一、复杂电路：

1、混联情况： 见图

2、两条以上支路含源支路的并联情况：见图 二、方程组

1、第一方程组（节点方程） II,,,入出

2、第二方程组（回路方程） = ,,,,IR,,

3、理论依据： jdS,0Edl,0,,,SL

4、符号法则：

（1）、关于电流：因为电流是双向标量，对未知电流，可以先任意哟定其正方向，

再由代数量的正负来判断电流的实际流向。

（2）、关于：先选择绕行方向，若绕行方向从电源负极到正极，则取“+”号，,,

否则取“-”号。同样，若,为未知时，可先假设极性，然后由代数量的正负判断实际极

性。

（3）、关于的正负：当绕行方向和电流的流向相同时，取“+”号，否则取“-”IR

号。即电压降为正，升为负。

5、独立方程数：

独立回路是指该回路中至少包含一条其他回路所不包含的新支路。

由网络拓扑学得：对P 条支路、N个节点、M 个回路，

pmn,，,1

6、基氏方程可以解决一切线形电路的计算问题。要求精炼，务必掌握。

三、实例求解：

例题见习题解：

巧列基尔霍夫方程

2012年3第28卷第3期 四川教育学院学报 月Mar(2012COLLEGE OF EDUCATION V01(28 JOURNAL OF SICHUAN

巧列基尔霍夫方程

张容

(四川教育学院物理与电子技术系，成都611130)宰 摘要:列基尔霍夫方程是解复

杂直流电路的关键，通过对基尔霍夫节点电流方程和回路电压方程的变形，并用容易懂记忆的规则，使列方程变得容易，难点教学变为易事，同时还进一步理解了其本质。

关键词:基尔霍夫方程;电流;电压

doi:10(3969,j(issn(1000—5757(2012。03(117

文献标志码:A 文章编号:1000-5757{2012}03--0117-02 中图分类号:0411

从负极进入电源时，其电动势前写“+”号，否则写 l引言

“一”号;当绕行方向与电阻的电流正方向相同时， 基尔霍夫方程包含两个方程，即节点电流方程 和回

该电阻的IR项前写“+”号，否则写“一”号。同时 路电压方程，也叫基尔霍夫第一定律和第二定

(2)式中也存在“双重正负号”问题，即:(1)当,为 律。节点电流方程比较简单，而回路电压方程在列

未知量时，也可先假设一个参考正、负极，若方程的时候较为麻烦一些，学生不易掌握，成为教学

列式解 中的一个难点。笔者经过多年教学，总结出基尔霍 方程后得到,>0，则表示,的真实

夫方程的简单列法，特别是判断回路电压方程正、负 性相同;若s

(2)I本身可正可负，取决

于其正方向与实际方向的关系。 号的方法，只要学生背诵16字的口诀即可掌握，方

3解决方法 便许多。

将(1)式变为:2现行教材处理基尔霍夫方程组的方法

(3) 目前大多数高校《电磁学》 课程一般采用由 (Y,^=(Y,f (I

梁灿彬主编、高教社出版的《电磁学》教材，教材中 即将流入某节点的电流之和(而不是代数和)

等式的左边，流出某节点的电流之和写在等式右边， 节点电流方程为: 写在

避免了前面的“?”号。在本质上来说，学生也好理 其中(1) 三(?t)=0

“?”号不可省略，当某支路电流的正方向指向 解，根据恒定电流的恒定条件:导体内各处的载流子 节点时用“一”，背离节点时用“+”，而每一t本身 尽管都在向前移动，但它们原来的位置又被后

载 续的 则仍为一可正可负的代数量，代数量的正负则反映

电流的实际方向，当不知‘的流向时，可沿电路先假 流子所占据，只要单位时间内从任一闭曲面的一

部分流出去的电荷等于从该面其他部分流进的电 设一个参考流向，若

荷，空间各点的电荷密度就不随时间变化，在节点处 列式解方程后得t>0，则表示

也是如此。 五的实际流向与参考流向相同;若t<0，则相反。>

回路电压方程为: 将(2)式变为:

(4)(2) 三(?占;)+三(?‘Ri)=0 三(?占‘)=三(?liR{)

方程(2)中的正、负号应由以下规则确定:任意选定 即表述为在电路的任何闭合回路中各段电压的代数 一个绕行回路的方向(叫做绕行方向)，写绕行方向 和等于零。在本质上学生也容易理解:当从电路的

?收稿日期:2011-11(28 作者简介:张容(1965一)，男，四川西充人，副教授，研究方

向:物理教学和物理教育。

117

万方数据

四川教育学院学报 2012年3月

任意一点出发，沿任意路径绕行，必然伴随着电位的 解:由(3)式可得: 升高和降低，当绕行一

M点的节点电流方程:j。+L=12 路时，由于周回到起点构成一个闭合电

电位的单值性，同一点电位是相等的，则 N点的节点电流方程:,2+厶=,5 电位升、降

取顺时针方向为回路绕行方向，由(4)式可得 的代数和必然为零。(4)式中正、负号由

回路电压方程为: 以下规则确定:任意选定一个绕行回路方向，当绕行 方向与‘的流向相同时，，iRi前取正号，相反时则取 一占1+，1R1+占2+，2R2一厶R3=0 负号;当

例2:求图2所示支路ab的电势差u止( 绕行方向若先遇见电源的正极时，则占;前 取

正号，若先遇见电源的负极则占i前取负号。概括 为16字的口诀“同向取正。反向取负;遇正取正，遇 负取负”。该确定“?”号的规则同样适用于求任意 崩产 图2一段复杂的含源恒定电流电路的电势差，只是将绕

行方向变为选定的巡行方向(a_b): 解:由(5)式可得:

(5) U曲=玑一仉=三(?占‘)+三(?‘Ri) ,=玑一巩

4应用举例 =占1+Iirl+‘R1一占2一，2r2一，2尺2+F3一，2r3

5结语例l:列出图l所示M、N点的节点电流方程和

回路电压方程。 综上所述:通过对基尔霍夫方程的变形，使列基

尔霍夫方程变得容易，学生掌握起来十分简便。同

时加深了对恒定电流的恒定条件及电势差的进一步

理解，从笔者的教学实践来看，原来需多次讲解，反

复练习才能过关的问题，现在一次讲解后，绝大多数

学生都掌握了，达到了事半功倍的教学效果。 参考文

献:

[1] 梁灿彬，秦光戎(电磁学(第二版)[M】(北京:高等教

育出版社，2004:141( 图1 Klrchhoff Ingeniously Llsthag Equation

ZHANG Roagof and Electronic of 61 1 Technology，Sichuan College Education，Chengdu 130，China) (Department Physics is the to the solution of d(C(drcnit(The deformation of Kirchhoff Kirchhoff Abstract:Listing equation key complex and and the return mute of mem- node dectric current equation voltage equation the印pficafion easily—undentood—and

to and ofizod rule make fist undemtand i协esoerlce( Call di伍cuhy easy，and equation teaching help

current;voltageKey words:Kirchhoff equation;electric

(责任编辑:刘春林责任校对:林子)

118

万方数据

巧列基尔霍夫方程

2012 28 3 3 第 卷 第 期四川教育学院学报年 月 Vol, 28JOURNAL OF SICHUAN COLLEGE OF EDUCATIONMar, 2012

巧列基尔霍夫方程

张容

, ( ，611130)四川教育学院 物理与电子技术系成都

，，: 摘 要列基尔霍夫方程是解复杂直流电路的关键通过对基尔霍夫节点电流方程和回路电压方程的变形并

，，，。 用容易懂记忆的规则使列方程变得容易难点教学变为易事同时还进一步理解了其本质

: ; ; 关键词基尔霍夫方程电流电压

doi: 10, 3969 / j, issn, 1000 ) 5757, 2012, 03, 117

:0411A1000-5757(2012)03-0117-02::中图分类号文献标志码文章编号

+ ，“”，引言号否从负极进入电源时其电动势前写

则写，基尔霍夫方程包含两个方程即节点电流方程 “ )” ; 号当绕行方向与电阻的电流正方向相，回路电压方程也叫基尔霍夫第一定律和第二定 同时IR + ) “”，“”该电阻的 项前写号否则写。，节点电流方程比较简单而回路电压方程在列 。号同时( 2) : ( 1) “”，式中也存在双重正负号问题即当 ，，程的时候较为麻烦一些学生不易掌握成为教学 ε 为 ，、，未知量时也可先假设一个参考正负极若列式解。，的一个难点笔者经过多年教学总结出基尔霍

,0 ，、 方程后得到 ε 则表示 ε 的真实正负与参考极，、方程的简单列法特别是判断回路电压方程正负

; , 0，。( 2) I ， 性相同若 ε 则相反本身可正可负取决，16 ，的方法只要学生背诵 字的口诀即可掌握方

。许多 。于其正方向与实际方向的关系

3 解决方法 现行教材处理基尔霍夫方程组的方法 ,1, ( 1) :将式变为 《》目前大多数高校电磁学课程一般采用由

、《》，灿彬主编高教社出版的电磁学教材教材中 II( 3)= Σ Σ 入出:点电流方程为 ( ) 即将流入某节点的电流之和而不是代数和写在

，等式的左边流出某节点的电流之和写在等式右边

= 0( 1)( ? I) ? “”。，Σ避免了前面的号在本质上来说学生也 i 好理 “ ? ”，: ，中号不可省略当某支路电流的正方向导体内各处的载流子解根据恒定电流的恒定条件

指向 ，尽管都在向前移动但它们原来的位置又被后续的

) ， + ，“”“”点时用背离节点时用而每 ，载流子所占据只要单位时间内从任一闭曲面的一I，一 本身 仍为一可正可负的代数量代数量的i 部分流出去的电荷等于从该面其他部分流进的电

，I，正负则反映 流的实际方向当不知 的流向时i ，，荷空间各点的电荷密度就不随时间变化在节点处

，可沿电路先假 一个参考流向若列式解方程后得 ; I 。的实际流向与参考流向相同若 ,0 ，。也是如此 则相反iI,0 ，则表示 ( 2) :将式变为 i :教材中回路电压方程为( ? ) + ( ? IR)ΣεΣ ( ? ) = ( ? IR)ΣεΣ i i i i i i = 0( 2)( 4)

( 2) : 、程中的正负号应由以下规则确定任意选定即表述为在电路的任何闭合回路中各段电压的代数

:+ I= I，M I升高和降低当绕行一周回到起点构成一个闭合电点的节点电流方程 1 4 2N :，，，点的节点电流方程 路时由于电位的单值性同一点电位是相等的则 I+ I= I 2 3 5、。( 4 ) 、电位升降的代数和必然为零式中正负号由 ( 4 ) ，取顺时针方向为回路绕行方向由式可 : ，以下规则确定任意选定一个绕行回路方向当绕行 :回路电压方程为 IIR，，方向与 的流向相同时前取正号相反时则取 i i i ) + IR+ + IR)I R= 0εε 1 1 1 2 2 2 2 3; ，负号当绕行方向若先遇见电源的正极时则 ε前 i 2: ab U,2 例 求图 所示支路 的电势差 ，ab 。取正号若先遇见电源的负极则 ε前取负号概括 i

16 “，;，为 字的口诀同向取正反向取负遇正取正

”。“ ?” 遇 负取负该确定号的规则同样适用

，于求任意 一段复杂的含源恒定电流电路的电势差 ( ab) :行方向变为选定的巡行方向只是将绕 ? 2图

: ( 5) :解由式可得

= U)U a h U= U)U ( 5)U= ( ? ) + ( ? IR)ΣεΣ aba babi i i

4 + Ir+ IR) )I r)I R+ )I r= 应用举例εεε 1 1 1 1 1 2 2 2 2 2 3 2 31: 1 M、N 例 列出图 所示 点的节点电流方程和 5 结语

。回路电压方程 : ，综上所述通过对基尔霍夫方程的变形使列

，。 尔霍夫方程变得容易学生掌握起来十分简便

时加深了对恒定电流的恒定条件及电势差的进一

，，，理解从笔者的教学实践来看原来需多次讲解

，，复练习才能过关的问题现在一次讲解后绝大多

，。学生都掌握了达到了事半功倍的教学效果

:参考文献

,1, ，, ( ) ,M,. : 梁灿彬秦光戎电磁学第二版北京高等 ，2004: 141, 育出版社 1图

Ingeniously Listing Kirchhoff Equation

ZHANG Rong

( Department Phof ysics and Electronic Technology，Sichuan College of Education，Chengdu 611130，China) Abstract: Listing Kirchhoff equation is the key to theso lution of complex d, c, circuit, The deformation of Kirchhoff

node eectrc current equaton and the return routevo tage equaton and thea ppcaton of easy )unde rstood) an d mem- liililiiil

orized rule can makel ist equation and difficulty teaching easy，andh elp to understanidts essence, Key words: Kirchhoff equation; electric current; v oltage

基尔霍夫方程

J. Math. Anal. Appl. 317(2006)

456–463

www.elsevier.com/locate/jmaa

Sign changing solutions of Kirchhoff type problems

via invariant sets of descent ?ow

Zhitao Zhang a , 1, Kanishka Perera b , ?, 2

a Academy of Mathematics and Systems Science, Institute of Mathematics, Chinese Academy of Sciences,

Beijing 100080, PR China

b Department of Mathematical Sciences, Florida Institute of Technology, Melbourne, FL 32901, USA

Received 11September 2004Available online 8September 2005

Submitted by P. Smith

Abstract

We obtain sign changing solutions of a class of nonlocal quasilinear elliptic boundary value problems using variational methods and invariant sets of descent ?ow.?2005Elsevier Inc. All rights reserved.

Keywords:Nonlocal problems; Kirchhoff’sequation; Variational methods; Invariant sets of descent ?ow

1. Introduction

In this paper we obtain sign changing solutions of the problem

?a +b Ω|?u |2Δu=f (x,u) in Ω,u =0on ?Ω,

(1.1)

where Ωis a smooth bounded domain in R n , a, b >0, and f (x,t) is locally Lipschitz continuous

in t ∈R , uniformly in x ∈, and subcritical:

*Corresponding author.

E-mail addresses:zzt@math.ac.cn(Z.Zhang), kperera@?t.edu(K.Perera). URL:http://my.?t.edu/~kperera/(K.Perera).

1Supported in part by the National Natural Science Foundation of China, Ky and Yu-Fen Fan Endowment of the AMS, Florida Institute of Technology, and the Humboldt Foundation. 2Supported in part by the National Science Foundation.

0022-247X/$–see front matter ?2005Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2005.06.102

Z. Zhang, K. Perera /J. Math. Anal. Appl. 317(2006)456–463457

f (x,t) C |t |p ?1+1

for some 2

<>

∞,

2n ,

n 3, n =1, 2,

(1.2)

where C denotes a generic positive constant.

This problem is related to the stationary analogue of the equation

u tt ?a +b |?u |2Δu=g(x,t)

Ω

(1.3)

proposed by Kirchhoff [11]as an extension of the classical D’Alembert’swave equation for free vibrations of elastic strings. Kirchhoff’smodel takes into account the changes in length of the string produced by transverse vibrations. Some early classical studies of Kirchhoff equations were Bernstein [5]and Poho?aev[14].However, Eq. (1.3)received much attention only after Lions [12]proposed an abstract framework to the problem. Some interesting results can be found, for example, in [4,6,10].More recently Alves et al. [2]and Ma and Rivera [13]obtained positive solutions of such problems by variational methods. Similar nonlocal problems also model several physical and biological systems where u describes a process which depends on the average of itself, for example the population density, see [1,3,7,8,17].We assume that

tf (x,t) 0and consider three cases:(i)4-sublinear case:p <4, (ii)asymptotically="" 4-linear="">

f (x,t)

=μuniformly in x,

|t |→∞bt lim

(iii)4-superlinear case:

?ν>4:νF(x,t) tf (x,t), |t |large ,

t

where F (x,t) =0f (x,s) ds , which implies

F (x,t) C |t |ν?1. By (1.2)and (1.5)(respectively(1.7)),

4 p <2?respectively><ν p=""><2?, so="" n="1," 2,="" or="" 3in="" (ii)and="">

Case (ii)leads us to the nonlinear eigenvalue problem

? u 2Δu=μu3in Ω,u =0on ?Ω,whose eigenvalues are the critical values of the functional

1

(Ω):u 4=1. I (u)= u 4, u ∈S :=u ∈H =H 0

Ω

(1.4)

(1.5)

(1.6)

(1.7)

(1.8)

(1.9)

(1.10)

458Z. Zhang, K. Perera /J. Math. Anal. Appl. 317(2006)456–463

We will see in the next section that I satis?esthe Palais–Smalecondition (PS)and that the ?rsteigenvalue μ1>0obtained by minimizing I has an eigenfunction ψ>0. We de?nea second eigenvalue μ1by

μ2:=inf

γ∈Γu ∈γ([0, 1])

max I (u),(1.11)

where Γis the class of paths γ∈C([0, 1], S) joining ±ψsuch that γ∪(?γ) is non-self-intersecting.

We are now ready to state our main result. Denote by 0<><λ2 ···the="" dirichlet="" eigen-values="" of="" ?δon="">

Theorem 1.1. Problem (1.1)has a positive solution, a negative solution, and a sign changing solution in the following cases :(i)p <>

aλ2

t , |t |small , 2

(ii)(1.5)and (1.12)hold and μ<>

(iii)(1.5)holds, μ>μ2is not an eigenvalue of (1.9), and

?λ>λ2:

F (x,t)

aλ12

t , 2

(iv)(1.6)and (1.13)hold.

F (x,t)

2. Variational setting

Lemma 2.1. I satis?es(PS), i.e., every sequence (uj ) in H such that I (uj ) is bounded and I (uj ) →0, called a (PS)sequence, has a convergent subsequence.

Proof. Since u j is bounded, for a subsequence, u j converges to some u weakly in H and strongly in L 4(Ω). Denoting by

3

P j v =v ?u j v u j (2.1)

Ω

(1.12)

|t |small , (1.13)

the projection of v ∈H onto the tangent space to S at u j , we have

1

I u j 2?u j ·?(uj ?u) =I (uj ) u 3(u?u) +(u), P (u?u) →0, j j j j j

4

Ω

Ω

(2.2)

so, passing to a subsequence, u j →0or u . 2

Since I is bounded from below and satis?es(PS),μ1:=inf I (u)

u ∈S

(2.3)

is achieved and hence positive. If ψis a minimizer, then so is |ψ|, so we may assume that ψ 0. Since ψis a nontrivial solution of (1.9),ψ>0in Ωand the interior normal derivative ?ψ>0on ?Ωby the strong maximum principle.

Z. Zhang, K. Perera /J. Math. Anal. Appl. 317(2006)456–463459

Recall that a function u ∈H is called a weak solution of (1.1)if

a +b u 2?u ·?v =f (x,u)v, ?v ∈H.

Ω

Ω

(2.4)

Weak solutions are the critical points of the C 2?0functional

a b 24

Φ(u)= u + u ?F (x,u), u ∈H.

24

Ω

(2.5)

They are also classical solutions if f is locally Lipschitz on ×R . Lemma 2.2. Φsatis?es(PS)in the following cases :(i)p <>

(ii)μis not an eigenvalue of (1.9), (iii)(1.6)holds.

Proof. As usual, it suf?cesto show that every (PS)sequence (uj ) of Φis bounded (see,e.g., Alves et al. [2,Lemma 1]).(i)We have

a b

u j 4=Φ(uj ) ? u j 2+F (x,u j ) C u j p +1(2.6)42

Ω

and hence (uj ) is bounded.

(ii)Suppose that ρj = u j →∞for a subsequence. Setting v j =u j /ρj and passing to a further subsequence, v j converges to some v weakly in H , strongly in L 4(Ω), and a.e. in Ω. Passing to the limit in

3v j (Φ (uj ), w) f (x,u j )

?v j ·?w ?w =(2.7)22)ρ1+a/bρ(a+bρbu 3j j j j

Ω

Ω

gives

Ω

?v ·?w ?μv3w =0

(2.8)

for each w ∈H , and passing to the limit with w =v j ?v shows that v =1, so μis an

eigenvalue of (1.9),contrary to assumption. (iii)The conclusion follows from

νν

?1b u j 4=νΦ(uj ) ?Φ (uj ), u j ??1a u j 242

+νF(x,u j ) ?uf (x,u) C +o u j . 2(2.9)

Ω

1, with the usual norm Let X =C 0

u X =max sup D αu(x) ,

0 |α| 1x ∈(2.10)

460Z. Zhang, K. Perera /J. Math. Anal. Appl. 317(2006)456–463

which is densely imbedded in H . By the elliptic regularity theory, the critical point set K :=

=Φ|X . A standard argument shows that Φ has the retracting {u ∈H :Φ (u)=0}?X . Let Φ

property (see,e.g., Dancer and Zhang [9,proof of Lemma 1]).Consider the initial value problem du

=?W (u),t >0, (2.11)u(0) =u 0∈X, where

(u)Φ

W (u)==u ?Au,

a +b u A =KG , G :X →L ∞(Ω), u →

f (x,u(x))

, a +b u (2.12)

and K =(?Δ)?1:L ∞(Ω)→X . Since f (x,t) is

locally Lipschitz in t , uniformly in , G is locally Lipschitz, and K is a bounded linear operator,

so W is locally Lipschitz on X . So a unique solution u(t,u 0) of (2.11)satisfying

t

u(t,u 0) =e ?t u 0+

e s Au (s,u 0) ds

(2.13)

exists in some maximal existence interval [0, T (u0)) , T (u0) ∞. Since

(u) 2 d Φdu Φu(t,u 0) =Φ(u),=? 0,

dt dt a +b u is nonincreasing along the orbits. Φ

(2.14)

if De?nition2.3. (Sun[15])We say that M ?X is an invariant set of descent ?owof Φ

u(t,u 0) :u 0∈M, t ∈0, T (u0) ?M. (2.15)Denote by

C(M)=u 0∈X :u(t,u 0) ∈M for some t ∈0, T (u0) ?M

(2.16)

the maximal subset of X retracted by an invariant set M . If M =C(M), we say that M is complete.

Lemma 2.4. (Sun[15])Let M be an invariant set.

achieves the in?mumat a critical point in M . (i)If inf Φ(M)>?∞and M is closed, then Φ

(ii)If M is open, then C(M)is an open complete invariant set, ?C(M)is a closed complete

invariant set, and inf Φ(?C(M)) inf Φ(?M). ?. Then P is Let P ={u ∈X :u 0}be the positive cone in X , which has nonempty interior P

a closed convex set and A(P) ?P by (1.4),so P is an invariant set (seeSun [16]).Combining this with (2.13)gives

u 0∈P

?

u(t,u 0) e ?t u 0,

(2.17)

?is also invariant. Similarly, the negative cone ?P and its interior ?P ?are invariant sets. so P

Z. Zhang, K. Perera /J. Math. Anal. Appl. 317(2006)456–463461

. Lemma 2.5. Let u be a nontrivial critical point of Φ

(i)If u ∈P (respectively ?P ) , then u is a positive (respectively negative ) solution of (1.1).

?) ∪?C(?P ?) , then u is a sign changing solution of (1.1). (ii)If u ∈?C(P

?Proof. (i)By the strong maximum principle, if u ∈P (respectively?P ), then, in fact, u ∈P ?). (respectively?P

?) ; the other case is similar. Since P ?is an open subset of C(P ?) , (ii)We suppose that u ∈?C(P

?. Since ?P ?is open and does not intersect C(P ?) , u ∈?. So u ∈u ∈/P /?P /P ∪(?P ) as in (i).23. Proof of Theorem 1.1

?) by Lemma 2.5. has nontrivial critical points in ±P and ?C(P It suf?cesto show that Φ(i)By (1.2),

b

u 4?C u p +1, Φ(u)(3.1)

4

is bounded from below. Let Y be the subspace of X spanned by the eigenfunctions of λ1so Φ

and λ2. By (1.2)and (1.12),

F (x,t) so

aλ2

t ?C |t |p , 2

u ∈Y ∩?Bδ,

(3.2)

a λb

?Φ(u)?1 u 2+ u 4+C u p <>

2λ24

(3.3)

?and where B δ={u ∈X : u <δ}with δ="">0suf?cientlysmall. Since Y ∩B δintersects ±P ?) by connectedness, it follows that the in?maof Φ?) are on ±P and ?C(P hence also ?C(P

negative and hence achieved at nontrivial critical points by (i)of Lemma 2.4. (ii)By (1.5),

F (x,t) so

bμ14

t +C, 4

(3.4)

b 44 Φ(u) u ?μ1u ?C ?C, 4

Ω

(3.5)

and the conclusion follows as in (i).(iii)By (1.2)and (1.13),

F (x,t) so

aλ12

t +C |t |p , 2

u ∈δ,

(3.6)

b b

u 4?C u p u 4, Φ(u)

48

(3.7)

and δ>0suf?cientlysmall, since p >4by (1.8).So 0is a strict local minimum of Φ

bδ4

8

(3.8)

462Z. Zhang, K. Perera /J. Math. Anal. Appl. 317(2006)456–463

is an invariant neighborhood of 0. Then C(U)is a complete invariant neighborhood of 0, ?C(U)

4

is a complete invariant set, and inf Φ(?C(U)) bδby (ii)of Lemma 2.4.

Fix ε∈(0, μ?μ2) . There is a γ∈Γsuch that max I (γ([0, 1])) <μ2+ε by="">

F (x,t)

b(μ?ε/2) 4

t ?C 4

(3.9)

by (1.5).Let

Z =tu :u ∈γ[0, 1]∪?γ[0, 1], t 0, which is a surface in H homeomorphic to R 2. For u ∈Z ,

?Φ(u)

b (μ?μ2?ε)a

u 4+ u 2+C →?∞as u →∞,

4(μ22

(3.10)

(3.11)

so C(U)∩Z is a bounded neighborhood of 0in Z . Denote by V its connected component

containing 0. By Lemma 2of Dancer and Zhang [9],?Vhas a connected component Σthat intersects each one-sided ray in Z through 0and hence contains some positive multiples of

?and intersects ?C(P ?) . Since Σ??C(U), it follows that ?C(U)∩(±P ) and ?C(U)∩±ψ∈±P

?) are nonempty closed invariant sets on which the in?maof Φ are positive and hence ?C(P

contain nontrivial critical points.

(iv)Since (1.7)holds and Y is a ?nite-dimensionalsubspace, Φ(u)→?∞as u →∞,

uniformly in Y , so the conclusion follows as in (iii)with Y in place of Z . References

[1]C.O. Alves, F.J.S.A. Correa, On existence of solutions for a class of problem involving a nonlinear operator, Comm.

Appl. Nonlinear Anal. 8(2001)43–56.

[2]C.O. Alves, F.J.S.A. Corrêa,T.F. Ma, Positive solutions for a quasilinear elliptic equation of Kirchhoff type, Comput.

Math. Appl. 49(2005)85–93.

[3]D. Andrade, T.F. Ma, An operator equation suggested by a class of nonlinear stationary problems, Comm. Appl.

Nonlinear Anal. 4(1997)65–71.

[4]A. Arosio, S. Panizzi, On the well-posedness of the Kirchhoff string, Trans. Amer. Math. Soc. 348(1996)305–330.[5]S. Bernstein, Sur une classe d’équationsfonctionnelles aux dérivéespartielles, Bull. Acad. Sci. URSS. Sér.Math.

[Izv.Akad. Nauk SSSR]4(1940)17–26.

[6]M.M. Cavalcanti, V .N. Domingos Cavalcanti, J.A. Soriano, Global existence and uniform decay rates for the

Kirchhoff–Carrierequation with nonlinear dissipation, Adv. Differential Equations 6(2001)701–730.

[7]M. Chipot, B. Lovat, Some remarks on nonlocal elliptic and parabolic problems, Nonlinear Anal. 30(1997)4619–

4627.

[8]M. Chipot, J.-F. Rodrigues, On a class of nonlocal nonlinear elliptic problems, RAIRO Modél.Math. Anal.

Numér.26(1992)447–467.

[9]E.N. Dancer, Z. Zhang, Fucik spectrum, sign-changing, and multiple solutions for semilinear elliptic boundary value

problems with resonance at in?nity,J. Math. Anal. Appl. 250(2000)449–464.

[10]P. D’Ancona,S. Spagnolo, Global solvability for the degenerate Kirchhoff equation with real analytic data, Invent.

Math. 108(1992)247–262.

[11]G. Kirchhoff, Mechanik, Teubner, Leipzig, 1883.

[12]J.-L. Lions, On some questions in boundary value problems of mathematical physics, in:Contemporary Develop-ments in Continuum Mechanics and Partial Differential Equations, Proc. Internat. Sympos., Inst. Mat., Univ. Fed. Rio de Janeiro, Rio de Janeiro, 1977, in:North-Holland Math. Stud., vol. 30, North-Holland, Amsterdam, 1978, pp. 284–346.

[13]T.F. Ma, J.E. Mu?ozRivera, Positive solutions for a nonlinear nonlocal elliptic transmission problem, Appl. Math.

Lett. 16(2003)243–248.

[14]S.I. Poho?aev,A certain class of quasilinear hyperbolic equations, Mat. Sb. (N.S.)96(1975)152–166,168.

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[15]J.X. Sun, On Some Problems about Nonlinear Operators, PhD thesis, Shandong University, Jinan, 1984.

[16]J.X. Sun, The Schauder condition in the critical point theory, Kexue Tongbao (EnglishEd.) 31(1986)1157–1162.[17]C.F. Vasconcellos, On a nonlinear stationary problem in unbounded domains, Rev. Mat. Univ. Complut. Madrid 5

(1992)309–318.

基尔霍夫方程论文

A note about positive solutions for an equation of Kirchhoff type

AndréL.M. Martinez a , Emerson V. Castelani b , ?, Jair da Silva c , Wesley V.I. Shirabayashi d

a

Universidade TecnológicaFederal do Paraná,86300-000CornélioProcópio,PR, Brazil

Colegiado de Matemática,Universidade Estadual do Oeste do Paraná,85819-110Cascavel, PR, Brazil c

Departamento de Matemática,Universidade Federal do Mato Grosso do Sul, 79070-900Campo Grande, MS, Brazil d

Departamento de Matemática,Universidade Estadual de Maringá,87020-900Maringá,PR, Brazil

b

a r t i c l e i n f o a b s t r a c t

A nonlinear boundary value problem related to an equation of Kirchhoff is considered. The existence of positive solutions is proved through alternative Leray–Schauder’stype com-bined with Krasnoselskii’s?xedpoint theorem. Numerical methods are presented and a result of local convergence is established.

ó2011Elsevier Inc. All rights reserved.

Keywords:

Krasnoselskii’s?xedpoint theorem Alternatives of Leray–SchauderNumerical solutions

1. Introduction

In this paper we consider the following nonlinear boundary value problem

(

000

àM eku 0k 22Tu ?q et Tf et ; u ; u T;

u e0T?u e1T?0;

e1T

where M :R t! R t; f :?0; 1 ?R ?R ! R and q :R t! R are continuous map. As cited in [10]this equation is related to

stationary states of the Kirchhoff equation [11]

Z u tt àc 0tc 1

L

!

j u x j dx u xx ?0;

2

e2T

and it is used in study of free vibrations in elastic strings. Still, in [10], the problem considered was an equation with re?ec-tion of argument of the type

àM

Z

1

j u es Tjds u 00?f ex ; u ex T; u eàx TT;

2

x 2?à1; 1 ;

à1

u eà1T?u e1T?0;

and results of existence and uniqueness were obtained using Brouwer’s?xedpoint theorem and a numerical method using ?nitedifference was presented, thereby complementing the studies presented in [9].

Several authors have studied variations of problems that are related to (2), sometimes for more theoretical purposes other times for most practical purposes [5–8].

In [13]we can ?nda study about the following equation

àM

Z

j u es Tjds u 00?f ex ; u ex TT;

2

x 2X ; u ?0on @X :e3T

X

?Corresponding author.

E-mail addresses:andrelmmartinez@yahoo.com.br(A.L.M.Martinez), emersonvitor@gmail.com(E.V.Castelani), jair@dmt.ufms.br(J.daSilva), wesleyvagner@gmail.com(W.V.I.Shirabayashi).

0096-3003/$-see front matter ó2011Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2011.07.025

A.L.M. Martinez et al. /Applied Mathematics and Computation 218(2011)2082–20902083

This study presents existence of positive solutions for the problem (3)by variational methods. Subsequently, a survey [12]was presented where theorems of existence and uniqueness for the same problem were established considering the cases:f =f (x ) and f =f (x , u ).

However, we can note that when f depends explicitly on ‘‘u 0’’little attention is paid. As known in the literature, the Kra-snoselskii’stheorem is an interesting tool to obtain existence results of positive solutions [2,4]. Thus the goal of this paper is to complement the studies cited above by establishing existence results for Eq. (1)as well as numerical methods.

In this sense we are presenting a massive study about this problem by using classical ?xedpoint theorems, that is, we are compiling two theorems in order to establish strong conditions of existence for (1), namely Krasnoselskii’stheorem and an alternative of Leray–Schauder’stype.

Essentially, in Section 2we demonstrate the existence of solutions for the above-mentioned problem by alternative Ler-ay–Schauder’stype. Since the existence was demonstrated we proved the existence of positive solutions using more restric-tive assumptions and Krasnoselskii’stheorem.

These results are very important in the theoretical sense because the existence is guaranteed with very reasonable hypotheses but, unfortunately, these results cannot generate an iterative method. Thus, in Section 4, we are presenting iter-ative algorithms and we demonstrate the local convergence of these algorithms by Banach’s?xedpoint theorem. For the theoretical purpose of this work we need to introduce the main tools. Theorem 1. Let E be a Banach space, C &E a closed and convex set, X an open set in C and p 2X . Then each completely continuous mapping T :! C has at least one the following properties:(A1)T has a ?xedpoint in (A2)There are u 2o X and k 2(0,1)such that u =k T(u)+(1àk )p.

Theorem 2. Let E be a Banach space and let K &E be a cone in E. Assume X 1, X 2are open subsets of E with 02X 1; 1&X 2, and let T :K \e2n X 1T! K be a completely continuous operator such that, either (a)k Tu k 6k u k , u 2K \o X 1, and k Tu k P k u k , u 2K \o X 2, or (b)k Tu k P k u k , u 2K \o X 1, and k Tu k 6k u k , u 2K \o X 2. Then T has a ?xedpoint in K \e2n X 1T.

The ?rsttheorem is well-known Leray–Schauderalternative and the second theorem is due Krasnoselskii, see [1,3]. 2. Existence of solutions

Let E ={u 2C 1[0,1];u (0)=u (1)=0}be Banach space equipped with the norm

k u k E ?k u 0k 1?max j u 0et Tj:

t 2?0; 1

We begin this section by pointing out that the solutions of (1)can be written as

u et T?

Z

1

f es ; u es T; u 0es TT

ds ; G et ; s Tq es T

20M k u k 2t e1às Tt 6s ;

s e1àt Ts 6t :

1

e4T

where G is the following function

G et ; s T?

&

We know that u is a solution of (1)if only if it is a ?xedpoint of the operator T :E ? E de?nedby

eTu Tet T?

Z

G et ; s Tq es T

f es ; u es T; u 0es TT

ds :

20M k u k 2

e5T

Next we show a result for existence of solutions for (1). For this purpose we need to de?nethe following conditions:(H 1) q (s ) f (s , u , v ) P 0, 8es ; u ; v T2?0; 1 ?R ?R .

(H 2) There exists positive constants a , A , B such that

A 6M k u 0k 226B ;

for k u k E 6a .

2084A.L.M. Martinez et al. /Applied Mathematics and Computation 218(2011)2082–2090

(H 3) Assume that a and A , de?nedin (H 2), satisfy :

a A

max j f es ; u ; v Tj<>

d 1es ; u ; v T2?0; 1 ??a ; a ??àa ; a

where

d 1?max

t

&Z

1

'

j u es ; t Tq es Tjds

and u :?0; 1 ??0; 1 ! R is de?nedby

@G et ; s T

u es ; t T??

@t

&

1às t 6s ; s

s 6t :

Theorem 3. Suppose that (H1)–(H3)hold. Then (1)has a nonnegative solution u ?2E such that k u ?k E 6a .

Proof. We shall employ Theorem 1with p =0and X ={u 2E ; k u k E

8 9

=Z 1Z 1< 0="" @g="" et="" ;="" s="" tf="" es="" ;="" u="" es="" t;="" u="" es="" tt100="" ds="" 6max="" q="" es="" tj="" u="" et="" tj6ketu="" tk="" ?max="" j="" u="" es="" ;="" t="" tq="" es="" tf="" es="" ;="" u="" es="" t;="" u="" 0es="" ttjds="">

2t t : 0; @t A 00 M k u k

2

Since k u k E 6a we have k u 0k 16a and, consequently, k u k 16a Thus we can apply (H 3) and we get 1

j u et Tj<>

A t

&Z

1

' A a

j u es ; t Tq es Tjds ?a :

d 1

Then we cannot have the conclusion (A 2) of Theorem 1, hence the conclusion (A 1) must hold. Therefore, there exists u ?2E such that k u ?k E 6a . Note that u ?is nonnegative by (H 1) and G (t , s ) P 0. h 3. Existence of positive solutions

In this section we establish the main result, that is, we are presenting a result of existence of positive solutions for (1). In this sense we need to introduce the following additional conditions:(H 4) There exists 0

min es ; u ; v T2?0; 1 ??0; b ??àb ; b f f es ; u ; v TgP

b B ; 2

where

d 2?

Z

1

w es Tq es Tds

and w :?0; 1 ! R is de?nedby

(

w es T?

s

s 61; 1às s P 1:(H 5) Suppose that b satis?es

b B a A

6:d 2d 1

Theorem 4. Suppose that (H1)–(H5)hold. Then (1)has a positive solution u ?2E, u is concave and b

K ?f u 2E :u ex TP 0; u is concave on ?0; 1 g:

Next, using (H 1) and the de?nitionof G , we have that T maps K into K . As in last result, T is completely continuous. We shall

employ Theorem 2. Thus, we can de?neX 1={u 2E ; k u k E

A.L.M. Martinez et al. /Applied Mathematics and Computation 218(2011)2082–20902085

i. if k u k E =a then k Tu k E 6a , ii. if k u k E =b then k Tu k E P b .

In fact, the proof of the ?rstcondition is analogous to proof of last theorem. To prove the second condition we note that if k u k 1?b then

keTu T0k 1?

where

M

1k u 0k 22

max f V 1; V 2g ;

V 1?V 2?

Thus

Z Z

1

sq es Tf es ; u es T; u 0es TTds ; e1às Tq es Tf es ; u es T; u 0es TTds :

&Z

10

1

keTu Tk 1P

B

1

' &Z 1'

1B b

w es Tq es Tf es ; u es T; u es TTds P w es Tq es Tds P b :

B 0d 2

Therefore, we have by Theorem 2that T has a ?xedpoint u ?2K with b

f es ; u ; v T?1t2u 2t2v 2; sin ep s T

; 2

M es T?1ts :q es T?

If we choose

es ; u ; v T2?0; 1 ?R ?R ;

a ?1;

A ?1; B ?2;

1; b ?2

and compute the constants

1; 21d 2?; d 1?

we have that (H 1)–(H 5) hold.

Now, we can note that the assumption (H 4) can be replaced by (H 4)’There exists 0

min es ; u ; v T2?m ; 1àm ??0; b ??àb ; b f f es ; u ; v TgP

b B ; d 2

where

d 2?

Z

1àm

w es Tq es Tds ;

m

àám 20; 1and w :?0; 1 ! R is de?nedby (

w es T?

s

s 61; 1às s P 1:Thus we can get the following result.

Theorem 6. Suppose that (H1)–(H3),(H4)’and (H5)hold. Then (1)has a positive solution u ?2E, u is concave and b

2086A.L.M. Martinez et al. /Applied Mathematics and Computation 218(2011)2082–2090

Proof. To show this result we need to verify that if u 2E and k u k E =b then k Tu k E P b . In fact, if k u 0k 1?b we have

keTu T0k E ?

where

M k u 0k 22

1

max f V 1; V 2g ;

V 1?V 2?

Thus

Z Z

1

sq es Tf es ; u es T; u 0es TTds ; e1às Tq es Tf es ; u es T; u 0es TTds :

&Z

1

1

keTu Tk E P

B

1

' &Z 1àm ' &Z 1àm '

11B b 0

w es Tq es Tf es ; u es T; u es TTds P w es Tq es Tf es ; u es T; u es TTds P w es Tq es Tds P b :

B m B m d 2

The rest of the proof is similar to last result. Therefore, the proof is complete. h 4. Iterative solutions

In this section we present an existence result for iterative solutions and consequently algorithms are given. To begin, we

de?ne

u k t1?T eu k T

Z 1

f es ; u k es T; u 0k es TTk t1

ds ; G et ; s Tq es Tu et T?

020M k u k k 2

e6T

and we shall show the existence of limit for the sequence in (6)by Banach’s?xedpoint theorem [1]. In this direction we

de?nesome conditions.

(S 1) There exists positive constants a , A and B such that

02

A 6M k u k 26B ;

for all u 2E with k u k E 6a .

(S 2) Assume that a and A , de?nedin (S 2), satisfy:

a A

max j f es ; u ; v Tj6;

a ; a ??àa ; a d es ; u ; v T2?0; 1 ??à1 where

d 1?max

t

&Z

1

'

j u es ; t Tq es Tjds :

(S 3) There exists k M >0such that j M (u ) àM (v ) j 6k M j u àv j , for all u , (S 4) There exists k f >0with

v 2[àa , a ],

j f es ; u ; u 0Tàf es ; v ; v 0Tj6k f max fj u àv j ; j u 0àv 0jg ;

??a and u 0; v 02?àa ; a . where s 2?0; 1 ; u ; v 2àa ; (S 5)

2k M a 2

B k f d 1A t<>

Theorem 7. Suppose that (S1)–(S5)hold. Then (1)has an iterative solution with k u k E 6a .

Proof. First, we show T :B [0,a ]? B [0,a ]where B [0,a ]={u 2E ; k u k E 6a }.In fact, if u 2B [0,a ]we have

8 9

= Z 1Z 1 < 0="" @g="" et="" ;="" s="" t="" @g="" et="" ;="" s="" tq="" es="" tf="" es="" ;="" u="" es="" t;="" u="" es="" tt100="" 6max="" j="" f="" es="" ;="" u="" ;="" u="" tj="" ds="" :ketu="" tk="" 1?max="" q="" es="" tds="" ;="" @t="" 22t="" s="" :="" @t="" 000="" 0m="" k="" u="" k="" 2m="" k="" u="" k="">

d 16d 1?a :

2d 1A 0M k u k 2

1

By hypothesis (S1)and (S2)we have

keTu Tk 16

a A

d 1

a A 1

A.L.M. Martinez et al. /Applied Mathematics and Computation 218(2011)2082–20902087

Therefore, T :B [0,a ]? B [0,a ].We shall show that T is a contraction by using Banach ?xedpoint theorem.

8 23 9

=Z 1< 00="" @g="" et="" ;="" s="" tf="" es="" ;="" u="" es="" t;="" u="" es="" ttf="" es="" ;="" v="" es="" t;="" v="" es="" tt04ds="" q="" es="" tketu="" àt="" v="" tk="" 1?max="" à="" ;="" t="" :="" @t="" m="" ek0k="" 2="" 0m="" k="" u="" 0k="">

8 9

< f="" es="" ;="" u="" es="" t;="" u="" 0es="" ttf="" es="" ;="" v="" es="" t;="" v="" 0es="" tt="">

à6d 1max ; 22s : 00 M k u k M k v k

2

2

A s

i d 1h 0200200

TàM k u k j f es ; u es T; u es TTjtM eku k Tmax j f es ; u es T; u es TTàf es ; v es T; v es TTjj max 6j M ekv 0k 2222s s A

! d 1a A 02

tB k f k u 0àv 0k 1:62k M jk v 0k 22àk u k 2j d 1A

Now we can note

6d 1

1

n o

0020

max j M ekv 0k 2Tf es ; u es T; u es TTàM k u k v es T; v es TTjf es ; 22

jk v

Then

02

k 2

à

k u 0k 22j

Z 1 Z

2200 6? e?v es T à?u es T Tds

1

j ev 0es Ttu 0es TTev 0es Tàu 0es TTj ds 6k u 0tv 0k 1k u 0àv 0k 1:

! !

a A d 1a A 0000000000

tB k f k u àv k 162k M e2a Teku àv k 1TtB k f k u àv k 1keTu àT v Tk 162k M k u tv k 1k u àv k 1

d 1d 1A A

2k M a 2B k f d 16tk u 0àv 0k 1:

A

d 1

Finally, as

2k M a 2B k f d 1

t2<1; a="">

we have from Banach ?xedpoint theorem the result. h

Motivated by the last result we can present the following algorithm:Algortihm 1:

1-De?nea uniformly spaced mesh {x j }.

2-Choose initial approximation u 0j ?u ex j T. 3-For k =1,2,3, ...

–Compute u k j by central-differences. –Compute u j k t1using

u k t1?T eu k T;

where the integrals are computed through trapezoidal rule.

–Test the convergence.

We can establish some modi?cationsin Algorithm 1as a way to achieve a better numerical results. More precisely we can compute the integrals through splines [14]. Thus we have the following algorithm. Algorithm 2:

1-De?nea uniformly spaced mesh {x j }.

2-Choose initial approximation u 0j ?u ex j T. 3-For k =1,2,3, ...

–Compute u 0j k by central-differences. –Compute u j k t1using

u k t1?T eu k T;

where the integrals are computed through cubic spline interpolation.

–Test the convergence.

2088A.L.M. Martinez et al. /Applied Mathematics and Computation 218(2011)2082–2090

In sequence we are presenting some examples in order to establish the effectiveness of the Algorithms 1and 2. In Tables

w k w k k +1

1–3, e k àu k k 1. Still, ‘‘It’’denotes ‘‘iteration’’.u denotes k u àu k 1where u is the exact solution and e denotes k u Example 8. For the ?rstexample we are considering

f et ; u ; u 0T?ep 2tu 2t2p 2Tu tt eu 0T2àM et T?1t2t :

t

2

;

For simplicity we de?neq (t ) =à1for all t 2[0,1].The exact solution is

u H et T?

sin p t

2

and the spaced mesh has n =10points. Example 9. In this example we have

f et ; u ; u 0T?4u 0à3u à3t t4à2e t à1;

1

M et T?t4t

and the exact solution

u H et T?t ee t à1à1T:

As before, we consider q (t ) =à1and n =10points for spaced mesh.

Example 10. Now, we are considering a classical function solution u w (t ) =t (t à1) 2and the problem is de?nedby:

f et ; u ; u 0T?

34

ee6àu Tt tt 4à2t 3tt 2à4tu 0eu àt 3Ttu 0e2t 2àt TT; 30302

M es T?1ts :

4

Again, we consider q (t ) =à1but n =20points for spaced mesh.

Table 1

Performance of Algorithms 1and 2in the Example 8. It 12351015

e k u (Alg.1)

9.8183809E à029.8984724E à029.9873824E à029.9130891E à029.9325251E à029.9314157E à02

e k (Alg.1)

3.5540297E à033.8565400E à032.4742332E à035.6049219E à042.7301223E à051.9550441E à06

e k u (Alg.2)

9.8259702E à029.8975140E à029.9703230E à029.9174853E à029.9311537E à029.9305760E à02

e k (Alg.2)

3.4703247E à033.5657322E à032.0194289E à034.0334824E à041.0890075E à055.6270784E à07

Table 2

Performance of Algorithms 1and 2in the Example 9. It 12510152030

e k u (Alg.1)

5.4284416E à012.8151610E à018.3756432E à018.9422046E à022.6154450E à022.5420703E à022.5694559E à02

e k (Alg.1)

6.0218720E à015.4169827E à019.9572360E à011.7723944E à014.3222377E à031.8427241E à031.2016155E à04

e k u (Alg.2)

5.3863780E à017.0328656E à019.4279440E à011.2105944E à011.2444909E à024.1020591E à034.2478119E à03

e k (Alg.2)

5.9798084E à017.8099875E à011.1854434E+02.2115774E à012.0089259E à024.1953717E à032.9407919E à05

Table 3

Performance of Algorithms 1and 2in the Example 10. It 12510

e k u (Alg.1)

2.7275169E à028.0718944E à034.4146441E à034.4141816E à03

e k (Alg.1)

1.2190900E à013.1153652E à025.4964838E à054.2285613E à10

e k u (Alg.2)

3.0751613E à024.8607874E à036.5283084E à046.5301780E à04

e k (Alg.2)

1.1185763E à013.0580363E à025.1153640E à051.3265517E à09

A.L.M. Martinez et al. /Applied Mathematics and Computation 218(2011)2082–2090Table 4

2089

We know that the integration by cubic spline can give better results than trapezoidal rule if we consider the same mesh. However, when we compute the integrals in Algorithms 1and 2, the results are equivalents in some cases, that is, even com-puting the integrals with cubic spline, the loss of accuracy using ?nitedifference do not allow a far superior performance of the Algorithm 2with respect to Algorithm 1.

Finally, we have that Examples 8–10are needed mainly to test the reliability of Algorithms 1and 2considering that in these examples we know the exact solution. Now, we can make an additional test. By using Theorem 4we have a solution to the Example 5but we do not know what it is. Let us apply the Algorithms 1and 2in this problem. For this purpose, we can consider the condition

k u k t1àu k k <>

as stopping criterion for the algorithms. The results for this example are presented in Table 4and Fig. 1. Acknowledgement

The authors are grateful to the referees for their valuable suggestions which improved this article. References

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