In the following paragraph of this paper, we address the elliptic two-dimensional static Klein-Gordon equation (SKGE),

$$ \nabla^{2}u(P) - k^{2}u(P) = 0, $$

(6)

where \(\nabla^{2}\) is the Laplace operator in the coordinates of the point *P* and *k* is a real constant parameter. The homogeneous boundary value problem

$$ M_{i}u(p) \equiv \alpha_{i}(P)\frac{\partial u(P)}{\partial n_{i}} + \beta_{i}(P)u(P) = 0,\quad P \in \Gamma_{i} $$

(7)

is set up for the nonhomogeneous equation

$$ \nabla^{2}u(P) - k^{2}u(P) = - f(P),\quad P \in \Omega, $$

(8)

where Ω represents a connected region in two-dimensional Euclidean space. In this paper, we focus on dealing with the SKGE on a rectangular region, whose boundary is \(\Gamma = \Gamma_{1} \cup \Gamma _{2} \cup \Gamma_{3} \cup \Gamma_{4}\), with

$$\begin{aligned}& \Gamma_{1} = \bigl\{ (x,y)\vert 0 \le x \le a,y = 0 \bigr\} , \\& \Gamma_{2} = \bigl\{ (x,y)\vert 0 \le y \le b,x = a \bigr\} , \\& \Gamma_{3} = \bigl\{ (x,y)\vert 0 \le x \le a,y = b \bigr\} , \end{aligned}$$

and

$$\Gamma_{4} = \bigl\{ (x,y)\vert 0 \le y \le b,x = 0 \bigr\} . $$

It is well known that the solution to the boundary value problem, (7) and (8), can be expressed as follows:

$$ u ( P ) = \iint_{\Omega } G ( P,Q ) f ( Q )\,d\Omega ( Q ). $$

(9)

In the above representation, the kernel \(G(P,Q)\) is said to be the Green’s function for the homogeneous problem corresponding to that of (7) and (8). *P* and *Q* in (9) are referred to as the field point and the source point, respectively.

For any source point \(Q \in \Omega \), the corresponding Green’s function, as a function of the coordinates of the field point *P*, can be expressed as [12, 15, 16]

$$ G ( P,Q ) = \frac{1}{2\pi } K_{0} \bigl( k\vert P - Q \vert \bigr) + R ( P,Q ), $$

(10)

where \(K_{0} ( k\vert P - Q \vert ) \) is the modified cylindrical Bessel function of the second kind of order zero, and \(R ( P,Q ) \) satisfies the homogeneous SKGE everywhere on Ω, regardless of the relative locations of *P* and *Q*. The singular component \(\frac{1}{2\pi } K_{0} ( k \vert P - Q \vert ) \) is the response at a field point *P* to a unit source placed at an arbitrary point *Q*. In the method of images, the regular component \(R ( P,Q ) \) of \(G(P,Q)\) is expressed as a sum of a number of unit sources and sinks placed at points \(Q_{1}^{*},Q_{2}^{*}, \ldots,Q_{m}^{*}\) outside the region. It yields the regular component

$$ R ( P,Q ) = \sum_{j = 1}^{m} \mp K_{0} \bigl( k \bigl\vert P - Q _{j}^{*} \bigr\vert \bigr) $$

(11)

of \(G(P,Q)\), a function satisfying the homogeneous SKGE at any point *P* in Ω.

As in paper [12], it is easy to construct Green’s function for the SKGE on the upper half-plane \(\Omega = \{ ( x,y) \mathpunct{\vert} - \infty < x < \infty,y > 0 \} \). In fact, the influence of the unit source at a point \(Q(\xi,\eta) \in \gamma \) represents the singular component

$$ \frac{1}{2\pi } K_{0} \bigl( k \sqrt{(x - \xi)^{2} + (y - \eta)^{2}} \bigr) $$

(12)

of the Green’s function. It can be compensated with a single unit sink placed at point \(Q^{*}(\xi, - \eta)\) located in the lower half-plane and symmetric to \(Q(\xi,\eta)\) about the boundary \(y=0\). The Green’s function for the upper half-plane is found to be

$$ G ( x,y;\xi,\eta) = \frac{1}{2\pi } \bigl[ K_{0} \bigl( k \vert z - \zeta \vert \bigr) - K_{0} \bigl( k\vert z - \bar{ \varsigma } \vert \bigr) \bigr] . $$

(13)

The Green’s function of the Dirichlet problem for the SKGE on the infinite strip \(\Omega = \{ ( x,y) \mathpunct{\vert} - \infty \le x \le \infty,0 \le y \le b \} \) is presented as

$$\begin{aligned} \tilde{G}_{0}^{ +} ( x,y;\xi,\eta) =& \frac{1}{2\pi } \sum_{n = - \infty }^{\infty } \bigl[ K_{0} \bigl( k\sqrt{(x - \xi)^{2} + (y - \eta + 2nb)^{2}} \bigr) \\ &{} - K_{0} \bigl( k\sqrt{(x - \xi)^{2} + (y + \eta - 2nb)^{2}} \bigr) \bigr]. \end{aligned}$$

(14)

To derive the Green’s function stated on the rectangular region \(\Omega = \{ ( x,y) \vert 0 \le x \le a, 0 \le y \le b \} \), we similarly use the method of images in the following paragraph.

Obviously, \(\tilde{G}_{0}^{ +} ( x,y;\xi,\eta) \) satisfies the Dirichlet conditions on the boundary fragment \(y=0\) and \(y=b\). However, it conflicts with the Dirichlet conditions on the boundary fragments \(x=0\) and \(x=a\). To construct the Green’s function satisfying all boundary conditions, we consider the compensation that results by introducing new sinks.

In fact, if we place the unit sources at the point set (Figure 1)

$$\begin{aligned} \tilde{S}_{0}^{ +} =& \bigl\{ ( \xi, - \eta), ( \xi,2b - \eta), ( \xi, - 2b + \eta), ( \xi,2b + \eta), ( \xi, - 2b - \eta), \\ & ( \xi,4b - \eta), \ldots, ( \xi,\eta - 2nb), ( \xi, - \eta + 2nb), \ldots \bigr\} \end{aligned}$$

(15)

in Ω, then the response to \(\tilde{S}_{0}^{ +}\) at field point \((x,y)\) is as follows:

$$\begin{aligned} \tilde{G}_{0}^{ +} ( x,y;\xi,\eta) =& \frac{1}{2\pi } \sum_{n = - \infty }^{\infty } \bigl[ K_{0} \bigl( k\sqrt{(x - \xi)^{2} + (y - \eta + 2nb)^{2}} \bigr) \\ &{} - K_{0} \bigl( k\sqrt{(x - \xi)^{2} + (y + \eta - 2nb)^{2}} \bigr) \bigr] \end{aligned}$$

(16)

and it is not difficult to show that \(\tilde{G}_{0}^{ +} ( x,y; \xi,\eta) \) satisfies the Dirichlet conditions on the boundary fragments \(y=0\) and \(y =b\) but conflicts with the Dirichlet conditions on the boundary fragments \(x=0\) and \(x=a\). Similar to the above discussion, the trace of the solution \(\tilde{G}_{0}^{ +} ( x,y; \xi,\eta) \) on the boundary line \(x=0\) can be compensated for with the unit sinks

$$\begin{aligned} \tilde{S}_{0,1}^{ +} =& \bigl\{ ( - \xi, - \eta), ( - \xi,2b - \eta), ( - \xi, - 2b + \eta), ( - \xi,2b + \eta), ( - \xi, - 2b - \eta), \\ & ( - \xi,4b - \eta), \ldots, ( - \xi,\eta - 2nb), ( - \xi, - \eta + 2nb), \ldots \bigr\} \end{aligned}$$

(17)

which represent the images of \(\tilde{S}_{0}^{ +}\) about \(x=0\). The response to these sinks at \((x,y)\) evidently is \(\tilde{G} _{0,1}^{ -} ( x,y;\xi,\eta) = - \frac{1}{2\pi } \tilde{G}_{0}^{ +} ( x,y; - \xi,\eta) \).

To compensate for the trace of \(\tilde{G}_{0}^{ +} ( x,y;\xi, \eta) \) on \(x=a\), we place a series unit sinks

$$\begin{aligned} \tilde{S}_{a,1}^{ +} =& \bigl\{ ( 2a - \xi ,- \eta), ( 2a - \xi ,2b - \eta), ( 2a - \xi ,- 2b + \eta), \\ & ( 2a - \xi ,2b + \eta), ( 2a - \xi ,- 2b - \eta), \\ & ( 2a - \xi ,4b - \eta), \ldots, ( 2a - \xi,\eta- 2nb), ( 2a - \xi ,- \eta + 2nb), \ldots \bigr\} \end{aligned}$$

(18)

which represent the images of \(\tilde{S}_{0}^{ +}\) about the line \(x=a\). The response to these sinks at \((x,y)\) evidently is

$$ \tilde{G}_{a,1}^{ -} ( x,y;\xi,\eta) = - \frac{1}{2 \pi } \tilde{G}_{0}^{ +} ( x,y;2a - \xi,\eta). $$

(19)

The functions \(\tilde{G}_{0,1}^{ -} ( x,y;\xi,\eta) \) and \(\tilde{G}_{a,1}^{ -} ( x,y;\xi,\eta) \) do not vanish on the boundary lines \(x=0\) and \(x=a\), and their traces can be properly compensated for by the set of unit sources

$$\begin{aligned} \tilde{S}_{0,2}^{ +} =& \bigl\{ ( - 2a + \xi, - \eta), ( - 2a + \xi,2b - \eta), ( - 2a + \xi, - 2b + \eta), ( - 2a + \xi,2b + \eta), \\ & ( - 2a + \xi, - 2b - \eta), ( - 2a + \xi,4b - \eta), \ldots, \\ & ( - 2a + \xi,\eta - 2nb), ( - 2a + \xi, - \eta + 2nb), \ldots \bigr\} \end{aligned}$$

(20)

and

$$\begin{aligned} \tilde{S}_{a,2}^{ +} =& \bigl\{ ( 2a + \xi, - \eta), ( 2a + \xi,2b - \eta), ( 2a + \xi, - 2b + \eta), ( 2a + \xi,2b + \eta), \\ &( 2a + \xi, - 2b - \eta), ( 2a + \xi,4b - \eta), \ldots, \\ & ( 2a + \xi, \eta - 2nb), ( 2a + \xi, - \eta + 2nb), \ldots \bigr\} . \end{aligned}$$

(21)

The responses to these at \(( x,y) \) are given as

$$ \tilde{G}_{0,2}^{ +} ( x,y; - 2a + \xi,\eta) = \frac{1}{2 \pi } \tilde{G}_{0}^{ +} ( x,y; - 2a + \xi,\eta) $$

(22)

and

$$ \tilde{G}_{a,2}^{ +} ( x,y;2a + \xi,\eta) = \frac{1}{2 \pi } \tilde{G}_{0}^{ +} ( x,y;2a + \xi,\eta). $$

(23)

Upon following the above described procedure, we properly place compensatory unit sources that alternate with compensatory unit sinks as follows:

$$\begin{aligned}& \begin{aligned}[b] \tilde{S}_{0,3}^{ +} ={}& \bigl\{ ( - 2a - \xi, - \eta), ( - 2a - \xi,2b - \eta), ( - 2a - \xi, - 2b + \eta), ( - 2a - \xi,2b + \eta), \\ & ( - 2a - \xi, - 2b - \eta), ( - 2a - \xi,4b - \eta), \ldots, \\ & ( - 2a - \xi,\eta - 2nb), ( - 2a - \xi, - \eta + 2nb), \ldots \bigr\} , \end{aligned} \end{aligned}$$

(24)

$$\begin{aligned}& \begin{aligned}[b] \tilde{S}_{a,3}^{ +} ={}& \bigl\{ ( 4a - \xi, - \eta), ( 4a - \xi,2b - \eta), ( 4a - \xi, - 2b + \eta), ( 4a - \xi,2b + \eta), \\ & ( 4a - \xi, - 2b - \eta), ( 4a - \xi,4b - \eta), \ldots, \\ & ( 4a - \xi,\eta - 2nb), ( 4a - \xi, - \eta + 2nb), \ldots \bigr\} \end{aligned} \end{aligned}$$

(25)

$$\begin{aligned}& \vdots \\& \begin{aligned}[b] \tilde{S}_{0,2n}^{ +} ={}& \bigl\{ ( - 2na + \xi, - \eta), ( - 2na + \xi,2b - \eta), ( - 2na + \xi, - 2b + \eta), \\ &( - 2na + \xi,2b + \eta), ( - 2na + \xi, - 2b - \eta), ( - 2na + \xi,4b - \eta), \ldots, \\ & ( - 2na + \xi,\eta - 2nb), ( - 2na + \xi, - \eta + 2nb), \ldots \bigr\} , \end{aligned} \end{aligned}$$

(26)

$$\begin{aligned}& \begin{aligned}[b] \tilde{S}_{a,2n}^{ +} ={}& \bigl\{ ( 2na + \xi, - \eta), ( 2na + \xi,2b - \eta), ( 2na + \xi, - 2b + \eta), \\ & ( 2na + \xi,2b + \eta), ( 2na + \xi, - 2b - \eta), \\ & ( 2na + \xi,4b - \eta), \ldots, ( 2na + \xi,\eta - 2nb), ( 2na + \xi, - \eta + 2nb), \ldots \bigr\} , \end{aligned} \end{aligned}$$

(27)

$$\begin{aligned}& \begin{aligned}[b] \tilde{S}_{0,2n + 1}^{ +} ={}& \bigl\{ ( - 2na - \xi, - \eta), ( - 2na - \xi,2b - \eta), ( - 2na - \xi, - 2b + \eta), \\ & ( - 2na - \xi,2b + \eta), ( - 2na - \xi, - 2b - \eta), ( - 2na - \xi,4b - \eta), \ldots, \\ & ( - 2na - \xi,\eta - 2nb), ( - 2na - \xi, - \eta + 2nb), \ldots \bigr\} , \end{aligned} \end{aligned}$$

(28)

$$\begin{aligned}& \begin{aligned}[b] \tilde{S}^{ +}_{a,2n + 1} ={}& \bigl\{ \bigl(2(n + 1)a - \xi, - \eta\bigr),\bigl(2(n + 1)a - \xi,2b - \eta\bigr), \\ & \bigl(2(n + 1)a - \xi, - 2b + \eta \bigr), \bigl(2(n + 1)a - \xi,2b + \eta \bigr), \\ & \bigl(2(n + 1)a - \xi, - 2b - \eta \bigr), \bigl(2(n + 1)a - \xi,4b - \eta \bigr), \ldots, \\ & \bigl(2(n + 1)a - \xi,\eta - 2nb \bigr), \bigl(2(n + 1)a - \xi, - \eta + 2nb \bigr), \ldots \bigr\} \end{aligned} \\& \vdots \end{aligned}$$

(29)

The structure is shown in Figure 2.

Hence, we express the Green’s function \(G ( x,y;\xi,\eta) \) stated on a rectangular region in the form

$$ G ( x,y;\xi,\eta) = \tilde{G}_{0}^{ +} ( x,y;\xi, \eta) + \sum_{n = 1}^{\infty } \bigl( \tilde{G}_{0,2n - 1}^{ -} + \tilde{G}_{a,2n - 1}^{ -} \bigr) + \sum _{n = 1}^{\infty } \bigl( \tilde{G}_{0,2n}^{ +} + \tilde{G}_{a,2n}^{ +} \bigr). $$

(30)

After a trivial substitution, the Green’s function of the SKGE stated on a rectangular region can be expressed as the double infinite series

$$\begin{aligned} G ( x,y;\xi,\eta) =& \frac{1}{2\pi } \sum _{n = - \infty } ^{\infty } \sum_{m = - \infty }^{\infty } \bigl[ K_{0} \bigl( k\sqrt{(x - \xi + 2ma)^{2} + (y - \eta + 2nb)^{2}} \bigr) \\ &{} - K_{0} \bigl( k \sqrt{(x - \xi - 2ma)^{2} + (y + \eta - 2nb)^{2}} \bigr) \bigr]. \end{aligned}$$

(31)