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Consider the following initial value problem for the autonomous system of ODEs \begin{equation}%%\label{eqn: } \begin{cases} %\vspace{3mm} x'(t)=y(t),\; t>0,\\ %\vspace{3mm} y'(t)=-\frac{\displaystyle x(t)}{\displaystyle\sqrt{1+c_1+x^2(t)}}\,y^2(t)-\big(1-\sqrt{1+c_1+x^2(t)}\big)\,x(t),\; t>0,\\ x(0)=0,\; y(0)=k, \end{cases} \end{equation} where $c_1$, $k$ are parameters. I'd like to find $c_1>-1$ and $k\in\mathbb{R}$ such that a nontrivial solution $(x(t),y(t))\not\equiv (0,0)$, $t>0$ exists and satisfies the additional condition $x(1)=0$. My idea essentially is "try and error", i.e. choose some $c_1$ and $k$ and use numerics to find $(x(t),y(t))$. However, I failed to find such $c_1$ and $k$ and this seems not a systematic approach and is not so efficient. Such problem belongs to the so-called inverse problems, controllability problems, or other kinds of problems? Or I may refer to some references or papers related to this kind of problems? Thanks!

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  • $\begingroup$ If you can post the explicit expressions for $f$ and $g$, it might be helpful. Also, it might help to solve both forwards from $t=0$ and backwards from $t=1$, but that's essentially a more elaborate version of trial and error. $\endgroup$ – Frits Veerman Jan 12 '16 at 12:45
  • $\begingroup$ Thanks. I've posted the explicit expression. Any idea to solve this problem is welcome. $\endgroup$ – LCH Jan 12 '16 at 14:03
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The best thing to do is to analyse the system using a dynamical systems perspective. That is, to (1) find all equilibria of the system, and (2) to classify the behaviour of these equilibria.

As you easily infer from the ODEs, the equilibria of the system are the solutions of the equations \begin{align} y &= 0,\\ x\left(1 - \sqrt{1+c_1+x^2}\right) &= 0. \end{align} the last equation is satisfied if either $x=0$ or \begin{equation} 1+c_1 + x^2 = 1 \;\Rightarrow x^2 = -c_1. \end{equation} So, for $c_1>0$, only the trivial equilibrium $(x,y)=(0,0)$ exists, whereas for $-1<c_1<0$, three equilibria exist, and the two nontrivial equilibria are found at $(x,y) = (\pm \sqrt{-c_1},0)$.

If we linearise the system around the trivial equilibrium, we see that the eigenvalues of the linearisation are given by $\lambda = \pm \sqrt{\sqrt{1+c_1}-1}$. Therefore, for $c_1>0$, the trivial equilibrium is a saddle; for $-1<c_1<0$, the trivial equilibrium is a centre. This is relevant for the following reason: you're looking for an orbit which starts somewhere on the $y$-axis, and which crosses the $y$-axis again at some specified later time ($t=1$). From the phase plane for $c_1>0$, it is immediately clear that no such orbit exists.

phase plane for $c_1>0$

However, for $-1<c_1<0$, any orbit found in between the two nontrivial equilibria is closed, i.e. periodic, and therefore crosses the $y$-axis infinitely many times.

phase plane for $c_1<0$

phase plane for $c_1<0$, zoomed in

This will definitely give you more insight in the possible values of $c_1$ and $k$.

Note 1: The phase plane figures are made with the java-applet PPlane.

Note 2: You can do a lot more analysis on this system, to get more explicit expressions, etc. However, as you're interested in numerics, I'll leave that for now.

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  • $\begingroup$ Thanks very much for the calculation using PPlane. I will read it in more details. So from the experiment using PPlane, such $c_1$ (as $-1<c_1<0$) and $k$ could exist? If so, I'd like to use mathematical analysis to show that such $c_1$ and $k$ exist for this problem. Then I will try to find numerical value of $c_1$ and $k$. $\endgroup$ – LCH Jan 12 '16 at 17:03
  • $\begingroup$ Such $c_1$ and $k$ could definitely exist. I suggest you give the analysis a go, and maybe post another question if you get stuck. $\endgroup$ – Frits Veerman Jan 12 '16 at 17:27
  • $\begingroup$ Thanks for introducing PPlane. Now I'm enjoying it and making the same plots as you did. I choose $k=0.005$ and $c_1=-0.7$, and obtain a solution $x(t)\not\equiv0$ satisfying $x(0)=0$ and at $t\approx5$, $x=0$ As the ODE system is autonomous, perhaps by suitably adjusting $c_1$ and $k$, we can find a solution $x(t)\not\equiv0$ satisfying $x(0)=x(1)=0$, right? $\endgroup$ – LCH Jan 12 '16 at 18:02
  • $\begingroup$ About the proof of the existence of $c_1$ and $k$, any idea? $\endgroup$ – LCH Jan 12 '16 at 18:30
  • $\begingroup$ I did some experiments on different choices of $c_1$ and $k$. However, it seems that I cannot find $c_1$ and $k$ such that $x(t^{\ast})=0$ for $t^{\ast}<t_1$, where $t_1\approx 3$. Does $t^{\ast}$ has a lower bound less than 3 (approximately)? This phenomenon seems interesting! $\endgroup$ – LCH Jan 12 '16 at 19:00
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$$\begin{cases} %\vspace{3mm} \frac{dx}{dt}=y(t),\; t>0,\\ %\vspace{3mm} \frac{dy}{dt}=-\frac{\displaystyle x(t)}{\displaystyle\sqrt{1+c_1+x^2(t)}}\,y^2(t)-\big(1-\sqrt{1+c_1+x^2(t)}\big)\,x(t),\; t>0,\\ x(0)=0,\; y(0)=k, \end{cases}$$ $$y\frac{dy}{dx}=-\frac{x}{\displaystyle\sqrt{1+c_1+x^2}}\,y^2-\big(1-\sqrt{1+c_1+x^2}\big)\,x,\;$$ With $\:Y=y^2\:$ : $$\frac{dY}{dx}=-2\frac{x}{\sqrt{1+c_1+x^2}}\,Y-2\big(1-\sqrt{1+c_1+x^2}\big)\,x$$ With $\:X=x^2\:$ : $$\frac{dY}{dX}=-\frac{1}{\displaystyle\sqrt{1+c_1+X}}\,Y-\big(1-\sqrt{1+c_1+X}\big)\;$$ This linear ODE is easy to solve. The solution is : $$Y(X)=2+c_1+X-2\sqrt{1+c_1+X}+Ce^{-2\sqrt{1+c_1+X}}$$

The equation of the trajectory is : $$y(x)=\pm\sqrt{2+c_1+x^2-2\sqrt{1+c_1+x^2}+Ce^{-2\sqrt{1+c_1+x^2}}}$$

In order to know the time at each position on the trajectory, one have to integrate :

$\frac{dx}{dt}=y(x)\:$ then $\:dt=\frac{dx}{dy(x)}$ $$t=\int \frac{dx}{ \sqrt{2+c_1+x^2-2\sqrt{1+c_1+x^2}+Ce^{-2\sqrt{1+c_1+x^2}}} }$$

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  • $\begingroup$ LCH: This is the type of analysis I was talking about. @JJaquelin: I believe there is a mistake in the solution of the ODE. The method is of course entirely correct. $\endgroup$ – Frits Veerman Jan 13 '16 at 11:42
  • $\begingroup$ You are right, there was a mistake in the solving of the linear ODE. it's corrected. Hoping that all is correct now. $\endgroup$ – JJacquelin Jan 13 '16 at 12:36

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