The Eigenfunction Expansion for a Dirichlet Problem with Explosive Factor

Abstract

We prove the eigenfunction expansion formula for a Dirichlet problem with explosive factor by two ways, first by standard method and second by proving a convergence in some metric space 𝐿₂(0,𝜋;𝜌(𝑥)).

1. Introduction

The solutions of many problems of mathematical physics are reduced to the spectral investigation of a differential operator or finding the expansion of arbitrary function, in series or integration, in terms of the eigenfunctions of a differential operator. The differential operator is called regular if it its domain is finite and its coefficients are continuous; otherwise it is called singular differential operator. The Sturm-Liouville theory occupies a central position of the spectral theory of regular operator. During the development of quantum mechanics, there was an increase in the interest of spectral theory of singular operator on which we will restrict our attention. The first basic role in the development of spectral theory of singular operator dates back to E. Ch. Tetchmarsh [1]. He gave a new approach in the spectral theory of singular differential operator of the second order by using contour integration. Also Levitan [2] gave a new method, he obtained the eigenfunction expansion in infinite interval by taking limit of a regular case. In the last time about twenty five or so years, due to the needs of mathematical physics, in particular, quantum mechanics, the question of solving various spectral problems with explosive factor has been arisen. These appeared also in the study of geophysics and electromagnetic field, see Alemov [3]. The spectral theory of differential operators with explosive factor is studied by A. N. Tekhanov, M. G. Krien, M. G. Gasimov, and others. In this paper, we find the eigenfunction expansion formula and prove its convergence for following version of a Dirichlet problem (1.2), (1.3). The introduction of the weight function 𝜌(𝑥) (1.4) as ± signs causes many analytical difficulties, see [4], because the problem is to be treated as two separated problems and so the formula of eigenfunctions expansion is written as two sums. In [5] the author considered the weight function of the form 𝜌(𝑥)=𝛼²1;;I𝑎m<𝛼𝑥≠≤0,𝜋0,≤𝑥≤𝑎<𝜋,(1.1) and the spectrum was both continuous and discreet so that the formula of eigenfunctions expansion obtained there was written as a summation and integration. We must notice that the result of this paper is a starting point in solving the inverse spectral problem which will be investigated later on.

Consider the following Dirichlet problem: −𝑦^+𝑞𝑦((𝑥0))𝑦==0,𝜆𝜌(𝑥𝑦)(𝑦𝜋,)=00≤,𝑥≤𝜋,((11..23)) where the nonnegative real function 𝑞(𝑥) has a second piecewise integrable derivatives on [0,𝜋], 𝜆 is a spectral parameter, and the weight function or the explosive factor 𝜌(𝑥) is of the form 𝜌(𝑥)=1−,1,0𝑎≤<𝑥𝑥≤≤𝑎𝜋.<𝜋,(1.4) In [4] the author proved that the eigenvalues 𝜆^±𝑛,𝑛=0,1,2,…, of problem (1.2)-(1.3) are real and the corresponding eigenfunctions 𝜑(𝑥,𝜆^±𝑛),𝜓(𝑥,𝜆^±𝑛) are orthogonal with weight function 𝜌(𝑥). We prove, here, the reality of these eigenfunctions under the condition that 𝑞(𝑥) is real. Indeed, let 𝜑(𝑥,𝜆) be the solution of the differential equation (1.2), 𝑥∈(0,𝑎) which satisfies the conditions𝜑(0,𝜆)=0,𝜑^(0,𝜆)=1(1.5) so that −𝜑^(𝑥,𝜆)+𝑞(𝑥)𝜑(𝑥,𝜆)=𝜆𝜑(𝑥,𝜆),(1.6) taking the complex conjugate we have −𝜑^(𝑥,𝜆)+𝑞(𝑥)𝜑(𝑥,𝜆)=𝜆𝜑(𝑥,𝜆).(1.7) By the aid of the uniqueness theorem, we have 𝜑(𝑥,𝜆)=𝜑(𝑥,𝜆). In a similar way, we can see that 𝜓(𝑥,𝜆)=𝜓(𝑥,𝜆) where 𝜓(𝑥,𝜆) is the solution of (1.2), 𝑥∈(𝑎,𝜋),  𝜓(𝜋,𝜆)=0,  𝜓′(𝜋,𝜆)=1, that is, the eigenfunctions of the problem (1.2)-(1.3) are real. As we know from [4], the eigenvalues of problem (1.2)-(1.3) coincide with the roots of the function Ψ(𝜆)=0, where Ψ(𝜆) is the Wronskian of the two solutions 𝜑(𝑥,𝜆), 𝜓(𝑥,𝜆) of (1.2)-(1.3)Ψ(𝜆)=𝑊[𝜑(𝑎,𝜆),𝜓(𝑎,𝜆)]=0.(1.8) In the following lemma, under the reality of 𝑞(𝑥), we prove the simplicity of the eigenvalues, that is, we prove that the roots of (1.8) are simple, in other cases for 𝑞(𝑥) is complex the roots of (1.8) may not be simple.

Lemma 1.1. Under the conditions stated in the introduction with respect to the problem (1.2)-(1.3), the eigenvalues of the problem (1.2)-(1.3) are simple.

Proof. We prove that Ψ̇(𝜆)≠0 where the dot means differentiation with respect to 𝜆. Let 𝜑(𝑥,𝜆) be the solution of the problem −𝑦^+𝑦(𝑞0()𝑥=)𝑦0=,𝜆𝑦,𝑦_(0≤𝑥≤𝑎),(0)=1,(1.9) and let 𝜓(𝑥,𝜆) be the solution of the problem −𝑦^+𝑦𝑞(𝜋(𝑥))=𝑦0=,−𝜆𝑦𝑦,_(𝑎<𝑥≤𝜋),(𝜋)=1.(1.10) From (1.9), we have −𝜑^(𝑥,𝜆)+𝑞(𝑥)𝜑(𝑥,𝜆)=𝜆𝜑(𝑥,𝜆).(1.11) Differentiating (1.11) with respect to 𝜆, we have −𝜑̇^(𝑥,𝜆)+𝑞(𝑥)𝜑̇(𝑥,𝜆)=𝜆𝜑̇(𝑥,𝜆)+𝜑(𝑥,𝜆).(1.12) Multiplying (1.11) by 𝜑̇(𝑥,𝜆) and (1.12) by 𝜑(𝑥,𝜆) and then subtracting the two results we have (𝑑/𝑑𝑥)[𝜑̇𝜑^−𝜑̇^𝜑]=𝜑²(𝑥,𝜆) from which, by integrating with respect to 𝑥, from 0 to 𝑎 and using the conditions in (1.10), we obtain 𝜑^𝑎,𝜆_𝑛𝜑̇𝑎,𝜆_𝑛−𝜑̇^𝑎,𝜆_𝑛𝜑𝑎,𝜆_𝑛=^𝑎0𝜑²𝑥,𝜆_𝑛𝑑𝑥.(1.13) In a similar way, from (1.10), we can write −𝜓^𝑎,𝜆_𝑛𝜓̇𝑎,𝜆_𝑛+𝜓̇^𝑎,𝜆_𝑛𝜓𝑎,𝜆_𝑛=−^𝜋𝑎𝜓²𝑥,𝜆_𝑛𝑑𝑥.(1.14) From (1.8), we have 𝜑𝑎,𝜆_𝑛𝜓^𝑎,𝜆_𝑛=𝜑^𝑎,𝜆_𝑛𝜓𝑎,𝜆_𝑛.(1.15) From which we have 𝜑^𝑎,𝜆_𝑛=𝜑𝑎,𝜆^𝑛𝜓^𝑎,𝜆_𝑛𝜓_𝜓𝑎,𝜆_𝑛,𝑎,𝜆_𝑛=𝜑^𝑎,𝜆_𝑛𝜓𝑎,𝜆_𝑛𝜑𝑎,𝜆_𝑛.(1.16) Substituting from (1.16) into (1.13) and (1.14), we can see by adding that Ψ̇𝜆_𝑛=𝑟_𝑛𝑎_𝑛,(1.17) where 𝑎_𝑛=^𝑎0𝜑²𝑥,𝜆_𝑛𝑑𝑥−1𝑟^2𝑛^𝜋𝑎𝜓²𝑥,𝜆_𝑛𝑑𝑥,𝑟_𝑛=𝜓𝑎,𝜆^𝑛𝜑𝑎,𝜆_𝑛,(1.18) and the numbers 𝑎_𝑛 are the normalization numbers of the eigenfunctions of problem (1.2)-(1.3). Following [4], we have 𝑟_𝑛≠0 and 𝑎_𝑛≠0 which complete the proof of lemma.

2. The Function 𝑅(𝑥,𝜉,𝜆)

We introduce the function 𝑅(𝑥,𝜉,𝜆) by 𝑅(𝑥,𝜉,𝜆)=−1Ψ(𝜆)𝜑𝜑((𝑥𝜉,,𝜆𝜆))𝜓𝜓((𝑥𝜉,,𝜆𝜆)),,𝑥𝜉≤≤𝑥𝜉,,(2.1) which is called the Green’s function of the nonhomogenous Dirichlet problem −𝑦^+𝑞(𝑥)𝑦=𝜆𝑦𝜌((0𝑥))=𝑦+𝑦(𝜌𝜋()𝑥=)𝑓0,(𝑥),0≤𝑥≤𝜋,(2.2) where 𝑓(𝑥)∈𝐿₁(0,𝜋) and 𝜌(𝑥) is defined by (1.4). The function 𝑅(𝑥,𝜉,𝜆) is, also, called the kernel of the resolvent 𝑅_𝜆=(𝐴−𝜆𝐼)⁻¹, where 𝐴≡−(𝑑²/𝑑𝑥²)+𝑞(𝑥),𝐷(𝐴)={𝑦(𝑥)∶∃𝑦^,𝑦(0)=𝑦(𝜋)=0}. In the following lemmas, we prove some essential properties of 𝑅(𝑥,𝜉,𝜆) which are useful in the forthcoming study of the eigenfunction expansion of the problem (1.2)-(1.3)

Lemma 2.1. Let 𝑓(𝑥) be any function belonging to 𝐿₁(0,𝜋), then the function 𝑦(𝑥,𝜆)=^𝜋0𝑅(𝑥,𝜉,𝜆)𝑓(𝜉)𝜌(𝜉)𝑑𝜉(2.3) is the solution of the nonhomogenous Dirichlet problem (2.2).

Proof. First we show that (2.3) satisfies the boundary conditions of (2.2). From (2.2) by using (1.9) and (1.10), we, respectively, have 𝑦(0)=−1Ψ(𝜆)^𝜋0𝑦(𝜋)=−1𝜑(0,𝜆)𝜓(𝜉,𝜆)𝜌(𝜉)𝑓(𝜉)𝑑𝜉=0,Ψ(𝜆)^𝜋0𝜑(𝜉,𝜆)𝜓(𝜋,𝜆)𝜌(𝜉)𝑓(𝜉)𝑑𝜉=0.(2.4) Secondly, we calculate the solution of (2.2) by the method of variation of parameters. We seek a solution of the nonhomogenous problem (2.2) in the form 𝑦(𝑥,𝜆)=𝐶₁𝜑(𝑥,𝜆)+𝐶₂𝜓(𝑥,𝜆),(2.5) where 𝜑(𝑥,𝜆),𝜓(𝑥,𝜆) are given together with their asymptotic formulas in [4]. By using the standard calculation, we find 𝐶₁(𝑥,𝜆)=−1Ψ(𝜆)^𝜋𝑥𝐶𝑓(𝜉)𝜌(𝜉)𝜓(𝜉,𝜆)𝑑𝜉,₂(𝑥,𝜆)=−1Ψ(𝜆)^𝑥0𝑓(𝜉)𝜌(𝜉)𝜑(𝜉,𝜆)𝑑𝜉.(2.6) Substituting from (2.6) into (2.5) and keeping in mind (2.1), we get the required formula (2.3).

Lemma 2.2. Under the conditions of Lemma 2.1, the function 𝑅(𝑥,𝜉,𝜆) satisfies the following formula: 𝑅(𝑥,𝜉,𝜆)=−1𝜆−𝜆_𝑛𝑎_𝑛𝜑𝑥,𝜆_𝑛𝜑𝜉,𝜆_𝑛+𝑅¹(𝑥,𝜉,𝜆),(2.7) where 𝑅₁(𝑥,𝜉,𝜆) is regular in the neighborhood of 𝜆=𝜆_𝑛 and 𝑎_𝑛=∫^𝜋0𝜌(𝜉)𝜑²(𝑥,𝜆_𝑛)𝑑𝑥.

Proof. So long as, from Lemma 1.1, the roots of the function Ψ(𝜆) are simple; it follows that the poles of the function 𝑅(𝑥,𝜉,𝜆) are simple. So that 𝑅(𝑥,𝜉,𝜆) can be represented in the form 𝑅(𝑥,𝜉,𝜆)=Res[𝑅(𝑥,𝜉,𝜆)]𝜆−𝜆_𝑛+𝑅₁(𝑥,𝜉,𝜆)(2.8) from (2.1); for 𝑥≤𝜉, we have Res_{𝜆=𝜆𝑛}[𝑅(𝑥,𝜉,𝜆)]=−𝜑𝑥,𝜆^𝑛𝜓𝜉,𝜆_𝑛Ψ̇𝜆_𝑛.(2.9) From (1.17) and (1.18), the relation (2.9) takes the form Res_{𝜆=𝜆𝑛}[𝑅(𝑥,𝜉,𝜆)]=−𝜑𝑥,𝜆^𝑛𝜑𝜉,𝜆_𝑛𝑎_𝑛.(2.10) Formula (2.7) is obtained by substituting from (2.10) into (2.8) we deduce the formula (2.7). We notice that in case of 𝜉≤𝑥, in a similar way. in order to prove the convergence of the eigenfunction expansion of the Dirichlet problem (1.2)-(1.3), we must write an equality for the function 𝑅(𝑥,𝜉,𝜆) and this, in turn, needs to extend the asymptotic formulas of 𝜑(𝑥,𝜆),𝜓(𝑥,𝜆) over all the interval [0,𝜋]. In [4], the asymptotic formulas for 𝜑(𝑥,𝜆),  𝜓(𝑥,𝜆) were deduced for 𝑥∈(0,𝑎) and (𝑎,𝜋), respectively. In the following lemma, we write this asymptotic formulas over all [0,𝜋] for both 𝜑(𝑥,𝜆) and 𝜓(𝑥,𝜆).

Lemma 2.3. The solutions 𝜑(𝑥,𝜆) and 𝜓(𝑥,𝜆) of the Dirichlet problem (1.2)-(1.3) have the following asymptotic formula: 𝜑(𝑥,𝜆)=⎧⎪⎪⎪⎨⎪⎪⎪⎩sin𝑠𝑥𝛽(𝑠𝑥)+○𝑒||𝑠²||^|Im𝑠|𝑥,0≤𝑥≤𝑎,𝑠𝛽+(𝑎○)[si𝑒n𝑠𝑎cosh𝑠(𝑎−_{|Im𝑠|𝑎+|ℜ𝑠|(𝑎−𝑥)}𝑥)−cos𝑠𝑎sinh𝑠(𝑎−𝑥)],𝜓(𝑥,𝜆)=⎧⎪⎪⎪⎨⎪⎪⎪⎩𝛼(𝑥)||𝑠²||,𝑎<𝑥≤𝜋,(2.11)𝑠𝛼+(𝑎)○[cos𝑒𝑠(𝑥−𝑎)sinh𝑠(_{|Im𝑠|(𝑥−𝑎)+|ℜ𝑠|(𝜋−𝑎)}𝜋−𝑎)−sin𝑠(𝑥−𝑎)cosh𝑠(𝜋−𝑎)],sinh𝑠(𝜋−𝑥)||𝑠²||,0≤𝑥≤𝑎,𝑠+○𝑒||𝑠²||^{|ℜ𝑠|(𝜋−𝑥)},𝑎≤𝑥≤𝜋,(2.12) where 𝛼(𝑥)=12^𝑥0𝑞(𝑡)𝑑𝑡,𝛽(𝑥)=12^𝜋𝑥𝑞(𝑡)𝑑𝑡,𝜆=𝑠².(2.13)

Proof. Following [4], the solutions 𝜑(𝑥,𝜆) and 𝜓(𝑥,𝜆) of the Dirichlet problem (1.2)-(1.3) have the representation 𝜑(𝑥,𝜆)=sin𝑠𝑥𝜓(𝑥,𝜆)=sinh𝑠(𝜋𝑠−𝑥+)○𝑒||𝑠²||^{|ℑ𝑠|𝑥},𝑥∈[0,𝑎],(2.14)𝑠+○𝑒||𝑠²||^{|ℜ𝑠|(𝜋−𝑥)},𝑥∈[𝑎,𝜋].(2.15) We can see, from [6], that the solution 𝑦(𝑥,𝑠) of the equation −𝑦^+𝑞(𝑥)𝑦=𝑠²𝑦0≤𝑥≤𝑎,  𝑦(0)=0 has the representation 𝑦(𝑥,𝑠)=𝑒^𝑖𝑠𝑥𝛼(𝑠𝑥)+○𝑠1²,𝑦^(𝑥,𝑠)=𝑒^𝑖𝑠𝑥𝑖𝛼(𝑥)+○1𝑠,(2.16) and the solution 𝑧(𝑥,𝑠) of the equation −𝑦^+𝑞(𝑥)𝑦=−𝑠²𝑦,𝑎<𝑥≤𝜋,𝑧(𝜋)=0 has the representation 𝑧(𝑥,𝑠)=𝑒^{𝑠(𝜋−𝑥)}𝛽(𝑠𝑥)+○𝑠1²,𝑧^(𝑥,𝑠)=𝑒^{𝑠(𝜋−𝑥)}−𝛽(𝑥)+○1𝑠,(2.17) where 𝛼(𝑥) and 𝛽(𝑥) are given by (2.13), so that, 𝜑(𝑥,𝜆) can be extended to (𝑎,𝜋) in terms of the two linearly independent solutions 𝑧(𝑥,𝑠),  𝑧(𝑥,−𝑠) as 𝜑(𝑥,𝜆)=𝑚₁𝑧(𝑥,𝑠)+𝑚₂𝑧(𝑥,−𝑠).(2.18) From the continuity of 𝜑(𝑥,𝜆) at 𝑥=𝑎 and by using the asymptotic relations, (2.14), of 𝜑(𝑥,𝜆) and (2.17) of 𝑧(𝑥,𝑠), the constants 𝑚₁, 𝑚₂ are calculated in the form 𝑚₁=𝑒^{𝑠(𝑎−𝑥)}𝛽(𝑥)𝑚2𝛽(𝑎)sin𝑠𝑎−𝑠cos𝑠𝑎+○𝑒𝑠^{|Im𝑠|𝑎2}₂,=𝑒^{𝑠(𝜋−𝑎)}12𝛽(𝑎)−sin𝑠𝑎−cos𝑠𝑎+○𝑒𝑠^|Im𝑠|𝑎.(2.19) Substituting from (2.19) into (2.18), we have, for 𝑥∈(𝑎,𝜋]𝜑(𝑥,𝜆)=𝛽(𝑥)𝑠𝛽(𝑎)[sin𝑠𝑎cosh𝑠(𝑎−𝑥)−cos𝑠𝑎sinh𝑠(𝑎−𝑥)]+○𝑒||𝑠²||^{|Im𝑠|𝑎+|Res|(𝑎−𝑥)}.(2.20) From (2.20) together with (2.14), the relation (2.11) is followed. The proof of the relation (2.12) is quite similar to the proof of (2.11). Indeed 𝜓(𝑥,𝜆),𝑥∈(𝑎,𝜋] is a linear combination of the two linearly independent solutions 𝑦(𝑥,𝑠),  𝑦(𝑥,−𝑠) as 𝜓(𝑥,𝜆)=𝑘₁𝑦(𝑥,𝑠)+𝑘₂𝑦(𝑥,−𝑠), where 𝑘₁=𝑒^{−𝑖𝑠𝑎}𝑘2𝛼(𝑎)[sinh𝑠(𝜋−𝑎)+𝑖cosh𝑠(𝜋−𝑎)]+○𝑒|𝑠|^{|Res|(𝜋−𝑎)}₂,=𝑒^𝑖𝑠𝑎2𝛼(𝑎)[−sinh𝑠(𝜋−𝑎)+𝑖cosh𝑠(𝜋−𝑎)]+○𝑒|𝑠|^{|Res|(𝜋−𝑎)},(2.21) and by using the asymptotic formulas (2.16) of 𝑦(𝑥,±𝑠) and (2.15) of 𝜓(𝑥,𝜆), we have, for 𝑥∈[0,𝑎], 𝜓(𝑥,𝜆)=𝛼(𝑥)+𝑠𝛼○(𝑎)𝑒[cos𝑠(𝑥−𝑎)sinh_{|Im𝑠|(𝑥−𝑎)+|Res|(𝜋−𝑎)}𝑠(𝜋−𝑎)−sin𝑠(𝑥−𝑎)cosh𝑠(𝜋−𝑎)]||𝑠²||.(2.22) The relation (2.20) together with (2.14) and the relation (2.22) together with (2.15) complete the proof of lemma.

The following inequality proves an inequality satisfied by 𝑅(𝑥,𝜉,𝜆).

Lemma 2.4. Under the conditions of Lemma 2.3, the resolvent 𝑅(𝑥,𝜉,𝜆) satisfies the following inequality: 𝑅(𝑥,𝜉,𝜆)=⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩○𝑒^{−|Im𝑠||2𝑎−𝑥−𝜉|}○𝑒|𝑠|,𝑥,𝜉∈[0,𝑎_{−|Res||𝑥+𝜉−2𝑎|}],○𝑒|𝑠|,𝑥,𝜉∈[𝑎,𝜋_{−|Im𝑠|(𝑎−𝑥)−|Res|(𝜉−𝑎)}],○𝑒|𝑠|,0≤𝑥≤𝑎<𝜉≤𝜋,_{−|Im𝑠|(𝑎−𝜉)−|Res|(𝑥−𝑎)}|𝑠|,0≤𝜉≤𝑎<𝑥≤𝜋.(2.23)

Proof. From (2.11) and (2.12), we have 𝜑(𝑥,𝜆)=⎧⎪⎪⎨⎪⎪⎩○𝑒^{−|Im𝑠|𝑥}○𝑒|𝑠|,0≤𝑥≤𝑎,_{−|Im𝑠|𝑎+|Res|(𝑎−𝑥)}𝜓(𝑥,𝜆)=⎧⎪⎪⎨⎪⎪⎩○𝑒|𝑠|_{−|Im𝑠|(𝑥−𝑎)+|Res|(𝜋−𝑎)},𝑎<𝑥≤𝜋,(2.24)○𝑒|𝑠|,0≤𝑥≤𝑎,_{−|Res|(𝜋−𝑥)}|𝑠|,𝑎<𝑥≤𝜋.(2.25) It can be easily seen that, for 𝑠∈Γ_𝑛, we have Ψ(𝜆)≥𝐶𝑒^{|Im𝑠|𝑎+|Res|(𝜋−𝑎)}|𝑠|,𝑠∈Γ^𝑛,(2.26) where 𝑠Γ_𝑛 is the quadratic contour, as defined in [4] Γ_𝑛=||Res||≤𝜋𝑎𝑛−14+2𝜋𝑎,||Im𝑠||≤𝜋𝜋−𝑎𝑛−14+2(𝜋𝜋−𝑎).(2.27) From (2.1) we have six possibilities, three of which for 𝑥≤𝜉 and the other three for 𝜉≤𝑥. Now for 𝜉≤𝑥 we have the following situation: (i) 0≤𝑥≤𝜉≤𝑎, (ii) 𝑎<𝑥≤𝜉≤𝜋, and (iii) 0≤𝑥≤𝑎≤𝜉≤𝜋. In cases (i), (ii), and (iii) by direct substitution from (2.24), (2.25), (2.26) into the first branch of (2.1), we obtain (i)𝑅(𝑥,𝜉,𝜆)=○𝑒^{−|Im𝑠||2𝑎−𝑥−𝜉|}(ii)𝑅(𝑥,𝜉,𝜆)=○𝑒|𝑠|_{−|Res||𝑥+𝜉−2𝑎|},0≤𝑥≤𝜉≤𝑎,(2.28)(iii)𝑅(𝑥,𝜉,𝜆)=○𝑒|𝑠|,𝑎_{−|Im𝑠|(𝑎−𝑥)−|Res|(𝜉−𝑎)}≤𝑥≤𝜉≤𝜋,(2.29)|𝑠|,0≤𝑥≤𝑎<𝜉≤𝜋.(2.30)In the case of 𝜉≥𝑥, we discuss (i*) 0≤𝜉≤𝑥≤𝑎, (ii*) 𝑎≤𝜉≤𝑥≤𝜋, and (iii*) 0≤𝜉≤𝑎≤𝑥≤𝜋.
Again by substituting (2.24), (2.25), and (2.26) into the second branch of (2.1), we get ii^∗i_∗𝑅(𝑥,𝜉,𝜆)=○𝑒|𝑠|^{−|Im𝑠||2𝑎−𝑥−𝜉|},0≤𝜉≤𝑥≤𝑎,(2.31)iii_∗𝑅(𝑥,𝜉,𝜆)=○𝑒|𝑠|^{−|Res||𝑥+𝜉−2𝑎|},𝑎≤𝜉≤𝑥≤𝜋,(2.32)𝑅(𝑥,𝜉,𝜆)=○𝑒|𝑠|^{−|Im𝑠|(𝑎−𝜉)−|Res|(𝑥−𝑎)},0≤𝜉≤𝑎<𝑥≤𝜋.(2.33) From (2.28) and (2.31), we have 𝑅(𝑥,𝜉,𝜆)=○𝑒^{−|Ims||2𝑎−𝑥−𝜉|}|𝑠|,𝑥,𝜉∈[0,𝑎],(2.34) and from (2.29) and (2.32), we have 𝑅(𝑥,𝜉,𝜆)=○𝑒^{−|Res||𝑥+𝜉−2𝑎|}|𝑠|,𝑥,𝜉∈[𝑎,𝜋].(2.35) From (2.30) and (2.33) together with (2.34) and (2.35), the lemma is proved. In the following lemma, we prove an integral formula which is satisfied by 𝑅(𝑥,𝜉,𝜆) and help in proving the eigenfunction expansion formula

Lemma 2.5. If the function 𝑓(𝑥) on [0,𝜋] has a second-order integrable derivatives and satisfies the Dirichlet condition 𝑓(0)=𝑓(𝜋)=0, then the following integral formula is true: ^𝜋0𝑅(𝑥,𝜉,𝜆)𝜌(𝜉)𝑓(𝜉)𝑑𝜉=−𝑓(𝑥)𝜆+^𝜋0𝑅(𝑥,𝜉,𝜆)𝜆−𝑓^(𝑥)+𝑞(𝑥)𝑓(𝜉)𝑑𝜉,(2.36) where 𝑅(𝑥,𝜉,𝜆) is the kernel of the resolvent of the nonhomogenous Dirichlet problem (2.2).

Proof. By the aid of (2.1), we have ^𝜋0𝑅(𝑥,𝜉,𝜆)𝜌(𝜉)𝑓(𝜉)𝑑𝜉=−1Ψ(𝜆)𝜓(𝑥,𝜆)^𝑥0𝜑(𝜉,𝜆)𝜌(𝜉)𝑓(𝜉)𝑑𝜉+𝜑(𝑥,𝜆)^𝜋𝑥𝜓(𝜉,𝜆)𝜌(𝜉)𝑓(𝜉)𝑑𝜉(2.3,7) where the functions 𝜑(𝑥,𝜆) and 𝜓(𝑥,𝜆) are the solutions of the homogenous Dirichlet problem (1.2)-(1.3), so that ^𝜋0𝑅(𝑥,𝜉,𝜆)𝜌(𝜉)𝑓(𝜉)𝑑𝜉=−1Ψ(𝜆)𝜓(𝑥𝜆,𝜆)^𝑥0𝜑^+𝜑(𝑥,𝜆)(𝜉,𝜆)+𝑞(𝜉)𝜑(𝜉,𝜆)𝑓(𝜉)𝑑𝜉𝜆^𝜋𝑥𝜓^(𝜉,𝜆)+𝑞(𝜉)𝜓(𝜉,𝜆)𝑓(𝜉)𝑑𝜉(2.38) from which we have ^𝜋0𝑅(𝑥,𝜉,𝜆)𝜌(𝜉)𝑓(𝜉)𝑑𝜉=1Ψ(𝜆)𝜓(𝑥𝜆,𝜆)^𝑥0𝜑^(𝜉,𝜆)𝑓(𝜉)𝑑𝜉+𝜑(𝑥,𝜆)𝜆^𝜋𝑥+1𝜓^(𝜉,𝜆)𝑓(𝜉)𝑑𝜉𝜆^𝜋0𝑅(𝑥,𝜉,𝜆)𝑞(𝜉)𝑓(𝜉)𝑑𝜉.(2.39) Integrating by parts twice the terms ∫^𝑥0 and ∫^𝜋𝑥, in (2.39), and then using the boundary conditions 𝑓(0)=𝑓(𝜋)=𝜑(0,𝜆)=0 and 𝑓(0)=𝑓(𝜋)=𝜓(𝜋,𝜆)=0, respectively, and keeping in mind (2.1), we deduce that 𝜓(𝑥,=𝜆)−Ψ^𝑥0𝜑(𝜆)^(𝑓𝜉,(𝑥𝜆))𝑓−(Ψ𝜉)(𝑑𝜆𝜉)+_𝜋0𝜑(𝑥,𝜆)^𝜋𝑥𝜓^(𝜉,𝜆)𝑓(𝜉)𝑑𝜉𝑅(𝑥,𝜉,𝜆)𝑓^(𝜉)𝑑𝜉.(2.40) Substituting from (2.40) into (2.39), we get the required result.

3. The Eigenfunctions Expansion Formula

We now construct and prove the eigenfunction expansion formula for the Dirichlet problem (1.2)-(1.3). Let 𝜆^+𝑛,𝑛=0,1,2,… and 𝜆^−𝑛,𝑛=0,1,2,… be the nonnegative and the negative eigenvalues of the problem (1.2)-(1.3), and let also 𝑎^+𝑛=^𝜋0𝜌(𝑥)𝜑²𝑥,𝜆^+𝑛𝑑𝑥,𝑎^−𝑛=^𝜋0𝜌(𝑥)𝜑²𝑥,𝜆^−𝑛𝑑𝑥(3.1) be the normalization numbers of the corresponding eigenfunctions 𝜑(𝑥,𝜆^±𝑛). We put 𝑣^±𝑘(𝑥)=𝜑𝑥,𝜆^±𝑘𝑎^±𝑘,𝑘=0,1,2,….(3.2) The set {𝑣^±𝑘(𝑥)}^∞𝑘=0 is an orthonormal system of eigenfunctions of the Dirichlet problem (1.2)-(1.3).

Theorem 3.1. Let f(x) be a second-order integrable derivatives on [0,𝜋] and satisfy the conditions 𝑓(0)=𝑓(𝜋)=0; then the following formula of eigenfunction expansion is true: 𝑓(𝑥)=^∞_𝑘=0𝑏^+𝑘𝑣^+𝑘(𝑥)+^∞_𝑘=0𝑏^−𝑘𝑣^−𝑘(𝑥),(3.3) where 𝑏^±𝑘=∫^𝜋0𝑣^±𝑘(𝜉)𝑓(𝜉)𝜌(𝜉)𝑑𝜉 and the series uniformly converges to 𝑓(𝑥),𝑥∈[0,𝜋].

Notice that, the expansion (3.3) can be written, more explicitly, in terms of 𝜑(𝑥,𝜆^±𝑘) as 𝑓(𝑥)=^∞_𝑘=01𝑎^+𝑘𝜓𝑥,𝜆^+𝑘^𝜋0𝜓𝜉,𝜆^+𝑘𝑓(𝜉)𝜌(𝜉)𝑑𝜉+^∞_𝑘=01𝑎^−𝑘𝜓𝑥,𝜆^−𝑘^𝜋0𝜓𝜉,𝜆^−𝑘𝑓(𝜉)𝜌(𝜉)𝑑𝜉(3.4) or in terms of 𝜓(𝑥,𝜆^±𝑘)𝑓(𝑥)=^∞_𝑘=01𝑟^+𝑘²𝑎^+𝑘+𝜑𝑥,𝜆^+𝑘^𝜋0𝜑𝜉,𝜆^+𝑘𝑓(𝜉)𝜌(𝜉)𝑑𝜉_∞_𝑘=01𝑟^−𝑘²𝑎^−𝑘𝜑𝑥,𝜆^−𝑘^𝜋0𝜑𝜉,𝜆^−𝑘𝑓(𝜉)𝜌(𝜉)𝑑𝜉,(3.5) where 𝑟^±𝑘 are defined by 𝜓(𝑥,𝜆^±𝑘)=𝑟^±𝑘𝜑(𝑥,𝜆^𝑘±,),  0≤𝑥≤𝜋.

Proof. We write (2.36) in the form ^𝜋0𝑅(𝑥,𝜉,𝜆)𝜌(𝜉)𝑓(𝜉)𝑑𝜉=−𝑓(𝑥)𝜆+𝑟(𝑥,𝜆),(3.6) where 𝑟(𝑥,𝜆)=^𝜋0𝑅(𝑥,𝜉,𝜆)𝜆−𝑓^(𝑥)+𝑞(𝑥)𝑓(𝜉)𝑑𝜉.(3.7) By the aid of Lemma 2.3 and the condition of the theorem imposed on 𝑞(𝑥), it can be easily seen that ||𝑟(𝑥,𝜆)||≤𝑀^𝑜||𝜆||^3/2,𝑠∈Γ_𝑛,(3.8) where 𝑀_𝑜 is constant which is independent of 𝑥,𝜉,𝜆 and the contour Γ_𝑛, as defined in [4], is given by (2.27). Let 𝜆=𝑠²; we denote by Γ^+𝑛 the upper half of the Γ_𝑛; let also Ł𝑛 denote the image of the contour Γ^+𝑛 under the transformation 𝜆=𝑠². We multiply both sides of (3.6) by 1/2𝜋𝑖 and integrating with respect to 𝜆 on the contour 𝐿_𝑛: 12𝜋𝑖_𝐿𝑛^𝜋0𝑅(𝑥,𝜉,𝜆)𝑓(𝜉)𝜌(𝜉)𝑑𝜉=−𝑓(𝑥)2𝜋𝑖_𝐿𝑛𝑑𝜆𝜆+21𝜋𝑖_𝐿𝑛𝑟(𝑥,𝜆)𝑑𝜆.(3.9) Among the poles of the function 𝑅(𝑥,𝜉,𝜆), as a function of 𝜆, lie only 𝜆^±𝑜,𝜆^±1,…,𝜆^±𝑛 inside 𝐿_𝑛. By using the residues formula and (2.10), we have 12𝜋𝑖_𝐿𝑛^𝜋0𝑅(𝑥,𝜉,𝜆)𝑓(𝜉)𝜌(𝜉)𝑑𝜉=^𝑛_𝑘=0Res_𝜆=𝜆^±𝑘^𝜋0=−_𝑛𝑅(𝑥,𝜉,𝜆)𝑓(𝜉)𝜌(𝜉)𝑑𝜉_𝑘=0𝜑𝑥,𝜆^+𝑘𝑎^+𝑘^𝜋0𝜑𝜉,𝜆^+𝑘𝜌(𝜉)𝑓(𝜉)𝑑𝜉−^𝑛_𝑘=0𝜑𝑥,𝜆^−𝑘𝑎^−𝑘^𝜋0𝜑𝜉,𝜆^−𝑘𝜌(𝜉)𝑓(𝜉)𝑑𝜉.(3.10) Further −𝑓(𝑥)2𝜋𝑖_𝐿𝑛𝑑𝜆𝜆=−𝑓(𝑥).(3.11) By using (3.8), we have ||||12𝜋𝑖_𝐿𝑛𝑟(𝑥,𝜆)𝑑𝜆||||≤𝑀^𝑜2𝜋_𝐿𝑛||𝑑𝜆||||𝜆||^3/2,(3.12) from which, by using the substitution 𝜆=𝑠², we have ||||12𝜋𝑖_𝐿𝑛𝑟(𝑥,𝜆)𝑑𝜆||||≤𝑀^𝑜𝜋_Γ^+𝑛||𝑑𝑠||𝑠²≤constant𝑛.(3.13) By substitution from (3.10), (3.11), and (3.13) into (3.9), we obtain |||||𝑓(𝑥)−_𝑛_𝑘=0𝑏^+𝑘𝑣^+𝑘+𝑏^−𝑘𝑣^−𝑘|||||≤constant𝑛,𝑥∈[0,𝜋]∀𝑛,(3.14) where 𝑏^±𝑘=^𝜋0𝜑𝜉,𝜆^±𝑘𝑎^±𝑘𝑓(𝜉)𝜌(𝜉)𝑑𝜉,𝑣^±𝑘=𝜑𝜉,𝜆^±𝑘𝑎^±𝑘,(3.15) which completes the uniform convergence of the series ∑^∞𝑘=0[𝑏^+𝑘𝑣^+𝑘+𝑏^−𝑘𝑣^−𝑘] to 𝑓(𝑥),𝑥∈[0,𝜋]. That is, 𝑓(𝑥)=^∞_𝑘=0𝑏^+𝑘𝑣^+𝑘(𝑥)+𝑏^−𝑘𝑣^−𝑘(𝑥).(3.16) It can be proved that the series (3.16) is not only uniformly but also absolutely convergent, to show this we use the asymptotic relations of 𝑣^±𝑘 and 𝑏^±𝑘 for 𝑛→∞. Following [4], we have 𝑎^+𝑛=𝑑¹𝑛²+○1𝑛³,𝑎^−𝑛=−𝑑²²𝑒^2𝑛𝑑2𝑒^−𝑑21𝑛²+○1𝑛³,(3.17) where 𝑑₁=𝑎³/2𝜋² and 𝑑₂=(𝜋−𝑎)/𝜋, from which we can write 𝑎^±𝑛=○1𝑛.(3.18) Using (3.18), (2.11) and (3.2), we deduce that ||𝑣^±𝑘||≤𝑀^±,∀𝑥∈[0,𝜋],andall𝑘,(3.19) where 𝑀^± are some constants. Further, arguing as in Lemma 2.4 and noticing that 𝑓(0)=𝑓(𝜋)=0, we have 𝑏^±𝑘=^𝜋0𝑣^±𝑘(𝑥)𝑓(𝑥)𝜌(𝑥)𝑑𝑥=1𝜆^±𝑘^𝜋0𝑓^(𝑥)+𝑞(𝑥)𝑓(𝑥)𝑣^±𝑘(𝑥)𝑑𝑥.(3.20) From [4], we have 𝜆^±𝑘=±𝑘²+○(1), and using (3.20) we have ||𝑏^±𝑘𝑣^±𝑘(𝑥)||≤costant𝑘²,𝑘⟶∞,(3.21) which complete the proof of absolute convergence of the series (3.16). It should be noted here that, in the proof of the absolute convergence of the series (3.16) we did not give the sum of the series as in the proof of uniform convergence (Theorem 3.1). In the following lemma, as a consequence of Theorem 3.1, we prove the Parsval’s identity which insures the convergence of the series (3.16) and helps in the proof of Theorem 3.3.

Lemma 3.2. Let 𝑓(𝑥) satisfy the conditions of Theorem 3.1; then the following Parsval’s identity holds true ^𝜋0𝜌(𝑥)||𝑓(𝑥)||²𝑑𝑥=^∞_𝑘=0||𝑏^+𝑘||²+||𝑏^−𝑘||²,(3.22) where 𝑏^±𝑘=^𝜋0𝑣^±𝑘(𝑥)𝑓(𝑥)𝜌(𝑥)𝑑𝑥.(3.23)

Proof. From Theorem 3.1, we have 𝑓(𝑥)=^∞_𝑘=0𝑏^+𝑘𝑣^+𝑘(𝑥)+𝑏^−𝑘𝑣^−𝑘(𝑥),(3.24) where 𝑏^±𝑘 are given by (3.22). Multiplying both sides of (3.24) by 𝑓(𝑥)𝜌(𝑥) and integrating with respect to 𝑥∈[0,𝜋], we have ^𝜋0𝜌(𝑥)||𝑓(𝑥)||²𝑑𝑥=^𝜋0∞_𝑘=0𝑏^+𝑘𝑣^+𝑘(𝑥)+𝑏^−𝑘𝑣^−𝑘(𝑥)𝑓(𝑥)𝜌(𝑥)𝑑𝑥.(3.25) By the aid of uniform convergence of the series (3.16), the integration and summation can be interchanged and we have ^𝜋0𝜌(𝑥)||𝑓(𝑥)||²𝑑𝑥=^∞_𝑘=0𝑏^+𝑘^𝜋0𝑣^+𝑘(𝑥)𝑓(𝑥)𝜌(𝑥)𝑑𝑥+^∞_𝑘=0𝑏^−𝑘^𝜋0𝑣^−𝑘(𝑥)𝑓(𝑥)𝜌(𝑥)𝑑𝑥,(3.26) where 𝜌(𝑥) and 𝑣^±𝑘(𝑥) are real (see introduction) which complete the proof of the lemma. In the following theorem, the validity of eigenfunction expansion and the Parsval’s identity can be extended to any function of 𝐿₂(0,𝜋;𝜌) but the convergence of the expansion will be in some weak sense, that is, in the metric sense of 𝐿₂(0,𝜋;𝜌).

Theorem 3.3. Suppose that 𝑓(𝑥) is any function from 𝐿₂(0,𝜋;𝜌); then the following Parsval’s identity (3.22) and the eigenfunction expansion (3.24) are true and the convergence of the series (3.16) to 𝑓(𝑥) is in the metric sense of the space 𝐿₂(0,𝜋;𝜌).

Proof. Let 𝑓(𝑥) be any function that belongs to 𝐿₂(0,𝜋;𝜌). It is known that the set of infinitely differential functions which vanish at the neighbourhood of the points 𝑥=0,  𝑥=𝜋 are dense in 𝐿₂(0,𝜋;𝜌), so that there exists a sequence {𝑓_𝑛(𝑥)} of finite smooth functions (and consequently, satisfy the conditions of the theorem) which converges to 𝑓(𝑥) in the metric of 𝐿₂(0,𝜋;𝜌); in equation notation this is can be written as ‖‖𝑓_𝑛(𝑥)−𝑓(𝑥)‖‖_𝐿2=^𝜋0𝜌(𝑥)||𝑓_𝑛(𝑥)−𝑓(𝑥)||²𝑑𝑥^1/2⟶0,as𝑛⟶∞.(3.27) By the last lemma, every function 𝑓_𝑛(𝑥) satisfies the parseval’s identity ^𝜋0𝜌(𝑥)||𝑓_𝑛(𝑥)||²𝑑𝑥=^∞_𝑘=0||𝑏^𝑘(𝑛)+||²+||𝑏^{𝑘(𝑛)−}||²,(3.28) where 𝑏^𝑘(𝑛)±=∫^𝜋0𝜌(𝑥)𝑓_𝑛(𝑥)𝑣^±𝑘(𝑥)𝑑𝑥,  𝑘=0,1,2,…
The identity (3.28) can be written as ‖‖𝑓_𝑛(𝑥)‖‖^2𝐿₂=‖‖𝑏^𝑘(𝑛)+‖‖^2𝑙₂+‖‖𝑏^{𝑘(𝑛)−}‖‖^2𝑙₂.(3.29) Consider the difference ‖‖𝑓_𝑛(𝑥)−𝑓_𝑚(𝑥)‖‖^2𝐿₂=‖‖𝑏^𝑘(𝑛)+−𝑏^𝑘(𝑚)+‖‖^2𝑙₂+‖‖𝑏^{𝑘(𝑛)−}−𝑏^{𝑘(𝑚)−}‖‖^2𝑙₂.(3.30) By the aid of (3.27), it follows that {𝑓_𝑛(𝑥)} is a fundamental sequence and hence by the completeness of 𝑙₂ the sequences {𝑏^𝑘𝑛±} are fundamentals, so, that there exists a limiting 𝑏^+𝑘 and 𝑏^−𝑘 such that ‖𝑏^𝑘(𝑛)+−𝑏^+𝑘‖^2𝑙₂→0 and ‖𝑏^{𝑘(𝑛)−}−𝑏^−𝑘‖^2𝑙₂→0; by using the continuity of the norm and passing to the limit as 𝑛→∞ in (3.29), we obtain ‖𝑓(𝑥)‖^2𝐿₂=‖‖𝑏^+𝑘‖‖^2𝑙₂+‖‖𝑏^−𝑘‖‖^2𝑙₂,(3.31) which is the Parsval’s identity. Now we prove the eigenfunction expansion formula by the help of Parsval’s identity. For any 𝑛, we have ^𝜋0𝜌(𝑥)|||||𝑓(𝑥)−^𝑛_𝑘=0𝑏^+𝑘=𝑣^+𝑘(𝑥)+𝑏^−𝑘𝑣^−𝑘(𝑥)|||||²𝑑𝑥_𝜋0𝜌(𝑥)𝑓(𝑥)−^𝑛_𝑘=0𝑏^+𝑘×𝑣^+𝑘(𝑥)+𝑏^−𝑘𝑣^−𝑘(𝑥)𝑓(𝑥)−^𝑛_𝑘=0𝑏^+𝑘𝑣^+𝑘(𝑥)+𝑏^−𝑘𝑣^−𝑘(𝑥)𝑑𝑥;(3.32) after calculation, we have ^𝜋0𝜌(𝑥)|||||𝑓(𝑥)−^𝑛_𝑘=0𝑏^+𝑘𝑣^+𝑘(𝑥)+𝑏^−𝑘𝑣^−𝑘(𝑥)|||||𝑑𝑥=^𝜋0𝜌(𝑥)||𝑓(𝑥)||²𝑑𝑥−^𝑛_𝑘=0||𝑏^+𝑘||²+||𝑏^−𝑘||²(3.33) from which, and by using Parseval identity (3.22), we have lim_𝑛→∞^𝜋0𝜌(𝑥)|||||𝑓(𝑥)−^𝑛_𝑘=0𝑏^+𝑘𝑣^+𝑘(𝑥)+𝑏^−𝑘𝑣^−𝑘(𝑥)|||||²𝑑𝑥⟶0.(3.34) So that, ∑^∞𝑘=0(𝑏^+𝑘𝑣^+𝑘(𝑥)+𝑏^−𝑘𝑣^−𝑘(𝑥))→𝑓(𝑥) in the metric of 𝐿₂(0,𝜋;𝜌), which completes the proof.