Journal of Inequalities in Pure and
Applied Mathematics
REARRANGEMENTS OF THE COEFFICIENTS OF ORDINARY
DIFFERENTIAL EQUATIONS
volume 5, issue 4, article 90,
2004.
SAMIR KARAA
Department of Mathematics and Statistics
Sultan Qaboos University
P.O. Box 36, Al-khod 123
Muscat, Sultanate of Oman.
EMail: skaraa@squ.edu.om
Received 12 March, 2004;
accepted 04 August, 2004.
Communicated by: D. Hinton
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
053-04
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Abstract
We establish extremal values of a solution y of a second-order initial value problem as the coefficients vary in a nonconvex set. These results extend earlier
work by M. Essen in particular by allowing a coefficient in the second derivative
expression.
2000 Mathematics Subject Classification: 26D15
Key words: Rearrangements, Nonconvex set, Extremal couples.
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
Oscillation and Nonoscillation Criteria . . . . . . . . . . . . . . . . . . . 8
3
Characterization of the Extremal Couples . . . . . . . . . . . . . . . . 12
4
Problem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
References
Rearrangements of the
Coefficients of Ordinary
Differential Equations
Samir Karaa
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1.
Introduction
Let L1+ (0, l) denote the set of all nonnegative functions from L1 (0, l). l is a
positive number. Let f ∈ L1+ (0, l) and µf its distribution function
µf (t) = |{x ∈ (0, l) : f (x) > t}|
for t ≥ 0,
where, here and below, |I| is the measure of the set I. Let f ∗ denote the decreasing rearrangement of f ,
f ∗ (x) = sup{t > 0 : µf (t) > x}.
It is known that f ∗ is nonnegative, right continuous and that [2]
Z t
Z t
(1.1)
f ds ≤
f ∗ ds, t ∈ [0, l],
0
Rearrangements of the
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Samir Karaa
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Z
l
Z
f ds =
(1.2)
0
l
f ∗ ds.
0
The increasing rearrangement of f is simply f ∗∗ defined by f ∗∗ (t) = f ∗ (l − t).
A crucial property of rearrangements is that if f and g are nonnegative with
f ∈ L1 (0, l) and g ∈ L∞ (0, 1) then
Z l
Z l
Z l
∗∗ ∗
(1.3)
f g ds ≤
f g ds ≤
f ∗ g ∗ ds.
0
0
0
We will say that f and g are equimeasurable or equivalently that f is a rearrangement of g if they have the same distribution function. We will denote this
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equivalence relation by f ∼ g. Let f0 be a member of L1+ (0, l) and C(f0 ) its
equivalence class for the relation ∼, i.e.,
C(f0 ) = {f ∈ L1+ (0, l), f ∗ = f0∗ }.
A function σ : [0, l] → [0, l] is measure-preserving if, for each measurable set
I ⊂ [0, l], σ −1 (I) is measurable and |σ −1 (I)| = |I|. Let Σ be the class of such
functions. According to Ryff [6], to each f ∈ L1+ (0, l) there corresponds σ ∈ Σ
such that f = f ∗ ◦ σ. In particular, we have
C(f0 ) = {f ∈
L1+ (0, l),
f=
f0∗
◦ σ, σ ∈ Σ}.
Let p and q be in L1+ (0, l) and consider the second-order differential equation
(1.4)
(p−1 (x)y 0 (x))0 + q(x)y(x) = 0,
y(0) = 1,
(p−1 y 0 )(0) = 0.
A solution of the equation is a function y such that y and y 0 are absolutely
continuous and the equation is satisfied almost everywhere. In the first part of
this paper we are interested in finding the supremum and the infimum of y(l)
when the couple (p, q) varies in the set C = C(f0 ) × C(g0 ), where g0 is also a
member of L∞
+ (0, l). Consider
1
Problem 1. Determine inf y(l), (p, q) ∈ C.
Problem 2. Determine sup y(l), (p, q) ∈ C.
To solve these problems, we shall use a kind of calculus of variations which
does not work in C; this class is not convex. Following Essen [3] and [4], and
1
The choice of p−1 instead of p is essential for the study of our problems.
Rearrangements of the
Coefficients of Ordinary
Differential Equations
Samir Karaa
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recalling that C(f0 ) and C(g0 ) are weakly relatively compact in L1 (0, l), we
introduce the set K = K(f0 ) × K(g) consisting of all weak limits of sequences
of C in [L1 (0, l)]2 . To simplify notations, we use the symbol ≺ introduced by
Hardy, Littlewood and Polya [5]. We say that f majorates g, written g ≺ f , if
Z x
Z x
∗
g dt ≤
f ∗ dt, x ∈ [0, l],
0
0
Z
l
∗
Z
g dt =
0
l
f ∗ dt.
0
We note that if g ≺ f (f and g are in L∞
+ (0, l)) then
ess sup g ≤ ess sup f,
ess inf f ≤ ess inf g.
The relations g ≺ f and f ≺ g imply that f ∼ g. In [7], it is shown that
K(f0 ) = {f ∈ L1+ (0, l), f ≺ f0 },
and K(f0 ) is the convex hull of C(f0 ). K(f0 ) is closed and weakly compact in
L1 (0, l). More generally, K(f0 ) is weakly compact in Lp (0, l) if f0 ∈ Lp+ (0, l),
1 ≤ p ≤ ∞. According to [1], C(f0 ) in the set of "∞-dimensional" extreme
points of K(f0 ). That is if f ∈ K(f0 ) − C(f0 ), then for any m ≥ 1, one can
find f1 , . . . , fm linearly independent in K(f0 ) and θ1 , . . . , θm ∈ (0, 1) such that
m
X
θi = 1,
i=1
The following result is given in [1].
m
X
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Samir Karaa
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θi fi = f.
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Proposition 1.1. Let h, g ∈ L1+ (0, l). Then the following are equivalent
(i) g ≺ f .
(ii) For all h ∈ L∞
+ (0, l),
Z x
Z
gh dt ≤
0
x
f h dt,
f dt.
0
(iii) For all h ∈ L∞
+ (0, l),
Z x
Z
∗ ∗
g h dt ≤
l
Z
g dt =
0
0
l
Z
∗ ∗
x
0
Z
∗ ∗
f h dt,
l
Z
g dt =
0
0
Rearrangements of the
Coefficients of Ordinary
Differential Equations
l
f dt.
0
(iv) We have
Z
l
F (g) dt =
0
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l
Z
Samir Karaa
F (f ) dt,
Contents
0
for all convex, nonnegative functions F such that F (0) = 0, F is Lipschitz.
As previously remarked we will consider the following problems
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Problem 3. Determine inf y(l), (p, q) ∈ K.
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Problem 4. Determine sup y(l), (p, q) ∈ K.
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Similar problems may be considered for the differential equation
(1.5)
Let then
(p−1 (x)y 0 (x))0 − q(x)y(x) = 0,
y(0) = 1,
(p−1 y 0 )(0) = 0.
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Problem 5. Determine inf y(l), (p, q) ∈ K.
Problem 6. Determine sup y(l), (p, q) ∈ K.
Proposition 1.2. Let y be the solution of (1.4) [resp. (1.5)]. Then
inf y(l) ≤ cos(Al) ≤ sup y(l),
resp.
inf y(l) ≤ cosh(Al) ≤ sup y(l),
1/2
where A = (||f0 ||L1 ||g0 ||L1 )
.
These estimates hold since the functions
p ≡ l−1 ||f0 ||L1
and
Rearrangements of the
Coefficients of Ordinary
Differential Equations
Samir Karaa
q ≡ l−1 ||g0 ||L1
are respectively members of K(f0 ) and K(g0 ).
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2.
Oscillation and Nonoscillation Criteria
To simplify this section, we assume that p, p−1 and q are in L∞
+ (0, l).
Lemma 2.1. If
Z
l
l
Z
q(x) dt ≤ 1,
p(x) dt
0
0
then a solution of (1.4) does not vanish in [0, l].
Proof. Let y0 be a solution of (1.4) vanishing in (0, l], and denote by a its smallest zero. We have
(2.1)
(p−1 (x)y00 (x))0 + q(x)y0 (x) = 0,
(p−1 y00 )(0) = 0,
y0 (a) = 0.
Multiplying (2.1) by y0 , we then integrate by parts to obtain
Z a
Z a
Z a
−1 0 2
2
2
p (y ) dx =
qy dx ≤ ymax
q dx,
0
0
a
0
Z
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0
|y | dx <
|ymax | ≤
0
21 Z
a
p dx
0
a
−1
0 2
p (y ) dx
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and then apply the inequality (y 0 and p are linearly independent)
Z
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12
.
0
By substitution
the bound for |ymax | into the first inequality and cancelling the
R a −1 0 of
2
term 0 p (y ) dx, the conclusion follows (by contradiction) since a ≤ l.
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Lemma 2.2. If
(2.2)
kpk∞ kqk∞ <
π 2
2l
,
then a solution of (1.4) does not vanish in [0, l].
Proof. Let y0 be as in the previous proof, so that λ0 = 1 is the first eigenvalue
of the problem
0
p−1 (x)y 0 (x) + λq(x)y(x) = 0, (p−1 y 0 )(0) = 0, y(a) = 0.
According to a variational principle,
Ra 0 2
R a −1
y (x) dx
p (x)y 0 (x)2 dx
−1
−1
0R
R0a
≤
kpk
kqk
inf
λ0 = inf
a
∞
∞
y(a)=0
y(a)=0
q(x)y(x)2 dx
y(x)2 dx
0
0
−1 2
−2
= kpk−1
∞ kqk∞ π (2a) .
Hence,
a2 ≥
π 2
2
−1
kpk−1
∞ kqk∞ ,
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Samir Karaa
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which contradicts (2.2).
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2
2
The proof shows that if kpk∞ kqk∞ = π /(2l) , then a solution of (1.4) may
vanish only at x = l. It is not difficult to show that this case holds only when p
and q are constants.
The following lemma gives sufficient conditions for oscillations.
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Lemma 2.3. Assume that p is nondecreasing, p−1 ∈ C 1 [0, l] and p(x) ≤ h−1 on
[0, l], where h is a positive constant. There exists a number H > 0 (depending
on h) such that if q ≥ H a.e. on (0, l) then every solution of (1.4) changes its
sign on (0, l).
Proof. Let z(x) = (l − x)2 (l + x)2 . Multiplying both sides in (1.4) by z(x) and
integrating over (0, l), we obtain
l
Z
y(x)[(p−1 z 0 )0 (x) + q(x)z(x)] dx = 0.
(2.3)
0
As p is nondecreasing we have for all x ∈ (0, l)
Rearrangements of the
Coefficients of Ordinary
Differential Equations
Samir Karaa
(p−1 z 0 )0 (x) = (p−1 )0 (x)z 0 (x) + p−1 (x)z 00 (x) ≥ p−1 (x)z 00 (x).
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Let ε be a positive number such that z 00 is positive on [l − ε, l]. Suppose that
y(x) ≥ 0 on [0, l]. Then (2.3) implies that
Z
l−ε
−1 0 0
y(x)[(p z ) (x) + q(x)z(x)] dx ≤ 0.
(2.4)
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Let
H > h max(−z 00 )(l − ε)−2 (l + ε)−2 .
[0,l]
Then,
(p−1 z 0 )0 (x) + q(x)z(x) ≥ hz 00 (x) + Hz(x) > 0
for all x ∈ (0, l − ε), which contradicts (2.4).
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Lemma 2.4. Any solution of (1.5) is positive and nondecreasing. Moreover, if
||p||L1 ||q||L1 < 1 then
y(l) ≤ (1 − ||p||L1 ||q||L1 )−1 .
Proof. Let y be a solution of (1.5). We have
Z x
0
y (x) = p(x)
q(t)y(t) dt,
0
which implies that y(x) ≥ 1 and y is nondecreasing. Therefore,
Z x
0
y (x) ≤ y(l)p(x)
q(t) dt.
Rearrangements of the
Coefficients of Ordinary
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Samir Karaa
0
Integrating both sides of the last inequality over (0, l), we get
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Z
y(l) − 1 ≤ y(l)
l
Z
p(t) dt
0
l
q(t) dt.
0
Hence,
y(l) ≤ (1 − ||p||L1 ||q||L1 )−1 .
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3.
Characterization of the Extremal Couples
The existence of extremal couples will be discussed at the end of this section.
We suppose that f0 , g0 ∈ L∞
+ (0, l) and f0 ≥ h where h is a positive constant.
Theorem 3.1. Assume that all solutions of (1.4) are positive when (p, q) varies
in K(f0 ) × K(g0 ). Let (p0 , q0 ) be an extremal couple for Problem 3 and y0 the
corresponding solution in (1.4). Then q0 = g0∗ and in the open set where
Z t
Z t
p0 (s) ds >
f0∗∗ (s) ds,
0
0
we have P 0 (t) = 0 where
Z l
y00 2 (t)
y00 (t)
−2
P (t) = 2
p0 (t)y0 (t) dt −
,
p0 (t)
(p0 y0 )(t)
t
Samir Karaa
t ∈ [0, l].
If f0 is bounded below by a positive constant then the above set is empty and
p0 = f0∗∗ , i.e., the infimum over the larger class K coincides with the infimum
over the smallest class C.
Theorem 3.2. Assume that all solutions of (1.4) are positive when (p, q) varies
in K(f0 ) × K(g0 ). Let (p0 , q0 ) be an extremal couple for Problem 4 and y0 the
corresponding solution in (1.4). Then q0 = g0∗∗ and in the open set where
Z t
Z t
p0 (s) ds <
f0∗ (s) ds,
0
Rearrangements of the
Coefficients of Ordinary
Differential Equations
0
we have P 0 (t) = 0 where P is as above. If f0 is far from zero then the above set
is empty and p0 = f0∗ , i.e. the supremum over the larger class K coincides with
the supremum over the smallest class C.
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Let ai and bi , (i = 1, 2), be positive numbers such that a1 < a2 and b1 < b2 .
Define the sets E and F by
Z l
∞
E = p ∈ L (0, l), a1 ≤ p ≤ a2 ,
p dx = A
0
and
Z l
∞
F = q ∈ L (0, l), b1 ≤ p ≤ b2 ,
q dx = B ,
0
where A and B are such that a1 l < A < a2 l and b1 l < B < b2 l. Then we have
Corollary 3.3. If AB ≤ 1, then inf y(l) when (p, q) varies in E × F is reached
by
(
a1 if x ∈ (0, α),
p0 (x) =
a2 if x ∈ (α, l),
Rearrangements of the
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Samir Karaa
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and
(
q0 (x) =
b2
if x ∈ (0, β),
b1 if x ∈ (β, l),
Rl
Rl
where α and β are chosen so that 0 p0 dx = A and 0 q0 dx = B. The supremum of y(l) over E × F is reached by p̄ = p∗0 and q̄ = q0∗∗ .
A counterexample. We show that Theorem 3.2 does not hold if the solutions
of (1.4) are allowed to vanish. Set l = 2π, and let p0 ≡ 1 in (0, l) and
(
0 if x ∈ (0, l0 ),
q0 (x) =
4 if x ∈ (l0 , l),
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where l0 = 3π/2. Then it is easily verified that the solution in (1.4) with (p, q) =
(p0 , q0 ) is
(
1
if x ∈ (0, l0 ),
y0 (x) =
cos 4(x − l0 ) if x ∈ (l0 , l).
Let p̄(x) ≡ q̄(x) ≡ 1 in (0, 2π). The corresponding solution in (1.4) is ȳ(x) =
cos x. We see that ȳ(l) > y0 (l) in spite of q̄ ≺ q0 . The assumption in Theorem 3.1 is also necessary.
Proofs of Theorems 3.1 and 3.2. Necessary conditions on p0 . By the change of
variable u = −y 0 /(py), i.e.,
(3.1)
y(x) = e−
Rx
0
Samir Karaa
pu dt
x ∈ [0, l],
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equation (1.4) is changed into
(3.2)
Rearrangements of the
Coefficients of Ordinary
Differential Equations
u0 − pu2 = q,
u(0) = 0.
The solution of (3.2) is written
Z t
Z t
u(t) =
q(s) exp
p(r)u(r) dr ds.
0
s
In view of (3.1), Problem 3 is equivalent to
Z l
maximising
pu dt subject to
0
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(p, q) ∈ K.
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Let p0 be an extremal function for the infimum problem and p an arbitrary member in K(f0 ). Define
pδ = (1 − δ)p0 + δp,
δ ∈ [0, 1].
We note that this type of variation is not possible in C(f0 ). Let uδ satisfy
u0δ − pδ u2δ = q0 ,
(3.3)
uδ (0) = 0.
Forming the difference of (3.3) and (3.3) with δ = 0, we have
u0δ
−
u00
= pδ (uδ − u0 )(uδ + u0 ) + δ(p −
p0 )u20 .
Therefore,
Rearrangements of the
Coefficients of Ordinary
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Samir Karaa
Z
(uδ − u0 )(t) = δ
Z t
t
2
(p − p0 )u0 exp
pδ (r)(uδ + u0 )(r) dr ds.
0
s
Writing pδ uδ − p0 u0 = pδ (uδ − u0 ) + (pδ − p0 )u0 and integrating over (0, l),
we obtain
Z t
Z l
Z l Z t
2
(pδ u − p0 u0 )dt =
pδ δ
(p − p0 )u0 exp
pδ (uδ + u0 ) dr ds dt
0
0
0
s
Z l
+ δ (p − p0 )u0 dt
0
Z l Z t
Z l
2
pδ exp
pδ (uδ + u0 ) dr dt ds
= δ (p − p0 )u0
s
s
0
Z l
+ δ (p − p0 )u0 dt.
0
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For Problem 3 the left-hand side is nonpositive. Dividing by δ and letting δ →
0+ brings
Z l
(3.4)
(p − p0 )(t)P (t) dt ≤ 0,
for all p ∈ K(f0 ),
0
where P is given in Theorem 3.1. If p0 is an extremal coefficient for Problem 4
then we find
Z l
(3.5)
(p − p0 )(t)P (t) dt ≥ 0,
for all p ∈ K(f0 ).
0
Let us first discuss (3.4). By Ryff’s characterization, there exists σ ∈ Σ such
that P = P ∗ ◦ σ. Substituting p = p∗0 ◦ σ into (3.4) we see that
Z l
Z l
Z l
Z l
∗ ∗
(3.6)
P p0 dt =
P p dt ≤
P p0 dt ≤
P ∗ p∗0 dt.
0
0
0
0
In the last step we used (1.3) which requires that P is nonnegative. This will be
proved later. As a result, equalities hold everywhere in (3.6) and we have
Z ∞ Z
Z ∞ Z
∗
(3.7)
p0 (t) dt ds =
p0 (t) dt ds
0
{P (t)>s}
0
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Samir Karaa
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{P ∗ (t)>s}
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for all s. As
|{P (t) > s}| = |{P ∗ (t) > s}|,
we know that
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Z
Z
p0 (t) dt ≤
{P (t)>s}
{P ∗ (t)>s}
p∗0 (t) dt
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for all s. It follows from (3.7) that
Z
Z
(3.8)
p0 (t) dt =
{P ∗ (t)>s}
{P (t)>s}
(3.9)
ess
inf
{P (t)>s}
p0 (t) ≥ ess
p∗0 (t) dt,
inf
{P (t)≤s}
p0 (t).
for all s. From (3.9) one deduces that if P is increasing on the interval I, then
p0 must be nondecreasing on this interval if we neglect a set of measure zero.
Similarly, if P is decreasing on some interval, p0 will be nonincreasing. If these
relations hold, we say that P and p0 are codependent.
We now return to the function P . We have P (0) = 0 and a straightforward
calculation yields
Z l
q0
0
0
−2
P (t) = q0 1 − 2 y0 y0
p0 (s)y0 (s) ds
p0
t
f0∗∗
that is nonnegative for all t ∈ (0, l). Choosing p =
in the variational
equation (3.4) and integrating by parts gives
Z l
Z l Z t
∗∗
∗∗
0≥
(f0 − p0 )P (t) dt =
(f0 − p0 ) ds d(−P (t)) ≥ 0.
0
0
We used the inequality
Z t
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Z
p0 ds ≥
0
0
Rearrangements of the
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Differential Equations
0
t
f0∗∗ ds,
t ∈ [0, l].
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Consequently,
Z
0
P (t)
t
(f0∗∗ − p0 ) ds = 0,
t ∈ [0, l],
0
and the second part of Theorem 3.1 is proved.
For the supremum problem we use the same arguments. If P = P ∗ ◦ σ,
where σ ∈ Σ, we choose p = p∗∗
0 ◦ σ in (3.5) to obtain
Z l
Z l
Z l
Z l
∗ ∗∗
(3.10)
P p0 dt =
P p dt ≥
P p0 dt ≥
P ∗ p∗∗
0 dt.
0
0
0
0
Thus, there is equality everywhere in (3.10) and
Z ∞ Z
Z ∞ Z
(3.11)
p0 (t) dt ds =
0
Since
{P (t)>s}
Samir Karaa
{P ∗ (t)>s}
0
Rearrangements of the
Coefficients of Ordinary
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∗∗
p0 (t) dt ds.
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Z
p∗∗ (t) dt ≤
{P ∗∗ (t)>s}
Z
p0 (t) dt,
{P (t)>s}
for all s, (3.11) implies that
Z
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Z
p0 (t) dt =
{P (t)>s}
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{P ∗ (t)>s}
p∗∗
0 (t) dt,
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ess
inf
{P (t)>s}
p0 (t) ≥ ess
inf
{P (t)≤s}
p0 (t),
for all s. In this case P and p0 are contra-dependent, i.e. if P is increasing (resp.
decreasing) on an interval I, p0 will be nonincreasing (resp. nondecreasing) on
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I. Choosing p = f0∗ in the variational equation (3.5) and arguing as above, we
prove the second part of Theorem 3.2.
Necessary conditions on q0 . Let q0 be an extremal function for Problem 3. For
q ∈ K(g0 ), we define
qδ = (1 − δ)q0 + δq,
δ ∈ [0, 1].
Let uδ be the solution of
u 0 − p 0 u 2 = qδ ,
(3.12)
u(0) = 0.
Forming the difference of (3.12) and (3.12) with δ = 0, calculations similar
to those of the preceding case allow us to derive the necessary conditions of
optimality
Z l
(q − q0 )(t)Q(t) dt ≤ 0
for all q ∈ K(g0 ),
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0
where
Q(t) =
l
Z
y02 (t)
p0 (s)y0−2 (s) ds.
t
We remark that Q(l) = 0 and
Q0 (t) = 2y0 y00
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Z
l
p0 (s)y0−2 (s) ds − p0
t
is nonpositive on (0, l). For Problem 4, q0 satisfies
Z l
(q − q0 )(t)Q(t) dt ≥ 0
for all q ∈ K(g0 ).
0
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Reasoning as above, we deduce that q0 and Q are codependent for the infimum
problem. The argument for characterizing p0 yields q0 = g0∗ . For the supremum
problem q0 and Q are contra-dependent and we get q0 = g0∗∗ which completes
the proofs.
Existence.
Let m0 denote the infimum of y(l) when (p, q) varies in K and (pn , qn ) a minimizing sequence in K. Let {un } be an associated sequence of solutions in the
Rl
differential equation (3.2) so that limn→∞ 0 pn un dt = m0 . Using weak∗ compactness, we find that (p0 , q0 ) ∈ K such that pn → p and qn → q weakly in
L∞ (0, l). From the expression of un , we see that
Rearrangements of the
Coefficients of Ordinary
Differential Equations
Samir Karaa
Z
un (t) ≤
l
qn (t)e−
Rl
0
pn un ds
dt ≤ ||g0 ||L1 e−m0 .
0
It follows from (3.2) that the sequence {u0n } is uniformly bounded in L∞ (0, l).
By Ascoli’s theorem, there exists a subsequence (we may assume that it is the
original sequence) such that un → u0 uniformly in [0, l]. It is easy to check
that u0 is the solution of (3.2) for (p, q) = (p0 , q0 ). The proof of the supremum
problem is quite the same.
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4.
Problem 6
Suppose that f0 , g0 ∈ L∞
+ (0, l) and f0 ≥ 1 over (0, l). The existence of extremal
couples for Problems 5 and 6 may be proved as above. Let
Z l
y00 2 (t)
y00 (t)
−2
P (t) = 2
p0 (s)y0 (s) ds −
,
p0 (t)
(p0 y0 )(t)
t
Z l
2
Q(t) = y0 (t)
p0 (s)y0 (s)−2 ds,
t ∈ [0, l].
t
Theorem 4.1. Let (p0 , q0 ) be the extremal couple for Problem 6, and y0 an
associated solution in (1.5). In the open set where
Z t
Z t
p0 ds >
f0∗∗ ds
0
0
resp.
Rearrangements of the
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Differential Equations
Samir Karaa
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Z
t
Z
q0 ds <
0
t
g0∗ ds,
0
we have P 0 (t) = 0, resp. Q0 (t) = 0.
Proof. By the change of variable u = y 0 /(py) equation (1.5) is changed into
u0 + pu2 = q,
u(0) = 0,
t ∈ [0, l].
We shall then study the equivalent problem
Z l
max
p u dt, (p, q) ∈ K.
0
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Let (p0 , q0 ) be the extremal couple for Problem 6. Arguing as above, we find
that p0 and q0 satisfy the conditions
Z l
(4.1)
(p − p0 )(t)P (t) dt ≥ 0
for all p ∈ K(f0 ),
0
Z
l
(q − q0 )(t)Q(t) dt ≤ 0
(4.2)
for all q ∈ K(g0 )n
0
where P and Q are given above. Unlike the preceding case, it is difficult here
to know the sign of P and Q. We shall then proceed as above: Let y1 be the
function defined by
Z l
y1 (t) = y0 (t)
p0 (s)y0−2 (s) ds, t ∈ [0, l].
t
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y1 is a solution of the differential equation
0
0
(p−1
0 (x)y (x)) − q0 (x)y(x) = 0,
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Differential Equations
x ∈ (0, l),
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y10 (l)
−1
but y1 (l) = 0 and
= −(y0 /p0 ) (l). Besides, it is easy to see that
0 for all t ∈ (0, l). Let
0
0
y0
y10
y0
y10
−
2,
η=−
+
2.
ξ=
y0 p0 y1 p0
y0 p0 y1 p0
y10 (t)
<
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Then, we have
(4.3)
ξ 0 = 2ξ η p0 ,
η 0 = p0 (ξ 2 + η 2 ) − q0 ,
−1 ,
Z l
2 = η(0).
ξ(0) =
p0 (s)y0−2 (s) ds
0
The key of deciding the sign of P and Q are the following relations
(4.4)
1
Q(t) = ξ(t)−1 ,
2
and
(4.5)
Samir Karaa
−1
1 q0 1
0
.
P (t) =
2 p0 ξ
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In fact, we have
(4.6)
Rearrangements of the
Coefficients of Ordinary
Differential Equations
ξQ = ξy0 y1 =
1
1
(y00 y1 − y0 y10 ) = ,
2p0 (t)
2
and
Z l
q0
0
P (t) = 2 y0 y0
p0 (s)y0−2 (s) ds − q0
p0
t
Z l
q0
0
−2
2y0 y0
p0 (s)y0 (s) ds − p0
=
p0
t
q0
= Q0 (t).
p0
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Relation (4.6) implies that ξ is positive and lim ξ(t) = ∞, t → l−. From (4.3)
it follows that lim sup η(t) ≥ 0, t → l−. Assume now that η changes its sign
on (0, l). Since η(0) > 0, there exists an interval [a, b] ⊂ [0, l) such that for
some c > 0, we have
η(t) ≤ η(a) < 0,
η(t) < 0,
t ∈ [a, a + c],
t ∈ [a, b),
η(b) = 0.
Since η is assumed negative on (a, b), ξ will be decreasing on this interval. (4.4)
and (4.5) imply that P and Q are both increasing on [a, b]. From (4.1) and (4.2)
we see that p0 is nonincreasing and q0 is nondecreasing on this interval. As a
result, we have
0 ≥ η(t) − η(a)
Z t
Z t
2
=
(p0 ξ − q0 ) +
p0 η 2
a
a
2
≥ (t − a) p0 (t)ξ (t) − q0 (t) + η(a)2 ,
t ∈ (a, a + c),
since ess inf (0,l) p0 (t) ≥ 1. Arguing as in [4], we arrive at the following contradiction: η(b) ≤ η(a) < 0. Hence, η is nonnegative and ξ is nondecreasing.
Taking p = f0∗∗ in the variational equation (4.1), we obtain
Z l
Z l Z t
∗∗
∗∗
0≤
(f0 − p0 )P (t) dt =
(f0 − p0 ) ds d(−P (t)) ≤ 0,
0
0
0
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and therefore
0
Z
P (t)
t
(f0∗∗ − p0 ) ds = 0,
t ∈ [0, l]
0
which proves the first part of Theorem 4.1. To complete the proof, we choose
q = g0∗ in (4.2).
Remark 1. For Problem 5, the arguments for deciding the sign of η on (0, l)
break down and the problem requires the development of other arguments.
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References
[1] A. ALVINO, G. TROMBETTI AND P.L. LIONS, On optimization problems
with prescribed rearrangements, Nonlinear Analysis, 13(2) (1989), 185–
220.
[2] C. BANDLE, Isoperimetric Inequalities and Applications, Pitman Monographs and Studies in Mathematics 7. Boston: Pitman, 1980.
[3] M. ESSEN, Optimization and α-Disfocality for ordinary differential equations, Canad. J. Math., 37(2) (1985), 310–323.
[4] M. ESSEN, Optimization and rearrangements of the coefficient in the operator d2 /dt2 − p(t)2 on a finite interval, J. Math. Anal. Appl., 115 (1986),
278–304.
Rearrangements of the
Coefficients of Ordinary
Differential Equations
Samir Karaa
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[5] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cambridge University Press, London/New York, 1934.
[6] J. RYFF, Orbits of L1 -functions under doubly stochastic transformation,
Trans. Amer. Math. Soc., 117 (1965), 92–100.
[7] J. RYFF, Majorized functions and measures, Nederl. Acad. Wekensch.
Indag. Math., 30(4) (1968), 431–437.
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