APPLICATIONS OF THE POINCARÉ INEQUALITY TO EXTENDED KANTOROVICH METHOD
DER-CHEN CHANG, TRISTAN NGUYEN, GANG WANG, AND NORMAN M. WERELEY
Received 3 February 2005; Revised 2 March 2005; Accepted 18 April 2005
We apply the Poincar´e inequality to study the extended Kantorovich method that was used to construct a closed-form solution for two coupled partial differential equations with mixed boundary conditions.
Copyright © 2006 Der-Chen Chang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
LetΩ⊂Rnbe a Lipschitz domain inRn. Consider the Dirichlet spaceH01(Ω) which is the collection of all functions in the Sobolev spaceL21(Ω) such that
H01(Ω)=
u∈L2(Ω) :u|∂Ω=0,uL2+ n k=1
∂u
∂xk
L2
<∞
. (1.1)
The famous Poincar´e inequality can be stated as follows: foru∈H01(Ω), then there exists a universal constantCsuch that
Ωu2(x)dx≤C n k=1
Ω
∂u
∂xk
2dx. (1.2)
One of the applications of this inequality is to solve the modified version of the Dirichlet problem (see, John [5, page 97]): find av∈H01(Ω) such that
(u,v)=
Ω
n
k=1
∂u
∂xk
∂v
∂xk dx=
Ωu(x)f(x)dx, (1.3)
Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2006, Article ID 32356, Pages1–21 DOI10.1155/JIA/2006/32356
where x=(x1,. . .,xn) with a fixed f ∈C( ¯Ω). Then the functionvin (1.3) satisfied the boundary value problem
Δv= −f, inΩ
v=0, on∂Ω. (1.4)
In this paper, we will use the Poincar´e inequality to study the extended Kantorovich method, see [6]. This method has been used extensively in many engineering problems, for example, readers can consult papers [4,7,8,11,12], and the references therein. Let us start with a model problem, see [8]. For a clamped rectangular boxΩ=n
k=1[−ak,ak], subjected to a lateral distributed load,ᏼ(x)=ᏼ(x1,. . .,xn), the principle of virtual dis- placements yields
n =1
a
−a
η∇4Φ−ᏼδΦDx=0, (1.5)
whereΦis the lateral deflection which satisfies the boundary conditions,ηis the flexural rigidity of the box, and
∇4= n k=1
∂4
∂x4k+
j=k
2 ∂4
∂x2j∂x2k. (1.6)
Since the domainΩis a rectangular box, it is natural to assume the deflection in the form Φ(x)=Φk1···kn(x)=
n =1
fkx, (1.7)
it follows that when fk2(x2)···fkn(xn) is prescribed a priori, (1.5) can be rewritten as a1
−a1
n
=2
a
−a
η∇4Φk1···kn−ᏼfkxdx δ fk1(x1)dx1=0. (1.8)
Equation (1.8) is satisfied when n =2
a
−a
η∇4Φk1···kn−ᏼfkxdx=0. (1.9)
Similarly, whenn=1,=mfk(x) is prescribed a priori, (1.5) can be rewritten as am
−am
n
=1,=m
a
−a
η∇4Φk1···kn−ᏼfk
x
dx δ fkm(xm)dxm=0. (1.10)
It is satisfied when
n =1,=m
a
−a
η∇4Φk1···kn−ᏼfkxdx=0. (1.11)
It is known that (1.9) and (1.11) are called the Galerkin equations of the extended Kan- torovich method. Now we may first choose
f20(x2)···fn0(xn)= n =2
c
x2 a2 −1
2
. (1.12)
ThenΦ10···0(x)=f11(x1)f20(x2)···fn0(xn) satisfies the boundary conditions Φ10···0=0, ∂Φ10···0
∂x =0 atx= ±a,x1∈
−a1,a1
, (1.13)
for=2,. . .,n. Now (1.9) becomes n
=2
c a
−a
∇4Φ10···0−ᏼ η
x2 a2 −1
2
dx=0, (1.14)
which yields
C4
d4f11
dx4 +C2
d2f11
dx2 +C0f11=B. (1.15)
After solving the above ODE, we can use f11(x1)n=3f0(x) as a priori data and plug it into (1.10) to find f21(x2). Then we obtain the function
Φ110···0(x)= f11
x1
f21
x2
f30
x3
···fn0 xn
. (1.16)
Continue this process until we obtainΦ1···1(x)= f11(x1)f21(x2)···fn1(xn) and there- fore completes the first cycle. Next, we use f21(x2)···fn1(xn) as our priori data and find f12(x1). We continue this process and expect to find a sequence of “approximate solu- tions.” The problem reduces to investigate the convergence of this sequence. Therefore, it is crucial to analyze (1.15). Moreover, from numerical point of view, we know that this sequence converges rapidly (see [1,2]). Hence, it is necessary to give a rigorous mathe- matical proof of this method.
2. A convex linear functional onH02(Ω) Denote
I[φ]=
Ω
|Δφ|2−2ᏼ(x)φ(x)dx (2.1)
forΩ⊂Rna bounded Lipschitz domain. Here x=(x1,. . .,xn). As usual, denote
D2φ=
⎡
⎢⎢
⎢⎣
∂2φ
∂x2
∂2φ
∂x ∂y
∂2φ
∂y ∂x
∂2φ
∂y2
⎤
⎥⎥
⎥⎦. (2.2)
ForΩ⊂R2, we define the Lagrangian functionLassociated toI[φ] as follows:
L:Ω×R×R2×R4−→R,
(x,y;z;X,Y;U,V,S,W)−→(U+V)2−2ᏼ(x,y)z, (2.3) whereᏼ(x,y) is a fixed function onΩwhich shows up in the integrand ofI[φ]. With the above definitions, we have
Lx,y;φ;∇φ;D2φ= |Δφ|2−2ᏼ(x,y)φ(x,y), (2.4) where we have identified
z←→φ(x,y), X←→∂φ
∂x, Y←→∂φ
∂y, U←→∂2φ
∂x2, V←→∂2φ
∂y2, S←→ ∂2φ
∂y ∂x, W←→ ∂2φ
∂x ∂y.
(2.5)
We also setH02(Ω) to be the class of all square integrable functions such that H02(Ω)=
ψ∈L2(Ω) :
|k|≤2
∂kψ
∂xk
L2
<∞,ψ|∂Ω=0, ∇ψ|∂Ω=0
. (2.6)
Fix (x,y)∈Ω. We know that
∇L(x,y;z;X,Y;U,V,S,W)=
−2ᏼ(x,y) 0 0 2(U+V) 2(U+V) 0 0T. (2.7) Because the convexity of the functionLin the remaining variables, then for all (z;X, Y;U, V,S,W) ∈R×R2×R4, one has
L x,y;z;X, Y;U,V,S,W!≥L(x,y;z;X,Y;U,V,S,W)−2ᏼ(x,y)z−z
+ 2(U+V)U−U+ V−V. (2.8) In particular, one has, withz=φ(x, y),
L x,y;φ;∇φ;D 2φ!≥Lx,y;φ;∇φ;D2φ+ 2Δφ∇φ− ∇φ
−2ᏼ(x,y)(φ−φ). (2.9)
This implies that
Δφ2−2ᏼ(x,y)φ≥Δφ2−2ᏼ(x,y)φ+ 2ΔφΔφ−Δφ−2ᏼ(x,y) φ−φ. (2.10) If instead we fix (x,y;z)∈Ω×R, then
L x,y;z;X, Y;U, V,S,W!≥L(x,y;z;X,Y ;U,V,S,W)
+ 2(U+V) U−U+ V−V. (2.11)
This implies that
Lx,y;φ;∇φ;D 2φ≥L(x,y;φ;∇φ;D2φ+ 2Δφ[∇φ− ∇φ] (2.12) Therefore,
|Δφ|2−2ᏼ(x,y)φ≥ |Δφ|2−2ᏼ(x,y)φ+ 2Δφ[Δφ−Δφ]. (2.13) Lemma 2.1. Suppose either
(1)φ∈H02(Ω)∩C4(Ω) andη∈Cc1(Ω); or (2)φ∈H02(Ω)∩C3( ¯Ω)∩C4(Ω) andη∈H02(Ω).
LetδI[φ;η] denote the first variation ofIatφin the directionη, that is,
δI[φ;η]=lim
ε→0
I[φ+εη]−I[φ]
ε . (2.14)
Then
δI[φ;η]=2
Ω
Δ2φ−ᏼ(x,y)η dx d y. (2.15) Proof. We know that
I[φ+εη]−I[φ]=2ε
Ω[ΔφΔη−ᏼη]dx d y+ε2
Ω(Δη)2dx d y. (2.16) Hence,
εI[φ;η]=2
Ω[ΔφΔη−ᏼη]dx d y. (2.17) If either assumption (1) or (2) holds, we can apply Green’s formula to a Lipschitz domain Ωto obtain
Ω(ΔφΔη)dx d y=
ΩηΔ2φdx d y+
∂Ω
"∂η
∂nΔφ−η ∂
∂nΔφ#dx d y, (2.18) where∂/∂n is the derivative in the direction normal to∂Ω. Since eitherη∈Cc1(Ω) or η∈H02(Ω), the boundary term vanishes, which proves the lemma.
Lemma 2.2. Letφ∈H02(Ω). Then
φH02(Ω)≈ ΔφL2(Ω). (2.19) Proof. The functionφ∈H02(Ω) implies that there exists a sequence{φk} ⊂C∞c (Ω) such that limk→∞φk=φinH02-norm. From a well-known result for the Calder ´on-Zygmund operator (see, Stein [10, page 77]), one has
∂2f
∂xj∂x
Lp≤CΔ fLp, j,=1,. . .,n (2.20)
for all f ∈Cc2(Rn) and 1< p <∞. HereCis a constant that depends onnonly. Applying this result to eachφk, we obtain
∂2φk
∂x2
L2(Ω),∂2φk
∂x ∂y
L2(Ω),∂2φk
∂y2
L2(Ω)≤CΔφk
L2(Ω). (2.21) Taking the limit, we conclude that
∂2φ
∂x2
L2(Ω), ∂2φ
∂x ∂y
L2(Ω),∂2φ
∂y2
L2(Ω)≤CΔφL2(Ω). (2.22) Applying Poincar´e inequality twice to the functionφ∈H02(Ω), we have
φL2(Ω)≤C1∇φL2(Ω)
≤C2
∂2φ
∂x2
L2(Ω)+ ∂2φ
∂x ∂y
L2(Ω)+∂2φ
∂y2
L2(Ω)
≤CΔφL2(Ω).
(2.23)
Hence,φL2(Ω)≤CΔφL2(Ω). The reverse inequality is trivial. The proof of this lemma
is therefore complete.
Lemma 2.3. Let{φk}be a bounded sequence inH02(Ω). Then there existφ∈H02(Ω) and a subsequence{φkj}such that
I[φ]≤lim infIφkj. (2.24)
Proof. By a weak compactness theorem for reflexive Banach spaces, and hence for Hilbert spaces, there exist a subsequence{φkj}of{φk}andφinH02(Ω) such thatφkj→φweakly inH02(Ω). Since
H02(Ω)⊂H01(Ω)⊂⊂L2(Ω), (2.25) by the Sobolev embedding theorem, we have
φkj−→φ inL2(Ω) (2.26)
after passing to yet another subsequence if necessary.
Now fix (x,y,φkj(x,y))∈R2×Rand apply inequality (2.13), we have
Δφkj2−2ᏼ(x,y)φkj(x,y)≥ |Δφ|2−2ᏼ(x,y)φkj(x,y) + 2ΔφΔφkj−Δφ. (2.27) This implies that
Iφkj≥
Ω
|Δφ|2−2ᏼ(x,y)φkjdx d y+ 2
ΩΔφ·
Δφkj−Δφdx d y. (2.28) Butφkj→φinL2(Ω), hence
Ω
|Δφ|2−2ᏼ(x,y)φkjdx d y−→
Ω
|Δφ|2−2ᏼ(x,y)φdx d y=I[φ]. (2.29)
Besidesφkj→φweakly inH02(Ω) implies that
ΩΔφ·
Δφkj−Δφdx d y−→0. (2.30) It follows that when taking limit
I[φ]≤lim inf
j Iφkj
. (2.31)
This completes the proof of the lemma.
Remark 2.4. The above proof uses the convexity ofL(x,y;z;X,Y;U,V,S,W) when (x,y;
z) is fixed. We already remarked at the beginning of this section that when (x,y) is fixed, L(x,y;z;X,Y;U,V,S,W) is convex in the remaining variables, including thez-variable.
That is, we are not required to utilize the full strength of the convexity ofLhere.
3. The extended Kantorovich method
Now, we shift our focus to the extended Kantorovich method for finding an approximate solution to the minimization problem
min
φ∈H02(Ω)I[φ] (3.1)
when Ω=[−a,a]×[−b,b] is a rectangular region in R2. In the sequel, we will write φ(x,y) (resp.,φk(x,y)) asf(x)g(y) (resp., fk(x)gk(y)) interchangeably as notated in Kerr and Alexander [8]. More specifically, we will study the extended Kantorovich method for the casen=2, which has been used extensively in the analysis of stress on rectangular plates. Equivalently, we will seek for an approximate solution of the above minimization problem in the formφ(x,y)= f(x)g(y) where f ∈H02([−a,a]) andg∈H02([−b,b]).
To phrase this differently, we will search for an approximate solution in the tensor product Hilbert spacesH02([−a,a])⊗$H02([−b,b]), and all sequences{φk},{φkj}involved hereinafter reside in this Hilbert space. Without loss of generality, we may assume that Ω=[−1, 1]×[−1, 1] for all subsequent results remain valid for the general case where Ω=[−a,a]×[−b,b] by approximate scalings/normalizing of thexandyvariables. As in [8], we will treat the special caseᏼ(x,y)=γ, that is, we assume that the loadᏼ(x,y) is distributed equally on a given rectangular plate.
To start the extended Kantorovich scheme, we first choose g0(y)∈H02([−1, 1])∩ C∞c (−1, 1), and find the minimizerf1(x)∈H02([−1, 1]) of the functional:
If g0
=
Ω
Δf g02−2γ f(x)g0(y)dx d y
=
Ω
g02(f)2+ 2f fg0g0+f2g02−2γ f g0
dx d y
= 1
−1(f)2dx 1
−1g02d y+ 2 1
−1
g02d y 1
−1(f)2dx +
1
−1
g02d y 1
−1f2dx−2γ 1
−1g0d y 1
−1f dx,
(3.2)
where the last equality was obtained via the integration by parts of f fandg0g0. Since g0has been chosen a priori; we can rewrite the functionalIas
J[f]=g02
L2
1
−1(f)2dx+ 2g02
L2
1
−1(f)2dx +g02L2
1
−1f2dx−2γ 1
−1g0(y)d y 1
−1f dx
(3.3)
for all f ∈H02([−1, 1]). Now we may rewrite (3.3) in the following form:
J[f]= 1
−1
C1(f)2+C2(f)2+C3f2+C4fdx
≡ 1
−1K(x,f,f,f)dx
(3.4)
withK:R×R×R×R→Rgiven by
(x;z;V;W)−→C1W2+C2V2+C3z2+C4z, (3.5) where
C1=g02
L2, C2=g02L2, C3=g02L2, C4= −2γ 1
−1g0(y)d y. (3.6) As long asg0≡0, as we have implicitly assumed, the Poincar´e inequality implies that
0< C1≤αC2≤βC3 (3.7)
for some positive constantsαandβ, independent ofg0. Consequently,K(x;z;V;W) is a strictly convex function in variablez,V,Wwhenxis fixed. In other words,Ksatisfies
K(x;z;V;W) −K(x;z;V;W)
≥∂K
∂z(x;z;V;W)(z−z) +∂K
∂V(x;z;V;W)(V−V) + ∂K
∂W(x;z;V;W)(W−W) (3.8) for all (x;z;V;W) and (x;z;V;W) in R4, and the inequality becomes equality at (x;z;V; W) only ifz=z, orV=V, orW=W.
Proposition 3.1. Letᏸ:R×R×R×R→Rbe a C∞ function satisfying the following convexity condition:
ᏸ(x;z+z;V+V;W+W)−ᏸ(x;z;V;W)
≥∂ᏸ
∂z(x;z;V;W)z+∂ᏸ
∂V(x;z;V;W)V+ ∂ᏸ
∂W(x;z;V;W)W (3.9)
for all (x;z;V;W) and (x;z+z;V+V;W+W)∈R4, with equality at (x;z;V;W) only ifz=0, orV=0, orW=0. Also, let
J[f]= β
αᏸx,f(x),f(x),f(x)dx, ∀f ∈H02(α,β). (3.10) Then
J[f +η]−J[f]≥δJ[f,η], ∀η∈C∞c (α,β) (3.11) and equality holds only ifη≡0. HereδJ[f,η] is the first variation ofJatf in the directionη.
Proof. Condition (3.9) means that at eachx, ᏸ(x;f+η;f+η;f+η)−ᏸ(x;f;f;f)
≥∂ᏸ
∂z(x;f;f;f)η(x) +∂ᏸ
∂V(x;f;f;f)η(x) + ∂ᏸ
∂W(x;f;f;f)η(x) (3.12) for allη∈C∞c (α,β) with equality only ifη(x)=0, orη(x)=0, orη(x)=0. Equivalently, the equality holds in (3.12) atxonly ifη(x)η(x)=0 orη(x)=0. In other words,
η(x) d dx
η2(x)=0. (3.13)
Integrating (3.12) gives J[f+η]−J[f]≥
β
α
"∂ᏸ
∂zη+∂ᏸ
∂Vη+ ∂ᏸ
∂Wη
#
dx=δJ[f,η]. (3.14) Now suppose there existsη∈Cc∞(α,β) such that (3.14) is an equality. Sinceᏸis a smooth function, this equality forces (3.12) to be a pointwise equality, which implies, in view of (3.13), that
η(x)d dx
η2(x)=0, ∀x. (3.15)
Ifη(x)≡0, thenη(x)=constant which implies thatη(x)≡0 (since η∈C∞c (α,β)).
This tells us thatη≡constant and conclude thatη≡0 on the interval (α,β).
Ifη(x)≡0, setU= {x∈(α,β) :η(x)=0}. ThenUis a non-empty open set which implies that there existx0∈Uand some open setᏻx0ofx0contained inU. Thenη(ξ)= 0 for allξ∈ᏻx0⊂U. Thus
d dx
η2=0 onᏻx0. (3.16)
Hence,η(ξ)≡constant onᏻx0. But this creates a contradiction becauseη(ξ)≡0 onᏻx0. Therefore,
J[f+η]−J[f]=δJ[f,η] (3.17) only ifη(x)≡0, as desired. This completes the proof of the proposition.
Corollary 3.2. LetJ[f] be as in (3.4). Then f1∈H02([−1, 1]) is the unique minimizer for J[f] if and only if f1solves the following ODE:
g02
L2
d4f
dx4 −2g02L2
d2f
dx2 +g02L2f =γ 1
−1g0d y. (3.18) Proof. Suppose f1 is the unique minimizer. Then f1 is a local extremum ofJ[f]. This implies thatδJ[f,η]=0 for allη∈H02([−1, 1]). Using the notations in (3.4), we have
0=δJ[f,η]
= 1
−1
"∂K
∂zη+∂K
∂Vη+ ∂K
∂Wη
# dx
= 1
−1
"∂K
∂z − d dx
∂K
∂V
+ d2 dx2
∂K
∂W
# η(x)dx
(3.19)
for allη∈H02([−1, 1]). This implies that
∂K
∂z − d dx
∂K
∂V
+ d2 dx2
∂K
∂W
=0, (3.20)
which is the Euler-Lagrange equation (3.18). This also follows fromLemma 2.1directly.
Conversely, assume f1solves (3.18). Then the above argument shows thatδJ[f,η]=0 for allη∈H02([−1, 1]). SinceKsatisfies condition (3.9) inProposition 3.1, we conclude that
Jf1+η−Jf1
≥δJf1,η, ∀η∈C∞c [−1, 1]. (3.21) This tells us thatJ[f1+η]≥J[f1] for allη∈Cc∞([−1, 1]) andJ[f1+η]> J[f1] ifη≡0.
Observe thatJ:H01([−1, 1])→Ras given in (3.4) is a continuous linear functional in theH02-norm. This fact, combined with the density ofC∞c ([−1, 1]) inH02([−1, 1]) (in the H02-norm), implies that
Jf1+η≥Jf1
, ∀η∈C∞c [−1, 1]. (3.22) This means that for allϕ∈H02([−1, 1]), we haveJ[ϕ]≥J[f1] and ifϕ≡ f1(almost ev- erywhere), then ϕ−f1≡0 and hence, J[ϕ]> J[f1]. Thus f1 is the unique minimum
forJ.
Reversing the roles of f andg, that is, fixing f0and findingg1∈H02to minimizeI[f0g]
overg∈H02([−1, 1]), we obtain the same conclusion by using the same arguments.
Corollary 3.3. Fix f0∈H02([−1, 1]). Theng1∈H02([−1, 1]) is the unique minimizer for J[g]=If0g=f02
L2
1
−1(g)2d y+ 2f02L2
1
−1(g)2d y +f02L2
1
−1g2d y−2γf0
L1
1
−1g d y
(3.23)
if and only ifg1solves the Euler-Lagrange equation f02
L2
d4g
d y4−2f02L2
d2g
d y2+f02L2g=2γ 1
−1f0(x)dx. (3.24) Now we search for the solutionf1∈H02([−1, 1]) in (3.18), that is,
g02
L2
d4f
dx4 −2g02L2
d2f
dx2 +g02L2f =2γ 1
−1g0(y)d y. (3.25) Rewrite the above ODE in the following form:
g02
L2
⎡
⎣
D−g2L2 g02
L2
2
+g2L2 g02
L2
−g4L2 g04
L2
⎤
⎦f =2γ 1
−1g0(y)d y, (3.26) whereD=d2/dx2.
Remark 3.4. In general wheng∈H2, that is,gneeds not satisfy the zero boundary con- ditions for function inH02, then the quantity
g02L2
g02
L2
−g04L2
g04
L2
(3.27) can take on any values. However, ifg∈H02andg0≡0, as proved below, this quantity is always positive.
Lemma 3.5. LetΩbe a Lipschitz domain inRn,n≥1. Letg∈H02(Ω) be arbitrary. Then
∇g2L2≤ gL2· ΔgL2, (3.28) and equality holds if and only ifg≡0.
Proof. Integration by parts yields
∇g2L2=
Ω∇g· ∇g dx= −
ΩgΔg dx +
∂Ωg∂g
∂ndσ= −
ΩgΔg dx. (3.29) By the Cauchy-Schwartz inequality, we have
∇g2L2≤ gL2· ΔgL2, (3.30) and the equality holds if and only if (see Lieb-Loss [9])
(i)|g(x)| =λ|Δg(x)|almost everywhere for someλ >0, (ii)g(x)Δg(x)=eiθ|g(x)| · |Δg(x)|.
Sincegis real-valued, (i) and (ii) imply
g(x)Δg(x)=λΔg(x)2. (3.31)
So,gmust satisfy the following PDE:
Δg−1
λg=0, (3.32)
whereg∈H02(Ω). But the only solution to this PDE isg≡0 (see, Evans [3, pages 300–
302]). This completes the proof of the lemma.
Remark 3.6. Ifn=1, one can solveg−λ−1g=0 directly without having to appeal to the theory of elliptic PDEs.
Proposition 3.7. The solutions of (3.18) and (3.24) have the same form.
Proof. Using eitherLemma 3.5in casen=1 to the above remark, we see that g2L2
g02
L2
−g4L2 g04
L2
>0 ifg0≡0. (3.33)
Hence the characteristic polynomial associated to (3.26) has two pairs of complex con- jugate roots as long asg0≡0. Apply the same arguments to the ODE in (3.24) and the
proposition is proved.
Remark 3.8. The statement inProposition 3.7was claimed in [8] without verification.
Indeed the authors stated therein that the solutions of (3.18) and (3.24) are of the same form because of the positivity of the coefficients on the left-hand side of (3.18) and (3.24).
As observed inRemark 3.4and proved inProposition 3.7, the positivity requirement is not sufficient. The fact that f0,g0∈H02must be used to conclude this assumption.
4. Explicit solution for (3.26)
We now find the explicit solution for (3.26), and hence for (3.18). Let r=gL2
g0
L2
, t=gL2
g0
L2
, ρ=
% t+r2
2 , κ=
% t−r2
2 .
(4.1)
Then from Proposition 3.7and its proof, the 4 roots of the characteristic polynomial associated to ODE (3.26) are
ρ+iκ, ρ−iκ, −ρ−iκ, −ρ+iκ. (4.2)
Thus the homogeneous solution of (3.26) is
fh(x)=c1cosh(ρx) cos(κx) +c2sinh(ρx) cos(κx)
+c3cosh(ρx) sin(κx) +c4sinh(ρx) sin(κx). (4.3)