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```									SCIENTIA
Series A: Mathematical Sciences, Vol. 15 (2007), 17–22
e
Valpara´ıso, Chile
ISSN 0716-8446
c Universidad T´cnica Federico Santa Mar´ 2007
e                      ıa

Maximum Principles For Some Elliptic Problems

M.M. Al-Mahameed

Abstract. In this paper we introduce a maximum principle for some semilinear
elliptic equations subject to mixed boundary conditions which may be used to
deduce bounds on important quantities in physical problems of iterest.

1. Introduction
2
In [6],maximum principles for the functions p = g(u) | u| + h(u) and
u
2
q = g(u) | u| + c f (s)g(s)ds, c ∈ R, which are deﬁned on solutions of the
0
semilinear partial diﬀerential equation ∆u + f (u) = 0 in some region Ω ⊂ Rn are
found using the classical maximum principle [3]. In [7], a maximum principle for the
function q at a critical point of u under some conditions on ∂Ω is introduced. In [2],
the following result is proved : Let u ∈ C 3 (Ω) be a solution of

∆u + f (u) = 0             in Ω ⊂ Rn ,

u=0                   on ∂Ω.
If the boudary ∂Ω has a nonnegative mean curvature, then the function

u
2
Φ = | u| + 2              f (s)g(s)ds
0

assumes its maximum at a point where u = 0. In [3], maximum principles are derived
for certain functions deﬁned for solutions of the equation

∆u + λρ(x)f (u) = 0

2000 Mathematics Subject Classiﬁcation. Primary 35B50, 35J60.
Key words and phrases. Maximum Principles, elliptic equations.
17
18                                       M.M. AL-MAHAMEED

in some region Ω ⊂ R2 subject to a mixed boundary condition.
In this paper we derive maximum principles for functions deﬁned for solutions of
the semilinear equation
(1.1)                                       ∆u + f (x, u) = 0
in some region Ω ⊂ Rn subject to a mixed boudary condition.
In order to motivate our work, let us ﬁrst look at the one dimensional problem

(1.2)                                       uxx + f (x, u) = 0.

If we multiply (1.2) by ux we get

1 2
(u )x + f (x, u)ux = 0,
2 x
that is

u
1 2
(1.3)                     u +           f (x, s)ds − H(x, u) = constant,
2 x
0

where H(x, u) satisﬁes:

u

Hx (x, u) =                     fx (x, s)ds.
0
Thus we conclude that the function

u

(1.4)                       p = u2 + 2
x                     f (x, s)ds − 2H(x, u)
0

is a constant, where u is a solution of (1.2). It is obvious that p satisﬁes a maximum
principle.
Let u be a solution of (1.1) . We look for a function p of the form

u
2
(1.5)                     p = | u| + 2                     f (x, s)ds − 2H(x, u),
0

where H(x, u) satisﬁes:

u

H,i(x,u) =                     f,i (x, s)ds.
0
MAXIMUM PRINCIPLES FOR SOME ELLIPTIC PROBLEMS                           19

Our goal is to ﬁnd conditions such that (1.5) satisﬁes a maximum principle.
Let us ﬁrst give the following lemma.
Lemma. Let u be a C 3 (Ω) solution of (1.1) with f ∈ C 1 (Ω × R), Ω ⊂ Rn ,
n 2.Then the function p deﬁned by (1.5) takes its maximum either on ∂Ω or at a
critical point of u.
Proof. By diﬀerentiating (1.5) we obtain

(1.6)                        p,i = 2u,j u,ij + 2f u,i

(1.7)             ∆p = p,ii = 2u,ij u,ij + 2u,j u,iij + 2f ∆u + 2f,i u,i .
Now we have

(1.8)                                      ∆u = −f,

(1.9)                                    u,iij = −f,j .
This allows us to rewrite (1.7) as

(1.10)                              ∆p = 2u,ij u,ij − 2f 2 .
From (1.6) and Schwarz’s inequality, it follows that

2
(1.11)        (p,i − 2f u,i )(p,i − 2f u,i ) = 4u,ji u,j u,ki u,k   4u,ij u,ij | u| .
Consequently, by (1.10) and (1.11) , we can write

Lk p,k
(1.12)                                ∆p +         2      0,
| u|
where

1
Lk = 2f u,k − p,k .
2
Hopf’s ﬁrst maximum principle [4] implies the lemma.

2. The Result and its Proof
We give our result by the following theorem.

Theorem. Let u be a C 3 (Ω) solution of the problem
∆u + f (x, u) = 0      in Ω ,
∂u
u=0       on Γ1 , ∂n = 0 , on Γ2 , Γ1 ∪ Γ2 = ∂Ω ,
20                               M.M. AL-MAHAMEED

∂u
where f ∈ C 1 (Ω ×R), Ω is a convex domain in R2 and ∂n denotes the outward
normal derivative. Then the function p deﬁned by (1.5) takes its maximum at a
critical point of u.
Proof. We will show that p cannot attain its maximum on ∂Ω unless it is
attained at a critical point of u which is on Γ2 .
Suppose that p takes its maximum at a point M ∈ Γ1 . Then M cant be a critical
∂u
point of u. Since u = 0 on Γ1 , we have| u| = ∂n and

∂p
(2.1)                            = 2un unn + 2f un ,
∂n
Where un denotes the outward normal derivative.By introducing normal coordinates
in the neighbourhood of the boundary , we can write

(2.2)                          ∆u = unn + kun = −f,
where k denotes the curvature of the boundary.Thus it follows that

∂p
(2.3)                                  = −2ku2 ,
n
∂n
∂p
and since Ω is convex , ∂n 0 at M . This contradicts Hopf’s second maximum
principle [5].
We now suppose that p takes its maximum at M ∈ Γ2 and that M is not a
critical point of u. Since ∂n = 0 on Γ2 , we have | u| = ∂u and
∂u
∂t

∂p
(2.4)                                   = 2ut utn ,
∂n
where ut denotes the tangential derivative of u. In terms of normal coordinates in the
neighbourhood of the boundary , we have

(2.5)                             utn = unt − kut ,
so that

∂p
= −2ku2 on Γ2 .
t
∂n
Thus we again have a contradiction of the second maximum principle when Ω is
convex. The lemma, and our calculations , gives the theorem.
Example. Let u ∈ C 3 (Ω) be a positive solution of the problem

∆u + 4u − (x2 + x2 ) exp(α2 − x2 − x2 ) = 0
1    2             1    2          inΩ,

u=0        on ∂Ω,
MAXIMUM PRINCIPLES FOR SOME ELLIPTIC PROBLEMS                                  21

where

Ω = {x = (x1 , x2 )\ |x| < α}

and

f (x, u) = 4u − (x2 + x2 ) exp(α2 − x2 − x2 ),
1    2             1    2

it follows from the theorem (2.1) that

u
2
| u| + 2                4s − (x2 + x2 ) exp(α2 − x2 − x2 )ds − 2H(x, u)
1    2             1    2
0

           u


max 2                 4s − (x2 + x2 ) exp(α2 − x2 − x2 )ds − 2H(x, u) 
1    2             1    2
Ω∪ ∂Ω
0

or

2
| u|                     max 4u2 − 2(x2 + x2 ) exp(α2 − x2 − x2 )u
1    2             1    2
Ω∪ ∂Ω

− 4u2 − 2(x2 + x2 ) exp(α2 − x2 − x2 )u
1    2             1    2

.
From the above inequality , we get

2                          2
| u|                  4(u2 − u2 ) + 2α2 eα uM ,
M

where uM is the maximum of u in Ω ∪ ∂Ω.

3. Concluding Remarks

1. One can prove the result of the lemma for the function
u
2
p = g(u) | u| + 2               f (x, s)g(s)ds − 2H(x, u) with suitable assumptions on g(u)
0
as in [5].
2. Theorem 2.1 is also valid for n > 2, [5].
3.One may give an extention of the maximum principle for a uniformly elliptic
equation Lu + f (x, u) = 0 under suitable assumptions, [5].
22                                    M.M. AL-MAHAMEED

References
[1] A. Greco,A Maximum principle for some second order elliptic semilinear
equations.Rend. Sem. Fac. Sci. Univ. Cagliari. 59(2)(1989), 147-154.
[2] L.E. Payne and I.Stakgold, Nonlinear problems in nuclear reactor analysis.Lecture Notes in math.
322(1972) 298-307.
[3] P.w.Schaefer and R.P. Sperb.Maximum principles and bounds in some inhomogeneous elliptic
boundary value problems.SIAM J.Math. Anal.8(1977), 871-878.
[4] M.H.Protter and Weinberger, Maximum principles in diﬀefential equations, Prentice-
Hall,Englewood Cliﬀs, N.J.,1984.
[5] M.H.Protter and Weinberger, Maximum principles in diﬀefential equations, Prentice-
Hall,Englewood Cliﬀs, N.J.,1984.
[6] M.M.Al-Mahameed, Maximum principles for semilinear elliptic partial diﬀerential equations, Far
East. j. applied Math.3(3)(1999), 287-292.
[7] M.M.Al-Mahameed ,A maximum principle for the P-function at a point where grad u = 0 in some
region Ω ⊂ Rn .J .Ins.Math.and Comp.Sci..14(1)(2001), 65-71.

Received 28 06 2006, revised 06 03 2007

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