0 引言

在理论物理和数学物理中，用带谱参数λ的Lax对去发现新的可积系统是一个有意义的研究领域.可积系统的另一个关键特征是它们具有无穷多守恒律^[1-2].具有Lax表示的可积系统可以用反散射变换方法^[3-4]、代数几何方法^[5-7]、Darboux方法^[8-9]及其他方法^[10-13]求解，尤其是起源于1882年Darboux对Sturm-Liouville方程进行研究时所提出的Darboux变换方法，它是一种获得演化方程精确解^[14-16]的非常有效的方法.

文中我们给出了一个与3×3矩阵谱问题相关的新的非线性演化方程族，并且通过考虑两个相关的线性谱问题，得到了这个方程族中前两个非线性演化方程的无穷多守恒律

$ \left\{ {\begin{array}{*{20}{l}} {{u_t} = 2{u_x} + 6{v_x},}\\ {{v_t} = - 2{u_{xxx}} + 16u{u_x} - 6{v_x}.} \end{array}} \right. $

(1)

$ \left\{ {\begin{array}{*{20}{l}} {{u_t} = {{\left( {6{u_{xx}} + 9{v_{xx}} - 24{u^2} - 36uv - 4u - 3v} \right)}_x},}\\ {{v_t} = {{\left( { - 3{u_{xxxx}} - 6{v_{xx}} + {u_{xx}} + 36u{u_{xx}} + 18u_x^2 - 32{u^3} - 4{u^2} - 18{v^2}} \right)}_x}.} \end{array}} \right. $

(2)

最后，通过引入3×3矩阵谱问题的规范变换，得到了方程(1)的Darboux变换，并由此给出一些精确解.

1 非线性演化方程族和无穷多守恒律

本节，我们首先引入一个以u，v为位势, λ为常谱参数的3×3矩阵谱问题

$ {\mathit{\boldsymbol{\psi }}_x} = \mathit{\boldsymbol{U\psi }};\mathit{\boldsymbol{\psi }} = \left( {\begin{array}{*{20}{l}} {{\psi _1}}\\ {{\psi _2}}\\ {{\psi _3}} \end{array}} \right);\mathit{\boldsymbol{U}} = \left( {\begin{array}{*{20}{l}} 0&u&{\lambda + v}\\ 1&1&{\;\;u}\\ 0&1&{\;\;0} \end{array}} \right). $

(3)

然后借助于零曲率方程得到与之相联系的非线性演化方程族.为了推导相对应的非线性演化方程，我们定义两组Lenard递推方程

$ \left\{ {\begin{array}{*{20}{l}} {\mathit{\boldsymbol{K}}{\mathit{\boldsymbol{s}}_j} = \mathit{\boldsymbol{J}}{s_{j + 1}},j \ge 0,{{\left. {{{\boldsymbol{s}}_j}} \right|}_{(u,v) = 0}} = 0,j \ge 1,}\\ {\mathit{\boldsymbol{K}}{{\mathit{\boldsymbol{\hat s}}}_j} = \mathit{\boldsymbol{ J}}{{\hat s}_{j + 1}},j \ge 0,{{\left. {{{\mathit{\boldsymbol{\hat s}}}_j}} \right|}_{(u,v) = 0}} = 0,j \ge 1,} \end{array}} \right. $

(4)

其中：初值s₀=(1, 0)^T, $ {\mathit{\boldsymbol{\hat s}}_0} = {(0, 1)^{\rm{T}}} $; 两个反称算子分别为(∂∂^－1=∂^－1∂=1),

$ \mathit{\boldsymbol{K}} = \left( {\begin{array}{*{20}{c}} { - {\partial ^3} + \partial u + u\partial + \frac{1}{3}\partial }&{\frac{2}{3}{\partial ^3} + 2\partial v + v\partial + \frac{2}{3}\partial u}\\ {\frac{2}{3}{\partial ^3} + 2v\partial + \partial v + \frac{2}{3}u\partial }&{{K_{22}}} \end{array}} \right);\mathit{\boldsymbol{J}} = \left( {\begin{array}{*{20}{c}} 0&{ - 3\partial }\\ { - 3\partial }&{4\partial } \end{array}} \right), $

(5)

$K_{22}=\frac{1}{3} \partial^{5}-\frac{2}{3} \partial^{3} u-\frac{2}{3} u \partial^{3}-\partial^{3}-\partial u \partial^{2}-\partial^{2} u \partial+2 \partial u^{2}+2 u^{2} \partial+\frac{4}{3} u \partial u-2 \partial v-2 v \partial $.从而我们可以唯一确定s_j和$ \hat{\boldsymbol{s}}_{j} $, 例如

$ {\mathit{\boldsymbol{s}}_1} = \left( {\begin{array}{*{20}{c}} { - \frac{1}{9}(4u + 3v)}\\ { - \frac{1}{3}u} \end{array}} \right);{{\mathit{\boldsymbol{\hat s}}}_1} = \left( {\begin{array}{*{20}{c}} {\frac{1}{{27}}\left( {6{u_{xx}} - 24{u^2} - 8u - 6v} \right)}\\ { - \frac{2}{9}(u + 3v)} \end{array}} \right). $

(6)

接下来，我们考虑静态零曲率方程

$ {\mathit{\boldsymbol{V}}_x} - \left[ {\mathit{\boldsymbol{U}},\mathit{\boldsymbol{V}}} \right] = 0;\mathit{\boldsymbol{V}} = \left( {\begin{array}{*{20}{l}} {{V_{11}}}&{{V_{12}}}&{{V_{13}}}\\ {{V_{21}}}&{{V_{22}}}&{{V_{23}}}\\ {{V_{31}}}&{{V_{32}}}&{{V_{33}}} \end{array}} \right), $

(7)

它等价于

$ \left\{ \begin{array}{l} {V_{11,x}} + {V_{12}} - u{V_{21}} - (\lambda + v){V_{31}} = 0,\\ {V_{12,x}} + {V_{12}} + {V_{13}} + u\left( {{V_{11}} - {V_{22}}} \right) - (\lambda + v){V_{32}} = 0,\\ {V_{13,x}} - u{V_{23}} + u{V_{12}} + (\lambda + v)\left( {{V_{11}} - {V_{33}}} \right) = 0,\\ {V_{21,x}} - {V_{21}} + {V_{22}} - {V_{11}} - u{V_{31}} = 0,\\ {V_{22,x}} + {V_{23}} - {V_{12}} + u{V_{21}} - u{V_{32}} = 0,\\ {V_{23,x}} - {V_{13}} - {V_{23}} + u\left( {{V_{22}} - {V_{33}}} \right) + (\lambda + v){V_{21}} = 0,\\ {V_{31,x}} + {V_{32}} - {V_{21}} = 0,\\ {V_{32,x}} + {V_{33}} - {V_{22}} + u{V_{31}} + {V_{32}} = 0,\\ {V_{33,x}} - {V_{23}} + u{V_{32}} + (\lambda + v){V_{31}} = 0. \end{array} \right. $

(8)

如果我们定义$ (a, b) = \sum\limits_{j = 0}^\infty {\left( {{a_j}, {b_j}} \right)} {\lambda ^{ - j}} $和

$ \begin{array}{l} {V_{11}} = \frac{1}{3}\left( {3\partial - 1} \right)a + \frac{1}{3}\left( {{\partial ^2} - 3\partial - 2u} \right)b;{V_{12}} = \frac{1}{3}\left( {\partial - 3{\partial ^2} + 3u} \right)a +\\ \frac{1}{3}\left( {3{\partial ^2} - {\partial ^3} + 2\partial u + 3u\partial + 6v} \right)b + 2\lambda b;\\ {V_{13}} = \frac{1}{3}\left( {2{\partial ^2} + 3v} \right)a + \frac{1}{3}\left( {{\partial ^4} - 2{\partial ^2}u - 3u{\partial ^2} - 3{\partial ^2} - 3{\partial u \partial} + 6{u^2} - 6v} \right)\\ b + \lambda (a - 2b);{V_{21}} = a + \partial b;\\ {V_{22}} = \frac{2}{3}a - \frac{2}{3}\left( {{\partial ^2} - 2u} \right)b;{V_{23}} = - \frac{1}{3}\left( {3{\partial ^2} + \partial - 3u} \right)a +\\ \frac{1}{3}\left( {{\partial ^3} + 3{\partial ^2} - 2\partial u - 3u\partial + 6v} \right)b + 2\lambda b;\\ {V_{31}} = 2b;{V_{32}} = a - \partial b;{V_{33}} = - \frac{1}{3}(3\partial + 1)a + \frac{1}{3}\left( {{\partial ^2} + 3\partial - 2u} \right)b. \end{array} $

(9)

把式(9)代入式(8)，可以得到

$ \mathit{\boldsymbol{K}}{\mathit{\boldsymbol{S}}_j} = \mathit{\boldsymbol{J}}{\mathit{\boldsymbol{S}}_{j + 1}},j \ge 0;\mathit{\boldsymbol{J}}{\mathit{\boldsymbol{S}}_0} = 0. $

(10)

易见方程JS₀=0有通解$ \boldsymbol{S}_{0}=\boldsymbol{\alpha}_{0} \boldsymbol{s}_{0}+\hat{\boldsymbol{\alpha}}_{0} \hat{\boldsymbol{s}}_{0} $，从而$\boldsymbol{S}_{j}=\left(a_{j}, b_{j}\right)^{\mathrm{T}}=\alpha_{0} \boldsymbol{s}_{j}+\hat{\boldsymbol{\alpha}}_{0} \hat{\boldsymbol{s}}_{j}$，满足递推方程(10), 这里的α₀和$ \hat{\alpha}_{0} $是任意常数.

令ψ满足谱问题(3)和辅谱问题

$ {\mathit{\boldsymbol{\psi }}_{{t_n}}} = {\mathit{\boldsymbol{V}}^{(n)}}\mathit{\boldsymbol{\psi }};{\mathit{\boldsymbol{V}}^{(n)}} = \left( {\begin{array}{*{20}{l}} {V_{11}^{(n)}}&{V_{12}^{(n)}}&{V_{13}^{(n)}}\\ {V_{21}^{(n)}}&{V_{22}^{(n)}}&{V_{23}^{(n)}}\\ {V_{31}^{(n)}}&{V_{32}^{(n)}}&{V_{33}^{(n)}} \end{array}} \right), $

(11)

其中：V_ij⁽ⁿ⁾=V_ij(a⁽ⁿ⁾，b⁽ⁿ⁾); $ \left( {{a^{(n)}}, {b^{(n)}}} \right) = \sum\limits_{l = 0}^n {\left( {{a_l}, {b_l}} \right)} {\lambda ^{n - l}} $.由式(3)和式(11)的相容性条件可得零曲率方程，U_tn－V_x⁽ⁿ⁾+[U，V⁽ⁿ⁾]=0，它等价于一个非线性演化方程族

$ {\left( {{u_{{t_n}}},{v_{{t_n}}}} \right)^{\rm{T}}} = \mathit{\boldsymbol{K}}{\mathit{\boldsymbol{S}}_n} = \mathit{\boldsymbol{J}}{\mathit{\boldsymbol{S}}_{n + 1}},n \ge 0. $

(12)

该方程族的前两个方程为

$ \left\{ {\begin{array}{*{20}{l}} {{u_{{t_0}}} = {\alpha _0}{u_x} + \frac{2}{3}{{\hat \alpha }_0}\left( {{u_x} + 3{v_x}} \right),}\\ {{v_{{t_0}}} = {\alpha _0}{v_x} - \frac{1}{3}{{\hat \alpha }_0}\left( {2{u_{xxx}} - 16u{u_x} + 6{v_x}} \right),} \end{array}} \right. $

(13)

$ \begin{array}{l} {u_{{t_1}}} = \frac{1}{{27}}{\alpha _0}\left( {6{u_{xxx}} + 9{v_{xxx}} - 48u{u_x} - 36{u_x}v - 36u{v_x} - 4{u_x} - 3{v_x}} \right) - \\ \frac{2}{{81}}{{\hat \alpha }_0}\left( {9{u_{xxxxx}} - 90u{u_{xxx}} - 9{u_{xxx}} - } \right.\\ \left. {225{u_x}{u_{xx}} + 180{u^2}{u_x} + 72u{u_x} + 4{u_x} + 9{v_{xxx}} + 135v{v_x} + 54{u_x}v + 54u{v_x} + 3{v_x}} \right);\\ {v_{{t_1}}} = \frac{1}{{81}}{{\hat \alpha }_0} \left( {6{u_{xxxxx}} - 18{v_{xxxxx}} + 180u{v_{xxx}} + 90{u_{xxv}}v - 12u{u_{xxx}}}\\ + 180{u_{xx}}{v_x} + 2{u_{xxx}} + 42{v_{xxx}} + 270{u_x}{v_{xx}} - \right.\\ \left. {144{u_x}{u_{xx}} + 252v{v_x} - 360{u^2}{v_x} - 720uv{u_x} - 288{u^2}{u_x} - 16u{u_x}} \right) -\\ \frac{1}{{27}}{\alpha _0}\left( {3{u_{xxxx}} + 6{v_{xxx}} - {u_{xxx}} - } \right.\\ \left. {36u{u_{xxx}} - 72{u_x}{u_{xx}} + 36v{v_x} + 96{u^2}{u_x} + 8u{u_x}} \right). \end{array} $

(14)

当α₀=0，$ \hat{\alpha}_{0}=3 $，t₀=t时，方程(13)约化为方程(1)；当α₀=27，$ \hat{\alpha}_{0}=0 $，t₁=t时，方程(14)约化为方程(2).

下面我们将推导式(13)和式(14)的无穷多守恒律.假设$\phi=\frac{\psi_{3, x}}{\psi_{3}} $, 由谱问题(3)可得Riccati方程

$ {\phi _{xx}} + 3\phi {\phi _x} - {\phi _x} + {\phi ^3} - {\phi ^2} - 2u\phi = {u_x} + v + \lambda . $

(15)

令λ=η³及$ \phi = \sum\limits_{j = - 1}^\infty {{\phi _j}} {\eta ^{ - j}}$, 把它们代入到方程(15)，并比较λ的同次幂系数，可得ϕ_j的表达式

$ \begin{array}{l} {\phi _{ - 1}} = 1;{\phi _0} = \frac{1}{3};{\phi _1} = \frac{2}{3}u + \frac{1}{9};{\phi _2} = - \frac{1}{3}{u_x} + \frac{2}{9}u + \frac{1}{3}v + \frac{2}{{81}};{\phi _3} = \frac{1}{9}{u_{xx}} - \frac{2}{9}{u_x} - \frac{1}{3}{v_x};\\ {\phi _j} = \frac{1}{3}\\ \left( {\sum\limits_{l = - 1}^{j - 1} {{\phi _l}{\phi _{j - 2 - l}}} - \sum\limits_{\begin{array}{*{20}{c}} {l + m + n = j - 2}\\ { - 1 \le l,m,n < j} \end{array}} {{\phi _l}{\phi _m}{\phi _n}} - 3\sum\limits_{l = - 1}^{j - 1} {{\phi _l}{\phi _{j - 2 - l,x}}} + {\phi _{j - 2,x}} - {\phi _{j - 2,xx}} + 2u{\phi _{j - 2}}} \right),\\ j \ge 3. \end{array} $

(16)

另外，由辅谱问题可以得到

$ \frac{{{\psi _{3,t}}}}{{{\psi _3}}} = V_{31}^{(n)}\left( {{\phi ^2} + {\phi _x} - \phi - u} \right) + V_{32}^{(n)}\phi + V_{33}^{(n)}. $

(17)

定义θ=V₃₁⁽ⁿ⁾(ϕ²+ϕ_x－ϕ－u)+V₃₂⁽ⁿ⁾ϕ+V₃₃⁽ⁿ⁾，由$ \frac{\partial}{\partial_{t}} \frac{\psi_{3, x}}{\psi_{3}}=\frac{\partial}{\partial_{x}} \frac{\psi_{3, t}}{\psi_{3}}$可知ϕ_t=θ_x.

对于方程(13)，当n=0时有$V_{31}^{(0)}=2 \hat{\alpha}_{0}$，V₃₂⁽⁰⁾=α₀，$ V_{33}^{(0)}=-\frac{1}{3} \alpha_{0}-\frac{2}{3} u \hat{\alpha}_{0} $.

把ϕ的展开式和上式代入到式(17)中，可得

$ \begin{array}{*{20}{c}} {\theta = 2{{\hat \alpha }_0}{\eta ^2} + \left( {{\alpha _0} - \frac{2}{3}{{\hat \alpha }_0}} \right)\eta + \sum\limits_{j = 1}^\infty {{\theta _j}} {\eta ^{ - j}};}\\ {{\theta _1} = {\alpha _0}\left( {\frac{2}{3}u + \frac{1}{9}} \right) + 2{{\hat \alpha }_0}\left( {\frac{2}{9}u + \frac{2}{3}v + \frac{1}{{81}}} \right);{\theta _j} = 2{{\hat \alpha }_0}\left( {\sum\limits_{l = - 1}^{j + 1} {{\phi _l}} {\phi _{j - l}} +{\phi _{j,x}} - {\phi _j}} \right)\\ + {\alpha _0}{\phi _j},j \ge 1.} \end{array} $

(18)

我们把ϕ和θ的展开式中的系数ϕ_j和θ_j叫作守恒密度及连带流.方程(13)的第一个守恒律为

$ {\left( {\frac{2}{3}u + \frac{1}{9}} \right)_{{t_0}}} = {\alpha _0}{\left( {\frac{2}{3}u + \frac{1}{9}} \right)_x} + 2{{\hat \alpha }_0}{\left( {\frac{2}{9}u + \frac{2}{3}v + \frac{1}{{81}}} \right)_x}. $

(19)

对于方程(14)，当n=1时，

$ \begin{array}{l} V_{31}^{(1)} = - \frac{2}{3}{\alpha _0}u - \frac{4}{9}{{\hat \alpha }_0}(u + 3v) + 2{{\hat \alpha }_0}\lambda ;\\ V_{32}^{(2)} = \frac{1}{9}{\alpha _0}\left( {3{u_x} - 4u - 3v} \right) + {\alpha _0}\lambda + \frac{1}{{27}}{{\hat \alpha }_0}\left( {6{u_{xx}} + 6{u_x} - 24{u^2} - 8u + 18{v_x} - 6v} \right);\\ V_{33}^{(1)} = \frac{1}{{27}}{\alpha _0}\left( {3{u_x} - 3{u_{xx}} + 6{u^2} + 4u + 9{v_x} + 3v} \right) + \\ \frac{1}{{81}}{{\hat \alpha }_0}\left( {36{u^2} - 18{u_{xxx}} - 12{u_{xx}} + 144u{u_x} + } \right.\\ \left. {6{u_x} + 8u - 18{v_{xx}} + 6v + 36uv} \right) - \frac{1}{3}\left( {{\alpha _0} + 2u{{\hat \alpha }_0}} \right)\lambda . \end{array} $

(20)

此时

$ \begin{array}{l} \theta = 2{{\hat \alpha }_0}{\eta ^5} + \frac{1}{3}\left( {3{\alpha _0} - 2{{\hat \alpha }_0}} \right){\eta ^4} + \frac{1}{{81}}\left( {2{{\hat \alpha }_0} + 9{\alpha _0}} \right){\eta ^2} + \frac{1}{{243}}\left( {2{{\hat \alpha }_0} + 6{\alpha _0}} \right)\eta + \sum\limits_{j = 1}^\infty {{\theta _j}} {\eta ^{ - j}};\\ {\theta _1} = \frac{1}{{81}}{\alpha _0}\left( {12{u_{xx}} + 18{v_{xx}} - 48{u^2} - 72uv - 8u - 6v - \frac{2}{9}} \right) -\\ \frac{2}{{243}}{{\hat \alpha }_0}\left( {18{u_{xxxx}} - 180u{u_{xx}} - 18{u_{xx}} + 120{u^3} + } \right.\\ \left. {72{u^2} + 8u + 18{v_{xx}} - 135u_x^2 + 135{v^2} + 108uv + 6v + \frac{4}{{27}}} \right);\\ {\theta _j} = \left[ {\frac{1}{{27}}{{\hat \alpha }_0}\left( {6{u_{xx}} + 6{u_x} + 18{v_x} - 24{u^2} + 4u + 30v} \right) + \frac{1}{9}{\alpha _0}\left( {2u - 3v + 3{u_x}} \right)} \right]{\phi _j} +\\ 2{{\hat \alpha }_0}\left( {\sum\limits_{l = - 1}^{j + 4} {{\phi _l}{\phi _{j + 3 - l}}} + {\phi _{j + 3,x}}} \right) + \\ \left( {{\alpha _0} - 2{{\hat \alpha }_0}} \right){\phi _{j + 3}} - \left[ {\frac{2}{3}{\alpha _0}u + \frac{4}{9}{{\hat \alpha }_0}(u + 3v)} \right]\left( {\sum\limits_{l = - 1}^{j + 1} {{\phi _l}{\phi _{j - l}}} + {\phi _{j,x}}} \right),j \ge 1. \end{array} $

(21)

方程(14)的第一个守恒律为

$ \begin{array}{l} \left( {\frac{2}{3}u + \frac{1}{9}} \right){t_1} = \frac{1}{{81}}{\alpha _0}{\left( {12{u_{xx}} + 18{v_{xx}} - 48{u^2} - 72uv - 8u - 6v - \frac{2}{9}} \right)_x} -\\ \frac{2}{{243}}{{\hat \alpha }_0}\left( {18{u_{xxxx}} - 180u{u_{xx}} - 18{u_{xx}} + } \right.\\ {\left. {120{u^3} + 72{u^2} + 8u + 18{v_{xx}} - 135u_x^2 + 135{v^2} + 108uv + 6v + \frac{4}{{27}}} \right)_x}. \end{array} $

(22)

2 方程(1)的Darboux变换和精确解

本节，我们将通过Darboux变换给出方程(1)的精确解.该方程所满足的Lax方程为矩阵谱问题(3)和如下辅谱问题

$ {\mathit{\boldsymbol{\psi }}_t} = {\mathit{\boldsymbol{V}}^{(0)}}\mathit{\boldsymbol{\psi }} = \left( {\begin{array}{*{20}{c}} { - 2u}&{2{u_x} + 6v + 6\lambda }&{ - 2{u_{xx}} + 6{u^2} - 6v - 6\lambda }\\ 0&{4u}&{ - 2{u_z} + 6v + 6\lambda }\\ 6&0&{ - 2u} \end{array}} \right)\mathit{\boldsymbol{\psi }}. $

(23)

首先我们假设ψ^(l)=(ψ₁^(l)，ψ₂^(l)，ψ₃^(l))^T，1≤l≤3，是方程(3)和(23)的3个解，由此可以定义一个基解矩阵Ψ=(ψ⁽¹⁾，ψ⁽²⁾，ψ⁽³⁾).接下来我们引入谱问题(3)和(23)的规范变换$ \mathit{\boldsymbol{ \boldsymbol{\varPsi} }} \to \mathit{\boldsymbol{ \boldsymbol{\widehat \varPsi} }}:\mathit{\boldsymbol{ \boldsymbol{\widehat \varPsi} }} = \mathit{\boldsymbol{T \boldsymbol{\varPsi} }} $.

其中：

$ \begin{array}{l} \mathit{\boldsymbol{T}} = \left( {\begin{array}{*{20}{c}} {{T_{11}}}&{{T_{12}}}&{{T_{13}}}\\ {{b_x} + c}&{\lambda + {c_x} + bu + c + d}&{\lambda b + {d_x} + bv + cu}\\ b&c&{\lambda + d} \end{array}} \right);\\ {T_{11}} = \lambda + {b_{xx}} + 2{c_x} + d - {b_x} - \frac{3}{2}b{b_x};\\ {T_{12}} = \lambda b + {c_{xx}} + 2{b_x}u + b{u_x} + {c_x} + 2{d_x} + bv + cu - \frac{3}{2}{b_x}c;\\ {T_{13}} = \lambda \left( {\frac{1}{2}{b_x} + c - b} \right) + {d_{xx}} + 2{b_x}v + b{v_x} + 2{c_x}u + c{u_x} + cv + b{u^2} - {d_x} - bv - \frac{3}{2}{b_x}d. \end{array} $

(24)

这里的b，c和d待定.通过计算可知det T是λ的三次多项式.令λ₁、λ₂、λ₃是3个已知的任意参数，并且是det T的3个根，即det T=(λ－λ₁)(λ－λ₂)(λ－λ₃).易知当λ=λ_i(i=1, 2, 3)时，$ \mathit{\boldsymbol{ \boldsymbol{\widehat \varPsi} }} $的列向量是线性相关的.因此，我们得到了下面的线性代数方程组

$ \begin{array}{l} {\lambda _i} + {b_{xx}} + 2{c_x} + d - {b_x} - \frac{3}{2}b{b_x} + \\ \left( {{\lambda _i}b + {c_{xx}} + 2{b_x}u + b{u_x} + {c_x} + 2{d_x} + bv + cu - \frac{3}{2}{b_x}c} \right) \sigma _1^{(i)} + \\ \left[ {{\lambda _i}\left( {\frac{1}{2}{b_x} + c - b} \right) + {d_{xx}} + 2{b_x}v + b{v_x} + 2{c_x}u + c{u_x} + cv + b{u^2} - {d_x} - bv - \frac{3}{2}{b_x}d} \right]\\ \sigma _2^{(i)} = 0, \end{array} $

(25)

$ {b_x} + c + \left( {{\lambda _i} + {c_x} + bu + c + d} \right)\sigma _1^{(i)} + \left( {{\lambda _i}b + {d_x} + bv + cu} \right)\sigma _2^{(i)} = 0, $

(26)

$ b + c\sigma _1^{(i)} + \left( {{\lambda _i} + d} \right)\sigma _2^{(i)} = 0, $

(27)

其中：

$ \sigma _1^{(i)} = \frac{{r_1^{(i)}\psi _2^{(1)}\left( {{\lambda _i}} \right) + r_2^{(i)}\psi _2^{(2)}\left( {{\lambda _i}} \right) + r_3^{(i)}\psi _2^{(3)}\left( {{\lambda _i}} \right)}}{{r_1^{(i)}\psi _1^{(1)}\left( {{\lambda _i}} \right) + r_2^{(i)}\psi _1^{(2)}\left( {{\lambda _i}} \right) + r_3^{(i)}\psi _1^{(3)}\left( {{\lambda _i}} \right)}};\sigma _2^{(i)} = \frac{{r_1^{(i)}\psi _3^{(1)}\left( {{\lambda _i}} \right) + r_2^{(i)}\psi _3^{(2)}\left( {{\lambda _i}} \right) + r_3^{(i)}\psi _3^{(3)}\left( {{\lambda _i}} \right)}}{{r_1^{(i)}\psi _1^{(1)}\left( {{\lambda _i}} \right) + r_2^{(i)}\psi _1^{(2)}\left( {{\lambda _i}} \right) + r_3^{(i)}\psi _1^{(3)}\left( {{\lambda _i}} \right)}}, $

(28)

而r_k⁽ⁱ⁾(i=1，2，3；k=1，2，3)是相关系数.通过计算，由式(27)可以推出式(26)，由式(26)可以推出式(25).如果合理选择参数λ_i和r_k⁽ⁱ⁾，使得式(27)所确定的系数行列式非零，那么b、c、d由式(27)唯一确定.

定理1  在线性变换$ \mathit{\boldsymbol{\hat \psi }} = \mathit{\boldsymbol{T\psi }} $下，式(3)和式(23)被转换成$\mathit{\boldsymbol{\hat \psi }}$的两个3×3矩阵谱问题

$ {{\mathit{\boldsymbol{\hat \psi }}}_x} = \mathit{\boldsymbol{\hat U\hat \psi }},{{\mathit{\boldsymbol{\hat \psi }}}_t} = {{\mathit{\boldsymbol{\hat V}}}^{(0)}}\mathit{\boldsymbol{\hat \psi }}, $

(29)

其中：$ \hat{\boldsymbol{U}}=\left(\boldsymbol{T}_{x}+\boldsymbol{T} \boldsymbol{U}\right) \boldsymbol{T}^{-1} $；$ \hat{\boldsymbol{V}}^{(0)}=\left(\boldsymbol{T}_{t}+\boldsymbol{T} \boldsymbol{V}^{(0)}\right) \boldsymbol{T}^{-1} $.

$ \mathit{\boldsymbol{\hat U}} = \left( {\begin{array}{*{20}{c}} 0&{\hat u}&{\lambda + \hat v}\\ 1&1&{\hat u}\\ 0&1&0 \end{array}} \right),{{\mathit{\boldsymbol{\hat V}}}^{(0)}} = \left( {\begin{array}{*{20}{c}} { - 2\hat u}&{2{{\hat u}_x} + 6\hat v + 6\lambda }&{ - 2{{\hat u}_{xx}} + 6{{\hat u}^2} - 6\hat v - 6\lambda }\\ 0&{4\hat u}&{ - 2{{\hat u}_x} + 6\hat v + 6\lambda }\\ 6&0&{ - 2\hat u} \end{array}} \right). $

(30)

新旧位势之间的变换公式为

$ \hat u = u + \frac{3}{2}{b_x};\hat v = v + \frac{3}{2}{b_{xx}} + 3{c_x} - 2{b_x} - 3b{b_x}. $

(31)

下面通过变换式(31)，我们构造方程(1)的精确解.易见方程(27)可以改写为

$ \left\{ {\begin{array}{*{20}{l}} {b + c\sigma _1^{(1)} + d\sigma _2^{(1)} = - {\lambda _1}\sigma _2^{(1)},}\\ {b + c\sigma _1^{(2)} + d\sigma _2^{(2)} = - {\lambda _2}\sigma _2^{(2)},}\\ {b + c\sigma _1^{(3)} + d\sigma _2^{(3)} = - {\lambda _3}\sigma _2^{(3)},} \end{array}} \right. $

(32)

由克莱姆法则得

$ b = {\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}_1}/\mathit{\boldsymbol{ \boldsymbol{\varDelta} }};c = {\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}_2}/\mathit{\boldsymbol{ \boldsymbol{\varDelta} }};d = {\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}_3}/\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}, $

(33)

其中：

$ \mathit{\boldsymbol{ \boldsymbol{\varDelta} }} = \left| {\begin{array}{*{20}{c}} 1&{\sigma _1^{(1)}}&{\sigma _2^{(1)}}\\ 1&{\sigma _1^{(2)}}&{\sigma _2^{(2)}}\\ 1&{\sigma _1^{(3)}}&{\sigma _2^{(3)}} \end{array}} \right|;{\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}_1} = \left| {\begin{array}{*{20}{c}} { - {\lambda _1}\sigma _2^{(1)}}&{\sigma _1^{(1)}}&{\sigma _2^{(1)}}\\ { - {\lambda _2}\sigma _2^{(2)}}&{\sigma _1^{(2)}}&{\sigma _2^{(2)}}\\ { - {\lambda _3}\sigma _2^{(3)}}&{\sigma _1^{(3)}}&{\sigma _2^{(3)}} \end{array}} \right|; \\ {\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}_2} = \left| {\begin{array}{*{20}{c}} 1&{ - {\lambda _1}\sigma _2^{(1)}}&{\sigma _2^{(1)}}\\ 1&{ - {\lambda _2}\sigma _2^{(2)}}&{\sigma _2^{(2)}}\\ 1&{ - {\lambda _3}\sigma _2^{(3)}}&{\sigma _2^{(3)}} \end{array}} \right|;{\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}_3} = \left| {\begin{array}{*{20}{c}} 1&{\sigma _1^{(1)}}&{ - {\lambda _1}\sigma _2^{(1)}}\\ 1&{\sigma _1^{(2)}}&{ - {\lambda _2}\sigma _2^{(2)}}\\ 1&{\sigma _1^{(3)}}&{ - {\lambda _3}\sigma _2^{(3)}} \end{array}} \right|. $

合理选择λ_i和r_k⁽ⁱ⁾使得式(32)所确定的方程的系数行列式Δ≠0.把式(33)代入式(31)，我们可以把Darboux变换的显式形式表示为

$ \begin{array}{l} \hat u = u + f(x,t,\lambda ) = u + \frac{3}{2}{\left( {{\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}_1}/\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}} \right)_x};\\ \hat v = v + h(x,t,\lambda ) = v + \frac{3}{2}{\left( {{\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}_1}/\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}} \right)_{xx}} + 3{\left( {{\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}_2}/\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}} \right)_x} - 2{\left( {{\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}_1}/\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}} \right)_x} - 3\left( {{\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}_1}/\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}} \right){\left( {{\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}_1}/\mathit{\boldsymbol{ \boldsymbol{\varDelta} }}} \right)_x}. \end{array} $

(34)

1) 取u=0，v=－λ时，式(3)和式(23)有基解矩阵

$ \mathit{\boldsymbol{ \boldsymbol{\varPsi} }} = \left( {\begin{array}{*{20}{c}} 0&{ - 1}&0\\ 0&1&{{{\rm{e}}^x}}\\ 1&{x - 6t}&{{{\rm{e}}^x}} \end{array}} \right). $

(35)

通过Darboux变换式(31)，我们得到方程(1)的一个精确解$\hat{u}=f(x, t, \lambda)$，$ \hat{v}=h(x, t, \lambda)-\lambda $.

2) 取$u=-\frac{1}{2} \lambda$，v=0(λ>0)时，式(3)和式(23)有基解矩阵(其中$A=\sqrt{\lambda}(x-6 t) $)

$ \mathit{\boldsymbol{ \boldsymbol{\varPsi} }} = \left( {\begin{array}{*{20}{c}} {\frac{1}{2}\lambda {{\rm{e}}^{x + 4\lambda t}}}&{\frac{1}{2}{{\rm{e}}^{ - 2\lambda t}}(2\sqrt \lambda \sin A - \lambda \cos A)}&{ - \frac{1}{2}{{\rm{e}}^{ - 2\lambda t}}(\lambda \sin A + 2\sqrt \lambda \cos A)}\\ {{{\rm{e}}^{x + 4\lambda t}}}&{ - \sqrt \lambda {{\rm{e}}^{ - 2\lambda t}}\sin A}&{\sqrt \lambda {{\rm{e}}^{ - 2\lambda t}}\cos A}\\ {{{\rm{e}}^{x + 4\lambda t}}}&{{{\rm{e}}^{ - 2\lambda t}}\cos A}&{{{\rm{e}}^{ - 2\lambda t}}\sin A} \end{array}} \right). $

(36)

通过Darboux变换式(31)，我们得到方程(1)的一个精确解$ \hat{u}=f(x, t, \lambda)-\frac{1}{2} \lambda $，$ \hat{v}=h(x, t, \lambda) $.

3) 取$ u=\frac{1}{2} \lambda $，v=－2λ(λ>0)时，式(3)和式(23)有基解矩阵(其中$ A=\sqrt{\lambda}(x-6 t) $)

$ \mathit{\boldsymbol{ \boldsymbol{\varPsi} }} = \left( {\begin{array}{*{20}{c}} { - \frac{1}{2}\lambda {{\rm{e}}^{x - 4\lambda t}}}&{\frac{1}{2}(\lambda - 2\sqrt \lambda ){{\rm{e}}^{A + 2\lambda t}}}&{\frac{1}{2}(\lambda + 2\sqrt \lambda ){{\rm{e}}^{ - A + 2\lambda t}}}\\ {{{\rm{e}}^{x - 4\lambda t}}}&{\sqrt \lambda {{\rm{e}}^{A + 2\lambda t}}}&{ - \sqrt \lambda {{\rm{e}}^{ - A + 2\lambda t}}}\\ {{{\rm{e}}^{x - 4\lambda t}}}&{{{\rm{e}}^{A + 2\lambda t}}}&{{{\rm{e}}^{ - A + 2\lambda t}}} \end{array}} \right). $

(37)

通过Darboux变换式(31)，我们得到方程(1)的一个精确解$\hat{u}=f(x, t, \lambda)+\frac{1}{2} \lambda$，$ \hat{v}=h(x, t, \lambda)-2 \lambda $.

4) 取u=0，v=0时，式(3)和式(23)有基解矩阵

$ \mathit{\boldsymbol{ \boldsymbol{\varPsi} }} = \left( {\begin{array}{*{20}{c}} {\left( {k_1^2 - {k_1}} \right){{\rm{e}}^{{\mathit{\Gamma }_1}}}}&{\psi _1^{(2)}}&{\psi _1^{(3)}}\\ {{k_1}{{\rm{e}}^{{\mathit{\Gamma }_1}}}}&{\psi _2^{(2)}}&{\psi _2^{(3)}}\\ {{{\rm{e}}^{{\mathit{\Gamma }_1}}}}&{{{\rm{e}}^{{\mathit{\Gamma }_2}}}\cos {\mathit{\Gamma }_3}}&{{{\rm{e}}^{{\mathit{\Gamma }_2}}}\sin {\mathit{\Gamma }_3}} \end{array}} \right), $

(38)

$ \psi _1^{(2)} = \frac{1}{9}{{\rm{e}}^{{\mathit{\Gamma }_2}}}\left[ {\left( {k_2^2 - k_3^2 - 3{k_2}} \right)\cos {\mathit{\Gamma }_3} - \left( {2{k_2}{k_3} - 3{k_3}} \right)\sin {\mathit{\Gamma }_3}} \right],\psi _2^{(2)} = \frac{1}{3}{{\rm{e}}^{{\mathit{\Gamma }_2}}}\\ \left( {{k_2}\cos {\mathit{\Gamma }_3} -{k_3}\sin {\mathit{\Gamma }_3}} \right), $

$ \psi _1^{(3)} = \frac{1}{9}{{\rm{e}}^{{\mathit{\Gamma }_2}}}\left[ {\left( {k_2^2 - k_3^2 - 3{k_2}} \right)\sin {\mathit{\Gamma }_3} + \left( {2{k_2}{k_3} - 3{k_3}} \right)\cos {\mathit{\Gamma }_3}} \right],\psi _2^{(3)} = \frac{1}{3}{{\rm{e}}^{^{{\mathit{\Gamma }_2}}}}\\ \left( {{k_2}\sin {\mathit{\Gamma }_3} + {k_3}\cos {\mathit{\Gamma }_3}} \right), $

$ \mathit{\Gamma } = {\left( {\frac{2}{{2 + 27\lambda + 3\sqrt {12\lambda + 81{\lambda ^2}} }}} \right)^{\frac{1}{3}}},{k_1} = \frac{1}{3}\left( {1 + \mathit{\Gamma } + \frac{1}{\mathit{\Gamma }}} \right),{k_2} = 1 - \frac{\mathit{\Gamma }}{2} - \frac{1}{{2\mathit{\Gamma }}},{k_3} = - \frac{{\sqrt 3 }}{2}\left( {\mathit{\Gamma } - \frac{1}{\mathit{\Gamma }}} \right), $

$ {\mathit{\Gamma }_1} = \frac{1}{3}\left( {1 + \mathit{\Gamma } + \frac{1}{\mathit{\Gamma }}} \right)x + \frac{2}{3}(\mathit{\Gamma } - 1)\left( {\mathit{\Gamma } - \frac{1}{{{\mathit{\Gamma }^2}}}} \right)t,{\mathit{\Gamma }_2} = \frac{1}{3}{k_2}x + \frac{2}{3}\left( {k_2^2 - k_3^2 - 3{k_2}} \right)t, $

$ {\mathit{\Gamma }_3} = \frac{1}{3}{k_3}x + \frac{2}{3}\left( {2{k_2}{k_3} - 3{k_3}} \right)t. $

通过Darboux变换式(31)，我们得到方程(1)的一个精确解$ \hat{u}=f(x, t, \lambda) $，$ \hat{v}=h(x, t, \lambda) $.

3 结论

本文我们引入了一个具有两个位势的3×3矩阵谱问题，通过该问题及其辅助谱问题得到了一个新的微分方程族.基于特征函数的Riccati方程，我们推导出该方程族前两个方程的无穷多守恒律，并借助于Darboux变换得到了第一个非平凡方程的一些显式解.关于这个非线性演化方程族的孤子解、Bäcklund变换、代数几何解和其他的一些性质，我们将在以后的文章中进行讨论.