MATLAB Project 2 Solution

NJIT Math 222, Fall 2021

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.

$y=\sin{x}-0.2\sin{4x}$
$y=\sin{x}-0.2\sin{4x}$

The direction field looks very similar to that of $\sin{x}$:

Direction field for $\frac{dx}{dt}=\sin{x}-0.2\sin{4x}$
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 $a>\frac14$.

$y=\sin{x}-0.4\sin{4x}$
$y=\sin{x}-0.4\sin{4x}$

We can see in the direction field regions where the direction of the flow is reversed compared to the previous one.

Direction field for $\frac{dx}{dt}=\sin{x}-0.4\sin{4x}$
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

$y=\sin{x}-0.8\sin{4x}$
$y=\sin{x}-0.8\sin{4x}$

We can also see that trajectories slow down near $x=\pm 2$ as seen from the slopes.

Direction field for $\frac{dx}{dt}=\sin{x}-0.8\sin{4x}$
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 $a=\frac{1}{4} \left(\frac{11}{\sqrt{6}}-\sqrt{\frac{2}{3}}\right)\approx 0.919$.

$y=\sin{x}-\sin{4x}$
$y=\sin{x}-\sin{4x}$

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 $\frac{dx}{dt}=\sin{x}-\sin{4x}$
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 $\frac{dx}{dt}=x - \frac{t x}{1+x^2}$
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.

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