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.
![$y=\sin{x}-0.2\sin{4x}$](/media/math222/a0p2_hu82b19cfae901783053296218b180a70e_313945_c0c1ae286c2ce8a5cbb285868574301c.webp)
The direction field looks very similar to that of $\sin{x}$:
![Direction field for $\frac{dx}{dt}=\sin{x}-0.2\sin{4x}$](/media/math222/a0p2df_hua7eb0a85ce888abec5ebc2dee0309db9_1703427_6b4d5a0d297739cac0d04d9a45088b04.webp)
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 $a>\frac14$.
![$y=\sin{x}-0.4\sin{4x}$](/media/math222/a0p4_hu87e54e59702c497d02757c06eaa24510_326109_3e2645d6772de94b380deaeb6ea7a474.webp)
We can see in the direction field regions where the direction of the flow is reversed compared to the previous one.
![Direction field for $\frac{dx}{dt}=\sin{x}-0.4\sin{4x}$](/media/math222/a0p4df_huf6ea2339d5dd29f4145303e44d886106_1686903_5e2dd8f3b6abe4c57e9c7e92c19e3616.webp)
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
![$y=\sin{x}-0.8\sin{4x}$](/media/math222/a0p8_hu20d0ae37c279a7115ebb865e1a578c2b_367138_593dfee0c9de7b2a47c0a9fb172ff4ef.webp)
We can also see that trajectories slow down near $x=\pm 2$ as seen from the slopes.
![Direction field for $\frac{dx}{dt}=\sin{x}-0.8\sin{4x}$](/media/math222/a0p8df_hudb9d34e1486b4109988820f1f71cea85_1612209_90665bfad04f78c53ea817f89b45b4b5.webp)
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 $a=\frac{1}{4} \left(\frac{11}{\sqrt{6}}-\sqrt{\frac{2}{3}}\right)\approx 0.919$.
![$y=\sin{x}-\sin{4x}$](/media/math222/a1_hueba2feb9601f8cf5c315e3e1e482acde_389992_56b430b486d6e6510e12ce233f03d460.webp)
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 $\frac{dx}{dt}=\sin{x}-\sin{4x}$](/media/math222/a1df_hu8a0152843c8e50b2b657f1f17fbf7da9_1713305_d841abc5661bfe85def32faafc042000.webp)
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 $\frac{dx}{dt}=x - \frac{t x}{1+x^2}$](/media/math222/matlab1problem2_hu70bffb798260cb2f69c8cbf57499f08b_1634681_d4eb9c3cd2ddc8a67614aac4490ea783.webp)
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.