# Math Insight

### Solving single autonomous differential equations using graphical methods

Math 1241, Fall 2019
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Due date: Nov. 29, 2019, 11:59 p.m.
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Total points: 3
1. Our goal is to use a graphical approach to solve the continuous dynamical system (or differential equation) \begin{align*} \diff{u}{t} &= 1-u^2/9\\ u(0) &= u_0 \end{align*} for various initial conditions $u_0$. (We could also have written it as $u'(t)=1-u(t)^2/9$.) By solving, we mean determine what $u(t)$ looks like as a function of time, though we won't find a formula for $u(t)$.
1. Two graphs. The graphical method for solving differential equations involves two different types of graphs. The first is a plot of the function defining the differential equation. Since our differential equation can be written as $u'(t)=f(u)$ for $f(u)=1-u^2/9$, we plot $f$, shown in the top panel, below.
Feedback from applet
Step 1: state variable axis:
Step 2: initial condition:
Step 3: equilibria:
Step 3: no overlap in vectors:
Step 3: reasonable relative vector lengths:
Step 3: valid vectors:
Step 4: equilibria:
Step 4: no overlap in vectors:
Step 4: reasonable relative vector angles:
Step 4: valid vectors:
Step 5: final values:
Step 5: initial conditions:
Step 5: speed profiles:

Our continuous dynamical system (or differential equation) has one state variable
, which we could also write as $u(t)$, to emphasize its value depends on time. The variable $u$ lives on the horizontal axis (or $u$-axis) of the top graph. (We could call this line the “state line” or “phase line” for the state variable $u$.) To remind us where the action occurs, we'll highlight the horizontal axis (or $u$-axis). In the above applet, change the step slider to 1. Move the points in the top panel so that the blue line is the $u$-axis.

The second graph is a graph of the solution $u(t)$ versus time $t$, which we'll draw in the bottom panel of the above applet. In this case, the value of the state variable $u$ is indicated by the vertical axis (which is now the $u$-axis). Move the blue points to highlight the vertical $u$-axis with the blue line.

The bottom panel has the $u$-axis rotated to be the vertical axis so that we can use the horizontal axis for time $t$. In this way, we'll plot the solution $u$ versus time in the bottom panel. You can think of the top plot with the graph of $f$ as an auxiliary plot that we'll use to help us draw the solution in the bottom plot.

In what follows, we'll go through the steps for sketching the solution to the differential equation using the above applet.

2. Initial condition. At first, we'll consider the initial condition $u_0=-2.5$. Change the applet to step=2 and move the points to indicate the initial condition on both graphs. In the top graph, everything we'll draw will be directly on the $u$-axis, so the initial condition must lie on the horizontal axis. In the bottom graph, the initial condition point must be on the $u$-axis, i.e., directly on the vertical axis (since the initial condition is the value of the state variable $u$ when $t=0$).
3. Initial rate of change. Our goal is to determine the qualitative shape of the function $u(t)$ based on information from the above differential equation, which is $u'(t)=f(u)=$
(enter formula for $f$). We know at time $t=0$, $u(0)=$
. But what happens as time increases? Should $u(t)$ be an increasing or decreasing function? Let's see if we can figure it out.

The rate of change of $u(t)$ at time $t=0$ is given by the derivative $u'(0)$. Since we don't know the function $u(t)$, we can't compute its derivative from a formula. However, the differential equation lets us look up the derivative directly. Since $u(0) =$
, we can calculate that $u'(0) = f(u(0)) = f(＿) =$
. (If rounding, include at least 3 significant digits in your response.)

At a more basic level, what is the sign of $u'(0)$?
Therefore, we learned an important fact about the solution $u(t)$: at $t=0$, the function $u(t)$ is
.

4. Solution velocity. Above, you calculated that, if we know $u=-2.5$, then the rate of change must be $u' = ＿$ so that $u(t)$ is ＿. Notice how this calculation did not depend on the fact that we were looking at the initial condition with $t=0$. The right hand side of the differentiation equation does not directly depend on the time $t$. If you know that $u(t)=-2.5$, the differential equation immediately tells you that, for this value of $t$, $u'(t) =$
.

From now on, we aren't even going to bother with the formula for $f(u)$. Since we are just looking for the qualitative shape of the solution $u(t)$, the graph of $f(u)$ is all we'll need to know about $f$. For example, you can simply look at the value of the function above $u=-2.5$ to determine that the sign of the function there is
so that $u(t)$ must be
whenever $u(t)$ happens to be equal to $-2.5$.

The next step is to represent graphically this rate of change (or velocity) of $u(t)$ using vectors on the phase line (the $u$-axis in the top graph). To show a visual aid, select the “show test point” box. A blue point appears in the top panel that you can move along the $u$-axis to change the value of $u$. In addition to showing the value of $f$, a horizontal vector (arrow) appears showing the direction that $u$ changes. If $u$ is increasing (i.e., $f(u)$ is positive), the vector points to the right; if $u$ is decreasing (i.e., $f(u)$ is negative), the vector points to the left. The length of the vector indicates how fast $u$ is changing, i.e., it indicates the magnitude of $u'(t)$. Your task is to draw vectors indicating how $u'(t)$ depends on the value of $u$; each vector will represent how $u(t)$ must be changing whenever the value of $u$ happens to be where you drew the vector.

Change the step slider to 3 to draw the vectors. A slider labeled by $n_v$ appears by which you can control the number of vectors. Change $n_v$ to 1 so that one vector appears. Move this vector so that it indicates how $u(t)$ must behave whenever it happens to be equal to $-2.5$. Put the base of the vector at $u=-2.5$. Since know that $u(t)$ must be increasing at this moment, draw the vector pointing toward the
. (You can move the test point to $-2.5$ to see what the vector should look like. You don't need to make your vector be exactly the same length as the vector attached to the test point.)

Set $n_v=2$ and draw another vector, this time around $u=0$. Compared to $u=-2.5$, the value of $f$ is
at $u=0$, so you should draw the vector
. Since the sign of $f$ is still
, the vector should still point toward the
. (If you like, use the test point to help visualize what the vector look like.)

Set $n_v=3$ and repeat this process for $u$ around $2$. The sign of $f$ is still
, so the vector should still point toward the
. But this time the vector should be
as the value of $f$ has gotten
.

Set $n_v=4$ and draw a vector around $u=3.5$. What's different this time? Since the sign of $f$ is
, you should draw the vector pointing
. Notice that the magnitude of $f(3.5)$ is small, so be sure to keep the vector
. (In particular, don't let the vector cross $u=3$, because on the left side of $u=3$, the sign of $f$ is
and vectors on the left side of $u=3$ should point to the
.)

Set $n_v=5$ and draw one more vector all the way to right, around $u=6$. At this point $f$ is so large and negative that it doesn't fit on the graph. This means the vector you draw should be
and pointing to the
. However, you'll need to draw the vector much shorter than the one shown from the test point, as you don't want it to touch the vector you drew around $u=3.5$. (The applet demands that the vectors do not overlap.)

To complete the picture with vectors, draw two more vectors at the left, one just to the left of $u=-3$ and one further to the left. (Increase $n_v$ to 6 and 7 to reveal these vectors.) Since the sign of $f$ in this region is
, these vectors should point to the
. The leftmost vector should be longest, as $f$ is
in magnitude there.

If $u$ happened to be $3$ or $-3$, what would be its velocity? Since $f(3)=f(-3)=$
, the velocity of $u(t)$ at these points would be
. These points are equilibria of the dynamical system. We don't usually draw velocity vectors there, but instead represent equilibria as points. In the applet though, draw those points as zero length vectors. Increase $n_v$ to 8 and 9 and move both endpoints of a vector to one of the equilibria. The vector will turn into a red point to represent the equilibrium.

5. Velocity of $u$ as a slope. The next step is to translate the left/right vectors you drew in the top panel into vectors illustrating the slope of the solution in the bottom panel. If you have the test point revealed, you can see the corresponding test point in the bottom panel. As you move the test point in the top panel left and right, the point in the bottom panel moves up and down so that both points are at the same value of $u$. In the bottom panel, you can see a vector pointing with slope given by the value of $f(u)$. Since $u'(t)$ is equal to this slope, the solution $u(t)$ must move in the direction of this vector whenever $u$ is equal to the value indicated by the test point.

Notice how, in the bottom panel, you can move the test point left and right to change the value of the time $t$. Nothing happens to the slope when you change the time, as the differential equation doesn't care about the time as long as it knows the value of the $u(t)$.

To begin, plot a slope vector in the bottom panel corresponding to the initial condition $u=-2.5$. Set step to 4, revealing a new $n_v$ slider for the number of the slope vectors. Change $n_v$ to 1 to reveal the first vector. A black vector appears that is anchored to the $u$ axis where $t=0$. Since the slope doesn't depend on time, the applet shows a series of “shadow vectors” at all later times. If those are confusing, ignore them for now and focus on the left black vector. Move it $u=-2.5$ and set its angle to correspond to the slope of $f(-2.5)$. If you like, you can move the test vector near the point to get an idea of the slope. But, the applet doesn't require you to exactly match the slope. Relative slopes between different points are more important than the actual value of the slope.

Next, plot a slope vector around $u=0$, corresponding to the second vector you drew on the phase line of the top graph. Set $n_v$ to 2 to reveal the vector and move it in the correct spot. Repeat for all seven vectors you drew above. For the equilibria at $u=-3$ and $u=3$, the slope is
. When you draw a vector with that slope, the applet also draws a red line to indicate that these horizontal arrows correspond to a constant solution, i.e., an equilibrium.

In the end, you should have vectors filling out the majority of the bottom panel. These vectors capture the behavior of the solution for any value of $u$.

6. Drawing the solution. Finally, we are ready to draw the solution to the differential equation for the initial condition $u(0)=-2.5$. We just need to interpret all the vectors that we've drawn and use them to draw a curve in the bottom panel starting at the initial condition point we drew at the beginning.

When $u(t) = -2.5$, you calculated that $u(t)$ must be
. The curve for $u(t)$ in the bottom panel must start by moving
. This corresponds to moving
in the top panel along the phase line.

As $u(t)$ increases what happens to the velocity $\diff{u}{t}$? According to the form of $f(u)$, for values of $u$ around $0$, you should have drawn an arrow with a
slope in the bottom panel and a
arrow in the top panel. As $u(t)$ increases from $-2.5$, it moves towards these vectors, so its velocity must be getting
.

At what point does $u$ reach its maximum speed? We don't know exactly at what time, but from the graph of $f$ (or the arrows we drew), we can see that the speed will be maximal at the moment when $u(t)=$
. Since $f(＿)=$
, $u(t)$ will be moving at the speed of $u'(t)=$
at the moment when $u(t)=＿$.

What happens once $u(t)$ passes $u(t)=0$? According to the form of $f(u)$, for values of $u$ around $2$, you should have drawn an arrow with a
slope in the bottom panel and a
arrow in the top panel. This means that as $u(t)$ passes $0$ it should begin to
.

The trickiest part is what happens after a long time. What must be the velocity of $u(t)$ if it were ever at $u=3$? The velocity is $f(3)=$
. What must be the velocity of $u$ if were ever past $u=3$? For $u \gt 3$, the sign of $f(u)$ is
, so $u(t)$ in that case must be
. Could it ever be possible for $u(t)$ to increase past $u=3$?
. On the other hand, can $u(t)$ ever stop before reaching $u=3$? For $-3 \lt u \lt 3$, the sign of $f(u)$ is
, so $u(t)$ must always be
as long as $u$ is in that interval. The solution $u(t)$ is boxed in. It must always be increasing and it can never cross $u=3$. Therefore, for large time $t$, $u(t)$ must slow down and get closer and closer to $u=$
. The slope vectors in the bottom panel illustrate this fact. The solution is going to get closer to the line of horizontal slope vectors at $u=3$, which are highlighted by the red line.

To use the applet to draw the solution, set step=5 to reveal the $n_c$ slider that controls the number of solution curves displayed. Set $n_c=1$ to reveal the first curve. Move the left point on the curve to the proper initial condition. Move the right point so that $u(t)$ approaches the correct value for large time $t$. Click the curve to change the shape of the curve. There are three possibilities for the shape: $u(t)$ will be either (a) always speeding up, (b) always slowing down, or (c) first speeding up and then slowing down. For the curve with initial condition $u(0)=-2.5$, we've discovered that $u(t)$ first speeds up as $u(t)$ approaches $u=0$ and then slows down to approach $u=3$. Click the curve so its shape reflects this speeding up and slowing down.

If all worked out well, the solution curve should be close to parallel to the slope vectors you drew. The applet isn't that picky, so they don't need to match exactly. The reason the applet shows all those shadow vectors is to help you see what the shape of the solution curve should be.

You might have noticed that while you were drawing the solution curve $u(t)$ in the bottom panel, the applet also drew a corresponding representation of the solution in the phase line of the top panel, using an arrow of the same color. This arrow doesn't indicate how $u(t)$ speeds up or slows down (as there is no representation of time in the phase line). But it does show where $u(t)$ starts, which direction it moves, and where it ends up as time gets large.

7. Other initial conditions. Now that you did all this work drawing the vectors in both the top and bottom panels, we quickly draw what solutions would look like if we had different initial conditions.

Plot a solution curve for the initial condition $u(0)=5$. Since the sign of $f(5)$ is
, $u(t)$ must start off
. As it cannot decrease past the equilibrium at $u=3$, the solution must slow down and approach that value for large time $t$. Set $n_c=2$ to draw the curve. The shape of the curve should reflect that the solution is always slowing down. (It never speeds up as it is always moving toward values of $f$ that are closer to zero.)

Plot a solution curve for the initial condition $u(0)=-3.5$. Since the sign of $f(-3.5)$ is
, $u(t)$ must start off
. Hence, the solution is moving
in the top panel and
in the bottom panel. As $u$
, it continues to
because it moves toward values of $u$ where $f(u)$ is further and further from zero. With this initial condition, the solution
without bound, heading toward $-\infty$. Set $n_c=3$ to draw the curve.

What if the initial condition was $u(0)=3$? Since $f(3)=$
, $u(t)$ is
. For this initial condition, the solution is $u(t)=$
. Such a ＿ solution is called an equilibrium. Usually, we'd ask you to draw the solution as a
line. In this case, the applet already drew if for you when you drew the constant slope vectors. The red line on the slope vectors is the equilibrium solution.

What about the initial condition $u(0)=-3$? Again, $f(-3)=$
, and $u(t)$ is
; $u(t)=-3$ is another equilibrium solution. If you drew the slope vectors correctly, there should be another red line at $u=-3$ corresponding to this equilibrium solution.

8. This last part is just for fun, but it will help you visualize the relationship between the solution plot in the bottom panel and the phase line (with the graph of $f$ on top of it) in the top panel. Even though we cannot represent time in the phase line plot, we can still visualize time with a movie. Set step=6 and either move time $t$ with the slider or click the triangle play button to start a move with $t$ moving forward. In the bottom panel, you can see a point on each solution curve moving with time. In synchrony, three points in the top panel show how each solution $u(t)$ can be visualized as a point moving along the phase line as time $t$ increases.

2. Consider the dynamical system \begin{align*} \diff{ w }{t} &= f(w)\\ w(0) & = w_0, \end{align*} where the function $f$ is graphed in the top panel, below, and $w_0$ is an initial condition. (The zeros of $f$ are $-8$, $3$, and $7$.)
Feedback from applet
Step 1: state variable axis:
Step 2: initial condition:
Step 3: equilibria:
Step 3: no overlap in vectors:
Step 3: reasonable relative vector lengths:
Step 3: valid vectors:
Step 4: equilibria:
Step 4: no overlap in vectors:
Step 4: reasonable relative vector angles:
Step 4: valid vectors:
Step 5: final values:
Step 5: initial conditions:
Step 5: speed profiles:

Repeat the procedure from the previous problem to (1) highlight the $w$-axis in both panels, (2), plot the initial condition $w_0=2$, (3) sketch vectors and equilibria in the top plot (suggestion: use three vectors in between $-8$ and $3$), (4) sketch slope vectors and equilibria in the bottom plot, and (5) sketch solutions for the initial condition $w_0=2$, as well as for two more initial conditions $w_0=4$ and $w_0=10$. (Remember, for the last step, you can click the curves to change their speed profiles.)

3. Consider the dynamical system \begin{align*} \diff{ z }{t} &= f(z) \end{align*} where the function $f$ is graphed to the right. Find all equilibria. (They are integers.)

The equilibria are $z=$

(Enter in increasing order, separated by commas.)

For each equilibrium, determine its stability by sketching the solution for initial conditions just above and below the equilibrium. For each equilibrium, enter stable or unstable. Write your answers in the same order as the equilibria and separate your answers by commas.

(For example, if there are four equilibria, the first two are stable, and the second two are unstable, you would enter stable, stable, unstable, unstable in the answer blank.

In the below applet, sketch solutions for the initial conditions $z_0 = -4.5$, $z_0 = -3.5$, $z_0 = 5.5$, and $z_0 = 6.5$. (Increase $n_c$ by one to reveal a new curve which you can move and click to change shape.) Plot the equilibria as horizontal lines. (Increase $n_e$ by one to reveal a new horizontal line, which you can move to the correct location for an equilibrium.) If an equilibrium is stable, draw the line as a solid line; if an equilibrium is unstable, draw the line as a dashed line. (Click the line to switch between solid and dashed.)

Feedback from applet
Equilibria:
Final values of curves:
Initial conditions of curves:
Speed profiles of curves:
Stability of equilibria: