Volume 9 (2008), Issue 2, Article 38, 5 pp.
NOTE ON AN INTEGRAL INEQUALITY APPLICABLE IN PDEs
V. ČULJAK
D EPARTMENT OF M ATHEMATICS
FACULTY OF C IVIL E NGINEERING
U NIVERSITY OF Z AGREB
K ACICEVA 26, 10 000 Z AGREB , C ROATIA
vera@master.grad.hr
Received 22 April, 2008; accepted 06 May, 2008
Communicated by J. Pečarić
A BSTRACT. 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.
Key words and phrases: Integral inequality, Nonlinear system of PDEs.
2000 Mathematics Subject Classification. Primary 26D15, Secondary 26D99.
1. I NTRODUCTION
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
0
then
(a) If p ≥ 2, the inequality
Z
b
p
Z
f dx ≤ Ap
(1.1)
a
f dx
a
holds for all Ap ≥ 2.
(b) If 1 ≤ p < 2, the inequality
Z b
Z
p
(1.2)
f dx ≤ Ap
a
a+b
2
a+b
2
f dx
a
holds for all Ap ≥ 4.
In this note we improve the optimal Ap for the case 1 < p < 2.
132-08
2
V. Č ULJAK
2. R ESULTS
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
(i) If a + b ≥ 0, then for 1 ≤ p, this inequality holds
Z b
Z a+b
2
p
(2.1)
f dx.
f dx ≤ 2
a
a
(ii) If a + b < 0 then
(a) If p ≥ 2, the inequality
Z b
Z
p
(2.2)
f dx ≤ Ap
a
(2.3)
a+b
2
f dx
a
holds for all Ap ≥ 2.
(b) If 1 < p < 2, the inequality
Z b
Z
p
f dx ≤ Ap
a
a+b
2
f dx
a
p−1
1+xmax
holds for all Ap ≥ 2 1+x
, where 0 < xmax ≤ 1 is the solution of
max
xp−1 (p − 2) + xp−2 (p − 1) − 1 = 0.
(2.4)
(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
Z 0
Z ϕ(x)
f dt =
f dt.
x
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 ϕ(x)
Z x+ϕ(x)
2
p
f dt.
f dt ≤ 2
x
x
In particular, for x = a,
Z
b
p
Z
f dt ≤ 2
a
a+b
2
f dt.
a
If we suppose that x + ϕ(x) < 0 for arbitrary x ∈ [a, 0], then we define a new function
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 38, 5 pp.
http://jipam.vu.edu.au/
N OTE ON AN I NTEGRAL I NEQUALITY
3
ψ : [a, 0] → R by
Z
x+ϕ(x)
2
ψ(x) = Ap
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
Z ϕ(x)
Z
p
f dt ≤ Ap
x
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))
1
x + ϕ(x)
1
x + ϕ(x)
= 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))
1
x + ϕ(x)
− Ap f (x)f (ϕ(x))
= Ap [f (ϕ(x)) − f (x)]f
2
2
+ f p (ϕ(x))f (x) + f p (x)f (ϕ(x)).
For p ≥ 1, if [f (ϕ(x)) − f (x)] ≤ 0, then
f (ϕ(x))ψ 0 (x)
1
x + ϕ(x)
≤ Ap [f (ϕ(x)) − f (x)]f
− Ap f (x)f (ϕ(x))
2
2
+ [f (ϕ(x))f (x) + f (x)f (ϕ(x))]
1
x + ϕ(x)
= 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.
that [f (ϕ(x)) − f (x)] > 0 then using the properties of ϕ, we can conclude that
If we suppose
x+ϕ(x)
f
≤ f (x) and we estimate f (ϕ(x))ψ 0 (x) as follows:
2
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
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 38, 5 pp.
http://jipam.vu.edu.au/
4
V. Č ULJAK
1
1
2
= − Ap f (x) + 2 − Ap f (ϕ(x))f (x)
2
2
1
1
2
≤ − Ap f (x) + 2 − Ap f 2 (ϕ(x))
2
2
1
≤ − (Ap − 4)f 2 (ϕ(x)).
2
0
So, ψ (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
0
Hence, ψ (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
1
1
p
p−1
≤ y − A p z − Ap y + z + y z
2
2
y p−1 1
y
p
= y − Ap z 1 +
+z 1+
2
z
z
1
y
y p−1
≤ 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
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 :
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.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 38, 5 pp.
http://jipam.vu.edu.au/
N OTE ON AN I NTEGRAL I NEQUALITY
5
pn −1
If we consider the sequence pn = 2 − n1 , then the limn→∞ 1+t1+t = 1, but we find that the
pn −1
point tmax where the function 1+t1+t achieves the maximum is a fixed point of the function
)n .
g(x) = (1 − 1+x
n
We use fixed point iteration to find the fixed point for the function g(x) = (1 − 1+x
)100 , by
100
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.
pn −1
When n → ∞, i.e. pn → 2, the point tmax where the function 1+t1+t achieves the maximum
is a fixed point of the function g(x) = e−(1+x) .
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 :
t0
t1
t2
t3
t4
t5
t6
t7
= 0.200000000000000,
= 0.301194211912202,
= 0.272206526577512,
= 0.280212642489384,
= 0.277978184195021,
= 0.278600009316777,
= 0.278426822683543,
= 0.278475046663319
If we consider the sequence pn = 1 +
for t → 0 + .
(c) For p = 1,
1
n
then limn→∞
1+tpn −1
1+t
• 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,
so, the best constant is A1 = 4.
=
2
,
1+t
and supt∈(0,1]
2
1+t
=2
R EFERENCES
[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].
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 38, 5 pp.
http://jipam.vu.edu.au/