In this section, we will construct periodic wave solutions for the potential Kadomtsev–Petviashvili equation and the (\(3+1\))-dimensional potential-YTSF equation by employing the Hirota method and the Riemann theta function, as well as property (2.5).

### Potential Kadomtsev–Petviashvili equation

The following (\(2+1\))-dimensional potential Kadomtsev–Petviashvili equation [26] is considered:

$$ u_{t} + \frac{3}{4}u_{x}^{2} + \frac{1}{4}u_{xxx} + \frac{3}{4}\partial_{x}^{ - 1}u_{yy} = 0, $$

(4.1)

which is transformed into the bilinear form

$$ \bigl[4D_{x}D_{t} + D_{x}^{4} + 3D_{y}^{2}\bigr]F \cdot F = 0, $$

(4.2)

under the dependent variable transformation \(u = 2(\ln F)_{x}\).

We introduce the Riemann theta function solution of Eq. (4.1) as

$$ F = \sum_{n = - \infty}^{\infty} e^{2\pi in\zeta + \pi in^{2}\tau}, $$

(4.3)

where \(n \in \mathcal {Z}\), \(\tau \in \mathcal {C}\), \(\operatorname{Im}\tau > 0\) and \(\zeta = kx + ly + \omega t\).

Substituting (4.3) into (4.2), we get

$$\begin{aligned} &G(D_{x},D_{y},D_{t})F \cdot F \\ &\quad = G(D_{x},D_{y},D_{t})\sum _{n = - \infty}^{\infty} e^{2\pi in\zeta + \pi in^{2}\tau} \sum _{m = - \infty}^{\infty} e^{2\pi im\zeta + \pi im^{2}\tau} \\ &\quad = \sum _{n = - \infty}^{\infty} \sum_{m = - \infty}^{\infty} G (D_{x},D_{y},D_{t})e^{2\pi in\zeta + \pi in^{2}\tau} e^{2\pi im\zeta + \pi im^{2}\tau} \\ &\quad = \sum_{n = - \infty}^{\infty} \sum _{m = - \infty}^{\infty} G \bigl[2\pi i(n - m)k,2\pi i(n - m)l,2\pi i(n - m)\omega \bigr]e^{2\pi i(n + m)\zeta + \pi i(n^{2} + m^{2})\tau} \\ &\quad = \sum_{p = - \infty}^{\infty} \sum _{n = - \infty}^{\infty} G\bigl[2\pi i(2n - p)k,2\pi i(2n - p)l,2 \pi i(2n - p)\omega \bigr]e^{\pi i(n^{2} + (p - n)^{2})\tau} e^{2\pi ip\zeta} \\ &\quad = \sum _{p = - \infty}^{\infty} \bar{G} (p)e^{2\pi ip\zeta}, \end{aligned}$$

where \(n + m = p\). Noting that

$$\begin{aligned} \bar{G}(p) =& \sum_{n = - \infty}^{\infty} G \bigl(2 \pi i(2n - p)k, 2\pi i(2n - p)l, 2\pi i(2n - p)\omega \bigr]e^{\pi i(n^{2} + (p - n)^{2})\tau} \\ =& \sum_{N = - \infty}^{\infty} G\bigl[2\pi i\bigl(2N - (p - 2)\bigr)k,2\pi i\bigl(2N - (p - 2)\bigr)l,2\pi i \times \bigl(2N - (p - 2) \bigr)\omega )\bigr]\\ &{}\times e^{\pi i((N + 1)^{2} + (p - N - 1)^{2})\tau} \\ =& \sum_{N = - \infty}^{\infty} G \bigl[2\pi i\bigl(2N - (p - 2)\bigr)k,2\pi i\bigl(2N - (p - 2)\bigr)l,2\pi i\bigl(2N - (p - 2)\bigr) \omega )\bigr] \\ &{}\times e^{\pi i(N^{2} + (p - N - 2)^{2})\tau}e^{2\pi i(p - 1)\tau} \\ =& \bar{G}(p - 2)e^{2\pi i(p - 1)\tau}, \end{aligned}$$

which indicates that if \(\bar{G}(0) = \bar{G}(1) = 0\), then

$$ \bar{G}(p) = 0,\quad p \in \mathcal {Z}. $$

(4.4)

Therefore, we may let

$$\begin{aligned} & \bar{G}(0) = \sum_{n = - \infty}^{\infty} \bigl[ - 64\pi^{2}n^{2}k\omega + 256\pi^{4}n^{4}k^{4} - 48\pi^{2}n^{2}l^{2} + \mu \bigr] e^{2\pi in^{2}\tau} = 0, \end{aligned}$$

(4.5)

$$\begin{aligned} & \bar{G}(1) = \sum_{n = - \infty}^{\infty} \bigl[ - 16\pi^{2}(2n - 1)^{2}k\omega + 16 \pi^{4}(2n - 1)^{4}k^{4} - 12\pi^{2}(2n - 1)^{2}l^{2} + \mu \bigr] \\ &\hphantom{ \bar{G}(1) =}{}\times e^{\pi i(n^{2} + (n - 1)^{2})\tau} = 0. \end{aligned}$$

(4.6)

Denote

$$\begin{aligned} &\Delta_{1}(n) = e^{2\pi in^{2}\tau},\qquad \Delta_{2}(n) = e^{\pi i(n^{2} + (n - 1)^{2})\tau}, \\ &A_{11} = - \sum_{n = - \infty}^{\infty} 64 \pi^{2}n^{2}k\Delta_{1}(n),\qquad A_{12} = \sum_{n = - \infty}^{\infty} \Delta_{1} (n), \\ &A_{21} = - \sum_{n = - \infty}^{\infty} 1 6 \pi^{2}(2n - 1)^{2}k\Delta_{2}(n),\qquad A_{22} = \sum_{n = - \infty}^{\infty} \Delta_{2} (n), \\ &B_{1} = - \sum_{n = - \infty}^{\infty} \bigl( 256\pi^{4}n^{4}k^{4} - 48 \pi^{2}n^{2}l^{2}\bigr)\Delta_{1}(n), \\ &B_{2} = - \sum_{n = - \infty}^{\infty} \bigl[16\pi^{4}(2n - 1)^{4}k^{4} - 12 \pi^{2}(2n - 1)^{2}l^{2}\bigr] \Delta_{2}(n). \end{aligned}$$

Then Eqs. (4.5) and (4.6) can be written as

$$A_{11}\omega + A_{12}\mu = B_{1},\qquad A_{21}\omega + A_{22}\mu = B_{2}. $$

By solving this system, we get

$$ \omega = \frac{B_{1}A_{22} - A_{12}B_{2}}{A_{11}A_{22} - A_{12}A_{21}},\qquad \mu = \frac{A_{11}B_{2} - A_{21}B_{1}}{A_{11}A_{22} - A_{12}A_{21}}. $$

(4.7)

Thus the periodic wave solution is given by

$$ u = 2(\ln F)_{x}, $$

(4.8)

where *ω* and *F* are determined by Eqs. (4.7) and (4.3), respectively.

Next, we will demonstrate that the soliton solution can be obtained as a limit of a periodic wave solution. From Eq. (4.3), we rewrite *F* as

$$ F = 1 + \alpha \bigl(e^{2\pi i\zeta} + e^{ - 2\pi i\zeta} \bigr) + \alpha^{4}\bigl(e^{4\pi i\zeta} + e^{ - 4\pi i\zeta} \bigr) + \cdots, $$

(4.9)

where \(\alpha = e^{\pi i\tau} \).

Setting

$$K = 2\pi ik, \qquad L = 2\pi il, \qquad \varOmega = 2\pi i\omega,\qquad \zeta ' = Kx + Ly + \varOmega t + \pi i\tau, $$

we get

$$\begin{aligned} F =& 1 + \alpha \bigl(e^{2\pi i\zeta} + e^{ - 2\pi i\zeta} \bigr) + \alpha^{4}\bigl(e^{4\pi i\zeta} + e^{ - 4\pi i\zeta} \bigr) + \cdots \\ =& 1 + e^{\zeta '} + \alpha^{2}\bigl(e^{ - \zeta '} + e^{2\zeta '}\bigr) + \alpha^{6}\bigl(e^{ - 2\zeta '} + e^{3\zeta '}\bigr) + \cdots \\ \to& 1 + e^{\zeta '},\quad \mbox{as } \alpha \to 0 . \end{aligned}$$

(4.10)

Thus, the periodic wave solution (4.8) turns to the soliton solution

$$ u = 2(\ln F)_{x},\qquad F = 1 + e^{\zeta '},\qquad \zeta ' = Kx + Ly + \varOmega t + \pi i\tau, $$

(4.11)

if we can prove that

$$ \varOmega \to - \frac{K^{3}}{4} - \frac{3L^{2}}{4K}. $$

(4.12)

In fact, it is easy to known that

$$\begin{aligned} &A_{11} = - 128\pi^{2}k\bigl(\alpha^{2} + 4 \alpha^{8} + \cdots \bigr),\qquad A_{12} = 1 + 2\alpha^{2} + 2\alpha^{8} + \cdots, \\ &A_{21} = - 32\pi^{2}k\bigl(\alpha + 9\alpha^{5} + \cdots \bigr),\qquad A_{22} = 2\alpha + 2\alpha^{5} + \cdots, \\ &B_{1} = 2\bigl(256\pi^{4}k^{4} - 48 \pi^{2}l^{2}\bigr)\alpha^{2} + 2\bigl(256 \pi^{4}2^{4}k^{4} - 48\pi^{2}2^{2}l^{2} \bigr)\alpha^{8} + \cdots, \\ &B_{2} = - 2\bigl(16\pi^{4}k^{4} - 12 \pi^{2}l^{2}\bigr)\alpha + 2\bigl(16\pi^{4}3^{4}k^{4} - 12\pi^{2}3^{2}l^{2}\bigr)\alpha^{5} + \cdots, \end{aligned}$$

which lead to

$$\begin{aligned} \begin{aligned} &B_{1}A_{22} - A_{12}B_{2} = 2\bigl(16\pi^{4}k^{4} - 12\pi^{2}l^{2} \bigr)\alpha + o(\alpha ),\\ & A_{11}A_{22} - A_{12}A_{21} = 32\pi^{2}k\alpha + o(\alpha ), \end{aligned} \end{aligned}$$

(4.13)

According to (4.7), we get

$$ \omega \to \pi^{2}k^{3} - \frac{3l^{2}}{4k},\quad \mbox{as } \alpha \to 0 , $$

(4.14)

which is equivalent to

$$\varOmega \to - \frac{K^{3}}{4} - \frac{3L^{2}}{4K},\quad \mbox{as } \alpha \to 0. $$

### (\(3+1\))-dimensional potential-YTSF equation

By a similar analysis process as in Sect. 4.1, we have

$$\begin{aligned} &\bar{G}(0) = \sum_{n = - \infty}^{\infty} \bigl[64 \pi^{2}n^{2}k\omega + 256\pi^{4}n^{4}k^{3}m - 48\pi^{2}n^{2}l^{2} + \lambda \bigr] e^{2\pi in^{2}\tau} = 0, \end{aligned}$$

(4.15)

$$\begin{aligned} &\bar{G}(1) = \sum_{n = - \infty}^{\infty} \bigl[16 \pi^{2}(2n - 1)^{2}k\omega + 16\pi^{4}(2n - 1)^{4}k^{3}m - 12\pi^{2}(2n - 1)^{2}l^{2} + \lambda \bigr] \\ &\hphantom{\bar{G}(1) =}{}\times e^{\pi i(n^{2} + (n - 1)^{2})\tau} = 0. \end{aligned}$$

(4.16)

Denote

$$\begin{aligned} &\Delta_{1}(n) = e^{2\pi in^{2}\tau},\qquad \Delta_{2}(n) = e^{\pi i(n^{2} + (n - 1)^{2})\tau},\\\ & A_{11} = \sum_{n = - \infty}^{\infty} 6 4\pi^{2}n^{2}k\Delta_{1}(n), \qquad A_{12} = \sum_{n = - \infty}^{\infty} \Delta_{1} (n),\\ &A_{21} = \sum _{n = - \infty}^{\infty} 1 6\pi^{2}(2n - 1)^{2}k\Delta_{2}(n),\qquad A_{22} = \sum _{n = - \infty}^{\infty} \Delta_{2} (n), \\ &B_{1} = - \sum_{n = - \infty}^{\infty} \bigl( 256\pi^{4}n^{4}k^{3}m - 48 \pi^{2}n^{2}l^{2}\bigr)\Delta_{1}(n), \\ & B_{2} = - \sum_{n = - \infty}^{\infty} \bigl[16\pi^{4}(2n - 1)^{4}k^{3}m - 12 \pi^{2}(2n - 1)^{2}l^{2}\bigr] \Delta_{2}(n). \end{aligned}$$

Then Eqs. (4.15) and (4.16) can be written as

$$A_{11}\omega + A_{12}\lambda = B_{1},\qquad A_{21}\omega + A_{22}\lambda = B_{2}. $$

By solving the system, we get

$$ \omega = \frac{B_{1}A_{22} - A_{12}B_{2}}{A_{11}A_{22} - A_{12}A_{21}},\qquad \lambda = \frac{A_{11}B_{2} - A_{21}B_{1}}{A_{11}A_{22} - A_{12}A_{21}}. $$

(4.17)

Thus, the periodic wave solution is given by

$$ u = \frac{3}{2}(\ln F)_{x}, $$

(4.18)

where *ω* and *F* are determined by Eqs. (4.17) and (4.3), respectively.

From Eq. (4.3), we rewrite *F* as

$$ F = 1 + \delta \bigl(e^{2\pi i\zeta} + e^{ - 2\pi i\zeta} \bigr) + \delta^{4}\bigl(e^{4\pi i\zeta} + e^{ - 4\pi i\zeta} \bigr) + \cdots, $$

(4.19)

where \(\delta = e^{\pi i\tau} \).

Setting

$$\begin{aligned} &K = 2\pi ik,\qquad L = 2\pi il,\qquad M = 2\pi im, \qquad \varOmega = 2\pi i \omega,\\ & \zeta' = Kx + Ly + Mz + \varOmega t + \pi i\tau, \end{aligned}$$

yields

$$\begin{aligned} F =& 1 + \delta \bigl(e^{2\pi i\zeta} + e^{ - 2\pi i\zeta} \bigr) + \delta^{4}\bigl(e^{4\pi i\zeta} + e^{ - 4\pi i\zeta} \bigr) + \cdots \\ =& 1 + e^{\zeta '} + \delta^{2}\bigl(e^{ - \zeta '} + e^{2\zeta '}\bigr) + \delta^{6}\bigl(e^{ - 2\zeta '} + e^{3\zeta '}\bigr) + \cdots \\ \to& 1 + e^{\zeta '},\quad \mbox{as } \delta \to 0 . \end{aligned}$$

(4.20)

Thus, if we can prove that

$$ \varOmega \to \frac{K^{2}M}{4} + \frac{3L^{2}}{4K},\quad \mbox{as } \delta \to 0, $$

(4.21)

the periodic wave solution (4.18) turns to the soliton solution

$$ u = \frac{3}{2}(\ln F)_{x},\qquad F = 1 + e^{\zeta '}, \qquad \zeta ' = Kx + Ly + Mz + \varOmega t + \pi i\tau. $$

(4.22)

In fact, it is easy to known that

$$\begin{aligned} &A_{11} = 128\pi^{2}k\bigl(\delta^{2} + 4 \delta^{8} + \cdots \bigr), \\ &A_{12} = 1 + 2\delta^{2} + 2\delta^{8} + \cdots, \qquad A_{21} = 32\pi^{2}k\bigl(\delta + 9 \delta^{5} + \cdots \bigr),\qquad A_{22} = 2\delta + 2 \delta^{5} + \cdots, \\ &B_{1} = 2\bigl(256\pi^{4}k^{3}m - 48 \pi^{2}l^{2}\bigr)\delta^{2} + 2\bigl(256 \pi^{4}2^{4}k^{3}m - 48\pi^{2}2^{2}l^{2} \bigr)\delta^{8} + \cdots, \\ &B_{2} = - 2\bigl(16\pi^{4}k^{3}m - 12 \pi^{2}l^{2}\bigr)\delta + 2\bigl(16\pi^{4}3^{4}k^{3}m - 12\pi^{2}3^{2}l^{2}\bigr)\delta^{5} + \cdots, \end{aligned}$$

which lead to

$$\begin{aligned} &B_{1}A_{22} - A_{12}B_{2} = 2\bigl(16 \pi^{4}k^{3}m - 12\pi^{2}l^{2}\bigr) \delta + o(\delta ),\\ & A_{11}A_{22} - A_{12}A_{21} = - 32\pi^{2}k\delta + o(\delta ). \end{aligned}$$

According to (4.17), we get

$$\omega \to - \pi^{2}k^{2}m + \frac{3l^{2}}{4k},\quad \mbox{as } \delta \to 0, $$

which is equivalent to

$$\varOmega \to \frac{K^{2}M}{4} + \frac{3L^{2}}{4K},\quad \mbox{as } \delta \to 0. $$