In this section, we first formulate the matrix Riemann–Hilbert
problem under the assumption that the solution $q(x,y)$ of
(1.1) exists and then we discuss the existence of the
solution $q(x,y)$ satisfying the boundary conditions. Using
equations (3.1) and the global relation
(3.6), after tedious but straightforward calculations, we formulate the following matrix Riemann–Hilbert
problem:

${M}_{-}(x,y,k)={M}_{+}(x,y,k)J(x,y,k),k\in \mathcal{\mathcal{L}},$(4.1)

where the oriented contours $\mathcal{\mathcal{L}}={L}_{1}\cup {L}_{2}\cup {L}_{3}\cup {L}_{4}$ are given by (cf. Figure 2)

${L}_{1}={D}_{1}\cap {D}_{2},{L}_{2}={D}_{2}\cap {D}_{3},$(4.2a)${L}_{3}={D}_{3}\cap {D}_{4},{L}_{4}={D}_{4}\cap {D}_{1},$(4.2b)

and the jump matrices are defined by

${J}_{1}(x,y,k)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill \frac{{b}_{2}(k)}{{a}_{1}(-k){a}_{3}(-k)}{e}^{2\theta (x,y,k)}\hfill & \hfill 1\hfill \end{array}\right),\mathrm{\hspace{1em}}k\in {L}_{1},$(4.3a)${J}_{2}(x,y,k)=\left(\begin{array}{cc}\hfill \frac{{a}_{2}(k)}{{a}_{1}(k){a}_{3}(-k)}\hfill & \hfill -\frac{{b}_{1}(-k)}{{a}_{1}(k)}{e}^{-2\theta (x,y,k)}\hfill \\ \hfill -\frac{{b}_{3}(k)}{{a}_{3}(-k)}{e}^{2(\theta (x,y,k)-{\omega}_{2}(k)L)}\hfill & \hfill 1\hfill \end{array}\right),\mathrm{\hspace{1em}}k\in {L}_{2},$(4.3b)${J}_{3}(x,y,k)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill -\frac{{b}_{2}(-k)}{{a}_{1}(k){a}_{3}(k)}{e}^{-2\theta (x,y,k)}\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right),\mathrm{\hspace{1em}}k\in {L}_{3},$(4.3c)${J}_{4}(x,y,k)={J}_{1}{J}_{2}^{-1}{J}_{3}=\left(\begin{array}{cc}\hfill 1\hfill & \hfill \frac{{b}_{3}(-k)}{{a}_{3}(k)}{e}^{-2(\theta (x,y,k)-{\omega}_{2}(k)L)}\hfill \\ \hfill \frac{{b}_{1}(k)}{{a}_{1}(-k)}{e}^{2\theta (x,y,k)}\hfill & \hfill \frac{{a}_{2}(-k)}{{a}_{1}(-k){a}_{3}(k)}\hfill \end{array}\right),\mathrm{\hspace{1em}}k\in {L}_{4}$(4.3d)

with $\theta (x,y,k)={\omega}_{1}(k)x+{\omega}_{2}(k)y$.
The matrix-valued functions ${M}_{\pm}$ are sectionally meromorphic
and defined as follows:

${M}_{+}(x,y,k)=\{\begin{array}{cc}(\frac{{\mu}_{2}^{(1)}}{{a}_{1}(-k)},{\mu}_{3}^{(12)})\hfill & \text{for}k\in {D}_{1},\hfill \\ ({\mu}_{3}^{(34)},\frac{{\mu}_{2}^{(3)}}{{a}_{1}(k)})\hfill & \text{for}k\in {D}_{3},\hfill \end{array}$$\mathrm{\hspace{1em}}{M}_{-}(x,y,k)=\{\begin{array}{cc}(\frac{{\mu}_{1}^{(2)}}{{a}_{3}(-k)},{\mu}_{3}^{(12)})\hfill & \text{for}k\in {D}_{2},\hfill \\ ({\mu}_{3}^{(34)},\frac{{\mu}_{1}^{(4)}}{{a}_{3}(k)})\hfill & \text{for}k\in {D}_{4}.\hfill \end{array}$

Note that $det{M}_{\pm}=1$ and ${M}_{\pm}=I+O(\frac{1}{k})$ as
$k\to \mathrm{\infty}$. Indeed, using the global relation (3.6), we find

${a}_{2}(k)-{b}_{1}(-k){b}_{3}(k){e}^{-2{\omega}_{2}(k)L}={a}_{1}(k){a}_{3}(-k),$

which implies that $det({J}_{2})=det({J}_{4})=1$.

The Riemann–Hilbert problem (4.1) can be
solved by a Cauchy-type integral equation. Let $\stackrel{~}{J}=I-J$.
Then equation (4.1) becomes

${M}_{+}(x,y,k)-{M}_{-}(x,y,k)={M}_{+}(x,y,k)\stackrel{~}{J}(x,y,k).$

Applying the Plemelj formula [9], the solution *M* of the Riemann–Hilbert problem (4.1)
can be expressed as

$M(x,y,k)=I+\frac{1}{2i\pi}{\int}_{\mathcal{\mathcal{L}}}{M}_{+}(x,y,{k}^{\prime})\stackrel{~}{J}(x,y,{k}^{\prime})\frac{d{k}^{\prime}}{{k}^{\prime}-k}.$

Expanding the integrand for *k*, we find

$M(x,y,k)=I-\frac{1}{2i\pi k}{\int}_{\mathcal{\mathcal{L}}}{M}_{+}(x,y,{k}^{\prime})\stackrel{~}{J}(x,y,{k}^{\prime})\mathit{d}{k}^{\prime}+O\left(\frac{1}{{k}^{2}}\right),k\to \mathrm{\infty}.$

Thus, we expand the solution *M* of the Riemann–Hilbert problem (4.1) as

$M(x,y,k)=I+\frac{{M}^{(1)}(x,y)}{k}+\frac{{M}^{(2)}(x,y)}{{k}^{2}}+O\left(\frac{1}{{k}^{3}}\right),k\to \mathrm{\infty}.$(4.4)

If we substitute the expansion (4.4) into the *x*-part of the Lax pair (2.1a), the $(2,1)$-component
at $O(1)$ yields

$i{q}_{x}(x,y)+{q}_{y}(x,y)=-4i{M}_{21}^{(1)}(x,y)$(4.5)

and the $(1,1)$-component at $O(\frac{1}{k})$ implies

${M}_{11x}^{(1)}(x,y)=-\frac{1}{8i}(\mathrm{cosh}q(x,y)-1)-\frac{1}{4}(i{q}_{x}(x,y)+{q}_{y}(x,y)){M}_{21}^{(1)}(x,y).$

Simplifying the above equation with (4.5), we obtain the
reconstruction formula for $q(x,y)$ in terms of the solution
of the Riemann–Hilbert problem

$\mathrm{cosh}q(x,y)=1-8i{M}_{11x}^{(1)}(x,y)-8{({M}_{21}^{(1)})}^{2}.$

Similarly, if we substitute the expansion (4.4) into the *y*-part of the Lax pair,
the solution $q(x,y)$ can be written equivalently as

$\mathrm{cosh}q(x,y)=1+8{M}_{11y}^{(1)}(x,y)+8{({M}_{21}^{(1)})}^{2}.$

Let us now state the existence theorem for
the elliptic sinh-Gordon equation in the semi-strip.

#### Theorem 4.1.

*Let the functions ${g}_{j}\mathit{}\mathrm{(}x\mathrm{)}$, ${h}_{j}\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\in}{H}^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^{\mathrm{+}}\mathrm{)}$ and ${f}_{j}\mathit{}\mathrm{(}y\mathrm{)}\mathrm{\in}{H}^{\mathrm{1}}\mathit{}\mathrm{(}\mathrm{[}\mathrm{0}\mathrm{,}L\mathrm{]}\mathrm{)}$ , $j\mathrm{=}\mathrm{0}\mathrm{,}\mathrm{1}$, be given
with the sufficiently small
${H}^{\mathrm{1}}$ norms. Let the functions ${a}_{j}\mathit{}\mathrm{(}k\mathrm{)}$, ${b}_{j}\mathit{}\mathrm{(}k\mathrm{)}$, $j\mathrm{=}\mathrm{1}\mathrm{,}\mathrm{2}\mathrm{,}\mathrm{3}$, be
given by (3.2), (3.3) and
(3.4) in Definitions 3.1, 3.2 and
3.3, respectively. Suppose that the spectral functions
${a}_{j}$ and ${b}_{j}$, $j\mathrm{=}\mathrm{1}\mathrm{,}\mathrm{2}\mathrm{,}\mathrm{3}$, satisfy the global relation (3.6).
Let $M\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{,}k\mathrm{)}$ be the solution of the following matrix
Riemann–Hilbert (RH) problem*

${M}_{-}(x,y,k)={M}_{+}(x,y,k)J(x,y,k),k\in \mathcal{\mathcal{L}},$(4.6)

*where $\mathrm{det}\mathit{}\mathrm{(}{M}_{\mathrm{\pm}}\mathrm{)}\mathrm{=}\mathrm{1}$, ${M}_{\mathrm{\pm}}\mathrm{=}I\mathrm{+}O\mathit{}\mathrm{(}\frac{\mathrm{1}}{k}\mathrm{)}$ as $k\mathrm{\to}\mathrm{\infty}$, the oriented contours $\mathcal{L}$ are defined in (4.2) and the jump matrices **J* are given in (4.3).

*Then the Riemann–Hilbert problem is uniquely solvable and the
function $q\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{)}$ defined by*

$i{q}_{x}+{q}_{y}=-4i\underset{k\to \mathrm{\infty}}{lim}k{M}_{21},\mathrm{cosh}q(x,y)=1-8i\underset{k\to \mathrm{\infty}}{lim}k{M}_{11x}-8\underset{k\to \mathrm{\infty}}{lim}{(k{M}_{21})}^{2}$(4.7)

*solves the elliptic sinh-Gordon equation (1.1) satisfying the boundary conditions*

$q(x,0)={g}_{0}(x),$$\mathrm{\hspace{1em}}{q}_{y}(x,0)={g}_{1}(x),$(4.8a)$q(0,y)={f}_{0}(y),$$\mathrm{\hspace{1em}}{q}_{x}(0,y)={f}_{1}(y),$(4.8b)$q(x,L)={h}_{0}(x),$$\mathrm{\hspace{1em}}{q}_{y}(x,L)={h}_{1}(x).$(4.8c)

#### Proof.

Using the dressing method and the vanishing lemma presented
in [7, 9], we can prove the unique solvability of the Riemann–Hilbert problem (4.6)
and then we can also verify that
the $q(x,y)$ defined in (4.7) solves the elliptic
sinh-Gordon equation (1.1).

The proof that $q(x,y)$ given in (4.7) satisfies the
boundary values is similar to the argument presented in
[14]. Here, we only state the proof of (4.8c).
We first define

${M}^{(L)}(x,k)=\{\begin{array}{cc}M(x,L,k){J}_{1}(x,L,k)\hfill & \text{for}k\in {D}_{1},\hfill \\ M(x,L,k)\hfill & \text{for}k\in {D}_{2},\hfill \\ M(x,L,k)F(x,k)\hfill & \text{for}k\in {D}_{3},\hfill \\ M(x,L,k){J}_{3}^{-1}(x,L,k)F(x,k)\hfill & \text{for}k\in {D}_{4},\hfill \end{array}$(4.9)

where

$F(x,k)=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill \hfill & \hfill -\frac{{b}_{2}(-k)}{{a}_{1}(k){a}_{3}(k)}{e}^{-2({\omega}_{1}(k)x+{\omega}_{2}(k)L)}\hfill \\ \hfill 0\hfill & \hfill \hfill & \hfill 1\hfill \end{array}\right).$

It is convenient to denote $M(x,L,k)$ and ${M}^{(L)}(x,k)$ for $k\in {D}_{j}$, $j=1,\mathrm{\dots},4$, by
${M}_{j}(x,L,k)$ and ${M}_{j}^{(L)}(x,k)$, respectively. Then, equations (4.6) and
(4.9) can be written as

${M}_{2}(x,L,k)={M}_{1}{J}_{1}(x,L,k),{M}_{2}(x,L,k)={M}_{3}{J}_{2}(x,L,k),$(4.10a)${M}_{4}(x,L,k)={M}_{3}{J}_{3}(x,L,k),{M}_{4}(x,L,k)={M}_{1}{J}_{4}(x,L,k),$(4.10b)

and

${M}_{1}^{(L)}(x,k)={M}_{1}(x,L,k){J}_{1}(x,L,k),$${M}_{2}^{(L)}(x,k)={M}_{2}(x,L,k),$(4.11a)${M}_{3}^{(L)}(x,k)={M}_{3}(x,L,k)F(x,k),$${M}_{4}^{(L)}(x,k)={M}_{4}(x,L,k){J}_{3}^{-1}(x,L,k)F(x,k).$(4.11b)

Combining (4.11) and (4.10), we find the jump conditions as

${M}_{2}^{(L)}(x,k)={M}_{1}^{(L)}(x,k),{M}_{3}^{(L)}(x,k)={M}_{2}^{(L)}(x,k){J}_{2}^{-1}(x,L,k)F(x,k),$${M}_{4}^{(L)}(x,k)={M}_{3}^{(L)}(x,k),{M}_{4}^{(L)}(x,k)={M}_{1}^{(L)}(x,k){J}_{2}^{-1}(x,L,k)F(x,k).$

Note that no jumps occur along the contours ${L}_{1}$ and ${L}_{3}$ and that
${J}_{2}^{-1}(x,L,k)F(x,k)={J}^{(L)}(x,k)$,
where ${J}^{(L)}(x,k)$ is given in (3.13). Thus,
we define

${M}^{(L)}(x,k)={M}_{+}^{(L)}(x,k),\mathrm{\hspace{1em}}k\in {D}_{1}\cup {D}_{2},$${M}^{(L)}(x,k)={M}_{-}^{(L)}(x,k),\mathrm{\hspace{1em}}k\in {D}_{3}\cup {D}_{4}$

and then
equations (4.10) are equivalent to the Riemann–Hilbert problem (3.12)
with the jump matrix ${J}^{(L)}(x,k)$.
Thus, evaluating (4.7) at $y=L$,
we prove that the function $q(x,y)$ defined in (4.7) satisfies the boundary values (4.8c).

In a similar way, equations (4.8a) and (4.8b) can be proved.
∎

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