Fifth Mississippi State Conference on Differential Equations and Computational Simulations, Electronic Journal of Differential Equations, Conference 10, 2003, pp 153–161. http://ejde.math.swt.edu or http://ejde.math.unt.edu ftp ejde.math.swt.edu (login: ftp)

A selfadjoint hyperbolic boundary-value problem ^{∗}

Nezam Iraniparast

Abstract

We consider the eigenvalue wave equation

u_{tt} − u_{ss} = λpu,

subject to u(s, 0) = 0, where u ∈ ℝ, is a function of (s, t) ∈ ℝ^{2}, with t ≥ 0. In the characteristic triangle T = {(s, t) : 0 ≤ t ≤ 1, t ≤ s ≤ 2 − t} we impose a boundary condition along characteristics so that

αu(t, t) − β \frac{∂u}{∂n_{1}} (t, t) = αu(1 + t, 1 − t) + β \frac{∂u}{∂n_{2}} (1 + t, 1 − t), 0 ≤ t ≤ 1.

The parameters α and β are arbitrary except for the condition that they are not both zero. The two vectors n_{1} and n_{2} are the exterior unit normals to the characteristic boundaries and \frac{∂u}{∂n_{1}} , \frac{∂u}{∂n_{2}} are the normal derivatives in those directions. When p ≡ 1 we will show that the above characteristic boundary value problem has real, discrete eigenvalues and corresponding eigenfunctions that are complete and orthogonal in L_{2}(T). We will also investigate the case where p ≥ 0 is an arbitrary continuous function in T.

1 Introduction

Consider the wave equation

u_{tt} − u_{ss} = λpu, (s, t) ∈ T := {(s, t) : 0 ≤ t ≤ 1, t ≤ s ≤ 2 − t},  (1.1)

where λ is a parameter, and p ≥ 0 is a continuous function in T. We impose the boundary conditions,

u(s, 0) = 0, 0 ≤ s ≤ 2,  (1.2)
αu(t, t) − β \frac{∂u}{∂n_{1}} (t, t) = αu(1 + t, 1 − t) + β \frac{∂u}{∂n_{2}} (1 + t, 1 − t), 0 ≤ t ≤ 1, (1.3)

where the parameters α and β are arbitrary with α^{2} + β^{2} ≠ 0. The two vectors n_{1} and n_{2} are the exterior unit normals to the characteristic boundaries and \frac{∂u}{∂n_{1}} , \frac{∂u}{∂n_{2}} are the normal derivatives in those directions. Our problem is a generalization of the problems studied by Kalmenov [6], and Kreith [7], where they consider the boundary conditions,

∗Mathematics Subject Classifications: 35L05, 35L20, 35P99.

Key words: Characteristics, eigenvalues, eigenfunctions, Green’s function, Fredholm alternative.

\begin{matrix}c \\ ◯ \end{matrix}2003 Southwest Texas State University.

Published February 28, 2003.

u(s, 0) = 0 = u(1, 1), 0 ≤ s ≤ 2
u(t, t) = u(1 + t, 1 − t), 0 ≤ t ≤ 1.

We note two facts. First, if we only prescribe the values of u along the characteristics, say, u(t, t) = f_{1}(t) and u(1 + t, 1 − t) = f_{2}(t) then we have a classical characteristic initial value problem,(see, e.g., Garabedian [1]) and equation (1.1) will have a solution for all values of λ. However, the conditions (1.2), (1.3) provide a boundary value problem with spectral properties. Second, if we set β = 0 in condition (1.3) we have the case of Kreith [7] . If in addition, we set p ≡ 1 we have the case of Kalmenov [6]. As a start we will let p ≡ 1 and use the technique of Kalmenov [6] to show that the equation, u_{tt} − u_{ss} = λu, (s, t) ∈ T subject to conditions (1.2), (1.3), is selfadjoint. Then we will study equation (1.1) subject to conditions (1.2), (1.3) by converting the problem into a nonhomogeneous eigenvalue integral equation and using the method of the Fredholm alternative. In this case we will assume that u along the characteristics is given. This will not weaken the problem because we still require that condition (1.2) be satisfied. As a result we will see that not all values of λ will produce a solution.

We would like to add here that the investigation into the spectral properties of the characteristic initial value problems for the wave equation has been conducted in different directions by several authors. In this context, beside the above mentioned references, the works of Haws [2], Kreith [8, 9] and the author [3, 4, 5] are the ones most closely related to the present work.

2 The selfadjoint problem

Consider the equation

u_{tt} − u_{ss} = λu, (s, t) ∈ T,  (2.1)

subject to the conditions (1.2), (1.3). Extend u as an odd function of t and define,

ũ(s, t) = \begin{cases}u(s, t) & if t > 0, (s, t) ∈ T \\ −u(s, −t) & if t < 0, (s, −t) ∈ T \end{cases}

and reflect T in t = 0 axis to obtain the rectangle

R = {(s, t) : |t| ≤ 1, |t| ≤ s ≤ 2 − |t|}.  (2.2)

Now ũ must satisfy the following conditions:

ũ_{tt} − ũ_{ss} = λũ, (s, t) ∈ R,  (2.3)
αũ(t, t) − β \frac{∂ũ}{∂n_{1}} (t, t) = αũ(1 + t, 1 − t) + β \frac{∂ũ}{∂n_{2}} (1 + t, 1 − t), |t| ≤ 1, (2.4)
ũ(s, t) = −ũ(s, −t), (s, t) ∈ R.  (2.5)

Using the change of variables, x = s − t, y = s + t and denoting ũ(s, t) = ũ(\frac{x+y}{2} , \frac{y−x}{2}) by Ũ(x, y), the rectangle R maps into S = \{(x, y) : 0 ≤ x ≤ 2, 0 ≤ y ≤ 2}, and we have,

−4Ũ_{x,y} = λŨ, (x, y) ∈ S,  (2.6)
αŨ(0, y) + βŨ_{x}(0, y) = αŨ(y, 2) + βŨ_{y}(y, 2),  (2.7)
Ũ(x, y) = −Ũ(y, x).  (2.8)

From this equation, we have Ũ_{x}(x, y) = −Ũ_{y}(y, x); therefore,

Ũ(0, y) = −Ũ(y, 0)  (2.9)
Ũ_{x}(0, y) = −Ũ_{y}(y, 0).  (2.10)

These two identities change the boundary condition (2.7) to,

−αŨ(y, 0) − βŨ_{y}(y, 0) = αŨ(y, 2) + βŨ_{y}(y, 2).  (2.11)

Now let Ũ(x, y) = ϕ(x)ψ(y), and plug it into equation (2.6), we have,

[2iϕ′(x)][2iψ′(y)] = λϕ(x)ψ(y)  (2.12)

which leads to

\frac{2iϕ′(x)}{ϕ(x)} = λ\frac{ψ(y)}{2iψ′(y)} = μ,

which in turn leads to two ODE equations:

ψ′(y) = −\frac{iλ}{2μ}ψ(y) = −\frac{iδ}{2} ψ(y), δ = \frac{λ}{μ}, μ ≠ 0,  (2.13)
ϕ′(x) = −\frac{iμ}{2}ϕ(x).  (2.14)

The effect of separation of variables on the boundary condition (2.11) will be

−αϕ(y)ψ(0) − βϕ(y)ψ′(0) = αϕ(y)ψ(2) + βϕ(y)ψ′(2).  (2.15)

Upon cancelling ϕ(y) here, we have the boundary-value problem

ψ′(y) = −\frac{iδ}{2} ψ(y), 0 ≤ y ≤ 2,  (2.16)
−αψ(0) − βψ′(0) = αψ(2) + βψ′(2).  (2.17)

We note that from (2.16) we have

ψ′(0) = −\frac{iδ}{2} ψ(0),  (2.18)
ψ′(2) = −\frac{iδ}{2} ψ(2).  (2.19)

These relations simplify the condition (2.17) to

ψ′(0) = −ψ(2),  (2.20)

The selfadjoint boundary-value problem (2.16), (2.20) is the same as the one in [6]. The eigenvalues and eigenfunctions of problem (2.16), (2.20) are

δ = (−2n − 1)π, ψ_{n}(y) = exp(\frac{(2n + 1)πiy}{2} ), n = 0, ±1, ±2, … (2.21)

Using (2.8) differently to write Ũ_{y}(x, y) = −Ũ_{x}(y, x),

Ũ(y, 2) = −Ũ(2, y)  (2.22)
Ũ_{y}(y, 2) = −Ũ_{x}(2, y).  (2.23)

The substitution of (2.22), (2.23) into the boundary condition (2.7) yields

αŨ(0, y) + βŨ_{x}(0, y) = −αŨ(2, y) − βŨ_{x}(2, y).  (2.24)

If we again use the separation of variables Ũ(x, y) = ϕ(x)ψ(y) in the equation (2.24) and proceed as for the case of the function ψ(y), we will have

ϕ′(x) = −\frac{iμ}{2}ϕ(x), 0 ≤ x ≤ 2,  (2.25)
−αϕ(0) − βϕ′(0) = αϕ(2) + βϕ′(2).  (2.26)

The condition (2.26) again can be replaced with

ϕ′(0) = −ϕ(2),  (2.27)

The eigenvalues and eigenfunctions of problem (2.25), (2.27) are

μ = (−2m − 1)π, ϕ_{m}(y) = exp(\frac{(2m + 1)πix}{2}), m = 0, ±1, ±2, … (2.28)

Both sets of eigenfunctions {ψ_{n}(y)} and {ϕ_{m}(x)} are complete and orthogonal in L_{2}(0, 2). The eigenfunctions and the eigenvalues of (2.6)–(2.8) will be

\begin{matrix}λ_{mn} = (2m + 1)(2n + 2)π^{2}, \\ Ũ_{mn}(x, y) = exp(\frac{(2m + 1)x}{2} + \frac{(2n + 1)y}{2} )πi, m, n = 0, ±1, ±2, … \end{matrix} (2.29)

To find the eigenfunctions of the problem (1.1)–(1.3), we introduce the functions

\begin{aligned}u_{mn}(x, y) & = Ũ_{mn}(x, y) − Ũ_{mn}(y, x) \\ & = exp(\frac{(2m + 1)x}{2} + \frac{(2n + 1)y}{2} )πi \\ & − exp(\frac{(2m + 1)y}{2} + \frac{(2n + 1)x}{2})πi, \end{aligned}  (2.30)

for m, n = 0, ±1, ±2, . . .. The set {u_{mn}(x, y)} is complete and orthogonal in L_{2}(T′) where T′ = {(x, y) : 0 ≤ y ≤ 2, 0 ≤ x ≤ y}.

Theorem 2.1 Problem (1.1)–(1.3) when p ≡ 1 is selfadjoint. It has eigenvalues (2.29) and eigenfunctions (2.30). The eigenfunctions are complete and orthogonal in L_{2}(T).

3 The non-selfadjoint problem

Now consider the problem (1.1)–(1.3) and make the change of variables x = s−t and y = s + t, to obtain

U_{xy} = γP (x, y)U(x, y), (x, y) ∈ T′,  (3.1)
T′ = {(x, y) : 0 ≤ y ≤ 2, 0 ≤ x ≤ y}  (3.2)
αU(0, y) + βU_{x}(0, y) = αU(y, 2) + βU_{y}(y, 2), 0 ≤ y ≤ 2  (3.3)
U(x, x) = 0, 0 ≤ x ≤ 2,  (3.4)

where U(x, y) = u(\frac{x+y}{2} , \frac{y−x}{2}), P (x, y) = p(\frac{x+y}{2} , \frac{y−x}{2}), and γ = −λ/4. Integrating (3.1) in T′ from 0 to ξ, we have

U_{y}(ξ, y) − U_{y}(0, y) = γ ∫_{0}^{ξ} P U dx.  (3.5)

The above equation when ξ = y and y = 2 will be

U_{y}(y, 2) − U_{y}(0, 2) = γ ∫_{0}^{y} P (x, 2)U(x, 2) dx.  (3.6)

Integrating equation (3.1) in T′ from y to η, we have

U_{x}(x, η) − U_{x}(x, y) = γ ∫_{y}^{η} P U dy.

When x = 0 and η = 2 this equation becomes

U_{x}(0, 2) − U_{x}(0, y) = γ ∫_{y}^{2} P (0, y)U(0, y) dy.  (3.7)

In (3.5) let ξ lie on the line y = x and integrate the equation from ξ to η in T′,

U(ξ, η) − U(ξ, ξ) − U(0, η) + U(0, ξ) = γ ∫_{ξ}^{η} ∫_{0}^{ξ} P Udxdy.  (3.8)

Since U(ξ, ξ) = 0 by the boundary condition (3.4), we have

U(ξ, η) = U(0, η) − U(0, ξ) + γ ∫_{ξ}^{η} ∫_{0}^{ξ} P U dx dy .  (3.9)

From (3.6) when y is replaced with η we have

U_{y}(η, 2) = U_{y}(0, 2) + γ ∫_{0}^{η} P (x, 2)U(x, 2) dx.  (3.10)

From (3.7), when y is replaced with η we have

U_{x}(0, η) = U_{x}(0, 2) − γ ∫_{η}^{2} P (0, y)U(0, y) dy.  (3.11)

158 A selfadjoint hyperbolic boundary-value problem EJDE/Conf/10 Finally from equation (3.9) we have,

U(η, 2) = U(0, 2) − U(0, η) + γ ∫_{η}^{2} ∫_{0}^{η} P U dx dy .  (3.12)

Now, we substitute the right hand side of the equations (3.10), (3.11), (3.12) into the boundary condition (3.3) with y replaced with η,

αU(0, η) + β(U_{x}(0, 2) − γ ∫_{η}^{2} P (0, y)U(0, y)dy)

= α(U(0, 2) − U(0, η) + γ ∫_{η}^{2} ∫_{0}^{η} P U dx dy) + β(U_{y}(0, 2)
+ γ ∫_{0}^{η} P (x, 2)U(x, 2)dx).

Placing αU(0, η) in right-hand side and combining,

\begin{matrix}2αU(0, η) + β(U_{x}(0, 2) + γ ∫_{η}^{2} P (0, y)U(0, y)dy) \\ = α(U(0, 2) + γ ∫_{η}^{2} ∫_{0}^{η} P U dx dy) + β(U_{y}(0, 2) \\ + γ ∫_{0}^{η} P (x, 2)U(x, 2)dx). \end{matrix}  (3.13)

Rewrite this equation with η replaced with ξ,

\begin{matrix}2αU(0, ξ) + β(U_{x}(0, 2) + γ ∫_{ξ}^{2} P (0, y)U(0, y)dy) \\ = α(U(0, 2) + γ ∫_{ξ}^{2} ∫_{0}^{ξ} P U dx dy) + β(U_{y}(0, 2) \\ + γ ∫_{0}^{ξ} P (x, 2)U(x, 2)dx). \end{matrix}  (3.14)

Subtract equation (3.14) from (3.13), 2α(U(0, η) − U(0, ξ)) + βγ(∫_{η}^{2} P (0, y)U(0, y)dy − ∫_{ξ}^{2} P (0, y)U(0, y)dy)

= αγ(∫_{η}^{2} ∫_{0}^{η} P Udxdy − ∫_{ξ}^{2} ∫_{0}^{ξ} P U dx dy)  (3.15)
+ βγ(∫_{0}^{η} P (x, 2)U(x, 2)dx − ∫_{0}^{ξ} P (x, 2)U(x, 2)dx).

Solve for U(0, η) − U(0, ξ) in (3.15), assuming α ≠ 0, and substitute in (3.9),

\begin{aligned}U(ξ, η) = & − \frac{β}{2α}γ(∫_{ξ}^{2} P (0, y)U(0, y)dy − ∫_{η}^{2} P (0, y)U(0, y)dy) \\ & + \frac{γ}{2} (∫_{η}^{2} ∫_{0}^{η} P U dx dy − ∫_{ξ}^{2} ∫_{0}^{ξ} P U dx dy) \\ & + \frac{β}{2α}γ(∫_{0}^{η} P (x, 2)U(x, 2)dx − ∫_{0}^{ξ} P (x, 2)U(x, 2)dx) \\ & + γ ∫_{ξ}^{η} ∫_{0}^{ξ} P U dx dy . \end{aligned}  (3.16)

Rewrite (3.16) in a compact form using the Green’s function, G(ξ, η; x, y) described in Figure 1,

\begin{matrix}U(ξ, η) = & γ(∫ ∫_{T′} G(ξ, η; x, y)P U(x, y) dx dy \\ − ∫_{0}^{2} g(ξ, η, x)(P (0, x)U(0, x) − P (x, 2)U(x, 2))dx), \end{matrix}  (3.17)

where

g(ξ, η, x) = \begin{cases}0 & if 0 ≤ x ≤ ξ \\ β/(2α) & if ξ ≤ x ≤ η \\ 0 & if η ≤ x ≤ 2 \end{cases}

Now, let G be the operator,

G[U] = ∫ ∫_{T′} G(ξ, η; x, y)P U(x, y) dx dy  (3.18)

defined on the Hilbert space of weighted square integrable functions H = L_{2}^{P} (T′). Assume the function U along the characteristics x = 0 and y = 0 in T′ is given, and denote,

f(ξ, η) = ∫_{0}^{2} g(ξ, η, x)(P (0, x)U(0, x) − P (x, 2)U(x, 2))dx.  (3.19)

Also, let Γ = 1/γ when γ ≠ 0. Then, (3.13) can be written as

G[U] = ΓU + f.  (3.20)

The operator G in (3.18), is the same as the one in [7], where it is shown to be selfadjoint in H. Denote the normalized eigenfunctions of G by E_{k}, and the inner product in H by ⟨., .⟩. Using the standard Fredholm alternative [10] we have the following theorem.

Theorem 3.1 Let Γ_{k} be the eigenvalues of GU = ΓU, where by the selfadjointness of G, are real and satisfy |Γ_{k}| → 0, as k → ∞ then, we have

[figure: x; 2; w; z; y; 2; w; z; 0; 0; 0; 0; 1/2; 1/2]

Figure 1: The Green’s function G(w, z; x, y)

1. Γ ≠ Γ_{k} for any integer k. A unique solution of (3.20) exists and is given in the form U = −\frac{f}{Γ} + ∑_{k} \frac{Γ_{k}}{Γ(Γ_{k}−Γ)}⟨f, E_{k}⟩E_{k}.

2. Γ = Γ_{m}, one of the eigenvalues of G, and Γ_{m} is not degenerate. If ⟨f, E_{m}⟩ ≠ 0 the equation (3.20) has no solution. If ⟨f, E_{m}⟩ = 0 then, (3.20) has infinitely many solutions

U = −\frac{f}{Γ_{m}} + ∑_{k≠m} \frac{Γ_{k}}{Γ_{m}(Γ_{k} − Γ_{m})}⟨f, E_{k}⟩E_{k} + cE_{m},

with c an arbitrary constant.

3. Γ_{m} is degenerate, Γ_{m_{1}} = Γ_{m_{2}} = ⋯ = Γ_{m_{j}}, for successive indices m_{1}, m_{2},… m_{j}, with some m_{i} = m, i = 1, 2, … j, where j is the multiplicity of Γ_{m}. Then, unless ⟨f, E_{i}⟩ = 0, i = 1, 2, …, j, the equation (3.20) has no solution. If however, these j solvability conditions are satisfied, the solution can be represented by,

U = −\frac{f}{Γ_{m}} + ∑_{k≠m_{i}} \frac{Γ_{k}}{Γ_{m}(Γ_{k} − Γ_{m})}⟨f, E_{k}⟩E_{k} + ∑_{i=1}^{j} c_{i}E_{m_{i}},

where c_{i}’s are arbitrary constants.

Remark When p ≡ 1 in problem (1.1)–(1.3), the problem is selfadjoint if α^{2} + β^{2} ≠ 0. When p is not necessarily 1, the case α ≠ 0, β = 0, reduces the problem to the selfadjoint problem of [7]. When α ≠ 0, β ≠ 0, Theorem 2 holds. When α = 0, no conclusion can be drawn.

References

[1] P. Garabedian. Partial differential equations. Second edition, Chelsea Publishing Co., New York, 1986. vxii+672 pp. ISBN: 0-8284-00325

[2] L. Haws. Symmetric Green’s functions for certain hyperbolic problems, Comput. Math. Appl. 21 (1991), no. 5, 65–78.

[3] N. Iraniparast. A method of solving a class of CIV boundary value problems, Canad. Math. Bull. 35 (1992), no. 3, 371–375.

[4] N. Iraniparast. A boundary value problem for the wave equation. Int. J. Math. Math. Sci. 22 (1999), no. 4, 835–845.

[5] N. Iraniparast. A CIV boundary value problem for the wave equation, Appl. Anal. 76 (2000), no. 3-4, 261–271.

[6] T. Kalmenov. On the Spectrum of a Selfadjoint Problem for the Wave Equation, Akad. Nauk. Kazakh. SSR, Vestnik 1 (1983), 63-66.

[7] K. Kreith. Symmetric Green Functions for a Class of CIV Boundary Value Problems, Canadian Math. Bull. 31 (1988), 272-279.

[8] K. Kreith. Mixed selfadjoint boundary conditions for the wave equation, Differential equations and its applications (Budapest, 1991), 219–226, Colloq. Math. Soc. Janos Bolyai, 62, North-Holland, Amsterdam, 1991.

[9] K. Kreith. Establishing hyperbolic Green’s functions via Leibniz’s rule, SIAM Rev. 33 (1991), no. 1, 101–105.

[10] I. Stakgold. Green’s Functions and Boundary Value Problems, John Wiley & Sons, New York, 1979.

Nezam Iraniparast Department of Mathematics Western Kentucky University E-mail: nezam.iraniparast@wku.edu