CHAPTER STSturm-Liouville ProblemsMany of the problems to be considered later will require an approximation of givenfunctions in terms of eigenfunctions of an ordinary differential operator. A well developedeigenvalue theory exists for so-called Sturm-Liouville differential operators, and we shallsummarize the results important for the solution of partial differential equations later on.However, in many applications only very simple and readily solved eigenvalue problemsarise which do not need the generality of the Sturm-Liouville theory. We shall consider suchproblems first.The simplest, but also constantly recurring, operator isLφ ≡ φ00(x) 0 < x < Ldefined on the vector space C2(0, L) of twice continuously differentiable functions on theinterval (0, L), or on some subspace M of C2(0, L). Henceforth M will denote the domainon which L is to be defined. In analogy to the matrix eigenvalue problemAφ = λφfor an n × n matrix A we shall consider the following problem:Find an eigenvalue µ and all eigenfunctions (= eigenvectors) φ(x) ∈ M which satisfy(ST.1) Lφ = µφ for all x ∈ (0, L).As in the matrix case the eigenvalue may be zero, real or complex, but the correspondingeigenfunction must not be the zero function. Note that if φ is an eigenvector then cφ forc 6= 0 is also an eigenvector.The domain on which L is defined has an enormous influence on the solvability of theeigenvalue problem. For example, if M = C2(0, L), then for any complex number µ theequationLφ − µφ ≡ φ00− µφ = 0has the two linearly independent solutionsφ1= e√µx1φ2= e−√µxfor µ 6= 0 andφ1= 1φ2= xfor µ = 0. Hence any number is an eigenvalue and has two corresponding eigenfunctions.On the other hand, if M = {f ∈ C2(0, L) ∩ C[0, L] : f(0) = f0(0) = 0} then for any µ theonly solution ofφ00− µφ = 0φ(0) = φ0(0) = 0is the zero solution. Hence there are no eigenvalues and eigenvectors in this case.The subspaces M of C2(0, L) of interest for applications are defined by the boundaryconditionsi) φ(0) = 0, φ(L) = 0ii) φ0(0) = 0, φ(L) = 0iii) φ(0) = 0, φ0(L) = 0iv) φ0(0) = 0, φ0(L) = 0v) αφ0(0) = φ(0), φ(L) = 0, α > 0vi) αφ0(0) = φ(0), φ0(L) = 0, α > 0vii) φ(0) = 0, βφ0(L) = −φ(L), β > 0viii) φ0(0) = 0, βφ0(L) = −φ(L), β > 0ix) αφ0(0) = φ(0), βφ0(L) = −φ(L), α, β > 0andx) φ(0) = φ(L), φ0(0) = φ0(L).We note that the first nine boundary conditions represent special cases of the general con-dition(St.2)α1φ(0) − α2φ0(0) = 0,β1φ(L) + β2φ0(L) = 02for real αi, βjsuch thatα1α2≥ 0 α21+ α226= 0β1β2≥ 0 β21+ β226= 0.The boundary condition x) is associated with periodic functions defined the line. In eachcase the subspace M will consist of those functions in C2(0, L) which are continuous on[0, L] and which satisfy the given boundary conditions.For several of these boundary conditions we can give explicitly the eigenvalues andeigenfunctions. To see what is involved let us look at the simple case ofφ00= µφφ(0) = φ(L) = 0.If µ = 0 then φ(x) = c1+ c2x, and the boundary conditions require c1= c2= 0 so thatφ(x) ≡ 0; hence µ = 0 is not an eigenvalue. For µ 6= 0 the differential equation has againthe general solutionφ(x) = c1e√µx+ c2e−√µx.The two boundary conditions requirec1+ c2= 0c1e√µL+ c2e−√µL= 0or in matrix formµ1 1e√µLe−√µL¶µc1c2¶=µ00¶.This system has a nontrivial solution (1, −1) if and only if the determinant f(µ) of thecoefficient matrix is zero. We needf(µ) ≡ e−√µL− e√µL= e−√µL(1 − e2√µL) = 0.This can be the case only if2√µL = 2nπi3for a non-zero integer n. Hence there are countably many eigenvaluesµn= −µnπL¶2with corresponding eigenfunctionφn(x) = c¡einπLx− e−inπLx¢Since eigenfunctions are determined only up to a multiplicative constant we can choosec =12iso thatφn(x) = sinnπLx.The above calculation would have been simpler had we known a priori that the eigenvalue hasto be real and non-negative. In that case complex numbers and functions can be avoidedas we shall see below. For the algebraic sign pattern of the coefficients in the boundaryconditions of all of the above ten eigenvalue problems this property is easy to establish.Theorem ST.1. The eigenvalues of (ST.1) for the boundary conditions i–x are real andnon-positive.Proof. If {µ, φ(x)} are an eigenvalue-eigenvector pair thenZL0φ00(x)φ(x)dx = µZL0φ(x)φ(x)dxIntegration by parts shows thatZL0φ00(x)φ(x)dx = φ0(x)φ(x)ZL0−ZL0φ0(x)φ0(x)dx.For each of the ten cases above the boundary terms either vanish or are real and non-positive.For example, if β16= 0 thenφ0(L)φ(L) = −β2β1φ0(L)φ0(L) ≤ 0.HenceµZL0|φ(x)|2dx ≤ 0which implies that µ is real and non-positive.Since the eigenvalue is real it follows that the real and imaginary part of any complexvalued eigenfunction must satisfy the eigenvalue equation. Hence the eigenfunctions maybe assumed to be real so that the conjugation in the integrals can be dropped.4Theorem ST.2. The eigenfunctions corresponding to distinct eigenvalues are orthogonalin L2(0, L).Proof. Let {µm, φm} and {µn, φn} be eigenvalues and eigenfunctions with µm6= µn. Then(µm− µn)ZL0φm(x)φn(x)dx =ZL0(φnφ00m− φmφ00n)dx.If we integrate by parts and use the boundary conditions we see that the right hand integralvanishes so thathφm, φni =ZL0φm(x)φn(x)dx = 0.The computation of the eigenvalues and eigenvectors for the above cases is straightfor-ward, at least in principle. Since µ ≤ 0 we shall writeµ = −λ2and solveφ00+ λ2φ = 0.We know that the general solution of this equation is(St.3) φ(x) = c1+ c2x if λ = 0(St.4) φ(x) = c1cos λx + c2sin λx if λ 6= 0.We now have to determine from the boundary conditions for what values of λ we can find anon-trivial solution. To introduce the required computations let us look at the simple casesii) and x) before tackling (St.2). A summary of eigenvalues and eigenvectors available inclosed form is given in Table St.1. We shall repeatedly have occasion to refer to this tableas we solve partial differential equations.ii) φ0(0) = φ(L) = 0We see by inspection that λ = 0 is not admissible because there is no non-zero eigen-function of the form (St.3). For λ 6= 0 the solution is given by (St.4). The boundaryconditions lead toλc2= 05c1cos λL + c2sin λL = 0which can be written in matrix form asµ0 λcos λL sin λL¶µc1c2¶=µ00¶.This linear system has a non-trivial