Initial Values, Existence and Uniqueness

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The idea

An initial-value problem combines $y'=f(t,y)$ with $y(t_0)=y_0$ . Geometrically, it asks for a solution curve through one point. Initial data select among solution curves, but do not automatically guarantee a solution for all future time.

Visual guide

VISUAL GUIDEThe same initial point can have several continuations

Departure at a = 0
Departure at a = 1
Departure at a = 2

For y′ = 2√|y| and y(0) = 0, a solution may stay zero until a chosen a ≥ 0, then follow (t − a)². These curves agree initially but depart at different times because the uniqueness hypothesis fails at y = 0.

Method and assumptions

Continuity of $f$ near the initial point ensures a local solution. Continuity of $f_y$ there is a convenient sufficient condition for local uniqueness. More generally a local Lipschitz condition in $y$ suffices. These are sufficient conditions: failure of a test is not itself proof of nonuniqueness.

Worked example: unique but not global

For $y'=y^2$ , $y(0)=1$ , both $f$ and $f_y=2y$ are smooth. The unique solution is $1/(1-t)$ , but it diverges at $t=1$ . Local uniqueness does not prevent finite-time blow-up.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

A local existence and uniqueness theorem applies at (t₀,y₀). What does it guarantee?

Every initial value gives the same solution

One solution on some interval around t₀

A solution for every real t

Hint 1 · Find a starting point

Pay attention to the word “local.”

Hint 2 · Take the next step

The interval can end when a solution blows up or leaves the theorem’s domain.

Show the reasoning

Answer: One solution on some interval around t₀

The theorem gives a unique solution nearby. Global existence requires additional information.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: multiple departures

For $y'=2\sqrt{|y|}$ and $y(0)=0$ , $y=0$ is a solution. So is a curve that stays zero until any $a\ge0$ and then follows $(t-a)^2$ . At the joining point both derivatives are zero. The right side is continuous, but is not locally Lipschitz in $y$ at zero.

Interpreting the result

For a linear equation $y'+p(t)y=q(t)$ , continuous coefficients on an interval give a unique solution throughout that interval. Divide by the coefficient of $y'$ only where it is nonzero.

Practice

Solve $y'=-y$ , $y(0)=2$ .
Where can the standard theorem apply to $t y'+y=1$ ?
Does a missing continuous $f_y$ prove multiple solutions?

Show worked solutions

$y=2e^{-t}$ , defined for every real $t$ .
On intervals avoiding $t=0$ , where division produces continuous coefficients.
No. It only makes this sufficient test inconclusive; inspect the equation directly.

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.

MAKE IT YOURS

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