Boundary value problems for the elliptic sine Gordon equation in a semi strip

Transcription

1 Boundary value problems for the elliptic sine Gordon equation in a semi strip Article Accepted Version Fokas, A. S., Lenells, J. and Pelloni, B. 3 Boundary value problems for the elliptic sine Gordon equation in a semi strip. Journal of Nonlinear Science, 3. pp ISSN 8 9 doi: Available at It is advisable to refer to the publisher s version if you intend to cite from the work. To link to this article DOI: Publisher: JNSA All outputs in CentAUR are protected by Intellectual Property Rights law, including copyright law. Copyright and IPR is retained by the creators or other copyright holders. Terms and conditions for use of this material are defined in the End User Agreement. CentAUR Central Archive at the University of Reading

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3 Boundary value problems for the elliptic sine-gordon equation in a semi-strip A.S. Fokas J. Lenells and B. Pelloni *Department of Applied Mathematics and Theoretical Physics Cambridge University Cambridge CB3 WA, UK. t.fokas@damtp.cam.ac.uk **Department of Mathematics Baylor University Waco, TX 76798, USA Jonatan Lenells@baylor.edu ***Department of Mathematics University of Reading Reading RG6 6AX, UK b.pelloni@reading.ac.uk September 4, Abstract We study boundary value problems posed in a semistrip for the elliptic sine-gordon equation, which is the paradigm of an elliptic integrable PDE in two variables. We use the method introduced by one of the authors, which provides a substantial generalization of the inverse scattering transform and can be used for the analysis of boundary as opposed to initial-value problems. We first epress the solution in terms of a matri Riemann-Hilbert problem whose jump matri depends on both the Dirichlet and the Neumann boundary values. For a well posed problem one of these boundary values is an unknown function. This unknown function is characterised in terms of the so-called global relation, but in general this characterisation is nonlinear. We then concentrate on the case that the prescribed boundary conditions are zero along the unbounded sides of a semistrip and constant along the bounded side. This corresponds to a case of the so-called linearisable boundary conditions, however a major difficulty for this problem is the eistence of non-integrable singularities of the function q y at the two corners of the semistrip; these singularities are generated by the discontinuities of the boundary condition at these corners. Motivated by the recent solution of the analogous problem for the modified Helmholtz equation, we introduce an appropriate regularisation which overcomes this difficulty. Furthermore, by mapping the basic Riemann-Hilbert problem to an equivalent modified Riemann-Hilbert problem, we show that the solution can be epressed in terms of a matri Riemann-Hilbert problem whose jump matri depends eplicitly on the width of the semistrip L, on the constant value d of the solution along the bounded side, and on the residues at the given poles of a certain spectral function denoted by h. The determination of the function h remains open.

4 y L S 3 Figure : The semistrip S Introduction A method for solving initial-boundary value problems for linear and integrable nonlinear PDEs was introduced in [3, 4] and developed by several authors, see the survey [5] and the references therein. This method has already been used for: a linear and integrable nonlinear evolution PDEs formulated on the half line and on a finite interval [4, 5,,, 7,,,, 3, 4, 7, 38, 39, 47]; b linear and integrable nonlinear hyperbolic PDEs [6, 4, 4]; c linear elliptic PDEs [, 3, 8, 9, 6, 8, 9, 44, 45, 46]. The aim of this paper is to implement this method in the case of the prototypical integrable nonlinear elliptic PDE, namely the celebrated elliptic sine-gordon equation. This equation was first analysed in [37] see also [7, 3]; simple boundary value problems for this equation, using the method of [3], have been considered in [4, 43]. For the case of nonlinear elliptic PDEs in cylindrical coordinates, see [3, 34]. We will consider the sine-gordon equation in the form q + q yy = sin q, q = q, y,. and we will analyze boundary value problems posed in the semi-infinite strip S = { < <, < y < L},. where L is a positive finite constant. The sides {y = L, < < }, { =, < y < L} and {y =, < < } will be referred to as side, and 3 respectively, see figure. Suppose that. is supplemented with appropriate, compatible boundary conditions on the boundary of the semistrip S, so that there eists a unique solution q, y. It will be shown in section that this solution can be epressed in terms of the solution of a matri Riemann- Hilbert RH problem with a jump on the union of the real and imaginary ais of the comple plane. The jump matri is epressed in terms of certain functions, called spectral functions, which will be denoted by {a j, b j }, j =,, 3. These functions can be uniquely characterized via the solution of certain linear Volterra integral equations, in terms of the Dirichlet and Neumann boundary values. Namely, {a, b }, {a, b } and {a 3, b 3 } are uniquely determined in terms of {q, L, q y, L}, {q, y, q, y} and {q,, q y, } respectively. However, for a well posed problem only a subset of these boundary values are prescribed as boundary conditions. Thus, in order to compute the spectral functions in terms of the given boundary conditions, one must first

5 determine the unknown boundary values, i.e. one must characterize the Dirichlet to Neumann map. The solution of this problem, which makes crucial use of the so-called global relation, yields in general a nonlinear map, see [6, 8, 5, 35, 36]. In the case of integrable nonlinear evolution PDEs, it has been shown in [7, 9, 3, 4, 33] that there eists a particular class of boundary conditions, called linearizable, for which it is possible to avoid the above nonlinear map. The main result of the present paper is the analysis of a particular case of linearizable boundary conditions for the sine-gordon equation on the semi-infinite strip. In particular, the following boundary conditions will be investigated in detail: q, L = q, =, < < ; q, y = d, < y < L,.3 where d is a finite constant. We assume that < d < π. These boundary conditions are discontinuous at the corners, and, L of the domain. This implies that q y, y has a non-integrable singularity at the two corners of the semistrip. Using an appropriate gauge transformation, which is motivated by the recent solution of the analogous problem for the modified Helmholtz equation [], see also Appendi A, we are able to overcome this difficulty and introduce well-defined spectral functions. Furthermore, we show that the basic Riemann-Hilbert problem can be mapped to a simpler Riemann-Hilbert problem whose jump matri, instead of depending on the si unknown spectral functions {a j, b j } 3, depends eplicitly on the given constant d, on the width L of the semistrip, and on the residues at the given poles of a certain spectral function, denoted by h. The rigorous analysis for the determination of h remains open. This result, as well as the analogous result valid for the elliptic version of the Ernst equation [3, 34], imply that the method of [3] provides a powerful tool for analyzing effectively a large class of interesting boundary conditions. Spectral analysis under the assumption of eistence In what follows we assume that. is supplemented with appropriate boundary conditions on the boundary of the semistrip S, compatible at the corners of the domain, so that the eistence of a unique, smooth solution q, y can be assumed. Furthermore, we assume the following: q, L, q y, L, q,, q y, L R +, q, L, q y, L, q,, q y, L R +,. q, y, q, y, yq, y, yq, y L [, L]. The sine-gordon equation is the compatibility condition of the following La pair [3] for the matri-valued function Ψ, y,, C: Ψ + Ω [σ 3, Ψ] = Q, y, Ψ,. Ψ y + ω [σ 3, Ψ] = iq, y, Ψ,.3 where Ω = i, ω = +,.4 3

6 , L, y, y, y, y Ψ Ψ, Ψ 3 Figure : The functions Ψ, Ψ and Ψ 3 Q, y, = i 4 cos q q iq y + i sin q q iq y i sin q cos q, q = q, y..5 Equations. and.3 can be written as the single equation d e Ω+ωy cσ 3 Ψ, y, = W, y,,.6 where the differential form W is given by W, y, = e Ω+ωy cσ 3 Q, y, Ψ, y, d + iq, y, Ψ, y, dy,.7 and σ 3 acts on a matri A by σ 3 A = [σ 3, A]. Remark. Note that Ω = Ω = Ω, ω = ω = ω.. Bounded and analytic eigenfunctions We define three solutions Ψ j, y,, j =,, 3, of.6 by where,y Ψ j, y, = I + e Ω+ωy cσ 3 W ξ, η,,.8 j,y j, y =, y,, y =, L, 3, y 3 =,..9 Since the differential form W is eact, the integral on the right hand side of.8 is independent of the path of integration. We choose the particular contours shown in figure. This choice implies the following inequalities on the contours:, y, y : ξ, 4

7 , y, y : ξ, η y, 3, y 3, y : ξ, η y. The first inequality above implies that the eponential appearing in the second first column of the right hand side of the equation defining Ψ is bounded and analytic for Im < Im >. Similar considerations are valid for Ψ and Ψ 3. Hence we denote the matrices Ψ j as follows: Ψ = Ψ, Ψ 34, Ψ = Ψ 4, Ψ, Ψ 3 = Ψ 3 3, Ψ 3, where the superscript denotes the union of the first and second quadrants of the comple plane, and similarly for the other superscripts. The function Ψ is analytic for Im > and it has essential singularities at = and = ; furthermore, Ψ = + O,, Im.. Similar considerations are valid for the column vectors Ψ 34, Ψ 3 3 and Ψ 3. The function Ψ is an analytic function in the entire comple plane, ecept at = and =, where it has essential singularities. In addition, Ψ 4 = Ψ =. Spectral functions + O,, + O,, Any two solutions Ψ, Ψ of.6 are related by an equation of the form We introduce the notations 3π arg π, π arg π.. Ψ, y, = Ψ, y, e Ω+ωy cσ 3 C.. S = Ψ, L,, S = Ψ,,, S 3 = Ψ,,..3 Then equation. implies the following equations: Ψ, y, = Ψ, y, e Ω+ωy cσ 3 e ω Lcσ3 S, R,.4 Ψ, y, = Ψ 3, y, e Ω+ωy cσ 3 S, C \ {},.5 Ψ, y, = Ψ 3, y, e Ω+ωy cσ 3 S3, R, R +..6 The notation R, R + means that the equation for the first column vector in.6 is valid for R, while the equation for the second vector is valid for R +. Equations.3-.6 suggest the following definitions: S = Φ,, Φ, = I e Ωξ cσ 3 Qξ, L, Φ ξ, dξ, 5

8 C +, C, < <,.7 S = Φ,, S 3 = Φ 3,, Φ y, = I i L y e ωη y cσ 3 Q, η, Φ η, dη, C, < y < L,.8 Φ 3, = I e Ωξ cσ 3 Qξ,, Φ3 ξ, dξ, C +, C, < <..9 The matri Q satisfies the symmetry properties Q = Q, Q = Q.. Hence the matrices Φ i, i =,.., 3, can be represented in the form A, B Φ =, A y, B, Φ B, A, = y, A3, B, Φ B y, A y, 3 = 3, B 3, A 3,, and therefore ai b S i = i b i a i, i =,, 3. The spectral functions {a, b }, {a, b } and {a 3, b 3 } are defined in terms of {q, L, q y, L}, {q, y, q, y} and {q,, q y, } respectively, through equations These functions have the following properties: a, b are analytic and bounded in C +. a a b b =, R. a = + O, b = O as, Im. a, b are analytic functions of for all C, ecept for essential singularities at = and =. a a b b =, C \ {}. a = + O, b = O as, 3π a 3, b 3 are analytic and bounded in C +. arg π. a 3 a 3 b 3 b 3 =, R. a 3 = + O, b3 = O as, Im. These properties follow from the analogous properties of the matri-valued functions Φ j, j =,, 3, from the condition of unit determinant, and from the large asymptotics of these functions. 6

9 .3 The global relation Evaluating equations.5 and.6 at =, y = L, we find and Eliminating Ψ 3, L, we obtain I = Ψ 3, L, e ω Lcσ3 S S = Ψ 3, L, e ω Lcσ3 S 3. e ω Lcσ3 S = S S 3.. The first column vector of this equation yields the following global relations: a = a a 3 b b 3, C +,.a b e ωl = a b 3 a 3 b, C +..b.4 The Riemann-Hilbert problem Equations.4-.6, relating the various analytic eigenfunctions, can be rewritten in a form that determines the jump conditions of a RH problem, with unitary jump matrices on the real and imaginary aes. This involves tedious but straightforward algebraic manipulations. The final form is M, y, = M +, y, J, y,, R ir,.3 where the matrices M ± and J are defined as follows: M + = Ψ, a 3 Ψ 3, arg M = Ψ, a Ψ, arg M + = a 3 Ψ3 3, Ψ34, arg M = a Ψ4, Ψ34, arg [, π ], [ π ], π, [ π, 3π ], [ 3π, π ],.4 J, y, = J α, y,, if arg = α, α =, π, π, 3π,.5 where, using the global relations., we find J π/ = J = a a a 3 b 3 a 3 e θ,y, e ωl b a e θ,y, b a a 3 e θ,y,, J 3π/ = 7, b a a 3 eθ,y,

10 and where J π = J 3π/ J J π/,.6 θ, y, = Ω + ωy..7 All the matrices J α have unit determinant: for J π/ and J 3π/ this is immediate, whereas for J we find where we have used the equation detj = a + e ωl b b 3 a a 3 = a a 3 a a 3 =, a = a a 3 b 3 b e ωl, R,.8 which is a consequence of the global relations.. J π/ Ψ, a Ψ Ψ, a 3 Ψ 3 J π Γ J a 3 Ψ3 3, Ψ34 a Ψ4, Ψ34 J 3π/ Figure 3: The bounded eigenfunctions and the Riemann-Hilbert problem The solution M, y, of this RH problem is a sectionally meromorphic function of. The possible poles of this function are generated by the zeros of the function a in the region {arg [ π, π]}, by the zeros of a 3 in the region {arg [, π ]}, and by the corresponding zeros of a, a 3. We assume The possible zeros of a in the region {arg π, π} are simple; these zeros are denoted j, j =,.., N.9 The possible zeros of a 3 in the region {arg, π } are simple; these zeros are denoted ζ j, j =,.., N 3 The residues of the function M at the corresponding poles can be computed using equations Indeed, equation.6 yields Ψ = a 3 Ψ b 3 e θ,y, Ψ 3, 8

11 hence Ψ 3 Res ζj a 3 = Ψ 3 ζ j ȧ 3 ζ j where ȧ 3 denotes the derivative of a 3 with respect to. Similarly, using.4, Ψ Res j a = Ψ ζ j ȧ 3 ζ j b 3 ζ j e θ,y,ζj,.3 = Ψ j Ψ j = ȧ j ȧ j b j e ωjl e θ,y,j..3 Using the notation [M] for the first column and [M] for the second column of the solution M of the RH problem.3, the equations.3 and.3 imply the following residue conditions: Res ζj [M, y, ] = e θ,y,ζj ȧ 3 ζ j b 3 ζ j [M, y, ζ j], < arg < π,.3 Res j [M, y, ] = e θ,y,j ȧ j b j e ωjl [M, y, j], Similar residue conditions are obtained in C by letting. π < arg < π. The inverse problem Rewriting the jump condition, we obtain M + M = M + M + J = M + I J M + M = M + J,.33 where J = I J. The asymptotic conditions.-. imply M, y, = I + M, y + O,, C \ R ir..34 Equation.33 and the condition.34 yield the following integral representation for the function M: M, y, = I + M +, y, J, y, πi Γ d, C \ Γ.35 where Γ = R ir. Equations.34 and.35 imply M = πi Using.34 in the first ODE in the La pair., we find Γ M +, y, J, y, d..36 i 4 [σ 3, M ] = i q iq y σ q iq y = M = lim 4 M,.37 where σ, σ 3 denote the usual Pauli matrices. In order to obtain an epression in terms of q rather than its derivatives, we consider the coefficient in. of the term. The, element of this coefficient yields cos q, y = + 4iM + M..38 9

12 3 Spectral theory assuming the validity of the global relation 3. The spectral functions The above analysis motivates the following definitions for the spectral functions. The spectral functions at the y = and y = L boundaries Definition 3. Given the functions q, L, q y, L satisfying conditions., define the map by S : {q, L, q y, L} {a, b } a b = [Φ, ], C +, where [Φ, ] denotes the first column vector of the unique solution Φ, of the Volterra linear integral equation Φ, = I e Ωξ cσ 3 Qξ, L, Φξ, dξ, 3. and Q, L, is given in terms of q, L and q y, L by equation.5. In what follows, we also assume that the function a may have N simple poles j in C +. Similarly for a 3. Proposition 3. The spectral functions a, b have the following properties. i a, b are continuous and bounded for Im, and analytic for Im >. ii a = + O, b = O as, Im. iii a = cos q,l + O, b = i sin q,l + O as, Im. iv a a b b =, R. v The map Q : {a, b } {q, L q y, L}, inverse to S, is given, cos q, L = + 4i lim M + lim M q y, L = iq, L + i lim M, where M is the solution of the following Riemann-Hilbert problem: The function { M+, C M, = + M, C is a sectionally meromorphic function of C.

13 M = I + O as, and M, = M +, J,, R, where J, = b a e Ω b a eω a a, R. 3. Let [M] i denote the i-th column vector of M, =,. The possible poles of M + occur at j, and the possible poles of M occur at j in C, and the associated residues are given by Res j [M, ] = Res j [M, ] = e Ωj ȧ j b j [M, j], e Ωj ȧ j b j [M, j]. 3.3 The spectral functions {a 3, b 3 } are defined similarly: Definition 3. Given the functions q,, q y,, satisfying conditions., define the map by S 3 : {q,, q y, } {a 3, b 3 } a3 b 3 = [Φ 3, ], C +, where [Φ 3, ] denotes the first column vector of the unique solution Φ 3, of the Volterra linear integral equation Φ, = I e Ωξ cσ 3 Qξ,, Φξ, dξ, 3.4 and Q,, is given in terms of q, and q y, by equation.5. Proposition 3. The spectral functions a 3, b 3 have the properties i-v of proposition 3., provided a is replaced by a 3, b is replaced by b 3, S is replaced by S 3 and L in replaced by in all epressions. The spectral functions at the = boundary Definition 3.3 Given the functions q, y, q, y, satisfying conditions., define the map S : {q, y, q, y} {a, b } by a b = [Φ, ], C +,

14 where [Φ, y] denotes the first column vector of the unique solution Φ, y of the Volterra linear integral equation Φy, = I i L y e ωη y cσ 3 Q, η, Φη, dη, 3.5 and Q, y, is given in terms of q, y and q, y by equation.5. Proposition 3.3 The spectral functions a, b have the following properties. i a, b are analytic functions of, ecept for essential singularities at = and =, bounded for Re. ii a = + O, b = O as, Re. iii a = cos q, + O, b = i sin q, + O as, Re. iv a a b b =, C. v The map Q : {a, b } {q, y q y, y}, inverse to S, is given by cos q, y = 4 lim M y lim M q, y = iq y, y + lim M, where M is the solution of the following Riemann-Hilbert problem: The function My, = is a sectionally meromorphic function of C. M = I + O as, and where J y, = { M+ y, Re M y, Re M y, = M + y, J y,, b a e ω b a eω a a ir, M satisfies appropriate residue conditions at the zeros of a. Proof of propositions , ir. The proof of properties i-iv follows from the discussion in Section.. In particular, property iii follows from the asymptotic behaviour at, which can be derived by analysing equations.-.3 see [4], and is given by Ψ = Ψ + O,, Ψ, y = cos q,y i sin q,y i sin q,y cos q,y. 3.6

15 To prove v, we note that the function Φ, given by.7 is the unique solution of the ODE Φ + Ω σ 3Φ = Q, L, Φ,, Φ, = I. lim Furthermore, Φ 3, given by.9 is the solution of the same ODE problem, with Q, L, replaced by Q,,. Similarly, Φ y, given by.8 is the unique solution of the ODE Φ y + ω σ 3Φ = iq, y, Φy,, ΦL, = I. The spectral analysis of the above ODEs yields the desired result. Regarding the rigorous derivation of the above results, we note the following: If {q, L, q y, L}, {q,, q y, } and {qy,, q y, } are in L, then the Volterra integral equations 3., 3.4 and 3.5 respectively, have a unique solution, and hence the spectral functions {a j, b j }, j =,.., 3, are well defined. Moreover, under the assumption. the spectral functions belong to H R, hence the Riemann-Hilbert problems that determine the inverse maps can be characterized through the solutions of a Fredholm integral equation, see [, 5]. QED 3. The Riemann-Hilbert problem Theorem 3. Suppose that a subset of the boundary values {q, L, q y, L}, {q,, q y, }, < <, and {qy,, q y, }, < y < L, satisfying., are prescribed as boundary conditions. Suppose that these prescribed boundary conditions are such that the global relations. can be used to characterize the remaining boundary values. Define the spectral functions {a j, b j }, j =,.., 3, by definitions Assume that the possible zeros { j } N j= of a and {ζ j } N3 j= of a 3 are as in assumption.9. Define M, y, as the solution of the following matri Riemann-Hilbert problem: The function M, y, is a sectionally meromorphic function of away from R ir. The possible poles of the second column of M occur at = ζ j, j =,..., N 3, in the first quadrant and at = j, j =,..., N, in the second quadrant of the comple plane. The possible poles of the first column of M occur at = j j =,..., N and = ζ j j =,..., N 3. The associated residue conditions satisfy the relations.3. M = I + O as, and M, y, = M +, y, J, y,, R ir, where M = M + for in the first or third quadrant, and M = M for in the second or fourth quadrant of the comple plane, and J is defined in terms of {a j, b j } by equations.6. 3

16 Then M eists and is unique, provided that the H norm of of the spectral functions is sufficiently small. Define q, y is terms of M, y, by q iq y = lim, 3.7 cos q, y = + 4i lim + lim. 3.8 Then q, y solves.. Furthermore, q, y evaluated at the boundary, yields the functions used for the computation of the spectral functions. Proof: Under the assumptions., the spectral functions are in H. In the case when a and a 3 have no zeros, the Riemann-Hilbert problem is regular and it is equivalent to a Fredholm integral equation. However, we have not been able to establish a vanishing lemma, hence we require a small norm assumption for solvability. If a and a 3 have zeros, the singular RH problem can be mapped to a regular one coupled with a system of algebraic equations []. Moreover, it follows from standard arguments, using the dressing method [48, 49], that if M solves the above RH problem and q, y is defined by , then q, y solves equation.. The proof that q evaluated at the boundary yields the functions used for the computation of the spectral functions follows arguments similar to the ones used in [3]. QED 4 Linearizable boundary conditions We now concentrate on the particular boundary conditions.3. We note that these boundary conditions are symmetric with respect to the line y = L. Hence, if q, y is a solution, so is q, L y. Assuming that the solution is unique, we can conclude that q, y = q, L y, < <, < y < L. 4. These boundary conditions are not compatible at the corners of the domain, and therefore introduce a discontinuity at each corner. It turns out that these discontinuities imply that if q, y is the solution of the resulting boundary value problem, then the function q y, = q y, L is not integrable near =. Similarly, q, y is not integrable near y = and y = L. Hence we cannot guarantee that the results of propositions hold. In particular, the spectral functions as given by.7-.9 and the resulting Riemann-Hilbert problem are not well defined. To overcome this lack of regularity, we will employ a gauge transformation to define a modified La pair. This transformation is motivated by the recent analysis of the linearised problem [], which we summarize in Appendi A. For the linear case, it can be shown that the behaviour of the boundary function q y, as is given by q y, d π. The contribution of this term can be eliminated by an appropriate gauge transformation. The advantage of the new La pair we define below by adapting the linear gauge transformation to the nonlinear setting, is that the spectral functions and the Riemann-Hilbert problem are well defined, indicating that in the nonlinear case, as in the linear, the singular behaviour introduced by the terms q y, and q, y is eliminated. 4

17 4. A new La pair The linearised version of the elliptic sine-gordon equation, namely the modified Helmholtz equation, with the boundary conditions.3, is discussed in Appendi A, where we show how, by incorporating appropriately in the differential form associated with the linear equation the term κ, y = 4 q y ξ, ydξ, 4. the spectral problem is regularized. Motivated by the linear analysis, we now introduce a new eigenfunction Φ via the gauge transformation cosh κ, y sinh κ, y gφ = Ψ, g, y := e κ,yσ =, 4.3 sinh κ, y cosh κ, y where Ψ denotes the solution of the La pair.-.3 and κ, y is given by 4.. Note that κ, y = 4 q y, y, κ y, y = 4 sin qξ, ydξ 4 q, y. The transformation matri g, y has unit determinant, and is chosen to satisfy the property g g = g g = q y, y. 4 q y, y Let the function Φ satisfy the La pair with Φ + Ω [σ 3, Φ] = V, y, Φ, 4.4 Φ y + ω [σ 3, Φ] = V, y, Φ, 4.5 V, y, = g g + g Q, y, g Ω g [σ 3, g] 4.6 = i coshκ coshκ iq+ 4 sinhκ +q +sinhκ iq 4 sinhκ +q sinhκ iq 4 coshκ + +coshκ iq 4, V, y, = g g y + g iq, y, g ω g [σ 3, g] 4.7 = R coshκ +coshκ iq 4 sinhκ sinhκ + R sin qdξ+iqy+sinhκ iq 4 sin qdξ+iqy+sinhκ iq 4 coshκ +coshκ iq 4 We now use the eigenfunctions determined by the La pair to define new spectral functions. Namely, in analogy with.7-.9, we define. S = ϕ,, ϕ, = I e Ωξ cσ 3 V ξ, L, ϕ ξ, dξ, 5

18 C +, C, < <, 4.8 S = ϕ,, S 3 = ϕ 3,, ϕ y, = I L y e ωη y cσ 3 V, η, ϕ η, dη, C, < y < L, 4.9 ϕ 3, = I e Ωξ cσ 3 V ξ,, ϕ 3 ξ, dξ, C +, C, < <. 4. Note that V does not involve q y, y, and V does not involve q, y, the terms respectively responsible, at least in the linear case, for the non-integrable behaviour. Note also that, since the symmetry relation. holds for V, V in place of Q, we can represent the matrices ϕ i in the form A, B ϕ =, A y, B, ϕ B, A, = y,, B y, A y, A3, B ϕ 3 = 3,, 4. B 3, A 3, and set ai b S i = i b i a i, i =,, The rest of the general construction of sections.3-3. is formally valid with the spectral functions a j, b j, j =,, 3 as defined by 4., ecept for the statement i-iii and v of propositions Symmetry conditions Given the boundary conditions.3, equations are written eplicitly as follows: ϕ, = I Ω ϕ y, = I e Ωξ cσ 3 cosh κ sinh κ ξ, Lϕ sinh κ cosh κ ξ, dξ, < <, C +, C, 4.3 L y e ωη y cσ 3 coshκ +coshκ id 4 sinhκ + R sin qdξ+sinhκ id 4 ϕ 3, = I Ω sinhκ + R sin qdξ sinhκ id 4 coshκ +coshκ id 4, ηϕ η, dη, < y < L, C, 4.4 e Ωξ cσ 3 cosh κ sinh κ ξ, ϕ sinh κ cosh κ 3 ξ, dξ, < <, C +, C

19 Using that κ, = κ, L, we can immediately conclude that A, B, = A3, B 3, = a = a 3, b = b 3, 4.6 where A i, B, a i, b i are as in In equations 4.3 and 4.5, the only dependence on is through Ω. Thus, since Ω = Ω, it follows that the vector functions A, B and A 3, B 3 satisfy the same symmetry properties. Hence, a j = a j, b j = b j, j =, 3, Im. 4.7 It turns out that the vector function A, B also satisfies a certain symmetry condition, as stated in the following proposition. Proposition 4. Let q, y be a sufficiently smooth function. Then the vector solution of the linear Volterra integral equation 4.4 satisfies the following symmetry conditions: A y, = A y, F B y, + F e ωy L B y, F e ωy L A y, F, B y, = B y, F A y, + F e ωy L A y, F e ωy L B y, F, where the function F is defined by Proof: Define a function φ y, by < y < L, C, 4.8 F = i + tan d. 4.9 ϕ y, = φ y, e ω σ3y L 4. where ϕ is defined by 4.4. It follows that φ satisfies the ODE where φ y = V φ, 4. φ L, = I, < y < L, V y, = V, y, We seek a nonsingular matri R, independent of y, such that ω, y, σ V y, = RV y, R

20 It can be verified that such a matri is given by F R = F, 4.4 where F is defined by 4.9. Replacing in equation 4. by, and using 4.3, we find the following equation: R φ y, = V y, R φ y, y, hence R φ y, = φ y, C, where C is a y-independent matri. Using the second of equations 4., it follows that C = R, and therefore φ y, = Rφ y, R. This equation and equation 4. imply ϕ y, = Rϕ y, e ω cσ 3 y L R. 4.5 The first column vector of this equation implies 4.8. QED Remark 4. Recalling that a = A,, and b = B,, equations 4.8 immediately imply the following important relations: a b = a F b + F e ωl b F e ωl a F, = b F a + F e ωl a F e ωl b F, In summary, the basic equations characterizing the spectral functions are: a the symmetry relations 4.6, 4.7 and 4.6; b the global relations.; c the conditions of unit determinant. C \ {}. 4.6 In the net lemma, we collect some important consequences of these conditions. For simplicity, we will use the notations f := f, ˆf := f. 8

21 Lemma 4. The spectral functions satisfy the following relations: where the function G is defined by a 3 â 3 b 3ˆb3 =, R, 4.7 â 3 b 3 a 3ˆb3 = G, R, 4.8 b = + e ωl â 3 b 3, R, 4.9 Lω G := F tanh d e ωl = i tan + e ωl Proof: Equation 4.7 is just the condition of unit determinant. Using the symmetry condition 4.6 to eliminate a and b from the global relations., we find a 3 [â ] = ˆb b 3, R, 4.3a b 3 [ e ωl + a ] = a3 b, R. 4.3b The equations 4.3 together with the equations obtained by letting in 4.3 are four equations which can be solved for the four functions {a, b, â, ˆb } with the result that a = + e ωl a 3 â 3 e ωl, R, 4.3a b = + e ωl â 3 b 3, R. 4.3b This proves 4.9. Replacing by / in 4.3 and using the symmetry 4.7, we find a 3 [ a ] = b b3, R, 4.33a b 3 [ ] e ωl + a = a3 b, R. 4.33b Consider the two equations 4.33 together with the two equations obtained by letting in We can eliminate a ± and b ± from these four equations by using the symmetry relations 4.6 as well as the symmetry relations obtained by letting in 4.6. The resulting four equations can then be solved for the four functions {a, b, â, ˆb } with the result that a = a 3F + b 3 â 3 F ˆb 3 e Lω a 3 b 3 F â 3 + ˆb 3 F F, 4.34a b = a 3 + b 3 F â 3 F ˆb 3 e Lω a 3 b 3 F â 3 F + ˆb 3 F. 4.34b Comparing 4.3a with 4.34a, we find 4.8. QED The functions a 3 and b 3 are defined by 4. only for in the upper half-plane. However, equation 4.9 implies that a 3 and b 3 can be analytically etended to the whole comple plane. Indeed, equation 4.9 provides the analytic continuation of a 3 into C : a 3 = b b 3 [ + e ωl ], C

22 Similarly, equation 4.9 with replaced with provides the analytic continuation of b 3 into C : b 3 = b a 3 [ + e ωl ], C Adopting these etended definitions of a 3 and b 3, analytic continuation implies that the relations and the global relations 4.3 are valid in the whole comple plane. Proposition 4. The spectral functions satisfy the equations as well as the global relations where the known function G is given by 4.3. a 3 â 3 b 3ˆb3 =, C, 4.37 â 3 b 3 a 3ˆb3 = G, C, 4.38 b = + e ωl â 3 b 3, C, 4.39 a 3 [â ] = ˆb b 3, C, 4.4a b 3 [ e ωl + a ] = a3 b, C, 4.4b 5 Spectral theory in the linearisable case In appendi A we give the solution of the linear case. In this case, the dependence on the unknown spectral function B is resolved by mapping the basic Riemann-Hilbert problem to an equivalent but simpler one. To define this mapping, in the linear case, it is convenient to employ the two equations A.9 and A.. For the nonlinear problem, we will use the following equations, which provide the nonlinear analogues of equations A.9 and A.: and b 3 a 3 h b 3 a 3 = G, C, 5. h b a 3 + b 3 e ωl a 3 h = + e ωl G, C, 5. h where the unknown function h is defined by h = a 3 b 3, C, 5.3 and the known function G is defined by 4.3. Note that from the above properties it follows is well defined at the zeros of the function h. Moreover, in the linear limit, that b3 a 3 b 3 B 3, a 3, h, tan d d, and equations 5. and 5. become equations A.9 and A..

23 Equation 5. and 5. are a direct consequence of Indeed, a 3 G = a 3 [a 3 b 3 a 3 b 3 ] = a 3 b 3 + b 3 [ + b 3 b 3 ] = b 3 [ a 3 b 3 ] b 3, which is equivalent to equation 5. in the first and second equations above we have used equations 4.38 and 4.37 respectively. Furthermore, b a 3 = + e ωl b 3 a 3 = + e ωl [ b 3 a 3 h G ], h which, using G = G, becomes equation 5. in the first and second equations above we have used the relation 4.39 and equation 5. with replaced by. The properties of the functions G and h We give a summary of some of the properties of the function G given by 4.3 and of the unknown function h given by 5.3. The set of poles of G, denoted by P G C, is given by P G = { e ωl + = } { = n = i πn + ± } L L + n + π n Z. 5.4 Note that G has no poles at = ±i since the term e ωl vanishes at these points. Proposition 5. The following statements hold: a G admits the symmetries G = G = G/, C. b G has essential singularities at and at and a countable number of simple poles on the imaginary ais accumulating at ±i and at. G has no other singularities. c Each of the functions ± G has a countable number of zeros. All these zeros lie on the imaginary ais and they accumulate only at ±i and at. d The set of zeros of the function G is the disjoint union of the set of zeros of a 3 b 3 and the set of zeros of a 3 + b 3. e The set of zeros of the function +G is the disjoint union of the set of zeros of a 3 +b 3 and the set of zeros of a 3 + b 3. f The set of zeros of the function G is the disjoint union of the set of zeros of h and the set of zeros of h. g The function h has a double pole at each point in the set P G C. The function h has a double pole at each point in the set P G C +. The functions h and h do not have any other poles.

24 Im Gi I I G = pole of G G = G = pole of G i pole of G. Re a b Figure 4: a The graph of G for on the imaginary ais when L = and d =. b The corresponding picture of the poles of G and the zeros of G ±. The poles and zeros accumulate at the origin and at infinity. Proof: The proof of a follows from the definition 4.3. The proof of b follows from the same definition and from equation 5.4. In order to prove c, we note that the function G is purely real for ir: Gi I = + I d tan tan L I, I R. I 4 I As I increases from to +, the argument L I 4 I increases from to. It follows that each of the functions + G and G has an infinite number of zeros on the imaginary ais, see figure 4. More precisely, if ip and ip are two consecutive poles of G on the positive imaginary ais, then, unless p < < p, there is eactly one zero of G and one zero of + G belonging to the interval ip, ip. If p < < p, then there are no zeros of + G in the interval ip, ip and there are two counted with multiplicity or no zeros of G in this interval depending on whether Gi = L tand/ is or >. Since the poles of G accumulate at and at ±i, the same is true for the zeros of ± G. This proves c. Taking the sum and difference of 4.37 and 4.38 we find a 3 ± b 3 â 3 ˆb 3 = ± G, C. 5.5 It follows that hh = G, C. 5.6

25 Equation 5.5 implies that either a 3 b 3 or â 3 + ˆb 3 vanishes whenever G =. Since all zeros of G are simple, the functions a 3 b 3 and â 3 + ˆb 3 cannot simultaneously vanish at one of these zeros. In order to prove d, it only remains to show that G vanishes whenever a 3 b 3 or â 3 + ˆb 3 does. Equation 5.5 suggests that this is true; however, it is conceivable that a zero pole of a 3 b 3 could coincide with a pole zero of â 3 + ˆb 3 in such a way that the product a 3 b 3 â 3 + ˆb 3 = G remains nonzero. We show now that this cannot occur. Suppose is a zero of a 3 b 3. The unit determinant condition 4.37 implies that a 3 and b 3 cannot simultaneously vanish. Thus, The global relations 4.4 yield a 3 = b 3. a = b, e ωl + a = b. 5.7 Using these equations to eliminate b and b from the determinant condition we find a a b b =, a e ωl a = e ωl. 5.8 If is a zero of a 3 b 3, then, by 4.7, so is /, and hence we also have a e ωl a = e ωl. 5.9 On the other hand, the symmetry condition 4.6 for a implies that a + e ωl a = a + e ωl a, C. 5. Indeed, if we use 4.6 to eliminate a ±/ from the left-hand side of 5. and then simplify, we find the right-hand side of 5.. Evaluating 5. at = and using 5.8 and 5.9 to eliminate a / and a from the resulting equation, we find a = a. 5. In view of 5.7, 5.8, and 5., the symmetry equation 4.6 for a evaluated at = reduces to = e ωl + F e ωl, i.e. G =. This shows that G = whenever a 3 b 3 =. A similar argument shows that G = also whenever a 3 + b 3 =. This proves d. The proof of e is similar. The statement f follows from d and e since h = a 3 b 3 a 3 + b 3. Since h is analytic in C + and h does not vanish at any point P G by f, equation 5.6 implies that h has a double pole at each point in the set P G C +. This proves g. QED 3

26 5. An equivalent Riemann-Hilbert problem Using the relations 4.6 and 4.3a, the jump matrices.6 become and J π/, y, = J, y, = + b3b3 a 3a 3 e ωl b 3 a 3 e θ,y, b 3 a 3 e ωl e θ,y, b a 3 e θ,y,, J 3π/, y, =, b a 3 e θ,y, J π = J 3π/ J J π/, e θ,y, = e Ω+ωy. 5. Let M, y, be defined by.4 with Ψ j replaced with Φ j, j =,, 3. Let D j denote the jth quadrant of the comple plane, { D j = C j π < arg < j π }, j =,..., 4, and let M j denote the restriction of M to D j. The jump matrices 5. involve the unknown spectral functions b, a 3, and b 3. We therefore seek matrices A j, y,, j =,..., 4, defined for D j, such that the functions { M j, y, } 4 by M j, y, = M j, y, A j, y,, D j, j =,..., 4, 5.3 satisfy a modified Riemann-Hilbert problem whose jump matrices involve only known functions. We would like A j to be bounded and analytic or at least meromorphic for D j. The requirement that A j is bounded in the j-th quadrant implies that A and A are upper triangular, while A 3 and A 4 are lower triangular. The requirement that A j has unit determinant implies that the diagonal elements of A j are d j and d j. The, components of equations 5.8 imply that d = d = d 3 = d 4. On the other hand, the four eponential factors e θ,y,, e θ,y, e ωl, e θ,y,, e θ,y, e ωl 5.4 are bounded in the first, second, third and fourth quadrant of the comple -plane, respectively. This suggests choosing the matrices A j in the following form: A = α e θ,y,, D, 5.5a A = α e ωl e θ,y,, D, 5.5b, 4

27 A 3 = A 4 =, D 3, 5.5c α 3 e θ,y,, D d α 4 e ωl e θ,y, where α j, j =,..., 4, is a scalar valued function of D j. Substituting 5.3 into the jump relations we find the equations M 4 = M J, M = M J π/, M 4 = M 3 J 3π/, 5.6 M 4 = M J, M = M J π/, M4 = M 3 J 3π/, 5.7 provided that the matrices J, J π/, J 3π/ satisfy the following equations: J A 4 = A J, J π/ A = A J π/, J 3π/ A 4 = A 3 J 3π/. 5.8 We analyse the first of equations 5.8. The, element of this equation is satisfied identically and the, element is a consequence of the, and, elements, as well as of the requirement that all matrices in 5.8 have unit determinant. Denoting the, and, components of J by e θ,y, Ũ and e θ,y, e ωl Ṽ respectively, we find that the, and, elements of the first of equations 5.8 yield b 3 a 3 α = Ũ 5.9 and b 3 a 3 + α 4 = Ṽ. 5. Comparing equation 5.9 with the identity 5. we find that a simple choice for the function α and hence for Ũ is α = b3 a and Ũ 3h = G h. Note that these functions are well defined on R since h does not have any real zero. However, with these choices the functions α and Ũ have i poles at the unknown zeros of h along the imaginary ais and ii poles at the known poles of G along the imaginary ais. To ensure that the poles in i are removable singularities we define a function G as follows: G = Gh This function takes value and eactly where G does. Indeed, if h =, then a 3 = ±b 3 and correspondingly, in view of 5.5, G = G = ±. The above discussion suggests that a suitable choice for the functions α and Ũ is α = b 3 a 3 h G h, Ũ G G = = G. 5. h 5

28 Similarly, equation 5. suggests α 4 = b 3 a 3 h + G h, Ṽ G G = = G. 5.3 h We net analyse the second of equations 5.8. The, element of this equation yields α e ωl + b a 3 = b 3 a 3 h G h + Ũ π/, where Ũ π/ e θ,y, denotes the, component of J π/. Using the identity 5., we find This suggests that we define α e ωl + e ωl b 3 a 3 h + eωl G h + G h = Ũ π/. α = b 3 a 3 h + G h, Ũ π/ = + e ωl G. 5.4 A similar analysis of the third of equations 5.8 yields α 3 = b 3 a 3 h G h, Ṽ 3π/ = + e ωl G, 5.5 where V 3π/ e θ,y, 3π denotes the, component of J. Note that the relations α = α 4 and α 3 = α are consistent with the symmetry.. We define the matrices A j, j =,..., 4, by equations 5.5 and Henceforth, we assume that, y lies in the interior of the semistrip. so that > and < y < L. Then the jth eponential factor in 5.4 has eponential decay as and for Īj. Thus, although the analysis of the linear problem suggests that the spectral functions a j, b j could have some minor growth as and caused by the jumps in the boundary data at the corners of the semistrip in the linear case this growth is logarithmic, see Appendi A, this ensures that the A j s are bounded as and as in the corresponding domains D j. In fact, since the A j s have removable singularities at the zeros of the function h along the imaginary ais, the only remaining difficulty is that the A j s have singularities at the known poles of G. To deal with these singularities, we add small indentations to the jump contour along the imaginary ais so that it passes to the right of the poles of G. Thus, instead of the four quadrants D j of the comple plane, we consider the deformed domains D j defined in such a way that all P G C + lie in D and all P G C lie in D 3, see Figure 5. We net determine the residue conditions at these poles. Let P G C + be a pole of α in D. In what follows, we use the notation M, y, = [M, y, ], [M, y, ] to denote the first and second column vector of a given matri M, y,. Then the relation M = M A implies that [ M, y, ] = [M, y, ], [ M, y, ] = α e θ,y, [M, y, ] + [M, y, ]. 6

29 J π/ D D J π J D 3 D4 J 3π/ Figure 5: The contour for the modified Riemann-Hilbert problem satisfied by M. The contour has small indentations bypassing the poles of G. Taking the residue of the second of these equations at, we find Res [ M, y, ] = Res α e θ,y, [M, y, ] = Res G + h e θ,y, [ M, y, ], where the residue of G at is known from the definition 4.3 whereas the number h remains unknown. Similarly, the relation M 3 = M 3 A 3 implies that [ M 3, y, ] = [M 3, y, ] + α 3 e θ,y, [M 3, y, ], [ M 3, y, ] = [M 3, y, ]. Taking the residue of the first of these equations at P G C, we find Res [ M 3, y, ] = Res α 3 e θ,y, [M 3, y, ] = Res G + h e θ,y, [ M 3, y, ], In summary, we have derived the following result. Theorem 5. The RH problem defined in theorem 3. and characterized by the jump matrices {J π/, J 3π/, J, J π } defined in equation 5., can be mapped to a new RH problem with the 7

30 following jump matrices: J π/, y, = + eωl Ge θ,y,, J 3π/, y, =, + e ωl Ge θ,y, e ωl Ge θ,y, J, y, = Ge ωl e θ,y,, 5.6 where the known function G is defined in 4.3. This is achieved by using the matrices 5.5, with α = b 3 a 3 h G h, α = b 3 a 3 h + G h, α 3 = α, α 4 = α. 5.7 where the functions h and G are defined by 5.3 and 5. respectively. The solution M of the new RH problem is a sectionally meromorphic function with simple poles at each point in the set P G given in 5.4. At these points the following residue conditions are valid: Res [ M, y, ] = Res G Res [ M, y, ] = Res G + + h e θ,y, [ y, ], P G C +, 5.8 e θ,y, [ M, y, ], P G C. 5.9 h The solution q, y, >, < y < L, of the boundary value problem determined by the boundary conditions.3 is given by q iq y, y = lim M, 5.3 cos q, y = + 4i lim M + lim M. 5.3 Proof: We only need to show how to represent the solution q, y of the boundary value problem in terms of the solution of the RH problem characterised by the jump matrices given by 5.6. Recall that if M, y, denotes the solution of the RH problem defined in theorem 3. then q, y is defined in terms of M by equations Using M j = A j M j in the j-th quadrant, j =,..., 4, by choosing in the first quadrant we obtain q iq y, y = lim M, [ cos q, y = + 4i lim M α e θ,y, M + lim ] M. Since e θ,y, decays eponentially as in the first quadrant, the term involving the unknown coefficient α does not contribute to the limit and we find

31 6 Conclusions and open problems We have analyzed the elliptic sine-gordon equation in a semistrip for general boundary data sections and 3 and in the particular case of a linearisable boundary value problem sections 4 and 5. The linearizable problem has the novelty that the function q y, possesses a nonintegrable singularity as while the function q, y possesses a non-integrable singularity as y. Motivated by the recent solution of an analogous problem for the modified Helmholtz equation presented in [], we have been able to bypass this problem by employing a suitable gauge transformation. Furthermore, we have shown that the RH problem characterizing the solution q, y can be mapped to a modified RH problem whose jump matri is determined only by the width L of the semistrip and the given constant value d of the boundary condition prescribed at = see theorem 5.. However, the modified RH problem also includes residue conditions at the points P G, where the set P G consists of a countable number of points on the imaginary ais. The formulation of these residue conditions requires the knowledge of h for P G, where h is an unknown meromorphic function defined in terms of the spectral functions. It remains an open problem to characterize the values of h for P G in terms of L and d alone; progress in this direction is likely to rely on the analyticity properties of h as well as on relations derived from the symmetry properties of the spectral functions, such as the relation 5.6, and the known structure of the poles of h. Acknowledgements This research was partially supported by EPSRC grant EP/E96/. ASF would like to epress his gratitude to the Guggenheim Foundation, USA. A The modified Helmholtz equation The basic differential form associated with the modified Helmholtz equation is given by W B, y, = e Ω ωy 4 u, y + u, y y u, y =, [ i u iu y iu d u iu y + iu ] dy, where Ω and ω are defined in.4. Indeed, it can be verified that dw B = e Ω ωy 4 [u + u yy u] d dy. Suppose that u y, y has non-integrable singularities at, and, L. In order to eliminate these singularities we consider the differential form W, y, = W B, y, d e Ω ωy κ, y, 9

32 y L Φ 4 Φ Φ 3 Figure 6: The contours of integration used to define Φ j, y,, j =, 3, 4. and choose κ in such a way that the term u y cancels. Noting that W = e Ω ωy 4 we choose κ as in 4.: Then W = e Ω ωy 4 where we have used { [iu + u y + u 4κ + 4Ωκ]d [u iu y + i u } + 4κ y 4ωκ]dy κ, y, = 4 u y ξ, ydξ. {[iu + u Ω u y ξ, ydξ]d + [iu y i u + uξ, ydξ ω κ y = 4 u yy ξ, ydξ = 4 We define Φ j, y,, j =, 3, 4, as the solutions of the equation with uξ, ydξ + u, y. A. } u y ξ, ydξ]dy A. d Φ j e Ω ωy = W, j =, 3, 4, A.3 Φ, y, =, Φ 3,, =, Φ 4, L, =, A.4 see Figure 6. The difference of any two of the above functions equals e Ω+ωy ρ, where ρ can be computed by evaluating the difference at any convenient point, y. Hence Φ 4, y, Φ, y, = e Ω+ωy e ωl B, C +, Φ 4, y, Φ 3, y, = e Ω+ωy B, C, A.5 Φ, y, Φ 3, y, = e Ω+ωy B 3, C +, where B = Φ, L,, B = Φ 4,,, B 3 = Φ,,, A.6 3

33 and the above choice for the domains of validity with respect to in equations A.5 will be justified below. Equations A.3 and the first of equations A.4 imply that Φ, y, = [ e Ω ξ iu ξ + u ] 4 Ω u y ξ, ydξ dξ, C +. A.7 Hence the first and the third of equations A.6 imply that B 3 = [ e Ωξ uξ, iu ξ ξ, + Ω 4 B = 4 [ e Ωξ uξ, L iu ξ ξ, L + Ω ξ ξ ξ u y ξ, dξ ] dξ, C +, A.8 u y ξ, Ldξ ] dξ, C +. A.9 In order to compute Φ 4,, we compute Φ 4 along the y-ais from, L to, : Φ 4,, = [ e ωη u, η ] iu η, η i + uξ, ηdξ ω u η ξ, ηdξ dη. 4 L The last integral of this equation is given by ω L [ ] e ωη η uξ, ηdξ dη = ω [e ωl uξ, Ldξ 4 4 L + ω e ωη uξ, ηdξ dη. 4 ] uξ, dξ A. Hence, using this equation, as well as the identity Ω + ω =, equation A. together with the second of equations A.6 yields B = L [ e ωη iu, η ] iu η, η + Ω uξ, ηdξ dη 4 + ω ] [e ωl uξ, Ldξ uξ, dξ, C. A. 4 Subtracting the second of equations A.5 from the sum of the other two equations in A.5 we find the global relation A. Eample Let e ωl B B 3 + B =, C +. A. u, = u, L =, < < ; u, y = d, < y < L. The epressions in A.8 and A. simplify as follows: B 3 = Ω e Ωξ f 3 ξdξ, f 3 ξ = 4 3 ξ u y ξ, dξ, C +, A.3

34 B = id e ωl + L Ω e ωη f ηdη, f η = 4 uξ, ηdξ, C. A.4 Note that ω±i =, thus the first term of the right hand side of equation A.4 has removable singularities at ±i. The overall symmetry u, y = u, L y implies Thus the global relation A. becomes B = B 3, C +. A.5 B = + e ωl B 3, C +. A.6 The only dependence of B 3 on is through Ω, which remains invariant under the transformation, thus B 3 = B 3, C +. A.7 The second term of the right hand side of the first of equations A.4 involves Ω and ω, which are invariant under the transformation. Thus we find therefore B id e ωl + B = B id = B id e ωl +, + e ωl, C. A.8 In summary, taking into account that B = B 3, it follows that the modified Helmholtz equation in the semistrip, with the boundary conditions.3, involves the two unknown spectral functions B 3 and B, defined in terms of the two unknown functions f and f 3 by equations A.3 and A.4. These two spectral functions satisfy the global relation A.6 as well as the symmetry relations A.7 and A.8. In what follows, we will show that the unknown functions B and B 3 yields a zero contribution to the representation of the solution u, y. In order to prove this fact, we need the following identities, which are a consequence of equations A.6-A.8: B 3 B 3 = G, G = id e ωl +, R, A.9 e ωl + B + e ωl B 3 = G, G = id Indeed, letting in the global relation A.6 we find B = + e ωl B 3, C. + e ωl, C. A. Using in the above equation the symmetry relation A.7 with, as well as the symmetry relation A.8, we find equation A.. Subtracting equation A. from the global relation A.6 we find equation A.9. The functions G, G have removable singularities at = ±i. 3

35 e Ω+ωy B 3 e Ω+ωy B 3 e Ω+ωy B Figure 7: The contour for the Riemann-Hilbert problem. The functions Φ j, y,, j =, 3, 4, define a Riemann-Hilbert problem with jumps on the real and negative imaginary ais, see figure 7. In order to map this Riemann-Hilbert problem to a problem with known jump conditions, we introduce the functions Φ j, y,, j =, 3, 4, through the following equations: Φ = Φ, C +, Φ 3 = Φ 3 e θ,y, B 3, π arg 3π, A. Φ 4 = Φ 4 + e θ,y, e ωl B 3, 3π arg π. It is shown in remark A. at the end of this appendi that the function B 3 has a logarithmic singularity as and. In particular, assuming that, y lies in the interior of the semistrip. so that > and < y < L, it follows that e θ,y, B 3 and e θ,y, e ωl B 3 are bounded and analytic for in the third and fourth quadrant of the plane, respectively. Using the definitions A. in equation A.5, we find Φ 4, y, Φ, y, = e θ,y, e ωl B 3 B 3, R +, A. Φ 4, y, Φ 3, y, = e θ,y, B + e ωl B 3, ir +, Φ, y, Φ 3, y, = e θ,y, B 3 B 3, R. Equations A.9 and A. imply that the jump conditions appearing in A. can be epressed in terms of the known functions G and G. Equation A.3 implies that the function Φ satisfies the equation This equation suggests that Φ ΩΦ = 4 Φ, y, = 4 [iu + u Ω u y ξ, ydξ + O 33 ] u y ξ, ydξ. A.3,. A.4