R E S E A R C H
Open Access
Computation of Green’s functions through
algebraic decomposition of operators
F Adrián F Tojo
**Correspondence:
[email protected] Departamento de Análise Matemática, Facultade de Matemáticas, Universidade de Santiago de Compostela, Santiago de Compostela, Spain
Abstract
In this article, we use linear algebra to improve the computational time for obtaining Green’s functions of linear differential equations with reflection (DER). This is achieved by decomposing both the ‘reduced’ equation (the ODE associated with a given DER) and the corresponding two-point boundary conditions.
MSC: 34Bxx; 47Lxx; 34Kxx
Keywords: Green’s functions; ODE; reflection; decomposition
1 Introduction
Differential operators with reflection have recently been of great interest, partly due to their applications to supersymmetric quantum mechanics [–] or topological methods applied to nonlinear analysis [].
In the last years, the works in this field have been related to either obtaining eigenvalues and explicit solutions of different problems [–], their qualitative properties [, ], or ob-taining the associated Green’s function [–]. In [], the authors described a method to derive the Green’s function of a differential equation with constant coefficients, reflection and two-point boundary conditions. This algorithm was implemented inMathematica (see []) in order to put it to a practical use. Unfortunately, it was soon observed that, although theoretically correct, there were severe limitations when computing the Green’s functions of problems of high order. In this respect, we have to point out that annth-order linear DER is reduced to a (n)th-order ordinary differential equation; see Theorem . and compare equations (.) and (.). This particularity posses a great challenge since the computational time increases greatly withn.
To sort this out, the best option is to go back from a (n)th-order problem totwo prob-lems of ordern. This procedure, compared to solving directly the order n, is much faster. Furthermore, it also happens that, in some cases, the decomposition provides two equiva-lent problems or a problem and its adjoint. In those cases, the improvement is even more notorious.
In the next section, we contextualize the problem with a brief introduction to differential equations with reflection and state some basic results concerning the Green’s function associated with them. In Section , we develop some theoretical results, which provide a way of decomposing the DER we are dealing with. Finally, in Section , we establish a
suitable decomposition for the boundary conditions, state criteria for self-adjointness of the decomposed problem, and provide examples to illustrate the theory.
2 Differential equations with reflection
In order to establish a useful framework to work with these equations, we consider the differential operator D, the pullback operator of the reflection ϕ(t) = –t, denoted by ϕ∗(u)(t) =u(–t), and the identity operator Id.
LetT ∈R+andI:= [–T,T]. We now consider the algebraR[D,ϕ∗] consisting of the linear operators of the form
L=
n
k=
akϕ∗+bk
Dk, (.)
wheren∈N,ak,bk∈R,k= , . . . ,n, which act as
Lu(t) =
n
k=
aku(k)(–t) + n
k=
bku(k)(t), t∈I,
on any functionu∈Wn,(I). The operation in the algebra is the usual composition of operators; we will omit the composition sign. We observe that Dkϕ∗ = (–)kϕ∗Dk for
k= , , . . . , which makes it anoncommutative algebra. We will consider, for convenience, the sumsnk=≡ksuch thatk∈ {, , . . .}, but taking into account that the coefficients ak,bkare zero for big enough indices.
Notice thatR[D,ϕ∗] is not a unique factorization domain. For instance, D– = (D+ )(D– ) = –ϕ∗D+ϕ∗.
LetR[D] be the ring of polynomials with real coefficients on the variableD. The follow-ing property is crucial for the obtainfollow-ing of a Green’s function.
Theorem .([], Theorem .) Take L be as defined in(.)and define
R:=
k
akϕ∗Dk+
l
(–)l+blDl∈R
D,ϕ∗. (.)
Then RL=LR∈R[D].
Remark . IfS:=RL=kn=ckDk, then
ck= ⎧ ⎨ ⎩
, kodd,
k
–
l=(–)l(alak–l–blbk–l) + (–)
k
(a
k
–b k
), keven.
This implies that thereducedoperatorRLhas only coefficients for the even powers of the derivative, so the equation is self-adjoint. If the boundary conditions are appropriate (we will clarify this statement in Theorem .), then the Green’s function is symmetric []. Observe thatc=a–b. Also, ifL=
n
i=(aiϕ∗+bi)Diwithan= orbn= , then we
two elements ofR[D,ϕ∗], we can get another element that has simpler terms in the sense of derivatives of lower order. This is quite difficult when computing the Green’s functions since, in this case, we could have one, many, or no solutions of our original problem []. The following example is quite illustrative.
Example . Consider the equation
x)(t) +x)(–t) =sint, t∈I.
This equation cannot have a solution since the left-hand side is an even function whereas the right-hand side is an odd function.
As we said before,S=RLis a usual differential operator with constant coefficients. Con-sider now the following problem:
Su(t) :=
n
k=
akuk)(t) =h(t), t∈I,
Bku:= n–
j=
αkjuj)(–T) +βkjuj)(T)
= , k= , . . . ,n.
(.)
The existence of Green’s functions for problems such as (.) is a classical result (see,e.g., []). Here we present it adapted to our framework.
Theorem . Assume that the following homogeneous problem has a unique solution:
Su(t) = , t∈I, Bku= , k= , . . . ,n.
Then there exists a unique function,calledGreen’s function,such that: (G) Gis defined on the squareI.
(G) The partial derivatives∂kG
∂tk exist and are continuous onIfork= , . . . ,n– . (G) ∂∂nt–n–G and
∂nG
∂tn exist and are continuous onI\{(t,t) :t∈I}. (G) The lateral limits ∂n–G
∂tn–(t,t+)and
∂n–G
∂tn–(t,t–)exist for everyt∈(a,b),and
∂n–G ∂tn–
t,t––∂
n–G
∂tn–
t,t+= an
.
(G) For eachs∈(a,b),the functionG(·,s)is a solution of the differential equation
Su= onI\{s}.
(G) For eachs∈(a,b),the functionG(·,s)satisfies the boundary conditionsBku= ,
k= , . . . ,n.
Furthermore,the function u(t) :=abG(t,s)h(s) ds is the unique solution of problem(.).
Now we can state the result that relates the Green’s function of a problem with reflection to the Green’s function of its associated reduced problem.
Theorem .([]) Let I= [–T,T].Consider the problem
Lu(t) =h(t), t∈I, Biu= , k= , . . . ,n, (.)
where L is defined as in(.),h∈L(I),and
Bku:= n–
j=
αkjuj)(–T) +βkjuj)(T)
, k= , . . . ,n.
Then,there exists R∈R[D,ϕ∗] (as in(.))such that S:=RL∈R[D],and the unique solu-tion of problem(.)is given byabRG(t,s)h(s) ds,where G is the Green’s function associ-ated with the problem
Su= , (.)
Bku= , k= , . . . ,n, (.)
BkRu= , k= , . . . ,n, (.)
assuming that it has a unique solution.
As stated in Section , Theorem . was implemented in Mathematica in []. We now proceed to describe some steps that could be added to the algorithm in order to improve it.
3 Decomposing the reduced equation
The computation of Green’s functions is prohibitive in computation time terms [], mostly for high-order equations, so it is necessary to find ways to palliate this problem. Our approach consists of decomposing our problem in order to deal with equations of lower order.
First, observe that from Remark . we know that the reduced equation has no deriva-tives of odd indices. For convenience, ifpis a real (complex) polynomial, then we denote byp–the polynomial with the same principal coefficient and opposite roots.
Lemma . Let n∈N,and let p(x) =nk=αkxkbe a real polynomial of ordern.Then
there is a complex polynomial q of order n such that p=αnqq–.Furthermore,ifp(x) =˜
n
k=αkxkhas no negative roots,then q is a real polynomial.
Proof First, observe thatpis a polynomial onx, and therefore, ifλis a root ofp, so is –λ. Hence, using the fundamental theorem of algebra, the first part of the result can be derived by separating the monomials that composepin two different polynomials with opposite roots.
Let us explicitly show that in the casep˜has no negative roots,qis a real polynomial. Take the change of variablesy=x. Then,p(x) =p(y), and, by the fundamental theorem˜ of algebra,
˜
p(y) =
n
k= αkyk
=αnyσ
y–λ· · ·y–λmy+λm+
· · ·y+λmy+μy+ν
for some integers σ,m, m,l and real numbersλ, . . . ,λm,ν, . . . ,νl,μ, . . . ,μl such that
λk> andνk>|μk|/ for everykin the appropriate set of indices. The terms of the form
y+μky+νkcorrespond to the pairs of complex roots of the polynomial. This means that
the discriminant =μ
k– νk< , that is,νk>|μk|/.
Hence,
p(x) =αnxσ
x–λ· · ·x–λmx+λm+
· · ·x+λmx+μx+ν
· · ·x+μlx+νl
.
Now we have
x–λk= (x+λk)(x–λk),
x+λk= (x+λki)(x–λki), and
x+μkx+νk
=x–xνk–μk+νk
x+xνk–μk+νk
for anykin the appropriate set of indices. Define
q(x) =xσ(x–λ)· · ·(x–λm)(x–λm+i)· · ·(x–λmi)
x–xν–μ+ν
· · ·x–xνl–μl+νl
and
q–(x) =xσ(x+λ)· · ·(x+λm)(x+λm+i)· · ·(x+λmi)
x+xν–μ+ν
· · ·x+xνl–μl+νl
.
We have thatp=αnqq–.
Observe that ifλis a root ofp, thenλis a root ofp. Hence, if˜ p˜has no negative roots, thenphas no roots of the formλ=aiwitha= . Thus,
p(x) =αnxσ
x–λ· · ·x–λmx+μx+ν
· · ·x+μlx+νl
,
q(x) =xσ(x–λ)· · ·(x–λm)
x–xν–μ+ν
· · ·x–xνl–μl+νl
,
q–(x) =xσ(x+λ)· · ·(x+λm)
x+xν–μ+ν
· · ·x+xνl–μl+νl
,
that is,qis a real polynomial.
Remark . Descartes’ rule of signs establishes that the number of positive roots (with multiple roots counted separately) of a real polynomial on one variable either is equal to the number of sign differences between consecutive nonzero coefficients or is less than it by an even number, provided that the terms of the polynomial are ordered by descending variable exponent. This implies that for a polynomialp(x) to have no negative roots, it suffices that all coefficients ofp(–x) are positive, that is,p(x) has positive even coefficients and negative odd coefficients.
The following lemma establishes a relation between the coefficients ofqandq–.
Lemma . Let n∈N,and let q(x) =nk=αkxkbe a complex polynomial.Then
q–(x) =
n
k=
(–)k+nαkxk.
Proof We proceed by induction. Forn= ,q(x) =α(x–λ). Clearly,qhas the rootλ, and q–(x) =α(x+λ) = (–)+αx+ (–)αλhas the root –λ.
Assume that the result is true for somen≥. Then, forn+ ,qis of the formq(x) = (x–λn+)r(x), wherer(x) =
n
k=αkxkis a polynomial of ordern, that is,
q(x) = (x–λn+)
n
k=
αkxk=xn++ n
k=
[αk––λn+αk]xk–λn+α.
Now,q–(x) = (x+λn+)r–(x). Since the formula is valid forn,
q–(x) = (x+λn+)r–(x) = (x+λn+)
n
k=
(–)k+nαkxk
=xn++
n
k=
(–)k+n+[αk––λn+αk]xk– (–)n+λn+α.
So the formula is valid forn+ as well.
Remark . The result can be directly proven by considering the last statement in Re-mark .. If we take a polynomialp(x) =a(x–λ)· · ·(x–λn), then the polynomialp(–x) has
exactly opposite roots. In fact,p(–x) =a(–x–λ)· · ·(–x–λn) = (–)na(x+λ)· · ·(x+λn).
It is easy to check that the coefficients ofp(–x) are precisely as described in Lemma . save for the factor (–)n.
This last lemma allows the computation of the polynomialsqandq–related to the poly-nomialRLon the variableDusing the formula given in Remark .. We will assume that RLis of order n, that is,a
n–bn= . Otherwise, the problem of computingqandq–would be the same, but these polynomials would be of lower order. Also, assume thatRL, con-sidered as a polynomial onD, has no negative roots in order forqto be a real polynomial. IfL=nk=(akϕ∗+bk)Dkandq(D) =Dn+
n–
k=αkDk, then
RL= n
k=
ckDk= (–)n
an–bnq(D)q–(D).
This relation establishes the following system of quadratic equations:
ck= k–
l= (–)l(a
lak–l–blbk–l) + (–)k
ak–bk
=an–bn
k–
l=
(–)l(αlαk–l) + (–)kαk
whereak,bk,αk= ifk∈ {/ , . . . ,n}andαn= . These arenequations withnunknowns
α, . . . ,αn. We present here the case ofn= to illustrate the solution of these equations.
Example . Forn= , we have that
RL=a–bD+–a+ aa+b– bb
D+a–b,
a–bq(D)q–(D) =
a–bD+α–α
a–bD+αa–b,
and the system of equations is
a–b=a–bα,
–a+ aa+b– bb=
a–bα–α
.
(.)
Before computing the solutions, let us state explicitly the limitations thatRL, considered as an order polynomial onD, that is,RL(x) =ax+bx+c, has no negative roots implies. There are two options:
(I) There are two complex roots, that is, =b– ac< . This is equivalent to ac> ∧ |b|< √acor, expressed in terms of the coefficients ofRL,
b–ab–a> and –a+aa+b–bb<
b –a
b –a
.
(II) There are two nonnegative roots, that is, =b– ac≥, and
–b+√b– ac/(a)≤.
This is equivalent to(a,c≥∧–b≥√ac)∨(a,c≤∧b≥√ac)or, expressed in terms of the coefficients ofRL,
b–a,b–a≥∧––a+ aa+b– bb
≥b –a
b –a
or
b–a,b–a≤∧––a+ aa+b– bb
≥b–ab–a.
Now, with these conditions, the solutions of the system of equations (.) are as follows. Case(I). We have two solutions:
α=
b–a b–a,
α=±
sign(a –b)
(b
–a)(b–a) – (–a+ aa+b– bb) a
–b
.
Case(II). We have four solutions depending on whether we chooseξ= orξ= –:
α=ξ
b –a b
α=±
ξsign(a –b)
(b
–a)(b–a) – (–a+ aa+b– bb) a
–b
.
These solutions provide well-defined real numbers by conditions (I) and (II).
4 Decomposing the boundary conditions
Now we consider the cases where the problem can be decomposed into two equations. We will try to identify those circumstances when problem (.)-(.)-(.) can be expressed as an equivalent factored problem of the form
Lu=y, Vju= , j= , . . . ,n, (.)
Ly=Rh, Vjy= , j= , . . . ,n, (.)
whereS=LL. If that where the case, then conditions (.)-(.) would be equivalent to
Vju= , VjLu= , j= , . . . ,n. (.)
In this case, the Green’s function of problem (.)-(.)-(.) can be expressed as
G(t,s) =
T
–T
G(t,r)G(r,s) dr,
whereGis the Green’s function associated with problem (.), andGis the one associ-ated with problem (.), assuming that both Green’s functions exist.
In order to guarantee that (.)-(.) and (.) are equivalent, let us establish the follow-ing definitions. Let
:= (αkj)jk=,...,=,...,nn–, Xn:=
u(T),u(T), . . . ,u(n–)(T),
:= (βkj)jk=,...,=,...,nn–, Xn:=
u(–T),u(–T), . . . ,u(n–)(–T).
Then the boundary conditions (.) can be expressed asXn+Xn= . In the same
way, (.) can be written as ()Xn+ ()Xn= for some matrices,,,∈ Mn(R). So, globally, the conditions on equation (.) can be expressed as
Xn+
Xn= . (.)
Now, assume thatLandVjare of the form
L=
n
l= clDl,
Vju= n–
k=
γjkuk)(–T) +δjkuk)(T)
=
n–
k=
γjk(–T)∗+δjkT∗
for somecl,γjk,δjk∈R,l,j,k= , . . . ,n, wherea∗denotes the pullback by the constanta.
Define now:= (γjk)j,k,:= (δjk)j,k∈Mn(R), and
= (djk)kj=,...,=,...,n–n–:=
⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
c c c · · · cn– cn · · ·
c c · · · cn– cn– cn · · ·
c · · · cn– cn– cn– cn · · ·
..
. ... ... . .. ... ... ... ... . .. ...
· · · c c c c · · · cn
⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ =
∈Mn×n(R),
where,∈Mn(R),is invertible (becausecn= ), andis invertible if and only if c= .
Now we are ready to start the calculations. We have that
(VjLu)j= n–
k=
γjk(–T)∗+δjkT∗
Dk
n
l= clDlu
j = n– k= n l=
γjkcl(–T)∗+δjkclT∗
Dk+lu
j = n– k=
k+n
m=k
γjkcm–k(–T)∗+δjkcm–kT∗
Dmu
j = n– k= n–
m=
γjkdkmu(m)(–T) +δjkdkmu(m)(T) j = n– k= γjkdkm
j,m
Xn+ m
k= δjkdkm
j,m
Xn=Xn+Xn.
Hence, we can write (.) in the form
Xn+
Xn= . (.)
Clearly, it is convenient to take=and=, that is,Vj=Bj,j= , . . . ,n.
Lemma . If andare invertible and=–+– –– – , then,taking
=, =, = Id, and =–– ,
condition(.)is equivalent to condition(.),and,therefore,problems(.)-(.)-(.) and(.)-(.)are equivalent.
The matrixAis invertible, and
=A
,
=A
.
Hence, conditions (.) and (.) are equivalent.
Analogously, we have a result whenandare invertible.
Lemma . Ifandare invertible and=– +– –– – , then,taking
=, =, = Id, and =– – ,
condition(.)is equivalent to condition(.),and,therefore,problems(.)-(.)-(.) and(.)-(.)are equivalent.
The following example illustrates this discussion explicitly.
Example . Consider the following problem:
u(t) +u(–t) +u(t) =h(t), t∈I,
u(–) –u() = , u(–) =u(), u(–) –u() = ,
(.)
whereh(t) =sint. Then, the operator we are studying isL=D+ϕ∗+ . If we takeR:= D+ϕ∗– , then we have thatRL=D, which admits a simple decomposition inR[D] as RL= (D)(D) =L
L.
The boundary conditions are
(–)∗– ∗Du= , (–)∗D– ∗Du= , (–)∗D– ∗u= .
Taking this into account, we add the conditions
=(–)∗– ∗DRu=u(–) –u()(),
=(–)∗D– ∗DRu=u()(–) –u()(),
=(–)∗D– ∗Ru=u()(–) –u().
Then our newreducedproblem, writing the boundary conditions in matrix form, is
u()(t) =f(t),
⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠
⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
u(–) u(–) u(–) u(–) u()(–) u()(–)
⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠
+
⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
–
–
–
–
–
–
⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠
⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
u() u() u() u() u()() u()()
⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠
= ,
wheref(t) =Rh(t) =h(t) +h(–t) –h(t) = –sint.
Now, we can check that we are working under the conditions of Lemma .. We have that== Id,== , and
==
⎛ ⎜ ⎝
–
–
–
⎞ ⎟ ⎠.
On the other hand,
= () =
⎛ ⎜ ⎝
⎞ ⎟ ⎠.
Thus, it is straightforward to check that
–+– –– – == ,
and therefore the hypotheses of Lemma . are satisfied. The conditionsVjare given by
the matrices= Id and=–– =. Hence, we know that this problem is equivalent to the factored system
u(t) =v(t), u(–) –u() = ,
u(–) =u(), u(–) –u() = ,
(.)
v(t) = –sint, v(–) –v() = ,
v(–) =v(), v(–) –v() = .
(.)
Thus, it is clear that
u(t) =
–
G(t,s)v(s) ds, v(t) =
–
G(t,s)f(s) ds,
whereG=Gare, respectively, the Green’s functions of (.) and (.). The Green’s func-tions of problems involving linear ordinary differential equafunc-tions with constant coeffi-cients and two-point boundary conditions can be computed with theMathematica note-books [] or []. Explicitly,
G(t,s) =
⎧ ⎨ ⎩
–(s–t)(s(t– ) +t– ), –≤s≤t≤,
Hence, the Green’s functionGfor problem (.) is given by
G(t,s) =
–
G(t,r)G(r,s) dr
= ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
s(t+ ) – s(t(t+ ) + )
+ st(t+ ) – s(t(t+ )– )
+ st(t(t(t+ ) + ) – ) – t
– t+ t– , – <t<s≤,
–s– s– (s(s+ )– )t
+ (s(s(s+ ) + ) – )st
+ s+ (s+ )t– (s(s+ ) + )t
+ (s+ )st– , –≤s≤t≤.
Therefore, using Theorem ., the Green’s function for problem (.) is
G(t,s) =RG(t,s) =∂ G
∂t (t,s) +G(–t,s) +G(t,s)
= ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
–(s– )t+ (s– )st+ (s– )t– (s– )s
– (s– s+ s+ s– s+ )t, –≤ |t| ≤s≤,
s(–(t– )) + s(t– )t– s(t– ) + (t– )t
+s(–t+ t– t+ t+ t+ ), –≤ |s|<t≤,
s(–(t+ )) – st(t+ ) – s(t+ ) – t(t+ )
–s(t+ t+ t– t– t– ), –≤ |s|< –t≤,
–(s+ )t– s(s+ )t+ (s+ )t+ s(s+ )
– (s+ s+ s+ s– s+ )t, –≤ |t| ≤–s≤.
Hence, the solution of problem (.) is given by
u(t) =
–
G(t,s)sin(s) ds
= –
– – t– t+ t+tsin()
+
t– t– cos() + sin(t) +cos(t).
Computationally, this procedure poses a big advantage: it is always easier to obtain the Green’s function for twonth-order problems than for one (n)th-order problem. Further-more, if the hypotheses of Lemma . are satisfied and we are able to obtain a factorization of the aforementioned kind usingqandq–in the place ofLandL, then we have an extra advantage: the differential equation given by q– is the adjoint equation of that given by qmultiplied by the factor (–)n. This fact, together with the following result (which can
Theorem . Consider an interval J= [a,b]⊂R,functionsσ,ai∈L1(J),i= , . . . ,n,real
numbersαij,βij,hi,i= , . . . ,n,j= , . . . ,n– ,a vector subspace D(Ln)⊂Wn,(J),the
oper-ator
Lnu(t) =au(n)(t) +a(t)u(n–)(t) +· · ·+an–(t)u(t) +an(t)u(t), t∈J,u∈D(Ln),
with a= ,and the problem
Lnu(t) =σ(t), t∈J, Ui(u) =hi, i= , . . . ,n, (.)
where
Ui(u) := n–
j=
αiju(j)(a) +βiju(j)(b)
, i= , . . . ,n.
Then,the associated adjoint problem is
L†nv(t) =
n
j=
(–)jan–j(t)u(j)(t), t∈J,v∈D
L†n, (.)
where
DL†
n
=
"
v∈Wn,(J) :b∗–a∗
n
j=
j–
i=
(–)(j–i–)(an–jv)j–i–u(i)
= ,u∈D(Ln) #
.
Furthermore,if G(t,s)is the Green’s function of problem(.),then that associated with problem(.)is G(s,t).
Hence, if we can decompose problem (.)-(.)-(.) in two adjoint problems of the form (.)-(.), then its Green’s function is
G(t,s) =
T
–T
G(t,r)G(r,s) dr=
T
–T
G(t,r)G(s,r) dr,
whereGis the Green’s function of (.), andG(t,s) =G(s,t) is that of (.). We note, though, that unless the operatorq–is the adjoint equation times (–)n, the boundary con-ditions may not be the adjoint ones.
Example . Consider the problem
u(–t) +u(t) +√u(–t) =f(t) :=et, t∈[–, ], u(–) =u(). (.)
TakingR=ϕ∗D+√ϕ∗– Id and composing problem (.) with this operator, we obtain the reduced problem
Problem (.) is equivalent to the factored system
u(t) +u(t) =v(t), u(–) =u(), (.)
–v(t) +v(t) = –Rf(t), v(–) =v() (.)
fort∈[–, ]. Observe that problem (.) is the adjoint problem of (.). Since the Green’s function of problem (.) is given by
G(t,s) :=
⎧ ⎨ ⎩
es–t+
e–, –≤s≤t≤,
es–t
e–, – <t<s≤,
and, therefore,G(s,t) is the Green’s function of problem (.), the Green’s function of problem (.) is
G(t,s) = –
–
G(t,r)G(s,r) dr=
⎧ ⎨ ⎩
–es–t++et–s
e– , –≤s≤t≤,
–es–t+ee––s+t+, – <t<s≤.
Finally, the Green’s function of problem (.) is
G(t,s) =RG(t,s) =∂G ∂t(–t,s) +
√
G(–t,s) –G(t,s)
=
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
e–s–t[(√–)(–e(s+t+))+es++et–√–]
(e–) , |t| ≤–s,
e–s–t[(√–)(–e(s+t))+es++et–(+√)e]
(e–) , |s|<t,
e–s–t[(√–)(–e(s+t+))+es+et+–√–]
(e–) , |s|< –t,
e–s–t[(√–)(–e(s+t))+es+et+–(+√)e]
(e–) , |t| ≤s.
Hence, the solution of problem (.) is
u(t)
:=–e
–t(–( +√)t+e(( +√)t+ √) +et(–t+e(t+√ – ) –√) +√ + )
(e– ) .
Competing interests
The author declares that they have no competing interests.
Acknowledgements
The author wants to acknowledge his gratitude to the anonymous referee for helping improve the manuscript, especially in the proof of Lemma 3.1. This work was partially supported by Consellería de Cultura, Educación e Ordenación Universitaria, Xunta de Galicia, Spain, project EM2014/032 and supported by FPU scholarship, Ministerio de Educación, Cultura y Deporte, Spain.
Received: 6 May 2016 Accepted: 31 August 2016
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