+1
2k(v1 v2)(T)k2+M
2kv1 v2k2
L2
M
2 eM2
kv1 v2k2
L2+ekv0
1 v0
2k2
L2;
whereh;iL2andkkL2denote the inner product of L2(0;T;H)and the corresponding
induced norm. Thus, if 0 <e<1=(2M);operator Qeis strongly monotone in L2(0;T;H)
with constant M
1
2 eM
.
Moreover, operator Qeis maximal monotone, i.e., for all g2L2(0;T;H)there exists
v2D(Qe)such that (M=2)v+Qev=g. Indeed, using the converse transformation u(t) =
eMtv(t), this equation becomes (P:2)e, with B0 and f(t) =eMtg(t), and we know that this
problem has a (unique) solution u2W2;2(0;T;H)(cf. Proposition 1), so v(t) =e Mtu(t)
belongs to D(Qe)and satisfies the above equation.
On the other hand, it is easily seen that ˜Bis Lipschitzian on L2(0;T;H), and for all
v1;v22L2(0;T;H)we have
M
2 L
kv1 v2k2
L2h˜Bv1 ˜Bv2;v1 v2iL2:
37
Consequently, if we choose M:=2L, then operator eBis monotone, hence maximal mono-
tone (since it is also Lipschitzian). Summarising, we see that ˜B+Qeis maximal monotone
and strongly monotone on D(Qe)L2(0;T;H)with constant L(1 4eL)for all 0 <e<
e0:=1=(4L). Therefore, ˜B+Qeis surjective, so in particular the equation Qev+˜Bv=
e 2Ltfhas a unique solution ve2W2;2(0;T;H)for all e2(0;e0). This means that, for
every e2(0;e0),ue(t) =e2Ltve(t)is the unique solution of problem (Pe). The proof of
Lemma 9 is complete.
By using classical arguments from the singular perturbation theory (see [ ?], [?]) we found
in [6] that problem (P:2)e(with the general condition u0(T) =uT) was regularly perturbed
of order zero with respect to the norm of C([0;T];H). Even if our present assumptions are
much weaker, we expect to have now the same situation, i.e., to have no boundary layer
function in the asymptotic expansion of order zero of ue:
ue(t) =u(t)+r0e(t);t2[0;T]: (3.4.41)
Here uis the solution of problem (P0), and r0edenotes the remainder of order zero. Obvi-
ously, r0eshould be the solution of the problem
(
e(r0e+u)00+r0
0e+Aue Au+Bue Bu30 in [0;T];
re(0) =0;(r0e+u)0(T) =u0
e(T) =0:(3.4.42)
Now let us state the main result of this section which confirms our guess above.
Theorem 21. Assume that (h1)0and(h2)0are satisfied and, in addition,
f2W1;1(0;T;H);u02D(A): (3.4.43)
Then, for every e>0small enough the problems (P:2)eand(P0)have unique solutions,
ue2W2;2(0;T;H)and respectively u2W1;¥(0;T;H), and the following estimates hold
kue ukC([0;T];H)=O(e1=4);kue ukL2(0;T;H)=O(e1=2): (3.4.44)
Proof. In what follows we will assume that e2(0;e0), where e0=1=(4L), as defined in
Lemma 9. In fact this choice does not restrict the generality since we are interested in small
values of e. From Lemmas 1 and 9 it follows the existence and uniqueness of the solutions
uandueto problems (P0)and(Pe), and
r0e+u=ue2W2;2(0;T;H);r0e2W1;¥(0;T;H):
38
In order to prove the estimates (3.4.44) let us show that
kr0ekL2(0;T;H)=O(e1=2);kr0
0ekL2(0;T;H)=O(1): (3.4.45)
To this purpose we denote se(t) =e Mtr0e(t). Taking into account the obvious equality
e Mtr0
0e=Mse+s0
e, we obtain from ( ??)
(
ee Mt(eMtse+u)00+s0
e+Mse+e Mt(Aue Au)+e Mt(Bue Bu)30;t2[0;T];
se(0) =0;(eMtse+u)0(T) =0
(3.4.46)
Taking the scalar product in Hof equation (3.4.46) 1andseand then integrating over [0;T],
we can derive
eZT
0h(eMtse+u)00;e Mtseidt+1
2kse(T)k2+Mksek2
L2
+ZT
0e MthAue Au;seidt+ZT
0e MthBue Bu;seidt=0:
(3.4.47)
Integrating by parts the first term of the left-hand side of (3.4.47) yields
eZT
0h(eMtse+u)0;e Mts0
e Me Mtseidt+Mksek2
L2+1
2kse(T)k2
+ZT
0e MthAue Au;seidt+ZT
0e MthBue Bu;seidt=0;
which implies (by using Cauchy-Schwarz and H ¨older)
eks0
ek2
L2+M(1 eM)ksek2
L2+1
2kse(T)k2+ZT
0e MthAue Au;seidt
ZT
0e MthBue Bu;seidt+e
C1ksekL2+C2ks0
ekL2
:(3.4.48)
Here and in what follows C1;C2;denote positive constants that depend on the given data
but are independent of e. By virtue of (hB)we have
ZT
0e MthBue Bu;seidtLZT
0e Mtkue ukk sekdt=Lksek2
L2:(3.4.49)
Let us choose M=2L. Then (3.4.48) and (3.4.49) lead us to
L(1 4eL)ksek2
L2+eks0
ek2
L2+1
2kse(T)k2+ZT
0e MthAue Au;seidtC1eksekL2+C2eks0
ekL2:
(3.4.50)
39
Choosing e1=minf1;e0;1=(8L)g, we easily deduce from (3.4.50) that for all e2(0;e1)
L
2ksek2
L2+eks0
ek2
L2+1
2kse(T)k2+ZT
0e MthAue Au;seidtC3e+L
4ksek2
L2+e
2ks0
ek2
L2
(3.4.51)
(where we have used the elementary inequality ab(a2+b2)=2). This implies (by using
the monotonicity of A)
ksekL2(0;T;H)=O(e1=2);ks0
ekL2(0;T;H)=O(1);kse(T)k=O(e1=2);
for all e2(0;e1). Since r0e(t) =e2Ltse(t)we obtain (3.4.45). Therefore, we deduce from
the obvious equality
kr0e(t)k2=2Zt
0hr0e(s);r0
0e(s)ids8t2[0;T];
thatkr0ekC([0;T];H)=kue ukC([0;T];H)=O(e1=4):
Remark 10. By(3.4.45) 2we also have u0
e!u0ase!0, weakly in L2(0;T;H):
Remark 11. Note that if A is the subdifferential of a proper, convex, lower semicontinu-
ous function j:H!( ¥;+¥], denoted A =¶j, then the existence of a unique strong
solution for problem (P0)holds under the weaker conditions f 2L2(0;T;H)and u 02H
withj(u0)<+¥(see Lemma 8). It is still not clear to us whether the weaker condition
j(u0)<+¥is enough to establish an existence result for problem (Pe)similar to Lemma
9. Answering this open question would allow us to derive a convergence result similar to
Theorem 21 in the case A =¶j,j(u0)<+¥.
Remark 12. In order to establish higher order asymptotic expansions for u ewe need more
regularity for both u eand u, the solution of problem (P0). Without going into details, we
think that a suitable framework would correspond to the case when A is the generator of a
C0-semigroup of linear operators and B is a smooth nonlinear operator. A boundary layer
of order one is expected to occur in this case due to the condition u0
e(T) =0or the general
one u0
e(T) =uT. A detailed analysis of such things will be the subject of a forthcoming
paper.
3.5 Applications
To illustrate Theorem 21 we are going to discuss some significant examples.
40
3.5.1 The nonlinear heat equation
Consider the heat (diffusion) equation
ut Du+b(u) =f(x;t);(x;t)2W(0;¥); (3.5.52)
with initial condition
u(x;0) =u0(x);x2W; (3.5.53)
and the nonlinear boundary condition
¶u
¶n+a(u)30 a.e. on G(0;¥); (3.5.54)
where WRnis a bounded domain with sufficiently smooth boundary G,nis the outer unit
normal to the boundary G,ais a maximal monotone graph in RR;such that 02D(a),
andb:R!Ris a Lipschitz continuous function.
For this example we consider the Hilbert space H1:=L2(W)equipped with the usual scalar
product and norm, and define the operators
A1:D(A1)H1!H1;D(A1) =fw2H2(W);¶w=¶n+a(w)30 a.e. on Gg;A1w= Dw;
andB1:H1!H1;B1w=bw:
Using these definitions we can express problem (3.5.52) (3.5.54) as the following
Cauchy problem in H1
(
u0(t)+A1u(t)+B1u(t) =f(t);t2[0;T];
u(0) =u0:(P1
0)
It is well-known the fact that operator A1is the subdifferential of the (proper, convex,
lower semicontinuous) function y:H1!( ¥;+¥];defined by
y(u) =8
<
:1
2R
WjÑuj2dx+R
Gh(u)dsifu2H1(W);h(u)2L1(G);
+¥ otherwise,
where h:R!( ¥;+¥]is a convex, lower semicontinuous function such that ¶h=a
(note that such a function hexists; see, e.g., [51, Proposition 1.3, p. 42]). On the other hand,
since bis a Lipschitz continuous function on R, it follows that B1is Lipschitzian on H1.
Therefore, from Lemma 8 we have
Lemma 10. Assume that the above conditions on W,a,bare fulfilled and f2L2(0;T;L2(W));u02
H1(W);h(u0)2L1(G). Then there exists a unique strong solution u 2W1;2(0;T;L2(W))\
L2(0;T;H2(W))to problem (P1
0).
41
Note that Lema 10 includes an additional regularity property, u2L2(0;T;H2(W)), due
to the specific form of problem (P1
0).
The elliptic regularization (in the sense above) of problem (P1
0)is the following
(
eu00+u0+A1u+B1u=f(t);t2[0;T];
u(0) =u0;u0(T) =0:(P1
e)
Taking into account Theorem 21, Lemma 10 and the particular form of this example,
we can state
Theorem 22. Assume that the above conditions on W,aandbare fulfilled and f 2
L2(0;T;L2(W));u02H2(W);¶u0=¶n+a(u0)30a.e. on G. Then, for every e>0small
enough, problem (P1
e)has a unique solution u e2W2;2(0;T;L2(W))\L2(0;T;H2(W))and
the following estimates hold
kue ukC([0;T];L2(W))=O(e1=4);kue ukL2(0;T;H1(W))=O(e1=2): (3.5.55)
Proof. From Theorem 21 we have (see (3.4.44))
kue ukC([0;T];L2(W))=O(e1=4);kue ukL2(0;T;L2(W))=O(e1=2):
In this case we can obtain the additional estimate
kÑue ÑukL2(0;T;L2(W))n=O(e1=2)
by using the fact that
hA1v A1w;v wiH1Z
WkÑv Ñwk2
Rndx8v;w2D(A1):
We have just to repeat the proof of Theorem 21 and exploit the above property of A1(see
especially (3.4.51)).
Remark 13. Theorem 22 can be extended to a more general case as follows. Let gbe
a maximal monotone graph in RRsuch that 02D(g). Thus gis the subdifferential
of a convex, lower semicontinuous function g :R!( ¥;+¥],g=¶g. Denote by gthe
cannonical extension of gto H 1and replace A 1byeA1:D(eA1)H1!H1;defined by
D(eA1):=n
w2H2(W);¶u
¶n+a(u)30a.e. on Go
\D(g);eA1(w) = Dw+gw:
It is well known that eA1is a maximal monotone operator, more precisely eA1=¶y1;where
42
y1:H1!( ¥;+¥]is defined by
y1(u) =8
<
:1
2R
WjÑuj2dx+R
Wg(u)dx+R
Gh(u)dsif u2H1(W);h(u)2L1(G);g(u)2L1(W);
+¥ otherwise,
Thus Theorem 22 can be readily extended to this more general case. The precise statement
of this extension is left to the reader.
Remark 14. Various examples of problems of the type (3.5.52) (3.5.54) withb0can be
found in [66]. Specifically, for different a0s problem (3.5.52) (3.5.54) describes different
processes such as: the thermostat control process, the enzyme diffusion process, etc.
3.5.2 The nonlinear telegraph system
Consider in DT:=f(x;t):x2[0;1]andt2[0;T]gthe telegraph system
(
ut+vx+r(x;u) =f1(x;t);
vt+ux+g(x;v) =f2(x;t);(3.5.56)
with initial conditions
u(x;0) =u0(x);v(x;0) =v0(x);0x1; (3.5.57)
and the following boundary conditions
u(0;t)2r0(v(0;t));u(1;t)2r1(v(1;t));0tT; (3.5.58)
where r0andr1are maximal monotone graphs in RR, and r;g:[0;1]R!Rare given
functions such that r(;x),g(;x)belong to L2(0;1)for all x2R, and there exists a positive
constant L>0 such that
jr(x;x) r(x;h)jLjx hj;jg(x;x) g(x;h)jLjx hj;8x;h2Rand a. a. x2(0;1):
Note that the case r0=g0=I(the identity function) corresponds to the usual Ohm’s law at
the ends of the electric line, x=0 and x=1.
For this example the natural framework is the space H2:=L2(0;1)2with the usual
scalar product and norm. Define the operators A2:D(A2)H2!H2by
D(A2) =n
(p;q)2H1(0;1)2; p(0)2r0(q(0));p(1)2r1(q(1))o
;A2(p;q) =
q0;p0
;
andB2:H2!H2;B2(p;q) = ( r(;p);g(;q)):
43
Using these notations, we can express problem (3.5.56) (3.5.58) as the following
Cauchy problem in H2
(
(u0;v0)+A2(u;v)+B2(u;v) = ( f1;f2);in[0;T];
u(0) =u0;v(0) =v0:(P2
0)
Obviously, operator B2is Lipschitz on H2, and operator A2is maximal monotone in H2
(cf. [39, Proposition 5.1.1]). Thus, taking into account Lemma 1 (see also [39, Proposition
5.1.1, p. 111]), we have
Lemma 11. Assume that the above conditions on r, g, r 0, r1are fulfilled. Then, for every
f1;f22W1;1(0;T;L2(0;1));u0;v02H1(0;1); u0(0)2r0(v0(0));u0(1)2r1(v0(1));
there exists a unique strong solution (u;v)of problem (P2
0)such that
u;v2W1;¥(0;T;L2(0;1));ux;vx2L¥(0;T;L2(0;1));
and u ;v satisfy (3.5.58) for all t2[0;T]:
The elliptic-like regularization of problem (P2
0)is given by the system
(
eutt+ut+vx+r(x;u) =f1(x;t);
evtt+vt+ux+g(x;v) =f2(x;t);inDT;(3.5.59)
subject to
u(x;0) =u0(x);v(x;0) =v0(x);
ut(x;T) =0;vt(x;T) =0;0x1;(3.5.60)
and the boundary conditions (3.5.58). An abstract form of this problem is the following
(
e(u00;v00)+(u0;v0)+A2(u;v)+B2(u;v) = ( f1;f2);t2[0;T];
u(0) =u0;v(0) =u0;u0(T) =0;v0(T) =0:(P2
e)
Under the above assumptions, for every e>0 small enough, problem (P2
e)has a unique
solution (ue;ve)2W2;2(0;T;L2(0;1)2)(see Lemma 9).
Summarizing, Theorem 21 can be formulated in this specific case as follows
Theorem 23. Assume that the above conditions on r, g, r 0, r1are fulfilled and, in addition,
f1;f22W1;1(0;T;L2(0;1));u0;v02H1(0;1); u0(0)2r0(v0(0));u0(1)2r1(v0(1)):
(3.5.61)
44
Then, for every e>0small enough problem (P2
e)has a unique solution
(ue;ve)2W2;2(0;T;L2(0;1)2)\L2(0;T;H1(0;1)2)
and the following estimates hold true
kue ukC([0;T];L2(0;1))=O(e1=4);kue ukL2(DT)=O(e1=2);
kve vkC([0;T];L2(0;1))=O(e1=4);kve vkL2(DT)=O(e1=2):(3.5.62)
Remark 15. Theorem ??remains true if operator A 2is replaced byeA2:D(A2)H2!H2
defined by
eA2(p;q) =
q0+˜r(;p);p0+˜g(;q)
;
where ˜r;˜g:[0;1]R!Rare given functions such that ˜r(;x),˜g(;x)belong to L2(0;1)
for all x2R, and x!˜r(x;x);x!˜g(x;x)are continuous and nondecreasing for a. a.
x2(0;1). This assertion is based on the fact that eA2is also maximal monotone in H 2(cf.
[39, Proposition 5.1.1, p. 111]).
Similar results can also be established in the case of algebraic-differential boundary
conditions of the form
u(0;t)2r0(v(0;t));vt(1;t) u(1;t)+r1(v(1;t))3e(t);0tT;
where r 0;r1are given maximal monotone graphs in RR, and e2L2(0;T):Such condi-
tions also occur in specific practical problems.
It is also worth pointing out that in a similar manner further types of boundary con-
ditions can be investigated, such as differential-differential, bilocal or periodic conditions
(for more information, see, e.g., [39, Chapter 5]).
Even more, our technique is applicable to systems consisting of n couples of telegraph
systems with unknowns u i;vi;i=1;2;:::;n, connected by various conditions at x =0and
x=1, as in the case of integrated circuits.
3.5.3 The nonlinear wave equation
Consider the wave equation
utt Du+h(ut) =f(x;t);(x;t)2W(0;¥); (3.5.63)
with initial conditions
u(x;0) =u0(x);ut(x;0) =v0(x);x2W; (3.5.64)
45
and the homogeneous Dirichlet boundary condition
u(x;t) =0 on G[0;¥); (3.5.65)
where WRnis a bounded domain with sufficiently smooth boundary G. Assume that
h:R!Ris Lipschitzian on R:
In this case, we consider as usual the Hilbert space (phase space) H3=H1
0(W)L2(W)
with the scalar product defined by
h(p1;q1);(p2;q2)i=Z
WÑp1Ñp2dx+Z
Wq1q2dx;
and define the operators A3:D(A3)H3!H3by
D(A3):=
H1
0(W)\H2(W)
H1
0(W);A3(p;q) = ( q; Dp);
andB3:H3!H3;B3(p;q) = ( 0;h(q)):
Obviously, problem (3.5.63) (3.5.65) can be expressed as a Cauchy problem in H3as
follows (
U0(t)+A3U(t)+B3U(t) =F(t);t2[0;T];
U(0) =U0;(P3
0)
where U(t):= (u(;t);ut(;t));F(t):= (0;f(;t));U0:= (u0;v0):
According to Theorem 7 and Lemma 11 above, we have
Lemma 12. Assume that the above conditions on on Wandhare fulfilled. Then, for every
f2W1;1(0;T;L2(W));u02H1
0(W)\H2(W);and v 02H1
0(W);there exists a unique strong
solution u of problem (P3
0)such that
u2W1;¥(0;T;H1
0(W))\W2;¥(0;T;L2(W));u(;t)2H1
0(W)\H2(W);8t2[0;T]:
The elliptic-like regularization of problem (P3
0)is
(
eU00+U0+A3U+B3U=F(t);t2[0;T];
U(0) =U0;U0(T) =0;(P3
e)
or, equivalently,
8
>>>><
>>>>: eutt+ut v=0;
evtt+vt Du+h(v) =finW[0;T];
u=v=0 onG[0;¥);
u(;0) =u0;v(;0) =v0;ut(;T) =0;vt(;T) =0 inW:
According to Lemma 9, if the above assumptions on Wandhare fulfilled, then for all e>0
46
sufficiently small, problem (P3
e)has a unique solution Ue:= (ue;ve)2W2;2(0;T;H1
0(W)
L2(W)):
In fact, ueis more regular. Indeed, from the equation
e(ue)tt+(ue)t ve=0
it follows that ue2W4;2(0;T;L2(W)). Even more, from the equation
e(ve)tt+(ve)t Due+h(ve) =f;
we derive Due2L2(0;T;L2(W))and hence ue2W4;2(0;T;L2(W))\W2;2(0;T;H1
0(W))\
L2(0;T;H2(W));showing that the effect of the regularization is really significant in this
case.
Taking into account Theorem 21 and the above comments, we can state
Theorem 24. Assume that the above conditions on Wandhare fulfilled. Then, for every
f2W1;1(0;T;L2(W));u02H1
0(W)\H2(W);v02H1
0(W);
and every e>0small enough, problem (P3
e)has a unique solution
(ue;ve)2
W4;2(0;T;L2(W))\W2;2(0;T;H1
0(W))\L2(0;T;H2(W))
W2;2(0;T;L2(W));
and the following estimates hold
kue ukC([0;T];H1
0(W))=O(e1=4);kue ukL2(0;T;H1
0(W))=O(e1=2);
kve utkC([0;T];L2(W))=O(e1=4);kve utkL2(0;T;L2(W))=O(e1=2):
Remark 16. Theorem 24 remains valid if his replaced by h+g, where his as above and
gis a maximal monotone graph in RR, with 02D(g):
It is also readily seen that a result similar to Theorem 24 above is valid if his a
Lipschitz continuous function depending on u. Other types of boundary conditions can also
be analyzed in a similar manner; see [66, p.241] for more details.
3.6 Elliptic regularization of the semilinear telegraph sys-
tem with nonlinear boundary conditions
The purpose of this section is to study a Lions type regularization of the semilinear tele-
graph system with nonlinear boundary conditions. An asymptotic expansion of order zero
for the solution of this regularization is established, including some boundary layer cor-
rections. Specifically, under some appropriate smoothness and compatibility conditions on
47
the data an estimate for the remainder term of the expansion is derived with respect to
theC([0;T];L2(0;1)2)norm. Our main theorem generalizes a result regarding the particu-
lar case of homogeneous linear boundary conditions reported by N. C. Apreutesei and B.
Djafari Rouhani [8].
More exactly, let us consider in DT:=
(x;t);x2(0;1);t2(0;T)
the telegraph
system(
ut+vx+g1(x;u) =f1(x;t);
vt+ux+g2(x;v) =f2(x;t);(3.6.66)
with the initial conditions
u(x;0) =u0(x);v(x;0) =v0(x);0x1; (3.6.67)
and the following nonlinear boundary conditions
u(0;t) =r1
v(0;t)
;u(1;t) =r2
v(1;t)
;0tT; (3.6.68)
where T>0 is a given time instant, u0;v0are given initial states, f1;f2:DT!R;r1;r2:
R!R, and g1;g2:[0;1]R!Rare given functions subject to the following assumptions:
(i1)for all K>0 there exists a positive constant lKsuch that
jri(x) ri(y)jlKjx yj;8jxj;jyjK;
and
ri(x) ri(y)
(x y)m0(x y)2;8x;y2R;i=1;2;
where m0is a positive constant;
gi(;x)2L2(0;1)for all x2R;i=1;2, and one of the conditions (i2),(i3)below
holds:
(i2)there exists a positive constant L>0 such that:
jgi(x;x) gi(x;h)jLjx hj;8x;h2Rand a. a. x2(0;1);i=1;2;
(i3)x!gi(x;x),i=1;2, are both nondecreasing functions for a. a. x2(0;1);and for
allK>0 there exists a positive constant LKsuch that
jgi(x;x) gi(x;h)jLKjx hj;a. a. x2(0;1);8jxj;jhjK;i=1;2:
Let us denote by P0the problem (3.6.66), (3.6.67) ;(3.6.68).
48
The elliptic regularization of the above problem, denoted by Peis :
8
>><
>>: e(utt;vtt)+(ut;vt)+(vx;ux)+
g1(;u);g2(;u)
= (f1;f2);inDT;
u(0;t) =r1
v(0;t)
;u(1;t) =r2
v(1;t)
;0tT;
u(;0);v(;0)
= (u0;v0);
u(;T);v(;T)
= (uT;vT);Pe
where uT;vTare given convenient functions.
In sub section 3.3.1, a similar regularization of the telegraph system subject to some
linear boundary conditions has been obtained. The case of nonlinear conditions is more
delicate since in this case operator Ais no longer linear, thus the abstract results ob-
tained in Section 3.1 cannot be directly applied. Note also that in Section 3.1 the problem
eu00(t) +u0(t) +Au(t) +Bu(t) =f(t);0tT;u(0) =u0;u0(T) =0 (see (P:2)e), has
been investigated, where Awas assumed to be a general maximal monotone operator. For
this type of condition at t=T;problem Peis regularly perturbed of order zero with respect
to the sup-norm of C([0;T];L2(0;1)2).
It is worth pointing out that the problem we consider in this Section is essentially
different (being a singularly perturbed one) and requires separate analysis.
First of all we state without proofs some results on the existence, uniqueness and reg-
ularity of the solutions of problems P0andPe.
In what follows we consider the Hilbert space H:=L2(0;1)2with its usual scalar prod-
uct and induced norm. In order to apply Theorem 7 and Remark 1, Chapter 1, to problem
P0we define the operators A:D(A)H!Hby
D(A) =n
(u;v)2H1(0;1)2; u(0) =r1(v(0));u(1) =r2(v(1))o
;A(u;v) = ( v0;u0)
andB:H!H,B(u;v) =
g1(;u);g2(;v)
8(u;v)2H.
It is not difficult to see that problem (3.6.66) (3.6.68) can be expressed as a Cauchy
problem in Has follows:
8
<
:w0(t)+A
w(t)
+B
w(t)
=F(t);0tT;
w(0) =w0;(3.6.69)
where w(t):= (u(;t);v(;t));F(t):= (f1(;t);f2(;t));w0:= (u0;v0).
According to [ ?], Proposition 5.1.1, p. 111, we have:
Lemma 13. Assume that r 1;r2are maximal monotone graphs in RR. Then A is a maximal
monotone operator and D (A)is dense in H.
In particular, assumption (i1)implies that r1andr2are maximal monotone graphs in
RR. Now, if (i2)is satisfied, then operator Bis Lipschitz on H. In the other case, i.e. if
hypotheses (i1)and(i3)hold, operator A+B:D(A)!His maximal monotone in H(cf.
49
[39, Proposition 5.1.1, p. 111]). Summarizing, if r1;r2satisfy (h1)andg1;g2verify either
(h2)or(h3), we can apply Theorem 7 and Remark 1 in order to derive the following result
regarding the existence, regularity and uniqueness for the solution (u;v)to problem P0:
Lemma 14. (see [39, Theorem 5.5.1]) Assume that (i1)holds and either (i2)or(i3)is
satisfied. Let T >0be fixed. Then, for every
f1;f22W1;1
0;T;L2(0;1)
;u0;v02H1(0;1)such that
u0(0) =r1
v0(0)
;u0(1) =r2
v0(1)
;(3.6.70)
there exists a unique strong solution (u;v)of problem P 0such that
(u;v)2W1;1
0;T;L2(0;1)2
;ux;vx2L¥
0;T;L2(0;1)
; (3.6.71)
and u ;v satisfy (3.6.68) for all t2[0;T]:
An abstract form of the elliptic-like regularization Peis the following:
8
<
: ew00(t)+w0(t)+A
w(t)
+B
w(t)
=F(t);0tT;
w(0) =w0;w(T) =wT;(3.6.72)
where w:= (ue;ve);wT= (uT;vT):
Letf2L2(0;T;H)andw0;wT2D(A):Recall that (see [ ?]) aH valued function w=w(t)
is said to be a strong solution of problem (3.6.72) if w2C
[0;T];H
\W2;2(0;T;H);w(t)
satisfies (3.6.72) 1for a. a. t2(0;T);w(0) =w0;w(T) =wT:
As far as the solution w= (ue;ve)of problem Peis concerned, the next result is also known
(see [7, Theorem 3.2] :
Lemma 15. Assume that (i1)and(i3)hold and, in addition, (f1;f2)2L2
DT2and(u0;v0);
(uT;vT)2D(A). Then for every e>0problem P ehas a unique strong solution (ue;ve)2
W2;2
0;T;L2(0;1)2
:
Remark 17. Obviously, under the assumptions of Lemma 15 above, the operator A +B is
maximal monotone (since B is maximal monotone and D (B) =H). In fact Lemma 15 holds
true if (i1)and(i3)are replaced by the condition that A +B:D(A)H!H is maximal
monotone.
In the other case, when (i1)and(i2)hold, Lemma 15 cannot be used. In this situation
we can prove the following result:
Lemma 16. Assume that (i1)and(i2)are satisfied. Then for all (f1;f2)2L2
DT2and
(u0;v0);(uT;vT)2D(A)there exits an e0>0(which depends on the Lipschitz constant L
of g 1;g2) such that problem P ehas a unique strong solution (ue;ve)2W2;2
0;T;L2(0;1)2
for all 0<e<e0.
50
Proof. Following an idea from the proof of Theorem 19 we multiply equation (3.6.72) 1by
e Mtwith M>0;a positive constant that will be chosen later. Denoting z=e Mtw, we get
from (3.6.72) that zverifies the following problem in [0;T]:
(
ee Mt
eMtz00+z0+Mz+e Mt
A
eMtz
+B
eMtz
=e MtF(t);
z(0) =w0;z(T) =zT:=e MTwT:(3.6.73)
Next, we are writing problem (3.6.73) in the form Qez+˜Bz=e MtF, where the operators
QeandeBare defined as follows: Qe:D(Qe)L2(0;T;H)!L2(0;T;H),˜B:L2(0;T;H)!
L2(0;T;H);
D(Qe):=n
z2W2;2(0;T;H);z(0) =w0;z(T) =zT;A(eMtz)2L2(0;T;H)o
;
Qe(z):= ee Mt
eMtz00+z0+M
2z+e MtA(eMtz)8z2D(Qe);
˜Bz:=e MtB
eMtz
+M
2z8z2L2(0;T;H):
As a consequence of Lemma 15 we get that D(Qe)6=/ 0. Indeed, z=e Mtw2D(Qe), where
w=weis the solution of problem Pewith B0:Ifz1;z22D(Qe)we have
Qez1 Qez2;z1 z2
L2 eZT
0h(eMtz1)00 (eMtz1)00;e Mt(z1 z2)idt
+M
2kz1 z2k2
L2M
2 eM2
kz1 z2k2
L2+ekz0
1 z0
2k2
L2;
whereh;iL2andkkL2denote the inner product of L2(0;T;H)and the corresponding
induced norm. Thus, for every 0 <e<1=(2M);operator Qeis strongly monotone in
L2(0;T;H).
In addition, operator Qeis maximal monotone, i.e., for all h2L2(0;T;H)there exists
z2D(Qe)such that (M=2)z+Qev=h. Indeed, using the substitution w(t) =eMtz(t), we are
lead to the original problem in w,Pe, with B0 and F(t) =eMth(t), and from Lemma 15 we
know that this problem has a unique strong solution w2W2;2(0;T;H), soz(t) =e Mtw(t)
belongs to D(Qe)and satisfies the above equation.
Moreover, one can see that ˜Bis Lipschitzian on L2(0;T;H), and, if we choose M:=2L,
then operatoreBis also monotone, hence maximal monotone. In conclusion, we obtain that
˜B+Qeis maximal monotone and strongly monotone on D(Qe)L2(0;T;H)with constant
L(1 4eL)for all 0 <e<e0:=1=(4L). Therefore, ˜B+Qeis surjective, so in particular the
equation (3.6.73) has a unique solution ze2W2;2(0;T;H)for all e2(0;e0). This means
that, for every e2(0;e0),we(t) =e2Ltze(t)is the unique solution of problem Pe. The proof
of Lemma 16 is complete.
Problem Peformulated above, under assumption (i1)and one of the assumptions (i2)
51
or(i3), is singularly perturbed with respect to the C
[0;T];L2(0;1)2
norm. Indeed, if we
suppose that its solution (ue;ve)!(u;v)ase!0 in this norm (here (u;v)denotes the
solution of problem P0), then from the condition
ue(;T);ve(;T)
= (uT;vT)we derive
that the equality (u;v)(;T) = ( uT;vT)for a. e. x2(0;1)is a necessary condition for the
convergence. Therefore (ue;ve)has a singular behavior with respect to ein a neighborhood,
[T d;T][0;1];of the setfTg[0;1]:This vicinity is called boundary layer.
Taking into account the above comments as well as the classical perturbation theory
(see [11] and [67] for details) we are going to derive formally an expansion of order zero of
the solution (ue;ve)in the form:
(ue;ve)(x;t) = ( u;v)(x;t)+(c1;c2)(x;t)+(s1e;s2e)(x;t);(x;t)2DT; (3.6.74)
where: t= (T t)=eis the stretched (fast) variable, (u;v)is the regular term (the solution
of problem P0),(c1;c2)is the corresponding correction (or boundary layer function), while
(s1e;s2e)denotes the remainder of order zero.
Let us impose to (ue;ve)given by (3.6.74) to satisfy formally the problem Pe:By
identifying the coefficients of e 1ande0we can find all the terms of the expansion (3.6.74).
We have to distinguish between the coefficients depending of tand those depending of the
fast variable t. Thus one can see that (u;v)satisfies the reduced problem P0, while (c1;c2)
satisfies the following problem:
8
<
:
c1tt;c2tt
(x;t)+
c1t;c2t
(x;t) = ( 0;0);(x;t)2(0;1)(0;¥);
c1;c2
(;0) =
uT u(;T);vT v(;T)
:(3.6.75)
Obviously, the boundary layer function c= (c1;c2)is introduced to compensate the sin-
gular behavior of (ue;ve)in a neighborhood of the set fTg[0;1]:So, from (3.6.75),
taking into account that (c1;c2)should be negligible far from the boundary layer, i.e.,
k(c1;c2)(;t)k!0 ast!¥;we deduce that
c1;c2
(x;t) =e t
uT(x) u(x;T);vT(x) v(x;T)
8t0;x2[0;1]: (3.6.76)
Finally, it is easy to see that the remainder term seshould satisfy:
8
>><
>>: e(se+w)tt+set+
ve v
x;
ue u
x
+
g1(;ue) g1(;u)
;
g2(;ve) g2(;v)
= (0;0)in(0;T);
se(;0) =
c1;c2
(;T=e);se(;T) = ( 0;0);(3.6.77)
where se:= (s1e;s2e);w= (u;v):Therefore, all the terms involved in our asymptotic ex-
pansion (3.6.74) are completely determined.
Summarizing what we have done so far, we can state the following concluding result:
52
Corolary 1. Assume that (i1)and either (i2)or(i3)are satisfied and, in addition:
f1;f22W1;1
0;T;L2(0;1)
;u0;v0;uT;vT2H1(0;1);
u0(0) =r1
v0(0)
;u0(1) =r2
v0(1)
; uT(0) =r1
vT(0)
;uT(1) =r2
vT(1)
:
(3.6.78)
Then, for every e>0(small enough if assumption (i2)holds) the problems P eand P 0have
unique strong solutions
ue;ve2W2;2
0;T;L2(0;1)
\L2
0;T;H1(0;1)
;
u;v2W1;¥
0;T;L2(0;1)
\L¥
0;T;H1(0;1)
;u(;t);v(;t)2D(A)8t2[0;T]:
(3.6.79)
Moreover, (ue;ve)admits an asymptotic expansion of the form (3.6.74) .
Remark 18. It should be pointed out that, under assumptions of Corollary 1, u (;T);
v(;T)2H1(0;1), thus by (3.6.76) , we get that
(c1;c2)(;t)2H1(0;1)28t0;s1e;s2e2W1;¥
0;T;L2(0;1)
\L2
0;T;H1(0;1)
:
In what follow we establish an estimate for the remainder se= (s1e;s2e)with respect
to the C
[0;T];L2(0;1)2
norm:
Theorem 25. Assume that all the assumptions of Corollary 1 hold. Then, for every e>0
small enough, the problem P ehas a unique strong solution w e= (ue;ve)2W2;2
0;T;L2(0;1)2
\
L2
0;T;H1(0;1)2
that admits an asymptotic expansion of the form (3.6.74) , where (u;v)
is the solution of problem P 0;(c1;c2)is defined by (3.6.76) , se= (s1e;s2e)satisfies (??), and
the following estimates hold
k
ue u c1;ve v c2
kC([0;T];H)=O(e1=4);k
ue u;ve v
kL2(DT)2=O(e1=2):
(3.6.80)
Proof. Throughout this proof we denote by C1;C2;… different positive constants which
depend on the data, but are independent of e:The following change of functions
bse= (bs1e;bs2e):= (s1e;s2e)+(r1e;r2e);
re= (r1e;r2e):= (1 t=T)
c1(;T=e);c2(;T=e)
inDT(3.6.81)
homogenizes the boundary condition in t=0 (see the first equality in (3.6.77) 2):
In order to establish the desired estimates ( ??) we are going to prove that, for e>0
small enough, the following estimates hold
k(s1e;s2e)kL2(0;T;H)=O(e1=2);k(s1et;s2et)kL2(0;T;H)=O(1): (3.6.82)
53
Let us denote se= (s1e;s2e);we= (ue;ve);c= (c1;c2);w= (u;v):Combining (3.6.74),
(3.6.76), (3.6.81), and Corollary 1, we infer that, for all e>0 (sufficiently small if assump-
tion(i2)holds),
bse+w
=we c+re2W2;2(0;T;H)\L2(0;T;H1(0;1)2);
bse2W1;¥(0;T;H)\L2
0;T;H1(0;1)2
:
Thus, one can see that bsesatisfies the problem
(
e(bse+w)00+bs0
e+A(we) A(w)+B(we) B(w) =he;t2[0;T];
bse(;0) =bse(;T) = ( 0;0);(3.6.83)
where he:= T 1c(;T=e). To achieve our goal, which is to obtain (3.6.82), let us multi-
ply (3.6.83) 1bye Mt;where the constant M>0 will be chosen later. Denote ese:=e Mtbse:
Taking into account the obvious equality e Mtbs0
e=Mese+es0
e, we infer that
8
>><
>>: ee Mt(eMtese+w)00+es0
e+Mese
+e Mt
A(we) A(w)
+
B(we) B(w)
=e Mthe;t2[0;T];
ese(;0) =ese(;T) = ( 0;0):(3.6.84)
If we take the scalar product in H=L2(0;1)2of equation ( ??)1byeseand integrate the
resulting equation over [0;T], we obtain
eZT
0h(eMtese+w)tt;e Mteseidt+Mkesek2
L2+ZT
0e MthA(we) A(w);eseidt
+ZT
0e MthB(we) B(w);eseidt=ZT
0e Mthhe;eseidt(3.6.85)
(we have denoted by kkL2the usual norm of L2(DT)2):
A new integration, this time by parts in the first term of (3.6.85) yields
eZT
0h(eMtese+w)t;e Mteset Me Mteseidt+ZT
0e MthA(we) A(w);eseidt
+ZT
0e MthB(we) B(w);eseidt+Mkesek2
L2=ZT
0e Mthhe;eseidt;
which implies
M(1 eM)kesek2
L2+ekesetk2
L2 T1e T2e
+
C1e+ke MthekL2
kesekL2+C2ekesetkL2;(3.6.86)
54
where we have denoted
T1e:=ZT
0e MthA(we) A(w);eseidt;T2e:=ZT
0e MthB(we) B(w);eseidt:(3.6.87)
Next, we are going to estimate T1e:First, let us point out that
ese=e Mt(we w+he);he= (h1e;h2e):=re c;
e Mt(A(we) Aw) =
(es2e eh2e)x;(es1e eh1e)x
;ehe:=e Mthe:(3.6.88)
Thus, we have
T1e=ZT
0Z1
0
(es2e eh2e)xes1e+(es1e eh1e)xes2e
dx dt
=ZT
0es1ees2e 1
0dt ZT
0Z1
0
es1eeh2ex+es2eeh1ex
dx dt :(3.6.89)
Now, by the boundary conditions (3.6.68) and (Pe)2;we infer that
T1e=ZT
0e Mt
r2(ve(1;t)) r2(v(1;t))+h1e(1;t)
es2e(1;t)
dt
+ZT
0e Mt
r1(ve(0;t)) r1(v(0;t))+h1e(0;t)
es2e(0;t)
dt
ZT
0Z1
0
es1eeh2ex+es2eeh1ex
dx dt :(3.6.90)
If we analyze the structure of c;he;reandhe;we can easily see that
kckL2
0;T;H1(0;1)2=O(e1=2);kehekL2
0;T;H1(0;1)2=O(e1=2);
krekL2
0;T;H1(0;1)2=O(ej);khekL2=O(ej);8j1:(3.6.91)
From estimate (3.6.91) 1;we deduce that
E1e:=ZT
0e Mt
r2(ve(1;t)) r2(v(1;t))+h1e(1;t)
es2e(1;t)
dt
=ZT
0e Mt
r2(v h2e+eMtes2e) r2(v)
es2e+eh1ees2e
(1;t)dt
C3e1=2kes2e(1;)kL2(0;T)
+ZT
0e Mt
r2(v h2e+eMtes2e) r2(v h2e)+r2(v h2e) r2(v)
es2e
(1;t)dt:
(3.6.92)
55
Using assumption (i1), we also have
ZT
0e Mt
r2(v h2e+eMts2e) r2(v h2e)
es2e+
r2(v h2e) r2(v)
es2e
(1;t)dt
m0kes2e(1;)k2
L2(0;T) C4keh2e(1;)kL2(0;T)kes2e(1;)kL2(0;T);
where the constant C4does not depend on e;since the arguments of r2;specifically
v
h2e
(1;t)andv(1;t), verifyj
v h2e
(1;t)jk1;jv(1;t)jk2a. e. t2(0;T)with con-
stants k1;k2which depend only on kvkL¥
0;T;H1(0;1);kvTkH1(0;1);kv(;T)kH1(0;1).
Finally, taking into account estimates ( ??)1and making use of the above inequality into
(3.6.92) we obtain that
E1e C5e1=2kes2e(1;)kL2(0;T)+m0kes2e(1;)k2
L2(0;T): (3.6.93)
Obviously, using similar computations one can obtain an analogous estimate for the second
integral term of (3.6.90). Thus, we derive
T1e C6e1=2
kesekL2+kes2e(1;)kL2(0;T)+kes2e(0;)kL2(0;T)
+m0
kes2e(1;)k2
L2(0;T)+kes2e(0;)k2
L2(0;T)
:(3.6.94)
It remains to estimate T2e. Let us consider in a first case that assumption (i2)holds. We have
T2e=ZT
0e MthB(we) B(w);eseidt=ZT
0e MthB(eMtese+w he) B(w);eseidt
LZT
0e Mtk he+eMtesekkesekdt
LZT
0khekkesekdt+kesek2
L2
C7e1=2kesekL2 Lkesek2
L2:
(3.6.95)
Now, let us choose M=2L:Combining, (3.6.86), (3.6.91), (3.6.94), and (3.6.95) leads to
L(1 4eL)kesek2
L2+ekes0
ek2
L2+m0
kes2e(1;)k2
L2(0;T)+kes2e(0;)k2
L2(0;T)
C1ekesekL2+C2ekes0
ekL2+C8e1=2kesekL2
+C6e1=2
kes2e(1;)kL2(0;T)+kes2e(0;)kL2(0;T)
:(3.6.96)
56
If, for example, 0 <e<minf1;1=(8L)g;then we can deduce from (3.6.96) that
L
2kesek2
L2+ekes0
ek2
L2+m0
kes2e(1;)k2
L2(0;T)+kes2e(0;)k2
L2(0;T)
C2ekes0
ekL2+C8e1=2kesekL2
+C6e1=2
kes2e(1;)kL2(0;T)+kes2e(0;)kL2(0;T)
C9e+L
4kesek2
L2+e
2kes0
ek2
L2+m0
2
kes2e(1;)k2
L2(0;T)+kes2e(0;)k2
L2(0;T)
;
where we have used the inequality ab(a2+b2)=2. This shows that for all 0 <e 1 we
have
kesekL2(0;T;H)=O(e1=2);kesetkL2(0;T;H)=O(1);
kes2e(1;)kL2(0;T)=O(e1=2);kes2e(0;)kL2(0;T)=O(e1=2):(3.6.97)
Since se=e2Ltese re;using ( ??) and ( ??), we obtain ( ??).
Now, let us suppose that assumption (i3)holds. Using the monotonicity of Bwe can see
that
ZT
0e MthB(we) B(w);eseidt=ZT
0e 2MthB(w+eMtese he) B(w he);eMteseidt
+ZT
0e 2MthB(w he) B(w);eMteseidt
ZT
0e MthB(w he) B(w);eseidt:
(3.6.98)
The local Lipschitz condition for g1;g2(see assumption (i3)) and (3.6.91) imply
ZT
0e 2MthB(w he) B(w);eMteseidt C10e1=2kesekL2: (3.20)
Indeed we have
kB
u h1e;v h2e
B(u;v)k2
H=k
g1(;u h1e) g1(;u);g2(;v h2e) g2(;v)
k2
H
=Z1
0
g1(;u h1e) g1(;u)2
dx+Z1
0
g2(;v h2e) g2(;v)2
dx:
From Corollary 1 we see that the solution (u;v)2L¥(DT)2;and since by assumption u0,
v0,uT,vT2H1(0;1);we obtain (by using the local Lipschitz condition on g1,g2) that there
exists a C10>0 depending on f1;f2;u0;v0;uT;vT, but independent of e, such that
kB
u h1e;v h2e
B(u;v)k2
HC10Z1
0h2
1edx+Z1
0h2
2edx
:
57
This implies estimate (3.20), as claimed. Thus, from (3.6.86), (3.6.94), (3.6.98) and (3.20)
we obtain
M(1 eM)kesek2
L2+ekes0
ek2
L2+m0
kes2e(1;)k2
L2(0;T)+kes2e(0;)k2
L2(0;T)
C1ekesekL2+C2ekes0
ekL2+C11e1=2kesekL2
+C6e1=2
kes2e(1;)kL2(0;T)+kes2e(0;)kL2(0;T)
:(3.6.99)
Taking for example M=1 in (3.6.99), we derive an estimate similar to (3.6.96) which
implies (3.6.97).
Summarizing all the information above, we can see that in both our cases sesatisfies
(3.6.82). Consequently, from the obvious equality
kse(;t)k2= 2ZT
thse(;s);set(;s)ids8t2[0;T];
we infer thatksekC
[0;T];H=O(e1=4):
Finally, (3.6.80) 2is a consequence of estimates (3.6.82) 1and (3.6.91) 1. This completes
the proof of the theorem.
Remark 19. From (3.6.82) we derive that s et!(0;0)weakly in L2(DT)2;i. e.,
(ue c1)t;(ve c2)t
!(ut;vt)
weakly in
L2(DT)2ase!0+.
Combining (3.6.91) and(3.6.97) implies that
k
ue u
(1;);
ve v
(1;)
kL2(0;T)=O(e1=2)
hence, taking into account assumption (h1)and boundary conditions (3.6.68) ,(Pe)2we also
obtain
k
ue u
(0;);
ve v
(0;)
kL2(0;T)=O(e1=2):
Remark 20. The estimate (3.6.80) 2shows that problem P eis regularly perturbed of order
zero with respect to the weaker norm L2(DT)2:
Remark 21. Similar results can be established in the case of nonlinear algebraic-differential
boundary conditions of the form
u(0;t) =r1
v(0;t)
;vt(1;t) u(1;t)+f0
v(1;t)
=e(t);0tT;
where r 1;f0are given monotone functions which satisfy appropriate assumptions, and e 2
L2(0;T):Such conditions also occur in specific practical problems.
58
It is also worth pointing out that in a similar manner further types of non linear bound-
ary conditions can be investigated, such as differential-differential conditions (for more
information, see, e.g., [39, Chapter 5] and [11, Chapter 5]).
Even more, our technique is applicable to systems consisting of n couples of telegraph
systems with unknowns u i;vi;i=1;2;:::;n, connected by various boundary conditions at
x=0and x =1. This case is relevant for integrated circuits.
3.7 Elliptic regularization of the semilinear heat equation
with nonlinear boundary conditions
In this Section we study an elliptic regularization of the following problem denoted by P0:
8
><
>:ut(x;t) Du(x;t)+b(u(x;t)) = f(x;t);(x;t)2WT=W(0;T];
¶u
¶n(x;t) =a(u(x;t));(x;t)2ST=¶W[0;T];
u(x;0) =u0(x);x2W;(P0)
where
(j1)WRnis a bounded open set with boundary ¶Wsufficiently smooth; nis the outer
unit normal to the boundary ¶W;T>0 is a given time;
(j2)f:WT!R;u0:W!Rare given functions, f2L2(WT);u02L2(W).
Concerning the nonlinear functions aandbwe assume that
(j33)a;b:R!Rare continuous and nondecreasing.
The elliptic regularization of problem P0, denoted by Pe, is the following:
8
>><
>>:eutt(x;t) ut(x;t)+Du(x;t) b(u(x;t)) = f(x;t);(x;t)2WT;
¶u
¶n(x;t) =a(u(x;t));(x;t)2ST;
u(x;0) =u0(x);u(x;T) =uT(x);x2W;Pe
where uT2L2(W):
We will see that these problems can be written in the abstract forms similar with those
considered in Section 3.4. Since in this case operator Ais no longer linear, the abstract
results obtained in Theorem 19 cannot be directly applied.
As a first step in our analysis we construct formal asymptotic expansions of the order
zero for the solution of the elliptic problem by employing the Vishik-Lyusternik method
(see [67]). Once our formal asymptotic expansions are determined, we may continue with
their validation. More precisely, we prove a result concerning the existence, uniqueness,
and regularity of the terms which occur in the previously determined asymptotic expansion.
All these guarantee the fact that this expansion is well defined. Moreover, they are also used
to derive estimates for the remainder.
59
Let us notice that problem Peis singularly perturbed with respect to C([0;T];L2(W))
norm. Indeed, if we suppose that u1e!X0ase!0 in this norm (here X0denotes the
solution of problem P0), then from u(x;T) =uT(x);x2Wwe derive that the equality
X0(x;T) =uT(x)for a. e. x2Wis a necessary condition for the convergence. Therefore
uehas a singular behavior with respect to ein a neighborhood [T d;T]Wof the set
fTgW:This vicinity is called boundary layer.
Taking into account the above comments as well as the classical perturbation theory
(see [67] for details) we are going to derive formally an expansion of the order zero of the
solution uein the form
u1e(x;t) =X0(x;t)+c0(x;t)+re(x;t);(x;t)2WT; (3.7.100)
where: t= (T t)=eis the stretched (fast) variable, X0is the regular term, c0is the corre-
sponding correction (or boundary layer function), while redenotes the remainder of order
zero.
Let us consider that uegiven by (3.7.100) satisfies formally Pe:By identifying the co-
efficients of e 1ande0we can find all the terms of (3.7.100). We have distinguish between
the coefficients depending of tand those depending of the fast variable t. Thus one can see
thatX0satisfies the reduced problem P0, while c0satisfies the following problem:
8
<
:c0tt(x;t)+c0t(x;t) =0;(x;t)2W(0;¥);
c0(x;0) =uT(x) X0(x;T);x2W:(3.7.101)
Obviously the boundary layer function c0is introduced to compensate the singular behavior
ofu1ein a neighborhood of our boundary layer WfTg:So, from (3.7.101) taking into
account that c0should be negligible far from the boundary layer, i. e.
kc0(;t)k!0 ast!¥;
wherekk denotes the usual norm of L2(W), we deduce
c0(x;t) = ( uT(x) X0(x;T))e t8t0;x2W: (3.7.102)
Finally, one can see that the remainder term satisfies the problem:
8
><
>:e(re+X0)tt ret+D(re+c0) b(u1e)+b(X0) =0 inWT;
¶re
¶n+a(u1e) a(X0)+¶c0
¶n=0 onST;
re(x;0) = c0(x;T=e);re(x;T) =0;x2W:(3.7.103)
Therefore, all the terms involved in our asymptotic expansion are completely deter-
mined.
60
We choose as our framework the Hilbert space H:=L2(W);endowed with the usual
scalar product, denoted h;i, and the corresponding induced norm, denoted kk:
It is not difficult to see that our problem Pecan be expressed as the following two-point
boundary problem in H:
8
<
:eu00 u0 Ju= f;t2(0;T);
u(0) =u0;u(T) =uT;(3.7.104)
where u(t) =u(;t);f(t) =f(;t);t2(0;T);b:D(b)H!H,
D(b) =fu2H;x!b(u(x))belongs to Hg;
b(u)
(x) =b(u(x))for a. a. x2W;8u2D(b);
(bis the canonical extension of btoH),J:D(J)H!H;
D(J) =fu2H2(W); ¶u=¶n=a(u)a. e. on ¶Wg\
D(b);
Ju= Du+b(u)8u2D(J):
It is well known that under assumptions (j1);(j3)operator J:D(J)!His maximal mono-
tone. In fact, Jis even cyclically monotone. More precisely, since bandaare the deriva-
tive of the convex functions j(x) =Rx
0b(y)dy;j0(x) =Rx
0a(y)dy;operator J=¶y, where
y:H!( ¥;¥];
y(u) =8
>>><
>>>:1
2R
WjÑuj2dx+R
Wj(u)dx+R
¶Wj0(u)ds;
ifu2H1(W);j(u)2L1(W);j0(u)2L1(¶W);
+¥;otherwise ;(3.7.105)
wherejjdenotes the euclidian norm of Rn(see, e.g., [59], p. 197). Note that Jis also
injective.
Therefore we have the following result, whose proof is known (see [7]):
Theorem 26. Assume that (j1);(j3)are satisfied and f 2L2(WT);u0;uT2D(J). Then,
problem P e;e>0, has a unique solution
ue2W2;2(0;T;L2(W)):
In the following we deal with problem P0. This can be represented as a Cauchy problem
61
in the same Hilbert space H:
8
<
:X0
0+JX0=f(t);t2(0;T);
X0(0) =u0;(3.7.106)
where X0(t):=X0(;t).
Therefore, we are able to prove a result concerning existence, uniqueness and regularity
of the solution X0of problem P0:
Theorem 27. Assume that (j1);(j3)are satisfied and
f2W1;1(0;T;L2(W));u02H2(W);b(u0)2L2(W);¶u0
¶n+a(u0) =0: (3.7.107)
Then, problem P 0has a unique solution
X02W1;¥(0;T;L2(W))\
W1;2(0;T;H1(W))\
L¥(0;T;H2(W)):
Proof. First we note that under assumptions (j1);(j3), operator Jis the subdifferential
of the proper convex lower semicontinuous function ydefined in (3.7.105). Therefore, if
f2L2(0;T;H);u02H1(W);j(u0)2L1(W);j0(u0)2L1(¶W);according to Theorem 9 our
problem has a unique solution X02W1;2(0;T;L2(W))TL2(0;T;H2(W))TL¥(0;T;H1(W)):
If, in addition f2W1;1(0;T;H)andu02D(J)we can use the classical existence theory (see
e.g.[19], p. 257) to infer that solution X0of problem P0belongs to W1;¥(0;T;H)TL¥(0;T;H2(W)):
Now, we are going to prove that X02W1;2(0;T;H1(W)):We start from the obvious
inequality
1
2d
dtkX0(;t+d) X0(;t)k2+kÑX0(;t+d) ÑX0(;t)k2
(L2(W))n
kf(;t+d) f(;t)kkX0(;t+d) X0(;t)k;
for a.a. 0tt+dT. Integration over [0;T d]the above inequality leads us to
ZT d
0kÑX0(;t+d) ÑX0(;t)k2
(L2(W))ndt1=2kX0(;d) u0k2
+ZT d
0kf(;t+d) f(;t)kkX0(;t+d) X0(;t)kdt;(3.7.108)
for all d2(0;T]. Since X02W1;¥(0;T;L2(W))andf2W1;1(0;T;L2(W))we have
kX0(;t+d) X0(;t)kK1d80t<t+dT;
ZT d
0kf(;t+d) f(;t)kdtK2d8d2(0;T];
62
where K1;K2are some positive constants. Thus, by (3.7.108), we obtain
ZT d
0kÑX0(;t+d) ÑX0(;t)k2
(L2(W))ndtK3d28d2(0;T];
from which it follows that X02W1;2(0;T;H1(W)).
Summarizing what we have done so far, we can state the following concluding result:
Corolary 2. Assume that (h1);(h3)are satisfied and, in addition
f2W1;1(0;T;L2(W));u0;uT2H2(W);b(u0);b(uT)2L2(W);
¶u0
¶n+a(u0) =0;¶uT
¶n+a(uT) =0:(3.7.109)
Then, problem P e;e>0, and P 0have unique solutions
ue2W2;2(0;T;L2(W));
X02W1;¥(0;T;L2(W))\
W1;2(0;T;H1(W))\
L¥(0;T;H2(W)):
Remark 22. It should be pointed out that, under assumptions of Corollary 2, X 0(;T)2
H1(W), thus by (3.7.102) , we get c 0(;t)2H1(W)8t0.
In the next result we establish estimates for the remainders rewith respect to the
C([0;T];L2(W))norm. These estimates validate completely our expansion (3.7.100).
Theorem 28. Suppose that assumptions (j1);(j3)and(3.7.109) of Corollary 2 are satisfied
anda0;b02L¥(R). Then, for every e>0, the solution u eof problem P e;i=1;2admits
an asymptotic expansion of the form (3.7.100) and the following estimates hold
krekC([0;T];L2(W))=O(e1=4);ku1e X0kL2(0;T;H1(W))=O(e1=2): (3.7.110)
In addition, u et!X0tweakly in L2(WT)ase!0+.
Proof. Throughout this proof we denote by C1;C2;… some positive constants which depend
on the data, but are independent of e:
From Corollary 2, (3.7.100) and (3.7.102) we derive re+X02W2;2(0;T;L2(W));re+
c02L2(0;T;H1(W)):
For the beginning, we take the scalar product in Hof equation (3.7.103) 1with re(;t),
and then integrate the resulting equation over [0;T]. Thus, by applying Green’s formula, we
63
obtain:
ekretk2
L2(WT)+kÑrek2
(L2(WT))n
+ZT
0Z
W
b(u1e) b(X0)
redxdt+ZT
0Z
¶W
a(u1e) a(X0)
redsdt
=1=2kre(;0)k2 ZT
0hÑc0;Ñreindxdt
eZT
0Z
WX0tretdxdt eZ
W
re+X0
t(x;0)re(x;0)dx;
whereh;indenotes the usual scalar product of Rn. Since aandbare nondecreasing func-
tions, it follows the inequality:
kÑrek2
(L2(WT))n+ekretk2
L2(WT)1=2kre(;0)k2
+ekretkL2(WT)kX0tkL2(WT)+kÑrek(L2(WT))nkÑc0k(L2(WT))n
+krekL2(WT)kb(X0+c0) b(X0)kL2(WT)
+krekL2(ST)ka(X0+c0) a(X0)kL2(ST)
+ekre(;0)kk
re+X0
t(;0)k:(3.7.111)
On the other hand, for every j2N, one can verify:
kc0kL2(WT)=O(e1=2);kÑc0k(L2(WT))n=O(e1=2);kre(;0)k=O(ej);
kb(X0+c0) b(X0)kL2(WT)=O(e1=2);
ka(X0+c0) a(X0)kL2(ST)=O(e1=2)(3.7.112)
(we have used that a0;b02L¥(R)):Now, (3.7.111), (3.7.112) and the trace theorem lead
us to the estimate:
1
2kÑrek2
(L2(WT))n+e
2kretk2
L2(WT)
C1e+C2e1=2
krekL2(WT)+kÑrek(L2(WT))n
:(3.7.113)
It is easily seen that krekL2(WT)C3kretkL2(WT);since re(;T) =0:Next, combining this
inequality with (3.7.113) one has
kretkL2(WT)=O(e 1=2): (3.7.114)
In what follows we are going to obtain an estimate for kretkL2(WT). Denoteere=e tre:
Thus, equation ( ??)1reads
ee t
re+X0
tt eret ere+D(ere+e tc0) e t(b(u1e) b(X0)) = 0 in(0;T):
64
If we multiply this equation by ereand then integrate the resulting equation over [0;T], we
get:
2eZT
0Z
We 2t(re+X0)tredxdt eZT
0Z
We 2t(re+X0)tretdxdt
eZ
W(re+X0)t(x;0)re(x;0)dx+1=2kre(;0)k2 kerek2
L2(WT)
kÑerek2
(L2(WT))n+ZT
0Z
¶W¶ere=¶neredsdt
ZT
0Z
We t
b(u1e) b(X0)
eredxdt+ZT
0Z
We tc0eredxdt=0;
from which we derive
kerek2
L2(WT)+ee 2Tkretk2
L2(WT)+kÑerek2
(L2(WT))n
1=2kre(;0)k2+2ekSetkL2(WT)krekL2(WT)
+ekX0tkL2(WT)kretkL2(WT)+kc0kL2(WT)krekL2(WT)
+ekre(;0)kk(re+X0)t(;0)k
+kb(X0+c0) b(X0)kL2(WT)kerekL2(WT)
+C4ka(X0+c0) a(X0)kL2(ST)(kerekL2(WT)+kÑerek(L2(WT))n)
1=2
ee 2Tkretk2
L2(WT)+kÑerek(L2(WT))n
+C5e+C6e1=2kerekL2(WT);(3.7.115)
where Se=re+X0(we have used Green’s formula, (3.7.114) and trace theorem).
Denoting Ee=kerekL2(WT)and taking into account estimate ( ??), we get E2
eC7e1=2Ee+
C8e;which implies
kerekL2(WT)=O(e1=2))krekL2(WT)=O(e1=2): (3.7.116)
This together with (3.7.115) leads us to
kÑrek(L2(WT))n=O(e1=2);kretkL2(WT)=O(1): (3.7.117)
Thus, from the obvious formula
kre(;t)k2= 2ZT
thre(;s);ret(;s)ids8t2[0;T];
we derivekrekC([0;T];L2(W))=O(e1=4):
Obviously, from (3.7.117) 2we obtainfretge>0converges to zero weakly in L2(WT).
65
Remark 23. As a consequence of (3.7.112) 1;2,(3.7.116) and(3.7.117) 1we can find
kue X0kL2(0;T;H1(W))=O(e1=2):
This estimate shows that problem P eis regularly perturbed of order zero with respect to the
weaker norm L2(0;T;H1(W)):
Next, if we assume that u T(x) =X0(x;T);x2W, then we obtain c 00. If, in addition,
X02W2;2(0;T;L2(W))(this happens under additional assumptions on f and u 0), then,
problem P eis regularly perturbed of order zero, and the following better estimates hold
kue X0kC([0;T];L2(W))=O(e3=4);kuet X0tkL2(WT)=O(e1=2);
kue X0kL2(0;T;H1(W))=O(e):
If X 02L¥(WT)and u T2L¥(W)(hence c 02L¥(W[0;¥))), then it is easily seen that
estimations (3.7.112) 2;3can be obtained under the following weaker assumptions on aand
b
a0;b02L¥
loc(R); (3.7.118)
and the conclusions of Theorem 28 hold.
Let us present a particular case in which (3.7.118) is enough: by the Sobolev-Kondrashov
theorem, if mp>n, then Wm;p(W)C(W)compactly. In particular, for m =p=2and
n3, we have X 02L¥(WT), and u T2C(W).
66
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