Question

In Problem 3, suppose the mass is released from an initial position $\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{0}}$with the initial velocity${\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$. Use a numerical solver to estimate an interval$\mathbf{a}\mathbf{\le }{\mathbf{x}}_{\mathbf{0}}\mathbf{\le }\mathbf{b}$for which the motion is oscillatory.

Short answer

The solution is an oscillatory one at$-0.8\le x_0\le 1.1$

Step-by-step solution

To estimate the interval

We have

$\frac{d^{2}x}{d{t}^2}+2x-x^2=0,x\left(0\right)=x_{0}and{x}^{\text{'}}\left(0\right)=1$

Our aim is to use a numerical solver, to find an interval$\mathrm{a}\le {\mathrm{x}}_0\le \mathrm{b}$, at which the motion is an oscillatory one, where$x_0$ is the initial position.

From Problem 3, We found that the motion is an oscillatory one at $x_0=1$, where$x_0=1$is below the equilibrium point$(x=0)$ . We need to find the maximum value $x_0$, for which the solution is an oscillatory one. We try different values for$x_0$until we reach a solution, for$x_0$, at which the solution is not oscillatory. And we find the following

$\mathrm{In}\left[1\right]:=\mathrm{NDSolve}\left[\left\{\mathrm{x}"\left[\mathrm{t}\right]+2\mathrm{x}\left[\mathrm{t}\right]-\mathrm{x}{\left[\mathrm{t}\right]}^2=0,\mathrm{x}\left[0\right]=1,{\mathrm{x}}^{\text{'}}\left[0\right]=1\right\},\mathrm{x}\left[\mathrm{t}\right],\{\mathrm{t},0,14\}\right]$

$\mathrm{Out}\left[1\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{Interpolating}\mathrm{Function}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,14.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]$

$\mathrm{In}\left[2\right]:=\mathrm{NDSolve}\left[\left\{\mathrm{x}"\left[\mathrm{t}\right]+2\mathrm{x}\left[\mathrm{t}\right]-\mathrm{x}{\left[\mathrm{t}\right]}^2=0,\mathrm{x}\left[0\right]=1.1,{\mathrm{x}}^{\text{'}}\left[0\right]=1\right\},\mathrm{x}\left[\mathrm{t}\right],\{\mathrm{t},0,14\}\right]$

$\mathrm{Out}\left[2\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{InterpolatingFunction}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,14.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]$

$\mathrm{In}\left[2\right]:=\mathrm{NDSolve}\left[\left\{\mathrm{x}"\left[\mathrm{t}\right]+2\mathrm{x}\left[\mathrm{t}\right]-\mathrm{x}{\left[\mathrm{t}\right]}^2=0,\mathrm{x}\left[0\right]=1.2,{\mathrm{x}}^{\text{'}}\left[0\right]=1\right\},\mathrm{x}\left[\mathrm{t}\right],\{\mathrm{t},0,14\}\right]$

$\mathrm{Out}\left[3\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{Interpolating}\mathrm{Function}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,5.16.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]$

The two sets of the initial condition

We plot them, yields

$\ln \left[5\right]:=$$\mathrm{Plot}[\{\mathrm{Evaluate}\left[\mathrm{x}\left[\mathrm{t}\right]/.\mathrm{\%}1\right],\mathrm{Evaluate}\left[\mathrm{x}\left[\mathrm{t}\right]/.\mathrm{\%}2\right],\mathrm{Evaluate}\left[\mathrm{x}\left[\mathrm{t}\right]/.\mathrm{\%}3\right]\},$

$\{\mathrm{t},0,5\},\mathrm{Plot}\mathrm{Legends}\to \{"{\mathrm{x}}_0=1","{\mathrm{x}}_0=1.1","{\mathrm{x}}_0=1.2"\}]$

Therefore, we have$b=1.1$

Above the equilibrium point

Using the same analysis, we try the following values for$X_0$.

$\mathrm{In}\left[6\right]:=\mathrm{NDSolve}\left[\left\{\mathrm{x}"\left[\mathrm{t}\right]+2\mathrm{x}\left[\mathrm{t}\right]-\mathrm{x}{\left[\mathrm{t}\right]}^2=0,\mathrm{x}\left[0\right]=-1,{\mathrm{x}}^{\text{'}}\left[0\right]=1\right\},\mathrm{x}\left[\mathrm{t}\right],\{\mathrm{t},0,14\}\right]$

$\mathrm{Out}\left[6\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{InterpolatingFunction}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,4.63.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]$

$\mathrm{In}\left[7\right]:=\mathrm{NDSolve}\left[\left\{\mathrm{x}"\left[\mathrm{t}\right]+2\mathrm{x}\left[\mathrm{t}\right]-\mathrm{x}{\left[\mathrm{t}\right]}^2=0,\mathrm{x}\left[0\right]=-0.9,{\mathrm{x}}^{\text{'}}\left[0\right]=1\right\},\mathrm{x}\left[\mathrm{t}\right],\{\mathrm{t},0,14\}\right]$

$\mathrm{Out}\left[7\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{Interpolating}\mathrm{Function}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,5.2.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]$

$\mathrm{Out}\left[8\right]=\{\{\mathrm{x}\left[\mathrm{t}\right]\to \mathrm{Interpolating}\mathrm{Function}\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,14.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]$

$\mathrm{In}\left[8\right]:=\mathrm{NDSolve}\left[\left\{\mathrm{x}"\left[\mathrm{t}\right]+2\mathrm{x}\left[\mathrm{t}\right]-\mathrm{x}{\left[\mathrm{t}\right]}^2=0,\mathrm{x}\left[0\right]=-0.8,{\mathrm{x}}^{\text{'}}\left[0\right]=1\right\},\mathrm{x}\left[\mathrm{t}\right],\{\mathrm{t},0,14\}\right]$

Plotting them, yields

$\begin{array}{l}\ln \left[5\right]:=\mathrm{Plot}[\{\mathrm{Evaluate}\left[\mathrm{x}\left[\mathrm{t}\right]/.\mathrm{\%}6\right],\mathrm{Evaluate}\left[\mathrm{x}\left[\mathrm{t}\right]/.\mathrm{\%}7\right],\mathrm{Evaluate}\left[\mathrm{x}\left[\mathrm{t}\right]/.\mathrm{\%}8\right]\},\\ \{\mathrm{t},0,5\},\mathrm{PlotLegends}\to \{"{\mathrm{x}}_0=-1","{\mathrm{x}}_0=0.9","{\mathrm{x}}_0=0.8"\}]\end{array}$

Final Answer

Therefore, we have$a=-0.8$ . Yields the interval, at which the solution is an oscillatory one to be$-0.8\le x_0\le 1.1$

The solution is an oscillatory one at$-0.8\le x_0\le 1.1$