MATLAB Project 2 Solution
NJIT Math 222
Problem 1
Consider the differential equation
\[ \frac{dx}{dt}= f(x) = \sin{x}-a \sin{4x} \]We will look at four different values \(a=0.2,\ a=0.4,\ a= 0.8,\ a= 1\).
Graphs below were drawn with Desmos and direction fields with the Slopes App.
The case \(\mathbf{a=0.2}\)
The graph of \(f(x)\) looks like a slightly wigglier version of \(\sin(x)\) and, more importantly, has the same zeros. Stable equilibrium are marked in blue and unstable in purple.
$y=\sin{x}-0.2\sin{4x}$
The direction field looks very similar to that of \(\sin{x}\):
Direction field for $\frac{dx}{dt}=\sin{x}-0.2\sin{4x}$
The case \(\mathbf{a=0.4}\)
Two new zeros have arisen on either side of \(x=0\). Note that the equilibrium at zero has become stable while the two new equilibria are unstable. In fact, these two new equilibria exist for all \(a>\frac14\).
$y=\sin{x}-0.4\sin{4x}$
We can see in the direction field regions where the direction of the flow is reversed compared to the previous one.
Direction field for $\frac{dx}{dt}=\sin{x}-0.4\sin{4x}$
The case \(\mathbf{a=0.8}\)
Not much has changed here compared with the previous case, however the curves come very close to the \(x\)-axis near \(x=\pm 2\). Watch this space, as things will change near here in the final case
$y=\sin{x}-0.8\sin{4x}$
We can also see that trajectories slow down near \(x=\pm 2\) as seen from the slopes.
Direction field for $\frac{dx}{dt}=\sin{x}-0.8\sin{4x}$
The case \(\mathbf{a=1}\)
Four new fixed points have arisen in pairs near \(x=\pm 2\). In fact, with some fancier math we can find that the critical value is \(a=\frac{1}{4} \left(\frac{11}{\sqrt{6}}-\sqrt{\frac{2}{3}}\right)\approx 0.919\).
$y=\sin{x}-\sin{4x}$
In the direction field, we can see new types of trajectories of the opposite slope near \(x=\pm 2\) in addition to four new fixed points.
Direction field for $\frac{dx}{dt}=\sin{x}-\sin{4x}$
Problem 2
Graph the direction field and some solutions for the non-autonomous equation
\[ \frac{dx}{dt}=x - \frac{t x}{1+x^2}. \]The direction field looks like this:
Direction field for $\frac{dx}{dt}=x - \frac{t x}{1+x^2}$
Solutions can either go to zero as \(t\to\infty\) or they can go to \(\pm\infty\). There is a pair of curves separating these two regimes of behavior, but I haven’t been able to figure out a formula for it.