NOTE ON AN INTEGRAL INEQUALITY
APPLICABLE IN PDEs
Note on an Integral Inequality
V. ČULJAK
Department of Mathematics
Faculty of Civil Engineering
University of Zagreb
Kaciceva 26, 10 000 Zagreb, Croatia
V. Čuljak
vol. 9, iss. 2, art. 38, 2008
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Contents
EMail: vera@master.grad.hr
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06 May, 2008
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Communicated by:
J. Pečarić
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2000 AMS Sub. Class.:
Primary 26D15, Secondary 26D99.
Key words:
Integral inequality, Nonlinear system of PDEs.
Abstract:
The article presents and refines the results which were proven in [1]. We give
a condition for obtaining the optimal constant of the integral inequality for the
numerical analysis of a nonlinear system of PDEs.
Received:
22 April, 2008
Accepted:
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Contents
1
Introduction
3
2
Results
4
Note on an Integral Inequality
V. Čuljak
vol. 9, iss. 2, art. 38, 2008
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Contents
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1.
Introduction
In [1] the following problem is considered and its application to nonlinear system of
PDEs is described.
Theorem A. Let a, b ∈ R, a < 0, b > 0 and f ∈ C[a, b], such that: 0 < f ≤ 1 on
[a, b], f is decreasing on [a, 0] and
Z 0
Z b
f dx =
f dx
a
Note on an Integral Inequality
V. Čuljak
vol. 9, iss. 2, art. 38, 2008
0
then
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(a) If p ≥ 2, the inequality
Z
(1.1)
a
b
f p dx ≤ Ap
Z
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a+b
2
f dx
holds for all Ap ≥ 2.
(b) If 1 ≤ p < 2, the inequality
Z b
Z
p
(1.2)
f dx ≤ Ap
a
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a
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a+b
2
f dx
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a
holds for all Ap ≥ 4.
In this note we improve the optimal Ap for the case 1 < p < 2.
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2.
Results
Theorem 2.1. Let a, b ∈ R, a < 0, b > 0 and f ∈ C[a, b], such that 0 < f ≤ 1 on
[a, b], f is decreasing on [a, 0] and
Z 0
Z b
f dx =
f dx.
a
0
Note on an Integral Inequality
(i) If a + b ≥ 0, then for 1 ≤ p, this inequality holds
Z b
Z a+b
2
p
f dx.
(2.1)
f dx ≤ 2
a
a
V. Čuljak
vol. 9, iss. 2, art. 38, 2008
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(ii) If a + b < 0 then
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(a) If p ≥ 2, the inequality
Z
b
p
Z
f dx ≤ Ap
(2.2)
a
a+b
2
f dx
a
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holds for all Ap ≥ 2.
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(b) If 1 < p < 2, the inequality
Z b
Z
p
(2.3)
f dx ≤ Ap
a
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a+b
2
f dx
a
p−1
max
holds for all Ap ≥ 2 1+x
, where 0 < xmax ≤ 1 is the solution of
1+xmax
(2.4)
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xp−1 (p − 2) + xp−2 (p − 1) − 1 = 0.
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(c) For p = 1 the inequality
Z
b
a+b
2
Z
f dx ≤ 4
(2.5)
f dx
a
a
holds.
Proof. As in the proof in [1], we consider two cases: (i) a + b ≥ 0 and (ii) a + b < 0.
(i) First, we suppose that a + b ≥ 0. Using the properties of the function f, we
conclude, for p ≥ 1, that:
Z b
Z b
Z 0
Z a+b
2
p
f dx.
f dx ≤
f dx = 2
f dx ≤ 2
a
a
a
a
The constant Ap = 2 is the best possible. To prove sharpness, we choose f = 1.
(ii) Now we suppose that a + b < 0. Let ϕ : [a, 0] → [0, b] be a function with the
property
0
Z
ϕ(x)
Z
f dt =
x
f dt.
ϕ(x)
p
Z
f dt ≤ 2
x+ϕ(x)
2
x
In particular, for x = a,
b
p
Z
f dt ≤ 2
a
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a+b
2
f dt.
a
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f dt.
x
Z
V. Čuljak
vol. 9, iss. 2, art. 38, 2008
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0
So, ϕ(x) is differentiable and ϕ(a) = b, ϕ(0) = 0.
For arbitrary x ∈ [a, 0], such that x + ϕ(x) ≥ 0, according to case (i) for p ≥ 1,
we obtain the inequality
Z
Note on an Integral Inequality
If we suppose that x + ϕ(x) < 0 for arbitrary x ∈ [a, 0], then we define a new
function
ψ : [a, 0] → R by
Z
ψ(x) = Ap
x+ϕ(x)
2
Z
ϕ(x)
f dt −
x
f p dt.
x
The function ψ is differentiable and
x + ϕ(x)
1
0
0
ψ (x) = Ap (1 + ϕ (x))f
− Ap f (x) − f p (ϕ(x))ϕ0 (x) + f p (x)
2
2
and ψ(0) = 0.
If we prove that ψ 0 (x) ≤ 0 then the inequality
Note on an Integral Inequality
V. Čuljak
vol. 9, iss. 2, art. 38, 2008
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Z
x
ϕ(x)
f p dt ≤ Ap
Z
x+ϕ(x)
2
f dt
x
holds.
Using the properties of the functions f , ϕ and the fact that f (ϕ(x))ϕ0 (x) =
−f (x), we consider f (ϕ(x))ψ 0 (x) and try to conclude that f (ϕ(x))ψ 0 (x) ≤ 0 as
follows:
f (ϕ(x))ψ 0 (x)
1
x + ϕ(x)
0
p
0
p
= f (ϕ(x)) Ap (1 + ϕ (x))f
− Ap f (x) − f (ϕ(x))ϕ (x) + f (x)
2
2
1
x + ϕ(x)
1
x + ϕ(x)
0
= Ap f (ϕ(x))f
+ Ap f (ϕ(x))ϕ (x)f
− Ap f (x)f (ϕ(x))
2
2
2
2
− f p (ϕ(x))ϕ0 (x)f (ϕ(x)) + f p (x)f (ϕ(x))
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1
x + ϕ(x)
x + ϕ(x)
1
= Ap f (ϕ(x))f
− Ap f (x)f
− Ap f (x)f (ϕ(x))
2
2
2
2
+ f p (ϕ(x))f (x) + f p (x)f (ϕ(x))
x + ϕ(x)
1
= Ap [f (ϕ(x)) − f (x)]f
− Ap f (x)f (ϕ(x))
2
2
+ f p (ϕ(x))f (x) + f p (x)f (ϕ(x)).
Note on an Integral Inequality
For p ≥ 1, if [f (ϕ(x)) − f (x)] ≤ 0, then
V. Čuljak
vol. 9, iss. 2, art. 38, 2008
f (ϕ(x))ψ 0 (x)
1
x + ϕ(x)
− Ap f (x)f (ϕ(x))
≤ Ap [f (ϕ(x)) − f (x)]f
2
2
+ [f (ϕ(x))f (x) + f (x)f (ϕ(x))]
x + ϕ(x)
1
= Ap [f (ϕ(x)) − f (x)]f
− (Ap − 2)f (x)f (ϕ(x)).
2
2
Then, obviously, ψ 0 (x) ≤ 0 for Ap − 2 ≥ 0.
If we suppose that [f(ϕ(x)) − f (x)] > 0 then using the properties of ϕ, we can
conclude that f x+ϕ(x)
≤ f (x) and we estimate f (ϕ(x))ψ 0 (x) as follows:
2
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f (ϕ(x))ψ 0 (x)
1
≤ Ap [f (ϕ(x)) − f (x)]f (x) − Ap f (x)f (ϕ(x)) + f p (ϕ(x))f (x) + f p (x)f (ϕ(x))
2
1
≤ Ap [f (ϕ(x)) − f (x)]f (x) − Ap f (x)f (ϕ(x)) + f (ϕ(x))f (x) + f (x)f (ϕ(x))
2
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1
1
1
1
2
2
= − Ap f (x) + 2 − Ap f (ϕ(x))f (x) ≤ − Ap f (x) + 2 − Ap f 2 (ϕ(x))
2
2
2
2
1
≤ − (Ap − 4)f 2 (ϕ(x)).
2
So, ψ 0 (x) ≤ 0 for Ap − 4 ≥ 0.
Now, we will consider the sign of f (ϕ(x))ψ 0 (x) for p = 1, p ≥ 2, and 1 < p < 2.
(a) For p ≥ 2, we try to improve the constant Ap ≥ 4 for the case a + b < 0 and
[f (ϕ(x)) − f (x)] > 0. We can estimate f (ϕ(x))ψ 0 (x) as follows:
f (ϕ(x))ψ 0 (x)
1
≤ Ap [f (ϕ(x)) − f (x)]f (x) − Ap f (x)f (ϕ(x)) + f p (ϕ(x))f (x) + f p (x)f (ϕ(x))
2
1
≤ Ap [f (ϕ(x)) − f (x)]f (x) − Ap f (x)f (ϕ(x)) + f 2 (ϕ(x))f (x) + f 2 (x)f (ϕ(x))
2
1
≤ f (x)[f (x) + f (ϕ(x))][2f (ϕ(x)) − Ap ].
2
Hence, ψ 0 (x) ≤ 0 for Ap ≥ 2.
(b) For 1 < p < 2, we can improve the constant Ap ≥ 4 for the case a + b < 0
and [f (ϕ(x)) − f (x)] > 0. We can estimate f (ϕ(x))ψ 0 (x) (for 0 < f (x) = y <
f (ϕ(x)) = z ≤ 1), as follows:
f (ϕ(x))ψ 0 (x)
1
≤ Ap [f (ϕ(x)) − f (x)]f (x) − Ap f (x)f (ϕ(x)) + f p (ϕ(x))f (x) + f p (x)f (ϕ(x))
2
Note on an Integral Inequality
V. Čuljak
vol. 9, iss. 2, art. 38, 2008
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y p−1 1
1
1
y
p
p−1
p
≤ y − Ap z − Ap y + z + y z = y − A p z 1 +
+z 1+
2
2
2
z
z
p−1
1
y
y
≤ yz − Ap 1 +
+1+
.
2
z
z
So, we conclude that ψ 0 (x) ≤ 0 if
1
p−1
− Ap (1 + t) + 1 + t ) < 0,
2
Note on an Integral Inequality
V. Čuljak
vol. 9, iss. 2, art. 38, 2008
for 0 < t = yz ≤ 1.
p−1
Therefore, for 1 < p < 2 the constant Ap ≥ 2 max0<t≤1 1+t
.
1+t
1+tp−1
The function 1+t is concave on (0, 1] and the point tmax where the maximum
is achieved is a root of the equation
tp−1 (p − 2) + tp−2 (p − 1) − 1 = 0.
Numerically we get the following values of Ap :
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for p = 1.01,
the constant Ap ≥ 3.8774,
for p = 1.99,
the constant Ap ≥ 2.0056,
for p = 1.9999, the constant Ap ≥ 2.0001.
pn −1
If we consider the sequence pn = 2 − n1 , then the limn→∞ 1+t1+t = 1, but we find
pn −1
that the point tmax where the function 1+t1+t achieves the maximum is a fixed point
of the function g(x) = (1 − 1+x
)n .
n
We use fixed point iteration to find the fixed point for the function g(x) = (1 −
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1+x 100
) ,
100
by starting with t0 = 0.2 and iterating tk = g(tk−1 ), k = 1, 2, ...7 :
t0
t1
t2
t3
t4
t5
t6
t7
= 0.200000000000000,
= 0.299016021496423,
= 0.270488141422931,
= 0.278419068898826,
= 0.276191402436672,
= 0.276815328895026,
= 0.276640438571483,
= 0.276689450339917.
Note on an Integral Inequality
V. Čuljak
vol. 9, iss. 2, art. 38, 2008
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1+tpn −1
1+t
When n → ∞, i.e. pn → 2, the point tmax where the function
achieves
−(1+x)
the maximum is a fixed point of the function g(x) = e
.
We use fixed point iteration to find the fixed point for the function g(x) =
e−(1+x) , by starting with t0 = 0.2 and iterating tk = g(tk−1 ), k = 1, 2, ...7 :
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t0
t1
t2
t3
t4
t5
t6
t7
= 0.200000000000000,
= 0.301194211912202,
= 0.272206526577512,
= 0.280212642489384,
= 0.277978184195021,
= 0.278600009316777,
= 0.278426822683543,
= 0.278475046663319
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If we consider the sequence pn = 1 +
2
supt∈(0,1] 1+t
= 2 for t → 0 + .
(c) For p = 1,
1
n
then limn→∞
1+tpn −1
1+t
=
2
,
1+t
and
• if [f (ϕ(x)) − f (x)] ≤ 0 then ψ 0 (x) ≤ 0 for A1 − 2 ≥ 0;
• if [f (ϕ(x)) − f (x)] > 0 then ψ 0 (x) ≤ 0 for A1 − 4 ≥ 0,
Note on an Integral Inequality
so, the best constant is A1 = 4.
V. Čuljak
vol. 9, iss. 2, art. 38, 2008
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References
[1] V. JOVANOVIĆ, On an inequality in nonlinear thermoelasticity, J. Inequal. Pure
Appl. Math., 8(4) (2007), Art. 105. [ONLINE: http://jipam.vu.edu.
au/article.php?sid=916].
Note on an Integral Inequality
V. Čuljak
vol. 9, iss. 2, art. 38, 2008
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