See the field before solving
Chapter 2 - Turn a rate rule into a picture
An equation (y'=f(t,y)) assigns one slope to every point ((t,y)). A direction field draws a small segment with that slope at each point. A solution curve does not invent its path; it threads through the assigned segments.
By the end of this chapter, you should be able to:
- read increasing, decreasing, and equilibrium behavior from a field;
- sketch solution curves through initial conditions;
- connect features of (f(t,y)) to visible features of the field; and
- identify what a numerical or symbolic answer must look like before calculating it.
Lab: move the initial condition
Try these moves before reading on:
- Choose (y'=-0.65y). Place the initial point above, on, and below (y=0).
- Choose the logistic equation. Find every horizontal row where the marks go flat.
- Choose (y'=\sin t-0.35y). Notice that equal heights can have different slopes at different times.
Read a field in four passes
1. Find zero-slope sets
Where (f(t,y)=0), the field is horizontal. If the equation is autonomous, (y'=f(y)), a horizontal zero-slope row often gives an equilibrium solution.
For (y'=y(4-y)), the zero-rate states are (y=0) and (y=4).
2. Mark the sign regions
- (y'>0): solution curves rise as (t) increases.
- (y'<0): solution curves fall as (t) increases.
- large (|y'|): solution curves look steep.
For (y'=y(4-y)), the rate is positive between 0 and 4 and negative above 4. Positive solutions move toward 4.
3. Ask what stays constant
If (f) depends only on (y), every mark on one horizontal row has the same slope. If (f) depends only on (t), every mark in one vertical column has the same slope. These repeated patterns identify structure before algebra does.
4. Thread, do not connect dots
A solution curve stays tangent to the field and remains smooth where (f) is smooth. It cannot turn at a point where the assigned slope is strictly positive or cross an equilibrium when uniqueness applies.
Field fingerprint. A direction field looks identical along every horizontal row but changes from one row to the next. What form is most likely?
Worked example: predict without solving
- Zero rate: \(2-y=0\) at \(y=2\). The row \(y=2\) is horizontal.
- Sign: above \(y=2\), \(y'<0\); below it, \(y'>0\). Arrows point toward 2 from both sides.
- Initial motion: \(y'(0)=2-5=-3\), so the curve starts downward with slope \(-3\).
- Long-run prediction: the rate weakens as \(y\) nears 2, so the curve falls and flattens toward \(y=2\).
- Shape check: differentiating the equation gives \(y''=-y'\). While the solution falls, \(y'<0\), so \(y''>0\). The curve is concave up.
Same structure, less support
For (y'=1-\tfrac12 y), (y(0)=-2):
- the equilibrium is (y=\underline{\hspace{2em}});
- the initial slope is (\underline{\hspace{3em}});
- the solution moves [toward / away from] the equilibrium; and
- while it rises, its concavity is [up / down].
Check the four predictions
The equilibrium is \(y=2\); the initial slope is \(2\); the solution moves toward 2; and \(y''=-\tfrac12 y'<0\), so it is concave down while rising.
What can go wrong?
The small segment at \((1,2)\) reports the slope a solution would have if it passed through \((1,2)\). It does not require every solution to pass through that point. A field encodes infinitely many possible initial-value problems at once.
A classical solution \(y(t)\) is a function of \(t\), so its graph cannot double back or become vertical. Very steep is possible; vertical is not a finite derivative.
When may solution curves cross?
If (f(t,y)) and its dependence on (y) are sufficiently well behaved near a point, one initial condition selects one local solution. Two distinct solution curves cannot cross there: the crossing point would give the same initial condition two different futures.
This is a conceptual use of uniqueness. You do not need the full theorem yet. Use the practical rule: for the smooth equations in the early chapters, never sketch two solution curves crossing.
Practice
- Practice - qualitative
2.1 For \(y'=y-3\), identify the equilibrium, sign regions, and long-run direction for initial values 1, 3, and 5.
- Practice - field fingerprint
2.2 Describe how the fields for \(y'=\cos t\), \(y'=\cos y\), and \(y'=\cos(t+y)\) differ in repeated pattern.
- Integrated - derivatives
2.3 A solution satisfies \(y'=y(1-y)\) and \(y(0)=1/4\). Predict monotonicity, limiting state, and the sign of \(y''\) before and after \(y=1/2\).
- Challenge - nonuniqueness
2.4 Verify that both \(y=0\) and \(y=(t/2)^2\) for \(t\ge0\) solve \(y'=\sqrt{y}\) with \(y(0)=0\). What field-sketching rule must you suspend at this nonsmooth point?
Spaced retrieval 1
Without looking back, translate: "A quantity moves toward 10 at a rate equal to twice its signed distance from 10." Check the sign by testing a state above 10.
Reveal after committing
\(y'=-2(y-10)=2(10-y)\). At \(y=12\), the rate is \(-4\), so the state moves down toward 10.