Difference between revisions of "Inverse scattering, full-line case"
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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: | ||
− | + | \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} | ||
− | + | \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 ) } & | + | \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:
|
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 |
Inverse scattering, full-line case. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Inverse_scattering,_full-line_case&oldid=55505