11
$\begingroup$

I have a second order differential equation, \begin{eqnarray} \dfrac{d^2 y}{d x^2} = H\left(x\right) \hspace{0.05ex}y \label{*}\tag{*} \end{eqnarray}

where, $\,H\left(x\right) = \dfrac{\mathop{\rm sech}\nolimits\left(x\right) \mathop{\rm sech}\nolimits\left(x\right)}{x + \ln\big(2\cosh\left(x\right)\big)}$. Plot of function $\,H\left(x\right) $ is shown below: enter image description here

I need to find solution of the equation $\eqref{*}$ for the boundary conditions

$$\begin{aligned} y\left(x\right)\bigg\rvert_{-\infty} &= 0, & \left.\dfrac{d\hspace{0.1ex}y\left(x\right)}{d\hspace{0.1ex}x}\right\rvert_{ -\infty} &= 0 \end{aligned} \label{**}\tag{**}$$

Obvious solution of the problem is $\,y=0$. But $\,y = x + \ln\big(2 \cosh\left(x\right)\big)\,$ also satisfies differential equation $\eqref{*}$, and satisfies boundary conditions $\eqref{**}$. Plot of $y\left(x\right)$ is shown below: enter image description here

As far as I know there cannot be two solution of the differential equation satisfying given boundary conditions. What am I missing here? Is uniqueness theorem not valid if the boundary conditions are applied at $\,\pm\infty$.

EDIT Thanks to the comment by Santiago, appearance contradiction is better seen:

Differential eq. $%\begin{align} y'\left(x\right) = y\left(x\right) %\end{align} $ with boundary condition $\displaystyle\lim_{x \to \infty}y\left(x\right) = 0$. There are infinitely many solution to this problem all of the form $y\left(x\right) = k\exp\left(x\right)$, where $k$ is some constant.

Post Edit Is it possible to generalize observation above that, boundary conditions at $\pm \infty$ may not yield unique solution?

$\endgroup$
  • 2
    $\begingroup$ Compare with $y'(x) = y(x)$ given that $y(-\infty) = y'(-\infty) = 0$. There are two solutions, $y \equiv 0$ and $y(x) = \exp x$. $\endgroup$ – Santiago May 3 '16 at 18:36
  • $\begingroup$ @Santiago, I see your point and completely agree with the example. $\endgroup$ – alekhine May 3 '16 at 18:53
  • $\begingroup$ @alekhine : This is the same for second order. Coming back to your example, with the simplest case $\quad H(x)=1\quad\to\quad y''(x)=y(x)\quad$ an infinity of solutions $\quad y=c\,e^x\quad$ any $c$ , satisfy the two conditions $y(-\infty)=y'(-\infty)=0$ . $\endgroup$ – JJacquelin May 4 '16 at 8:32
  • $\begingroup$ Another exemple with finite boundary conditions : The EDO : $\quad x^2y''(x)=6y(x)\quad$ with two conditions $\quad y(0)=y'(0)=0\quad$ has an infinity of solutions : $\quad y(x)=c\,x^3\quad$ $\endgroup$ – JJacquelin May 4 '16 at 8:56
  • 1
    $\begingroup$ Your last example is not surprising since $1/x^2$ is not locally lipschitz (not even $C^0$). The cauchy-lipschitz (or picard lindelöf) theorem guarantee the uniqueness and existence of solutions of $y'=f(x,y)$ if $f$ is locally lipschitz. But there is another theorem for which $f$ is only supposed continuous and then you have existence of a solution but not uniqueness. See "peano existence theorem". $\endgroup$ – Renart May 13 '16 at 12:24
0
$\begingroup$

Consider the constant coefficient first order linear dynamical system of dimension $n$ \begin{equation}{\bf{x}}'=A{\bf{x}},\end{equation} where $A$ is an $n\times n$ constant matrix and ${\bf{x}}$ is an $n$-dimensional vector. If $A$ has $k$ eigenvalues with positive real parts, then system above has a $k$ dimensional space of solutions satisfying $\displaystyle{\lim_{t\rightarrow-\infty}{\bf{x}}=0}$.

More generally, we now consider a class of equations which the system (*) in the question is part of. More precisely, we consider systems of the form \begin{equation}{\bf{x}}'=A(t){\bf{x}},\end{equation} where $A$ is $t$-dependent and such that the limit $A^{-\infty}\equiv \displaystyle{\lim_{t\rightarrow -\infty} A(t)}$ exists. Assume also that $A(t)$ approaches $A^{-\infty}$ exponentially fast as $t\rightarrow -\infty$. Then the solutions of the system above behave like the solutions of the constant coefficient system $$ {\bf{x}}'=A^{-\infty}{\bf{x}}, $$ as $t\rightarrow -\infty$, which in particular implies that If $A^{-\infty}$ has $k$ eigenvalues with positive real parts, then the non-constant system above has a $k$ dimensional space of solutions satisfying $\displaystyle{\lim_{t\rightarrow -\infty}{\bf{x}}=0}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.