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1st-Order Eqs, Existence and Uniqueness of Solutions

Given:

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

Questions:

Is there a solution? (Existence)
If so, is that solution unique? (Uniqueness)

Conceptual Overview: The fundamental questions of ODE theory address whether an initial value problem (IVP) is well-posed: existence ensures at least one trajectory passes through $(t_0, y_0)$, while uniqueness ensures trajectories do not branch or intersect at that point.

Let's start with linear eqs.

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

We know that if an integrating factor exists, we can (in principle) solve it.

Integrating Factor

\[ \mu(t) = e^{\int p(t) \, dt} \]

1st requirement: this MUST exist

so $ \int p(t) \, dt $ must exist

$ p(t) $ MUST be continuous at least on some interval of $ t $

Connection to Integration: By the Fundamental Theorem of Calculus, continuity of the coefficient function $ p(t) $ on an open interval $ I $ containing $ t_0 $ guarantees that the antiderivative $ \int p(t) \, dt $ exists and is differentiable, ensuring the integrating factor $ \mu(t) $ is well-defined and strictly positive on $ I $.

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Existence of Solutions for First-Order Linear ODEs

The eq can be written as

\[ \frac{d}{dt} \bigl[ \mu(t) y \bigr] = \mu(t) g(t) \]

then

\[ \mu(t) y = \int \mu(t) g(t) \, dt + C \]

2nd requirement:

$ \int \mu(t) g(t) \, dt $ must exist.

So $ g(t) $ must be continuous AND $ p(t) $ and $ g(t) $ are continuous on some common interval.

Explanation of Requirements: The integrating factor is $ \mu(t) = e^{\int p(t)\,dt} $. For $ \mu(t) $ to exist and be differentiable, $ p(t) $ must be continuous (the 1st requirement). For the right-hand side to be integrated, the product $ \mu(t)g(t) $ must be integrable, which is guaranteed when $ g(t) $ is also continuous.

So, for $ y' + p(t)y = g(t) $, $ y(t_0) = y_0 $, a solution exists on some interval of $ t $ on which BOTH $ p(t) $ and $ g(t) $ are continuous AND containing the initial $ t_0 $.

Now, is the solution unique?

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Uniqueness of Solutions: Can Solutions Branch?

For example, can this happen?

Coordinate plot showing two solution curves y_1 and y_2 branching from the same initial point (t_0, y_0) — Figure: Coordinate plot showing two solution curves $ y_1 $ and $ y_2 $ branching from the same initial point $ (t_0, y_0) $ Show Details

A coordinate system with vertical axis $ y $ and horizontal axis $ t $. An initial value point is marked at $ (t_0, y_0) $, from which two distinct solution curves, labeled $ y_1 $ and $ y_2 $, branch upward and to the right for $ t > t_0 $, posing the question of whether solutions to a first-order initial value problem can fail to be unique.

Example: Linear First-Order Initial Value Problem

Consider the linear first-order differential equation and initial condition:

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

Suppose that two solutions exist: $ y_1, y_2 $.

Let's define:

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

difference between them as a function of $ t $

Strategy for Proving Uniqueness: To prove that the solution is unique, we define the difference function $ u(t) = y_2(t) - y_1(t) $. If both satisfy $ y(t_0) = y_0 $, then $ u(t_0) = y_0 - y_0 = 0 $. By substituting into the differential equation, we show that $ u(t) \equiv 0 $ for all $ t $, meaning $ y_1(t) = y_2(t) $.

If $ y_1 $ and $ y_2 $ are solutions, then:

\[ \begin{aligned} y_1' + p(t)y_1 &= g(t) \\[4pt] y_2' + p(t)y_2 &= g(t) \end{aligned} \]

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Uniqueness Proof: Homogeneous Differential Equation

Subtract:

$$(y_2' - y_1') + \bigl(p(t)y_2 - p(t)y_1\bigr) = 0$$

Factoring out $ p(t) $:

$$\underbrace{(y_2' - y_1')}_{u'} + p(t)\underbrace{(y_2 - y_1)}_{u} = 0$$

Substitution: Define the difference function $ u(t) = y_2(t) - y_1(t) $. Differentiating both sides with respect to $ t $ yields $ u'(t) = y_2'(t) - y_1'(t) $, transforming the difference between the two candidate solutions into a first-order linear homogeneous differential equation.

$$u' + p(t)u = 0$$

$$u(t_0) = 0$$

We can solve this for $ u $

Linear and/or separable

$$\vdots$$

$$\begin{aligned} u(t) &= C e^{-\int p(t)\,dt} \\[6pt] u(t) &= u(t_0) e^{-\int p(t)\,dt} \\[6pt] &= 0 \quad \text{because } u(t_0) = 0 \end{aligned}$$

This means $ y_2 - y_1 = 0 $ so $ y_2 = y_1 \to \underline{\mathbf{one}} \text{ solution only} $.

Conclusion: Since the initial condition satisfies $ u(t_0) = y_2(t_0) - y_1(t_0) = y_0 - y_0 = 0 $, the only solution to the homogeneous equation is identically zero: $ u(t) \equiv 0 $ for all $ t $. Therefore, $ y_2(t) = y_1(t) $, establishing the uniqueness of the solution.

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Existence & Uniqueness Theorem for Linear First-Order ODEs

For linear eq. $ y' + p(t)y = g(t) $, $ y(t_0) = y_0 $:

A solution exists and is unique on some interval of $ t $ on which BOTH $ p(t) $ and $ g(t) $ are continuous and containing $ t_0 $.

Conceptual Breakdown: Linear Existence & Uniqueness Theorem

Unlike nonlinear equations, a first-order linear differential equation written in standard form $ y' + p(t)y = g(t) $ is guaranteed to have a unique solution on the entire continuous interval where both coefficient functions $ p(t) $ and $ g(t) $ are continuous and which contains the initial time $ t_0 $.

Example

$ \ln(t)y' + y = \cot(t), \quad y(2) = 3 $

Write the equation in standard form $ y' + p(t)y = g(t) $ by dividing by $ \ln(t) $:

\[ y' + \underbrace{\frac{1}{\ln(t)}}_{p(t)} y = \underbrace{\frac{\cot(t)}{\ln(t)}}_{g(t)} \]

Domain Analysis of $ p(t) $ and $ g(t) $

For $ p(t) = \frac{1}{\ln(t)} $: $ \ln(t) $ is defined only for $ t > 0 $, and the denominator cannot be zero, so $ \ln(t) \neq 0 \implies t \neq 1 $. Hence, $ p(t) $ is continuous on $ (0, 1) \cup (1, \infty) $.
For $ g(t) = \frac{\cot(t)}{\ln(t)} $: In addition to $ t > 0 $ and $ t \neq 1 $, $ \cot(t) = \frac{\cos(t)}{\sin(t)} $ is discontinuous wherever $ \sin(t) = 0 $, which occurs at integer multiples of $ \pi $ ($ t = \pi, 2\pi, 3\pi, \dots $).

Continuity of the coefficient functions:

$ p(t) $ continuous on: $ (0, 1), \; (1, \infty) $
$ g(t) $ continuous on: $ (0, 1), \; (1, \pi), \; (\pi, 2\pi), \; (2\pi, 3\pi), \; \dots $

The interval where BOTH are continuous AND containing $ t_0 = 2 $ is $ (1, \pi) $.

Why $ (1, \pi) $?

The initial condition is given at $ t_0 = 2 $. Since $ 1 < 2 < \pi \approx 3.14159 $, the point $ t_0 = 2 $ falls strictly inside the open subinterval $ (1, \pi) $, where neither $ \ln(t) $ nor $ \sin(t) $ vanishes.

So, solution exists and is unique on $ 1 < t < \pi $.

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Properties of Linear Equations: General Solutions

Another nice property of linear eqs. is that the general solution contains ALL possible solutions.

\[ y' = y \]

$ y = 0 $ is a solution

Solve as separable

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

\[ \frac{1}{y}\,dy = dt \]

$ (y \neq 0) $

Division by Zero Caution: Dividing both sides by $ y $ requires the assumption $ y \neq 0 $. At this stage, the constant solution $ y(t) = 0 $ is temporarily excluded from the separable calculation.

Intermediate Steps: Integrating both sides gives $ \int \frac{1}{y}\,dy = \int dt \implies \ln|y| = t + k_1 \implies |y| = e^{k_1}e^t \implies y = \pm e^{k_1}e^t $. Letting $ C = \pm e^{k_1} $ yields $ y = C e^t $.

\[ y = C e^t \]

Note $ y = 0 $ is included ($ C = 0 $)

linear eqs. ALWAYS behave like this

not always true for nonlinear eqs.

Linear vs. Nonlinear Completeness: For any first-order linear differential equation $ y' + p(t)y = g(t) $, the general solution family accounts for every single solution. By contrast, nonlinear equations often possess "singular solutions" that cannot be obtained by choosing any value for the arbitrary integration constant $ C $.

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Nonlinear Equations: Existence and Uniqueness

For nonlinear eqs,

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

existence : $ f(t, y) $ is continuous on some interval containing $ t_0 $

(for example, think of how you solve $ y' = y^3 $)

Peano's Existence Theorem: Continuity of $ f(t, y) $ on a domain containing $ (t_0, y_0) $ is sufficient to guarantee the existence of at least one local solution. However, continuity alone does not guarantee uniqueness.

uniqueness is harder

consider $ y' = y^{1/3} $

easy to solve as a separable

\[ \begin{aligned} y^{-1/3} \, dy &= dt \\[6pt] \int y^{-1/3} \, dy &= \int dt \\[6pt] \frac{3}{2} y^{2/3} &= t + c \\[6pt] y^{2/3} &= \frac{2}{3} t + c \end{aligned} \]

Counterexample to Uniqueness: For the initial value problem with $ y(0) = 0 $ (where $ c = 0 $), solving gives $ y(t) = \pm \left(\frac{2}{3}t\right)^{3/2} $. Notice that the trivial solution $ y(t) \equiv 0 $ is also a valid solution with the same initial value. Uniqueness fails here because the partial derivative $ \frac{\partial f}{\partial y} = \frac{1}{3y^{2/3}} $ is discontinuous at $ y = 0 $, violating the conditions of the Picard–Lindelöf theorem.

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Non-Uniqueness of Solutions: $ y' = y^{1/3} $

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

Note $ y = 0 $ is a solution to $ y' = y^{1/3} $

but in the solution above, $ y = \left(\frac{2}{3}t + c\right)^{3/2} $, if $ y(0) = 0 $,

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

two solutions out of $ y(0) = 0 $?

Concept Explanation: Failure of Uniqueness

This example illustrates why the Picard–Lindelöf (Existence and Uniqueness) Theorem requires continuity of the partial derivative $ \frac{\partial f}{\partial y} $.

Here, $ f(t, y) = y^{1/3} $, which is continuous at $ y = 0 $.
However, $ \frac{\partial f}{\partial y} = \frac{1}{3}y^{-2/3} = \frac{1}{3y^{2/3}} $, which approaches $ \infty $ as $ y \to 0 $ and is undefined at $ y = 0 $.
Because $ f $ is not Lipschitz continuous around $ y = 0 $, uniqueness fails, giving multiple solutions: the trivial solution $ y(t) = 0 $ and the power solution $ y(t) = \left(\frac{2}{3}t\right)^{3/2} $ for $ t \ge 0 $.

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Slope Field for $ y' = y^{1/3} $ demonstrating Non-Uniqueness at Origin

Figure: Slope field and solution curves for differential equation y prime equals y to the one-third power demonstrating multiple solutions at the origin Show Details

Slope field for the autonomous first-order differential equation $ y' = y^{1/3} $. The vector field slopes increase with $ |y| $ and are horizontal along the line $ y = 0 $. At the initial value $ y(0) = 0 $, two distinct solutions branch forward for $ t \ge 0 $: the solid blue line representing the equilibrium solution $ y(t) = 0 $, and the dashed dark-blue curve representing the secondary branch $ y(t) = \left(\frac{2}{3}t\right)^{3/2} $. Solid red and purple curves illustrate trajectories with initial conditions $ y(0) = 0.25 $ and $ y(0) = -0.1 $, respectively.

Summary of Solution Curves and Initial Conditions

Differential equation solution curves and initial conditions illustrated in the slope field
Curve Label	Solution Formula $ y(t) $	Initial Condition	Trajectory Behavior
Equilibrium	$ y(t) = 0 $	$ y(0) = 0 $	Constant zero solution along the $ t $-axis
Secondary Branch	$ y(t) = \left(\frac{2}{3}t\right)^{3/2} $	$ y(0) = 0 $	Branches away from $ y = 0 $ for $ t \ge 0 $ (Non-uniqueness)
Upper Solution	$ y(t) = \left(\frac{2}{3}t + 0.25^{2/3}\right)^{3/2} $	$ y(0) = 0.25 $	Distinct positive trajectory passing through $ (0, 0.25) $
Lower Solution	$ y(t) = -\left(\frac{2}{3}t + 0.1^{2/3}\right)^{3/2} $	$ y(0) = -0.1 $	Distinct negative trajectory passing through $ (0, -0.1) $

Analysis of Non-Uniqueness (Picard–Lindelöf Theorem Failure)

Consider the initial value problem $ y' = f(t, y) = y^{1/3} $ with initial value $ y(0) = 0 $.

Existence: The function $ f(t, y) = y^{1/3} $ is continuous everywhere on $ \mathbb{R}^2 $, so by Peano's Existence Theorem, at least one solution exists.
Failure of Uniqueness: The partial derivative with respect to $ y $ is: \[ \frac{\partial f}{\partial y} = \frac{1}{3} y^{-2/3} = \frac{1}{3y^{2/3}} \] As $ y \to 0 $, $ \frac{\partial f}{\partial y} \to \infty $. Thus, $ f(t, y) $ is not Lipschitz continuous in any neighborhood containing $ y = 0 $.

Because the Lipschitz condition is violated at the origin, uniqueness is not guaranteed. As depicted, both the constant equilibrium solution $ y(t) = 0 $ and the non-trivial curve $ y(t) = \left(\frac{2}{3}t\right)^{3/2} $ satisfy the initial condition $ y(0) = 0 $ for $ t \ge 0 $.

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Sensitivity of Slope and Partial Derivatives

What happens here is that for a tiny change in $ y $, the slope can change dramatically near $ (0, 0) $.

To track the change of slope of $ y' = f(t, y) $ with respect to $ y $, we look at $ \frac{\partial f}{\partial y} $.

For example:

\[ y' = y^{1/3} \]

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

If $ y $ is small, small changes in $ y $ can make $ \frac{\partial f}{\partial y} $ big.

Conceptual Note on Existence and Uniqueness:

By the Picard–Lindelöf theorem, uniqueness of solutions to an initial value problem $ y' = f(t, y) $ typically requires $ \frac{\partial f}{\partial y} $ to be continuous (or Lipschitz continuous in $ y $). When $ y = 0 $, $ \frac{\partial f}{\partial y} = \frac{1}{3 y^{2/3}} $ is unbounded, which can lead to non-unique solutions through the point $ (0, 0) $.

(continue next time)

Curve Label	Solution Formula \( y(t) \)	Initial Condition	Trajectory Behavior
Equilibrium	\( y(t) = 0 \)	\( y(0) = 0 \)	Constant zero solution along the \( t \)-axis
Secondary Branch	\( y(t) = \left(\frac{2}{3}t\right)^{3/2} \)	\( y(0) = 0 \)	Branches away from \( y = 0 \) for \( t \ge 0 \) (Non-uniqueness)
Upper Solution	\( y(t) = \left(\frac{2}{3}t + 0.25^{2/3}\right)^{3/2} \)	\( y(0) = 0.25 \)	Distinct positive trajectory passing through \( (0, 0.25) \)
Lower Solution	\( y(t) = -\left(\frac{2}{3}t + 0.1^{2/3}\right)^{3/2} \)	\( y(0) = -0.1 \)	Distinct negative trajectory passing through \( (0, -0.1) \)

1st-Order Eqs, Existence and Uniqueness of Solutions

Let's start with linear eqs.

Existence of Solutions for First-Order Linear ODEs

Uniqueness of Solutions: Can Solutions Branch?

Example: Linear First-Order Initial Value Problem

Uniqueness Proof: Homogeneous Differential Equation

Existence & Uniqueness Theorem for Linear First-Order ODEs

Example

Properties of Linear Equations: General Solutions

Nonlinear Equations: Existence and Uniqueness

Non-Uniqueness of Solutions: \( y' = y^{1/3} \)

Slope Field for \( y' = y^{1/3} \) demonstrating Non-Uniqueness at Origin

Summary of Solution Curves and Initial Conditions

Analysis of Non-Uniqueness (Picard–Lindelöf Theorem Failure)

Sensitivity of Slope and Partial Derivatives

Table of Contents

Let's start with linear eqs.

Existence of Solutions for First-Order Linear ODEs

Uniqueness of Solutions: Can Solutions Branch?

Example: Linear First-Order Initial Value Problem

Example

Example Derivation

Key Takeaways

Nonlinear Equations: Existence and Uniqueness

Summary of Solution Curves and Initial Conditions

Analysis of Non-Uniqueness (Picard–Lindelöf Theorem Failure)