The Ordinary Differential Equations Project

Subsection 2.2.1 Direction Fields

Consider the differential equation for a simple harmonic oscillator that we developed in Section 1.1,

$$m{x}^{″}+kx=0.$$

If we assume that $k$ and $m$ are both equal to one and let $x^{\prime }=v\text{,}$ we can rewrite this equation as the first order system,

$$\begin{array}{rl}{x}^{\prime }& =v,\\ v^{\prime }& =-x.\end{array}$$

The direction field is relatively easy to understand. After plotting only few vectors, we can very quickly see that the vectors are tangent to circles centered at the origin (Figure 2.2.2). Since the solutions to the undamped harmonic oscillator $x^{″}+x=0$ are of the form

$$x(t)=A\cos t+B\sin t$$

for arbitrary constants $A$ and $B\text{,}$ this should not be too surprising.

The system

$$\begin{array}{rl}{x}^{\prime }& =x\\ y^{\prime }& =y\end{array}$$

gives us a direction field where the vectors point away from the origin (Figure 2.2.4).

The system

$$\begin{array}{rl}{x}^{\prime }& =-x,\\ y^{\prime }& =-y.\end{array}$$

gives us a direction field where the vectors point towards the origin (Figure 2.2.5).

The system

$$\begin{array}{rl}{x}^{\prime }& =-x,\\ y^{\prime }& =-5y.\end{array}$$

also gives us a direction field where the vectors point towards the origin; however, we shall soon see that there are important differences between this direction field and the direction field of the previous system (Figure 2.2.5).

Subsection 2.2.2 Modified Predator-Prey System

Subsection 2.2.3 A Competing Species Model

Suppose that $x=x(t)$ and $y=y(t)$ are two populations, competing for the same resources, are governed by the following system of differential equations,

$$\begin{array}{rl}\frac{dx}{dt}& =x(1-x)-\alpha xy\\ \frac{dy}{dt}& =y(1-y)-\alpha xy.\end{array}$$

If we study how the two populations interact, we will discover two very different cases depending on the value of the parameter $\alpha \text{.}$

If we let $\alpha =2\text{,}$ then there is a strong interaction between the two species and our system becomes

$$\begin{array}{rl}\frac{dx}{dt}& =x(1-x)-2xy\\ \frac{dy}{dt}& =y(1-y)-2xy.\end{array}$$

We are interested in what happens as $t\to \mathrm{\infty }\text{.}$ First we will examine what happens when $dx/dt=0$ and $dy/dt=0\text{:}$

$$\begin{array}{rl}\frac{dx}{dt}& =x(1-x-2y)\\ \frac{dy}{dt}& =y(1-y-2x).\end{array}$$

If $dx/dt=0\text{,}$ then

$$\begin{array}{}\text{(2.2.3)}& x& =0\text{(2.2.4)}& y& =(1-x)/2.\end{array}$$

Similarly, if $dy/dt=0\text{,}$ we have

$$\begin{array}{}\text{(2.2.5)}& y& =0\text{(2.2.6)}& y& =1-2x.\end{array}$$

The lines in equations (2.2.3)–(2.2.6) are called nullclines. In general, if we are given the system

$$\begin{array}{rl}\frac{dx}{dt}& =f(x,y)\\ \frac{dy}{dt}& =g(x,y),\end{array}$$

then the values of $x$ and $y$ satisfying $dx/dt=f(x,y)=0$ are called the $x$-nullclines of the system and the values of $x$ and $y$ satisfying the equation $dy/dt=g(x,y)=0$ are called the $y$-nullclines of the system.

In our example, we can plot the $x$ and $y$-nullclines to help us understand our system (Figure 2.2.9). Of course, we have an equilibrium solution whenever an $x$-nullcline intersects with a $y$-nullcline. Thus, the equilibrium points for our particular system are $(0,0)\text{,}$ $(1,0)\text{,}$ $(0,1)\text{,}$ and $(1/3,1/3)\text{.}$ Furthermore, we can choose a representative point in each region to find how the direction field is oriented. If we are given the initial condition $(1.4,1.3)$ for example, we can see that

$$\begin{array}{rl}\frac{dx}{dt}& =-4.2\\ \frac{dy}{dt}& =-4.03.\end{array}$$

Since both of these numbers are negative, we can see that our initial trajectory is headed down and to the left—slightly more to the left than down. However, we have no guarantee that the trajectory will continue in this direction.

Before we proceed further with our analysis, let us determine what happens on the nullclines themselves. That is, we will examine the case when $dx/dt=0$ or $dy/dt=0\text{.}$ If $dx/dt=0\text{,}$ then any solution crossing this $x$-nullcline must be moving vertically. There can be no right or left movement at this point in the phase portrait. We can indicate this fact along with the direction of vertical movement by drawing a vertical slash with an arrow on the $x$-nullcline (Figure 2.2.9). For example, the point $(0.5,0.25)$ lies on the $x$-nullcline and

$$\frac{dy}{dt}(0.5,0.25)<0.$$

Therefore, the trajectory that crosses the $R$-nullcline at $(0.5,0.25)$ is moving down. As we move along the $x$-nullcline, the direction of this arrow can only change when we cross an $y$-nullcline.

Similarly, there can be no vertical motion on an $y$-nullcline. We will indicate the direction of horizontal motion by drawing a horizontal line with an arrow pointing to the right if $dx/dt>0$ on the $y$-nullcline and an arrow pointing to the left if $dx/dt<0$ on the $y$-nullcline.

We can also determine the basic direction of the solution curve by checking what happens at a point in each of the regions bounded by the nullclines. For example, at the point $(0.1,0.1)$ we find that

$$\begin{array}{rl}\frac{d}{dt}x(0.1,0.1)& =0.7\\ \frac{d}{dt}y(0.1,0.1)& =0.7.\end{array}$$

Thus, the general direction of any solution curve in this region is up and right (Figure 2.2.9).

What happens to the initial condition $(1.4,1.3)\text{?}$ We see three possible scenarios if we follow the nullclines for large values of $t\text{.}$

• Only species $x$ survives and species $y$ becomes extinct.

• Only species $y$ survives and species $x$ becomes extinct.

• There are essentially equal numbers of species $x$ and $y\text{.}$

There is no reason why our nullclines should be limited to straight lines. The system

$$\begin{array}{rl}{x}^{\prime }& =y-x^2\\ y^{\prime }& =x-2\end{array}$$

has an $x$-nullcline $y=x^2$ and a $y$-nullcline $x=2\text{.}$ These two nullclines intersect at $(2,4)\text{,}$ and the system has a single equilibrium solution. The nullclines divide the plane into four basic regions (Figure 2.2.11). By choosing a point in each of these regions and determining the direction of the slope field at that point, we can decide the direction of the slope field in the entire region.

The only equilibrium solution of our system occurs at $(2,4)\text{.}$

Subsection 2.2.4 Plotting Direction Fields with Sage

Subsection 2.2.5 A Summary of Phase Plane Analysis

• Step 1. Draw the curves where $f(x,y)=0\text{.}$ These curves are called the $x$-nullclines. When a solution curve $\mathbf{x}(t)$ lies on one of these curves, $dx/dt=0\text{,}$ draw vertical slash marks on the $x$-nullclines to remind yourself that a trajectory crossing the nullcline can only do so if it is moving in a vertical direction at the instant of crossing.

• Step 2. Draw the curves where $g(x,y)=0\text{.}$ These curves are called the $y$-nullclines. When $\mathbf{x}(t)$ lies on one of these curves, $dg/dt=0\text{.}$ Draw horizontal slash marks on the $y$-nullclines to remind yourself that a trajectory crossing the nullcline can only do so if it is moving in a horizontal direction at the instant of crossing.

• Step 3. Label the points where the $x$ and $y$-nullclines intersect. These intersections are the equilibrium points. If $\mathbf{x}(t)$ is ever at one of these points, then both $dx/dt$ and $dy/dt$ vanish. This means that the trajectory stays at the equilibrium point for all time. If our system is going to tend towards a steady state, then $\mathbf{x}(t)$ will approach on of the equilibrium points as $t\to \mathrm{\infty }\text{.}$

• Step 4. Label the regions of the $xy$-plane where $dx/dt<0$ and where $dx/dt>0\text{.}$ These regions are always separated by $x$-nullclines. Likewise, label the regions where $dy/dt$ is positive and negative.

• Step 5.Go back and put arrows on the vertical hash marks of the $x$ nullclines. These arrows indicate whether the motion across the nullcline is up or down. The arrows are up on the parts of the $x$-nullclines that are in the $dy/dt>0$ region, and down on those parts of the $x$-nullclines in the $dy/dt<0$ regions. Likewise, draw arrows on the horizontal slash marks of the $y$-nullclines. These arrows are pointing right on the parts of the $y$-nullclines in the $dx/dt>0$ regions and left point on the parts in the $dx/dt<0$ regions.

• Step 6.

  • If $dx/dt>0$ and $dy/dt>0\text{,}$ then both $x(t)$ and $y(t)$ are increasing and the trajectory moves up and right.

  • If $dx/dt>0$ and $dy/dt<0\text{,}$ the trajectory moves down and right.

  • If $dx/dt<0$ and $dy/dt>0\text{,}$ the trajectory moves up and left.

  • If $dx/dt<0$ and $dy/dt<0\text{,}$ the trajectory moves down and left.

Subsection 2.2.6 Important Lessons

• The righthand side of the system

  $$\begin{array}{rl}\frac{dx}{dt}& =f(x,y)\\ \frac{dy}{dt}& =g(x,y)\end{array}$$

  can be viewed as a vector field, $(f(x,y),g(x,y))\text{,}$ which can be plotted in the $x,y$-plane.

• Competition between two competing species can be modeled with the system

  $$\begin{array}{rl}\frac{dx}{dt}& =\alpha x(1-\frac{x}{M})-\beta xy\\ \frac{dy}{dt}& =\gamma y(1-\frac{y}{N})-\delta xy.\end{array}$$

• We can use nullclines and phase plane analysis to sketch solution curves for the system

  $$\begin{array}{rl}\frac{dx}{dt}& =f(x,y)\\ \frac{dy}{dt}& =g(x,y).\end{array}$$

Subsection 2.2.9 Plotting Nullclines with Sage

    

        
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x, y, t = var('x y t')
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F = [x + y, -2*x + y]
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P = desolve_system_rk4(F,[x, y],ics=[0,0.55,0],ivar=t,end_points=10,step=0.01)
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Q = [ [j,k] for i,j,k in P]
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p = line(Q, axes_labels=['$x(t)$','$y(t)$'], thickness=2)
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n = sqrt(F[0]^2 + F[1]^2)
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F_unit = [F[0]/n, F[1]/n] #set all vectors in the vector field to be same length
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p += plot_vector_field(F_unit, (x,-4,4), (y,-4,4), axes_labels=['$x(t)$','$y(t)$'], xmax = 4, xmin = -4, ymax = 4, ymin = -4, aspect_ratio=1)
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p

    
    
    Evaluate (Sage)
    

        

            
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xxxxxxxxxx
 
1
x, y, t = var('x y t')
2
F = [x + y, -2*x + y]
3
P = desolve_system_rk4(F,[x, y],ics=[0,0.55,0],ivar=t,end_points=10,step=0.01)
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Q = [ [j,k] for i,j,k in P]
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p = line(Q, axes_labels=['$x(t)$','$y(t)$'], thickness=2)
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n = sqrt(F[0]^2 + F[1]^2)
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F_unit = [F[0]/n, F[1]/n] #set all vectors in the vector field to be same length
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p += plot_vector_field(F_unit, (x,-4,4), (y,-4,4), axes_labels=['$x(t)$','$y(t)$'], xmax = 4, xmin = -4, ymax = 4, ymin = -4, aspect_ratio=1)
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p += implicit_plot(F[0], (x,-4,4), (y,-4,4), color="green")
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p += implicit_plot(F[1], (x,-4,4), (y,-4,4), color="red")
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p

    
    
    Evaluate (Sage)
    

        

            
                Language:
                
                    Sage
                    Gap
                    GP
                    HTML
                    Macaulay2
                    Maxima
                    Octave
                    Python
                    R
                    Singular
                
            
        
    
    




    

    

        

        Messages