MA354 Lecture: Graphical Slns to 1D Autonomous ODEs
Homework: 1D autonomous ODEs
Sketching a solution by drawing slope field: (HW 1d, 1e)
Sketching a solution by analyzing the phase line: (HW 3, 7 11e)
Methods for Finding Solutions to Diff Eqs
 Find an explicit solution.
- can always be done for linear
- often for separable
- often for exact
- sometimes others with special substitutions
 Using the slope field.
 Using numerical approximations (Euler’s, etc.)
Using Slope Fields (Direction Fields or Vector Fields)
Recall a differential equation f ' x, y  specifies the slope of f at any point (x,y). The
solution to the differential equation is the family of functions that have that slope.
dy
 g  x, y  (first order because highest
dx
derivative is 1), the slope field of dy/dx is an array of slope marks in the phase space,
usually of uniform length where the slope of a mark at (x,y) is g(x,y) and indicates the
slope of the tangent to the solution at (x,y).
Given a first-order differential equation
Worksheet Example
dy
 x2 .
Consider
dx



At the point (1,1) the slope would be 12 = 1.
At the point (5,1) the slope would be 52 = 25.
At the point (-3,1) the slope would be (-3)2 = 9.
Note the derivative does not depend on y!
You can sketch this by hand … or use Mathematica.
StreamPlot[{1,x^2},{x,-2,2},{y,-2,2}]
DSolve[y'[x] = = x*x,y[x],x]
Mathematica says: {{y[x] x3/3+C[1]}}
You can find the solution by using the fact that each mark indicates the slope of the
tangent. Can see the solutions by plotting the direction field: Without a computer, this
would be very tedious... we can make this more efficient by identifying how the tangent
changes as a function of x and y. This is particularly easy to do when the tangent only
depends upon one variable.
Autonomous ODEs: Graphical Solutions
An autonomous system is an ordinary differential equation not depending on the
independent variable. (the one we are taking the derivative with respect to)
Example: y´=y2 (doesn’t depend on x)
Counter-example: y´=x2 (non-autonomous function)
Some definitions:
Equilibrium values: All values of y such
dy
 0 that are called equilibrium values or
dx
dy
 g ( y ) , equilibrium values are the zeros of g(y). They are the
dx
constant functions which are solutions to the diff eqn.
rest points. Given
Phase line: A phase line is a plot on the y axis that shows the equilibrium values along
with the intervals where y’ and y’’ are positve and negative. We can use the phase line to
determine qualitative characteristics of the solution curves. (A phase line is a line through
phase space.)
Recall: Phase space: A multi-dimensional space in which all possible states of a system
are represented, with each possible state of the system corresponding to one unique point
in the phase space.
Drawing Phase Lines
Step 1: Plot equilibrium values on the real line.
Step 2: Identify signs of y’ within intervals. (represent this with directed arrows)
Step 3: Find y’’, zeros of y’’, and identify signs of y’ within intervals.
* Just like Calculus when we wanted to plot a function! *
Example 1 on page 428
dy
 ( y  1)( y  2)
dx
Making the phase line.
Equilibrium points: when dy/dx = 0: The equilibrium points are y=-1 and y=2.
Phase line: ------- (-1) -------------- (2) ---------------Signs of the derivative:
y’
+
-------------- (-1) -------------y’’
(2)
+
--------------
Now find y’’:
y’ = y2-y-2
y’’ = 2y*y’-1*y’ = (2y-1)*y’=(2y-1)*(y+1)*(y-2)
zeros are: -1, ½, 2
Signs of the second derivative:
y’
+
-------------- (-1) -------------y’’
+
(1/2)
--------------
(2)
+
-------------+
Sketching Solution Curves
Draw the horizontal lines which are equilibrium values for y. They may be stable or
unstable, but they are lines where y won’t change. They are constant functions which
satisfy the diff eqn.
Slope and Concavity of the Solution
y’’<0
y’’>0
y’<0
negative slope
concave down
negative slope
concave up
y’<0
positive slope
concave down
positive slope
concave up
Can see the solutions by plotting the direction field:
PlotVectorField[{1,(y+1)*(y-2)},{x,-15,5},{y,-2,5}]
Stability of the Equilibria:
If solutions near an equilibrium value tend to get closer to that value as x increases, then
it is an asymptotically stable equilibrium. If solutions move away from the equilibrium as
x increases, then it is an unstable equilibrium. An equilibrium can be semistable if it is
stable if approached from one direction and unstable if approached from another direction.
Examples – page 433
dy
 ( y)
dx
Step 1: Plot equilibrium values on the real line. (y=0)  not stable
Step 2: Identify signs of y’ within intervals. (<0, >0)
Step 3: Find y’’, zeros of y’’, and identify signs of y’ within intervals.
 y’’ = y’ = y,
 no additional zeros
 (<0, >0)
1a:
dy
 ( y  2)( y  3)
dx
Find equilibrium values and identify their stability.
2:
Step 1: Plot equilibrium values on the real line.
Step 2: Identify signs of y’ within intervals.
Step 3: Find y’’, zeros of y’’, and identify signs of y’ within intervals.
Homework: 1d, 1e, 3, 6
Next time: 12.1 look at autonomous systems that are 2-dim
1-dimensional:
dy
 g ( y)
dx
yields a phase line
2-dimensional:
dy1
 g ( y1 , y 2 )
dx
dy 2
 f ( y1 , y 2 )
dx
yields a phase plane (y1, y2)
equivalently:
dy
 g ( x, y )
dt
dx
 f ( x, y )
dt
(y1)
interpretation: we might have y1 and
y2 as spatial variables and x as time
yields a phase plane (x, y)
Another example using phase lines to sketch a solution.
Logistic Growth
dP
 r (M  P) P
dt
1. find rest points
2. find signs of P’ over
intervals
3. repeat steps for P’’
y’
y’’
(-)
--------------
zeros of P’: P=M and P=0
circle the rest points on the # line
draw little arrows!
  stable
 unstable
 semi-stable
P’ = rMP-rP2
P’’ = rM(P’)-[2rP(P’)]=P’(rM-2rP)=rP’*(M-2P)
zeros are zeros of P’: 0 and <
and also: P=M/2
(0)
(+)
------------(+) (M/2) (-)
(M)
(-)
------------(+)
Note that y’’ has a local maximum at M/2; this shows that the maximum rate of growth
occurs when the population is M/2. This has been used to find M in real populations.
When the change in the rate of growth is zero, P  M/2.