---
abstract: 'The nonlocal Darboux transformation of the two - dimensional stationary Schrödinger equation is considered and its relation to the Moutard transformation is established. It is shown that a particular case of the nonlocal Darboux transformation provides the Moutard transformation. New examples of exactly solvable two - dimensional stationary Schrödinger operators with smooth potentials are obtained as an application of the nonlocal Darboux transformation.'
---

**[A. G. Kudryavtsev]{}**

Institute of Applied Mechanics,

Russian Academy of Sciences, Moscow 119991, Russia

kudryavtsev\_a\_g@mail.ru

PACS: 02.30.Jr, 02.30.Ik, 03.65.Ge

Introduction
============

The two - dimensional stationary Schrödinger equation is a model of physical phenomenon in nonrelativistic quantum mechanics [@Landau], acoustics [@Morse] and tomography [@Kak]. The case of fixed energy for two - dimensional Schrödinger equation is of interest for the theory of two - dimensional integrable nonlinear systems [@Veselov1984], [@Novikov2010] and for the multidimentional inverse scattering theory [@Grinevich2000] .

Consider the two - dimensional stationary Schrödinger equation in the form $$\label{eq1}
\left( \Delta -{\it u}\left( x,y \right) \right) Y \left( x,y
\right) =0$$

where $\Delta$ is the two - dimensional Laplace operator $$\Delta= {\frac {\partial ^{2}}{\partial {x}^{2}}}
+{\frac {\partial ^{2}}{\partial {y}^{2}}}$$

The Schrödinger equation is related with the Fokker-Planck equation

$$\label{eq2}
{\it W_{xx}}+{\it W_{yy}}+{\frac {\partial }{\partial x}} \left(
2\,{ \it h_{x}}\,W \right) +{\frac {\partial }{\partial y}} \left(
2\,{\it h_{y}}\,W \right) =0$$

by substitution [@Risken1989]

$$\label{eq3}
Y \left( x,y \right) =W \left( x,y \right) {e^{h \left( x,y
\right) }}$$

if condition $$\label{eq4}
{\it u}=-\Delta {\it h} +{{\it h_{x}}}^{2}+{{\it h_{y}}}^{2}$$ holds.

The Fokker-Planck equation has the conservation low form that yields a pair of potential equations

$$\label{eq5}
{\it W_{x}}+ 2\,{ \it h_{x}}\,W  -{\it Q_{y}} =0$$

$$\label{eq6}
{\it W_{y}}+ 2\,{ \it h_{y}}\,W  +{\it Q_{x}} =0$$

In the paper [@Kudryavtsev2012] the special form of Darboux transformation for the potential equations , was considered. As the potential variable is a nonlocal variable for the Schrödinger equation that provides the nonlocal Darboux transformation for the Schrödinger equation. It was shown that this nonlocal transformation is a useful tool for obtaining exactly solvable two - dimensional stationary Schrödinger operators. The consideration in the paper [@Kudryavtsev2012] is restricted by the simple case $h=0$. In the present paper the case of arbitrary $h$ is considered and relation of the nonlocal Darboux transformation with the Moutard [@Moutard1878] transformation is established.

The nonlocal Darboux transformation and its relation with the Moutard transformation
====================================================================================

For completeness we outline some results of the paper [@Kudryavtsev2012]. Let us consider linear operator corresponding to the system of equations ,

$$\hat L\left( h \left( x ,y \right)  \right) \, {\bf F}=
\begin{pmatrix} { 2\,{ \it h_{x}}+\frac {\partial }{\partial x} }  &
{ -\frac {\partial }{\partial y} } \\ {{2\,{ \it h_{y}}+\frac
{\partial }{\partial y}}}  & {\frac {\partial }{\partial x}}
\end{pmatrix}\,
\begin{pmatrix} F_1 \\ F_2 \end{pmatrix}$$

Consider Darboux transformation in the form

$$\hat L_D \, {\bf F}=
\begin{pmatrix} { r_{11}-a_{11}\,\frac {\partial }{\partial x}-b_{11}\,\frac {\partial }{\partial y}  }  &
{  r_{12}-a_{12}\,\frac {\partial }{\partial x}-b_{12}\,\frac
{\partial }{\partial y} } \\ { r_{21}-a_{21}\,\frac {\partial
}{\partial x}-b_{21}\,\frac {\partial }{\partial y} }  & {
r_{22}-a_{22}\,\frac {\partial }{\partial x}-b_{22}\,\frac
{\partial }{\partial y} }
\end{pmatrix} \,
\begin{pmatrix} F_1 \\ F_2 \end{pmatrix}$$

If linear operators $\hat L$ and $\hat L_{D}$ hold the intertwining relation

$$\label{eq7}
\left( \hat L\left( h \left( x ,y \right) + s \left( x ,y
\right) \right)\hat L_{D} - \hat L_{D} \hat L\left( h \left( x ,y
\right) \right) \right) \, {\bf F}= 0$$

for any $ {\bf F} \in  \mathcal{F} \supset Ker\left( \hat L\left(
h \right)\right)$ where $Ker\left( \hat L\left( h
\right)\right)=\{{\bf F}:{\hat {L}}\left( h \right){\bf F}=0\}$, then for any ${\bf F_s}\in Ker\left( \hat L\left( h
\right)\right)$ the function $\tilde {\bf F} \left( x,y \right)=
\hat L_{D} {\bf F_s} \left( x,y \right)$ is a solution of the equation ${\hat {L}}\left( {\tilde {h}} \right) \tilde {\bf F} =0$ with new potential $\tilde h = h+s$.

If one consider equation on the set ${\mathcal{F}}$ of arbitrary functions, then treating $F_1, F_2$ and each of its derivatives as independent variables the following equation can be obtained

$$\label{eq8}
{\left({\frac {\partial \left( h+s \right)}{\partial x}}
\right)}^{2}+{\left({\frac {\partial \left( h+s \right)}{\partial
x}} \right)}^{2}={\left({\frac {\partial h }{\partial x}}
\right)}^{2}+{\left({\frac {\partial  h }{\partial x}}
\right)}^{2}$$

This is strong limitation for the new potential $\tilde h = h+s$.

In the paper [@Kudryavtsev2012] it was proposed to consider equation on the following domain: $${\mathcal{F}}_0=\{\,{\bf F}: {F_1}_{x}+ 2\,{ \it h_{x}}\, F_1  -
{F_2}_{y} =0\}$$

Taking into account this dependance of $F_1, F_2$ derivatives, the equations for $s, r_{ij}, a_{ij}, b_{ij}$ can be obtained.

It should be noted that instead of restriction to solutions of equation domain ${\mathcal{F}}_0$ can be defined by restriction to solutions of equation . Because the system of equations , has $x,y$ interchange with change of $Q$ sign symmetry, one can simply use this symmetry in resulting formulae.

In the paper [@Kudryavtsev2012] the case $h=0$ was considered. Now let us consider the case of arbitrary $h$.

When solving equation on the domain ${\mathcal{F}}_0$ the special situation arise in the case $s=-2h$ that corresponds to $\tilde h = -h$. In this case one can obtain from the equation the following Darboux transformation

$$\label{eq9}
\hat L_D = {e^{2\,h \left( x,y \right) }}
\begin{pmatrix} { 0 }  &
{ 1 } \\ { - 1 }  & { 0 }
\end{pmatrix} \,$$

By the formula with $\tilde h = -h$ we obtain for the new Schrödinger potential

$$\label{eq10}
\tilde {u} \left( x ,y \right) = u\left( x ,y \right) + 2\,\Delta
\, h\left( x,y \right)$$

Consider $$\label{eq11}
Y_h\left( x ,y \right)=e^{-h \left( x,y \right) }$$

According to the formula , $Y_h$ is a solution of the Schrödinger equation with potential $u$.

Then we get another form of the formula

$$\label{eq12}
\tilde {u} \left( x ,y \right) = u\left( x ,y \right) - 2\,\Delta
\, \ln  \left( Y_h \left( x,y \right)  \right)$$

This formula coincides with the formula of Moutard transformation [@Moutard1878] for the potential of the Schrödinger equation.

From the formula and relation

$$\label{eq13}
\tilde {Y} \left( x,y \right) =\tilde {W} \left( x,y \right) {e^{
\tilde {h} \left( x,y \right) }}$$

we have for the new solutions of the Fokker-Planck and Schrödinger equations

$$\label{eq14}
\tilde {W} \left( x ,y \right) = {e^{2\,h \left( x,y \right) }}Q
\left( x,y \right)$$

$$\label{eq15}
\tilde {Y} \left( x ,y \right) = {e^{h \left( x,y \right) }}Q
\left( x,y \right)$$

One can express $W, Q, h$ by equations , , trough $Y, \tilde {Y}, Y_h$ and substitute to the system of equations , . The result is

$$\label{eq16}
{\frac {\partial }{\partial x}} \left( {\it Y_h} \left( x,y
\right) {\tilde {Y}} \left( x,y \right)  \right) + \left( {\it
Y_h} \left( x,y \right)  \right) ^{2}{\frac {\partial }{\partial
y}} \left( {\frac {{ \it Y} \left( x,y \right) }{{\it Y_h} \left(
x,y \right) }} \right)=0$$

$$\label{eq17}
{\frac {\partial }{\partial y}} \left( {\it Y_h} \left( x,y
\right) {\tilde {Y}} \left( x,y \right)  \right) - \left( {\it
Y_h} \left( x,y \right)  \right) ^{2}{\frac {\partial }{\partial
x}} \left( {\frac {{ \it Y} \left( x,y \right) }{{\it Y_h} \left(
x,y \right) }} \right)=0$$

These formulae coincide with the formulae of the Moutard transformation for the solution of the Schrödinger equation.

Thus the case $\tilde h = -h$ of the nonlocal Darboux transformation provides the Moutard transformation.

The twofold application of the Moutard transformation can be effective for obtaining nonsingular solvable potentials for the Schrödinger equation [@Tsarev2008]. Now we give some formulae of the twofold Moutard transformation in the form convenient for later use.

Let us consider a solvable Schrödinger potential $u$ and select two solutions $Y_{1}, Y_{2}$ of the Schrödinger equation with this potential. Then according to we choose $Y_{h1}=Y_{1}$ and perform the Moutard transformation by the change of $h_1$ sign. From we have for the result of $Y_{2}$ transformation ${\tilde {Y}}_{2}=Q_{12}/Y_{h1} $, where $Q_{12}$, according to , , and , satisfies the following system of equations: $$\label{eq18}
 {\frac {\partial Y_{2}}{\partial x}}  \, Y
_{1}  -   Y_{2} \, {\frac {\partial Y_{1}}{\partial x}}
 -{\frac {\partial Q_{12}}{\partial y}} =0, \,\,\,\,
{\frac {\partial Y_{2}}{\partial y}} \, Y_{1} -Y_{2}  {\frac
{\partial Y_{1}}{\partial y}} +{\frac {\partial Q_{12}}{\partial
x} } =0$$

Then we choose $Y_{h2}={\tilde {Y}}_{2}$ and perform the Moutard transformation by the change of $h_2$ sign. From for $h=-h_2$ taking into account we get

$$\begin{gathered}
 \label{eq19}
\tilde {\tilde {u}}=u + 4 \, {\left( {Q_{12}} \right) ^{-1}}
\left( {\frac {\partial {Y_{2}}}{\partial y}}  {\frac {\partial
{Y_{1}}}{\partial x}} - {\frac {\partial {Y_{2}}}{\partial x}}
{\frac {\partial {Y_{1}}}{\partial y}} \right)
\\
+ 2 \, {\left( {Q_{12}} \right) ^{-2}}\left( \left( {\frac
{\partial {Y_{2}}}{\partial x}}  {Y_{1}} -{Y_{2}} {\frac {\partial
{Y_{1}}}{\partial x}} \right) ^{2}+ \left(  {\frac {\partial
{Y_{2}}}{\partial y}} {Y_{1}} -{Y_{2}} {\frac {\partial
{Y_{1}}}{\partial y}} \right) ^{2} \right)\end{gathered}$$

For nonsingular $u, Y_{1}, Y_{2}$ one can see from this formula that singularity of $\tilde {\tilde {u}}$ can arise from $Q_{12}$ zeros. Note that $Q_{12}$ is defined by to within arbitrary constant. In some cases, this constant allows to make $Q_{12}$ of constant signs.

According to , for $Y=Y_h$ one has simple solution $\tilde {Y}=1/Y_h$. Therefore $\tilde {\tilde
{Y}}=1/Y_{h2}=Y_{1}/Q_{12} $ is an example of solution for the Schrödinger equation with potential $\tilde {\tilde {u}}$.

Now let us return to the consideration of the equation . In the case $s$ not equal to $-2h$ from the equation on the domain ${\mathcal{F}}_0$ we obtain the following Darboux transformation

$$\label{eq20}
\hat L_D = {e^{- s \left( x,y \right) }}
\begin{pmatrix} { R_{1}-\frac {\partial }{\partial y}  }  &
{  R_{2} } \\ { -{s}_{x}-R_{2} }  & { -{s}_{y}+R_{1} -\frac
{\partial }{\partial y}}
\end{pmatrix} \,$$

where $R_{1} =F_{1}/F, \, R_{2} =F_{2}/F$ $$\begin{gathered}
 \label{eq21}
F =  2s_{{x}} \left( s_{{x}}+2h _{{x}} \right) +2s_{{y}} \left(
s_{{y}}+2h_{{y}} \right)
\\
\shoveleft{
F_{1} =
 \left( s_{{x}}+2h_{{x}} \right)  \left( -2 \left(
s_{{{\it xy}}}+h_{{{\it xy}}} \right) + \left( s_{{y}}-2h_{{y}}
\right) s_{{x}} \right) }
\\
\shoveright{+ \left(s_{{y}} +2h_{{y}} \right)  \left( s_{{{\it
xx}}}-s_{{{\it yy} }}-2h_{{{\it yy}}}+ \left( s_{{y}}-2h_{{y}}
\right) s_{{y}}
 \right)
}
\\
\shoveleft{
F_{2} =
 s_{{x}}s_{{{\it xx}}}+2\,s_{{y}} \left( s_{{{\it
xy}}}+h_{{{ \it xy}}} \right) -s_{{x}} \left( s_{{{\it
yy}}}+2h_{{{\it yy}}}
 \right)
 }
\\
 - \left( s_{{x}}+2h_{{x}} \right){s_{{x}}}^{2} - \left( s_{{y}} +2
h_{{y}}\right) s_{{x}}s_{{y}}\end{gathered}$$

and $s$ satisfies the following system of two nonlinear differential equations: $$\begin{gathered}
 \label{eq22}
 \left( {s_{{x}}}^{3}+2\,{s_{{x}}}^{2}h_{{x}}+ \left( 2\,s_{{y}}h_{{y}
}+{s_{{y}}}^{2} \right) s_{{x}} \right) s_{{{\it xxx}}}
\\
\shoveleft{+ \left(
 \left( s_{{y}}-2\,h_{{y}} \right) {s_{{x}}}^{2}+ \left( -4\,h_{{y}}h_
{{x}}+2\,h_{{x}}s_{{y}} \right)
s_{{x}}+{s_{{y}}}^{3}-4\,{h_{{y}}}^{2} s_{{y}} \right) s_{{{\it
xxy}}} }
\\
\shoveleft{+ \left( {s_{{x}}}^{3}+6\,{s_{{x}}}^{2} h_{{x}}+ \left(
8\,{h_{{x}}}^{2}+2\,s_{{y}}h_{{y}}+{s_{{y}}}^{2}
 \right) s_{{x}}+8\,h_{{x}}s_{{y}}h_{{y}}+4\,h_{{x}}{s_{{y}}}^{2}
 \right) s_{{{\it xyy}}}  }
\\
\shoveleft{ + \left(  \left( 2\,h_{{y}}+s_{{y}} \right) {s
_{{x}}}^{2}+ \left( 4\,h_{{y}}h_{{x}}+2\,h_{{x}}s_{{y}} \right)
s_{{x}
}+4\,h_{{y}}{s_{{y}}}^{2}+{s_{{y}}}^{3}+4\,{h_{{y}}}^{2}s_{{y}}
 \right) s_{{{\it yyy}}}  }
\\
\shoveleft{ + \left( 2s_{{y}}h_{{y}}+{s_{{y}}}^{2}-{s_{{ x}}}^{2}
\right) {s_{{{\it xx}}}}^{2}  +  2\left(  \left( -2s_{{
y}}+h_{{y}} \right) s_{{x}}+2h_{{y}}h_{{x}}-h_{{x}}s_{{y}}
 \right)s_{{{\it xx}}} s_{{{\it xy}}}  }
\\
\shoveleft{ +  2\left(
-h_{{x}}s_{{x}}+2{h_{{y}}}^{2}+s_{{y}}h_{{y}} \right) s_{{{\it
xx}}} s_{{{\it yy}}} + 4\left( s_{{y}}h_{{y}}-h_{{x}}s_{{x}
}-2{h_{{x}}}^{2} \right) {s_{{{\it xy}}}}^{2}  }
\\
\shoveleft{+2\left(-{h_{{{\it xx}}}s_{{x}}}^{2} + 2\left(
-h_{{{\it yy}}}h_{{x}}-s_{{y}}h_{{{\it xy}}}+h_{{{ \it
xy}}}h_{{y}} \right) s_{{x}}\right) s_{{{\it xx}}}  }
\\
\shoveleft{-2\left(h_{{{\it yy}}}{s_{{y}}}^{2}+2
 \left( h_{{{\it yy}}}h_{{y}}+h_{{{\it xy}}}h_{{x}} \right) s_{
{y}} \right) s_{{{\it xx}}}  }
\\
\shoveleft{ + \left(
 \left( -6\,h_{{y}}-4\,s_{{y}} \right) s_{{x}}-10\,h_{{x}}s_{{y}}-12\,
h_{{y}}h_{{x}} \right) s_{{{\it xy}}} s_{{{\it yy}}}  }
\\
\shoveleft{ + \left(  \left( -4\,h_{{{\it xx} }}-4\,h_{{{\it yy}}}
\right) s_{{y}}-4\,h_{{{\it yy}}}h_{{y}}-4\,h_{{{ \it xy}}}h_{{x}}
\right) s_{{x}} s_{{{\it xy}}}  }
\\
\shoveleft{+ \left( \left( -12\,h_{{{\it yy}}}h_{{x}}+12
\,h_{{{\it xy}}}h_{{y}} \right) s_{{y}}-8\,h_{{x}}h_{{{\it
yy}}}h_{{y} }-8\,{h_{{x}}}^{2}h_{{{\it xy}}} \right) s_{{{\it
xy}}}  }
\\
\shoveleft{+ \left( -4\,{h
_{{y}}}^{2}+{s_{{x}}}^{2}+2\,h_{{x}}s_{{x}}-4\,s_{{y}}h_{{y}}-{s_{{y}}
}^{2} \right) {s_{{{\it yy}}}}^{2}  }
\\
\shoveleft{+ \left(  \left( 4\,h_{{{\it yy}}}+2 \,h_{{{\it xx}}}
\right) {s_{{x}}}^{2}+ \left( -8\,h_{{{\it xy}}}h_{{y
}}+8\,h_{{{\it yy}}}h_{{x}}-4\,s_{{y}}h_{{{\it xy}}} \right)
s_{{x}} \right) s_{{{\it yy}}}  }
\\
\shoveleft{+\left( -2 \,h_{{{\it yy}}}{s_{{y}}}^{2}+ \left(
-8\,h_{{{\it xy}}}h_{{x}}-8\,h_{ {{\it yy}}}h_{{y}} \right)
s_{{y}}-8\,h_{{x}}h_{{{\it xy}}}h_{{y}}-8\, h_{{{\it
yy}}}{h_{{y}}}^{2} \right) s_{{{\it yy}}}  }
\\
\shoveleft{-{s_{{x}}}^{6}-6\,h_ {{x}}{s_{{x}}}^{5}-3\left(
4{h_{{x}}}^{2}+2s_{{y}}h_{{y}}+{s_ {{y}}}^{2} \right)
{s_{{x}}}^{4}-4 \left( 2{h_{{x}}}^{3}+6h_{{x}}
s_{{y}}h_{{y}}+3h_{{x}}{s_{{y}}}^{2} \right) {s_{{x}}}^{3}  }
\\
\shoveleft{- \left( 3{s_{{y}}}^{4}+12h_{{y}}{s_{{y}}}^{3}+12\left(
{h_{{y}}}^{2}+{h_{{x}}}^{2} \right) {s_{{y}}}^{2}- 2\left(
-12{h_{{x}}}^{2}h_{{ y}}+h_{{{\it yyy}}}+h_{{{\it xxy}}} \right)
s_{{y}}\right) {s_{{x}}}^{2}  }
\\
\shoveleft{+4\left(h_{{{\it yyy}}}h_{{y}}+h_{{{\it yy}}}h_{{{\it
xx}}}+{h_{{{\it yy}}}}^{2}+ h_{{{\it xyy}}}h_{{x}} \right)
{s_{{x}}}^{2}  }
\\
\shoveleft{-6\left( h_{{x}}{s_
{{y}}}^{4}+4h_{{x}}h_{{y}}{s_{{y}}}^{3}+4h_{{x}}{h_{{y}}}^{2}{s_
{{y}}}^{2}\right)s_{{x}}  }
\\
\shoveleft{-4\left( h_{{{\it yy}}}h_{{{\it xy}}}-h_{{x}}h_{{{ \it
yyy}}}+h_{{{\it xy}}}h_{{{\it xx}}}-h_{{{\it xxy}}}h_{{x}}
 \right) s_{{y}}s_{{x}}  }
\\
\shoveleft{+\left(8\,h_{{x}}h_{{{\it
yyy}}}h_{{y}}+8\,{h_{{x}}}^{2}h_{{ {\it xyy}}}-8\,h_{{{\it
xy}}}h_{{{\it yy}}}h_{{y}}+8\,h_{{x}}{h_{{{ \it yy}}}}^{2} \right)
s_{{x}}-{s_{{y}}}^{6}-6\,h_{{y}}{s_{{y}}}^{5}  }
\\
\shoveleft{- 12\,{h_{{y}}}^{2}{s_{{y}}}^{4}  + 2\left( h_{{{\it
yyy}}}-4{h_{{y}}} ^{3}+h_{{{\it xxy}}} \right) {s_{{y}}}^{3}+
4\left( h_{{{\it xxy} }}h_{{y}}+2h_{{{\it yyy}}}h_{{y}}+h_{{{\it
xyy}}}h_{{x}} \right) {s_{{y}}}^{2}  }
\\
+ \left( -8\,h_{{x}}h_{{{\it xy}}}h_{{{\it yy}}}+8\,h_{{{ \it
yyy}}}{h_{{y}}}^{2}+8\,h_{{x}}h_{{{\it xyy}}}h_{{y}}+8\,{h_{{{\it
xy}}}}^{2}h_{{y}} \right) s_{{y}}=0\end{gathered}$$

$$\begin{gathered}
 \label{eq23}
 \left(  \left( 2\,h_{{y}}+s_{{y}} \right) {s_{{x}}}^{2}+ \left( 4\,h_
{{y}}h_{{x}}+2\,h_{{x}}s_{{y}} \right)
s_{{x}}+4\,h_{{y}}{s_{{y}}}^{2}
+{s_{{y}}}^{3}+4\,{h_{{y}}}^{2}s_{{y}} \right) s_{{{\it xxx}}}
\\
\shoveleft{+
 \left( -{s_{{x}}}^{3}-6\,{s_{{x}}}^{2}h_{{x}}+ \left( -2\,s_{{y}}h_{{
y}}-8\,{h_{{x}}}^{2}-{s_{{y}}}^{2} \right)
s_{{x}}-4\,h_{{x}}{s_{{y}}} ^{2}-8\,h_{{x}}s_{{y}}h_{{y}} \right)
s_{{{\it xxy}}}  }
\\
\shoveleft{+ \left( \left( s_{{y}}-2\,h_{{y}} \right)
{s_{{x}}}^{2}+ \left( -4\,h_{{y}}h_{{x}}+2 \,h_{{x}}s_{{y}}
\right) s_{{x}}+{s_{{y}}}^{3}-4\,{h_{{y}}}^{2}s_{{y}}
 \right) s_{{{\it xyy}}}   }
\\
\shoveleft{ + \left( -{s_{{x}}}^{3}-2\,{s_{{x}}}^{2}h_{{x} }+
\left( -{s_{{y}}}^{2}-2\,s_{{y}}h_{{y}} \right) s_{{x}} \right)
s_{ {{\it yyy}}}   }
\\
\shoveleft{+ \left(  \left( -2\,s_{{y}}-4\,h_{{y}} \right)
s_{{x}}-4 \,h_{{y}}h_{{x}}-2\,h_{{x}}s_{{y}} \right) {s_{{{\it
xx}}}}^{2}   }
\\
\shoveleft{+ \left(
10\,h_{{x}}s_{{x}}-6\,s_{{y}}h_{{y}}+2\,{s_{{x}}}^{2}+
8\,{h_{{x}}}^{2}-4\,{h_{{y}}}^{2}-2\,{s_{{y}}}^{2} \right)
s_{{{\it xy }}}s_{{{\it xx}}}   }
\\
\shoveleft{+ \left(
2\,s_{{x}}h_{{y}}+2\,h_{{x}}s_{{y}}+4\,h_{{y}}h_{{x}}
 \right) s_{{{\it yy}}} s_{{{\it xx}}}   }
\\
\shoveleft{ +\left(2\,h_{{{\it xy}}}{s_{{x}}}^{2}+ \left(
 \left( -2\,h_{{{\it xx}}}+2\,h_{{{\it yy}}} \right) s_{{y}}+12\,h_{{{
\it xy}}}h_{{x}}-4\,h_{{{\it xx}}}h_{{y}}+8\,h_{{{\it yy}}}h_{{y}}
 \right) s_{{x}}\right)s_{{{\it xx}}}   }
\\
\shoveleft{ +\left(-2\,h_{{{\it xy}}}{s_{{y}}}^{2}+ \left(
4\,h_{{{\it yy }}}h_{{x}}-4\,h_{{{\it xy}}}h_{{y}} \right)
s_{{y}}+8\,h_{{x}}h_{{{ \it yy}}}h_{{y}}+8\,{h_{{x}}}^{2}h_{{{\it
xy}}} \right) s_{{{\it xx}}}    }
\\
\shoveleft{ + 4\left(
s_{{x}}h_{{y}}+2h_{{y}}h_{{x}}+h_{{x}}s_{{y}}
 \right) {s_{{{\it xy}}}}^{2}+  2\left( h_{{x}}s_{{x}}-{s_
{{y}}}^{2}+{s_{{x}}}^{2}+2{h_{{y}}}^{2}+s_{{y}}h_{{y}}
 \right) s_{{{\it yy}}}s_{{{\it xy
}}}     }
\\
\shoveleft{ + 4\left(h_{{{\it xy}}}h_{{y}}  -h_{{{\it
yy}}}h_{{x}}\right) s_{{x}}s_{{{\it xy }}}-4\left( h_{{{\it
xx}}}+h_{{{\it yy }}} \right) {s_{{y}}}^{2}s_{{{\it xy }}}   }
\\
\shoveleft{+ 4\left( 3h_{{{\it xy}}}h_{{x}}-2h_{{{ \it
xx}}}h_{{y}}+h_{{{\it yy}}}h_{{y}} \right) s_{{y}}s_{{{\it xy }}}
}
\\
\shoveleft{+\left( 8\,h_{{x}}h_ {{{\it xy}}}h_{{y}}+8\,h_{{{\it
yy}}}{h_{{y}}}^{2} \right) s_{{{\it xy }}}+ \left(
2\,h_{{y}}+2\,s_{{y}} \right) s_{{x}}{s_{{{\it yy}}}}^{2}    }
\\
\shoveleft{+
 \left( 2\,h_{{{\it xy}}}{s_{{x}}}^{2}+ \left(  \left( 6\,h_{{{\it yy}
}}+2\,h_{{{\it xx}}} \right) s_{{y}}+4\,h_{{{\it
yy}}}h_{{y}}+4\,h_{{{ \it xx}}}h_{{y}} \right) s_{{x}}-2\,h_{{{\it
xy}}}{s_{{y}}}^{2}
 \right) s_{{{\it yy}}}     }
\\
\shoveleft{ + \left( -2\,h_{{{\it yyy}}}-2\,h_{{{\it xxy}}}
 \right) {s_{{x}}}^{3}+ \left( -4\,h_{{x}}h_{{{\it yyy}}}-8\,h_{{{\it
xxy}}}h_{{x}}-4\,h_{{y}}h_{{{\it xyy}}} \right) {s_{{x}}}^{2}
}
\\
\shoveleft{+ \left(
 \left( -2\,h_{{{\it yyy}}}-2\,h_{{{\it xxy}}} \right) {s_{{y}}}^{2}+
 \left( -4\,h_{{{\it xxy}}}h_{{y}}-4\,h_{{{\it yyy}}}h_{{y}}+4\,{h_{{{
\it yy}}}}^{2}+4\,h_{{{\it yy}}}h_{{{\it xx}}} \right)
s_{{y}}\right) s_{{x}}     }
\\
\shoveleft{+\left(-8\,h_{{ x}}h_{{{\it xy}}}h_{{{\it
yy}}}-8\,h_{{x}}h_{{{\it xyy}}}h_{{y}}-8\,{h _{{x}}}^{2}h_{{{\it
xxy}}}+8\,h_{{{\it yy}}}h_{{y}}h_{{{\it xx}}}
 \right) s_{{x}}     }
\\
\shoveleft{ + \left( -4\,h_{{y}}h_{{{\it xyy}}}-4\,h_{{{\it yy}}}h
_{{{\it xy}}}-4\,h_{{{\it xxy}}}h_{{x}}-4\,h_{{{\it xy}}}h_{{{\it
xx}} } \right) {s_{{y}}}^{2}     }
\\
+ \left( -8\,{h_{{y}}}^{2}h_{{{\it xyy}}}-8\,h_
{{y}}h_{{x}}h_{{{\it xxy}}}+8\,h_{{x}}{h_{{{\it
xy}}}}^{2}-8\,h_{{{ \it xx}}}h_{{y}}h_{{{\it xy}}} \right) s_{{y}}
=0\end{gathered}$$

By the formula with $\tilde h = h+s$ we obtain for the new Schrödinger potential

$$\label{eq24}
\tilde {u} = u  -\Delta s+2\,h_{{x}}s_{{x}}+{s_{{x}}}^{2}+2\,s_{{
y}}h_{{y}}+{s_{{y}}}^{2}$$

From the formula we have for the new solution of the Fokker-Planck equation

$$\label{eq25}
\tilde {W}  = e^{- s}\left( R_{1}\,W- \frac {\partial W}{\partial
y}  + R_{2}\,Q \right)$$

From the formulae , and we have for the new solution of the Schrödinger equation

$$\label{eq26}
\tilde {Y}  =  \left( R_{{1}}  +h_{{y}} \right) Y  -\frac
{\partial Y}{\partial y}   +{e^{h}}R_{{2}}\, Q$$

Where $Q$, according to , , and , satisfies the following system of equations $$\label{eq27}
 {\frac {\partial Y}{\partial x}}  \, Y
_{{h}}  -   Y \, {\frac {\partial Y_{{ h}}}{\partial x}}
 -{\frac {\partial Q}{\partial y}} =0$$

$$\label{eq28}
  {\frac {\partial Y}{\partial y}} \, Y
_{{h}} -Y  {\frac {\partial Y_{{h}}}{\partial y}} +{\frac
{\partial Q}{\partial x} } =0$$

The Darboux transformation is a nonlocal transformation as the potential variable $Q$ is a nonlocal variable connected with $Y$ by the system , .

Application of the nonlocal Darboux transformation 
===================================================

In the simple case $h=0$ the system of equations , for $B=\exp \left( -s \right)$ has the form

$$\begin{gathered}
 \label{eq29}
 - \left( 2\,B{\it
B_y}\,{\it B_{xy}}+B{\it B_x}\, \left( {\it B_{xx}}-{\it B_{yy}}
 \right) +{\it B_x} \left( {{\it B_x}}^{2}+{{\it B_y}}^{2}
 \right)  \right)  \left( {\it B_{xx}}+{\it B_{yy}} \right)
\\
\shoveright{ +B \left( {{\it B_x}}^{2}+{{\it B_y}}^{2} \right)
{\frac {\partial }{
\partial x}} \left( {\it B_{xx}}+{\it B_{yy}} \right) =0}
\\
\shoveleft{ - \left( 2\,B{\it B_x}\,{\it B_{xy}}-B{\it B_y}\,
\left( {\it B_{xx}}-{\it B_{yy}}
 \right) +{\it B_y}\, \left( {{\it B_x}}^{2}+{{\it B_y}}^{2}
 \right)  \right)  \left( {\it B_{xx}}+{\it B_{yy}} \right)}
 \\
 +B \left( {{\it B_x}}^{2}+{{\it B_y}}^{2} \right) {\frac
{\partial }{
\partial y}} \left( {\it B_{xx}}+{\it B_{yy}} \right)
 =0\end{gathered}$$

These equations were considered in the paper [@Kudryavtsev2012]. If $B$ is a solution of equations then $1/B$ and $CB$, where $C$ is an arbitrary constant, are solutions as well. It is obvious from the form of equations that any solution of the Laplace equation $\Delta B =
0$ is the solution of these equations.

The system of equations can be integrated. In the case $\Delta B$ is not zero one can obtain

$$\label{eq30}
 \left(  {B} ^{2}-K \right)  \left( {\it B_{xx}}+{\it B_{yy}} \right) -2\,B
  \left(  {{\it B_x}}^{2}+{{\it B_y}}^{2} \right)=0$$

or in the other form

$$\label{eq31}
{\frac { {B} ^{4} \Delta  \left(  1/B \right)  }
{\Delta B}}=-K$$

where $K$ is an arbitrary constant. It is obvious that if $B$ is a solution of equation with constant $K_B$ then $1/B$ is a solution of this equation with constant $1/K_B$.

According to equation , the initial potential $u$ of the Schrödinger equation corresponding to $h=0$ is $u=0$. The new potential of Schrödinger equation corresponds to $s=-\ln
\left( B \right)$ and is given by

$$\label{eq32}
\tilde {u}  =  \Delta B/B$$

Note that according to formula $B$ is an example of solution for the Schrödinger equation with new potential $\tilde
{u}$.

The solution $B_L$ of the Laplace equation provide $\tilde {u}=0$ . Taking $B=1/B_L$ one obtains nontrivial $\tilde {u}$, but potentials of this kind have singularities:

$$\tilde {u} =2\, \left( {\it B_L}  \right) ^{-2}\left( \left( {
\frac {\partial {\it B_L} }{\partial x}} \right) ^ {2}+\left(
{\frac {\partial {\it B_L} }{\partial y}} \right) ^{2}\right)$$

Note that this formula coincides with the formula for the Moutard transformation of the potential where $u=0, Y_h=B_L$.

Let us consider the following ansatz for $B$

$$B \left( x,y \right) =F \left( {x}^{2}+{y}^{2} \right)$$

This ansatz provides the following solution of equation

$$\label{eq33}
B_r \left( x,y \right) =-{\frac { \sqrt {K}\left(  \left(
{x}^{2}+{y}^{2} \right) ^{{\it C_1}}-{\it C_2}
 \right) }{ \left( {x}^{2}+{y}^{2} \right) ^{{\it C_1}}+{\it
C_2}}}$$

where ${\it C_1},{\it C_2}$ are arbitrary constants.

The solution $B_r$ provides by the formula

$$\label{eq34}
{\tilde {u}} =-8\,{\frac {{\it C_2}\,{{\it C_1}}^{2} \left(
{x}^{2}+{y}^{2} \right) ^{ {\it C_1}-1}}{ \left(  \left(
{x}^{2}+{y}^{2} \right) ^{{\it C1}}+{\it C_2} \right) ^{2}}}$$

This solvable potential is nonsingular if ${\it C_1} \ge 1, {\it
C_2}>0$. For ${\it C_1}=1, 2, 3, 4$ this potential is the special case of potentials considered in [@Kudryavtsev2012].

Consider for example ${\it C_1} =3/2, {\it C_2}=1$. In this case we have

$$\label{eq35}
B ={\frac { \left( {x}^{2}+{y}^{2} \right) ^{3/2}-1  }{ \left(
{x}^{2}+{y}^{2} \right) ^{3/2}+1}}$$

Consider two solutions of the Laplace equation $$Y_{L1} ={\frac {x}{{x}^{2}+{y}^{2}}}, \,\,\, Y_{L2} ={\frac
{y}{{x}^{2}+{y}^{2}}}$$

From , with $Y_h=1$ we have for $Y=Y_{L1}$ and $Y=Y_{L2}$ $$Q_{L1} =-{\frac {y}{{x}^{2}+{y}^{2}}}+{\it C_{L1}}, \,\,\, Q_{L2}
={\frac {x}{{x}^{2}+{y}^{2}}}+{\it C_{L2}}$$

where ${\it C_{L1}},{\it C_{L2}}$ are arbitrary constants.

From with $h=0$ and $B$ from we have for $Y=Y_{L1}, Q=Q_{L1}$ and $Y=Y_{L2}, Q=Q_{L2}$

$$\label{eq36}
{\tilde {Y}}_{L1} = {\frac {-2\,xy \left( 7\, \left(
{x}^{2}+{y}^{2} \right) ^{3/2}+1 \right) }{ \left( {x}^{2}+{y}^{2}
\right) ^{2} \left(  \left( {x}^{2}+{y}^{2} \right) ^{3/2}+1
\right) }}+{\frac {{\it C_{L1}} \, x \left( 5\, \left(
{x}^{2}+{y}^{2} \right) ^{3/2}-1 \right) }{ \left( {x}^{2}+{y}
^{2} \right) \left( \left( {x}^{2}+{y}^{2} \right) ^{3/2}+1
\right) }}$$

$$\label{eq37}
{\tilde {Y}}_{L2} = {\frac { \left( {x}^{2}-{y}^{2} \right) \left(
7\, \left( {x}^{ 2}+{y}^{2} \right) ^{3/2}+1 \right) }{ \left(
{x}^{2}+{y}^{2} \right) ^{2} \left(  \left( {x}^{2}+{y}^{2}
\right) ^{3/2}+1 \right) }}+{ \frac {{\it C_{L2}} \, x \left( 5\,
\left( {x}^{2}+{y}^{2} \right) ^{3/2}- 1 \right) }{ \left(
{x}^{2}+{y}^{2} \right)  \left(  \left( {x}^{2}+{y }^{2} \right)
^{3/2}+1 \right) }}$$

These functions are examples of solutions for the Schrödinger equation with solvable potential $$\label{eq38}
{\tilde {u}} =-18\,{\frac {\sqrt {{x}^{2}+{y}^{2}}}{ \left( \left(
{x}^{2}+{y}^{2} \right) ^{3/2}+1 \right) ^{2}}}$$

Now the Moutard transformation can be applied to the Schrödinger equation with potential . A single Moutard transformation can provide singular solvable potential. To obtain new nonsingular solvable potential for the Schrödinger equation from the solvable potential , let us apply twofold Moutard transformation . Let us consider $u$ from and select two solutions $Y_{1}, Y_{2}$ of the Schrödinger equation equal to functions , with $C_{L1}=C_{L2}=0$. From the equation we obtain

$$\label{eq39}
Q_{12}=
 {\frac {49\, \left( {x}^{2}+{y}^{2} \right) ^{3/2}+1}{ 2 \, \left( {x}
^{2}+{y}^{2} \right) ^{2} \left(  \left( {x}^{2}+{y}^{2} \right)
^{3/2 }+1 \right) }}+C/2$$

where $C$ is an arbitrary constant. From the equation we obtain

$$\label{eq40}
\tilde {\tilde {u}}= {\frac {-2\, r \left(
441+9\,{C}^{2}{r}^{8}+2\,Cr \left( 392\,{r}^{6}+ 49\,{r}^{3}+8
\right)  \right) }{ \left( C{r}^{4} \left( {r}^{3}+1
 \right) +49\,{r}^{3}+1 \right) ^{2}}}$$

where $r=\sqrt {{x}^{2}+{y}^{2}}$.

This solvable potential is nonsingular if $C \ge 0$. The example of solution for the Schrödinger equation with potential is $$\label{eq41}
\tilde {\tilde {Y}}=Y_{1}/Q_{12}= {\frac { -4\,xy\left( 7\, \left(
{x}^{2}+{y}^{2} \right) ^{3/2}+1 \right) }{ C \left(
{x}^{2}+{y}^{2} \right) ^{2} \left(  \left( {x}^{ 2}+{y}^{2}
\right) ^{3/2}+1 \right) +49\, \left( {x}^{2}+{y}^{2}
 \right) ^{3/2}+1}}$$

The solutions of equations , can be obtained not only for $h=0$. For example let us consider $h(x,y)=H(y)$ and use the ansatz $s(x,y)=-2\,H(y)+S(x)$. That reduces the equations , to the ordinary differential equation $$\label{eq42}
S_{{x}}S_{{{\it xxx}}}-{S_{{{\it xx}}}}^{2}-{S_{{x}}}^{4}=0$$

The solution of the equation is $$\label{eq43}
S(x)=\ln  \left( {\frac {\exp \left( {\it C_1}\,x \right)-{\it
C_2}}{\exp \left( {\it C_1}\,x \right) +{ \it C_2}}} \right) +{\it
C_3}$$ where ${\it C_1},{\it C_2},{\it C_3}$ are arbitrary constants.

Thus from any solvable potential $$\label{eq44}
u_H=-{\frac {d^{2} H \left( y \right)}{d{y}^{2}}} + \left( {\frac
{d H \left( y \right)}{dy}}  \right) ^{2}$$ we obtain the new solvable potential $$\label{eq45}
{\tilde {u}}_{H}=u_H + 2\,{\frac {d^{2} H \left( y
\right)}{d{y}^{2}}} + {\frac {2\,{\it C_2}\,{{\it C_1}}^{2} \exp
\left({\it C_1}\,x\right)}{ \left( \exp \left({\it
C_1}\,x\right)-{\it C_2} \right) ^{2}}}$$

Consider $H(y)=-\ln \left( \sin \left( y \right) \right)$ and choose the solution in the form $S(x)=\ln  \left(
\tanh \left( p \left( x-{\it x_0} \right)  \right) \right)$ where $p,{\it x_0}$ are arbitrary constants. In this case the initial solvable potential is $u_H=-1$ and $$\label{eq46}
{\tilde {u}}_{H}=-1+2\, \left( \sin \left( y \right)  \right)
^{-2}+2\,{p}^{2}\left( \sinh \left( p \left( x-{\it x_0} \right)
\right) \right) ^{-2}$$ This potential has singularities. To obtain new nonsingular solvable potential for the Schrödinger equation from the solvable potential , let us apply Moutard transformation. For the solution $\sin \left( x \right)$ of the Schrödinger equation with potential $u_H=-1$ by the formulae , we obtain $Q= C -\cos \left( y \right)
\cos \left( x \right)$ where $C$ is an arbitrary constant. The formula provides the solution of the Schrödinger equation with potential $$\label{eq47}
{\tilde {Y}}_p ={\frac { \left( p \left( C - \cos \left( y \right)
\cos \left( x \right) \right) -\cos \left( y \right) \sin \left( x
\right) \tanh \left( p \left( x-{\it x_0} \right) \right) \right)
}{\sin \left( y \right) \tanh \left( p \left( x-{\it x_0} \right)
\right) }}$$ We choose $Y_h={\tilde {Y}}_p$ and perform the Moutard transformation for the potential . The new solvable potential is $$\label{eq48}
-1+ {\frac { 2 \, \left( {\it f_1}   \left( \cosh \left( p \left(
x-{\it x_0} \right)  \right)  \right) ^{-2}+{\it f_2} +{\it f_3}
\tanh \left( p \left( x-{\it x_0} \right)  \right) \right) }{
\left( p \left(C -\cos \left( y \right) \cos \left( x \right)
\right) -\cos \left( y \right) \sin \left( x \right) \tanh \left(
p \left( x-{\it x_0} \right) \right) \right) ^{2}}}$$ where $$\begin{gathered}
{\it f_1} =  -{p}^{2} \left( {p}^{2}+1 \right)  \left(C -\cos
\left( y \right) \cos \left( x \right)  \right)
^{2}+{p}^{2}{C}^{2}-\left( {p} ^{2}+1 \right)  { \cos \left( y
\right)  } ^{2}-{ \sin \left( x \right) } ^{2}
\\
\shoveleft{ {\it f_2} = -2\,{p}^{2}C \cos \left( y \right) \cos
\left( x \right) +1+ \left( {p}^{2}-1 \right){ \cos \left( x
\right)  } ^{2}  + \left( {p}^{2}+1 \right)  { \cos \left( y
\right)  } ^{2}
 }
\\
\shoveleft{ {\it f_3} = 2\,p \, \sin \left( x \right)  \left( \cos
\left( x \right) -C\cos \left( y \right)  \right)
 }
\\\end{gathered}$$ This potential is obviously nonsingular for $p>0, \, C>1/p+1$. The function $1/{\tilde {Y}}_p$ is an example of solution for the Schrödinger equation with potential .

Results and Discussion
======================

In the past decades substantial progress in the description of the exact solutions for nonlinear partial differential equations and linear partial differential equations with variable coefficients has taken place. The useful tool for solving one - dimensional Schrödinger equation is the Darboux transformation [@Matveev1991]. Some generalizations of the Darboux transformation for two - dimensional case were proposed, including operators of second order in derivatives [@Andrianov1995], [@Ioffe2004]. Examples of solvable two – dimensional models were obtained in the framework of these generalized Darboux transformations (see for example [@Andrianov2012], [@Bardavelidze2013] and references therein). Some examples of exactly solvable two - dimensional stationary Schrödinger operators with smooth rational potentials decaying at infinity were obtained in the paper [@Tsarev2008] by application of the Moutard transformation which is a two-dimensional generalization of the Darboux transformation. In the past years progress was made in the symmetry group analysis of differential equations by extending the spaces of symmetries of a given partial differential equations system to include nonlocal symmetries [@Ibragimov1991], [@Bluman2010]. In the present paper the inclusion of nonlocal variables in the Darboux transformation for the 2D stationary Schrödinger equation is considered.

In the paper [@Kudryavtsev2012] the Fokker-Planck equation associated with the two - dimensional stationary Schrödinger equation was considered. The Fokker-Planck equation has the conservation low form that yields a pair of potential equations. The special form of the Darboux transformation for these potential equations was introduced. As the potential variable is a nonlocal variable that provides the nonlocal Darboux transformation for the Schrödinger equation. It was shown that this nonlocal transformation is a useful tool for obtaining exactly solvable two - dimensional stationary Schrödinger operators. The consideration in the paper [@Kudryavtsev2012] is restricted by the simple case $h=0$. In the present paper the case of arbitrary $h$ is considered and relation of the nonlocal Darboux transformation with the Moutard [@Moutard1878] transformation is established. It is shown that the special case $\tilde h = -h$ of the nonlocal Darboux transformation provides the Moutard transformation. New examples of exactly solvable two - dimensional stationary Schrödinger operators with smooth potentials are obtained as an application of the nonlocal Darboux transformation.

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