Content-Type: multipart/mixed; boundary="-------------0311090851728" This is a multi-part message in MIME format. ---------------0311090851728 Content-Type: text/plain; name="03-493.comments" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="03-493.comments" Classification Codes: Primary 34L40; Secondary 34L30. Affiliations: Royal Institute of Technology in Stockholm (KTH), Sweden e-mails: hoppe@math.kth.se jorgen@math.kth.se laptev@math.kth.se ---------------0311090851728 Content-Type: text/plain; name="03-493.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="03-493.keywords" Fourth Order Differential Operators. Factorization. Follytons. ---------------0311090851728 Content-Type: application/x-tex; name="hlo.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="hlo.tex" % % This file was prepared using % Latex2E % % % \documentclass[12pt]{amsart} %\usepackage{draftcopy} %\usepackage{euler} \usepackage{euscript} \usepackage{multicol} \usepackage{amssymb} \usepackage{amsmath} \usepackage{times} \sloppy \newcommand{\im}{\mbox{\upshape Im\ }} %\newcommand{\det}{\mbox{\upshape det\ }} \newcommand{\re}{\mbox{\upshape Re\ }} \newcommand{\tr}{\mbox{\upshape tr\ }} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}{Proposition}[section] \newtheorem{lemma}{Lemma}[section] \numberwithin{equation}{section} \begin{document} \title[Follytons and the Removal of Eigenvalues] {Follytons\\ and the Removal of Eigenvalues\\ for Fourth Order Differential Operators} \author[Hoppe, Laptev and \"Ostensson]{J. Hoppe, A. Laptev and J. \"Ostensson} \begin{abstract} A non-linear functional $Q[u,v]$ is given that governs the loss, respectively gain, of (doubly degenerate) eigenvalues of fourth order differential operators $L = \partial^4 + \partial\,u\,\partial + v$ on the line. Apart from factorizing $L$ as $A^{*}A + E_{0}$, providing several explicit examples, and deriving various relations between $u$, $v$ and eigenfunctions of $L$, we find $u$ and $v$ such that $L$ is isospectral to the free operator $L_{0} = \partial^{4}$ up to one (multiplicity 2) eigenvalue $E_{0} < 0$. Not unexpectedly, this choice of $u$, $v$ leads to exact solutions of the corresponding time-dependent PDE's. \end{abstract} \email{hoppe@math.kth.se, laptev@math.kth.se, jorgen@math.kth.se} %\thanks{} \subjclass{Primary 34L40; Secondary 34L30.} \maketitle %\setcounter{section}{-1} \section{Factorization of the operator $L = \partial^4 + \partial\,u\,\partial + v$.} Let us assume that $u$ and $v$ are real-valued functions and $u, v\in \mathcal{S}\left(\mathbb{R}\right)$, where $\mathcal{S}\left(\mathbb{R}\right)$ denotes the Schwarz class of rapidly decaying functions. Let $L$ be a linear fourth order selfadjoint operator in $L^2\left(\mathbb{R}\right)$ % \begin{equation}\label{L} L := \partial^4 + \partial\,u\,\partial + v \end{equation} % defined on functions from the Sobolev class $H^4(\mathbb{R})$. This operator is bounded from below and we assume that its lowest eigenvalue $E_0<0$ is of double multiplicity and therefore there exist two orthogonal in $L^2(\mathbb{R})$ eigenfunctions $\psi_+$ and $\psi_-$ satisfying the equation % \begin{equation}\label{groundstate} L\,\psi = E_{0}\,\psi. \end{equation} % As shown in the appendix, the Wronskian % \begin{equation} \label{wronsk} W(x):= \psi_{+}(x)\,\psi_{-}'(x) - \psi_{-}(x)\,\psi_{+}'(x) \end{equation} % is necessarily non-vanishing, $W(x)\not= 0$, $x \in \mathbb{R}$. Let us try to factorize $L-E_{0}$ as % \begin{equation} \label{factorization} A^{*} A = \left(- \partial^2 - f \partial + g - f'\right) \left(- \partial^2 + f \partial + g\right), \end{equation} % with $f$ and $g$ real-valued. Clearly, % \begin{equation} \label{odefg} \left\{ \begin{array}{ll} f' + f^2 + 2g &= - u\\ g^2 - (fg + g')' &= v - E_{0}. \end{array} \right. \end{equation} % Instead of discussing these non-linear differential equations directly, let us express $f, g, u$ and $v$ in terms of the functions $\psi_{+}$, $\psi_{-}$. Straightforwardly, one finds that since $\psi_{+}$ and $\psi_{-}$ are eigenfunctions of $A^{*} A$ with eigenvalue $0$, we have $A \psi_{+} = A \psi_{-} = 0$, which implies % \begin{equation} \label{eqnfg} \left\{ \begin{array}{rl} f\,W &= W'\\ -g\,W &= \psi_{+}'\,\psi_{-}'' - \psi_{+}''\,\psi_{-}' =: W_{12}, \end{array} \right. \end{equation} % while $\left(L - E_{0}\right) \psi_{+} = \left(L - E_{0}\right) \psi_{-} = 0$ implies % \begin{equation} \label{eqnuv} \left\{ \begin{array}{rl} u\,W &= 2 \, W_{12} - W'' + \epsilon\\ \left(v - E_{0}\right)\,W&= u\,W_{12} + W_{12}'' -W_{23}, \end{array} \right. \end{equation} % where $\epsilon$ is an integration constant and % \begin{equation} \label{W23} W_{23} := \psi_{+}''\,\psi_{-}''' - \psi_{+}'''\,\psi_{-}'' \end{equation} % is expressible in terms of $W$ and $W_{12}$ via % \begin{equation} \label{W23relation} W\,W_{23} = W_{12}'\,W'-W_{12}\,W'' + W_{12}^2. \end{equation} % Equations \eqref{eqnfg} say that % \begin{equation} \label{fg} f = \frac{W'}{W}, \quad g = - \frac{W_{12}}{W}. \end{equation} % Since $u\,W + W'' - 2\,W_{12}$ vanishes at infinity, $\epsilon$ has to be $0$, and one finds, using equations \eqref{eqnuv}-\eqref{W23relation}, that % \begin{alignat}{4} \label{eqnu} u &= \frac{2\,W_{12} - W''}{W} \\ \label{eqnv} v - E_{0} &= \frac{W_{12}^2}{W^2} + \left(\frac{W_{12}'}{W}\right)'. \end{alignat} % Note that % \begin{equation} \label{darbouxL} \tilde{L} := A A^{*} + E_{0} = L + 4\,\partial\,f'\,\partial + 2\,f\,g' -f\,f''+f''' \end{equation} % will be isospectral to $L$, apart from $E_{0}$, which has been removed. To see why $E_{0}$ is not an eigenvalue of $\tilde{L}$, let us for simplicity assume that $u, v \in C_{0}^{\infty}\left(\mathbb{R}\right)$, say that supp$\,u$, supp$\,v \subset (-c,c)$. Then, % \begin{equation*} \begin{array}{rl} \psi_{+}(x) = \alpha_{1}\,e^{-\kappa x}\,\cos\,(\kappa x) + \beta_{1}\, e^{-\kappa x}\,\sin\,(\kappa x)\\ \psi_{-}(x) = \alpha_{2}\,e^{-\kappa x}\,\cos\,(\kappa x) + \beta_{2}\, e^{-\kappa x}\,\sin\,(\kappa x) \end{array}, \quad x > c, \end{equation*} % where $E_{0} = - 4\,\kappa^4$, $k>0$. This implies % \begin{equation*} \begin{array}{rl} W(x) &= \kappa\,e^{-2\kappa x}\,\left(\alpha_{1}\,\beta_{2} - \beta_{1}\,\alpha_{2}\right)\\ W_{12}(x) &= 2\,\kappa^3\,e^{-2\kappa x}\,\left(\alpha_{1}\,\beta_{2} - \beta_{1}\,\alpha_{2}\right) \end{array}, \quad x > c. \end{equation*} % (note that the bracket does not vanish, since $\psi_{+}$ and $\psi_{-}$ are linearly independent.) This (and a similar investigation at the other end) implies that \begin{equation*} f(x) = \mp 2\,\kappa, \quad g(x) = -2\,\kappa^2, \quad \mbox{for } \pm\,x >c. \end{equation*} Since $\tilde{L}\psi = E_{0}\,\psi$ implies $A^{*} \psi = 0$, we obtain \begin{equation*} \psi'' - 2\kappa\,\psi' + 2\kappa^2 \psi = 0, \quad x > c. \end{equation*} It clearly follows that $\psi$ cannot be in $L^{2}\left(\mathbb{R}\right)$ unless it vanishes identically. \smallskip Before giving some explicit examples, let us make some comments concerning the problem of actually finding $f$ and $g$, or $\psi_{+}$ and $\psi_{-}$, when $u$ and $v$ are given. Instead of solving the non-linear system \eqref{odefg}, or the spectral problem \eqref{groundstate}, one may also try to solve the Hirota-type equation which follows from \eqref{eqnu}, \eqref{eqnv} % \begin{equation} \label{hirota} 4\,\left(v-E_{0}\right) = \left(\frac{W''}{W} + u\right)^2 + 2\,\left(\frac{W'''}{W} + u' + u\,\frac{W'}{W}\right)', \end{equation} % and which for $u \equiv 0$ reads % \begin{equation*} \label{hirota2} 4\,\left(v-E_{0}\right)\,W^{2} = 2\,\left(W''''\,W - W'''\,W'\right) + W''^{2}. \end{equation*} % Once $W\,(\neq 0)$ is obtained, $f$ and $g$ can be given by the equations \eqref{fg}. With $f$ and $g$ defined in this way \cite{Hoppe1}, equation \eqref{odefg} is satisfied and the factorization \eqref{factorization} is valid. \smallskip Note also the following: the functions $\psi_{+}$ and $\psi_{-}$ are solutions of $A\,\psi = 0$, i.e. % \begin{equation*} -\psi'' + f\,\psi' + g\,\psi = 0. \end{equation*} % By writing % \begin{equation*} \psi_{\pm} = \sqrt{W}\,\phi_{\pm} , \end{equation*} % one finds that $\phi_{+}\,\phi_{-}' - \phi_{+}'\,\phi_{-} =1$ and that $\phi_{\pm}$ are (oscillating) solutions of the equation in Liouville form % \begin{equation*} -\phi'' + \left(g +\frac{3}{4}\left(\frac{W'}{W}\right)^2 - \frac{1}{2}\,\frac{W''}{W}\right)\,\phi = 0, \end{equation*} % i.e. associated to a second order self-adjoint diffential operator. \section{Addition and removal of eigenvalues.} Although adding and removing eigenvalues may be thought to be a procedure that can be read both ways (symmetrically), the steps involved are actually quite different in both cases (in particular, it is not yet clear, which conditions on $u$ and $v$ allow for the addition of a doubly degenerate eigenvalue below the spectrum of $\partial^4 + \partial\,u\partial + v$). Let us therefore 'summarize' them separately, in both cases starting from a given operator \begin{equation*} L_{n} := \partial^4 + \partial\,u_{n}\partial + v_{n},\quad n \in \mathbb{N}, \end{equation*} and the equation \eqref{hirota} with $u, v$ replaced by $u_{n}, v_{n}$. This equation shall be referred to as \eqref{hirota}$_{n}$.\\ \textbf{Removal of eigenvalues:}\\ \textit{1}. Solve \eqref{hirota}$_{n}$ $\left(\mbox{with } E_{0} \rightarrow E_{0}^{(n)} = -4 \kappa_{n}^4\right)$ for $W_{n} := W$ $\left( \rightarrow 0 \mbox{ at infinity}\right)$ and define $W_{12}^{(n)}$ as $\frac{1}{2}\left(W_{n}\,u_{n} + W_{n}''\right)$, as is natural in accordance with equation \eqref{eqnu}$_{n}$. Alternatively, if $\psi_{\pm}^{(n)}$ are known, calculate $W_{n}$ and $W_{12}^{(n)}$ via their definitions, i.e. as \begin{equation*} \begin{array}{rl} W_{n} &= \psi_{+}^{(n)}\,{\psi_{-}^{(n)}}' - \psi_{-}^{(n)}\,{\psi_{+}^{(n)}}' \\ W_{12}^{(n)} & = {\psi_{+}^{(n)}}'\,{\psi_{-}^{(n)}}'' - {\psi_{+}^{(n)}}''\,{\psi_{-}^{(n)}}' . \end{array} \end{equation*} \textit{2}. Define $f_{n}$ and $g_{n}$ according to \eqref{fg}$_{n}$, thus solving the system \eqref{odefg}, and obtaining the factorization \begin{equation*} L_{n} =A_{n}^{*} A_{n} - 4\kappa_{n}^4. \end{equation*} \textit{3}. The operator \begin{equation*} \tilde{L}_{n} =A_{n} A_{n}^{*} - 4\kappa_{n}^4 =: L_{n-1} \end{equation*} will then be isospectral to $L_{n}$ apart form the lowest eigenvalue $E_{0}^{(n)} = - 4\kappa_{n}^4$ (of multiplicity 2), which has been removed.\\ \textbf{Addition of eigenvalues:}\\ \textit{1}. Solve \eqref{hirota}$_{n}$ $\left(\mbox{with } E_{0} \rightarrow E_{0}^{(n+1)} = -4 \kappa_{n+1}^4\right)$ for $\hat{W}_{n+1} := W \sim e^{\pm 2 \kappa_{n+1} x}$, as $x \rightarrow \pm \infty$, i.e. $\hat{W}_{n+1}$ diverging at infinity and non-vanishing for finite $x$. (As mentioned above, conditions on $u_{n}$, $v_{n}$ ensuring the existence of $\hat{W}_{n+1}$ are still unclear.) \\ \textit{2}. Define $W_{n+1} := \frac{1}{\hat{W}_{n+1}}$, which will then solve the (more complicated looking) equation \begin{alignat}{4} \label{addeqn1} &40\,\frac{W'^{4}}{W^{4}}-2\,\frac{W''''}{W}+14\,\frac{W'''\,W'}{W^{2}}+ 13\,\frac{W''^{2}}{W^{2}}-64\,\frac{W''\,W'^{2}}{W^{3}}+\\ \notag &2u''+ u^2-2u'\,\frac{W'}{W}+2u\left(\frac{W'^{2}}{W^{2}}- 2\left(\frac{W'}{W}\right)'\right) = 16 \kappa^4 + 4 v \end{alignat} (with $u$, $v$ $\rightarrow$ $u_{n}$, $v_{n}$ and $\kappa \rightarrow \kappa_{n+1}$). In fact, \eqref{addeqn1} is equivalent to % \begin{alignat*}{4} &-2f'''+6ff''+7f'^2-8f'f^2+f^4+2u\left(f^2-2f'\right)-2u'f+u^2+2u'' \\ \notag &= 4v+16\kappa^4 \end{alignat*} $\left(\mbox{via } f = \frac{W_{n+1}'}{W_{n+1}} =:f_{n+1},\,u, v \rightarrow u_{n}, v_{n} \mbox{ and } \kappa \rightarrow \kappa_{n+1}\right)$ that arises in the factorization of $L_{n+1}$. \\ \textit{3}. Write \begin{equation*} L_{n} = A_{n+1} A_{n+1}^{*} - 4\kappa_{n+1}^4 \end{equation*} (implying $2\,g_{n+1} := 3\,f_{n+1}' - f_{n+1}^2 - u_{n}$). \\ \textit{4}. Then, \begin{equation*} L_{n+1} := A_{n+1}^{*} A_{n+1} - 4\kappa_{n+1}^4, \end{equation*} will be isospectral to $L_{n}$ apart from having one additional (doubly degenerate) eigenvalue $E_{0}^{(n+1)}$ below the spectrum of $L_{n}$. \section{A non-linear functional $Q$ and a system of PDE's associated with the operator $L$.} As observed 100 years ago \cite{Sch}, the operator $L = \partial^4 + \partial\,u\,\partial + v$ has a unique 4'th root in the form %of a %pseudodifferential operator $L^{1/4} := \partial + \sum_{k=1}^{\infty} l_{k}(x) \partial^{-k}$. %For any positive integer $m$ we may decompose %the operator $L^{m/4} := \left(L^{1/4}\right)^{m}$ into a positive and %a negative part % %\begin{equation*} %L^{m/4} = \sum_{k=0}^{m} P_{k}(x)\,\partial^{k} + \sum_{k=1}^{\infty} %p_{k}(x)\,\partial^{-k} = \left(L^{m/4}\right)_{+} + \left(L^{m/4}\right)_{-}. %\end{equation*} % Define $M$ to be the positive (differential operator) part of any %such integer power of $L^{1/4}$. Then it is well known, that % \begin{equation*} L_{t} = \left[L, M\right], \end{equation*} % where $L_{t}$ is the operator defined by $ L_{t}\,\varphi = \partial\,u_{t}\,\partial\,\varphi + v_{t}\,\varphi, $ consistently defines evolution equations (for $u = u(x, t)$, $v = v(x, t)$) that have infinitely many conserved quantities (i.e. functionals of $u$ and $v$, and their spatial derivatives, that do not depend on $t$). We shall make use of this by letting % \begin{equation*} M := 8\left(L^{3/4}\right)_{+} = 8\,\partial^3 + 6\,u\,\partial + 3\,u', \end{equation*} % and focusing on the quantity % \begin{equation} \label{functionalQ} Q[u, v] := \int_{\mathbb{R}} \left(48\,v^2 + \frac{5}{4}\,u^4 - 12\,u^2\,v - 40\,u\,v'' - 13\,u\,u'^2 + 9\,u''^2\right) \,dx. \end{equation} % This quantity does not change when $u$ and $v$ evolve according to % \begin{equation} \label{KdVtype} \left\{ \begin{array}{rl} u_{t} &= 10\,u''' + 6\,u\,u' - 24\,v' \\ v_{t} & = 3\left(u''''' + u\,u''' + u'\,u''\right) - 8\,v''' - 6\,u\,v'. \end{array} \right. \end{equation} % Formula \eqref{darbouxL} for $\tilde{L} = \partial^{4} +\partial\,\tilde{u}\,\partial + \tilde{v}$ implies that % \begin{equation} \label{darbouxL2} \left\{ \begin{array}{rl} \tilde{u} - u &= 4\,f'\\ \tilde{v} - v &= 2\,f\,g' -f\,f''+ f'''. \end{array} \right. \end{equation} % By using the asymptotic properties of $f$ and $g$ ($f \rightarrow \mp 2\,\kappa$, $g \rightarrow -2\kappa^{2}$, as $x \rightarrow \pm \infty)$, one can show that % \begin{equation} \label{deltaQ} \delta Q := Q[\tilde{u},\tilde{v}] - Q[u, v] = - 32\,\kappa^7\,\frac{2^{9}}{7}. \end{equation} % $\left(\mbox{making } \frac{7}{2^9\,\sqrt{2}}\,\delta Q = -\,2\,\left(4\,\kappa^{4}\right)^{7/4}\right)$. This result is similar to that for Schr\"{o}dinger operators \cite{BenLoss} and reflects the loss of a doubly degenerate eigenvalue $E_{0} = -4\,\kappa^{4}$, when going from $L$ to $\tilde{L}$. The constant in the right hand side of \eqref{deltaQ} is related to the semiclassical constant appearing in the trace formula for a fourth order differential operator considered in \cite{Ost}. \\ The proof of \eqref{deltaQ}, just as the derivation of \eqref{functionalQ}, involves very lengthy calculations. When deriving \eqref{deltaQ} one uses \eqref{odefg} and \eqref{darbouxL2} to write the expression for $\delta Q$ as an integral of terms involving only the functions $f$ and $g$, and their spatial derivatives. The crucial step is to note that the integrand is a pure derivative of $x$, i.e. $\delta Q = \int {\mathbb{Q}}\,' \, dx$ for some function $\mathbb{Q}$, which makes it possible to evaluate the integral solely from the limits of $f$ and $g$ at infinity. Thus, to compute $\delta Q$, one selects the terms in $\mathbb{Q}$ which are free of derivatives, as those are the only ones that contibute. The terms in $\mathbb{Q}$ still containing derivatives, for instance the ones quadratic in $g$ and linear in $f$, \begin{alignat*}{4} \int &\bigg((96-48)\,g^2\,f''' - 2\cdot96\,f\,g'\,g'' - 8\cdot12\,g''\,f'\cdot2g -4\cdot40\,f'''\,g^2 \\ &+ 160\,g''\,g'\,f - 16\cdot26\,f''\,g'\,g - 16\cdot13\,g'^2\,f'\bigg)\,dx, \end{alignat*} %which contributes to $\mathbb{Q}$ by %\begin{equation*} %-112\,f''\,g^2 - 192\,f'\,g\,g' - 16\,f\,g'^2 %\end{equation*} give zero. \section{Some examples.} \textit{Example 1.} The operator % \begin{equation*} L = \partial^4 - 5\,\partial^2 + \partial\,\frac{12}{\cosh^2 x}\,\partial - \frac{6}{\cosh^2 x} = A^{*} A - 4 \end{equation*} % with % \begin{equation*} A = - \partial^2 - 3\,\tanh x\,\partial - 2 \end{equation*} % has 2 linearly independent eigenfunctions with eigenvalue $E_{0} = -4$, % \begin{equation*} \psi_{+}(x) = \frac{1}{\cosh^2 x}, \quad \psi_{-}(x) = \frac{\sinh x}{\cosh^2 x}. \end{equation*} % One can easily check that $A\,\psi_{\pm} = 0$ and that $u, v$ are reflectionless, as % \begin{equation*} \tilde{L} = A A^{*} - 4 = \partial^4 - 5\,\partial^2 \end{equation*} % (note that $\psi_{+}$ and $\psi_{-}$ have different fall-off behaviour at $\infty$ and that $W(x) = \cosh^{-3} x$). \\ \textit{Example 2.} The operator % \begin{equation*} L = \partial^4 + 16\,\partial\,\frac{1}{\cosh^2 x}\,\partial + \frac{40}{\cosh^4 x} - \frac{88}{\cosh^2 x} = A^{*} A - 64 \end{equation*} % with % \begin{equation*} A = - \partial^2 - 4\,\tanh x\,\partial - 8 + \frac{2}{\cosh^2 x} \end{equation*} % has 2 linearly independent eigenfunctions with eigenvalue $E_{0} = -64$, % \begin{equation*} \psi_{+}(x) = \frac{\cos 2x}{\cosh^2 x}, \quad \psi_{-}(x) = \frac{\sin 2x}{\cosh^2 x}. \end{equation*} % One easily verifies that $A\,\psi_{\pm} = 0$, and that % \begin{equation*} \tilde{L} = A A^{*} - 4 = \partial^4 - \frac{40}{\cosh^2 x}. \end{equation*} % A computation gives that \begin{equation*} Q = \frac{2^{10}}{7}\cdot 4 \cdot 687, \quad \tilde{Q} = 2^{10} \cdot 100, \quad \delta Q = -\frac{2^{21}}{7} \left(= - 32\,(\kappa = 2)^7\,\frac{2^9}{7}\right). \end{equation*}\\ \textit{Example 3.} The operator % \begin{equation*} L = \partial^4 + \left(45\,\Psi^4 - 40\,\Psi^2\right) = A^{*} A - 4 \end{equation*} % with % \begin{equation*} W = \Psi^{2} := \frac{1}{\cosh^{2} x}, \quad W_{12} = 2\,\Psi^2 - 3\,\Psi^4 \end{equation*} % and % \begin{equation*} A = -\partial^2 - 2\,\tanh x\,\partial - 2 + 3\,\Psi^2 \end{equation*} % has a doubly degenerate eigenvalue $E_{0} = -4$. One easily verifies, that % \begin{equation*} \tilde{L} = \partial^4 - 8\,\partial\,\Psi^2\,\partial + 25\,\Psi^4-16\,\Psi^2. \end{equation*} % \\ \textit{Example 4.} The operator % \begin{equation*} L = \partial^4 - \partial^2 + 4\,\partial\,\frac{1}{\cosh^2 x} \,\partial + \frac{6}{\cosh^{2} x} - \frac{8}{\cosh^{4} x} = A^{*} A \end{equation*} % with % \begin{equation*} A = -\partial^2 - \,\tanh x\,\partial - \frac{1}{\cosh^{2} x} = \partial \left(-\partial - \tanh x\right) \end{equation*} % has a unique ground-state $E_{0} = 0$ with eigenfunction % \begin{equation*} \psi (x) = \frac{1}{\cosh x}. \end{equation*} % The second solution of $A\,\psi = 0$ is $\psi = \tanh x \not\in L^{2}\left(\mathbb{R}\right)$. One easily verifies, that % \begin{equation*} \tilde{L} = \partial^4 - \partial^2. \end{equation*} %\\ \textit{Example 5.} For any $k > 0$, the operator % \begin{equation*} L = \partial^4 + \partial\,u\,\partial + v \end{equation*} % with % \begin{equation*} \left\{ \begin{array}{rl} u(x) &= 2\left(1 + \frac{2}{k}\right)\,\Psi^{2}\left(\frac{x}{k}\right)\\ v(x) &= - 4\left(1 +\frac{1}{k} - \frac{1}{k^3}\right) \Psi^{2}\left(\frac{x}{k}\right) + \left(1 - \frac{1}{k}\right)\left(1 + \frac{5}{k} + \frac{6}{k^2}\right)\Psi^{4}\left(\frac{x}{k}\right), \end{array} \right. \end{equation*} % where % \begin{equation*} \Psi(x) := \frac{1}{\cosh x}, \end{equation*} % has a doubly degenerate ground-state, $E_{0} = -4$, with eigenfunctions % \begin{equation*} \psi_{\pm}^{(k)}(x) = e^{\pm i x} \left(\frac{1}{\cosh \frac{x}{k}}\right)^{k}. \end{equation*} % \section{Follytons.} In order to find $u$ and $v$ such that $L = A^{*} A + E_{0}$ is 'conjugate' to the free operator $\tilde{L} = \partial^{4} =: L_{0}$ one has to solve \eqref{odefg} with $u = v = 0$. Eliminating $g$ and writing $E_{0} = - 4\kappa^{4}$ one obtains the ODE % \begin{equation*} 2\,f''' + 6\,f\,f'' + 7\,f'^2 + 8\,f'\,f^2 + f^4 = 16\,\kappa^{4}. \end{equation*} % One may reduce the order by taking $f$ as the independent variable, and $F(f) := f'$ as the dependent one, yielding % \begin{equation*} 2\left(F''\,F^2 + F'^2\,F\right) + 6\,F\,F'\,f + 7\,F^2 + 8\,F\,f^2 + f^4 = 16\,\kappa^4, \end{equation*} % but both forms seem(ed) to be too difficult to solve. By using \eqref{hirota}, however, it takes the form % \begin{equation*} 16\kappa^4\,W^2 = 2\left(W''''\,W - W'''\,W'\right) + W''^2, \end{equation*} % in which it is easier to see the solution \cite{Hoppe1} % \begin{equation*} \hat{W} = \mbox{const } \cdot \left(\sqrt{2} + \cosh\left(2\,\kappa\,x\right)\right). \end{equation*} % Correspondingly, % \begin{equation*} \hat{f} := \frac{\hat{W}'}{\hat{W}} = 2\,\kappa \, \frac{\sinh\left(2\,\kappa\,x\right)}{\sqrt{2} + \cosh\left(2\,\kappa\,x\right)}. \end{equation*} % As interchanging $A^{*}$ and $A$ (as far as $f$ is concerned) only changes the sign of $f$, % \begin{equation*} f (x) = -2\,\kappa \, \frac{\sinh\left(2\,\kappa\,x\right)}{\sqrt{2} + \cosh\left(2\,\kappa\,x\right)}. \end{equation*} % The Wronskian of the two ground-states $\psi_{\pm}$ (of $L = \partial^4 + \partial\,u\,\partial + v$, conjugate to $L_{0} = \partial^4$) is simply the inverse of $\hat{W}$, i.e. (choosing the constant in $\hat{W}$ to be $1$), % \begin{equation*} W(x) = \frac{1}{\sqrt{2} + \cosh\left(2\,\kappa\,x\right)} =: \chi\left(2\,\kappa\,x\right). \end{equation*} % The function $g$ is given by % \begin{equation*} g = \frac{1}{2} \left(3\,f' - f^2\right) = -2\,\kappa^2 \left(1 + \sqrt{2}\,W - 2\,W^2\right). \end{equation*} % Insertion into equation \eqref{odefg} yields the reflectionless 'potentials' % \begin{equation} \label{folly1} \left\{ \begin{array}{rl} u_{\kappa} &= 16\,\kappa^2 \left(\sqrt{2}\,W - W^2\right)\\ v_{\kappa} &= 16\,\kappa^4 \left(\sqrt{2}\,W - 12\,W^2 + 16\,\sqrt{2}\,W^3 - 8\,W^4\right) \end{array} \right. \end{equation} % with $L = \partial^4 + \partial\,u_{\kappa}\,\partial + v_{\kappa}$ having exactly one doubly degenerate negative eigenvalue $-4\,\kappa^4$. While in most other examples we scaled $\kappa$ to be equal to $1$ it is, in this case (due to the appearence of $2 \kappa$ in $W$) easiest to choose $\kappa = \frac{1}{2}$, i.e. to take % \begin{equation} \label{folly2} \left\{ \begin{array}{rl} u &= 4 \left(\sqrt{2}\, \chi - \chi^{2}\right)\\ v &= \left(\sqrt{2}\,\chi - 12\,\chi^2 + 16\,\sqrt{2}\,\chi^3-8\chi^4\right) \end{array} \right. \end{equation} % and, when needed, use formulas like % \begin{equation*} \begin{array}{rl} \chi '' &= \chi\,\left(1 - 3\,\sqrt{2}\,\chi + 2\,\chi^2\right)\\ \chi'^2 &= \chi^2\,\left(1 - 2\,\sqrt{2}\,\chi + \chi^2\right)\\ \chi''' &= \chi'\,\left(1 - 6\,\sqrt{2}\,\chi + 6\,\chi^2\right)\\ \chi'''' &= \chi\,\left(1 - 15\,\sqrt{2}\,\chi + 80\,\chi^2 - 60\,\sqrt{2}\,\chi^{3} + 24\,\chi^4\right) . \end{array} \end{equation*} % (Note that redefining $\chi$ by a factor of $-\sqrt{2}$ would make all the coefficients positive (integers)). These formulas are useful when checking that $u(x + 4\,t)$ and $v(x + 4\,t)$, with $u$ and $v$ given by \eqref{folly2}, are exact solutions of the non-linear system of PDE's \eqref{KdVtype} (just as $u_{\kappa}(x + 16\kappa^2 \,t), v_{\kappa}(x + 16\kappa^2\,t)$). \appendix \section{$W\not= 0$.} We shall prove here that the Wronskian type function defined in \eqref{wronsk} never equals zero. \medskip \begin{theorem}\label{Wronsk} Let $\psi_\pm$ be two orthonormal eigenfunctions of the operator \eqref{L} corresponding to the lowest eigenvalue $E_0$ of double multiplicity. Then % \begin{equation}\label{wronsk1} W[\psi_+,\psi_-](x):= \psi_{+}(x)\,\psi_{-}'(x) - \psi_{-}(x)\,\psi_{+}'(x) \not= 0, \quad x\in \mathbb{R}. \end{equation} % \end{theorem} % \medskip \noindent In order to prove this result we need a simple auxiliary statement. % \medskip \begin{lemma}\label{overdet} Let $E_0$ be the lowest eigenvalue of the operator $L$ and let $\psi\in L^2(\mathbb{R})$ be a solution of the equation \eqref{groundstate} satisfying $\psi(x_0) = \psi^{'}(x_0)= 0$ for some $x_0\in \mathbb{R}$. Then $\psi(x) \equiv 0$. \end{lemma} % \noindent \textit{Proof}. Indeed, the function $$ \tilde \psi(x) = \begin{cases} \psi(x), & \text{if \(x\leq x_0\),} \\ -\psi(x), & \text{if \(x\geq x_0\),} \end{cases} $$ is linear independent with $\psi$. Since $\tilde\psi(x_0) = \tilde\psi^{'}(x_0)= 0$ we obtain $(L \tilde\psi, \tilde\psi) = E_0\|\tilde\psi\|^2$. Then $\tilde\psi$ is also an eigenfunction of the operator $L$ with the eigenvalue $E_0$. Consider now the linear combination $$ \psi_1(x) = \tilde\psi^{''}(x_0)\psi(x) - \psi^{''}(x_0)\tilde\psi(x). $$ Obviously $\psi_1(x_0)=\psi_1{'}(x_0)=\psi_1{''}(x_0)=0$, $\psi_1\in L^2\left(\mathbb{R}\right)$ and $\psi_1$ satisfies the fourth order differential equation $L\psi_1 = E_0\psi_1$. Being overdetermined, $\psi_1\equiv 0$ which also implies $\psi\equiv 0$. $\Box$ \medskip \noindent {\it Remark}. In Lemma~\ref{overdet} the conditions $\psi(x_0) = \psi^{'}(x_0)= 0$ split the problem for the operator $L$ in $L^2(\mathbb{R})$ into two Dirichlet boundary value problems on semiaxes $L^2((x_0,\infty))$ and $L^2((-\infty, x_0))$. Therefore, the lowest eigenvalue moves up. \medskip \noindent {\it Proof of Theorem~\ref{Wronsk}}. \noindent {\it a.} Let $\psi_\pm$, be two orthonormal eigenfunctions corresponding to the lowest eigenvalue $E_0$ of the operator $L$. The functions $\psi_+$ and $\psi_-$ cannot vanish at the same point. Indeed, assume that they do. Then there is a point $x_0$ such that $\psi_+(x_0)=\psi_-(x_0) = 0$. If in addition we assume that say $\psi_+^{'}(x_0)=0$, then by Lemma~\ref{overdet} $\psi_+\equiv 0$ and we obtain a contradiction. Therefore we can assume that $\psi_\pm^{'}(x_0) \not=0$. Introduce a new function $$ \psi_2(x) = \psi_-^{'}(x_0) \psi_+(x) - \psi_+^{'}(x_0)\psi_-(x). $$ It is a non-trivial eigenfunction of the equation \eqref{groundstate} satisfying $\psi_2(x_0)=\psi_2^{'}(x_0)=0$. By using Lemma~\ref{overdet} again we find that $\psi_2\equiv0$ which cannot be true because $\psi_+$ and $\psi_-$ are linear independent. \smallskip \noindent {\it b.} Consider now the following pair of complex functions $$ \Psi_{\pm}(x) = \psi_+(x) \pm i\psi_-(x) =: \psi(x) e^{\pm i \phi(x)}. $$ By using {\it a}. we observe that $\psi$ never vanishes, $\psi(x)\not= 0$, $x\in \mathbb{R}$. Besides $$ W[\Psi_+, \Psi_-] = (\psi_+ +i\psi_-) (\psi_+ -i\psi_-)^{'} - (\psi_+ +i\psi_-)^{'} (\psi_+ -i\psi_-) $$ $$ = -2i \,W[\psi_+,\psi_-] = -2i \,\phi^{'}\psi^2. $$ Thus, in order to prove Theorem \ref{Wronsk} it remains to prove that $\phi^{'}\not=0$. Assume that there is $x_0$ such that $\phi^{'}(x_0) = 0$ and consider $$ \Phi(x) = e^{-i\phi(x_0)}\Psi_+(x) - e^{i\phi(x_0)}\Psi_-(x). $$ Clearly $\Phi(x_0)=\Phi^{'}(x_0)=0$ and by using Lemma~\ref{overdet} we obtain $\Phi\equiv0$ which contradicts the linear independency of the functions $\Psi_\pm$. \noindent The proof is complete. $\Box$ \medskip \noindent {\it Acknowledgments.} The authors would like to thank H. Kalf, E. Langmann, A. Pushnitski and O. Safronov for useful discussions, as well as the ESF European programme SPECT and the EU Network: ``Analysis \& Quantum" for partial support. \begin{thebibliography}{99} % \bibitem{BenLoss} R. Benguria, M. 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