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Existence & Uniqueness (Continued)

Linear Equations

linear :\[ y' + p(t)y = g(t) \quad y(t_0) = y_0 \]

unique solution on interval where \( p(t) \) and \( g(t) \) are continuous and containing \( t_0 \)

Nonlinear Equations

nonlinear :\[ y' = f(t, y) \quad y(t_0) = y_0 \]

last time :\[ y' = y^{1/3} \quad y(0) = 0 \]

\[ y = \left(\frac{2}{3} t\right)^{3/2} \]

but \( y = 0 \) for all \( t \) is also a solution

Solution is not unique out of \( y(0) = 0 \)

Why does uniqueness fail? According to the Picard-Lindelöf Existence and Uniqueness Theorem for nonlinear ODEs \( y' = f(t, y) \), uniqueness is guaranteed near \( (t_0, y_0) \) only if both \( f \) and \( \frac{\partial f}{\partial y} \) are continuous. Here, \( f(t, y) = y^{1/3} \), so \( \frac{\partial f}{\partial y} = \frac{1}{3} y^{-2/3} = \frac{1}{3 y^{2/3}} \), which blows up and is discontinuous at \( y = 0 \). Hence, multiple solutions can emerge from \( y(0) = 0 \).

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Conditions for Unique Solutions in Nonlinear Equations

What conditions do nonlinear equations have to satisfy to guarantee a unique solution?

For linear equations, we defined:

\[ u(t) = y_2 - y_1 \]

We showed that \( u(t) = 0 \) (unique solution) as long as \( p(t) \) and \( g(t) \) are continuous.

Coordinate graph showing two candidate solutions y_1 and y_2 with separation u(t) — Figure: Coordinate graph showing two candidate solutions \( y_1 \) and \( y_2 \) with separation \( u(t) \) Show Details

A graph on the \( (t, y) \) plane illustrating two candidate solution curves, \( y_1 \) and \( y_2 \), with a bracket denoting their pointwise difference \( u(t) = y_2 - y_1 \).

Key Strategy for Uniqueness: If two distinct solutions \( y_1 \) and \( y_2 \) satisfy the same initial conditions \( y_1(t_0) = y_2(t_0) = y_0 \), their difference \( u(t) = y_2(t) - y_1(t) \) starts at \( u(t_0) = 0 \). Showing that \( u(t) \equiv 0 \) for all \( t \) proves that \( y_1(t) \equiv y_2(t) \), ensuring uniqueness.

Generalizing to Nonlinear Equations

Let's do that for nonlinear:

\[ y' = f(t, y), \quad y(t_0) = y_0 \]

Let \( y_1 \) and \( y_2 \) be two solutions.

Define \( u(t) = y_2 - y_1 \).

\[ y_1' = f(t, y_1) \quad \text{because } y_1 \text{ is a solution} \]

\[ y_2' = f(t, y_2) \]

Subtract:

\[ y_2' - y_1' = f(t, y_2) - f(t, y_1) \]

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Applying the Mean-Value Theorem

We will use the Mean-Value Theorem on the right:

MVT:

\[ \frac{f(b) - f(a)}{b - a} = f'(c), \quad a < c < b \]

Right side of \( y_2' - y_1' = f(t, y_2) - f(t, y_1) \):

Using MVT

\[ \frac{f(t, y_2) - f(t, y_1)}{y_2 - y_1} = \frac{\partial f}{\partial y}(t, y^*) \]

\( \frac{\partial f}{\partial y}(t, y^*) \) — depends explicitly on \( t \) only

where \( y_1 < y^* < y_2 \)

Context & Clarification: Applying the Single-Variable Mean-Value Theorem with respect to the second variable \( y \) while treating \( t \) as fixed yields an intermediate value \( y^* \) between \( y_1(t) \) and \( y_2(t) \).

\( y_2' - y_1' = f(t, y_2) - f(t, y_1) \) becomes:

\[ u'(t) = \frac{\partial f}{\partial y}(t, y^*) \cdot u(t) \]

(where \( u(t) = y_2 - y_1 \))

\[ u'(t) - \frac{\partial f}{\partial y}(t) \cdot u(t) = 0 \]

is linear! matching standard linear form: \( y' + p(t)y = g(t) \)

Significance: Because \( u(t) \) satisfies a linear homogeneous first-order differential equation with initial condition \( u(t_0) = y_2(t_0) - y_1(t_0) = 0 \), the only solution is \( u(t) \equiv 0 \). This implies \( y_1(t) = y_2(t) \), completing the uniqueness proof of the Picard-Lindelöf theorem.

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Uniqueness Condition & The Nonlinear Existence-Uniqueness Theorem

\( -\frac{\partial f}{\partial y} \) is acting like \( p(t) \)

Concept Context: To establish uniqueness, consider the difference \( u(t) = y_1(t) - y_2(t) \) between two solutions satisfying \( u(t_0) = 0 \). By linearization / Mean Value Theorem, \( u'(t) \approx \frac{\partial f}{\partial y} u(t) \), which transforms into a homogeneous linear differential equation \( u' - \frac{\partial f}{\partial y} u = 0 \) where \( -\frac{\partial f}{\partial y} \) plays the role of the coefficient \( p(t) \).

\[ \mu(t) = e^{-\int \frac{\partial f}{\partial y} \, dt} \]

integrating factor

\( u(t_0) = 0 \)

\[ u(t) = u(t_0) e^{\int \frac{\partial f}{\partial y} \, dt} \]

\( \longrightarrow \) will be \( 0 \) for all \( t \) as long as \( \frac{\partial f}{\partial y} \) is continuous

For nonlinear \( y' = f(t, y) \), \( y(t_0) = y_0 \)

A unique solution is guaranteed on some interval wherever both \( f(t, y) \) and \( \frac{\partial f}{\partial y} \) are continuous and containing \( y(t_0) = y_0 \).

let's revisit:

\[ y' = y^{1/3}, \quad y(0) = 0 \]

\( f(t, y) = y^{1/3} \) continuous for all \( (t, y) \)

\( \frac{\partial f}{\partial y} = \frac{1}{3} y^{-2/3} = \frac{1}{3y^{2/3}} \) not continuous for \( y = 0 \)

Conclusion: Since \( \frac{\partial f}{\partial y} \) is undefined and fails to be continuous at the initial condition \( y = 0 \), the hypothesis of the Existence and Uniqueness Theorem is not satisfied. Hence, uniqueness fails, confirming why multiple distinct solutions (such as \( y(t) = 0 \) and \( y(t) = \pm \left(\frac{2}{3}t\right)^{3/2} \)) can originate from \( y(0) = 0 \).

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Existence and Uniqueness: Region of Continuity

Coordinate plane of t versus y showing valid initial condition regions away from y = 0 and non-uniqueness along y = 0 . — Figure: Coordinate plane of \( t \) versus \( y \) showing valid initial condition regions away from \( y = 0 \) and non-uniqueness along \( y = 0 \). Show Details

A coordinate plane with horizontal axis \( t \) and vertical axis \( y \) enclosed in a dashed rectangular boundary. Arrows indicate that initial conditions are valid ('ok here') in the open regions where \( y \neq 0 \). A highlighted red line along the horizontal axis \( y = 0 \) is labeled 'not ok here', indicating the failure of the existence and uniqueness conditions where \( f \) or \( \frac{\partial f}{\partial y} \) is discontinuous, leading to non-unique solutions.

Initial condition \( y(0) = 0 \) is NOT at a location where BOTH \( f \) and \( \frac{\partial f}{\partial y} \) are continuous.

\( \rightarrow \) solution MAY NOT be unique

Concept: Picard–Lindelöf Theorem

For an initial value problem \( y' = f(t, y) \) with \( y(t_0) = y_0 \), uniqueness is guaranteed if both \( f(t, y) \) and \( \frac{\partial f}{\partial y} \) are continuous on a rectangle enclosing \( (t_0, y_0) \). When \( \frac{\partial f}{\partial y} \) fails to be continuous along \( y = 0 \) (e.g., for \( y' = y^{1/3} \) or \( y' = y^{2/3} \)), uniqueness often breaks down, allowing multiple solutions through the same point.

Autonomous Diff. Eqs.

1st-order eqs. of the form

\[ y' = f(y) \]

No explicit independent variable \( t \) appears on the right-hand side.
Rate of change of \( y \) depends on its own size.

Concept: Autonomous Systems

An ordinary differential equation is termed autonomous when time \( t \) does not appear explicitly. Such equations describe time-invariant physical processes where the dynamics depend solely on the current state of the system.

in a lot of applications

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

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Logistic Population Model

As a population model, it does not model environmental limitations.

An Improved Model

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

where \( r, a \) are constants.

Note on the term \( (r - ay) \):

No longer just constant \( k \)
Growth rate is NOT constant

Concept Explanation: In simple exponential growth (\( \frac{dy}{dt} = ky \)), the relative per capita growth rate is the constant \( k \). In this improved model, the per capita growth rate \( (r - ay) \) decreases linearly as the population \( y \) grows, accounting for crowding and resource competition.

Equilibrium Solutions

Note:

\( y' = 0 \) when \( y = 0 \)
\( y' = 0 \) when \( y = \frac{r}{a} \) “carrying capacity” (environmental limit)

Standard Logistic Growth Model

Rewrite:

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

\( r \): “intrinsic growth rate”

\( K = \frac{r}{a} \): carrying capacity

Logistic Growth Model

Algebraic Step: Factoring \( r \) out of \( (r - ay) \) produces \( r\left(1 - \frac{a}{r}y\right) = r\left(1 - \frac{y}{K}\right) \) by setting \( K = \frac{r}{a} \). When \( 0 < y < K \), the population increases (\( y'> 0 \)). When \( y > K \), mortality exceeds births due to environmental limits, causing the population to decrease (\( y' < 0 \)).

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Long-Term Behavior and Initial Conditions

We want to know: as \( t \to \infty \) will \( y \to 0 \) or \( y \to \frac{r}{a} \)?

Does the initial condition matter?

Example Analysis

For example, let's look at

\[ \frac{dy}{dt} = 3\left(1 - \frac{y}{3}\right)y = f(y) \]

\( f(y) = 0 \to \text{critical values/points} \)

(\( y \) values where \( y' = 0 \))

Here, the critical values are \( y = 0 \), \( y = 3 \).

Concept Explanation: Setting \( f(y) = 0 \) determines the equilibrium solutions (steady states) where \( \frac{dy}{dt} = 0 \). At these points, \( y(t) \) remains constant for all time. Here, factoring \( 3\left(1 - \frac{y}{3}\right)y = 0 \) directly yields roots \( y = 0 \) and \( y = 3 \).

Phase Line Analysis

Let's draw a phase line:

Phase line diagram for \frac{dy}{dt} = 3\left(1 - \frac{y}{3}\right)y showing critical points at y = 0 and y = 3 with directional arrows. — Figure: Phase line diagram for \( \frac{dy}{dt} = 3\left(1 - \frac{y}{3}\right)y \) showing critical points at \( y = 0 \) and \( y = 3 \) with directional arrows. Show Details

A horizontal phase line diagram for the autonomous differential equation \( \frac{dy}{dt} = f(y) = 3\left(1 - \frac{y}{3}\right)y \). The critical equilibrium values are indicated at \( y = 0 \) and \( y = 3 \). Above the axis, the signs of \( f(y) \) are labeled: negative (\( - \)) for \( y < 0 \), zero (\( 0 \)) at \( y = 0 \), positive (\( + \)) for \( 0 < y < 3 \), zero (\( 0 \)) at \( y = 3 \), and negative (\( - \)) for \( y > 3 \). Directional arrows along the line indicate decreasing motion to the left for \( y < 0 \), increasing motion to the right for \( 0 < y < 3 \), and decreasing motion to the left for \( y > 3 \).

\( f(y) = y' \)

\( f(y) > 0 \to y' > 0 \) increasing (to right)

\( f(y) < 0 \to y' < 0 \) decreasing (to left)

Reading the Phase Line:

For \( y < 0 \): \( f(y) < 0 \implies y' < 0 \) (arrows point left toward \( -\infty \)).
For \( 0 < y < 3 \): \( f(y)> 0 \implies y' > 0 \) (arrows point right toward \( y = 3 \)).
For \( y > 3 \): \( f(y) < 0 \implies y' < 0 \) (arrows point left toward \( y=3 \)).
Conclusion: \( y = 0 \) is an unstable equilibrium point, while \( y = 3 \) is a stable equilibrium (attractor / carrying capacity).

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Sketch Some Solutions

Phase line and sketched solution curves in the (t, y) -plane — Figure: Phase line and sketched solution curves in the \( (t, y) \)-plane Show Details

A vertical phase line on the left is paired with solution curves in the \( (t, y) \)-plane on the right. On the phase line, equilibrium levels are marked at \( y = 3 \) and \( y = 0 \). Arrows indicate that the flow moves downward toward \( y = 3 \) for \( y > 3 \), upward toward \( y = 3 \) for \( 0 < y < 3 \), and downward away from \( y = 0 \) for \( y < 0 \). In the coordinate system, the dashed horizontal line at \( y = 3 \) represents an asymptotically stable equilibrium. Three representative solution trajectories are plotted: one decreasing asymptotically toward \( y = 3 \) from above, an S-shaped trajectory starting just above \( y = 0 \) that passes through a labeled inflection point before flattening toward \( y = 3 \), and a decreasing trajectory in the region \( y < 0 \) that bends downward.

\( y' = 0 \rightarrow \text{equilibrium solutions} \)

Equilibrium Solutions: Setting the derivative \( y' = 0 \) determines the constant solutions \( y(t) = c \). Here, the equilibrium solutions are \( y = 3 \) (an asymptotically stable sink) and \( y = 0 \) (an unstable source).

Inflection Points: For an autonomous equation \( y' = f(y) \), differentiating yields \( y'' = f'(y)y' = f'(y)f(y) \). An inflection point in a solution curve occurs where \( y'' = 0 \), marking a transition in concavity as the solution transitions toward the carrying capacity \( y = 3 \).