Initial value problem for a system of ODEs with two parameters

Question

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!

Answer 1

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.

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.

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.

Answer 2

$$\begin{cases} %\vspace{3mm} \frac{dx}{dt}=y(t),\; t>0,\\ %\vspace{3mm} \frac{dy}{dt}=-\frac{\displaystyle x(t)}{\displaystyle\sqrt{1+c_1+x^2(t)}}\,y^2(t)-\big(1-\sqrt{1+c_1+x^2(t)}\big)\,x(t),\; t>0,\\ x(0)=0,\; y(0)=k, \end{cases}$$ $$y\frac{dy}{dx}=-\frac{x}{\displaystyle\sqrt{1+c_1+x^2}}\,y^2-\big(1-\sqrt{1+c_1+x^2}\big)\,x,\;$$ 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 $\:dt=\frac{dx}{dy(x)}$ $$t=\int \frac{dx}{ \sqrt{2+c_1+x^2-2\sqrt{1+c_1+x^2}+Ce^{-2\sqrt{1+c_1+x^2}}} }$$