The hyperbolic Ernst equation in a triangular domain

The collision of two plane gravitational waves in Einstein’s theory of relativity can be described mathematically by a Goursat problem for the hyperbolic Ernst equation in a triangular domain. We use the integrable structure of the Ernst equation to present the solution of this problem via the solution of a Riemann–Hilbert problem. The formulation of the Riemann–Hilbert problem involves only the prescribed boundary data, thus the solution is as effective as the solution of a pure initial value problem via the inverse scattering transform. Our results are valid also for boundary data whose derivatives are unbounded at the triangle’s corners—this level of generality is crucial for the application to colliding gravitational waves. Remarkably, for data with a singular behavior of the form relevant for gravitational waves, it turns out that the singular integral operator underlying the Riemann–Hilbert formalism can be explicitly inverted at the boundary. In this way, we are able to show exactly how the behavior of the given data at the origin transfers into a singular behavior of the solution near the boundary.


Introduction
Half a century after Einstein presented his theory of relativity, Ernst made the remarkable discovery that, in the presence of one space-like and one time-like Killing vector, the entire solution of the vacuum Einstein field equations reduces to solving a single equation for a complex-valued function E of two variables [10]. This single equation, now known as the (elliptic) Ernst equation, has proved instrumental in the study and construction of stationary axisymmetric spacetimes, cf. [21].
It later became clear that a similar reduction of Einstein's equations is possible also in the presence of two space-like Killing vectors, a situation relevant for the description of two colliding plane gravitational waves [4]. In this case the associated equation is known as the hyperbolic Ernst equation and can be written in the form where the Ernst potential E(x, y) is a complex-valued function of the two real variables (x, y) and subscripts denote partial derivatives. The problem of finding the nonlinear interaction of two plane gravitational waves following their collision has a distinguished history going back to the work of Khan and Penrose [19], Szekeres [36], Nutku and Halil [32], and Chandrasekhar and coauthors [4,5]; see the monograph [17] for further references and historical remarks. In terms of the Ernst potential, this collision problem reduces to a Goursat problem for equation (1.1) in the triangular region D defined by (see Fig. 1) (1.2) More precisely, the problem can be formulated as follows (see [17] and the "Appendix"): Given complex-valued functions E 0 (x), x ∈ (1, 1), and E 1 (y), y ∈ [0, 1), find a solution E(x, y) of the hyperbolic Ernst equation (1.1) in D such that E(x, 0) = E 0 (x) for x ∈ [0, 1) and E(0, y) = E 1 (y) for y ∈ [0, 1). (1.3) In this paper, we use the integrable structure of equation (1.1) and Riemann-Hilbert (RH) techniques to analyze the Goursat problem (1.3). We present four main results, denoted by Theorem 1-4: • Theorem 1 is a solution representation result: Assuming that the given data satisfy the following conditions for some n ≥ 2: x α E 0x , y α E 1y ∈ C([0, 1)) for some α ∈ [0, 1), Re E 0 (x) > 0 for x ∈ [0, 1), Re E 1 (y) > 0 for y ∈ [0, 1), (1.4) and the Goursat problem (1.3) has a solution (in the precise sense specified in Definition 3.1), we give a representation formula for this solution. This formula is given in terms of the solution of a corresponding RH problem whose formulation only involves the given boundary data. • Theorem 2 is a uniqueness result: Assuming that the given data satisfy the conditions (1.4) for some n ≥ 2, we show that the solution of the Goursat problem (1.3) is unique, if it exists. • Theorem 3 is an existence and regularity result: Assuming that the given data satisfy the conditions (1.4) for some n ≥ 2, we show that there exists a unique solution E of the problem (1.3) whenever the associated RH problem has a solution, and this E has the same regularity as the given data. In the case of collinearly polarized waves, this yields existence for general data; for noncollinearly polarized waves, a small-norm assumption is also needed. • Theorem 4 provides exact formulas for the singular behavior of the solution E near the boundary for data satisfying (1.4).
We emphasize that the assumptions (1.4) allow for functions E 0 (x) and E 1 (y) whose derivatives blow up as x and y approach the origin. This level of generality is necessary for the application to gravitational waves. Indeed, in order for the problem (1.3) to be relevant in the context of gravitational waves, it turns out that the solution should obey the conditions (see [17] and the "Appendix") lim x↓0 x α |E x (x, y)| = m 1 Re E 1 (y) √ 1 − y for each y ∈ [0, 1), (1.5a) where m 1 and m 2 are real constants such that m 1 , m 2 ∈ [1, √ 2) and α = 1/2. Remarkably, for data with a singular behavior at the origin of the form given in (1.4), the singular integral operator underlying the RH formalism can be explicitly inverted in the limit of small x or y. This leads to the characterization of the boundary behavior given in Theorem 4. In particular, it implies the following important conclusion for the collision of gravitational waves: A solution E(x, y) of the Goursat problem for (1.1) fulfills (1.5) iff the boundary data are such that lim x↓0 x α |E 0x (x)| and lim y↓0 y α |E 1y (y)| lie in the interval [1, √ 2). The assumptions Re E 0 (x) > 0 and Re E 1 (y) > 0 in (1.4) are natural because in the context of gravitational waves the real part of the Ernst potential is automatically strictly positive. The assumption E 0 (0) = E 1 (0) in (1.4) expresses the compatibility of the boundary values at the origin. If E is a solution of (1.1), then so is aE + ib for any choice of the real constants a and b. Thus, since E(0, 0) = 0 as a consequence of the assumption Re E 0 (x) > 0, there is no loss of generality in assuming that E(0, 0) = 1.
The analysis of a boundary or initial-boundary value problem for an integrable equation is usually complicated by the fact that not all boundary values are known for a well-posed problem cf. [13]. This issue does not arise for (1.3) which is a Goursat problem. This means that the presented solution is as effective as the solution of the initial value problem via the inverse scattering transform for an equation such as the KdV or nonlinear Schrödinger equation.
Despite its great importance in the context of gravitational waves, there are few results in the literature on the Goursat problem (1.3). In fact, rather than solving a given initial or boundary value problem, most of the literature on the Ernst equation has dealt with the generation of new exact solutions via solution-generating techniques, cf. [17,21,22]. Solving an initial or boundary value problem is much more difficult than generating particular solutions. In fact, even if a large class of particular solutions are known, the problem of determining which of these solutions satisfies the given initial and boundary conditions remains a highly nonlinear problem, often as difficult as the original problem. As noted by Griffiths [17, p. 210], "What would be much more significant would be to find a practical way to determine the solution in the interaction region for an arbitrary set of initial conditions." Regarding the problem of determining the interaction of two colliding plane waves from arbitrary initial conditions, important first progress was made in a series of papers by Hauser and Ernst, see [18]. Their approach is based on the so-called Kinnersley H -potential [20] rather than on equation (1.1). In terms of the 2 × 2-matrix valued Kinnersley potential H (r , s), the problem of determining the spacetime metric in the interaction region can be formulated as a Goursat problem in the triangular region Hauser and Ernst were able to relate the solution of this problem to the solution of a homogeneous Hilbert problem. The analysis of [18] relies, at least implicitly, on the fact that equation (1.7) admits the Lax pair (see Eq. (3.1) in [18]) where P(r , s, τ ) is a 2 × 2-matrix valued eigenfunction and τ ∈ C is the spectral parameter. The authors of [15] have addressed the Goursat problem in the triangle for the equation 2(s − r )g rs + g r − g s + (r − s)(g r g −1 g s + g s g −1 g r ) = 0, (1.8) where g(r , s) is a 2 × 2-matrix valued function.
Eq. (1.8) reduces to the scalar equation which is related to Eq. (1.1) by the change of variables y = (r +1)/2 and x = (1−s)/2. Through a clever series of steps, the authors of [15] express the solution of (1.8) in terms of the solution of a RH problem.
Our approach here is inspired by the works [23,25,31] on the elliptic Ernst equation. We have also drawn some inspiration from [15] and [18], although in contrast to these references, we analyze equation (1.1). Two further differences between the present work and [15] are: (i) It is assumed in [15] that the solution is C 2 on all of up to and including the non-diagonal part of the boundary. However, as explained above (see equation (1.5)), the Ernst potentials relevant for gravitational waves have boundary values E(x, 0) and E(0, y) whose derivatives are not continuous (actually unbounded) at the origin. Here we allow for such singularities in E x (x, 0) and E y (0, y). These singularities transfer, in general, into singularities of the associated eigenfunction solutions of the Lax pair, and the rigorous treatment of all these singularities was one of the main challenges of the present work. (ii) The normalization condition for the RH problem derived in [15] involves the solution itself; hence the solution representation is not effective. We circumvent this problem by defining the eigenfunctions on a Riemann surface S (x,y) with branch points at x and 1 − y. The Riemann surface S (x,y) is dynamic in the sense that it depends on the spatial point (x, y). This dependence on (x, y) creates some technical difficulties which we handle by introducing a map F (x,y) from S (x,y) to the standard Riemann sphere which takes the two moving branch points to the two fixed points −1 and 1. After transferring the RH problem to the Riemann sphere in this way, we can analyze it using techniques from the theory of singular integral equations.
In the traditional implementation of the inverse scattering transform, the two equations in the Lax pair are treated separately-usually the spatial part of the Lax pair is first used to define the scattering data and the temporal part is then used to determine the time evolution. The Goursat problem (1.3) does not fit this pattern, so a different approach is required; this is one reason why the solution of the problem (1.3) has proved elusive. Actually, the approach in [15] was one of the first implementations of a general framework for the analysis of boundary value problems for integrable PDEs now known as the unified transform or Fokas method [11] (see also [3,7,12,13,34]). In this method the two equations in the Lax pair are analyzed simultaneously rather than separately [14]. The ideas of this method play an important role also in this paper.
It is an interesting open problem to investigate whether existence and uniqueness results for (1.3) can be obtained also via functional analytic techniques. As was explained already in Chapter IV of Goursat's original treatise [16], existence and uniqueness results for Goursat problems for linear hyperbolic PDEs can be established by means of successive approximations and Riemann's method (see also [6]). It is possible to extend these ideas to prove existence theorems also for certain nonlinear Goursat problems [35,38]. However, even in the linear case, these theorems tend to assume that {E, E x , E y , E xy } are all continuous [6,16,35], or at least that the boundary values are Lipschitz [38]. These conditions fail for the assumptions (1.4) relevant for gravitational waves.
We recently became aware of some relatively recent works in the physics literature which also address the initial value problem for two colliding plane gravitational waves [1,2,33]. Compared with these papers, our work is more mathematical in character and there appears to be little overlap.
Let us finally point out that many exact solutions describing colliding plane gravitational waves are known (see e.g. [5,9,32,37]) and that there is a growing literature on colliding gravitational waves which are not necessarily plane (see e.g. [26,27]).

Organization of the paper
We begin by establishing some notation in Sect. 2. Our main results (Theorems 1-4) are stated in Sect. 3.
In Sect. 4, as preparation for the general case, we analyze the special case in which the colliding waves have collinear polarization. In this case, the problem reduces to a problem for the so-called Euler-Darboux equation. We prove a theorem for this equation (Theorem 5) which is analogous to Theorem 1-4.
In Sect. 5, we discuss the Lax pair of equation (1.1) and analyze the spectral data as well as the uniqueness of the solution of the corresponding RH problem.
In Sect. 6, we present the proofs of Theorem 1-4. Section 7 contains two short examples and the "Appendix" contains some background on the origin of the Goursat problem (1.3) in the context of colliding gravitational waves.

Notation
We introduce notation that will be used throughout the paper.
We let D denote the triangular region defined in (1.2) and displayed in Fig. 1. Given δ > 0, we let D δ denote the slightly smaller triangular region obtained by removing a narrow strip along the diagonal of D as follows (see Fig. 2): The interiors of D and D δ will be denoted by int D and int D δ , respectively. The Riemann sphere will be denoted byĈ = C ∪ {∞}.

The Riemann surface S (x,y)
For each (x, y) ∈ D, we let S (x,y) denote the Riemann surface consisting of all points P := (λ, k) ∈ C 2 such that together with two points ∞ + = (1, ∞) and ∞ − = (−1, ∞) at infinity and a branch point x ≡ (∞, x) which make the surface compact. The surface S (x,y) is two-sheeted in the sense that to each k ∈Ĉ\{x, 1− y}, there correspond exactly two values of λ. We 1−λ is a biholomorphism from the two-sheeted Riemann surface S (x,y) to the Riemann sphereĈ = C ∪ {∞}. It maps the branch points x and 1 − y to z = −1 and z = 1, respectively, and the upper (lower) sheet to the outside (inside) of the unit circle introduce a branch cut in the complex k-plane from x to 1− y and, for k ∈Ĉ\[x, 1− y], we let k + and k − denote the corresponding points on the upper and lower sheet of S (x,y) , respectively. By definition, the upper (lower) sheet is characterized by λ → 1 (λ → −1) as k → ∞. Writing λ(x, y, P) for the value of λ corresponding to the point P ∈ S (x,y) , we have where the sign of the square root in (2.3) is chosen so that λ(x, y, k + ) has positive real part.

The map F (x,y)
For each point (x, y) ∈ D, S (x,y) is a compact genus zero Riemann surface with branch points at k = x and k = 1 − y. In order to fix the locations of these branch points, we introduce a new variable z by and let F (x,y) : S (x,y) →Ĉ be the map that sends P to z, i.e., For each (x, y) ∈ D, F (x,y) is a biholomorphism (i.e. a bijective holomorphic function whose inverse is also holomorphic) from S (x,y) toĈ which maps the two branch points x and 1 − y to z = −1 and z = 1, respectively, see Fig. 3.

The contours 6 and 0
For each (x, y) ∈ D, we let 0 ≡ 0 (x, y) denote the shortest path from 0 + to 0 − in S (x,y) , and we let 1 ≡ 1 (x, y) denote the shortest path from 1 − to 1 + in S (x,y) . More precisely, where, for a subset S of the complex plane, we use the notation S ± = {k ± ∈ S (x,y) | k ∈ S} to denote the sets in the upper and lower sheets of S (x,y) which project onto S, see Fig. 4. We write := 0 ∪ 1 for the union of 0 and 1 .
Given (x, y) ∈ D, we let 0 ≡ 0 (x, y) and 1 ≡ 1 (x, y) denote two clockwise nonintersecting smooth contours in the complex z-plane which encircle the real intervals and respectively, but which do not encircle zero, see Fig. 5. We let ≡ (x, y) denote the union := 0 ∪ 1 of 0 and 1 .

Boundary values and function spaces
Let ⊂ C be a piecewise smooth oriented contour. For an analytic function m : C\ → C which extends continuously to from either side, we denote the boundary values of m from the left and right sides of by m + and m − , respectively. Given a subset S ⊂ R n , n ≥ 1, we let C(S) denote the space of complex-valued continuous functions on S. If S is open, we define C n (S) as the space of complex-valued functions Σ0 Σ1 The map F (x,y) sends the contours 0 and 1 onto the two real intervals F (x,y) ( 0 ) and F (x,y) ( 1 ), respectively Re z Im z The contour in the complex z-plane is the union of the loops 0 and 1 which encircle the intervals F (x,y) ( 0 ) and F (x,y) ( 1 ) respectively on S which are n times continuously differentiable, i.e., all partial derivatives of order ≤ n exist and are continuous. By B(X , Y ), we denote the space of bounded linear maps from a Banach space X to another Banach space Y equipped with the standard operator norm; if X = Y , we write B(X ) ≡ B(X , X ).

Main results
We adopt the following notion of a C n -solution of the Goursat problem (1.3).
We next state the four main results of the paper (Theorem 1-4), which all address different aspects of the Goursat problem (1.3).
In the formulation of Theorem 1-4, it is assumed that n ≥ 2 is an integer and that E 0 (x), x ∈ [0, 1), and E 1 (y), y ∈ [0, 1), are two complex-valued functions satisfying the assumptions in (1.4) for a fixed α ∈ [0, 1). The first theorem provides a representation formula for the solution in terms of the given boundary data via a RH problem.
Theorem 1 (Representation formula) If E(x, y) is a C n -solution of the Goursat problem for (1.1) in D with data {E 0 , E 1 }, then this solution can be expressed in terms of the boundary values E 0 (x) and E 1 (y) by 21 1 + (m(x, y, 0)) 11 + (m(x, y, 0)) 21 ,

2)
and the jump matrix v(x, y, z) is defined as follows: Let 0 and 1 be the unique solutions of the linear Volterra integral equations where U 0 and V 1 are defined by Theorem 2 establishes uniqueness of the C n -solution. Theorem 3 establishes existence of a C n -solution-in the collinear case, for general data; otherwise under a small-norm assumption.
Theorem 3 (Existence and regularity) For each δ > 0, the following three existence and regularity results hold: Then there exists a C n -solution of the Goursat problem for

Remark 3.2 Part (a)
of Theorem 3 shows that the solution E(x, y) exists and has the same regularity as the given data as long as the associated RH problem has a solution. By taking δ > 0 arbitrarily small, we see that the same statement holds also in all of D.
Theorem 4 establishes explicit formulas for the singular behavior of the solution near the boundary in terms of the given data.
Theorem 4 (Boundary behavior) Let α ∈ (0, 1) and n ≥ 2 be an integer. Let E(x, y) be a C n -solution of the Goursat problem for (1.1) in D with data {E 0 , E 1 }. Let m 1 , m 2 ∈ C denote the values of these functions at the origin, i.e., Then the solution E(x, y) has the following behavior near the boundary: In particular,

Remark 3.3 Theorem 4 yields
. In particular, the behavior of E x (x, 0) and E y (0, y) at the origin fully determines whether the functions E x (x, y) and E y (x, y) have the appropriate singular behavior near the edges ∂ D ∩ {x = 0} and ∂ D ∩ {y = 0}.

Collinearly polarized waves
Before turning to the general case, it is useful to first consider the special case in which the Ernst potential E is strictly positive. In the context of gravitational waves, this corresponds to the important situation when the two colliding waves have collinear polarization, see [17].

The Euler-Darboux equation
which is a version of the Euler-Darboux equation [29]. Since (4.1) is a linear equation, we can, without loss of generality, assume that V is real-valued and that V (0, 0) = 0.

Remark 4.1 (Linear limit)
In addition to being a reformulation of (1.1) in the special case of collinearly polarized waves, equation (4.1) can also be viewed as the linearized version of (1.1). Indeed, substituting E(x, y) considering the terms of O( ), we see that (4.1) is the linear limit of (1.1).
The analysis of the Euler-Darboux equation (4.1) presented in this section serves two purposes. First, it is used to prove the part of Theorem 3 regarding existence in the collinearly polarized case. Second, it turns out that the more difficult case of noncollinearly polarized solutions can be analyzed following steps which are conceptually very similar to-but technically more difficult than-those involved in the analysis of the collinear case. In fact, the analysis of (1.1) presented in later sections strongly relies on the insight gained in this section.
We are interested in the following Goursat problem for (4.1) in the triangle D: We introduce a notion of C n -solution of this problem as follows. 1), and V 1 (y), y ∈ [0, 1), be real-valued functions and α ∈ [0, 1). We define a function V : D → R to be a C n -solution of the Goursat problem The following theorem establishes the unique existence of a solution of the Goursat problem for (4.1) in D. It also provides a representation for the solution in terms of the boundary data and characterizes the singular behavior near the boundary.

Theorem 5 (Solution of the Euler-Darboux equation in a triangle
Then there exists a unique C n -solution V (x, y) of the Goursat problem for (4.1) in D with data {V 0 , V 1 }. Moreover, this solution is given in terms of the boundary values V 0 (x) and V 1 (y) by where m(x, y, z) is the unique solution of the scalar RH problem Furthermore, if α ∈ (0, 1) is such that the functions x α V 0x and y α V 1y are continuous on [0, 1) and then the solution V (x, y) has the following behavior near the boundary:

Remark 4.3 The scalar RH problem (4.4) has the unique solution
Hence the solution V (x, y) can be expressed in terms of v by Collapsing the contour in (4.9) onto the intervals in (2.5) and changing variables from z to k leads to the following representation for the solution in terms of Abel type integrals: for (x, y) ∈ D. Formulas analogous to (4.10) for equation (4.1) have been derived in [18] and [15].

Remark 4.4
The representation (4.10) can be found more directly by formulating a RH problem for on S (x,y) with jump across . This is essentially the approach adopted in [15]. The representation (4.10) has the advantage that it is explicit in its dependence on V 0 and V 1 , but it has the disadvantage that the integrands are singular at some of the endpoints of the integration intervals. These singularities complicate the verification that V satisfies the appropriate regularity and boundary conditions, especially in the situation relevant for gravitational waves where V 0x and V 1y are singular at the origin. For the nonlinear equation (1.1), this becomes a serious complication. For this reason, we have formulated the RH problems in Theorem 1 and Theorem 5 in terms of the contour (which avoids the problematic endpoints of the intervals in (2.5)) rather than in terms of a contour running along the real axis. However, the representation (4.10) allows for applying more classical techniques. This approach is used in [28] to compute an asymptotic expansion of the solution near the diagonal of D.

Remark 4.5
In [36] there was derived an alternative integral formula for the solution of the Goursat problem for the Euler-Darboux equation by applying Riemann's classical method [6,16]. Whereas the representation (4.10) relies on Abel integrals, the expression of [36] is given in terms of the Legendre function P −1/2 of order −1/2.

Remark 4.6
In order to emphasize the analogy between (1.1) and its linearized version (4.1), we will use the same symbols in this section for the various linearized quantities as we use elsewhere for the corresponding quantities of the nonlinear problem. Many quantities which are matrices in the noncollinear case reduce to scalar quantities in the collinear case. For example, in other sections will denote a 2 × 2-matrix valued eigenfunction, but in this section is a scalar-valued eigenfunction.

Proof of Theorem 5
The proof of Theorem 5 is divided into three parts. In the first part, we prove uniqueness and establish the solution representation formula (4.3). In the second part, we prove existence. In the third part, we consider the boundary behavior.

Proof of uniqueness and of (4.3)
Let V 0 (x), x ∈ [0, 1), and V 1 (y), y ∈ [0, 1) be real-valued functions satisfying (4.2) for some n ≥ 2 and α ∈ [0, 1). Suppose that V (x, y) is a C n -solution of the Goursat problem for (4.1) in D with data {V 0 , V 1 }. We will show that V (x, y) can be expressed in terms of V 0 and V 1 by (4.3). Equation (4.1) admits the Lax pair where (x, y, k) is an eigenfunction, λ = λ(x, y, k) is defined by (2.2), and k is a complex spectral parameter. Indeed, using the relations it is straightforward to check that the compatibility condition xy = yx of (4.11) is equivalent to (4.1).
The occurrence of λ in (4.11) implies that the spectral parameter is naturally considered as an element of the Riemann surface S (x,y) . Thus, we will henceforth view (x, y, ·) as a function defined on S (x,y) and write (x, y, P) for the value of at P = (λ, k) ∈ S (x,y) . We emphasize, however, that the partial derivatives x (x, y, P) and λ x (x, y, P) (resp. y (x, y, P) and λ y (x, y, P)) are still computed with (y, k) (resp. (x, k)) held fixed (and λ allowed to change). The basic idea in what follows is to write (4.11) in the differential form d = W , where W denotes the one-form W = λV x dx + 1 λ V y dy, and then define a solution of (4.11) by Since the one-form W is closed, the integral on the right-hand side is independent of path. However, since W in general is singular on the boundary of D, we need to be more careful when defining . We therefore choose to define using the specific contour which consists of the horizontal segment from (0, 0) to (x, 0) followed by the vertical segment from (x, 0) to (x, y) (see the left half of Fig. 6), that is, we define , the integrals on the right-hand side of (4.12) are welldefined. The next lemma establishes several properties of . (a) can be alternatively expressed using the contour consisting of the vertical segment from (0, 0) to (0, y) followed by the horizontal segment from (0, y) to (x, y) (see the right half of Fig. 6): obeys the symmetries (e) For each (x, y) ∈ D, (x, y, P) extends continuously to an analytic function of Proof Let (x, y) ∈ D. In order to prove (a), we need to show that the expression Hence, we can write (4.14) as Since V is a solution of (4.1), the Lax pair compatibility condition (λV Hence Fubini's theorem implies that the expression in (4.15) vanishes. This proves (a). Moreover, if k ∈Ĉ\[0, 1], then it follows from (4.12) and (4.13) that is a continuous function of (x, y) ∈ D and a C n -function of (x, y) ∈ int D, which proves (b).
The assumption x α V x ∈ C(D) implies that the right-hand side of (4.16) is a continuous function of (x, y) ∈ D. Similarly, we see that y α y (x, y, k + ) and are continuous functions of (x, y) ∈ D. This proves (c).
The symmetries in (d) are a consequence of the symmetries (4.17) and the definition (4.12) of .
To prove (e), we note that λ( This shows that the values of on the upper and lower sheets of S (x,y) fit together across the branch cut; hence extends to an analytic function of P ∈ S (x,y) \ . This proves (e).

Remark 4.9
The point P in (4.21) belongs to S (x,y) whereas the maps (x, 0, ·) and (0, y, ·) are defined on S (x,0) and S (0,y) , respectively. The interpretation of equation (4.21) therefore deserves a comment of clarification: If (x, y) and (x,ỹ) are two points in D and F is a map from S (x,y) to some space X , then F naturally induces a mapF from We sometimes, as in (4.21) (and also in (3.5)), identify these two maps and simply write F forF.
where the values of (0, y, P) and λ(x , y, P) in (4.22) are to be interpreted as in Remark 4.9. Since defines an analytic map U → C for each x ∈ [0, x], the map (4.22) is also analytic for P ∈ U . Since U was arbitrary, this establishes the desired statement for the second map in (4.21); the proof for the first map is similar.
Let 0 , 1 , and ∞ denote the three open components ofĈ\ chosen so that (see Fig. 7) (4.23) Fig. 7 The domains 0 , 1 , and ∞ in the complex z-plane We have showed that if V (x, y) is a C n -solution of the Goursat problem for (4.1) in D with data {V 0 , V 1 }, then V (x, y) can be expressed in terms of V 0 and V 1 by (4.3). This also proves that the solution V is unique if it exists, and completes the first part of the proof.

Proof of existence
The second part of the proof is devoted to proving existence. Let us therefore suppose that V 0 (x), x ∈ [0, 1), and V 1 (y), y ∈ [0, 1) are real-valued functions satisfying (4.2) for some n ≥ 2. We will construct a solution V (x, y) of the associated Goursat problem as follows: Using the given data V 0 and V 1 , we define 0 (x, P) and 1 (x, P) by (4.6). Then we define the jump matrix v by (4.5) and let m(x, y, z) denote the unique solution of the RH problem (4.4). Finally, we show that the function V (x, y) defined in terms of m(x, y, 0) via (4.3) constitutes a C n -solution of the Goursat problem in D with data {V 0 , V 1 }. The proof proceeds through a series of lemmas. (4.25) (c) For each x ∈ [0, 1), 0 (x, P) extends continuously to an analytic function of For each x ∈ (0, 1), 0x (x, P) is an analytic function of P ∈ S (x,0) except for a simple pole (at most) at the branch point k = x. (f) For each x 0 ∈ (0, 1) and each compact subset K ⊂Ĉ\[0, Proof If we note that 0 (x, P) is analytic at the points 1 ± ∈ S (x,0) for each x ∈ [0, 1), the properties (a)-(d) follow immediately by setting y = 0 in Lemma 4.7. Moreover, It remains to prove ( f ). Fix x 0 ∈ (0, 1) and let K be a compact subsetĈ\[0, x 0 ]. The function λ(x, 0, ·) is bounded on S (x,0) except for a simple pole at k = x. Hence, for where the right-hand side tends to zero as x 2 → x 1 because V 0x ∈ L 1 ((0, x 0 )). This shows that the map (4.26) is where ξ lies between x and x + h. As h → 0, the right-hand side goes to zero. Hence (4.26) is differentiable as a map (0, x 0 ) → L ∞ (K ) and the derivative satisfies 0k (x, k)).
The map 1 on (0, 1). Hence the map This proves ( f ) and completes the proof of the lemma.
In the same way that we constructed the eigenfunction 0 (x, k) of the x-part, we can construct an eigenfunction 1 (y, k) of the y-part.
Recall from the definition in Sect. 2 that the contour ≡ (x, y) consists of two nonintersecting clockwise loops 0 and 1 which encircle the intervals F (x,y) ( 0 ) and F (x,y) ( 1 ) respectively, but which do not encircle the origin. We are free to choose 0 and 1 as long as these requirements are met. It turns out to be convenient to choose 0 and 1 independent of (x, y). However, we see from (2.5) that the intervals F (x,y) ( 0 ) and F (x,y) ( 1 ) get arbitrarily close to the origin as (x, y) approaches the diagonal edge x + y = 1 of D. Hence we cannot take independent of (x, y) for all (x, y) ∈ D. However, if we restrict ourselves to points (x, y) which lie in the slightly smaller triangle D δ , δ > 0, defined in (2.1), then we can choose independent of (x, y).
Thus, fix δ ∈ (0, 1) and choose > 0 so small that F (x,y) ( 0 ) and F ( where v(x, y, z) is given by (4.5). We will show that Since δ > 0 can be chosen arbitrarily small, this will complete the proof of the theorem.
Consider the family of scalar RH problems given in (4.4) parametrized by the two parameters (x, y) ∈ D δ . For each (x, y) ∈ D δ , the unique solution of (4.4) is given by Lemma 4. 13 The map (x, y) → v(x, y, ·) is continuous from D δ to L ∞ ( ) and C n from int D δ to L ∞ ( ). Moreover, the three maps are continuous from D δ to L ∞ ( ).
Proof The map (x, y) → v(x, y, ·) is continuous from D δ to L ∞ ( ) and C n from int D δ to L ∞ ( ) as a consequence of part ( f ) of Lemmas 4.11 and 4.12. Furthermore, The other two maps in (4.32) are treated in a similar way. These symmetries imply that m(x, y, 0) − m(x, y, z −1 ) and m(x, y,z) satisfy the same RH problem as m(x, y, z). Hence, by uniqueness, (4.33) holds. This proves (a). For each z ∈Ĉ\ , the map Hence properties (b) and (c) follow immediately from (4.31) and Lemma 4.13.
Given a contour γ ⊂ C, we use the notation N (γ ) to denote an open tubular neighborhood of γ . We extend the definition (4.5) of v to a tubular neighborhood N ( ) = N ( 0 ) ∪ N ( 1 ) of as follows, see Fig. 9: v(x, y, z(x, y, P)) = 0 (x, P), We define functions f 0 (x, y, z) and f 1 (x, y, z) for (x, y) ∈ D δ by Moreover, we let n 0 (x, y, z) and n 1 (x, y, z) denote the functions given by and is analytic for z ∈Ĉ\{−1, ∞} with simple poles at z = −1 and z = ∞. Equation Differentiating (4.36) with respect to x and y and evaluating the resulting equations and v x (x, y, z) + z x x, y, (4.41) Using the first equations in (4.40) and (4.41) in (4.39), we conclude that f 0 is analytic across 1 and has the following jump across 0 : Consequently, n 0 is analytic across . Furthermore, by Lemma 4.11, 0x (x, F −1 (x,y) (z)) is analytic for z ∈Ĉ\{−1} with at most a simple pole at z = −1. It follows that n 0 satisfies (a). The proof of (b) is similar and relies on the second equations in (4.40) and (4.41).

Equation (4.24) suggests that we define a function (x, y, P) for (x, y) ∈ D δ and
Proof The analyticity structure of n 0 established in Lemma 4.15 implies that there exists a function C(x, y) independent of z such that We determine C(x, y) by evaluating (4.46) at z = 0. By Lemma 4.15 (d), this gives C(x, y) = −2V x (x, y). It follows that Note that we did not exclude that n 0 is free of singularities. In this case we have C = −2V x = 0 by Lemma 4.15.
Differentiating (4.44) with respect to x and using (4.43) and (4.47), we find, for  Proof The function V (x, y) = − 1 2 m(x, y, 0) is real-valued by (4.33). Moreover, by part (b) of Lemma 4.14, the map (x, y) → m(x, y, 0) is continuous from D δ to C and is C n from int D δ to C.
By Lemma 4.16, it satisfies the Lax pair equations (4.45). Since n ≥ 2, it follows that 0 = xy (x, y, P) − yx (x, y, P) is similar. By definitions (4.29) and (4.5) of V and v, we have Lemma 4.11, so using Cauchy's formula to compute the contributions from z = 0 and z = ∞, we find This completes the proof of the lemma. Since δ > 0 was arbitrary, it also completes the proof of existence.
, we find where the square roots have positive (negative) real part for |z| > 1 (|z| < 1). Thus where the square root has a branch cut along the interval [ √ y ] and the branch is fixed so that the root has positive real part for z < 0. Hence This proves (4.8a); the proof of (4.8b) is similar. Thus the proof of Theorem 5 is complete.

Lax pair and eigenfunctions
In this section we introduce a Lax pair for (1.1) and define appropriate eigenfunctions in preparation for the proofs of Theorems 1-4.

Lax pair
The hyperbolic Ernst equation (1.1) admits the Lax pair where k is the spectral parameter, the function (x, y, k) is a 2 × 2-matrix valued eigenfunction, and the 2 × 2-matrix valued functions U(x, y, k) and V(x, y, k) are defined as follows: with λ given by (2.2). This Lax pair is easily obtained from the Lax pair for the elliptic Ernst equation [30,31] by making small modifications (the same Lax pair is used in [33]).
We write (5.1) in terms of differential forms as d = W , (5.2) where W is the closed one-form As in Sect. 4, we will view the map (x, y, ·) as being defined on the Riemann surface S (x,y) and write (x, y, P) for the value of at P = (λ, k) ∈ S (x,y) .

Lemma 5.1 (Solution of the x-part) The eigenfunction 0 (x, P) defined via the Volterra integral equation (3.3) has the following properties:
For each x ∈ [0, 1), 0 (x, P) extends continuously to an analytic function of P ∈ S (x,0) \ 0 . (d) The value of 0 at P = ∞ + is given by The determinant of 0 is given by is a continuous map [0, x 0 ) → L ∞ (K ) and a C n -map (0,

Remark 5.2
Some of the properties listed in Lemma 5.1 can be found in [33].
Proof We first use successive approximations to show that the integral equation The function λ(x, 0, k + ) is analytic for k ∈Ĉ\[x, 1]; in particular, it is a bounded function of k ∈ K for each fixed x ∈ [0, 1). In view of the assumptions (1.4), this implies where the function C(x) is bounded on each compact subset of [0, 1). Thus Hence the series converges absolutely and uniformly for k ∈ K and x in compact subsets of [0, 1) to a continuous solution 0 (x, k + ) of (5.8). The fact that x → 0 (x, k + ) ∈ C n ((0, 1)) follows from differentiating x → ( j) 0 (x, k + ) and applying estimates similar to (5.10) to the derivative. Differentiating (with respect to k) under the integral sign in (5.9), we see that k → ( j) 0 (x, k + ) is analytic on int K for each j; the uniform convergence then proves that k → 0 (x, k + ) is analytic on int K . A similar argument applies to the integral equation defining 0 (x, k − ). We conclude that the functions 0 (x, k + ) and 0 (x, k − ) are well-defined for x ∈ [0, 1) and k ∈Ĉ\[0, 1] and are analytic functions of k ∈Ĉ\[0, 1] for each fixed x.
We next show uniqueness. Assume that˜ 0 is another solution of the Volterra equation (5.8) such that x → 0 (x, k ± ) is continuous on [0, 1), respectively, and let = 0 −˜ 0 . Then is a solution of the homogeneous equation Iterating this yields Hence, as in the proof of existence, we get the estimate which yields = 0. This proves (a).
The symmetries (4.17) of λ show that Hence σ 3 0 (x, k − )σ 3 and σ 1 0 (x,k + )σ 1 satisfy the same Volterra equation as 0 (x, k + ). By uniqueness, all three functions must be equal. This proves (b). We next show that 0 (x, k ± ) can be continuously extended across the branch cut to an analytic function on S (x,0) \ 0 . Since U 0 (x, k ± ) has continuous boundary values on the interval (x, 1), the above argument (applied with a K that reaches up to the boundary) shows that 0 (x, k ± ) also has continuous boundary values on (x, 1). Moreover, since the boundary functions (x, 0, (k + i0) + ) and (x, 0, (k − i0) − ) satisfy the same integral equation, so by uniqueness they are equal: Hence the values of 0 on the upper and lower sheets of S (x,0) fit together across the branch cut (x, 1), showing that 0 extends to an analytic function of P ∈ S (x,0) \ 0 ∪ {1} . But λ(x, 0, P) is bounded in a neighborhood of the branch point 1, hence the possible singularity of 0 (x, P) at this point must be removable. This shows that 0 satisfies (c).
Since λ(x, y, ∞ + ) = 1, 0 (x, ∞ + ) satisfies the equation This equation has the two linearly independent solutions Hence there exists a constant matrix A such that We determine A by evaluating this equation at x = 0 and using that E 0 (0) = 1 and 0 (0, ∞ + ) = I . This yields (5.5) and proves (d).
The proof of (e) relies on the general identity where B = B(x) is a differentiable matrix-valued function taking values in G L(n, C). We find This relation is valid at least for small x because 0 (0, k ± ) = I is invertible. In fact, since Re E 0 (x) > 0 for x ∈ [0, 1) by assumption (1.4), it extends to all of [0, 1) and we infer that, for P ∈ S (x,0) \ 0 , where C(P) ∈ C is independent of x. Evaluation at x = 0 gives C(P) = 1. This proves (e). It remains to prove ( f ). Fix x 0 ∈ (0, 1) and let K be a compact subset ofĈ\[0, x 0 ]. The function λ(x, 0, ·) is bounded on S (x,0) except for a simple pole at k = x. Hence, where the right-hand side tends to zero as where C(x 0 ) is chosen as in the proof of (a). This shows that the map (5.7) is continuous where ξ lies between x and x + h. As h → 0, the right-hand side goes to zero. Hence (5.7) is differentiable as a map (0, x 0 ) → L ∞ (K ) and the derivative satisfies is C ∞ from (0, x 0 ) to L ∞ (K ) and E 0 is C n on (0, 1). Hence the map This proves ( f ) and completes the proof of the lemma.
(d) The value of 1 at P = ∞ + is given by The determinant of 1 is given by 14) (f) For each y 0 ∈ (0, 1) and each compact subset K ⊂Ĉ\[1 − y 0 , 1], Proof The proof is similar to that of Lemma 5.1.

Uniqueness
The where the two functions c 0 (x) > 0 and c 1 (y) > 0 are independent of z. The function However, since W in general is singular on the boundary of D, we need to be more careful with the definition. We therefore instead define as the solution of (x, y, k + ) = 0 (x, k + ) + are continuous on D.  The same type of successive approximation argument already used in the proof of Lemma 5.1 shows that the Volterra equation (6.6) has a unique solution for each fixed x ∈ (0, 1) and each k ∈Ĉ\[0, 1], and that this solution (x, y, P) extends continuously to an analytic function of P ∈ S (x,y) \ . This proves (b). In order to prove (a), it remains to deduce the alternative representation (6.2). Note that y = V by definition and Since E is a solution of the Goursat problem, we have and, moreover, x (x, 0, k + ) = U (x, 0, k + ). Now a straightforward calculation shows Thus the function˜ = x −U is the unique solution of the Volterra integral equatioñ (x, y, k + ) = y 0 V˜ (x, y , k + )dy giving˜ = 0. This implies x = U . Consequently, , defined by (6.1), is an eigenfunction for the Lax pair equations (5.1). The difference between (6.1) and (6.2) is given by and (V ) x = (U ) y is the compatibility condition for the Lax pair. Hence the two representations (6.1) and (6.2) are equal. This proves (a).
Proof Let U be an open set in S (x,y) \ 1 . Multiplying (6.6) by (x, 0, P) −1 from the right, we find where the values of (x, 0, P) and λ(x, y , P) in (6.8) are to be interpreted as in Remark 4.9. Since is an analytic map U → C for each y , so is V(x, y , ·). It follows that the solution (x, y, P) (x, 0, P) −1 of (6.8) also is analytic for P ∈ U . This establishes the desired statement for the first map in (6.7); the proof for the second map is similar.
We have showed that if E(x, y) is a C n -solution of the Goursat problem for (1.1) in D with data {E 0 , E 1 }, then E(x, y) can be expressed in terms of the function m defined in (6.9) via equation (3.1). By Lemma 5.4, this function m(x, y, z) is the unique solution of the RH-problem (3.2) whose formulation involves only the values E 0 (x ) and E 1 (y ) for 0 ≤ x ≤ x and 0 ≤ y ≤ y. As a consequence, the value of the solution E at (x, y) is uniquely determined by the values E 0 (x ) and E 1 (y ) for 0 ≤ x ≤ x and 0 ≤ y ≤ y, if it exists. This completes the proofs of Theorem 1 and 2.
(6.11) We next list some facts about Cauchy integrals that we will use throughout the proof. If h ∈ L 2 ( ), then the Cauchy transform Ch is defined by We denote the nontangential boundary values of C f from the left and right sides of by C + f and C − f respectively. Then C + and C − are bounded operators on L 2 ( ) and C + − C − = I . Let w(x, y, z) = v(x, y, z) − I . We define the operator C w : where C = C − B(L 2 ( )) . We henceforth assume that the RH problem (3.2) has a solution for all (x, y) ∈ D δ or, equivalently, that The theory of singular integral equations then implies that the solution of the RH problem (3.2) is given by (see e.g. [8] or [24, Proposition 5.8]) m = I + C(μw), (6.15) where the 2 × 2-matrix valued function μ(x, y, ·) is defined by Equation ( is continuous from D δ to L ∞ ( ) and C n from int D δ to L ∞ ( ). Moreover, the three maps (x, y) → x α w x (x, y, ·), (x, y) → y α w x (x, y, ·), (x, y) → x α y α w xy (x, y, ·), (6.18) are continuous from D δ to L ∞ ( ).
Proof For N ≥ 0, let C N (K ) denote the Banach space of functions on K with continuous partial derivatives of order ≤ N equipped with the usual norm By part ( f ) of Lemma 5.1 the map is continuous for any compact set K not intersecting . Moreover, assuming as (x , y ) → (x, y) by uniform continuity of f ∈ C(K ) on the compact set K . It follows that the composed map also is continuous. A similar argument shows that is continuous. Recalling the definition (3.5) of v, this shows that the map (6.17) is continuous from D δ to L ∞ ( ). If a sequence of holomorphic functions f n converges uniformly on an open set then the sequence of derivatives f n converges uniformly on compact subsets of . Fix N ≥ n and let K be a compact subset of S (x,0) \ 0 . Then part ( f ) of Lemma 5.1 implies that the map is C n . On the other hand, the map is C n . Indeed, the map is continuous because as (x , y ) → (x, y) by uniform continuity of f on the compact set K . Moreover, the map has a continuous partial derivative with respect to x because as h → 0 by the mean-value theorem and the uniform continuity of the first partial derivatives of f . Similar arguments show that all partial derivatives of order ≤ n exist and are continuous. We conclude that the composed map built from (6.21) and (6.22) is C n . A similar argument shows that is C n . Recalling the definition (3.5) of v, this shows that the map (6.17) is C n as a map from int D δ to L ∞ ( ). If z ∈ 0 , we have where d dx F −1 (x,y) (z) denotes the derivative of the k-projection of F −1 (x,y) (z), which is given by Thus part ( f ) of Lemma 5.1 and of Lemma 5.3 imply that (x, y) → x α w x (x, y, ·) is a continuous map D δ → L ∞ ( ). The maps (x, y) → y α w y (x, y, ·) and (x, y) → x α y α w xy (x, y, ·) can be treated similarly.
Proof In view of the definition of μ, the map (6.23) is given by , y, ·)).
We note that the map is smooth by the estimate and that the linear map is bounded. Since (6.23) can be viewed as a composition of maps of the form (6.17), (6.25), and (6.26) together with the smooth inversion map I − C w → (I − C w ) −1 , it follows that (6.23) is continuous D δ → L 2 ( ) and C n from int D δ to L 2 ( ). Similarly, (x, y) → x α μ x (x, y, ·) can be viewed as composition of the continuous maps (6.25), (6.26), I − C w → (I − C w ) −1 , (6.17), and (6.24), and is hence continuous. The maps (x, y) → y α μ y (x, y, ·) and (x, y) → x α y α μ xy (x, y, ·) can be treated analogously.

Lemma 6.8
The function E(x, y) defined in (6.40) has the following properties: Re E(x, y) > 0 for (x, y) ∈ D δ .
The (21)-entry of this equation reads It follows that E(x, y) satisfies (1.1) for (x, y) ∈ int D δ . This completes the proof of the lemma. Simple algebra now shows that (I − C w ) =˜ . where f (y) is a function of y. Since λ(0, y, 0) = ∞, we can determine f (y) by solving the equation 1y (y, 0) = 1 2Re E 1 (y) E 1y (y) 0 0 E 1y (y) 1 (y, 0), which is a consequence of (5.1). This gives the desired statement.
The following lemma completes the proof of Theorem 4. where the derivation of (6.68) employs Lemma 6.12 and Lemma 6.13 as well as the residue calculation Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.