Mathematics.

ordinary differential equations

Existence and Uniqueness of ODE Solutions

Differential Equations30 minDifficulty7 out of 10

You should know: first order differential equation

Overview

Before solving an initial value problem y′ = f(x,y), y(x₀) = y₀, it is worth asking whether a solution exists at all, and if so whether it is the only one. The Picard–Lindelöf theorem (also called the Picard existence theorem) answers both questions: if f(x,y) is continuous and Lipschitz continuous in y near (x₀,y₀), then a unique solution exists on some (possibly small) interval around x₀. Continuity alone (Peano's existence theorem) guarantees existence but not uniqueness, and the interval of existence can be much shorter than the interval where f is defined — solutions can 'blow up' in finite time even when f itself is perfectly smooth everywhere.

Intuition

Continuity of f alone only guarantees the direction field is well-defined at every point, which is enough to guarantee SOME curve threads through it (Peano's theorem) — but continuity doesn't prevent two different curves from touching at (x₀,y₀) and splitting apart afterward, as happens when f is 'too steep' in y (like y^{1/3} near y=0). The Lipschitz condition is a bound on how fast f can change with y — it prevents nearby solution curves from ever being forced to converge to exactly the same point and then diverge, ruling out the branching that causes non-uniqueness. Separately, even a perfectly well-behaved (Lipschitz, even polynomial) f can cause a solution to escape to infinity in finite x — existence is only guaranteed locally, near x₀, not necessarily for all x.

Formal Definition

Definition

The Picard–Lindelöf theorem gives sufficient conditions for a unique local solution to an initial value problem:

y′=f(x,y),y(x0)=y0y' = f(x,y), \qquad y(x_0) = y_0
Initial value problem (IVP)
∣f(x,y1)−f(x,y2)∣≤L ∣y1−y2∣(Lipschitz condition in y, near (x0,y0))|f(x,y_1) - f(x,y_2)| \le L\,|y_1 - y_2| \quad \text{(Lipschitz condition in } y\text{, near } (x_0,y_0))
Lipschitz condition
  ⟹  ∃ ε>0 such that a UNIQUE solution y(x) exists on (x0−ε, x0+ε)\implies \exists\, \varepsilon > 0 \text{ such that a UNIQUE solution } y(x) \text{ exists on } (x_0-\varepsilon,\ x_0+\varepsilon)
Conclusion: local existence and uniqueness

Worked Examples

  1. This is separable: dy/y² = dx.

    ∫dyy2=∫dx  ⟹  −1y=x+C\int \frac{dy}{y^2} = \int dx \implies -\frac{1}{y} = x + C

  2. Apply y(0)=1 to find C, then solve for y.

    −1=0+C  ⟹  C=−1  ⟹  −1y=x−1  ⟹  y=11−x-1 = 0 + C \implies C=-1 \implies -\frac{1}{y} = x - 1 \implies y = \frac{1}{1-x}

  3. f(x,y)=y² is smooth (polynomial, hence Lipschitz in y) everywhere, so Picard–Lindelöf guarantees a unique LOCAL solution — but the solution itself has a vertical asymptote at x=1.

    y=11−x→∞ as x→1−y = \frac{1}{1-x} \to \infty \text{ as } x \to 1^-

Answer: y = 1/(1-x), which exists uniquely only on (-∞, 1) despite f(x,y)=y² being defined and smooth for all real x, y — a finite-time blowup (verified: y'=1/(1-x)^2=y^2 ✓).

Practice Problems

Difficulty 6/10

For y′ = y², y(0) = 2, find the solution and its interval of existence.

Difficulty 6/10

Is f(x,y) = y² Lipschitz in y on the strip |y| ≤ 5? Is it Lipschitz on all of ℝ?

Difficulty 7/10

Explain why y′ = y^{1/3} with y(0) = 1 (not y(0)=0) does NOT exhibit the non-uniqueness seen at y(0)=0.

Common Mistakes

Common Mistake

Assuming that because f(x,y) is smooth (infinitely differentiable), a solution to y′=f(x,y) must exist for all x.

Smoothness only gives LOCAL existence and uniqueness; the maximal interval of existence can still be finite, as in y′=y², y(0)=1, which blows up at x=1 despite f(x,y)=y² being a polynomial.

Common Mistake

Concluding that failure of the Lipschitz condition at a point means no solution exists there.

Peano's theorem only needs continuity to guarantee existence; a failure of the Lipschitz condition threatens UNIQUENESS, not existence — as seen in y′=y^{1/3}, y(0)=0, which has solutions, just more than one.

Quiz

The Picard–Lindelöf theorem guarantees a unique solution to y′=f(x,y), y(x₀)=y₀ when:
The solution y=1/(1-x) to y′=y², y(0)=1 illustrates that:
Non-uniqueness of solutions to y′=y^{1/3}, y(0)=0 occurs because:

Summary

  • The Picard–Lindelöf theorem guarantees a unique local solution to y′=f(x,y), y(x₀)=y₀ when f is continuous and Lipschitz continuous in y.
  • Continuity alone (Peano's theorem) guarantees existence of some solution but not uniqueness.
  • Even smooth, globally-defined f can produce solutions that blow up in finite time, so existence is generally only local, not global.
  • Failure of the Lipschitz condition (e.g. an unbounded ∂f/∂y) can allow multiple distinct solutions through the same initial point, as in y′=y^{1/3}, y(0)=0.

References

  1. WebsiteWikipedia — Picard–Lindelöf theorem
  2. BookBoyce, W. & DiPrima, R. Elementary Differential Equations, 11th ed., Ch. 2.