Math 555: Differential Equations I |
Consider a first order differential equation (FODE) of the form
$$
\frac{dy}{dt} = f(t,y).
$$
At any point $(t_0,y_0)$ in the domain of $f$, the value $f(t_0,y_0)$ prescribes
the instantaneous rate of change of the solution curve passing through the point.
We may plot a short line segment with this prescribed slope at the point. The
solution curve of the differential equation passing through this point must then be
tangent to the line segment.
We may repeat this process at various points in the plane to get a picture of how
solution curves must behave. The result is a slope field corresponding to
the DE. As we will see in the examples below, a slope field can also give us insight
into obstructions to solutions having certain behavior.
Example 1. As a first example, consider the first order differential equation
$$
\frac{dy}{dt} = 2 - y.
$$
We begin by plotting the slope field. (Click the "Run" button below to run the code
and view the slope field.)
Next, we add the equilibrium solution to the graph. Recall that the equilibrium solution occurs when the derivative is identically zero; that is, the solution is not changing. For this DE it is easy to find the equilibrium solution by simple inspection. In general, you might have to do some work to find an equilibrium solution--provided one even exists.
Finally, we add a couple of solution curves that have distinctly different behavior:
one with initial point $(0,3)$ above the equilibrium, and one with initial point
$(0,1)$ below the equilibrium.
We use ${\tt Sage}$ to solve the initial value problems and plot both curves on the
same setof axes as the slope field and equilibrium solution.
${\tt Sage}$ not only solve the DE with the given initial condition, it saves the solution as a callable symbolic function. The ${\tt \% display\ latex}$ command tells ${\tt Sage}$ to format the output in $\LaTeX$ math typesetting.
Example 2.
Next, consider the first order differential equation
$$
\frac{dy}{dt} = \dfrac{t^2}{(2y - 6)}.
$$
First, notice that this DE only makes sense when $y \neq 3$. This does not necessarily
mean that the solution curves cannot pass through points for which $y = 3$. It means
that the rate of change of the solution curves at these points is undefined. As we know,
derivatives can be undefined at sharp corners or cusps, or points of discontinuity: jumps,
holes, vertical asymptotes. But smooth curves can also have undefined derivatives at
certain points if they are not the graph of a function. Think: parametric curves from
Calculus II. In such situations it means that the tangent line is vertical, but the curve
is otherwise well-behaved. So what's going on with this example?
Let's see what the slope field looks like.
Next we use ${\tt Sage}$ to solve the initial value problem to find the solution curve passing through the origin.
Notice that this expression can not be written as a single function $y = y(t)$.
There are two solutions for $y$: one with a positive square root, and one with a negative square root. But both "solution functions" do not correspond to graphs passing through the origin. Instead, they should be regarded together to form two halves of a single smooth solution curve that is not the graph of a function.
Only the darker blue portion of this curve corresponds to the solution function of the initial value problem. The concatenation of the darker blue and cyan curves does form a single smooth solution curve to the IVP. In practice it is important to consider both points-of-view.
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