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Lesson 3: Existence and Uniqueness of Solutions

Estimated time: 25-30 minutes

Learning Objectives

By the end of this lesson, you will be able to:

State the Existence and Uniqueness Theorem for first-order IVPs
Check the hypotheses of the theorem for a given IVP
Identify situations where existence or uniqueness may fail
Understand the concept of the interval of existence

Why Existence and Uniqueness Matter

Before spending effort solving an IVP, it is wise to ask two fundamental questions:

Existence: Does a solution exist at all?
Uniqueness: If a solution exists, is it the only one?

If uniqueness fails, the same initial condition could lead to multiple different futures -- a serious problem in modeling physical systems where we expect deterministic behavior.

The Picard-Lindelöf Theorem

Existence and Uniqueness Theorem (Picard-Lindelöf): Consider the IVP

dy/dx = f(x, y), y(x₀) = y₀

If f(x, y) and ∂f/∂y are both continuous in some rectangle R containing the point (x₀, y₀), then there exists an interval I containing x₀ on which a unique solution y = φ(x) to the IVP exists.

        Key Points About the Theorem
        Continuity of f alone guarantees existence (Peano's theorem).
Continuity of ∂f/∂y additionally guarantees uniqueness.
The theorem is local: it guarantees a solution near (x0, y0), not necessarily for all x.
If the hypotheses fail at a point, the theorem says nothing -- a solution might still exist, but we cannot be sure.

      

Applying the Theorem

Example 1: Checking Hypotheses

Does the IVP dy/dx = x² + y, y(0) = 1 have a unique solution near (0, 1)?

Solution:

Step 1: Identify f(x, y) = x² + y.

Step 2: Check continuity of f. Polynomials are continuous everywhere. ✔

Step 3: Compute ∂f/∂y = 1. This is continuous everywhere. ✔

Conclusion: Both conditions are satisfied in any rectangle around (0, 1). The IVP has a unique solution on some interval containing x = 0.

Example 2: Where Uniqueness Fails

Consider the IVP dy/dx = y^1/3, y(0) = 0.

Solution:

Step 1: f(x, y) = y^1/3 is continuous for all y. ✔ (existence is guaranteed)

Step 2: ∂f/∂y = (1/3)y^-2/3 = 1/(3y^2/3).

At y = 0, this is undefined (not continuous). ✘

Conclusion: The uniqueness hypothesis fails at the initial point. Indeed, this IVP has multiple solutions:

y₁(x) = 0 (the trivial solution)

y₂(x) = (2x/3)^3/2 for x ≥ 0

Both satisfy the DE and the initial condition y(0) = 0.

Example 3: Checking a Rational Function

Does the IVP dy/dx = y/(x - 1), y(0) = 2 have a unique solution near x = 0?

Solution:

f(x, y) = y/(x - 1) is continuous as long as x ≠ 1.

∂f/∂y = 1/(x - 1) is continuous as long as x ≠ 1.

Since (0, 2) is not near x = 1, both conditions hold in a rectangle around (0, 2) that stays away from x = 1.

Conclusion: A unique solution exists on some interval containing x = 0 (at least on an interval within (-∞, 1)).

Interval of Existence

The theorem guarantees a solution exists on some interval, but that interval may not extend to all of (-∞, ∞). A solution can cease to exist if it:

Blows up (goes to ±∞ in finite time), or
Reaches a point where f or ∂f/∂y is discontinuous.

Example 4: Finite-Time Blow-Up

Solve dy/dx = y², y(0) = 1.

Solution: This is separable: dy/y² = dx. Integrating: -1/y = x + C. From y(0) = 1: C = -1. So y = 1/(1 - x).

As x → 1^-, y → +∞. The solution exists only on (-∞, 1). Despite f(x,y) = y² being smooth everywhere, the solution blows up in finite time.

Check Your Understanding

1. For the IVP dy/dx = sin(xy), y(0) = 3, does the Existence and Uniqueness Theorem guarantee a unique solution?

Answer: f(x,y) = sin(xy) is continuous everywhere. ∂f/∂y = x cos(xy) is also continuous everywhere. Both hypotheses are satisfied, so yes, a unique solution is guaranteed near (0, 3).

2. For dy/dx = sqrt(y), y(0) = 0, does uniqueness hold?

Answer: f(x,y) = y^1/2 is continuous for y ≥ 0. But ∂f/∂y = 1/(2y^1/2) is not continuous at y = 0. Uniqueness is not guaranteed. Indeed, both y = 0 and y = x²/4 satisfy the IVP.

3. For dy/dx = ln(y), y(1) = 1, does the theorem apply?

Answer: f(x,y) = ln(y) is continuous for y > 0. ∂f/∂y = 1/y is continuous for y > 0. Since y₀ = 1 > 0, both conditions hold. Yes, a unique solution exists near (1, 1).

4. Why does the solution to dy/dx = y², y(0) = 1 not exist for all x?

Answer: The solution y = 1/(1-x) blows up to infinity as x approaches 1. The theorem only guarantees a local solution; global existence depends on the specific behavior of the equation.

5. State the two conditions needed for the Picard-Lindelöf theorem to guarantee a unique solution.

Answer: (1) f(x, y) must be continuous in a rectangle containing (x₀, y₀). (2) ∂f/∂y must be continuous in that same rectangle.

Key Takeaways

The Existence and Uniqueness Theorem requires continuity of both f and ∂f/∂y near the initial point.
Continuity of f alone guarantees existence; adding continuity of ∂f/∂y ensures uniqueness.
When ∂f/∂y fails to be continuous (e.g., at y = 0 for f = y^1/3), multiple solutions may pass through the same initial point.
Even when a unique solution exists, it may only be defined on a finite interval (blow-up in finite time).
Always check the hypotheses before applying the theorem.

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