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PAGE 1

Autonomous Diff. Eqs. (continued)

Review from Last Time

last time:

\[ y' = 3\left(1 - \frac{y}{3}\right)y = 3y - y^2 = f(y) \]

equilibrium: \( y' = 0 \implies y = 0,\; y = 3 \)

Phase Line Analysis

phase line:

Sign analysis of f(y) and flow direction across intervals determined by equilibrium solutions
Interval / Point	Sign of \( f(y) \)	Flow Direction	Stability
\( y < 0 \)	\( - \) (negative)	Left (\( \leftarrow \))	Decreasing away from 0
\( y = 0 \)	\( 0 \)	Stationary	Unstable Source
\( 0 < y < 3 \)	\( + \) (positive)	Right (\( \rightarrow \))	Increasing toward 3
\( y = 3 \)	\( 0 \)	Stationary	Stable Sink / Carrying Capacity
\( y > 3 \)	\( - \) (negative)	Left (\( \leftarrow \))	Decreasing toward 3

Solution Curves in the \( (t, y) \) Plane

Vertical phase line and solution curves plotted against time t — Figure: Vertical phase line and solution curves plotted against time \( t \) Show Details

Vertical phase line aligned with coordinate axes representing time \( t \) horizontally and \( y \) vertically. Equilibrium solutions are shown at \( y = 0 \) and \( y = 3 \) (dashed line). Solutions with \( y(0) > 3 \) decrease asymptotically toward \( y = 3 \). Solutions with initial values in \( 0 < y(0) < 3 \) form characteristic S-shaped logistic curves that begin near \( y = 0 \), accelerate through an inflection point, and decelerate toward \( y = 3 \). Solutions with \( y(0) < 0 \) decrease rapidly away from \( y = 0 \).

Observation:

If \( y(0) \) is near \( y = 0 \), \( y' \) is initially small positive, increases, then eventually decreases and is near zero near \( y = 3 \).

Notice there is an inflection point on solutions that start near \( y = 0 \).

That's where \( y' \) reaches a maximum.

To find the inflection pt between equilibrium solutions:

How to locate the inflection point:

Recall that an inflection point occurs where the second derivative \( y'' = 0 \). By the chain rule:

\[ y'' = \frac{d}{dt}(y') = \frac{d}{dt}[f(y)] = f'(y) \cdot y' = f'(y) \cdot f(y) \]

Between the equilibrium solutions, \( f(y) \neq 0 \). Therefore, \( y'' = 0 \) precisely when \( f'(y) = 0 \). Since \( f(y) = 3y - y^2 \), differentiating yields \( f'(y) = 3 - 2y = 0 \implies y = \frac{3}{2} \). At this height, the rate of increase \( y' \) attains its absolute maximum.

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Autonomous Differential Equations: Equilibrium Stability & Concavity

\( y' = f(y) \)

\( \frac{df}{dy} = 0 \)how slope changes with respect to \( y \)

here, \( f(y) = 3y - y^2 \)

\[ \frac{df}{dy} = 3 - 2y = 0 \to y = \frac{3}{2} \]

Phase portrait and solution curves in the (t, y) -plane for y' = 3y - y^2 — Figure: Phase portrait and solution curves in the \( (t, y) \)-plane for \( y' = 3y - y^2 \) Show Details

Graph of solution curves \( y(t) \) versus time \( t \) for the autonomous differential equation \( y' = 3y - y^2 \). Horizontal lines indicate the equilibria at \( y = 3 \) and \( y = 0 \), as well as the inflection threshold at \( y = \frac{3}{2} \). Solutions starting above \( y = 3 \) decrease asymptotically toward \( y = 3 \). Solutions between \( y = 0 \) and \( y = 3 \) increase toward \( y = 3 \), switching from concave up to concave down as they cross the dashed line \( y = \frac{3}{2} \). Solutions starting below \( y = 0 \) curve downward away from the line \( y = 0 \), confirming that \( y = 3 \) is asymptotically stable while \( y = 0 \) is unstable.

\( \leftarrow \) At \( y = \frac{3}{2} \): every time solution goes through it changes concavity

Concept Explanation: Inflection Points in Autonomous Equations

Using the chain rule, the second derivative is \( y'' = \frac{d}{dt}(y') = \frac{d}{dt}[f(y)] = f'(y) \frac{dy}{dt} = f'(y) f(y) \). Inflection points occurred where \( y'' = 0 \). Besides equilibrium points where \( f(y) = 0 \), this occurs when \( f'(y) = \frac{df}{dy} = 0 \). For \( f(y) = 3y - y^2 \), setting \( f'(y) = 3 - 2y = 0 \) yields \( y = \frac{3}{2} \). When a solution crosses \( y = \frac{3}{2} \), the derivative \( f'(y) \) changes sign, producing a change in concavity from concave up to concave down.

all solutions near the equilibrium \( y = 3 \) want to converge onto it \( \to \) the equilibrium is asymptotically stable

\( y = 0 \) is the opposite situation \( \to \) unstable

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Phase Line Analysis and Semi-Stable Equilibria

Let's look at

\[ \frac{dy}{dt} = y(y-1)^2 \]

Equilibrium: \( y = 0 \), \( y = 1 \) \( (y' = 0) \)

Equilibrium points occur where the rate of change is zero, \( \frac{dy}{dt} = 0 \). Setting \( y(y-1)^2 = 0 \) yields the critical values \( y = 0 \) and \( y = 1 \).

Phase Line Analysis

Tracking the sign of \( y' \) across intervals partitioned by the equilibria:

Sign analysis of the derivative y' along the phase line intervals
Interval / Point	\( y < 0 \)	\( y = 0 \)	\( 0 < y < 1 \)	\( y = 1 \)	\( y > 1 \)
Sign of \( y' \)	\( - \)	\( 0 \)	\( + \)	\( 0 \)	\( + \)
Direction	Decreasing (\( \leftarrow \))	Equilibrium	Increasing (\( \rightarrow \))	Equilibrium	Increasing (\( \rightarrow \))

Phase portrait and solution curves in the (t, y) plane for \frac{dy}{dt} = y(y-1)^2 — Figure: Phase portrait and solution curves in the \( (t, y) \) plane for \( \frac{dy}{dt} = y(y-1)^2 \) Show Details

Vertical phase line and solution curves in the \( (t, y) \) plane for \( \frac{dy}{dt} = y(y-1)^2 \). The vertical phase line on the left marks equilibria at \( y = 0 \) and \( y = 1 \), with downward flow for \( y < 0 \) and upward flow for both \( 0 < y < 1 \) and \( y > 1 \). In the coordinate plane, solution curves for \( y(0) < 0 \) diverge downward away from \( y = 0 \). Curves with initial values in \( 0 < y(0) < 1 \) increase monotonically toward the dashed horizontal asymptote \( y = 1 \). Curves with \( y(0) > 1 \) increase and diverge away from \( y = 1 \), showing that \( y = 1 \) is semi-stable and \( y = 0 \) is unstable.

The solution \( y = 1 \) is said to be semi-stable (solutions on one side approach and solutions on the other side run away).

Semi-Stability Note: Because the factor \( (y-1)^2 \) is squared, it is strictly positive for all \( y \neq 1 \). Therefore, the sign of \( y' \) does not change as \( y \) passes through \( 1 \); \( y' > 0 \) on both sides. Hence, solutions below \( y = 1 \) approach it asymptotically, while solutions above \( y = 1 \) diverge away from it.

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Exact Differential Equations

In calculus 3, we studied level curves \( f(x, y) = C \) (constant)

Coordinate graph of level curves f(x, y) = C_1 and f(x, y) = C_2 in the xy -plane. — Figure: Coordinate graph of level curves \( f(x, y) = C_1 \) and \( f(x, y) = C_2 \) in the \( xy \)-plane. Show Details

Hand-drawn coordinate graph showing axes \( x \) and \( y \) with two nested level curves opening downward. Pointers indicate the outer curve as \( f(x, y) = C_1 \) and the inner curve as \( f(x, y) = C_2 \), visually illustrating contour lines of constant function values \( C \).

\( y \) is an implicit function of \( x \)

if we differentiate \( f(x, y) = C \) with respect to \( x \)

\[ \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx} = 0 \tag{1} \]

Concept Note: Multivariable Chain Rule

Differentiating \( f(x, y(x)) = C \) with respect to \( x \) uses the chain rule: \( \frac{df}{dx} = \frac{\partial f}{\partial x}\frac{dx}{dx} + \frac{\partial f}{\partial y}\frac{dy}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx} = 0 \), because the derivative of a constant is \( 0 \).

we also remember that if the partial derivatives exist then

\[ \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right) = \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right) \tag{2} \]

Theorem Note: Clairaut's Theorem (Symmetry of Mixed Partials)

If the mixed partial derivatives are continuous on an open domain, the order of differentiation does not matter: \( \frac{\partial^2 f}{\partial x\partial y} = \frac{\partial^2 f}{\partial y\partial x} \). This symmetry is the foundational condition for an ordinary differential equation to be exact.

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Exact Differential Equations

Let \(\frac{\partial f}{\partial x} = M(x,y)\) \(\frac{\partial f}{\partial y} = N(x,y)\)

then 1 becomes \( M(x,y) + N(x,y)\frac{dy}{dx} = 0 \)

\(\rightarrow\) Sometimes expressed as \( M(x,y)\,dx + N(x,y)\,dy = 0 \)

2 becomes \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \)

Condition for Exactness (Clairaut's Theorem):

Equation (2) arises from the equality of mixed partial derivatives: \(\frac{\partial M}{\partial y} = \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial N}{\partial x}\). If this equality holds on a simply connected domain, the differential equation is exact.

this kind of differential eq. is called exact diff. eq.

its solution came from the level curve \( f(x,y) = C \)

such that

\[ \frac{\partial f}{\partial x} = M \quad \text{and} \quad \frac{\partial f}{\partial y} = N \]

in the context of calculus 3, this means we are working with a potential function \( f(x,y) \) in a conservative vector field

Calculus 3 Connection:

A vector field \(\mathbf{F}(x, y) = M(x, y)\mathbf{i} + N(x, y)\mathbf{j}\) is conservative if \(\mathbf{F} = \nabla f\). The condition \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\) corresponds to \(\text{curl}(\mathbf{F}) = 0\), guaranteeing the existence of a potential function \(f\).

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Example: Testing for Exactness

Given differential equation:

\[ (3x^2 + 2y^2) + (4xy + 6y^2)y' = 0 \]

Note on Standard Form: This first-order differential equation has the form \( M(x, y) + N(x, y)y' = 0 \), where \( M(x, y) = 3x^2 + 2y^2 \) and \( N(x, y) = 4xy + 6y^2 \).

looks exact but might not be

\[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \]

must be true for \( M + Ny' = 0 \) to be exact.

Criterion for Exactness: If a potential function \( f(x, y) \) exists such that \( \frac{\partial f}{\partial x} = M \) and \( \frac{\partial f}{\partial y} = N \), then by Clairaut's theorem on mixed partial derivatives, \( \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y} \), requiring \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).

check:

\[ \left. \begin{aligned} \frac{\partial M}{\partial y} &= \frac{\partial}{\partial y}(3x^2 + 2y^2) = 4y \\[8pt] \frac{\partial N}{\partial x} &= \frac{\partial}{\partial x}(4xy + 6y^2) = 4y \end{aligned} \right\} \text{ match} \]

So, the eq. is exact

and the solution is \( f(x, y) = c \)

such that

\[ \begin{aligned} \frac{\partial f}{\partial x} &= M \\[6pt] \frac{\partial f}{\partial y} &= N \end{aligned} \]

Next Step: To find the potential function \( f(x, y) \), integrate \( M(x, y) \) with respect to \( x \) treating \( y \) as constant: \( f(x, y) = \int (3x^2 + 2y^2)\,dx = x^3 + 2xy^2 + g(y) \), then differentiate with respect to \( y \) and equate to \( N(x, y) \) to determine \( g(y) \).

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Solving Exact Equations: Integrating with Respect to \( x \)

Given partial derivatives:

\[ \frac{\partial f}{\partial x} = 3x^2 + 2y^2 \quad - \text{①} \]

\[ \frac{\partial f}{\partial y} = 4xy + 6y^2 \quad - \text{②} \]

goal: find \( f(x, y) = c \)

pick one to integrate

here, pick ① to integrate with respect to \( x \) (nothing wrong with choosing to integrate ② with respect to \( y \))

\[ f(x, y) = \int \frac{\partial f}{\partial x} \, dx = \int (3x^2 + 2y^2) \, dx \]

Integration notes:

variable is \( x \)
\( y \) is treated as constant

\[ = x^3 + 2y^2 x + h(y) \]

Component \( h(y) \):

some function of \( y \) which disappears when differentiated with respect to \( x \)

Concept Explanation: Function of Integration \( h(y) \)

When performing partial integration with respect to \( x \), the traditional constant of integration \( C \) is generalized to an arbitrary function \( h(y) \). This is because \( y \) is held fixed during partial differentiation with respect to \( x \), so \( \frac{\partial}{\partial x}[h(y)] = 0 \).

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Exact Differential Equations: Determining \( h(y) \)

Partial of the above with respect to \( y \) must match \( N \) because:

\[ \frac{\partial f}{\partial y} = N = 4xy + 6y^2 \]

Differentiating the expression for \( f(x, y) \) with respect to \( y \):

\[ \frac{\partial}{\partial y} \left( x^3 + 2y^2 x + h(y) \right) = 4xy + \frac{dh}{dy} \]

Comparing the differentiated potential function \( 4xy + \frac{dh}{dy} \) with \( N = 4xy + 6y^2 \), the \( 4xy \) terms cancel, directly isolating \( \frac{dh}{dy} \).

So,

\[ \frac{dh}{dy} = 6y^2 \]

so, \[ h(y) = 2y^3 \]

Integrating \( 6y^2 \) with respect to \( y \) gives \( 2y^3 \). The constant of integration can be omitted here as it is absorbed into the general constant \( C \).

Solution (implicit) : \( f(x, y) = C \)

\[ x^3 + 2xy^2 + 2y^3 = C \]