Math Insight

Solving single autonomous differential equations using graphical methods

Elementary dynamical systems
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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. The plot of $f$. 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 below.

Our continuous dynamical system (or differential equation) has one state variable $u$, 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 above graph. Highlight the horizontal axis thickly to remind us this is where the action occurs.

At first, we'll consider the initial condition $u_0=-2$. Draw a point on the $u$-axis at the location $u=-2$. This point shows the initial state of the system, at time $t=0$.

2. A plot of the solution. The second graph will be a graph of the solution $u(t)$ versus time $t$. The axes for this plot are shown below. In this case, the value of the state variable $u$ is indicated by the vertical axis (which is now the $u$-axis). Highlight the vertical axis thickly to indicate this vertical axis plays the same role as the horizontal axis, above.

Plot the initial condition $u(0)=u_0 = -2$ as a point on the solution graph. Since this point corresponds to $t=0$, plot the point directly on the vertical axis.

3. Now, let the action begin. We know that at time $t=0$, $u(0)=-2$. But what happens as time increases? Should $u(t)$ be an increasing or decreasing function? The rate of change is given by the derivative $u'(0)$. We don't know the function $u(t)$, so we can't compute its derivative that way. But, we have the differential equation that tells us what $u'(t)$ is. The differential equation tells us that the derivative $u'(0)$ must be $f(u(0))=f(-2)$. What is $u'(0)$? What is its sign? Therefore, is $u(t)$ increasing or decreasing at $t=0$?

4. You probably computed $f(-2)$ from the formula. Such precise information is overkill for this approach. Instead, use the graph of $f$ to estimate $f(-2)$. The graph indicates $f(-2)$ is positive; $u(t)$ must be increasing at $t=0$. For $t$ slightly larger than 0, $u(t)$ must become larger than $-2$. We'll indicate this change in two different ways on the two graphs.

On the plot of $f$, draw an arrow pointing right from the initial condition point you drew. $u(t)$ is moving rightward along the $u$ axis in that plot.

On the solution plot, start to draw the curve $u(t)$. Given that $u'(0)$ is positive, the curve begins with a positive slope. Draw a little bit of the curve $u(t)$ starting at $t=0$, making it slope upward. You can ignore the scale on the $t$ axis, so don't worry about the exact value of the slope.

5. As $u(t)$ increases from $-2$ toward $-1$, what happens to the velocity $\diff{u}{t}$? Since the velocity is $\diff{u}{t}=f(u)$, you just need to know what happens to $f(u)$ as $u$ increases from $-2$ to $-1$. Does $u(t)$ speed up, slow down, stop, or turn around?
6. At what point does $u(t)$ reach its maximum speed? (When is $f$ maximal?) At what value of $u$ does the velocity $f(u)$ finally slow down to a halt? Sketch the results on both the graph of $f$ and on the solution plot. On the graph of $f$, draw arrows showing movement of $u$ along the horizontal axis. Longer arrows indicate faster speed. On the solution plot, sketch the curve $u(t)$. Its slope should reflect it speeding up and slowing down. As $t$ gets very large, what happen to $u(t)$? The function $u(t)$ approaches what value for large $t$? Make sure your solution graph reflects this fact.

7. Repeat this process for the initial conditions $u_0=4$ and $u_0=-3.1$. In each case, draw leftward or rightward movement using arrows along the horizontal axis of the graph of $f$. Then, draw different curves in the solution plot, starting with $u=u_0$ when $t=0$, with slopes corresponding to values of $f$. In each case, does $u(t)$ approach a steady value? Or does $u(t)$ shoot upward or downward off the graph.

8. Repeat the process for $u_0=-3$ and $u_0=3$. What is the velocity $\diff{u}{t}$? Does $u(t)$ change or stay constant for these special initial conditions?

A constant solution is an equilibrium. How can you find the equilibria from the graph of $f$? In the solution plot, equilibria are horizontal lines.

2. Consider the dynamical system \begin{align*} \diff{ z }{t} &= f(z)\\ z(0) & = z_0, \end{align*} where the function $f$ is graphed to the right and $z_0$ is an initial condition. For each of the following initial conditions, sketch the graph of the solution $z(t)$. (The zeros of $f$ are $-5$, $3$, and $7$.)
1. $z_0 = -7.5$
2. $z_0 = -3.4$
3. $z_0 = 4.2$
4. $z_0=9.4$

3. Consider the dynamical system \begin{align*} \diff{ y }{t} &= 4 \left(y + 4\right) \left(y - 1\right) \left(y - 9\right)\\ y(0) & = y_0, \end{align*} where $y_0$ is an initial condition. Sketch the graph of the function $f(y) = 4 \left(y + 4\right) \left(y - 1\right) \left(y - 9\right)$. Use the graph to sketch the solution $y(t)$ for each of the following initial conditions: $y_0 = -7.6$, $y_0 = -4$, $y_0 = -1.0$, $y_0 = 1$, $y_0 = 2.6$, $y_0 = 9$, and $y_0=9.3$.

4. Consider the dynamical system \begin{align*} \diff{ v }{t} &= f(v) \end{align*} where the function $f$ is graphed to the right. Find all equilibria. (They are integers.) For each equilibrium, determine its stability by sketching the solution for initial conditions just above and below the equilibrium.