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Qualitative behavior of solutions to differential equations

In earlier work, we have used the tangent line to the graph of a function f at a point a to approximate the values of f near a.

Introduction

In earlier work, we have used the tangent line to the graph of a function \(f\) at a point \(a\) to approximate the values of \(f\) near \(a\). The usefulness of this approximation is that we need to know very little about the function; armed with only the value \(f(a)\) and the derivative \(f'(a)\), we may find the equation of the tangent line and the approximation \[\begin{aligned}\end{aligned}\].

Remember that a first-order differential equation gives us information about the derivative of an unknown function. Since the derivative at a point tells us the slope of the tangent line at this point, a differential equation gives us crucial information about the tangent lines to the graph of a solution. We will use this information about the tangent lines to create a slope field for the differential equation, which enables us to sketch solutions to initial value problems with the goal of understanding the solutions qualitatively. That is, we would like to understand the basic nature of solutions, such as their long-range behavior, without precisely determining the value of a solution at a particular point.

In the following Preview Activity, we consider a certain differential equation and explore how we can use the equation to find tangent lines to provide a kind of road map for an approximate solution.

Exploration
Exploration

Slope fields

Preview Activity shows that we can sketch the solution to an initial value problem if we know an appropriate collection of tangent lines. We can use the differential equation to find the slope of the tangent line at any point of interest, and hence generate and plot such a collection.

Let's continue looking at the differential equation \(\frac{dy}{dt} = t-2\). If \(t=0\), this equation says that \(dy/dt = 0-2=-2\). Note that this value holds regardless of the value of \(y\). We will therefore sketch tangent lines for several values of \(y\) and \(t=0\) with a slope of \(-2\), as shown in Figure. These tangent lines then provide a kind of road map for a solution to follow, starting from a certain initial condition.

Let's continue in the same way: if \(t=1\), the differential equation tells us that \(dy/dt = 1-2=-1\), and this holds regardless of the value of \(y\). We now sketch tangent lines for several values of \(y\) and \(t=1\) with a slope of \(-1\) in Figure.

Similarly, we see that when \(t=2\), \(dy/dt = 0\) and when \(t=3\), \(dy/dt=1\). We may therefore add to our growing collection of tangent line plots to achieve Figure.

In Figure, we begin to see the solutions to the differential equation emerge. For the sake of even greater clarity, we add more tangent lines to provide the more complete picture shown at right in Figure.

Figure is called a slope field for the differential equation. It allows us to sketch solutions just as we did in the preview activity. We can begin with the initial value \(y(0) = 1\) and start sketching the solution by following the tangent line. Whenever the solution passes through a point at which a tangent line is drawn, that line is tangent to the solution. This principle leads us to the sequence of images in Figure.

In fact, we can draw solutions for any initial value. Figure shows solutions for several different initial values for \(y(0)\).

Just as we did for the equation \(\frac{dy}{dt} = t-2\), we can construct a slope field for any differential equation of interest. The slope field provides us with visual information about how we expect solutions to the differential equation to behave.

Equilibrium solutions and stability

As our work in Activity demonstrates, first-order autonomous equations may have solutions that are constant. These are simple to detect by inspecting the differential equation \(dy/dt = f(y)\): constant solutions necessarily have a zero derivative, so \(dy/dt = 0 = f(y)\).

For example, in Activity, we considered the equation \(\frac{dy}{dt} = f(y)=-\frac12(y-4)\). Constant solutions are found by setting \(f(y) = -\frac12(y-4) = 0\), which we immediately see implies that \(y = 4\).

Values of \(y\) for which \(f(y) = 0\) in an autonomous differential equation \(\frac{dy}{dt} = f(y)\) are called equilibrium solutions of the differential equation.

Summary

  • A slope field is a plot created by graphing the tangent lines of many different solutions to a differential equation.

  • Once we have a slope field, we may sketch the graph of solutions by drawing a curve that is always tangent to the lines in the slope field.

  • Autonomous differential equations sometimes have constant solutions that we call equilibrium solutions. These may be classified as stable or unstable, depending on the behavior of nearby solutions.

Practice (2)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Consider the differential equation \[\begin{aligned}\end{aligned}\].

    1. Sketch a slope field on the axes at right.

    2. Sketch the solutions whose initial values are \(y(0)= -4, -3, \ldots, 4\).

    3. What do your sketches suggest is the solution whose initial value is \(y(0) = -1\)? Verify that this is indeed the solution to this initial value problem.

    4. By considering the differential equation and the graphs you have sketched, what is the relationship between \(t\) and \(y\) at a point where a solution has a local minimum?

    జవాబు వెల్లడి చేయండి

    1. Sketching the slopes at individual points by computing \(\frac{dy}{dt}\vert_{(t,y)}\) at various values of \((t,y)\), we see the following slope field.

    2. By following the slope field so that we pass through the points \((0,-4)\), \((0,-3)\), \(\ldots\), \((0,4)\), we see the collection of curves in the figure above.

    3. It looks like the solution that passes through \((0,-1)\) is a straight line with slope \(m = 1\), and thus has the formula \(y(t) = t-1\). To verify this function is a solution, we compute both \(\frac{dy}{dt}\) and \(t-y\). Doing so, we see \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\]. Thus, \(y(t) = t-1\) is indeed a solution to the differential equation.

    4. From both the differential equation and the graphs we have sketched, it appears that \(t\) and \(y\) are equal at any point where a solution has a local minimum. This is because when \(t = y\), \(\frac{dy}{dt} = t - y = 0\), and having the derivative be zero is required for any local minimum on a differentiable function.

  2. Consider the differential equation \[\begin{aligned}\end{aligned}\].

    1. Sketch a slope field on the axes at right.

    2. Sketch the solutions whose initial values are \(y(0)= -4, -3, \ldots, 4\).

    3. What do your sketches suggest is the solution whose initial value is \(y(0) = -1\)? Verify that this is indeed the solution to this initial value problem.

    4. By considering the differential equation and the graphs you have sketched, what is the relationship between \(t\) and \(y\) at a point where a solution has a local minimum?

    జవాబు వెల్లడి చేయండి

    1. Sketching the slopes at individual points by computing \(\frac{dy}{dt}\vert_{(t,y)}\) at various values of \((t,y)\), we see the following slope field.

    2. By following the slope field so that we pass through the points \((0,-4)\), \((0,-3)\), \(\ldots\), \((0,4)\), we see the collection of curves in the figure above.

    3. It looks like the solution that passes through \((0,-1)\) is a straight line with slope \(m = 1\), and thus has the formula \(y(t) = t-1\). To verify this function is a solution, we compute both \(\frac{dy}{dt}\) and \(t-y\). Doing so, we see \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\]. Thus, \(y(t) = t-1\) is indeed a solution to the differential equation.

    4. From both the differential equation and the graphs we have sketched, it appears that \(t\) and \(y\) are equal at any point where a solution has a local minimum. This is because when \(t = y\), \(\frac{dy}{dt} = t - y = 0\), and having the derivative be zero is required for any local minimum on a differentiable function.

Symbols used here

f'(x),\ \frac{dy}{dx}
derivative
Instantaneous rate of change; slope of the graph.
e
Euler's number
2.71828…, the base whose exponential is its own derivative.
\lim_{x \to a} f(x)
limit
The value f(x) approaches as x approaches a.
\int f(x)\,dx,\ \int_a^b
integral
Antiderivative (indefinite) or signed area from a to b (definite).
y',\ y''
first and second derivative of y
Prime notation for derivatives with respect to x (or t).
C,\ C_1,\ C_2
arbitrary constants
Constants of integration fixed by initial conditions.
\frac{\partial u}{\partial t},\ \nabla^2 u
partial derivative in time, Laplacian
Rate of change in time; sum of second partials (the diffusion operator).

How to: Qualitative behavior of solutions to differential equations

  1. What is a slope field?
  2. How can we use a slope field to obtain qualitative information about the solutions of a differential equation?
  3. What are stable and unstable equilibrium solutions of an autonomous differential equation?

Questions people ask

What is a differential equation?

An equation whose unknown is a function, relating it to its own derivatives. "The rate of growth is proportional to the population" is y′ = ky, and solving it means finding y as a function of time.

Why does the solution have arbitrary constants?

Integrating loses information: many functions share the same derivative. An n-th order equation has n constants, fixed by n initial or boundary conditions.

మీ సొంత ప్రయత్నించండి

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

ఇంకా Differential Equations