3.1 The solution of the model RH problem on the z-plane

Now, transform the model RH problem (3.0.11) from the $\lambda $ plane into z plane for $z=-\lambda ^2$ by taking the lower half $\lambda $ -plane onto $\mathbb C\setminus (-\infty ,0)$ . As a result, the model RH problem $S^{\infty }(\lambda )$ is converted into the RH problem on the z-plane, denoted as $S^{\infty }(z)$ which satisfies the following jump conditions (see also Figure 6)

(3.1.1) $$ \begin{align} S^{\infty}_+(z)=S^{\infty}_-(z)\begin{cases} \begin{aligned} &\begin{pmatrix} 0 & e^{x\Omega_0+\Delta_0}\\ e^{-x\Omega_0-\Delta_0}& 0 \end{pmatrix}, &&z\in[-\eta_1^2,0],\\ &\begin{pmatrix} i & 0\\ 0& i \end{pmatrix}, &&z\in[-\eta_2^2,-\eta_1^2]\cup[-\eta_4^2,-\eta_3^2],\\ &\begin{pmatrix} 0 & e^{x\Omega_1+\Delta_1}\\ e^{-x\Omega_1-\Delta_1}& 0 \end{pmatrix}, &&z\in[-\eta_3^2,-\eta_2^2],\\ &\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}, &&z\in[-\infty,-\eta_4^2].\\ \end{aligned} \end{cases} \end{align} $$

Furthermore, $S^{\infty }(z)\to \begin {pmatrix} 1&1 \end {pmatrix}$ as $z\to \infty $ . The last jump matrix in the jump condition (3.1.1) is generated by the symmetry $S^{\infty }(-\lambda )=S^{\infty }(\lambda ) \begin {pmatrix} 0&1\\ 1&0 \end {pmatrix}$ , which also changes the diagonal jump matrices in the $\lambda $ -plane into off-diagonal ones in the z-plane, see [Reference Zhao and Fan37].

Figure 6. The jump contour for $ S^{\infty }(z) $ and the associated jump matrices.

Figure 7. The Riemann surface $\hat {\mathcal {S}}$ and its basis $\{\hat {a}_j ,\hat {b}_j\},~j=1,2$ of circles.

Introduce the Riemann surface of genus two (see Figure 7) as

$$ \begin{align*} \hat{\mathcal{S}}:=\{(z,y)|y^2=z(z+\eta_1^2)(z+\eta_2^2)(z+\eta_3^2)(z+\eta_4^2)\}, \end{align*} $$

and define the cycle basis $\{\hat a_j,\hat b_j\}$ for $j=1,2$ . Further, denote $\hat R(z)=\sqrt {z(z+\eta _1^2)(z+\eta _2^2)(z+\eta _3^2)(z+\eta _4^2)}$ and define the normalized holomorphic differential vector $\hat \omega =(\hat \omega _1, \hat \omega _2)^{T}$ such that $\oint _{a_j}\hat {\omega }_k=\delta _{jk}$ for $j=1,2$ where $\delta _{jk}=0$ and $1$ for $j\neq k$ and $j=k$ respectively. The associated period matrix $\hat \tau _{ij}=\oint _{\hat b_j}\hat \omega _i~(i,j=1,2)$ and $\hat {\tau }_j=\hat {\tau } e_j$ , where $e_j$ is the j-th column of $2\times 2$ unit matrix for $j=1,2$ . Thus the Abel map $\hat {J}(z)=\int _{\infty }^z \hat \omega $ has the following properties

(3.1.2) $$ \begin{align} \begin{aligned} &\hat{J}_+(z)+\hat{J}_-(z)=0,&&z\in(-\infty,\eta_4^2],\\ &\hat{J}_+(z)-\hat{J}_-(z)=-e_1-e_2,&&z\in[-\eta_4^2,-\eta_3^2],\\ &\hat{J}_+(z)+\hat{J}_-(z)=-\hat{\tau}_1,&&z\in[-\eta_3^2,-\eta_2^2],\\ &\hat{J}_+(z)-\hat{J}_-(z)=-e_2,&&z\in[-\eta_2^2,-\eta_1^2],\\ &\hat{J}_+(z)+\hat{J}_-(z)=-\hat{\tau}_2,&&z\in[-\eta_1^2,0].\\ \end{aligned} \end{align} $$

Introduce the Riemann-Theta function:

(3.1.3) $$ \begin{align} \Theta(z;\tau)=\sum_{\vec{n}\in\mathbb{Z}^2}e^{2\pi i\left(\vec{n}^T z+\frac{1}{2}\vec{n}^T\tau \vec{n}\right)},\quad z\in\mathbb{C}^2, \end{align} $$

which satisfies the following properties:

(3.1.4) $$ \begin{align} \Theta(z+e_j;\hat{\tau})=\Theta(z-e_j;\hat{\tau})=\Theta(z;\hat{\tau}),\quad \Theta(z+\hat{\tau}_j;\hat{\tau})=e^{2\pi i(-z_j-\hat{\tau}_{jj}/2)}\Theta(z;\hat{\tau}), \end{align} $$

Furthermore, define

$$ \begin{align*} \hat\gamma(z)=\left(\frac{(z+\eta_1^2)(z+\eta_3^2)}{(z+\eta_2^2)(z+\eta_4^2)}\right)^{\frac{1}{4}}. \end{align*} $$

Notice that the solution of model RH problem (3.1.1) on the $\lambda $ -plane has at most fourth root singularities at $\pm \eta _j,~j=1,\cdots ,4$ , and this condition transforms into the solution of RH problem on the z-plane with singularities at $-\eta ^2_{1}$ and $-\eta ^2_{3}$ less than $1/4$ . Furthermore, the zeros of $\hat {\gamma }(z)$ is $-\eta _1^2$ and $-\eta _3^2$ and the solution of the RH problem has the form

(3.1.5) $$ \begin{align} \frac{\Theta(\pm \hat{J}(z)-\frac{\Omega}{2\pi i}+\hat{d};\hat{\tau})}{\Theta(\pm \hat{J}(z)+\hat{d};\hat{\tau})},\quad \text{with}\quad \Omega=(x\Omega_1+\Delta_1,x\Omega_0+\Delta_0)^T. \end{align} $$

It suffices to choose the parameter $\hat {d}$ , which makes the zeros of denominator at $-\eta _1^2$ and $-\eta _3^2$ , to nail down the singularities. On the other hand, we have

$$ \begin{align*} \begin{aligned} \hat{J}(\infty)&=0,\ \hat{J}(-\eta_4^2)=-\frac{e_1+e_2}{2},\ \hat{J}(-\eta_3^2)=\frac{\hat{\tau}_1}{2}-\frac{e_1+e_2}{2},\\ \hat{J}(-\eta_2^2)&=\frac{\hat{\tau}_1}{2}-\frac{e_2}{2},\ \hat{J}(-\eta_1^2)=\frac{\hat{\tau}_2}{2}-\frac{e_2}{2},\ \hat{J}(0)=\frac{\hat{\tau}_1}{2}. \end{aligned} \end{align*} $$

The Riemann constant $\hat {\mathcal {K}}=\frac {\hat {\tau }_1+\hat {\tau }_2}{2}-\frac {e_1}{2}$ , and further $\hat {J}(-\eta _1^2)+\hat {J}(-\eta _3^2)=\hat {\mathcal {K}}$ , which implies $\hat {d}=0$ . Consequently the solution of the model RH problem (3.1.1) on the z-plane is

(3.1.6) $$ \begin{align} \hat{S}^{\infty}(z)=\hat\gamma(z)\frac{\Theta(0;\hat\tau)}{\Theta(\frac{\Omega}{2\pi i};\hat\tau)}\left(\begin{array}{cc} \frac{\Theta(\hat{J}(z)-\frac{\Omega}{2\pi i};\hat\tau)}{\Theta(\hat{J}(z);\hat\tau)} & \frac{\Theta(-\hat{J}(z)-\frac{\Omega}{2\pi i};\hat\tau)}{\Theta(-\hat{J}(z);\hat\tau)} \end{array} \right). \end{align} $$

In order to get the expansion of the Abel map $\hat {J}(z)$ as $z\to \infty $ , we also need to transform the scalar RH problem for $g(\lambda )$ on the $\lambda $ -plane into the case on the z-plane, that is

$$ \begin{align*} \begin{aligned} &g_+(z)-g_-(z)=2i\sqrt{z}, &&z\in (-\eta_4^2,-\eta_3^2)\cup (-\eta_2^2,-\eta_1^2),\\ &g_+(z)+g_-(z)=0, &&z\in(-\infty,-\eta_4^2),\\ &g_+(z)+g_-(z)=\Omega_0, &&z\in(-\eta_1^2,0),\\ &g_+(z)+g_-(z)=\Omega_1, &&z\in(-\eta_3^2,-\eta_2^2),\\ &g(z)={\mathcal{O}\left(\frac{1}{\sqrt{z}}\right),} &&z\to\infty. \end{aligned} \end{align*} $$

Consequently, the formula of $g(z)$ is

$$ \begin{align*} g(z)=\hat{R}(z)\left(\int_{-\eta_3^2}^{-\eta_2^2}\frac{\Omega_1}{\hat R_+(\mu)(\mu-z)}d\mu+\int_{-\eta_1^2}^{0}\frac{\Omega_0}{\hat R_+(\mu)(\mu-z)}d\mu +\left(\int_{-\eta_4^2}^{-\eta_3^2}+\int_{-\eta_2^2}^{-\eta_1^2}\right)\frac{2i\sqrt{\mu}}{\hat R_+(\mu)(\mu-z)}d\mu\right). \end{align*} $$

In order to keep consistence with the asymptotic condition, it follows that

$$ \begin{align*} \int_{\hat a_1}{\Omega_1}{\hat\omega_j}+\int_{\hat a_2}{\Omega_0}{\hat\omega_j}+4i\left(\int_{-\eta_4^2}^{-\eta_3^2}+\int_{-\eta_2^2}^{-\eta_1^2}\right) \sqrt{z}{\hat\omega_j}=0,\quad j=1,2, \end{align*} $$

and $4i\left (\int _{-\eta _4^2}^{-\eta _3^2}+\int _{-\eta _2^2}^{-\eta _1^2}\right )\sqrt {z}{\hat \omega _j}=2\pi \mathrm {res}_{\infty }\sqrt {z}{\hat \omega _j}$ for $j=1,2$ , which indicates that $\mathrm {res}_{\infty }\sqrt {z}{\hat {\omega }}=-\frac {1}{2\pi }\begin {pmatrix} \Omega _1 &\Omega _0 \end {pmatrix}^T=-i\frac {\Omega _x}{2\pi i}$ . As $z\to \infty $ , expand $\hat {S}^{\infty }_1(z)$ as

$$ \begin{align*} \hat{S}^{\infty}_1(z)=1+\frac{\hat{S}^{\infty}_{11}}{z^{1/2}}+\mathcal{O}\left(\frac{1}{z}\right), \end{align*} $$

with

(3.1.7) $$ \begin{align} \hat{S}^{\infty}_{11}=-i\left[\nabla\log\left(\Theta\left(\frac{\Omega}{2\pi i}\right);\hat\tau\right)-\nabla\log(\Theta(0;\hat\tau))\right]\cdot \frac{ \Omega_x}{2\pi i}. \end{align} $$

Recall that $z=-\lambda ^2$ , hence the WKB expansion of solution $\hat S_1^{\infty }(z)$ on the parameter $\lambda $ is

(3.1.8) $$ \begin{align} \hat{S}^{\infty}_1(z)=1-\frac{1}{\lambda}\left[\nabla\log\left(\Theta\left(\frac{\Omega}{2\pi i}\right);\hat\tau\right)-\nabla\log(\Theta(0);\hat\tau)\right]\cdot \frac{ \Omega_x}{2\pi i}+\mathcal{O}\left(\frac{1}{\lambda^2}\right). \end{align} $$

Here we choose $i\lambda $ , since the map $\lambda \to z=-\lambda ^2$ from the upper $\lambda $ -plane to $\mathbb C\setminus {(-\infty ,0)}$ .

So far, we obtain the expression of $\hat {S}^{\infty }_1(z)$ and its expansion for $z\to \infty $ . To reconstruct the potential $u(x)$ , one needs to concentrate on the error RH problem of $Y(\lambda )$ which in fact contributes the error term $\mathcal {O}(x^{-1})$ in the asymptotic behavior of $u(x)$ for $x\to +\infty $ . The techniques for error estimation are illustrated in [Reference Girotti, Grava, Jenkins and McLaughlin21], although they differ slightly from those in this paper. Now, to analyze the error RH problem for $Y(\lambda )$ , we need to transform the vector-form solution into a matrix form (see Remark 3.7). To ensure logical coherence, we will reconstruct $u(x)$ at the end of subsection 3.2 (Remark 3.9) using equation (3.1.8), leveraging the connection between the z-plane and the $\lambda $ -plane. This allows us to complete the proof of Theorem 3.8 from the perspective of the z-plane.

3.2 The solution of the model RH problem on the $\lambda $ -plane

We have derived the solution of model RH problem $\hat S^{\infty }(z)$ on the z-plane; nevertheless, we also need to get the solution on $\lambda $ -plane. Moreover, notice that the transformation $z=-\lambda ^2$ can be considered as a holomorphic map between the Riemann surfaces $\mathcal {S}$ and $\hat {\mathcal {S}}$ in Figure 4 and Figure 7. In general, suppose that $\mathcal {S}$ is a Riemann surface of genus $2g-1$ for $g\in \mathbb {Z}_{+}$ , that is, $\mathcal {S}=\{(\lambda ,y)|y^2=\Pi _{j=1}^{2g}(\lambda ^2-\eta _j^2)\},$ and Riemann surface $\hat {\mathcal {S}}$ on the z-plane, that is, $\hat {\mathcal {S}}=\{(z,y)|y^2=z\Pi _{j=1}^{2g}(z+\eta _j^2)\}$ . Define the holomorphic map $\varphi :\mathcal {S}\to \hat {\mathcal {S}}$ with $-\lambda ^2\to z$ , and according to the Riemann-Hurwitz formula, it follows that the genus of Riemann surface $\hat {\mathcal {S}}$ is g. Indeed, we just transform the solution on the z-plane into $\lambda $ -plane, especially, the normalized holomorphic differentials, basic cycles and period matrix. In the following, we will directly illustrate the model solution of $S^{\infty }(\lambda )$ and then develop the equivalence between the two solutions.

Similarly, define the normalized holomorphic differentials $\omega _j~(j=1,2,3)$ associated with Riemann surface $\mathcal {S}$ by $\omega =(\omega _1, \omega _2, \omega _3)^T=A^{-1}\tilde {\omega }$ , where $\tilde {\omega }=(\tilde {\omega }_1, \tilde {\omega }_2, \tilde {\omega }_3)^{T}$ with $\tilde \omega _j=\frac {\zeta ^{j-1}d\zeta }{R(\zeta )}~(j=1,2,3)$ and $A=(a_{ij})_{3\times 3}$ with $a_{ij}=\oint _{a_j}\tilde \omega _i$ . Define the period matrix $\tau =(\tau _{ij})_{3\times 3}$ with $\tau _{ij}=\oint _{b_j}\omega _i$ and recall that the cycles are defined in Fig.4. Indeed, it follows from the parity of $\tilde \omega _j$ and the symmetries of a-cycles that $a_{11}=a_{13},a_{21}=-a_{23}$ and $a_{31}=a_{33}$ . Moreover, it is straightforward to check that

(3.2.1) $$ \begin{align} \omega_1=A_{11} \tilde \omega_1+A_{12}\tilde \omega_2+A_{13}\tilde \omega_3,\ \omega_2=A_{21} \tilde \omega_1+A_{23}\tilde \omega_3,\ \omega_3=A_{11} \tilde \omega_1-A_{12}\tilde \omega_2+A_{13}\tilde \omega_3, \end{align} $$

where $A_{i,j}$ is the $(i,j)$ element of the matrix $A^{-1}$ . It follows that $\omega _1+\omega _3$ and $2\omega _2$ have some nice symmetries on the Riemann surface $\mathcal {S}$ and the period matrix $\tau $ has the following properties:

(3.2.2) $$ \begin{align} \tau_{11}=\tau_{33},\tau_{12}=\tau_{23}. \end{align} $$

Now, define the Jacobi map

(3.2.3) $$ \begin{align} \check{J}(\lambda)=\int_{\eta_4}^{\lambda} \check\omega:=\int_{\eta_4}^{\lambda} \begin{pmatrix} \omega_1+\omega_3\\ 2\omega_2 \end{pmatrix}, \end{align} $$

and the corresponding period matrix is

(3.2.4) $$ \begin{align} \check\tau= \begin{pmatrix} \tau_{11}+\tau_{31}& \tau_{12}+\tau_{32}\\ 2\tau_{21}& 2\tau_{22} \end{pmatrix}. \end{align} $$

If one wants to use the above Jacobi map to construct the solution of model RH problem (3.0.11), it suffices to show that the imaginary part of $\check \tau $ is positive definite. In fact, it follows from the Riemann bilinear identity that $\tau _{ij}=\tau _{ji}$ and the properties of $\tau $ in (3.2.2) that $\check \tau $ is a principal minor of $\tau $ up to congruent transformations, which indicates that the imaginary part of $\check \tau $ is also positive definite. Consequently, the quotient space we choose is not $\mathbb C^3\setminus \{\tilde {e}_j,\tau _j\}$ for $j=1,2,3$ but $\mathbb C^2\setminus \{e_j,\check \tau _j\}$ for $j=1,2$ , where $\tilde {e}_j$ is the j-th column of the $3\times 3$ identity matrix and $\check \tau _j$ is the j-th column of the matrix $\check {\tau }$ . Indeed, we have

(3.2.5) $$ \begin{align} \begin{aligned} &\oint_{b_1}\check\omega=\check\tau_1,\ \oint_{b_2}\check\omega=\check\tau_2,\ \oint_{b_3} \check\omega= \begin{pmatrix} \tau_{13}+\tau_{33}\\ 2\tau_{23} \end{pmatrix}=\check\tau_1,\\ &\oint_{a_1}\check\omega=e_1,\ \oint_{a_2}\check\omega=2e_2,\ \oint_{b_3} \check\omega=e_1. \end{aligned} \end{align} $$

Furthermore, it follows that the Jacobi map $\check {J}(z)$ satisfies

(3.2.6) $$ \begin{align} \begin{aligned} &\check{J}_{+}(z)+\check{J}_{-}(z)=0, && z \in \Sigma_3, \\ &\check{J}_{+}(z)+\check{J}_{-}(z)=-e_1, && z \in \Sigma_1, \\ &\check{J}_{+}(z)+\check{J}_{-}(z)=-e_1-2 e_2, && z \in \Sigma_2, \\ &\check{J}_{+}(z)+\check{J}_{-}(z)=-2 e_1-2 e_2, && z \in \Sigma_4, \\ &\check{J}_{+}(z)-\check{J}_{-}(z)=-\check\tau_1, && z \in\left(\eta_2, \eta_3\right), \\ &\check{J}_{+}(z)-\check{J}_{-}(z)=-\check\tau_2, && z \in\left(-\eta_1, \eta_1\right), \\ &\check{J}_{+}(z)-\check{J}_{-}(z)=-\check\tau_1, && z \in\left(-\eta_3,-\eta_2\right), \end{aligned} \end{align} $$

and the Jacobi map on the branch points is half periods, that is,

(3.2.7) $$ \begin{align} \begin{aligned} \check{J}(\eta_4)&=0,\ \check{J}(\eta_3)=-\frac{\check\tau_1}{2},\ \check{J}(\eta_2)=-\frac{\check\tau_1}{2}-\frac{e_1}{2},\ \check{J}(\eta_1)=-\frac{\check\tau_2}{2}-\frac{e_1}{2},\\ \check{J}(-\eta_4)&=0,\ \check{J}(-\eta_3)=-\frac{\check\tau_1}{2},\ \check{J}(-\eta_2)=-\frac{\check\tau_1}{2}-\frac{e_1}{2},\ \check{J}(-\eta_1)=-\frac{\check\tau_2}{2}-\frac{e_1}{2}. \end{aligned} \end{align} $$

On the other hand, based on the formula (3.2.3), for $\lambda \in \mathbb C\setminus \mathbb R$ , one has

(3.2.8) $$ \begin{align} \check{J}(-\lambda)=\check{J}(\lambda)+e_1+e_2, \end{align} $$

which indicates that $\check {J}(\infty _+)=\frac {e_1+e_2}{2}$ . Now, the solution of the model RH problem for $S^{\infty }(\lambda )$ on the $\lambda $ -plane can be written down. Initially, suppose that $\check {\gamma }(\lambda )=\left (\frac {(\lambda ^2-\eta _1^2)(\lambda ^2-\eta _3^2)}{(\lambda ^2-\eta _2^2) (\lambda ^2-\eta _4^2)}\right )^{\frac {1}{4}}$ and similarly assume that

$$ \begin{align*} S^{\infty}(\lambda)=\check{c}\check{\gamma}(\lambda)\begin{pmatrix} \frac{\Theta(\check{J}(\lambda)-\check{d}+\frac{\Omega}{2\pi i};\check\tau )}{\Theta(\check{J}(\lambda)-\check{d};\check\tau )}& \frac{\Theta(-\check{J}(\lambda)-\check{d}+\frac{\Omega}{2\pi i};\check\tau )}{\Theta(-\check{J}(\lambda)-\check{d};\check\tau )} \end{pmatrix}, \end{align*} $$

where we demote the period matrix $\check {\tau }$ on the $\lambda $ -plane comparing to $\hat \tau $ on the z-plane. In fact, it will be proven in the Lemma 3.10 that $\check \tau =\hat \tau $ . Moreover, the parameters $\check {d}$ and $\check {c}$ should be determined. Note that we require the solution of RH problem on the $\lambda $ -plane to have at most fourth rootsingularities at branch points $\pm \eta _j$ for $j=1,\cdots ,4$ , which implies that the zeros of $\Theta (\check {J}(\lambda )-\check {d};\check \tau )$ and $\Theta (-\check {J}(\lambda )-\check {d};\check \tau )$ both only lies at $\pm \eta _{1}$ and $\pm \eta _{3}$ . Thus, it follows from the zeros of the Riemann-Theta function are odd half periods that $\check {d}=\frac {e_1+e_2}{2}$ , and combining with $\check {J}(\infty _+)=\frac {e_1+e_2}{2}$ shows that $\check {c}=\frac {\Theta (0;\check \tau )}{\Theta (\frac {\Omega }{2\pi i};\check \tau )}$ . Recall that $\Omega =(x\Omega _1+\Delta _1,x\Omega _0+\Delta _0)^T$ , thus $S^{\infty }(\lambda )$ exactly satisfies the jump conditions in (3.0.11). Now, we claim that the function $\Theta (\check {J}(\lambda ))$ has precise four simple zeros on the Riemann surface $\mathcal {S}$ , see [Reference Bertola4].

Proof. Integrate $d\ln \vartheta $ along the boundary of the Riemann surface $\mathcal {S}$ denoted by $\delta \mathcal {S}$ as follows

$$ \begin{align*} \begin{aligned} \frac{1}{2\pi i}\oint_{\delta\mathcal S}\frac{d\vartheta(\lambda)}{\vartheta(\lambda)}&=\frac{1}{2\pi i} \sum_{j=1}^{3}\left(\int_{\eta_4}^{\eta_4+a_j}+\int_{\eta_4+a_j}^{\eta_4+a_j+b_j}+\int_{\eta_4+a_j+b_j}^{\eta_4+b_j}+\int_{\eta_4+b_j}^{\eta_4}\right)d\ln \vartheta(\lambda)\\ &=\frac{1}{2\pi i} \sum_{j=1}^{3}\left(\int_{\eta_4}^{\eta_4+a_j}-\int_{\eta_4+b_j}^{\eta_4+a_j+b_j}+\int_{\eta_4+b_j}^{\eta_4}-\int_{\eta_4+a_j+b_j}^{\eta_4+a_j}\right)d\ln \vartheta(\lambda)\\ &=\int_{\eta_4}^{\eta_4+a_1}(\omega_1+\omega_3)+\int_{\eta_4}^{\eta_4+a_2}2\omega_2+\int_{\eta_4}^{\eta_4+a_3}(\omega_1+\omega_3)=4, \end{aligned} \end{align*} $$

where we have used the fact that $d\ln \vartheta (\lambda +b_{1})=d\ln \vartheta (\lambda +b_{3})=-2\pi i d(\check {J}(\lambda ))_1+d\ln \vartheta (\lambda )$ , $d\ln \vartheta (\lambda +b_2)=-2\pi i d(\check {J}(\lambda ))_2+d\ln \vartheta (\lambda )$ and $d(\check {J}(\lambda ))_1=\omega _1+\omega _3,d(\check {J}(\lambda ))_2=2\omega _2$ . Similar to the above computation, integrate $\check {J}(\lambda )d\ln \vartheta (\lambda )$ along the boundary of Riemann surface $\delta \mathcal {S}$ in the following:

$$ \begin{align*} \begin{aligned} \frac{1}{2\pi i}\oint_{\delta\mathcal S}(\check{J}(\lambda))_kd\ln \vartheta(\lambda)&=\frac{1}{2\pi i} \sum_{j=1}^{3}\left(\int_{\eta_4}^{\eta_4+a_j}+\int_{\eta_4+a_j}^{\eta_4+a_j+b_j}+\int_{\eta_4+a_j+b_j}^{\eta_4+b_j}+\int_{\eta_4+b_j}^{\eta_4}\right)(\check{J}(\lambda))_kd\ln \vartheta(\lambda)\\ &=\frac{1}{2\pi i} \sum_{j=1}^{3}\left(\int_{\eta_4}^{\eta_4+a_j}-\int_{\eta_4+b_j}^{\eta_4+a_j+b_j}+\int_{\eta_4+b_j}^{\eta_4}-\int_{\eta_4+a_j+b_j}^{\eta_4+a_j}\right)(\check{J}(\lambda))_kd\ln \vartheta(\lambda)\\ &=\frac{1}{2\pi i} \sum_{j=1}^{3}\int_{\eta_4}^{\eta_4+a_j}\left((\check{J}(\lambda))_kd\ln \vartheta(\lambda)-((\check{J}(\lambda))_k+\hat\tau_{kj})(d\ln \vartheta(\lambda)-2\pi i d(\check{J}(\lambda))_j)\right)\\ &\quad +\frac{1}{2\pi i} \sum_{j=1}^{3}\int_{\eta_4}^{\eta_4+a_j}\left((\check{J}(\lambda))_kd\ln \vartheta(\lambda)-((\check{J}(\lambda))_k+2\pi i \delta_{jk})d\ln \vartheta(\lambda)\right)\\ &=\oint_{a_1}(\check{J}(\lambda))_k(\omega_1+\omega_3)+\oint_{a_2}2(\check{J}(\lambda))_k\omega_2+\oint_{a_3} (\check{J}(\lambda))_k(\omega_1+\omega_3)-\frac{\hat\tau_{kk}}{2}+(d_0)_k. \end{aligned} \end{align*} $$

Finally, according to Lemma 3.4, it is immediate that both $\Theta (\check {J}(\lambda )-\check {d})$ and $\Theta (- \check {J}(\lambda )-\check {d})$ have four simple zeros, that is, $\pm \eta _{1}$ and $\pm \eta _{3}$ . So the solution to the model RH problem (3.0.11) is given by the following theorem.

Theorem 3.5. Define $\check {J}(\lambda )=\int _{\eta _4}^{\lambda }\begin {pmatrix} \omega _1+\omega _3\\ 2\omega _2 \end {pmatrix}$ , where $\omega _j~(j=1,2,3)$ as defined previously are normalized holomorphic differentials on Riemann surface $\mathcal {S}$ , then the corresponding period matrix $\check \tau $ is defined in (3.2.4), and let $\check {d}=\frac {e_1+e_2}{2},\Omega =\begin {pmatrix} x\Omega _1+\Delta _1,x\Omega _0+\Delta _0 \end {pmatrix}^T$ , then the vector valued funtion

(3.2.9) $$ \begin{align} S^{\infty}(\lambda)=\check{\gamma}(\lambda)\frac{\Theta(0;\check\tau)}{\Theta(\frac{\Omega}{2\pi i};\check\tau)} \begin{pmatrix} \frac{\Theta(\check{J}(\lambda)-\check{d}+\frac{\Omega}{2\pi i};\check\tau)}{\Theta(\check{J}(\lambda)-\check{d};\check\tau)}& \frac{\Theta(-\check{J}(\lambda)-\check{d}+\frac{\Omega}{2\pi i};\check\tau)}{\Theta(-\check{J}(\lambda)-\check{d};\check\tau)} \end{pmatrix}, \end{align} $$

solves the RH problem (3.0.11).

Remark 3.7. To finish the proof of Theorem 1.1, one should consider the error estimation of the potential $u(x)$ for $x\to +\infty $ . However, to keep the length of the paper manageable, we omit this step since the detailed discussions are made in [Reference Girotti, Grava, Jenkins and McLaughlin21]. Still, it is necessary to introduce some notations to make our results complete.

Consider the $1$ -form $dp=\frac {\lambda ^4+\alpha \lambda ^2+\beta }{R(\lambda )}d\lambda $ and the Abelian integral $p(\lambda )=\int _{\eta _4}^{\lambda }dp$ , which satisfied

(3.2.10) $$ \begin{align} \begin{aligned} &p_+(\lambda)+p_-(\lambda)=0, && \lambda\in\Sigma_{1,2,3,4}, \\ &p_+(\lambda)-p_-(\lambda)=\Omega_0, && \lambda\in[-\eta_1,\eta_1], \\ &p_+(\lambda)-p_-(\lambda)=\Omega_1, && \lambda\in[-\eta_3,-\eta_2]\cup[\eta_2,\eta_3], \end{aligned} \end{align} $$

and

(3.2.11) $$ \begin{align} p(-\lambda)=-p(\lambda), \quad \lambda\in\mathbb{C}\setminus\mathbb{R}. \end{align} $$

Further define

(3.2.12) $$ \begin{align} P^{\infty}(\lambda):=\frac{1}{2} \begin{pmatrix} (1+\frac{p(\lambda)}{\lambda})S_1^{\infty}+\frac{1}{\lambda}S_{1x}^{\infty} & (1-\frac{p(\lambda)}{\lambda})S_2^{\infty}+\frac{1}{\lambda}S_{2x}^{\infty} \\ (1-\frac{p(\lambda)}{\lambda})S_1^{\infty}-\frac{1}{\lambda}S_{1x}^{\infty} & (1+\frac{p(\lambda)}{\lambda})S_2^{\infty}-\frac{1}{\lambda}S_{2x}^{\infty} \end{pmatrix}, \end{align} $$

which satisfies the following matrix-valued RH problem with same jump matrices as $S^{\infty }(\lambda )$ and

(3.2.13) $$ \begin{align} \begin{aligned} &P^{\infty}(\lambda)=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \mathcal{O}\left(\frac{1}{\lambda}\right),\quad \lambda\to\infty, \\ &P^{\infty}(\lambda) \mathrm{~is~analytic~for~} \lambda\in\mathbb{C}\setminus[-\eta_4,\eta_4] \mathrm{ ~with~a~singularity~at~} \lambda=0. \end{aligned} \end{align} $$

One can proof that $\det P^{\infty }(\lambda )\equiv 1$ for $\lambda \in \mathbb {C}$ .

Now, let the solution of the error vector-valued RH problem defined by

(3.2.14) $$ \begin{align} \mathcal{E}(\lambda)=S(\lambda)(P(\lambda))^{-1}, \end{align} $$

where the global parametrix $P(\lambda )$ is given by

(3.2.15) $$ \begin{align} P(\lambda)=\begin{cases} P^{\infty}(\lambda), & \lambda\in\mathbb{C}\setminus\cup_{j=1,2,3,4}B_{\rho}^{\pm\eta_j}, \\ P^{\diamond}(\lambda), & \lambda\in B_{\rho}^{\diamond}, \end{cases} \end{align} $$

where $\diamond $ traverses all $8$ branch points, that is, $\pm \eta _j$ , $j=1,2,3,4$ and $B_{\rho }^{\diamond }$ denotes the open disc of radius $\rho $ centered at $\diamond $ . We claim that the local parametrix $P^{\pm \eta _j}$ near $\lambda =\pm \eta _j$ , $j=1,2,3,4$ can be described by the modified Bessel function which will finally contribute the term $\mathcal {O}(x^{-1})$ in the asymptotic result of $u(x)$ for $x\to +\infty $ .

Therefore, one can see that

(3.2.16) $$ \begin{align} Y(\lambda)=\mathcal{E}(\lambda)P(\lambda)e^{-xg(\lambda)\sigma_3}f(\lambda)^{-\sigma_3} =\left(\begin{pmatrix}1 & 1\end{pmatrix} + \frac{\mathcal{E}_1(x)}{x\lambda} +\mathcal{O}\left(\frac{1}{\lambda^2}\right)\right)P(\lambda)e^{-xg(\lambda)\sigma_3}f(\lambda)^{-\sigma_3}. \end{align} $$

Recall that the potential $u(x)$ can be reconstructed from the solution of $Y(\lambda )$ as following

$$ \begin{align*} u(x)=2 \frac{\mathrm{d}}{\mathrm{d} x}\left[\lim _{\lambda \rightarrow \infty} {\lambda}\left(Y_1(\lambda; x)-1\right)\right]. \end{align*} $$

Theorem 3.8. As $x\to +\infty $ , the potential function $u(x)$ is subject to the following asymptotic expression

(3.2.17) $$ \begin{align} u(x)=-\left(2\alpha+{\sum_{j=1}^4\eta_j^2}+2\partial_x^2\log\left(\Theta\left(\frac{\Omega}{2\pi i};\check\tau\right)\right)\right)+\mathcal{O}\left(\frac{1}{x}\right), \end{align} $$

where $\alpha $ is associated to a second kind Abel differential and can be calculated in Remark 3.1.

Proof. After a series of deformations of the original RH problem, we have

$$ \begin{align*} Y_1(\lambda)=\left(S_1^{\infty}(x;\lambda)+\frac{(\mathcal{E}_1(x))_1}{x\lambda} +\mathcal{O}\left(\frac{1}{\lambda^2}\right)\right)\frac{e^{-xg(\lambda)}}{f(\lambda)}, \end{align*} $$

where $(\mathcal {E}_1(x))_1$ is the first entry of the vector $\mathcal {E}_1(x)$ given by equation (3.2.16). Moreover, reminding the expression of $g(\lambda )$ in (3.0.1), it follows that

(3.2.18) $$ \begin{align} e^{-xg(\lambda)}=1-\frac{x}{\lambda}\left(\alpha+\frac{\eta_1^2+\eta_2^2+\eta_3^2+\eta_4^2}{2}\right)+\mathcal{O}\left(\frac{1}{\lambda^2}\right). \end{align} $$

Similarly, from the formula of $f(\lambda )$ , one has

(3.2.19) $$ \begin{align} f(\lambda)=1+\frac{f_1}{\lambda}+\mathcal{O}\left(\frac{1}{\lambda^2}\right). \end{align} $$

In addition, recall the linear equations in (3.0.3), and change $\tilde \omega _j$ into $\omega _j~(j=1,2,3)$ , it follows from the Riemann Bilinear relations [Reference Bertola4] that

$$ \begin{align*} 2\pi i\ \sum \mathrm{res}_{\infty_{\pm}}\frac{\omega_1}{z}=\Omega_1,\ 2\pi i\ \sum \mathrm{res}_{\infty_{\pm}}\frac{\omega_2}{z}=\Omega_2,\ 2\pi i\ \sum \mathrm{res}_{\infty_{\pm}}\frac{\omega_3}{z}=\Omega_1. \end{align*} $$

Incorporate with the symmetry in (3.2.8), which implies that as $\lambda \to \infty $

$$ \begin{align*} J(\lambda)=\frac{e_1+e_2}{2}-\frac{\Omega_x}{\lambda}+\mathcal{O}\left(\frac{1}{\lambda^2}\right). \end{align*} $$

Moreover, the expanding of $S_1^{\infty }(\lambda )$ as $\lambda \to \infty $ is

$$ \begin{align*} \begin{aligned} {S}^{\infty}_1(\lambda)&=1-\frac{1}{\lambda}\left[\nabla\log\left(\Theta\left(\frac{\Omega}{2\pi i};\check\tau\right)\right)-\nabla\log(\Theta(0;\check\tau))\right]\cdot \frac{ \Omega_x}{2\pi i}+\mathcal{O}\left(\frac{1}{\lambda^2}\right),\\ &=1-\frac{1}{\lambda}\partial_x\log\left(\Theta\left(\frac{\Omega}{2\pi i};\check\tau\right)\right)+\mathcal{O}\left(\frac{1}{\lambda^2}\right). \end{aligned} \end{align*} $$

Thus, it is obtained that

$$ \begin{align*} Y_1(\lambda)=1-\frac{1}{\lambda}\left[f_1+x\left(\alpha+\frac{1}{2}\sum_{j=1}^4\eta_j^2\right)+\partial_x\log\left(\Theta\left(\frac{\Omega}{2\pi i};\check\tau\right)\right)+\frac{(\mathcal{E}_1(x))_1}{x}\right]+\mathcal{O}\left(\frac{1}{\lambda^2}\right). \end{align*} $$

Consequently, based on the relationship between $u(x)$ and $Y(\lambda )$ , we have

$$ \begin{align*} u(x)=-\left(2\alpha+{\sum_{j=1}^4\eta_j^2}+2\partial_x^2\log\left(\Theta\left(\frac{\Omega}{2\pi i};\check\tau\right)\right)\right) +\mathcal{O}\left(\frac{1}{x}\right).\\[-40pt] \end{align*} $$

Remark 3.9. We now complete the proof of Theorem 3.8 on the z-plane as follows. Similar to the proof in Theorem 3.8, we have

(3.2.20) $$ \begin{align} Y_1(\lambda)= \left(\hat S_1^{\infty}(x;-\lambda^2)+\frac{(\mathcal{E}_1(x))_1}{x\lambda}+\mathcal{O}\left(\frac{1}{\lambda^2}\right)\right)\frac{e^{-xg(\lambda)}}{f(\lambda)}, \end{align} $$

where $S_1^{\infty }(x;-\lambda ^2)$ is given by (3.1.8). Consequently, combining (3.1.8), (3.2.18) with (3.2.19), it gives that

(3.2.21) $$ \begin{align} \begin{aligned} Y_1(\lambda) &=1-\frac{1}{\lambda}\left(f_1+x\left(\alpha+\frac{\sum_{j=1}^4\eta_j^2}{2}\right)\right.\\ &\quad\left.+\left(\nabla\log\left(\Theta\left(\frac{\Omega}{2\pi i}\right);\hat\tau\right)-\nabla\log(\Theta(0;\hat\tau))\right)\cdot \frac{ \Omega_x}{2\pi i} +\frac{(\mathcal{E}_1(x))_1}{x}\right)+\mathcal{O}\left(\frac{1}{\lambda^2}\right), \end{aligned} \end{align} $$

which indicates that for $x\to +\infty $ the genus two KdV soliton gas potential behaves

(3.2.22) $$ \begin{align} u(x)=-\left(2\alpha+{\sum_{j=1}^4\eta_j^2}+2\partial_x^2\log\left(\Theta\left(\frac{\Omega}{2\pi i};\hat\tau\right)\right)\right)+\mathcal{O}\left(\frac{1}{x}\right). \end{align} $$

So far, we almost complete the construction of the soliton gas potential in the regime $x\to +\infty $ . However, there is a minor flaw, which lies in proving the equivalence of the period matrices $\check \tau $ and $\hat \tau $ .

Proof. Recalling the definition of $\tilde \omega _j$ and changing the variable $\lambda $ into $i\sqrt {z}$ , we have the following local expressions

$$ \begin{align*} \tilde \omega_1=\frac{d\lambda}{R(\lambda)}=\frac{i}{2}\frac{dz}{\hat R(z)},\ \tilde \omega_2=\frac{\lambda d\lambda}{R(\lambda)}=-\frac{1}{2}\frac{\sqrt{z}dz}{\hat R(z)},\ \tilde \omega_3=\frac{\lambda^2 d\lambda}{R(\lambda)}=-\frac{i}{2}\frac{z dz}{\hat R(z)}. \end{align*} $$

It follows that after the holomorphic transformation $z=-\lambda ^2$ , the original holomorphic differential $\tilde \omega _{2}$ is not a holomorphic differential in $\hat {\mathcal {S}}$ . Fortunately, by the expressions of $\omega _j$ in terms of $\tilde {\omega }_j$ in (3.2.1), $\omega _1+\omega _2$ and $2\omega _{2}$ are irrelated to $\tilde {\omega }_2$ . In particular,

$$ \begin{align*} \begin{aligned} &\oint_{\hat a_1}(\omega_1+\omega_3)\rvert_{\hat{\mathcal{S}}}=\oint_{ a_1}(\omega_1+\omega_3)\rvert_{\mathcal{S}}=\oint_{ a_3}(\omega_1+\omega_3)\rvert_{\mathcal{S}}=1, \quad \oint_{\hat a_2}2\omega_2\rvert_{\hat{\mathcal{S}}}=\oint_{ a_2}\omega_2\rvert_{\mathcal{S}}=1,\\ &\oint_{\hat a_1}2\omega_2\rvert_{\hat{\mathcal{S}}}=\oint_{ a_1}2\omega_2\rvert_{\mathcal{S}}=\oint_{ a_3}2\omega_2\rvert_{\mathcal{S}}=0, \quad \oint_{\hat a_2}(\omega_1+\omega_3)\rvert_{\hat{\mathcal{S}}}=\frac{1}{2}\oint_{ a_1}(\omega_1+\omega_3)\rvert_{\mathcal{S}}=0. \end{aligned} \end{align*} $$

By the Riemann Bilinear relationship, it indicates that $(\omega _1+\omega _3)\rvert _{\hat {\mathcal {S}}}$ and $2\omega _2\rvert _{\hat {\mathcal {S}}}$ are the normalized holomorphic differentials on $\hat {\mathcal {S}}$ and the corresponding period matrix are equivalence.

3.3 Behavior of the genus two KdV soliton gas potential for $x\to -\infty $

The behavior of the genus two KdV soliton gas potential is quite simple since the jump matrices of the RH problem $Y(\lambda )$ are all converge to identity matrix exponentially when $x\to -\infty $ . To be precise, there exist a fixed constant $c\in \mathbb {R}_+$ such that for $x\to -\infty $ , the boundary behavior of the potential $u(x)$ is given by

(3.3.1) $$ \begin{align} u(x)=\mathcal{O}(e^{-c|x|}). \end{align} $$

4.1 Modulated one-phase wave region

Introduce the critical value $\xi _{crit}^{(1)}$ that is defined by (4.1.12) below, and consider the constraint

$$ \begin{align*} \eta_1^2<\xi<\xi_{crit}^{(1)}. \end{align*} $$

Refine the contours $\Sigma _{1}$ and $\Sigma _2$ as illustrated in Figure 8, by splitting them as follows:

(4.1.1) $$ \begin{align} \Sigma_{1_{\alpha_1}} = (\eta_1, \alpha_1), \quad \text{and} \quad \Sigma_{2_{\alpha_1}} = (-\alpha_1, -\eta_1), \end{align} $$

where $\alpha _1$ is a function of $\xi $ to be determined in (4.1.8) below.

Figure 8. The solid lines represent $\Sigma _{1_{\alpha _1}}$ and $\Sigma _{2_{\alpha _1}}$ , where $\eta _1 < \alpha _1 < \eta _2$ .

Similarly, introduce the g function, denoted as $g_{\alpha _1}$ , which satisfies the following scalar RH problem:

(4.1.2) $$ \begin{align} \begin{aligned} &g_{\alpha_{1,+}}(\lambda)+g_{\alpha_{1,-}}(\lambda)+8\lambda^3-8\xi\lambda=0, &&\lambda\in \Sigma_{1_{\alpha_{1}}}\cup \Sigma_{2_{\alpha_{1}}},\\ &g_{\alpha_{1,+}}(\lambda)-g_{\alpha_{1,-}}(\lambda)={\Omega_{\alpha_{1}}}, &&\lambda\in [-\eta_1,\eta_1],\\ &g_{\alpha_{1}}(\lambda)=\mathcal{O}\left(\frac{1}{\lambda}\right), && \lambda\to \infty. \end{aligned} \end{align} $$

To further deform the RH problem (4.0.1), it is required that the $g_{\alpha _1}$ function satisfies the following properties:

(4.1.3) $$ \begin{align} \begin{aligned} &g_{\alpha_1}(\lambda)+4\lambda^3-4\xi\lambda=(\lambda\pm\alpha_{1})^{\frac{3}{2}}, &&\lambda\to\pm\alpha_{1},\\ &\operatorname{\mathrm{Re}}\left(g_{\alpha_{1}}(\lambda)+4\lambda^3-4\xi\lambda\right)>0, &&\lambda \in (\alpha_{1},\eta_2)\cup (\eta_3,\eta_4),\\ &\operatorname{\mathrm{Re}}\left(g_{\alpha_{1}}(\lambda)+4\lambda^3-4\xi\lambda\right)<0, &&\lambda \in (-\eta_2,-\alpha_{1})\cup (-\eta_4,-\eta_3),\\ &-i(g_{\alpha_1,+}(\lambda)-g_{\alpha_1,-}(\lambda)) \text{ is real-valued and monotonically increasing},&& \lambda\in\Sigma_{1_{\alpha_{1}}},\\ &-i(g_{\alpha_1,+}(\lambda)-g_{\alpha_1,-}(\lambda)) \text{ is real-valued and monotonically decreasing}, && \lambda\in\Sigma_{2_{\alpha_{1}}}.\\ \end{aligned} \end{align} $$

The $g_{\alpha _1}$ can be derived from its derivative $ g^{\prime }_{\alpha _1} $ according to the uniqueness of $ g_{\alpha _1} $ . Here the $ g^{\prime }_{\alpha _1}d\lambda $ can be viewed as the second kind Abel differential on Riemann surface of genus one.

Define

(4.1.4) $$ \begin{align} g^{\prime}_{\alpha_1}(\lambda)=-12\lambda^2+4\xi+12\frac{Q_{\alpha_1,2}(\lambda)}{R_{\alpha_1}(\lambda)}-4\xi\frac{Q_{\alpha_1,1}(\lambda)}{R_{\alpha_1}(\lambda)}, \end{align} $$

with

$$ \begin{align*} R_{\alpha_1}(\lambda)=\sqrt{(\lambda^2-\eta_1^2)(\lambda^2-\alpha_{1}^2)}, \end{align*} $$

which is analytic for $\mathbb C\setminus \left (\Sigma _{1_{\alpha _1}}\cup \Sigma _{2_{\alpha _1}}\right )$ , and takes positive real value for $\lambda>\alpha _{1}$ . From the definition of $g_{\alpha _1}'(\lambda )$ in (4.1.4), we can derive the expression of $g_{\alpha _1}(\lambda )$ :

(4.1.5) $$ \begin{align} g_{\alpha_{1}}(\lambda)=-4\lambda^3+4\xi\lambda+12\int_{\alpha_{1}}^{\lambda}\frac{Q_{\alpha_1,2}(\zeta)}{R_{\alpha_1}(\zeta)}d\zeta-4\xi\int_{\alpha_{1}}^{\lambda}\frac{Q_{\alpha_1,1}(\zeta)}{R_{\alpha_1}(\zeta)}d\zeta. \end{align} $$

On the other hand, suppose that

(4.1.6) $$ \begin{align} Q_{\alpha_1,1}(\lambda)=\lambda^2+c_{\alpha_1,1},\quad Q_{\alpha_1,2}(\lambda)=\lambda^4-\frac{1}{2}(\alpha_1^2+\eta_1^2)\lambda^2+c_{\alpha_1,2}, \end{align} $$

where

(4.1.7) $$ \begin{align} c_{\alpha_1,1}=-\alpha_1^2+\alpha_1^2\frac{E(m_{\alpha_1})}{K(m_{\alpha_1})},\quad c_{\alpha_1,2}=\frac{1}{3} \alpha_1^2\eta_1^2+\frac{1}{6}(\alpha_1^2+\eta_1^2)c_{\alpha_1,1},\quad m_{\alpha_1}=\frac{\eta_1}{\alpha_1}. \end{align} $$

Here $K(m_{\alpha _1})$ and $E(m_{\alpha _1})$ are the first and the second kind complete elliptic integral, respectively, that is, $K(m)=\int _{0}^{\frac {\pi }{2}} \frac {d\vartheta }{\sqrt {1-m^2\sin \vartheta ^2}}$ and $E(m)=\int _{0}^{\frac {\pi }{2}} {\sqrt {1-m^2\sin \vartheta ^2}}{d\vartheta }$ . The $Q_{\alpha _1,1}$ and $Q_{\alpha _1,2}$ are determined by the conditions in (4.1.2), and the first property in (4.1.3) about the behavior near $\pm \alpha _{1}$ implies that the parameter $\alpha _1$ is determined by

(4.1.8) $$ \begin{align} \xi=3\frac{Q_{\alpha_1,2}(\pm \alpha_{1})}{Q_{\alpha_1,1}(\pm \alpha_{1})}=\frac{1}{2}(\alpha_{1}^2+\eta_1^2)+(\alpha_{1}^2-\eta_1^2)\frac{K(m_{\alpha_1})}{E(m_{\alpha_1})}, \end{align} $$

which states that the $\alpha _1$ is modulated by $\xi $ . Indeed, rewrite (4.1.8) as Whitham velocity

(4.1.9) $$ \begin{align} \xi=\frac{x}{4t}=\frac{\eta_1^2}{2}\left[1+\frac{1}{m_{\alpha_1}^2}+2\left(\frac{1}{m_{\alpha_1}^2}-1\right)\frac{K(m_{\alpha_1})}{E(m_{\alpha_1})}\right]:=\frac{\eta_1^2}{2}W(m_{\alpha_1}). \end{align} $$

It follows that $\partial _{\alpha _1}W(m_{\alpha _1})>0$ for $\eta _1<\alpha _{1}<+\infty $ , since the Whitham equation associated with the Whitham velocity (4.1.9) is strictly hyperbolic [Reference Levermore30]. Alternatively, one obtains the inequality $\partial _{m_{\alpha _{1}}}W(m_{\alpha _{1}})<0$ by direct calculation:

$$ \begin{align*}\begin{aligned} \partial_{m_{\alpha_1}} W(m_{\alpha_1}) = & -\frac{2}{m_{\alpha_1}^3} - \frac{4K(m_{\alpha_1})}{m_{\alpha_1}^3 E(m_{\alpha_1})} \\ & + 2\frac{1 - m_{\alpha_1}^2}{m_{\alpha_1}^2} \frac{ \left( \frac{E(m_{\alpha_1})}{m_{\alpha_1}(1 - m_{\alpha_1}^2)} - \frac{K(m_{\alpha_1})}{m_{\alpha_1}} \right) E(m_{\alpha_1}) - \frac{E(m_{\alpha_1}) - K(m_{\alpha_1})}{m_{\alpha_1}} K(m_{\alpha_1}) }{E(m_{\alpha_1})^2} \\ = & \frac{2K(m_{\alpha_1}) \left[ (1 - m_{\alpha_1}^2) K(m_{\alpha_1}) - 2(2 - m_{\alpha_1}^2) E(m_{\alpha_1}) \right]}{E(m_{\alpha_1})^2 m_{\alpha_1}^3} < 0, \end{aligned} \end{align*} $$

where the first equality follows from

(4.1.10) $$ \begin{align} K'(m)=\frac{E(m)}{m(1-m^2)}-\frac{K(m)}{m},\quad E'(m)=\frac{E(m)-K(m)}{m}, \end{align} $$

and the last inequality is given by $K(m)<\frac {E(m)}{\sqrt {1-m^2}}$ . Therefore the chain rule gives

(4.1.11) $$ \begin{align} \partial_{\alpha_1}W(m_{\alpha_1})=\frac{\eta_1^2}{2}\partial_{m_{\alpha_{1}}}W(m_{\alpha_{1}})\partial_{\alpha_1}m_{\alpha_1}>0,\quad \eta_1<\alpha_1<+\infty. \end{align} $$

According to the expansions of elliptic functions, it is immediate that

$$ \begin{align*} \begin{aligned} \lim_{\alpha_1\to\eta_1}\xi=\eta_1^2, \ m_{\alpha_1}\to1,\ \text{and}\ \lim_{\alpha_1\to+\infty}\xi=+\infty, \ m_{\alpha_1}\to0. \end{aligned} \end{align*} $$

Define

(4.1.12) $$ \begin{align} \xi_{crit}^{(1)}:=3\frac{Q_{\alpha_1,2}(\eta_2)}{Q_{\alpha_1,1}(\eta_2)}=\frac{1}{2}(\eta_1^2+\eta_2^2)+(\eta_2^2-\eta_1^2)\frac{K(m_{\eta_{2}})}{E(m_{\eta_{2}})}, \end{align} $$

where $m_{\eta _{2}}=\frac {\eta _1}{\eta _2}$ . Consequently, (4.1.8) defines $\alpha _{1}$ as a monotone increasing function of $\xi $ in $[\eta _1^2, \xi _{crit}^{(1)}]$ by the implicit function theorem.

If the relationship (4.1.8) is established, we can verify the first condition in (4.1.3). Indeed, together with (4.1.7), rewrite the function $g^{\prime }_{\alpha _{1}}(\lambda )$ as

$$ \begin{align*} \begin{aligned} g^{\prime}_{\alpha_1}(\lambda)&=-12\lambda^2+4\xi+12\frac{Q_{\alpha_1,2}(\lambda)-Q_{\alpha_1,2}(\alpha_1)}{R_{\alpha_1}(\lambda)}-4\xi\frac{Q_{\alpha_1,1}(\lambda)-Q_{\alpha_1,1}(\alpha_1)}{R_{\alpha_1}(\lambda)}\\ &=-12\lambda^2+4\xi+12\frac{(\lambda^2-\alpha_1^2)}{R_{\alpha_{1}}(\lambda)}\left[\lambda^2-\left(\frac{\eta_1^2-\alpha_{1}^2}{2}+\frac{\xi}{3}\right)\right]. \end{aligned} \end{align*} $$

It follows that $g^{\prime }_{\alpha _1}(\lambda )$ behaves like $(\lambda \pm \alpha _{1})^{\frac {1}{2}}$ as $\lambda \to \pm \alpha _{1}$ , and $(\lambda \pm \eta _{1})^{-\frac {1}{2}}$ as $\lambda \to \pm \eta _{1}$ , which is coincided with (4.1.3). Before verifying the other conditions in (4.1.3), we will determine ${\Omega }_{\alpha _{1}}$ in the jump condition (4.1.2) and introduce the following Riemann surface of genus one related to $R_{\alpha _1}(\lambda )$ :

$$ \begin{align*} \mathcal{S}_{\alpha_1}:=\{(\lambda,\eta)|\eta^2=(\lambda^2-\eta_1^2)(\lambda^2-\alpha_1^2)\}, \end{align*} $$

with the A, B-cycles defined in Figure 9. Define

$$ \begin{align*} \omega_{\alpha_1}=-\frac{ \alpha_1 d\zeta}{4K(m_{\alpha_1})R_{\alpha_1}(\zeta)}, \end{align*} $$

as the normalized holomorphic differential on $\mathcal {S}_{\alpha _1}$ with $\oint _{A}\omega _{\alpha _1}=1$ and $\oint _{B}\omega _{\alpha _1}=\tau _{\alpha _1}$ . Reminding the definition of $g_{\alpha _{1}}(\lambda )$ and jump conditions in (4.1.2), it implies that

(4.1.13) $$ \begin{align} \begin{aligned} &24\int_{\alpha_{1}}^{\eta_1}\frac{Q_{\alpha_1,2}(\zeta)}{R_{\alpha_1}(\zeta)}d\zeta-8\xi\int_{\alpha_{1}}^{ \eta_1}\frac{Q_{\alpha_1,1}(\zeta)}{R_{\alpha_1}(\zeta)}d\zeta={\Omega}_{\alpha_{1}},\\ &\int_{-\eta_1}^{\eta_1}\frac{Q_{\alpha_1,2}(\zeta)}{R_{\alpha_1}(\zeta)}d\zeta=\int_{-\eta_1}^{\eta_1}\frac{Q_{\alpha_1,1}(\zeta)}{R_{\alpha_1}(\zeta)}d\zeta=0.\\ \end{aligned} \end{align} $$

In fact, $\frac {Q_{\alpha _1,2}(\lambda )}{R_{\alpha _1}(\lambda )}d\lambda $ and $\frac {Q_{\alpha _1,1}(\lambda )}{R_{\alpha _1}(\lambda )}d\lambda $ are the normalized Abel differentials of the second kind and by using the Riemann Bilinear relations [Reference Bertola4], it follows from (4.1.13) that

(4.1.14) $$ \begin{align} \begin{aligned} {\Omega}_{\alpha_{1}}&=-12\int_{B}\frac{Q_{\alpha_1,2}(\zeta)}{R_{\alpha_1}(\zeta)}d\zeta+4\xi\int_{B}\frac{Q_{\alpha_1,1}(\zeta)}{R_{\alpha_1}(\zeta)}d\zeta\\ &=2\pi i (4\xi \mathrm{res}_{\lambda=\pm \infty}\lambda^{-1} \omega_{\alpha_1}-4 \mathrm{res}_{\lambda=\pm \infty}\lambda^{-3} \omega_{\alpha_1}) =2\pi i \alpha_1 \frac{\alpha_1^2+\eta_{1}^2-2\xi}{K(m_{\alpha_{1}})}\in i\mathbb R. \end{aligned} \end{align} $$

Figure 9. The Riemann surface $\mathcal {S}_{\alpha _1}$ of genus one and its basis of cycles.

In order to recover the potential function $u(x,t)$ , it is necessary to formulate $\partial _x(tg^{\prime }_{\alpha _{1}}(\lambda ))$ and $\partial _x(t{\Omega }_{\alpha _{1}})$ . Thus we have the following lemma.

Lemma 4.2. According the representation of $g^{\prime }_{\alpha _{1}}(\lambda )$ in (4.1.4) and ${\Omega }_{\alpha _{1}}$ in (4.1.13), the following two identities hold

(4.1.15) $$ \begin{align} \begin{aligned} \partial_x(tg^{\prime}_{\alpha_{1}}(\lambda))&=1-\frac{Q_{\alpha_1,1}(\lambda)}{R_{\alpha_{1}}(\lambda)},\\ \partial_x(t{\Omega}_{\alpha_{1}})&=-\frac{\pi i\alpha_{1}}{K(m_{\alpha_1})}. \end{aligned} \end{align} $$

Proof. The equation (4.1.4) shows that

$$ \begin{align*} \partial_x(tg^{\prime}_{\alpha_{1}}(\lambda))=1-\frac{Q_{\alpha_{1},1} (\lambda)}{R_{\alpha_{1}}(\lambda)}+\partial_{\alpha_1} \left(12t\frac{Q_{\alpha_1,2}(\lambda)}{R_{\alpha_{1}}(\lambda)} -x\frac{Q_{\alpha_1,1}(\lambda)}{R_{\alpha_{1}}(\lambda)}\right)\partial_{x}{\alpha_{1}}. \end{align*} $$

It suffices to show that $\partial _{\alpha _1}\left (12t\frac {Q_{\alpha _1,2}(\lambda )}{R_{\alpha _{1}}(\lambda )}d\lambda -x\frac {Q_{\alpha _1,1}(\lambda )}{R_{\alpha _{1}}(\lambda )}d\lambda \right )\equiv 0$ . Indeed, the term $12t\frac {Q_{\alpha _1,2}(\lambda )}{R_{\alpha _{1}}(\lambda )}d\lambda -x\frac {Q_{\alpha _1,1}(\lambda )}{R_{\alpha _{1}}(\lambda )}d\lambda $ is a holomorphic differential and its integral along the A-cycle is zero. Thus by the Riemann bilinear relation, it is identically zero. On the other hand, from the equation (4.1.13), it can be derived that

$$ \begin{align*} \partial_x(t{\Omega}_{\alpha_{1}})=-\partial_x\left(\oint_Bt(g^{\prime}_{\alpha_{1}}(\lambda)+12\lambda^2-4\xi)\right)=\oint_B\frac{Q_{\alpha_1,1}(\lambda)}{R_{\alpha_{1}}(\lambda)}=-\frac{\pi i\alpha_{1}}{K(m_{\alpha_1})}. \end{align*} $$

By the same way in discussing the asymptotic behavior of $u(x,0)$ as $x\to +\infty $ in Section 3, introduce

$$ \begin{align*} T_{\alpha_{1}}(\lambda)=Y(\lambda){e^{tg_{\alpha_{1}}(\lambda)\sigma_3}}f_{\alpha_{1}}(\lambda)^{\sigma_3}, \end{align*} $$

where $f_{\alpha _{1}}(\lambda )$ satisfies the following scalar RH problem:

(4.1.16) $$ \begin{align} \begin{aligned} &f_{\alpha_{1},+}(\lambda)f_{\alpha_{1},-}(\lambda)={r(\lambda)},&&\lambda\in(\eta_{1},\alpha_{1}),\\ &f_{\alpha_{1},+}(\lambda)f_{\alpha_{1},-}(\lambda)=\frac{1}{r(\lambda)},&&\lambda\in(-\alpha_{1},-\eta_{1}),\\ &\frac{f_{\alpha_{1},+}(\lambda)}{f_{\alpha_{1},-}(\lambda)}=e^{{\Delta}_{\alpha_{1}}},&&\lambda\in [-\eta_1,\eta_1],\\ &f_{\alpha_{1}}(\lambda)=1+\mathcal{O}\left(\frac{1}{\lambda}\right),&&\lambda\to\infty. \end{aligned} \end{align} $$

It is immediate that the function $f_{\alpha _{1}}(\lambda )$ is formulated as

(4.1.17) $$ \begin{align} \begin{aligned} f_{\alpha_{1}}(\lambda)=&\exp\left(\frac{R_{\alpha_{1}}(\lambda)}{2\pi i}\left[ \int_{\eta_1}^{\alpha_1}\frac{\log{r(\zeta)}}{R_{\alpha_{1},+}(\zeta)(\zeta-\lambda)}d\zeta +\int_{-\alpha_1}^{-\eta_1}\frac{\log\frac{1}{r(\zeta)}}{R_{\alpha_{1},+}(\zeta)(\zeta-\lambda)}d\zeta +\int_{-\eta_1}^{\eta_1}\frac{{\Delta}_{\alpha_{1}}}{R_{\alpha_{1}}(\zeta)(\zeta-\lambda)}d\zeta\right]\right) \end{aligned}, \end{align} $$

in which the normalization condition indicates that $ {\Delta }_{\alpha _{1}}=\frac {\alpha _{1}}{K(m_{\alpha _{1}})}\int _{\eta _1}^{\alpha _1}\frac {\log {r(\zeta )}}{R_{\alpha _{1},+}(\zeta )}d\zeta $ . Thus the row vector $T_{\alpha _{1}}(\lambda )$ obeys the following RH problem:

(4.1.18) $$ \begin{align} \begin{aligned} T_{\alpha_{1},+}(\lambda)=T_{\alpha_{1},-}(\lambda) \begin{cases} \begin{aligned} &\begin{pmatrix} e^{t(g_{\alpha_{1},+}(\lambda)-g_{\alpha_{1},-}(\lambda))} \frac{f_{\alpha_{1},+}(\lambda)}{f_{\alpha_{1},-}(\lambda)} & -i \\ 0 & e^{-t(g_{\alpha_{1},+}(\lambda)-g_{\alpha_{1},-}(\lambda))} \frac{f_{\alpha_{1},-}(\lambda)}{f_{\alpha_{1},+}(\lambda)} \end{pmatrix} , & \lambda \in \Sigma_{1_{\alpha_{1}}},\\ & \begin{pmatrix} e^{t(g_{\alpha_{1},+}(\lambda)-g_{\alpha_{1},-}(\lambda))} \frac{f_{\alpha_{1},+}(\lambda)}{f_{\alpha_{1},-}(\lambda)} & 0 \\ i & e^{-t(g_{\alpha_{1},+}(\lambda)-g_{\alpha_{1},-}(\lambda))} \frac{f_{\alpha_{1},-}(\lambda)}{f_{\alpha_{1},+}(\lambda)} \end{pmatrix}, & \lambda \in \Sigma_{2_{\alpha_{1}}}, \\ & \begin{pmatrix} e^{t {\Omega}_{\alpha_{1}}+{\Delta}_{\alpha_{1}}} & 0 \\ 0 & e^{-t {\Omega}_{\alpha_{1}}-{\Delta}_{\alpha_{1}}} \end{pmatrix}, & \lambda \in\left[-\eta_1, \eta_1\right], \\ & \begin{pmatrix} 1 & \frac{-ir(\lambda)}{ f^2(\lambda)}e^{-2t(g(\lambda)+4\lambda^3-4\xi\lambda)} \\ 0 & 1 \end{pmatrix}, & \lambda \in[\alpha,\eta_2]\cup\Sigma_{3}, \\ & \begin{pmatrix} 1 & 0 \\ {ir(\lambda)f^2(\lambda)}e^{2t(g(\lambda)+4\lambda^3-4\xi\lambda)} & 1 \end{pmatrix}, & \lambda \in[-\eta_2,\alpha]\cup\Sigma_{4},\\ & \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, & \lambda \in[\eta_2,\eta_3]\cup[-\eta_3,-\eta_2], \end{aligned} \end{cases} \\ \end{aligned} \end{align} $$

and

$$ \begin{align*} T_{\alpha_{1}}(\lambda)= \begin{pmatrix} 1&1 \end{pmatrix}+\mathcal{O}\left(\frac{1}{\lambda}\right),\quad \lambda\to\infty. \end{align*} $$

Similarly, factorize the RH problem (4.1.18) by introducing

(4.1.19) $$ \begin{align} S_{\alpha_{1}}(\lambda)= \begin{cases} T_{\alpha_{1}}(\lambda)\begin{pmatrix} 1 & 0 \\ \frac{if_{\alpha_{1}}^2(\lambda)}{\hat r(\lambda)}e^{2t(g_{\alpha_{1}}(\lambda)+4\lambda^3-4\xi\lambda)} & 1 \end{pmatrix}, & \mathrm{inside~the~contour}~\mathcal{C}_{1,\alpha_{1}}, \\ T_{\alpha_{2}}(\lambda)\begin{pmatrix} 1 & \frac{-i}{\hat r(\lambda)f_{\alpha_{1}}^2(\lambda)}e^{-2t(g_{\alpha_{1}}(\lambda)+4\lambda^3-4\xi\lambda)} \\ 0 & 1 \end{pmatrix}, & \mathrm{inside~the~contour}~\mathcal{C}_{2,\alpha_{1}}, \\ T_{\alpha_{2}}(\lambda), & \mathrm{elsewhere}, \end{cases} \end{align} $$

where $\mathcal {C}_{j,\alpha _{1}}$ are around $\Sigma _{j_{\alpha _{1}}}$ for $j=1$ and $j=2$ , respectively. Moreover, the contours and jump matrices for $S_{\alpha _{1}}$ are depicted in Figure 10.

Figure 10. The contours and the jump matrices for $ S_{\alpha _{1}}(\lambda ) $ : the gray entries in the matrices vanish exponentially as $ t \to +\infty $ , and the gray contours also vanish as $ t \to +\infty $ .

In order to transform the RH problem for $S_{\alpha _{1}}(\lambda )$ into a model problem, the other properties of $g_{\alpha _{1}}(\lambda )$ in (4.1.3) should be illustrated.

Lemma 4.3. The following inequalities are established:

(4.1.20) $$ \begin{align} \begin{aligned} &\operatorname{\mathrm{Re}}\left(g_{\alpha_{1}}(\lambda)+4\lambda^3-4\xi\lambda\right)>0, &&\lambda\in\Sigma_{3}\cup(\alpha_1,\eta_2)\cup\mathcal{C}_{2,\alpha_{1}}\setminus\{-\eta_1,-\alpha_1\},\\ &\operatorname{\mathrm{Re}}\left(g_{\alpha_{1}}(\lambda)+4\lambda^3-4\xi\lambda\right)<0, &&\lambda\in\Sigma_{4}\cup(-\eta_2,-\alpha_1)\cup\mathcal{C}_{1,\alpha_{1}}\setminus\{\eta_1,\alpha_1\}.\\ \end{aligned} \end{align} $$

Proof. Recall that

$$ \begin{align*} g^{\prime}_{\alpha_{1}}(\lambda)+12\lambda^2-4\xi=12\frac{(\lambda^2-\alpha_1^2)}{R_{\alpha_{1}}(\lambda)}\left[\lambda^2-\left(\frac{\eta_1^2-\alpha_{1}^2}{2}+\frac{\xi}{3}\right)\right]. \end{align*} $$

For $\lambda \in (\eta _1,\alpha _{1})$ , one has

(4.1.21) $$ \begin{align} g^{\prime}_{\alpha_{1},+}(\lambda)+12\lambda^2-4\xi=12\frac{i\sqrt{\alpha_1^2-\lambda^2}}{\sqrt{\lambda^2-\eta_1^2}}\left[\lambda^2-\left(\frac{\eta_1^2-\alpha_{1}^2}{2}+\frac{\xi}{3}\right)\right], \end{align} $$

which indicates that $g_{\alpha _{1},+}$ is purely imaginary and the normalization condition in (4.1.13) implies the right hand side of the equation (4.1.21) has a nonnegative root in $[0,\eta _1]$ . Thus ${-i(g^{\prime }_{\alpha _{1},+}(\lambda )+12\lambda ^2-4\xi )>0}$ and together with Cauchy-Riemann equation, it follows that

$$ \begin{align*} \operatorname{\mathrm{Re}}(g_{\alpha_{1},+}(\lambda)+4\lambda^3-4\xi\lambda)<0, \end{align*} $$

for $\lambda \in \mathbb {C}^{+}\cap \mathcal {C}_{1,\alpha _{1}}\setminus \{\eta _1,\alpha _1\}$ . Similarly, the signature of $\operatorname {\mathrm {Re}}\left (g_{\alpha _{1}}(\lambda )+4\lambda ^3-4\xi \lambda \right )$ on $\lambda \in \mathcal {C}_{2,\alpha _{1}}\setminus \{-\alpha _1,-\eta _1\}$ can also be proved.

For $\lambda \in (\alpha _{1},\eta _2)\cup \Sigma _{3}$ , one has

(4.1.22) $$ \begin{align} g^{\prime}_{\alpha_{1}}(\lambda)+12\lambda^2-4\xi=12\frac{\sqrt{\lambda^2-\alpha_1^2}}{\sqrt{\lambda^2-\eta_1^2}}\left[\lambda^2-\left(\frac{\eta_1^2-\alpha_{1}^2}{2}+\frac{\xi}{3}\right)\right]. \end{align} $$

Since the nonnegative root of the right hand side of the equation (4.1.21) lives in $[0,\eta _1]$ , it is immediate that $( g^{\prime }_{\alpha _{1}}(\lambda )+12\lambda ^2-4\xi )>0$ . Consequently, by the definition of $g_{\alpha _{1}}$ in (4.1.5), it follows $\operatorname {\mathrm {Re}}\left (g_{\alpha _{1}}(\lambda )+4\lambda ^3-4\xi \lambda \right )>0$ . In a similar way, the inequality on $(-\eta _2,-\alpha _{1})\cup \Sigma _{4}$ can be obtained.

As a result, as $t\to +\infty $ , the jump matrices of the RH problem for $S_{\alpha _{1}}(\lambda )$ restricted on the gray contours in Figure 10 converge to identity matrix exponentially outside the points $\pm \alpha _{1}$ and $\pm \eta _1$ . So we arrive at the model RH problem for $S^{\infty }_{\alpha _{1}}(\lambda )$ as follows:

(4.1.23) $$ \begin{align} {S}_{\alpha_{1},+}^{\infty}(\lambda)={S}_{\alpha_{1},-}^{\infty}(\lambda) \begin{cases}{\begin{pmatrix} e^{t {\Omega_{\alpha_{1}}}+{\Delta}_{\alpha_{1}}} & 0 \\ 0 & e^{-t {\Omega_{\alpha_{1}}}-{\Delta}_{\alpha_{1}}} \end{pmatrix}}, & \lambda \in[-\eta_1, \eta_1], \\ {\begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix}}, & \lambda \in \Sigma_{1_{\alpha_1}}=(\eta_1,\alpha_{1}), \\ {\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}}, & \lambda \in \Sigma_{ 2_{\alpha_1}}=(-\alpha_{1},-\eta_{1}),\end{cases} \end{align} $$

and

$$ \begin{align*} S^{\infty}_{\alpha_{1}}(\lambda)\to\begin{pmatrix} 1&1 \end{pmatrix}+\mathcal{O}\left(\frac{1}{\lambda}\right),\quad \lambda\to\infty, \end{align*} $$

whose solution can be expressed explicitly by

(4.1.24) $$ \begin{align} S^{\infty}_{\alpha_{1}}(\lambda)=\gamma_{\alpha_{1}}(\lambda)\frac{\vartheta_3(0;2\tau_{\alpha_{1}})}{\vartheta_3(\frac{t{\Omega_{\alpha_{1}}}+{\Delta_{\alpha_{1}}}}{2\pi i};2\tau_{\alpha_{1}})}\begin{pmatrix} \frac{\vartheta_3\left(2J_{\alpha_{1}}(\lambda)+\frac{t\Omega_{\alpha_{1}}+{\Delta_{\alpha_{1}}}}{2\pi i}-\frac{1}{2};2\tau_{\alpha_{1}}\right)}{\vartheta_3\left(2J_{\alpha_{1}}(\lambda)-\frac{1}{2};2\tau_{\alpha_{1}}\right)} & \frac{\vartheta_3\left(-2J_{\alpha_{1}}(\lambda)+\frac{t\Omega_{\alpha_{1}}+{\Delta_{\alpha_{1}}}}{2\pi i}-\frac{1}{2};2\tau_{\alpha_{1}}\right)}{\vartheta_3\left(-2J_{\alpha_{1}}(\lambda)-\frac{1}{2};2\tau_{\alpha_{1}}\right)} \end{pmatrix}, \end{align} $$

where $\gamma _{\alpha _{1}}(\lambda )=\left (\frac {\lambda ^2-\eta _{1}^2}{\lambda ^2-\alpha _{1}^2}\right )^{\frac {1}{4}}$ , $J_{\alpha _{1}}(\lambda )=\int _{\alpha _{1}}^\lambda \omega _{\alpha _1}$ and $\tau _{\alpha _{1}}=\oint _{B}\omega _{\alpha _1}$ . More precisely, $J_{\alpha _{1},+}-J_{\alpha _{1},-}=-\tau _{\alpha _{1}}$ for $\lambda \in [-\eta _{1},\eta _{1}]$ , $J_{\alpha _{1},+}+J_{\alpha _{1},-}=0~\mod (\mathbb {Z})$ for $\lambda \in \Sigma _{ 2_{\alpha _1}}\cup \Sigma _{ 1_{\alpha _1}}$ and $J_{\alpha _{1}}(\infty )=-\frac {1}{4}$ . Thus it is immediate that (4.1.24) solves the RH problem (4.1.23).

Theorem 4.4. For $\xi =\frac {x}{4t}$ , in the region $\eta _1^2<\xi <\xi _{crit}^{(1)}$ , the large-time asymptotic behavior of the solution to the KdV equation with genus two soliton gas potential is described by

(4.1.25) $$ \begin{align} u(x,t)=\alpha_{1}^2-\eta_{1}^2-2\alpha_{1}^2\frac{E(m_{\alpha_{1}})}{K(m_{\alpha_{1}})}-2{\partial_x^2}\log\vartheta_3\left(\frac{\alpha_{1}}{2K(m_{\alpha_{1}})}(x-2(\alpha_{1}^2+\eta_{1}^2)t+\phi_{\alpha_{1}});2\tau_{\alpha_{1}}\right)+\mathcal{O}\left(\frac{1}{t}\right), \end{align} $$

where

$$ \begin{align*} \phi_{\alpha_{1}}=\int_{\alpha_{1}}^{\eta_{1}}\frac{\log r(\zeta)}{R_{\alpha_{1},+}(\zeta)}\frac{d\zeta}{\pi i}, \end{align*} $$

and the parameter $\alpha _{1}$ is determined by (4.1.8). Alternatively, the expression (4.1.25) can also be rewritten as

(4.1.26) $$ \begin{align} u(x,t)=\alpha_{1}^2-\eta_{1}^2-2\alpha_{1}^2 \mathrm{dn}^2\left(\alpha_{1}(x-2(\alpha_{1}^2+\eta_{1}^2)t+\phi_{\alpha_{1}}) +K(m_{\alpha_{1}}); m_{\alpha_{1}}\right)+\mathcal{O}\left(\frac{1}{t}\right), \end{align} $$

where $\mathrm {dn}(z; m)$ is the Jacobi elliptic function with modulus $m_{\alpha _{1}}=\frac {\eta _{1}}{\alpha _{1}}$ .

Proof. By the same way as the analysis of the genus two KdV soliton gas potential $u(x,0)$ for $x\to +\infty $ , take

$$ \begin{align*} Y_1(\lambda)=\left(S_{\alpha_{1},1}^{\infty}(\lambda) +\frac{(\mathcal{E}_{\alpha_1,1}(x,t))_1}{t\lambda}+\mathcal{O}\left(\frac{1}{\lambda^2}\right)\right) e^{-tg_{\alpha_{1}}(\lambda)}f_{\alpha_{1}}(\lambda)^{-1}, \end{align*} $$

in which the $f_{\alpha _{1}}(\lambda )$ has the asymptotics

$$ \begin{align*} f_{\alpha_{1}}(\lambda)=1+\frac{f_{\alpha_{1}}^{(1)}(\alpha_{1},\eta_{1})}{\lambda}+\mathcal{O}\left(\frac{1}{\lambda^2}\right), \end{align*} $$

with

$$ \begin{align*} f_{\alpha_{1}}^{(1)}(\alpha_{1},\eta_{1})=\int_{\alpha_{1}}^{\eta_{1}}\frac{\zeta^2\log r(\zeta)}{R_{\alpha_{1}}(\zeta)}\frac{d\zeta}{\pi i}-{\Delta_{\alpha_{1}}}\int_{-\eta_{1}}^{\eta_{1}}\frac{\zeta^2}{R_{\alpha_{1}}(\zeta)}\frac{d\zeta}{2\pi i}, \end{align*} $$

and the term $\frac {(\mathcal {E}_{\alpha _1,1}(x,t))_1}{t\lambda }$ is the first entry of the error vector corresponding to the modulated one-phase wave region. It is noted that the local parametrix near $\pm \alpha _1$ and $\pm \eta _1$ can be described by the Airy function and the modified Bessel function, respectively, and both of them contribute the error term $\mathcal {O}(t^{-1})$ in the asymptotic behavior of potential $u(x,t)$ .

The derivative of term $e^{-tg_{\alpha _{1}}(\lambda )}$ on x is

$$ \begin{align*}\partial_xe^{-tg_{\alpha_{1}}(\lambda)}=-\frac{1}{\lambda}\left[\frac{\alpha_{1}^2+\eta_{1}^2}{2}+\alpha_{1}^2\left(\frac{E(m_{\alpha_{1}})}{K(m_{\alpha_{1}})}-1\right)\right]+\mathcal{O}\left(\frac{1}{\lambda^2}\right). \end{align*} $$

So the function $S_{\alpha _{1},1}^{\infty }(\lambda )$ behaves

$$ \begin{align*} S_{\alpha_{1},1}^{\infty}(\lambda)=1+\frac{1}{\lambda}\left[\nabla\log\left(\vartheta_3\left(\frac{t\Omega_{\alpha_{1}}+{\Delta_{\alpha_{1}}}}{2\pi i};2\tau_{\alpha_{1}}\right)\right)-\nabla\log(\vartheta_3(0;2\tau_{\alpha_{1}}))\right] \frac{ \alpha_{1}}{2K(m_{\alpha_{1}})} +\mathcal{O}\left(\frac{1}{\lambda^2}\right), \end{align*} $$

where $\nabla $ stands for the derivative of $\vartheta _3$ . In particular, one has

$$ \begin{align*} \partial_xS_{\alpha_{1},1}^{\infty}(\lambda)=\frac{1}{\lambda}\nabla^2\log\left(\vartheta_3\left(\frac{t\Omega_{\alpha_{1}}+{\Delta_{\alpha_{1}}}}{2\pi i};2\tau_{\alpha_{1}}\right)\right)\frac{ \alpha_{1}}{2K(m_{\alpha_{1}})}\left(\frac{\partial_x(t\Omega_{\alpha_{1}}+{\Delta_{\alpha_{1}}})}{2\pi i}\right)+\mathcal{O}\left(\frac{1}{\lambda^2}\right). \end{align*} $$

Together with (4.1.15) and $\partial _x{\Delta _{\alpha _{1}}}=\frac {\partial _{\alpha _{1}}{\Delta _{\alpha _{1}}}\partial _{\xi }\alpha _{1}}{4t}$ , it follows that

$$ \begin{align*} \partial_xS_{\alpha_{1},1}^{\infty}(\lambda)=-\frac{1}{\lambda}\left[\partial^2_x\log\left(\vartheta_3\left(\frac{t\Omega_{\alpha_{1}}+{\Delta_{\alpha_{1}}}}{2\pi i};2\tau_{\alpha_{1}}\right)\right)+\mathcal{O}\left(\frac{1}{t}\right)\right] +\mathcal{O}\left(\frac{1}{\lambda^2}\right). \end{align*} $$

Combining all the above expansions with the fact that $\partial _x\alpha _{1}\sim \frac {1}{t}$ for $t\to +\infty $ , it is sufficient to show the large-time asymptotics of $u(x,t)$ in (4.1.25) for $\eta _{1}^2<\xi <\xi _{crit}^{(1)}$ .

4.2 Unmodulated one-phase wave region

The equality (4.1.8) shows that $\alpha _{1}(\xi _{crit}^{(1)})=\eta _2$ . Thus when $\xi _{crit}^{(1)}<\xi <\xi _{crit}^{(2)}$ , where the parameter $\xi _{crit}^{(2)}$ is determined by (4.3.13), the jump matrices on $\Sigma _{3,4}$ are still exponentially decreasing to identity matrix for $t \to +\infty $ . The large-time behavior of $u(x,t)$ in this region is expressed by an unmodulated one-phase Jacobi elliptic wave. For convenience, we just need to modify the subscripts in Subsection 4.1, such as $ g_{\eta _{2}} $ , $R_{\eta _{2}}(\lambda )$ , $ \mathcal {S}_{\eta _{2}} $ , $\Omega _{\eta _{2}}$ and $\Delta _{\eta _{2}}$ , which are defined by replacing $ \alpha _{1} $ with $ \eta _{2} $ . In particular, the equation (4.1.5) becomes

$$ \begin{align*} g_{\eta_{2}}(\lambda)=-4\lambda^3+4\xi\lambda+12\int_{\eta_{2}}^{\lambda}\frac{Q_{\eta_{2},2} (\zeta)}{R_{\eta_{2}}(\zeta)}d\zeta-4\xi\int_{\eta_{2}}^{\lambda}\frac{Q_{\eta_{2},1}(\zeta)} {R_{\eta_{2}}(\zeta)}d\zeta, \end{align*} $$

with $R_{\eta _2}(\lambda ):=\sqrt {(\lambda ^2-\eta _1^2)(\lambda ^2-\eta _2^2)}$ , which satisfies the following RH problem:

$$ \begin{align*} \begin{aligned} &g_{\eta_{2,+}}(\lambda)+g_{\eta_{2,-}}(\lambda)+8\lambda^3-8\xi\lambda=0, &&\lambda\in \Sigma_{1}\cup \Sigma_{2},\\ &g_{\eta_{2,+}}(\lambda)-g_{\eta_{2,-}}(\lambda)={\Omega_{\eta_{2}}}, &&\lambda\in [-\eta_1,\eta_1],\\ &g_{\eta_{2}}(\lambda)=\mathcal{O}\left(\frac{1}{\lambda}\right), && \lambda\to \infty. \end{aligned} \end{align*} $$

Moreover, the function $g_{\eta _{2}}(\lambda )$ obeys the similar conditions in (4.1.3) as follows

$$ \begin{align*} \begin{aligned} &g_{\eta_2}(\lambda)+4\lambda^3-4\xi\lambda=(\lambda\pm\eta_2)^{\frac{3}{2}}, &&\lambda\to\pm \eta_2,\\ &\operatorname{\mathrm{Re}}\left(g_{\eta_2}(\lambda)+4\lambda^3-4\xi\lambda\right)>0, &&\lambda \in \Sigma_3,\\ &\operatorname{\mathrm{Re}}\left(g_{\eta_2}(\lambda)+4\lambda^3-4\xi\lambda\right)<0, &&\lambda \in \Sigma_4,\\ &-i(g_{\eta_2,+}(\lambda)-g_{\eta_2,-}(\lambda)) \text{ is real-valued and monotonically increasing},&& \lambda\in\Sigma_{1},\\ &-i(g_{\eta_2,+}(\lambda)-g_{\eta_2,-}(\lambda)) \text{ is real-valued and monotonically decreasing}, && \lambda\in\Sigma_{2}.\\ \end{aligned} \end{align*} $$

Consequently, the RH problem for $Y(\lambda )$ can be transformed into a model problem for $S^{\infty }_{\eta _{2}}(\lambda )$ whose solution is

$$ \begin{align*}S^{\infty}_{\eta_2}(\lambda)=\gamma_{\eta_2}(\lambda)\frac{\vartheta_3(0;2\tau_{\eta_2})}{\vartheta_3(\frac{t{\Omega_{\eta_2}}+{\Delta_{\eta_2}}}{2\pi i};2\tau_{\eta_2})}\begin{pmatrix} \frac{\vartheta_3\left(2J_{\eta_2}(\lambda)+\frac{t\Omega_{\eta_2}+{\Delta_{\eta_2}}}{2\pi i}-\frac{1}{2};2\tau_{\eta_2}\right)}{\vartheta_3\left(2J_{\eta_2}(\lambda)-\frac{1}{2};2\tau_{\eta_2}\right)} & \frac{\vartheta_3\left(-2J_{\eta_2}(\lambda)+\frac{t\Omega_{\eta_2}+{\Delta_{\eta_2}}}{2\pi i}-\frac{1}{2};2\tau_{\eta_2}\right)}{\vartheta_3 \left(-2J_{\eta_2}(\lambda)-\frac{1}{2};2\tau_{\eta_2}\right)} \end{pmatrix}. \end{align*} $$

In particular, the model problem satisfies the jump conditions as illustrated in Figure 11.

Figure 11. The jump contour for $ S^{\infty }_{\eta _2}(\lambda ) $ and the associated jump matrices.

Thus for $\xi _{crit}^{(1)}<\xi <\xi _{crit}^{(2)}$ , the following theorem holds.

Theorem 4.5. For $\xi =\frac {x}{4t}$ , in the region $\xi _{crit}^{(1)}<\xi <\xi _{crit}^{(2)}$ , the large-time asymptotic behavior of the solution to the KdV equation with genus two soliton gas potential is described by

(4.2.1) $$ \begin{align} u(x,t)=\eta_{2}^2-\eta_{1}^2-2\eta_{2}^2 \mathrm{dn}^2\left(\eta_{2}(x-2(\eta_{2}^2+\eta_{1}^2)t+\phi_{\eta_{2}})+K(m_{\eta_{2}}); m_{\eta_{2}}\right)+\mathcal{O}\left(\frac{1}{t}\right), \end{align} $$

where the modulus of the Jacobi elliptic function is $m_{\eta _{2}}=\frac {\eta _{1}}{\eta _{2}}$ and

$$ \begin{align*} \phi_{\eta_{2}}=\int_{\eta_{2}}^{\eta_{1}}\frac{\log r(\zeta)}{R_{\eta_{2},+}(\zeta)}\frac{d\zeta}{\pi i}. \end{align*} $$

4.3 Modulated two-phase wave region

This section considers the case that the parameter $\xi $ obeys

$$ \begin{align*} \xi_{crit}^{(2)}<\xi<\xi_{crit}^{(3)}, \end{align*} $$

where the parameters $\xi _{crit}^{(2)}$ and $\xi _{crit}^{(3)}$ are determined by (4.3.16) and (4.3.13) below. Define

(4.3.1) $$ \begin{align} \Sigma_{3_{\alpha_2}}=(\eta_3,\alpha_2),\ \text{and}\ \Sigma_{4_{\alpha_2}}=(-\alpha_2,-\eta_3), \end{align} $$

see Figure 12, where $\alpha _{2}$ is defined in (4.3.8) below.

Figure 12. The solid lines represent $\Sigma _{3_{\alpha _2}}$ and $\Sigma _{4_{\alpha _2}}$ , where $\eta _3 < \alpha _2 < \eta _4$ .

In addition, we introduce $g_{\alpha _{2}}(\lambda )$ which is subject to the following scalar RH problem:

(4.3.2) $$ \begin{align} \begin{aligned} &g_{\alpha_{2},+}(\lambda)+g_{\alpha_{2},-}(\lambda)+8 \lambda^3-8 \xi \lambda=0, && \lambda \in \Sigma_{1} \cup \Sigma_2 \cup \Sigma_{3_{\alpha_{2}}} \cup \Sigma_{4_{\alpha_{2}}}, \\ &g_{\alpha_{2},+}(\lambda)-g_{\alpha_{2},-}(\lambda)={\Omega}_{\alpha_{2},0}, && \lambda \in\left[-\eta_1, \eta_1\right], \\ &g_{\alpha_{2},+}(\lambda)-g_{\alpha_{2},-}(\lambda)={\Omega}_{\alpha_{2},1}, && \lambda \in\left[\eta_2, \eta_3\right]\cup\left[-\eta_3,-\eta_2\right], \\ &g_{\alpha_{2}}(\lambda)=\mathcal{O}\left(\frac{1}{\lambda}\right), && \lambda \rightarrow \infty. \end{aligned} \end{align} $$

It remains to show that the function $g_{\alpha _{2}}(\lambda )$ satisfies the following properties:

(4.3.3) $$ \begin{align} \begin{aligned} &g_{\alpha_{2}}(\lambda)+4\lambda^3-4\xi\lambda=(\lambda\pm\alpha_{2})^{\frac{3}{2}}, &&\lambda\to\pm\alpha_{2},\\ &Re\left(g_{\alpha_{2}}(\lambda)+4\lambda^3-4\xi\lambda\right)>0, &&\lambda \in (\alpha_{2},\eta_4),\\ &Re\left(g_{\alpha_{2}}(\lambda)+4\lambda^3-4\xi\lambda\right)<0, &&\lambda \in (-\eta_4,-\alpha_{2}),\\ &-i(g_{\alpha_{2},+}(\lambda)-g_{\alpha_{2},-}(\lambda)) \text{ is real-valued and monotonically increasing},&& \lambda\in\Sigma_{3_{\alpha_{2}}}\cup\Sigma_{1},\\ &-i(g_{\alpha_{2},+}(\lambda)-g_{\alpha_{2},-}(\lambda)) \text{ is real-valued and monotonically decreasing}, && \lambda\in\Sigma_{4_{\alpha_{2}}}\cup\Sigma_{2}.\\ \end{aligned} \end{align} $$

Further, introduce

(4.3.4) $$ \begin{align} g^{\prime}_{\alpha_{2}}(\lambda)=-12\lambda^2+4\xi+12\frac{Q_{\alpha_{2},2}(\lambda)}{R_{\alpha_{2}}(\lambda)}-4\xi\frac{Q_{\alpha_{2},1}(\lambda)}{R_{\alpha_{2}}(\lambda)}, \end{align} $$

with

$$ \begin{align*} R_{\alpha_{2}}(\lambda)=\sqrt{(\lambda^2-\eta_1^2)(\lambda^2-\eta_2^2)(\lambda^2-\eta_3^2)(\lambda^2-\alpha_{2}^2)}, \end{align*} $$

where $R_{\alpha _{2}}$ is analytic for $\mathbb C\setminus \Sigma _{\{1,2, 3_{\alpha _2},4_{\alpha _2}\}}$ with positive real value for $\lambda>\alpha _{2}$ . More precisely, we have

(4.3.5) $$ \begin{align} Q_{\alpha_{2},1}=\lambda^4+b_{\alpha_{2},1} \lambda^2+c_{\alpha_{2},1},\ Q_{\alpha_{2},2}=\lambda^6-\frac{1}{2}(\eta_{1}^2+\eta_{2}^2+\eta_{3}^2+\alpha_{2}^2)\lambda^4+b_{\alpha_{2},2} \lambda^2+c_{\alpha_{2},2},\ \end{align} $$

where the constants $b_{\alpha _{2},1},b_{\alpha _{2},2}$ and $c_{\alpha _{2},1},c_{\alpha _{2},2}$ can be determined by

(4.3.6) $$ \begin{align} \int_{-\eta_{1}}^{\eta_{1}}\frac{Q_{\alpha_{2},j}(\zeta)}{R_{\alpha_{2}}(\zeta)}d\zeta=0,\quad \int_{\eta_{2}}^{\eta_{3}}\frac{Q_{\alpha_{2},j}(\zeta)}{R_{\alpha_{2}}(\zeta)}d\zeta=0,\quad j=1,2. \end{align} $$

Lemma 4.7. The following identities are established

(4.3.7) $$ \begin{align} \partial_x(tg^{\prime}_{\alpha_{2}}(\lambda))=1-\frac{Q_{\alpha_2,1}(\lambda)}{R_{\alpha_{2}}(\lambda)},\ \partial_{x}\left(t\Omega_{\alpha_{2},1}\right)=\oint_{b_1}\frac{Q_{\alpha_{2},1}(\zeta)}{R_{\alpha_{2}}(\zeta)}d\zeta,\ \partial_{x}\left(t\Omega_{\alpha_{2},0}\right)=\oint_{b_2}\frac{Q_{\alpha_{2},1}(\zeta)}{R_{\alpha_{2}}(\zeta)}d\zeta. \end{align} $$

Proof. From the definition of the function $g^{\prime }_{\alpha _{2}}(\lambda )$ in (4.3.4), it is seen that

$$ \begin{align*} \partial_x(tg^{\prime}_{\alpha_{2}}(\lambda))=1-\frac{Q_{\alpha_{2},1} (\lambda)}{R_{\alpha_{2}}(\lambda)}+\partial_{\alpha_{2}}\left(12t\frac{Q_{\alpha_{2},2} (\lambda)}{R_{\alpha_{2}}(\lambda)}-x\frac{Q_{\alpha_{2},1}(\lambda)}{R_{\alpha_{2}}(\lambda)}\right) \partial_x{\alpha_{2}}, \end{align*} $$

and since $\partial _{\alpha _{2}}\left (12t\frac {Q_{\alpha _{2},2}(\lambda )}{R_{\alpha _{2}}(\lambda )} d\lambda -x\frac {Q_{\alpha _{2},1}(\lambda )}{R_{\alpha _{2}}(\lambda )}d\lambda \right )$ doesn’t have singularities at $\pm \alpha _{2}$ and $\infty $ , the term $12t\frac {Q_{\alpha _{2},2}(\lambda )}{R_{\alpha _{2}}(\lambda )} d\lambda -x\frac {Q_{\alpha _{2},1}(\lambda )}{R_{\alpha _{2}}(\lambda )}d\lambda $ is a holomorphic differential 1-form. Simultaneously, it is normalized to zero on the $a_{\alpha _2,j}~(j=1,2,3)$ -cycles, thus by Riemann bilinear relation one has

$$ \begin{align*} \partial_{\alpha_{2}}\left(12t\frac{Q_{\alpha_{2},2}(\lambda)}{R_{\alpha_{2}}(\lambda)} -x\frac{Q_{\alpha_{2},1}(\lambda)}{R_{\alpha_{2}}(\lambda)}\right)=0. \end{align*} $$

On the other hand, it is obvious that

$$ \begin{align*} \Omega_{\alpha_{2},1}=-\oint_{b_1}(g'(\zeta)+12\zeta^2-4\xi)d\zeta,\ \Omega_{\alpha_{2},0}=-\oint_{b_2}(g'(\zeta)+12\zeta^2-4\xi)d\zeta. \end{align*} $$

So it follows that

$$ \begin{align*} \begin{aligned} &\partial_{x}\left(t\Omega_{\alpha_{2},1}\right)=-\partial_x\left(t\oint_{b_1}(g'(\zeta)+12\zeta^2-4\xi)d\zeta\right)=\oint_{b_1}\frac{Q_{\alpha_{2},1}(\zeta)}{R_{\alpha_{2}}(\zeta)}d\zeta,\\ &\partial_{x}\left(t\Omega_{\alpha_{2},0}\right)=-\partial_x\left(t\oint_{b_2}(g'(\zeta)+12\zeta^2-4\xi)d\zeta\right)=\oint_{b_2}\frac{Q_{\alpha_{2},1}(\zeta)}{R_{\alpha_{2}}(\zeta)}d\zeta. \end{aligned}\\[-34pt] \end{align*} $$

In particular, the function $g_{\alpha _{2}}(\lambda )+4\lambda ^3-4\xi \lambda $ is endowed with the behavior $\mathcal {O}((\lambda \pm \alpha _{2})^{\frac {3}{2}})$ as $\lambda \to \pm \alpha _{2}$ if and only if $\xi $ and $\alpha _{2}$ satisfy the following relationship:

(4.3.8) $$ \begin{align} \xi=3\frac{Q_{\alpha_2,2}(\pm \alpha_{2})}{Q_{\alpha_2,1}(\pm \alpha_{2})}. \end{align} $$

Also, $g_{\alpha _{2}}(\lambda )+4\lambda ^3-4\xi \lambda =\mathcal {O}((\lambda \pm \alpha _{2})^{\frac {3}{2}})$ as $\lambda \to \pm \alpha _{2}$ implies that $g^{\prime }_{\alpha _{2}}(\alpha _{2})+12\alpha _{2}^2-4\xi =0$ , thus the equation (4.3.8) is established. Conversely, if the equation (4.3.8) holds, the equation (4.3.4) can be rewritten as

(4.3.9) $$ \begin{align} \begin{aligned} g^{\prime}_{\alpha_{2}}(\lambda)+12\lambda^2-4\xi&=12\frac{Q_{\alpha_{2},2}(\lambda)-Q_{\alpha_{2},2}(\alpha_{2})}{R_{\alpha_{2}}(\lambda)}-4\xi\frac{Q_{\alpha_{2},1}(\lambda)-Q_{\alpha_{2},1}(\alpha_{2})}{R_{\alpha_{2}}(\lambda)}\\ &=12\frac{(\lambda^2-\alpha_{2}^2)(\lambda^2-\lambda_{1}^2)(\lambda^2-\lambda_{2}^2)}{R_{\alpha_{2}}(\lambda)}, \end{aligned} \end{align} $$

where $\lambda _{1}\in (0,\eta _{1})$ and $\lambda _{2}\in (\eta _{2},\eta _{3})$ due to the normalization condition (4.3.6). Consequently, it follows $g^{\prime }_{\alpha _{2}}(\lambda )+12\lambda ^2-4\xi =\mathcal {O}(\sqrt {\lambda -\alpha _{2}})$ as $\lambda \to \alpha _{2}$ . In addition, the expression of $g_{\alpha _{2}}(\lambda )$ is

(4.3.10) $$ \begin{align} g_{\alpha_{2}}(\lambda)=-4\lambda^3+4\xi\lambda+12\int_{\alpha_{2}}^{\lambda}\frac{Q_{\alpha_{2},2}(\zeta)}{R_{\alpha_{2}}(\zeta)}d\zeta-4\xi\int_{\alpha_{2}}^{\lambda}\frac{Q_{\alpha_{2},1}(\zeta)}{R_{\alpha_{2}}(\zeta)}d\zeta. \end{align} $$

It is immediate that $g_{\alpha _{2}}(\lambda )$ satisfies the first condition in (4.3.3).

As in the case of the modulated one-phase wave region, the Whitham equation (4.3.8) is strictly hyperbolic; see [Reference Levermore30]. By following the approach developed in [Reference Grava and Tian25], we conclude that $\alpha _2$ is a monotonically increasing function of $\xi $ . More precisely, we introduce the following Abelian differentials of the second kind, referred to as the quasi-momentum and quasi-energy differentials:

$$\begin{align*}dp=\frac{Q_{\alpha_2,1}(\lambda)}{R_{\alpha_2}(\lambda)}\,d\lambda, \qquad dq=\frac{Q_{\alpha_2,2}(\lambda)}{R_{\alpha_2}(\lambda)}\,d\lambda, \end{align*}$$

and define

$$\begin{align*}d\varphi=\bigl(g^{\prime}_{\alpha_2}(\lambda)+12\lambda^2-4\xi\bigr)\,d\lambda =12\,dq-4\xi\,dp. \end{align*}$$

Proof. By equation (4.3.9), we obtain

$$ \begin{align*} \partial_{\xi}(d\varphi) =-\left( \frac{2\lambda_1\,\partial_{\xi}\lambda_1}{\lambda^2-\lambda_1^2} +\frac{2\lambda_2\,\partial_{\xi}\lambda_2}{\lambda^2-\lambda_2^2} +\frac{\alpha_2\,\partial_{\xi}\alpha_2}{\lambda^2-\alpha_2^2} \right)d\varphi. \end{align*} $$

On the other hand, by Lemma 4.7, we have $ \partial _{\xi }(d\varphi )=-4\,dp $ . Comparing the two expressions yields the identity

(4.3.11) $$ \begin{align} 4\frac{dp}{d\varphi} = \frac{2\lambda_1\,\partial_{\xi}\lambda_1}{\lambda^2-\lambda_1^2} +\frac{2\lambda_2\,\partial_{\xi}\lambda_2}{\lambda^2-\lambda_2^2} +\frac{\alpha_2\,\partial_{\xi}\alpha_2}{\lambda^2-\alpha_2^2} = \frac{Q_{\alpha_2,1}(\lambda)}{3(\lambda^2-\alpha_2^2)(\lambda^2-\lambda_1^2)(\lambda^2-\lambda_2^2)}. \end{align} $$

Since $Q_{\alpha _2,1}(\lambda )$ is even, has zeros in $(0,\eta _1)$ and $(\eta _2,\eta _3)$ for $\lambda>0$ , and satisfies $Q_{\alpha _2,1}(\lambda )\sim \lambda ^4$ as $\lambda \to \infty $ , we conclude that $Q_{\alpha _2,1}(\alpha _2)>0$ . Taking the residue at $\lambda =\alpha _2$ in (4.3.11), it follows that

(4.3.12) $$ \begin{align} \partial_{\xi}\alpha_2 = \frac{Q_{\alpha_2,1}(\alpha_2)}{6\alpha_2(\alpha_2^2-\lambda_1^2)(\alpha_2^2-\lambda_2^2)}>0. \end{align} $$

Since $dp$ is the normalized Abelian differential of the second kind, we have $Q_{\alpha _2,1}(\alpha _2)>0$ . As a result, $ \alpha $ is a monotone increasing function of $ \xi $ with $ \alpha _{2} \in (\eta _{3},\eta _{4}) $ . Thus, we define

(4.3.13) $$ \begin{align} \xi_{crit}^{(2)}=3\frac{Q_{\alpha_2,2}(\eta_{3})}{Q_{\alpha_2,1}(\eta_{3})},\quad \xi_{crit}^{(3)}=3\frac{Q_{\alpha_2,2}(\eta_{4})}{Q_{\alpha_2,1}(\eta_{4})}. \end{align} $$

More precisely, note that (4.3.8) also depends on the Riemann surface $ R_{\alpha _2}(\lambda ) $ , which is given by

$$\begin{align*}\xi=3\frac{Q_{\alpha_2,2}(\pm\alpha_2;\eta_1,\eta_2,\eta_3,\alpha_2)}{Q_{\alpha_2,1}(\pm{\alpha_2};\eta_1,\eta_2,\eta_3,\alpha_2)} \end{align*}$$

and the critical values $ \xi _{crit}^{(2)} $ and $ \xi _{crit}^{(3)} $ can be rewritten as

(4.3.14) $$ \begin{align} \xi_{crit}^{(2)}&=\lim_{\epsilon\to0}3\frac{Q_{\alpha_2,2}(\pm(\eta_3+\epsilon);\eta_1,\eta_2,\eta_3,\eta_3+\epsilon)}{Q_{\alpha_2,1}(\pm(\eta_3+\epsilon);\eta_1,\eta_2,\eta_3,\eta_3+\epsilon)},\\\xi_{crit}^{(3)}&=3\frac{Q_{\alpha_2,2}(\pm\eta_4;\eta_1,\eta_2,\eta_3,\eta_4)}{Q_{\alpha_2,1}(\pm{\eta_4};\eta_1,\eta_2,\eta_3,\eta_4)}.\nonumber \end{align} $$

By the genuine nonlinearity condition in (4.3.12), for any $ \epsilon>0 $ , it follows that

$$\begin{align*}3\frac{Q_{\alpha_2,2}(\pm(\eta_3+\epsilon);\eta_1,\eta_2,\eta_3,\eta_3+\epsilon)}{Q_{\alpha_2,1}(\pm(\eta_3+\epsilon);\eta_1,\eta_2,\eta_3,\eta_3+\epsilon)}<\xi_{crit}^{(3)}. \end{align*}$$

If the limit exists as $ \epsilon \to 0 $ , then the inequality $ \xi _{crit}^{(3)}<\xi _{crit}^{(2)} $ is established. In fact, consider the Riemann surface

$$\begin{align*}\mathcal{S}_{\alpha_2}^{(\epsilon)}:=\{(\lambda,\eta) \mid \eta^2=(\lambda^2-\eta_1^2)(\lambda^2-\eta_2^2)(\lambda^2-\eta_3^2)(\lambda^2-(\eta_3+\epsilon)^2)\}. \end{align*}$$

As $ \epsilon \to 0 $ , the algebraic curve associated with $ \mathcal {S}_{\alpha _2}^{(\epsilon )} $ develops two nodes, causing the genus of the limiting Riemann surface to degenerate to $ 1 $ . By the expansion given in [Reference Grava27, Reference Fay18], the limit of (4.3.14) exists, yielding

(4.3.15) $$ \begin{align} \begin{aligned} \xi_{crit}^{(2)} &= 3\lim_{\epsilon\to0}\frac{Q_{\alpha_2,2}(\pm(\eta_3+\epsilon);\eta_1,\eta_2,\eta_3,\eta_3+\epsilon)}{Q_{\alpha_2,1}(\pm(\eta_3+\epsilon);\eta_1,\eta_2,\eta_3,\eta_3+\epsilon)}\\ &= 3\frac{\int_{-R_{\eta_2}(\eta_3)}^{R_{\eta_2}(\eta_3)}\frac{Q_{\eta_2,2}(\zeta)}{R_{\eta_2}(\zeta)}d\zeta}{\int_{-R_{\eta_2}(\eta_3)}^{R_{\eta_2}(\eta_3)}\frac{Q_{\eta_2,1}(\zeta)}{R_{\eta_2}(\zeta)}d\zeta}, \end{aligned} \end{align} $$

where $ R_{\eta _2}(\lambda ) = \sqrt {(\lambda ^2-\eta _1^2)(\lambda ^2-\eta _2^2)} $ and

$$\begin{align*}Q_{\eta_2,1}(\lambda) = \lambda^2 + c_{\eta_2,1}, \quad Q_{\eta_2,2}(\lambda) = \lambda^4 - \frac{1}{2}(\eta_2^2+\eta_1^2)\lambda^2 + c_{\eta_2,2}, \end{align*}$$

with

$$\begin{align*}c_{\eta_2,1} = -\eta_2^2 + \eta_2^2\frac{E(m_{\eta_2})}{K(m_{\eta_2})}, \quad c_{\eta_2,2} = \frac{1}{3} \eta_2^2\eta_1^2 + \frac{1}{6}(\eta_2^2+\eta_1^2)c_{\eta_2,1}, \quad m_{\eta_2} = \frac{\eta_1}{\eta_2}. \end{align*}$$

Moreover, using the formula in [Reference Grava27], we obtain that

$$\begin{align*}\int_{-R_{\eta_2}(\eta_3)}^{R_{\eta_2}(\eta_3)}\frac{Q_{\eta_2,1}(\zeta)}{R_{\eta_2}(\zeta)}d\zeta = -4\left(\oint_A\frac{d\zeta}{R_{\eta_2}(\zeta)}\right)^{-1} \oint_A \frac{d\zeta}{R_{\eta_2}(\zeta)} \frac{R_2(\eta_3)}{\zeta^2-\eta_3^2}, \end{align*}$$

and

$$\begin{align*}\int_{-R_{\eta_2}(\eta_3)}^{R_{\eta_2}(\eta_3)}\frac{Q_{\eta_2,2}(\zeta)}{R_{\eta_2}(\zeta)}d\zeta = -\frac{4}{3} \left(-R_{\eta_2}(\eta_3) + \frac{\eta_1^2+\eta_2^2}{2} \left(\oint_A\frac{d\zeta}{R_{\eta_2}(\zeta)}\right)^{-1} \oint_A \frac{d\zeta}{R_{\eta_2}(\zeta)} \frac{R_2(\eta_3)}{\zeta^2-\eta_3^2} \right), \end{align*}$$

where $ A $ is the contour shown in Figure 9. Consequently, the expression for $ \xi ^{(2)}_{crit} $ simplifies to

(4.3.16) $$ \begin{align} \xi^{(2)}_{crit} = \frac{\eta_1^2+\eta_2^2}{2} - \frac{\int_{0}^{\eta_1}\frac{d\zeta}{R_{\eta_2}(\zeta)}}{\int^{\eta_1}_{0}\frac{d\zeta}{R_{\eta_2}(\zeta)(\zeta^2-\eta_3^2)}} = \frac{\eta_1^2+\eta_2^2}{2} + \eta_3^2{\frac{K(m_{\eta_2})}{\Pi(\kappa,m_{\eta_2})}}, \end{align} $$

where $ \Pi (\kappa ,m_{\eta _2}) := \int _{0}^{\frac {\pi }{2}} \frac {d\theta }{(1-\kappa \sin ^2\theta )\sqrt {1-m_{\eta _2}^2\sin ^2\theta }} $ is the complete elliptic integral of the third kind, with $ \kappa = \frac {\eta _1^2}{\eta _3^2} $ and $ m_{\eta _2} = \frac {\eta _1}{\eta _2} $ .

Notably, it is found that

$$\begin{align*}\xi_{crit}^{(2)} - \xi_{crit}^{(1)} = \eta_3^2{\frac{K(m_{\eta_2})}{\Pi(\kappa,m_{\eta_2})}} - (\eta_2^2-\eta_1^2)\frac{K(m_{\eta_2})}{E(m_{\eta_2})}. \end{align*}$$

Since $ {\Pi (\kappa ,m)\to \frac {E(m)}{1-m^2}} $ , as $\kappa \to m^2$ , and let

$$ \begin{align*}\Delta_{\xi}(\alpha)=K(m_{\eta_2})\,\left(\frac{\alpha^2}{\Pi\left(\frac{\eta_1^2}{\alpha^2};m_{\eta_2}\right)}-\frac{\eta_2^2-\eta_1^2}{E(m_{\eta_2})}\right)\,. \end{align*} $$

Notice that $\Delta _{\xi }(\eta _2)=0$ , which means that $\xi _{crit}^{(2)} = \xi _{crit}^{(1)}$ as $\eta _3=\eta _2$ , and by $ \partial _{\kappa } \Pi (\kappa ,m_{\eta _2})> 0 $ for $ 0 < \kappa < 1 $ , we conclude that $\partial _{\alpha }\Delta _{\xi }(\alpha )>0$ and $ \xi ^{(2)}_{crit}> \xi ^{(1)}_{crit} $ for $\eta _3>\eta _2$ .

On the other hand, based on the expression of the function $g_{\alpha _{2}}(\lambda )$ in (4.3.10), the inequalities like (4.1.20) are obtained in the following lemma.

Lemma 4.9. For $\xi _{crit}^{(2)}<\xi <\xi _{crit}^{(3)}$ , the inequalities below hold

(4.3.17) $$ \begin{align} \begin{aligned} &\operatorname{\mathrm{Re}}\left(g_{\alpha_{2}}(\lambda)+4\lambda^3-4\xi\lambda\right)>0, &&\lambda\in(\alpha_2,\eta_4)\cup\mathcal{C}_{2,\alpha_{2}}\setminus\{-\eta_1,-\eta_2,-\eta_3,-\alpha_2\},\\ &\operatorname{\mathrm{Re}}\left(g_{\alpha_{2}}(\lambda)+4\lambda^3-4\xi\lambda\right)<0, &&\lambda\in(-\eta_4,-\alpha_2)\cup\mathcal{C}_{1,\alpha_{2}}\setminus\{\eta_1,\eta_2,\eta_3,\alpha_2\},\\ \end{aligned} \end{align} $$

where the contours $\mathcal {C}_{j,\alpha _{2}}$ for $j=1,2$ are depicted in Figure 13.

Figure 13. The jump contours for $ S_{\alpha _{2}}(\lambda ) $ and the associated jump matrices: the gray terms in the matrices vanish exponentially as $ t \to +\infty $ , and the gray contours also vanish as $ t \to +\infty $ . Here $(V_T)_{12}=-\frac {ir(\lambda )}{ f^2(\lambda )}e^{-2t(g(\lambda )+4\lambda ^3-4\xi \lambda )}$ and $(V_T)_{21}=ir(\lambda )f^2(\lambda )e^{2t(g(\lambda )+4\lambda ^3-4\xi \lambda )}$ .

Proof. Similarly, recall that

$$ \begin{align*} g^{\prime}_{\alpha_{2}}(\lambda)+12\lambda^2-4\xi=12\frac{(\lambda^2-\alpha_{2}^2)(\lambda^2-\lambda_{1}^2)(\lambda^2-\lambda_{2}^2)}{R_{\alpha_{2}}(\lambda)}, \end{align*} $$

with $\lambda _{1}\in [0,\eta _{1}]$ and $\lambda _{2}\in [\eta _{2},\eta _{3}]$ . As a result, it can be derived that

$$ \begin{align*} g^{\prime}_{\alpha_{2},+}(\lambda)+12\lambda^2-4\xi= \begin{cases} \begin{aligned} &12\frac{\sqrt{\lambda^2-\alpha_{2}^2}(\lambda^2-\lambda_{1}^2)(\lambda^2-\lambda_{2}^2)} {\sqrt{(\lambda^2-\eta^2_{1})(\lambda^2-\eta^2_{2})(\lambda^2-\eta^2_{3})}}, &&\lambda\in(\alpha_{2},\eta_{4}),\\ &12\frac{i\sqrt{\alpha_{2}^2-\lambda^2}(\lambda^2-\lambda_{1}^2)(\lambda^2-\lambda_{2}^2)} {\sqrt{(\lambda^2-\eta^2_{1})(\lambda^2-\eta^2_{2})(\lambda^2-\eta^2_{3})}}, &&\lambda\in(\eta_{3},\alpha_{2}),\\ &12\frac{i\sqrt{\alpha_{2}^2-\lambda^2}(\lambda^2-\lambda_{1}^2)(\lambda_{2}^2-\lambda^2)} {\sqrt{(\lambda^2-\eta^2_{1})(\eta^2_{2}-\lambda^2)(\eta^2_{3}-\lambda^2)}}, &&\lambda\in(\eta_{1},\eta_{2}), \end{aligned} \end{cases} \end{align*} $$

which implies that $\operatorname {\mathrm {Re}}\left (g_{\alpha _{2}}(\lambda )+4\lambda ^3-4\xi \lambda \right )>0$ for $\lambda \in (\alpha _{2},\eta _{4})$ and $\text {Im}\left (g^{\prime }_{\alpha _{2},+}(\lambda )+12\lambda ^2-4\xi \right )>0$ for $\lambda \in \Sigma _{1}\cup \Sigma _{3_{\alpha _{2}}}$ . Together with the Cauchy-Riemann equation, it follows that $\operatorname {\mathrm {Re}}\left (g_{\alpha _{2}}(\lambda )+4\lambda ^3-4\xi \lambda \right )<0$ for $\lambda \in \mathcal {C}_{1,\alpha _{2}}\setminus \{\eta _1,\eta _2,\eta _3,\alpha _2\}$ . The other cases in (4.3.17) can also be proved by the similar way.

To deform the RH problem for $Y(x,t;\lambda )$ into the model problem, introduce

$$ \begin{align*} T_{\alpha_{2}}(\lambda)=Y(\lambda){e^{tg_{\alpha_{2}}(\lambda)\sigma_3}}f_{\alpha_{2}}(\lambda)^{\sigma_3}, \end{align*} $$

where $f_{\alpha _{2}}(\lambda )$ is subject to the following scalar RH problem:

(4.3.18) $$ \begin{align} \begin{aligned} &f_{\alpha_{2},+}(\lambda)f_{\alpha_{2},-}(\lambda)={r(\lambda)},&&\lambda\in\Sigma_{\{1,3_{\alpha_{2}}\}},\\ &f_{\alpha_{2},+}(\lambda)f_{\alpha_{2},-}(\lambda)=\frac{1}{r(\lambda)},&&\lambda\in\Sigma_{\{2,4_{\alpha_{2}}\}},\\ &\frac{f_{\alpha_{2},+}(\lambda)}{f_{\alpha_{2},-}(\lambda)}=e^{\Delta_0,\alpha_{2}},&&\lambda\in [-\eta_1,\eta_1],\\ &\frac{f_{\alpha_{2},+}(\lambda)}{f_{\alpha_{2},-}(\lambda)}=e^{\Delta_1,\alpha_{2}},&&\lambda\in [\eta_2,\eta_3]\cup[-\eta_3,-\eta_2],\\ &f(\lambda)=1+\mathcal{O}\left(\frac{1}{\lambda}\right),&&\lambda\to\infty. \end{aligned} \end{align} $$

In a similar manner, the function $ f_{\alpha _{2}}(\lambda ) $ can be derived akin to (3.0.4). The normalization condition then indicates the values of $\Delta _{j,\alpha _{2}},~j=0,1$ , which correspond to (3.2.19). All the calculations here are omitted for simplicity.

In the same way, open lenses of the RH problem for $T_{\alpha _{2}}$ by the transformation

(4.3.19) $$ \begin{align} S_{\alpha_{2}}(\lambda)= \begin{cases} T_{\alpha_{2}}(\lambda)\begin{pmatrix} 1 & 0 \\ \frac{if_{\alpha_{1}}^2(\lambda)}{\hat r(\lambda)}e^{2t(g_{\alpha_{1}}(\lambda)+4\lambda^3-4\xi\lambda)} & 1 \end{pmatrix}, & \mathrm{inside~the~contour}~\mathcal{C}_{1,\alpha_{2}}, \\ T_{\alpha_{2}}(\lambda)\begin{pmatrix} 1 & \frac{-i}{\hat r(\lambda)f_{\alpha_{1}}^2(\lambda)}e^{-2t(g_{\alpha_{1}}(\lambda)+4\lambda^3-4\xi\lambda)} \\ 0 & 1 \end{pmatrix}, & \mathrm{inside~the~contour}~\mathcal{C}_{2,\alpha_{2}}, \\ T_{\alpha_{2}}(\lambda), & \mathrm{elsewhere}. \end{cases} \end{align} $$

The jump matrices and contours are illustrated in Figure 13. As $t\to +\infty $ , the gray contours in Figure 13 vanish exponentially according to Lemma 4.9. Once again for $t\to +\infty $ , we arrive at the model problem for $S_{\alpha _{2}}^{\infty }(\lambda )$ below

(4.3.20) $$ \begin{align} {S}_{\alpha_{2},+}^{\infty}(\lambda)={S}_{\alpha_{2},-}^{\infty}(\lambda) \begin{cases}{\begin{pmatrix} e^{t {\Omega_{0,\alpha_{2}}}+{\Delta}_{0,\alpha_{2}}} & 0 \\ 0 & e^{-t {\Omega_{0,\alpha_{2}}}-{\Delta}_{0,\alpha_{2}}} \end{pmatrix}}, & \lambda \in[-\eta_1, \eta_1], \\ \begin{pmatrix} e^{t {\Omega_{1,\alpha_{2}}}+{\Delta}_{1,\alpha_{2}}} & 0 \\ 0 & e^{-t {\Omega_{1,\alpha_{2}}}-{\Delta}_{1,\alpha_{2}}} \end{pmatrix}, & \lambda \in[\eta_2, \eta_3]\cup[-\eta_3, -\eta_2], \\ {\begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix}}, & \lambda \in \Sigma_{1}\cup\Sigma_{3_{\alpha_1}}, \\ {\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}}, & \lambda \in \Sigma_{2}\cup\Sigma_{ 4_{\alpha_1}},\end{cases} \end{align} $$

and

$$ \begin{align*} {S}_{\alpha_{2}}^{\infty}(\lambda)=\begin{pmatrix} 1&1 \end{pmatrix}+\mathcal{O}\left(\frac{1}{\lambda}\right),\quad \lambda\to\infty. \end{align*} $$

Furthermore, as the solution of the model problem for $S^{\infty }(\lambda )$ in (3.2.9), the solution of $S_{\alpha _{2}}^{\infty }(\lambda )$ can be derived directly by

(4.3.21) $$ \begin{align} S_{\alpha_{2}}^{\infty}(\lambda)=\gamma_{\alpha_{2}}(\lambda)\frac{\Theta(0;\hat{\tau}_{\alpha_{2}})}{\Theta\left(\frac{\Omega_{\alpha_{2}}}{2\pi i};\hat{\tau}_{\alpha_{2}}\right)} \begin{pmatrix} \frac{\Theta\left(J_{\alpha_{2}}(\lambda)-d_{\alpha_{2}}+\frac{\Omega_{\alpha_{2}}}{2\pi i};\hat{\tau}_{\alpha_{2}}\right)}{\Theta\left(J_{\alpha_{2}}(\lambda)-d_{\alpha_{2}};\hat{\tau}_{\alpha_{2}}\right)}& \frac{\Theta\left(-J_{\alpha_{2}}(\lambda)-d_{\alpha_{2}}+\frac{\Omega_{\alpha_{2}}}{2\pi i};\hat{\tau}_{\alpha_{2}}\right)}{\Theta\left(-J_{\alpha_{2}}(\lambda)-d_{\alpha_{2}};\hat{\tau}_{\alpha_{2}}\right)} \end{pmatrix}, \end{align} $$

where $\gamma _{\alpha _{2}}(\lambda )=\left (\frac {(\lambda ^2-\eta _1^2)(\lambda ^2-\eta _3^2)}{(\lambda ^2-\eta _2^2)(\lambda ^2-\alpha _{2}^2)}\right )^{\frac {1}{4}} $ and $\Omega _{\alpha _{2}}=\begin {pmatrix} {t\Omega _{\alpha _{2},1}+{\Delta _{\alpha _{2},1}}}&{t\Omega _{\alpha _{2},0}+{\Delta _{\alpha _{2},0}}} \end {pmatrix}^T$ . Before computing the expression of large-time asymototics of $u(x,t)$ in this region, introduce the corresponding Riemann surface of genus three with ${a}_{\alpha _{2},j} ,{b}_{\alpha _{2},j}$ -cycles for $j=1,2,3$ , which is depicted in Figure 14 and

$$ \begin{align*}\mathcal{S}_{\alpha_{2}}:=\{(\lambda,\eta)|\eta^2=(\lambda^2-\eta_1^2)(\lambda^2-\eta_2^2) (\lambda^2-\eta_3^2)(\lambda^2-\alpha_{2}^2)\}. \end{align*} $$

Figure 14. The Riemann surface $\mathcal {S}_{\alpha _{2}}$ of genus three and its basis $\{{a}_{\alpha _{2},j} ,{b}_{\alpha _{2},j}\}~(j=1,2,3)$ of circles.

The normalized holomorphic differentials associated with the Riemann surface $\mathcal {S}_{\alpha _{2}}$ are denoted as $\omega _{\alpha _{2},j}~(j=1,2,3)$ . Suppose the period matrix of $\omega _{\alpha _{2},j}$ is $\tau _{\alpha _{2}}:=(\tau _{\alpha _{2},ij})_{3\times 3}$ with the symmetry $\tau _{\alpha _{2},11}=\tau _{\alpha _{2},33},\tau _{\alpha _{2},12}=\tau _{\alpha _{2},23}$ like the case in (3.2.2). Similarly, define the Jacobi map $J_{\alpha _{2}}(\lambda )$ as

(4.3.22) $$ \begin{align} J_{\alpha_{2}}(\lambda)=\int_{\alpha_{2}}^{\lambda} \hat\omega_{\alpha_{2}}:=\int_{\alpha_{2}}^{\lambda} \begin{pmatrix} \omega_{\alpha_{2},1}+\omega_{\alpha_{2},3}\\ 2\omega_{\alpha_{2},2} \end{pmatrix}, \end{align} $$

and the corresponding period matrix is

(4.3.23) $$ \begin{align} \hat\tau_{\alpha_{2}}= \begin{pmatrix} \tau_{\alpha_{2},11}+\tau_{\alpha_{2},31}& \tau_{\alpha_{2},12}+\tau_{\alpha_{2},32}\\ 2\tau_{\alpha_{2},21}& 2\tau_{\alpha_{2},22} \end{pmatrix}. \end{align} $$

In fact, the Jacobi map $J_{\alpha _{2}}(\lambda )$ satisfies similar properties in (3.2.6) and (3.2.7) just by replacing $\eta _{4}$ with $\alpha _{2}$ . In addition, by the symmetry of $J_{\alpha _{2}}(\lambda )$ like that in (3.2.8), it is derived that $d_{\alpha _{2}}=d=\frac {e_2+e_1}{2}$ . According to the second and third jump conditions in (4.3.2), it follows

(4.3.24) $$ \begin{align} \begin{aligned} & {\Omega}_{\alpha_{2},1}=24 \int_{\alpha_{2}}^{\eta_3} \frac{Q_{\alpha_{2},2}(\zeta)}{R_{\alpha_{2}}(\zeta)} d \zeta-8 \xi \int_{\alpha_{2}}^{\eta_3} \frac{Q_{\alpha_{2},1}(\zeta)}{R_{\alpha_{2}}(\zeta)} d \zeta, \\ & {\Omega}_{\alpha_{2},0}={\Omega}_{\alpha_{2},1}+24 \int_{\eta_{2}}^{\eta_1} \frac{Q_{\alpha_{2},2}(\zeta)}{R_{\alpha_{2}}(\zeta)} d \zeta-8 \xi \int_{\eta_{2}}^{\eta_1} \frac{Q_{\alpha_{2},1}(\zeta)}{R_{\alpha_{2}}(\zeta)} d \zeta. \end{aligned} \end{align} $$

Moreover, since the reconstruction formula involves the derivative with respect to x, the following lemma is necessary.

Thus for $\xi _{crit}^{(2)}<\xi <\xi _{crit}^{(3)}$ , the following theorem holds.

Theorem 4.10. For $\xi =\frac {x}{4t}$ , in the region $\xi _{crit}^{(2)}<\xi <\xi _{crit}^{(3)}$ , the large-time asymptotic behavior of the solution to the KdV equation with genus two soliton gas potential is described by

(4.3.25) $$ \begin{align} u(x,t)=-\left(2b_{\alpha_{2},1}+{\sum_{j=1}^3\eta_j^2+\alpha_{2}^2}+2\partial_x^2\log\left(\Theta\left(\frac{\Omega_{\alpha_{2}}}{2\pi i};\hat \tau_{\alpha_{2}}\right)\right)\right)+\mathcal{O}\left(\frac{1}{t}\right), \end{align} $$

where the parameter $\alpha _{2}$ is determined by (4.3.8).

Proof. Recall that

$$ \begin{align*} Y_1(\lambda)=\left(S_{\alpha_{2},1}^{\infty}(\lambda) +\frac{(\mathcal{E}_{\alpha_2,1}(x,t))_1}{t\lambda}+\mathcal{O}\left(\frac{1}{\lambda^2}\right)\right) e^{-tg_{\alpha_{2}}(\lambda)}f_{\alpha_{2}}(\lambda)^{-1}, \end{align*} $$

in which the function $f_{\alpha _{2}}(\lambda )$ behaves

$$ \begin{align*} f_{\alpha_{2}}(\lambda)=1+\frac{f_{\alpha_{2}}^{(1)}(\alpha_{2})}{\lambda}+\mathcal{O}\left(\frac{1}{\lambda^2}\right), \end{align*} $$

with

$$ \begin{align*} f_{\alpha_{2}}^{(1)}(\alpha_{2})=\left(\int_{\eta_1}^{\eta_{2}}+\int_{\eta_3}^{\alpha_{2}}\right)\frac{\zeta^4\log{r(\zeta)}}{R_{\alpha_{2}}(\zeta)} \frac{d\zeta}{\pi i}+\Delta_{\alpha_{2},0}\int_{-\eta_1}^{\eta_1}\frac{\zeta^4}{R_{\alpha_{2}}(\zeta)} \frac{d\zeta}{2\pi i}+\Delta_{\alpha_{2},1}\int_{\eta_2}^{\eta_3}\frac{\zeta^4}{R(\zeta)} \frac{d\zeta}{\pi i}, \end{align*} $$

and the term $\frac {(\mathcal {E}_{\alpha _2,1}(x,t))_1}{t\lambda }$ is the first entry of the error vector subject to the modulated two-phase wave region. Similar to the modulated one-phase case, one can conclude that the local parametrix near $\pm \alpha _2$ and $\pm \eta _j$ ( $j=1,2,3$ ) can be described by the Airy function and the modified Bessel function, respectively, and both of them contribute the error term $\mathcal {O}(t^{-1})$ in the asymptotic behavior of potential $u(x,t)$ for $t\to +\infty $ .

The derivative of the term $e^{-tg_{\alpha _{2}}(\lambda )}$ has the asymptotics

$$ \begin{align*} \partial_xe^{-tg_{\alpha_{1}}(\lambda)}=-\frac{1}{\lambda}\left[\frac{\eta_{1}^2+\eta_{2}^2+\eta_{3}^2+\alpha_{2}^2}{2}+b_{\alpha_{2},1}\right]+\mathcal{O}\left(\frac{1}{\lambda^2}\right). \end{align*} $$

In addition, from Lemma 4.7, it follows from Riemann Bilinear relations [Reference Bertola4] that

$$ \begin{align*} \sum \mathrm{res}_{\infty_{\pm}}\frac{\omega_{\alpha_{2},1}}{\lambda}=\frac{\partial_x\left(t\Omega_{\alpha_{2},1}\right)}{2\pi i},\ \sum \mathrm{res}_{\infty_{\pm}}\frac{\omega_{\alpha_{2},2}}{\lambda}=\frac{\partial_x(t\Omega_{\alpha_{2},0})}{2\pi i},\ \sum \mathrm{res}_{\infty_{\pm}}\frac{\omega_{\alpha_{2},3}}{\lambda}=\frac{\partial_x(t\Omega_{\alpha_{2},1})}{2\pi i}. \end{align*} $$

Recalling $J_{\alpha _{2}}(\infty )=(e_1+e_2)/2$ , the Jacobi map $J_{\alpha _{2}}(\lambda )$ has the asymptotics

$$ \begin{align*} J_{\alpha_{2}}(\lambda)=\frac{e_1+e_2}{2}-\frac{\begin{pmatrix} \partial_x(t\Omega_{\alpha_{2},1})&\partial_x(t\Omega_{\alpha_{2},0}) \end{pmatrix}^T}{\lambda}+\mathcal{O}\left(\frac{1}{\lambda^2}\right),\quad \lambda\to\infty. \end{align*} $$

Since $\partial _x{\Delta _{\alpha _{2}}}=\mathcal {O}\left (\frac {1}{t}\right )$ as $t\to +\infty $ , the Jacobi map $J_{\alpha _{2}}(\lambda )$ can be rewritten as

$$ \begin{align*}J_{\alpha_{2}}(\lambda)=\frac{e_1+e_2}{2}-\frac{\partial_x(\Omega_{\alpha_{2}})}{2\pi i \lambda}+\mathcal{O}\left(\frac{1}{\lambda^2}\right)+\mathcal{O}\left(\frac{1}{t}\right). \end{align*} $$

Thus the expansion of the function $S_{\alpha _{2},1}^{\infty }(\lambda )$ as $\lambda \to \infty $ is expressed by

$$ \begin{align*} \begin{aligned} {S}^{\infty}_{\alpha_{2},1}(\lambda)&=1-\frac{1}{\lambda}\left[\nabla\log\left(\Theta\left(\frac{\Omega_{\alpha_{2}}}{2\pi i};\hat\tau_{\alpha_{2}}\right)\right)-\nabla\log(\Theta(0;\hat\tau_{\alpha_{2}}))\right]\cdot \frac{ \partial_x(\Omega_{\alpha_{2}})}{2\pi i}+\mathcal{O}\left(\frac{1}{t}\right)+\mathcal{O}\left(\frac{1}{\lambda^2}\right),\\ &=1-\frac{1}{\lambda}\partial_x\log\left(\Theta\left(\frac{\Omega_{\alpha_{2}}}{2\pi i};\hat\tau_{\alpha_{2}}\right)\right)+\mathcal{O}\left(\frac{1}{t}\right)+\mathcal{O}\left(\frac{1}{\lambda^2}\right). \end{aligned} \end{align*} $$

Therefore, the asymptotics $\partial _x f_{\alpha _{2}}^{(1)}=\mathcal {O}\left (\frac {1}{t}\right )$ for $t\to +\infty $ and all the formulae above result in the large-time asymptotic behavior of $u(x,t)$ as

$$ \begin{align*} u(x,t)=-\left(2b_{\alpha_{2},1}+{\sum_{j=1}^3\eta_j^2+\alpha_{2}^2}+2\partial_x^2\log\left(\Theta\left(\frac{\Omega_{\alpha_{2}}}{2\pi i};\hat \tau_{\alpha_{2}}\right)\right)\right)+\mathcal{O}\left(\frac{1}{t}\right),\quad t\to +\infty.\\[-34pt] \end{align*} $$

4.4 Unmodulated two-phase wave region

For $\xi> \xi _{crit}^{(3)}$ , the large-time behavior of $u(x,t)$ is described by an unmodulated two-phase Riemann-Theta function. Similar to the case in Section 4.2.1, we only need to modify relevant notations, such as $ g_{\alpha _{2}} $ , $ R_{\alpha _{2}}(\lambda ) $ , $ \mathcal {S}_{\alpha _{2}} $ , $ \Omega _{\alpha _{2}} $ , and $ \Delta _{\alpha _{2}} $ in Section 4.3 by replacing $\alpha _{2}$ with $\eta _{4}$ , for example $R_{\eta _4}:=\sqrt {(\lambda ^2-\eta _4^2)(\lambda ^2-\eta _2^2)(\lambda ^2-\eta _3^2)(\lambda ^2-\eta _4^2)}$ . Similarly, one can verify that $ g_{\eta _{4}} $ and $ f_{\eta _{4}} $ can still deform the RH problem for $ Y(x,t;\lambda ) $ into a model problem $ S_{\eta _4}^{\infty }(\lambda ) $ . Indeed, the model problem $ S_{\eta _4}^{\infty }(\lambda ) $ is also similar to $ S^{\infty }(\lambda ) $ in (3.0.11) with the same jump contours, but the diagonal matrices are replaced by $ e^{(t{\Omega }_{j,\eta _{4}} + {\Delta }_{j,\eta _{4}})\sigma _3} $ for $j=0,1$ . We omit all the details here for brevity. Thus for $\xi> \xi _{crit}^{(3)}$ , the following theorem holds.

Theorem 4.11. For $\xi =\frac {x}{4t}$ , in the region $\xi _{crit}^{(3)}<\xi $ , the large-time asymptotic behavior of the solution to the KdV equation with genus two soliton gas potential is described by

(4.4.1) $$ \begin{align} u(x,t)=-\left(2b_{\eta_{4},1}+{\sum_{j=1}^4\eta_j^2}+2\partial_x^2\log\left(\Theta\left(\frac{\Omega_{\eta_{4}}}{2\pi i};\hat \tau_{\eta_{4}}\right)\right)\right)+\mathcal{O}\left(\frac{1}{t}\right), \end{align} $$

where $\hat \tau _{\eta _{4}}=\hat {\tau }$ in (3.2.4) and $b_{\eta _{4}}, \Omega _{\eta _{4}}=\begin {pmatrix} {t\Omega _{\eta _{4},1}+{\Delta _{\eta _{4},1}}}&{t\Omega _{\eta _{4},0}+{\Delta _{\eta _{4},0}}} \end {pmatrix}^T$ are defined by (4.3.6) and (4.3.24), respectively.