PAGE 1

2.4 (Continued)

Nonlinear First-Order ODEs

\[ \frac{dy}{dt} = f(t, y) \quad y(t_0) = y_0 \]

A solution is unique within \( t_0 - h < t < t_0 + h \) with in the rectangle \( \alpha < t < \beta \), \( \gamma < y < \delta \) where \( f \) and \( \frac{\partial f}{\partial y} \) are BOTH continuous.

Example

\[ y' = \frac{\ln(2y - t^2)}{t + 1} + (y - 2)^{1/3} \]

Let \( f(t, y) = \frac{\ln(2y - t^2)}{t + 1} + (y - 2)^{1/3} \).

\( f \) is continuous on \( 2y - t^2 > 0 \), \( t \neq -1 \).

\[ \frac{\partial f}{\partial y} = \frac{2}{(2y - t^2)(t + 1)} + \frac{1}{3}(y - 2)^{-2/3} \]

\( \frac{\partial f}{\partial y} \) is continuous on \( 2y - t^2 \neq 0 \), \( t \neq -1 \), \( y \neq 2 \).

PAGE 2

\( 2y - t^2 = 0 \rightarrow y = \frac{1}{2}t^2 \) (parabola), we must be above it (\( 2y - t^2 \neq 0 \), \( 2y - t^2 > 0 \)).

Then exclude \( t = -1 \), \( y = 2 \).

A coordinate graph showing a dashed parabola y = \frac{1}{2}t^2 , a vertical dashed line at t = -1 , and a horizontal dashed line at y = 2 . — Figure: A coordinate graph showing a dashed parabola \( y = \frac{1}{2}t^2 \), a vertical dashed line at \( t = -1 \), and a horizontal dashed line at \( y = 2 \).

The initial condition \( y(t_0) = y_0 \) determines where the unique solution is contained.

PAGE 3

2.5 Autonomous Diff. Eqs.

\[ \frac{dy}{dt} = f(y) \]

no \( t \) explicitly

\( \frac{dy}{dt} \) only depends on itself (\( y \)), so "autonomous".

Often used to model population dynamics.

Simple exponential growth:

\[ \frac{dy}{dt} = ry \]

growth rate (\( r > 0 \))

if \( r < 0 \) → exponential decay

Notice autonomous diff. eqs. are always separable, but let's focus on the qualitative behavior in 2.5.

\[ \frac{dy}{dt} = ry \longleftrightarrow y = Ce^{rt} \]

not very interesting

Figure: A coordinate graph with t as the horizontal axis and y as the vertical axis, showing three upward-sloping exponential curves.

PAGE 4

But population doesn't grow exponentially forever → run out of resources, for example.

Modify model:

\[ \frac{dy}{dt} = h(y)y \]

rate depends on \( y \), not constant anymore

Model the slow down of growth as \( y \) becomes "large".

\[ \frac{dy}{dt} = (r - ay)y \]

as \( y \to \frac{r}{a} \), \( y' \to 0 \)

Logistic growth

(models a limit the environment puts on the population)

Rewrite:

\[ \frac{dy}{dt} = r \left( 1 - \frac{y}{K} \right) y \]

Intrinsic growth

\( r \): intrinsic growth rate

PAGE 5

Equilibrium Solutions in Population Dynamics

We see \[ \frac{dy}{dt} = 0 \] when \( y = 0 \) and \( y = K \).

\( y = 0 \): no population
\( y = K \): limiting population (carrying capacity)

Equilibrium Solutions

These solutions where \( y' = 0 \) are called equilibrium solutions.

(where \( f(y) = 0 \rightarrow \) critical points of \( f(y) \))

Investigation Goals

We want to investigate:

As \( t \to \infty \), which equilibrium solution do solutions converge to?
Does it depend on initial condition?

PAGE 6

Example: Quadratic Growth Model

For example, \[ \frac{dy}{dt} = (2 - 3y)y = f(y) = -3y^2 + 2y \] This is a parabola opening down.

Critical points: \( y = 0 \), \( y = \frac{2}{3} \)

Graph of \( f(y) \) vs. \( y \)

Figure: Graph of f(y) = -3y^2 + 2y showing a parabola opening down with roots at y=0 and y=2/3.

\( f(y) > 0 \): growth. Growth rate max at \( y = \frac{1}{3} \).

\( f(y) > 0 \) but the rate of growth slows down as it approaches \( y = \frac{2}{3} \).

\( f(y) < 0 \): rate \( < 0 \) (losing population).

Stability Analysis

Solutions want to leave the equilibrium solution \( y = 0 \).

Solutions want to converge onto the equilibrium solution \( y = \frac{2}{3} \).

Phase Line

PAGE 7

Stability of Solution Curves

Figure: Phase line and t-y graph showing curves diverging from y=0 and converging toward y=2/3.

Uniqueness Property

Can solution curves cross \( y = \frac{2}{3} \)?

NO, otherwise the uniqueness property is violated.

Slope and Concavity Analysis

Up, \( f > 0 \) → positive slope
\( y = \frac{1}{3} \) → max of \( f(y) \) (inflection point)
Below: concave up
Above: concave down

Solution diverge from \( y = 0 \) → \( y = 0 \) is unstable

Solution converge onto \( y = \frac{2}{3} \) → \( y = \frac{2}{3} \) is asymptotically stable