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The Boundary Value Problem for a Static 2D Klein–Gordon Equation in the Infinite Strip and in the Half-Plane∗

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The Boundary Value Problem for a Static 2D Klein–Gordon Equation in the Infinite Strip and in the Half-Plane∗ The Boundary Value Problem for a Static 2D Klein{Gordon Equation in the Infinite Strip and in the Half-Plane∗ Dmitry Muraveyy International Laboratory of Quantitative Finance, National Research University Higher School of Economics, Moscow, Russia October 3, 2018 Abstract We provide explicit formulas for the Green function of an elliptic PDE in the infinite strip and the half-plane. They are expressed in elementary and special functions. Proofs of uniqueness and existence are also given. 1 Motivation This research is motivated by the multi-asset American option pricing problem(see e.g. [2]). Us- ing the standard arbitrage theory framework it can be shown that the option prices are bounded by solutions of elliptic PDE problems. These problems have specific features: the domains are unbounded and the boundary functions are not smooth and do not vanish at infinity. We reduce the original PDE determining the price to a static Klein{Gordon equation (SKGE) in an infinite strip (for the alternative dual options) and in the half plane (the exchange and basket options). The boundary conditions of the corresponding problems are bounded H¨olderfunctions. This leads to the following questions: i. Can we explicitly solve the boundary problems for the SKGE? ii. If this is the case, can we construct computer- friendly representations? iii. Is the obtained solution classical? iv. Is it unique? arXiv:1504.02428v1 [math.AP] 8 Apr 2015 To our knowledge, these aspects of the mentioned problems have not been studied in the liter- ature. Even the answers to the two last questions seem not be obvious. The majority of the known results deal with bounded domains and one can not directly apply available theorems to the equations in the infinite strip and in the half-plane. Also, if the boundary conditions are not twice differentiable, then the smoothness of the solution is not clear. In this paper we give answers to all four questions. Our results are as follows: i. The boundary value problem in the infinite strip and in the half-plane allows a closed form solution. ii. Solutions can be represented in terms of elementary functions or in terms of special func- tions. ∗I am grateful to Prof. Yu.M. Kabanov and to Dr. M. Zhitlukhin for important discussions and help [email protected] 1 iii. For all bounded boundary functions with H¨olderproperty the solution is classical. iv. The solution is unique. We could find in the literature only a few related works. The paper [4] contains closed form formulas of boundary problem for the Laplace equation in the infinite strip. In the recent paper [6] there is a variety of Green functions for the homogeneous Poisson problems for SKGE in some unbounded domains, including the strip and the half-plane. The paper is organized as follows. In Section 2 we obtain explicit formulas for the Green function in the case of infinite strip and formulate the main theorem on existence, uniqueness and closed form of the solution. In Section 3 we address the case of half-plane domain. All proofs are given in Section 4. 2 Problem in the infinite strip 2.1 Existence and uniqueness theorem π 2 2 Let Π = R × [0; π] = (x; y) 2 R ; y 2 [0; π] . Let LV = ∆V − r V be an operator acting on the twice differentiable functions V = V (x; y) defined in the interior of Ππ. Here ∆ is the Laplacian and r 2 R. We consider the boundary value problem 8 π < LV (x; y) = 0; (x; y) 2 int Π ; V (x; 0) = '(x); x 2 R; (2.1) : V (x; π) = 0; x 2 R: The following theorem claims that under certain assumptions the problem (2.1) admits a unique classical solution and provides an explicit form for it. Theorem 2.1. Let Hλ be the space of H¨olderfunctions of order λ > 0 and let ' be a bounded function from Hλ.Then the solution of the problem (2.1) exists in the classical sense, is unique, and allows the representation Z V (x; y) = '(u)Gπ(x − u; y)du; (2.2) R with the Green function 1 p 1 X k sin ky 2 2 Gπ(x; y) = δ(x)Θ(−y) + p e−|xj k +r : (2.3) π 2 2 k=1 k + r where δ(x) is the Dirac delta-function and Θ(x) = I(0;1)(x). π Corollary 2.1. The Green function G∆ for the Laplace equation ∆V = 0 (i.e. for r = 0) has the representation (see [4]): 1 sin y Gπ (x; y) = δ(x)Θ(−y) + : (2.4) ∆ 2π cosh x − cos y Theorem 2.2. The Green function Gπ given by (2.3) has the integral representation p Z 1 2 π jxj J0(jxjr t − 1) sinh jxjt G (x; y) = δ(x)Θ(−y) + sin y 2 dt (2.5) 2π 1 (cosh xt − cos y) where J0(z) is the Bessel function of zero order. 2 π Corollary 2.2. The Green function (2.3) can be represented in terms of the Green function G∆ as follows: p Z 1 J (r t2 − x2)t Gπ(x; y) = Gπ (x; y) − r Gπ (t; y) 1 p dt; (2.6) ∆ ∆ 2 2 jxj t − x where J1(z) is the Bessel function of first order. Remark 2.1. The problem LV = 0 with boundary conditions V (x; 0) = 0, V (x; π) =' ~(x) can be reduced to the problem (2.1 by the substitutiony ~ = π − y. Using the linearity of L we obtain the solution of the problem LV = 0 with the boundary conditions V (x; 0) = '(x) and V (x; π) =' ~(x) as the sum of solutions of the problems in which one of the boundary conditions is a function equal to zero. 2.2 Green function: construction In this subsection we derive the formulas (2.3) for the Green function. To this aim we introduce the interaction potential P π(x; y; u) defined as the limit π π P (x; y; u) = lim P" (x; y; u); "!0 π The function P" (x; y; u) is the solution in the distribution sense (see, e.g. [7]) of the boundary value problem 8 π LP" = 0; < π −"(x−u) P" jy=0 = e Θ(x − u); (2.7) π : P" jy=π = 0: We define the Green function as the partial derivative in x of P π: @P π(x; y; u) G(x − u; y) = : (2.8) @x π Let us consider the Fourier transform in x of the potential P" : Z iξx π v"(y; ξ; u) = e P" (x; y; u)dx; R It solves, as a function of y, the two-point problem 8 d2v > " − (ξ2 + r2)v = 0; <> dy2 " iξu (2.9) v (0; ξ; u) = e ; > " "−iξ :> v"(π; ξ; u) = 0: The solution has the form eiξu sinh (π − y)pr2 + ξ2 v"(y; ξ; u) = : (−iξ + ") sinh πpr2 + ξ2 π Making the inverse transform, we obtain the explicit formula for the potential P" : p Z eiξ(u−x) sinh (π − y) r2 + ξ2 π 1 P" (x; y; u) = dξ: (2.10) 2π p 2 2 R (−iξ + ") sinh π r + ξ The integrand is an analytical function in the whole complex plane except zeros of functions p ξ + i" = 0; sinh π ξ2 + r2 = 0; 3 Figure 1: Integration loop for the strip case that is except the points ± p 2 2 ξ = −i"; ξk = ± k + r : Integrating along the loop depicted in Figure 1 and applying the Jordan lemma we obtain the infinite sum representation of (2.10) 1 π X + P" (x; y; u) = iΘ(x − u) Res Φ"(ξk ; x; y) k=1 1 ! X − +iΘ(u − x) −Res Φ"(−i"; x; y) − Res Φ"(ξk ; x; y) : k=1 Here Φ"(ξ; x; y) is integrand function in (2.10). Computing the residuals, we get p ik sin(ky)e∓(x−u) r2+k2 sinh ((π − y)r) Res Φ (ξ±; x; y) = ∓ ; Res Φ (−i"; x; y) = : " k π(r2 + k2) " sinh(πr) Letting " ! 0 and calculating the partial derivative @P π=@x, we can obtain the result. 2.3 The general elliptic operator Now we extend our formula for the strip Πl of width l and more general operator 8 2 2 2 2 @ V @ V 2 @ V @V @V 2 l > σ + 2ρσ1σ2 + σ + α1 + α2 − r V = 0; (x; y) 2 int Π ; < 1 @x2 @x@y 2 @y2 @x @y V jy=0 = '(x); x 2 R; > : V jy=π = 0; x 2 R; 4 where the coefficient ρ satisfies the condition jρj < 1. The solution of the problem (2.3) also allows the representation V = ' ∗ G^l where the Green function G^l is given as follows: 1 2 α2y −R(k)js(x;y)j πσ2 − 2 +βs(x;y) X πy e G^l(x; y) = δ(s(x; y))Θ(−y) + e 2σ2 k sin ; (1 − ρ2)σ2l2 l R(k) 1 k=0 where ρσ1 1 ρα2 α1 s(x; y) = y − x; β = 2 − + ; σ2 2(1 − ρ )σ1 σ2 σ1 s 2 2 2 2 2 2 1 α1 ρα1α2 α2 2 2 4(1 − ρ )σ2π k R(k) = 2 2 − 2 + 2 + 4(1 − ρ )r + 2 : 2(1 − ρ )σ1 σ1 σ1σ2 σ2 l The proof of uniqueness and existence of the solution and the construction of the Green function is a straightforward extension of arguments given in Section 4 for the Laplace operator.
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