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Let $q ( x ) \in L _ { 1,1 } : = \left\{ q : \int _ { - \infty } ^ { \infty } ( 1 + | x | ) | q ( x ) | d x < \infty , q = \overline { q } \right\}$, where the bar stands for complex conjugation. Consider the (direct) scattering problem:
 
Let $q ( x ) \in L _ { 1,1 } : = \left\{ q : \int _ { - \infty } ^ { \infty } ( 1 + | x | ) | q ( x ) | d x < \infty , q = \overline { q } \right\}$, where the bar stands for complex conjugation. Consider the (direct) scattering problem:
  
<table class="eq" style="width:100%;"> <tr><td style="width:94%;text-align:center;" valign="top"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i130/i130050/i1300502.png"/></td> <td style="width:5%;text-align:right;" valign="top">(a1)</td></tr></table>
+
\begin{equation}
 +
\tag{a1}
 +
\mathrm{l}u_- - k^2 u_- := (-u'' + q(x) - k^2) u_- = 0,
 +
\end{equation}
  
 
\begin{equation*} x \in \mathbf{R} : = ( - \infty , \infty ), \end{equation*}
 
\begin{equation*} x \in \mathbf{R} : = ( - \infty , \infty ), \end{equation*}
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\begin{equation} \tag{a3} \text{l}u _ { + } -  { k } ^ { 2 } u _ { + } = 0 , x \in \mathbf{R}, \end{equation}
 
\begin{equation} \tag{a3} \text{l}u _ { + } -  { k } ^ { 2 } u _ { + } = 0 , x \in \mathbf{R}, \end{equation}
  
<table class="eq" style="width:100%;"> <tr><td style="width:94%;text-align:center;" valign="top"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i130/i130050/i13005018.png"/></td> <td style="width:5%;text-align:right;" valign="top">(a4)</td></tr></table>
+
\begin{equation}
 +
\tag{a4}
 +
u_{+} = \left\{
 +
\begin{array} { ll }
 +
t_{+}(k) e^{-ikx}, & x \xrightarrow{\qquad\qquad\qquad\qquad\qquad\qquad } +\infty, \\
 +
e^{-ikx} + r_{+}e^{ikx}, & x \xrightarrow{\qquad\qquad\qquad\qquad\qquad\qquad } -\infty.
 +
\end{array} \right. \end{equation}
  
 
One proves that $t _ { - } ( k ) = t _ { + } ( k ) : = t ( k )$, $t ( - k ) = \overline { t ( k ) }$, $k \in \mathbf{R}$, where the bar stands for complex conjugation, $r _ { \pm } ( - k ) = \overline { r _ { \pm } ( k ) }$. The matrix
 
One proves that $t _ { - } ( k ) = t _ { + } ( k ) : = t ( k )$, $t ( - k ) = \overline { t ( k ) }$, $k \in \mathbf{R}$, where the bar stands for complex conjugation, $r _ { \pm } ( - k ) = \overline { r _ { \pm } ( k ) }$. The matrix
  
\begin{equation*} \left( \begin{array} { c c } { t ( k ) } &amp; { r _ { - } ( k ) } \\ { r _ { + } ( k ) } &amp; { t ( k ) } \end{array} \right) = S ( k ) \end{equation*}
+
\begin{equation*}  
 +
\left(  
 +
\begin{array} { c c }  
 +
{ t ( k ) } & { r _ { - } ( k ) } \\  
 +
{ r _ { + } ( k ) } & { t ( k ) }  
 +
\end{array}  
 +
\right) = S ( k )  
 +
\end{equation*}
  
 
is called the $S$-matrix (cf. [[Scattering matrix|Scattering matrix]]). Conservation of energy implies $| t ( k ) | ^ { 2 } + | r ( k ) | ^ { 2 } = 1$.
 
is called the $S$-matrix (cf. [[Scattering matrix|Scattering matrix]]). Conservation of energy implies $| t ( k ) | ^ { 2 } + | r ( k ) | ^ { 2 } = 1$.
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Let $f ( x , k )$ and $g ( x , k )$ be the solutions to (a1) satisfying the conditions:
 
Let $f ( x , k )$ and $g ( x , k )$ be the solutions to (a1) satisfying the conditions:
  
\begin{equation*} f ( x , k ) = e ^ { i k x } + o ( 1 ) , x \rightarrow \infty , \end{equation*}
+
\begin{equation*}  
 +
f ( x , k ) = e ^ { i k x } + o ( 1 ) , \quad x \rightarrow \infty ,  
 +
\end{equation*}
  
\begin{equation*} g ( x , k ) = e ^ { - i k x } + o ( 1 ) , x \rightarrow - \infty. \end{equation*}
+
\begin{equation*}  
 +
g ( x , k ) = e ^ { - i k x } + o ( 1 ) , \quad x \rightarrow - \infty.  
 +
\end{equation*}
  
 
Then
 
Then
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where
 
where
  
\begin{equation*} a ( - k ) = \overline { a ( k ) } , b ( - k ) = \overline { b ( k ) }, \end{equation*}
+
\begin{equation*}  
 +
a ( - k ) = \overline { a ( k ) } , \quad b ( - k ) = \overline { b ( k ) },  
 +
\end{equation*}
  
\begin{equation*} | a ( k ) | ^ { 2 } = 1 + | b ( k ) | ^ { 2 } , r _ { - } ( k ) = \frac { b ( k ) } { a ( k ) } , r _ { + } ( k ) = - \frac { b ( - k ) } { a ( k ) }, \end{equation*}
+
\begin{equation*}  
 +
| a ( k ) | ^ { 2 } = 1 + | b ( k ) | ^ { 2 } , \quad
 +
r _ { - } ( k ) = \frac { b ( k ) } { a ( k ) } , \quad
 +
r _ { + } ( k ) = - \frac { b ( - k ) } { a ( k ) },  
 +
\end{equation*}
  
 
\begin{equation*} t ( k ) = \frac { 1 } { \alpha ( k ) }. \end{equation*}
 
\begin{equation*} t ( k ) = \frac { 1 } { \alpha ( k ) }. \end{equation*}

Latest revision as of 04:43, 15 February 2024


Let $q ( x ) \in L _ { 1,1 } : = \left\{ q : \int _ { - \infty } ^ { \infty } ( 1 + | x | ) | q ( x ) | d x < \infty , q = \overline { q } \right\}$, where the bar stands for complex conjugation. Consider the (direct) scattering problem:

\begin{equation} \tag{a1} \mathrm{l}u_- - k^2 u_- := (-u'' + q(x) - k^2) u_- = 0, \end{equation}

\begin{equation*} x \in \mathbf{R} : = ( - \infty , \infty ), \end{equation*}

\begin{equation} \tag{a2} u _ { - } = \left\{ \begin{array} { l } { e ^ { - i k x } + r _ { - } ( k ) e ^ { - i k x } } & {x \xrightarrow{\qquad\qquad\qquad\qquad\qquad\qquad }-\infty ,} \\ { t _{-} ( k ) e ^ { i k x } ,} & {x \xrightarrow{\qquad\qquad\qquad\qquad\qquad\qquad }+\infty .} \end{array} \right. \end{equation}

The coefficients $r ( k )$ and $t ( k )$ are called the reflection and transmission coefficients. One can prove that $t_- ( k )$ is analytic in $\mathbf{C} _ { + } : = \{ k : \operatorname { Im } k > 0 \}$ except at a finite number of points $i k_j$, $1 \leq j \leq J$, $k_ j > 0$, which are simple poles of $t ( k )$.

Problem (a1)–(a2) describes scattering by a plane wave $e ^ { i k x }$ falling from $- \infty$ and scattered by the potential $q ( x )$.

One can also consider the scattering of the plane wave falling from $+ \infty$:

\begin{equation} \tag{a3} \text{l}u _ { + } - { k } ^ { 2 } u _ { + } = 0 , x \in \mathbf{R}, \end{equation}

\begin{equation} \tag{a4} u_{+} = \left\{ \begin{array} { ll } t_{+}(k) e^{-ikx}, & x \xrightarrow{\qquad\qquad\qquad\qquad\qquad\qquad } +\infty, \\ e^{-ikx} + r_{+}e^{ikx}, & x \xrightarrow{\qquad\qquad\qquad\qquad\qquad\qquad } -\infty. \end{array} \right. \end{equation}

One proves that $t _ { - } ( k ) = t _ { + } ( k ) : = t ( k )$, $t ( - k ) = \overline { t ( k ) }$, $k \in \mathbf{R}$, where the bar stands for complex conjugation, $r _ { \pm } ( - k ) = \overline { r _ { \pm } ( k ) }$. The matrix

\begin{equation*} \left( \begin{array} { c c } { t ( k ) } & { r _ { - } ( k ) } \\ { r _ { + } ( k ) } & { t ( k ) } \end{array} \right) = S ( k ) \end{equation*}

is called the $S$-matrix (cf. Scattering matrix). Conservation of energy implies $| t ( k ) | ^ { 2 } + | r ( k ) | ^ { 2 } = 1$.

Let $f ( x , k )$ and $g ( x , k )$ be the solutions to (a1) satisfying the conditions:

\begin{equation*} f ( x , k ) = e ^ { i k x } + o ( 1 ) , \quad x \rightarrow \infty , \end{equation*}

\begin{equation*} g ( x , k ) = e ^ { - i k x } + o ( 1 ) , \quad x \rightarrow - \infty. \end{equation*}

Then

\begin{equation*} f ( x , k ) = e ^ { i k x } + \int _ { x } ^ { \infty } A _ { + } ( x , y ) e ^ { i k y } d y, \end{equation*}

\begin{equation*} g ( x , k ) = e ^ { - i k x } + \int _ { - \infty } ^ { x } A _ { - } ( x , y ) e ^ { - i k y } d y, \end{equation*}

where $A _ { \pm } ( x , y )$ are the kernels which define the transformation operators. One has

\begin{equation*} f ( x , k ) = b ( k ) g ( x , k ) + a ( k ) g ( x , - k ), \end{equation*}

\begin{equation*} g ( x , k ) = - b ( - k ) f ( x , k ) + a ( k ) f ( x , - k ), \end{equation*}

where

\begin{equation*} a ( - k ) = \overline { a ( k ) } , \quad b ( - k ) = \overline { b ( k ) }, \end{equation*}

\begin{equation*} | a ( k ) | ^ { 2 } = 1 + | b ( k ) | ^ { 2 } , \quad r _ { - } ( k ) = \frac { b ( k ) } { a ( k ) } , \quad r _ { + } ( k ) = - \frac { b ( - k ) } { a ( k ) }, \end{equation*}

\begin{equation*} t ( k ) = \frac { 1 } { \alpha ( k ) }. \end{equation*}

The function $a ( k )$ is analytic in $\mathbf{C} _ { + } : = \{ k : \operatorname { Im } k > 0 \}$ and has finitely many simple zeros all of which are at the points $i k_j$, $1 \leq j \leq J$, $a ( i k _ { j } ) = 0$, $\dot { a } ( i k _ { j } ) \neq 0$, $\dot { a } : = d a / d k $.

If $k = i k_j$, then $f ( x , i k_{ j} ) \in L ^ { 2 } ( \mathbf{R} )$,

\begin{equation*} - f ^ { \prime \prime } ( x , i k _ { j } ) + q ( x ) f ( x , i k _ { j } ) + k ^ { 2 _ j } f ( x , i k _ { j } ) = 0, \end{equation*}

\begin{equation*} \int _ { - \infty } ^ { \infty } | f ( x , i k _ { j } ) | ^ { 2 } d x = ( m _ { j } ^ { + } ) ^ { - 2 }, \end{equation*}

\begin{equation*} \int _ { - \infty } ^ { \infty } | g ( x , i k _ { j } ) | ^ { 2 } d x = ( m _ { j } ^ { - } ) ^ { - 2 }. \end{equation*}

The numbers $- k _ { j } ^ { 2 }$ are the eigenvalues of the operator $- d ^ { 2 } / d x ^ { 2 } + q ( x )$ in $L ^ { 2 } ( \mathbf{R} )$. They are called the bound states.

The scattering data are the values

\begin{equation*} S : = \{ r _ { + } ( k ) , i k _ { j } , ( m _ { j } ^ { + } ) ^ { 2 } : \forall k > 0,1 \leq j \leq J \}. \end{equation*}

The inverse scattering problem (ISP) consists of finding $q ( x ) \in L _ { 1,1 }$ from $\mathcal{S}$.

The inverse scattering problem has at most one solution in the class $L _ { 1 , 1}$. This solution can be calculated by the following Marchenko method:

1 Define

\begin{equation} \tag{a5} F + ( x ) = \sum _ { j = 1 } ^ { J } ( m _ { j } ^ { + } ) ^ { 2 } e ^ { - k _ { j } x } + \frac { 1 } { 2 \pi } \int _ { - \infty } ^ { \infty } r _ { + } ( k ) e ^ { i k x } d k \end{equation}

and solve the following Marchenko equatio for $A _ { + } ( x , y )$:

\begin{equation*} A _ { + } ( x , y ) + F _ { + } ( x + y ) + \int _ { x } ^ { \infty } A ( x , t ) F _ { + } ( t , y ) d t = 0, \end{equation*}

\begin{equation*} y \geq x. \end{equation*}

If the data $\left\{ r _ { + } ( k ) , i k _ { j } , ( m _ { j } ^ { + } ) ^ { 2 } : 1 \leq j \leq J \right\}$ correspond to a $q \in L _ { 1 , 1} $, then equation (a5) is uniquely solvable in $L ^ { 1 } ( x , \infty )$ for every $x > - \infty$.
2 If $A _ { + } ( x , y )$ is found, then $q ( x ) = - 2 d A _ { + } ( x , x ) / d x$.

The main result [a7] is the characterization property for the scattering data: In order that $\mathcal{S} : = \left\{ r _ { + } ( k ) , i k _ { j } , ( m _ { j } ^ { + } ) ^ { 2 } : 1 \leq j \leq J , k _ { j } > 0 , m _ { j } ^ { + } > 0 , k > 0 \right\}$ be the scattering data corresponding to a $q ( x ) \in L _ { 1,1 } ( \mathbf{R} )$, it is necessary and sufficient that the following conditions hold:

i) $r ( - k ) = \overline { r ( k ) }$ for $k > 0$, the function $r ( k )$ for $k \neq 0$ is continuous,

\begin{equation*} | r _ { + } ( k ) | \leq 1 - c k ^ { 2 } ( 1 + k ^ { 2 } ) ^ { - 1 }, \end{equation*}

where $c = \text{const} > 0$, and $r_+ ( k ) = O ( 1 / k )$ as $k \rightarrow \pm \infty$.

ii) The function

\begin{equation*} R _ { + } ( x ) : = \frac { 1 } { 2 \pi } \int _ { - \infty } ^ { \infty } r _ { + } ( k ) e ^ { i k x } d k \end{equation*}

is absolutely continuous and

\begin{equation*} \int _ { s } ^ { \infty } | R _ { + } ^ { \prime } ( x ) | ( 1 + | x | ) d x < \infty \end{equation*}

for every $s > - \infty$.

iii) Denote

\begin{equation*} a ( z ) : = \prod _ { j = 1 } ^ { J } \frac { z - i k _ { j } } { z + i k _ { j } } \operatorname { exp } \left\{ - \frac { 1 } { 2 \pi i } \int _ { - \infty } ^ { \infty } \frac { \operatorname { ln } ( 1 - | r _ { + } ( k ) | ^ { 2 } ) } { k - z } d k \right\} . \end{equation*}

The function $a ( z )$ is continuous in $\overline { \mathbf{C} } _ { + }$ and

\begin{equation*} \operatorname { lim } _ { k \rightarrow 0 } k \alpha ( k ) [ r _ { + } ( k ) + 1 ] = 0. \end{equation*}

iv) The function

\begin{equation*} R _ { - } ( x ) : = - \frac { 1 } { 2 \pi } \int _ { - \infty } ^ { \infty } r _ { + } ( - k ) \frac { a ( - k ) } { a ( k ) } e ^ { - i k x } d k \end{equation*}

is absolutely continuous and

\begin{equation*} \int _ { s } ^ { \infty } ( 1 + | x | ) | R _ { - } ^ { \prime } ( x ) | d x < \infty \end{equation*}

for every $s > - \infty$.

A similar result holds for the data

\begin{equation*} \{ r _ { - } ( k ) , i k _ { j } , ( m _ { j } ^ { - } ) ^ { 2 } : 1 \leq j \leq J , \forall k > 0 \} \end{equation*}

and the potential $q ( x )$ can be obtained by the Marchenko method, $q ( x ) = - 2 d A _ { - } ( x , x ) / d x$.

In [a2] the above theory is generalized to the case when $q ( x )$ tends to a different constants as $x \rightarrow + \infty$ and $x \rightarrow - \infty$.

In [a5] a different approach to solving the inverse scattering problem is described for

\begin{equation*} q \in L _ { 1,2 } : = \left\{ q : q = \overline { q } , \int _ { - \infty } ^ { \infty } ( 1 + x ^ { 2 } ) | q ( x ) | d x < \infty \right\}. \end{equation*}

The approach in [a5] is based on a trace formula.

If $q ( x ) = 0$ for $x < x _ { 0 } < \infty$, then the reflection coefficient $\{ r_+ ( k ) : \forall k > 0 \}$ alone, without the knowledge of $i k_j$ and $( m _ { j } ^ { + } ) ^ { 2 }$, determines $q ( x )$ uniquely. A simple proof of this and similar statements, based on property $C$ for ordinary differential equations (cf. Ordinary differential equations, property $C$ for), is given in [a10].

An inverse scattering problem for an inhomogeneous Schrödinger equation is studied in [a5].

The inverse scattering method is a tool for solving many evolution equations (cf. also Evolution equation) and is used in, e.g., soliton theory [a7], [a1], [a3], [a6] (cf. also Korteweg–de Vries equation; Harry Dym equation).

Methods for adding and removing bound states are described in [a5]. They are based on the Darboux–Crum transformations and commutation formulas.

A large bibliography can be found in [a4].

References

[a1] M. Ablowitz, H. Segur, "Solutions and inverse scattering transform" , SIAM (1981)
[a2] A. Cohen, T. Kappeler, "Scattering and inverse scattering for step-like potentials in the Schrödinger equation" Indiana Math. J. , 34 (1985) pp. 127–180
[a3] F. Calogero, A. Degasperis, "Solutions and the spectral transform" , North-Holland (1982)
[a4] K. Chadan, P. Sabatier, "Inverse problems in quantum scattering" , Springer (1989)
[a5] P. Deift, E. Trubowitz, "Inverse scattering on the line" Commun. Pure Appl. Math. , 32 (1979) pp. 121–251
[a6] L. Faddeev, L. Takhtadjian, "Hamiltonian methods in the theory of solutions" , Springer (1986)
[a7] V. Marchenko, "Sturm–Liouville operators and applications" , Birkhäuser (1986)
[a8] A.G. Ramm, "Multidimensional inverse scattering problems" , Longman/Wiley (1992)
[a9] A.G. Ramm, "Inverse problem for an inhomogeneous Schrödinger equation" J. Math. Phys. , 40 : 8 (1999) pp. 3876–3880
[a10] A.G. Ramm, "Property C for ODE and applications to inverse problems" A.G. Ramm (ed.) P.N. Shivakumar (ed.) A.V. Strauss (ed.) , Operator Theory and Applications , Fields Inst. Commun. , 25 , Amer. Math. Soc. (2000) pp. 15–75
How to Cite This Entry:
Inverse scattering, full-line case. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Inverse_scattering,_full-line_case&oldid=55505
This article was adapted from an original article by A.G. Ramm (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article