An initial value problem (IVP) pairs an equation $y' = f(t, y)$ with a starting value $y(t_0) = y_0$. In Direction Fields and Euler's Method you traced a curve from a point by following the slopes. It is natural to assume exactly one curve leaves each point. Usually that is true, but not always, and the exceptions are worth knowing.
Key idea
Linear case. For $y' + p(t)y = g(t)$, if $p$ and $g$ are continuous on an open interval $I$ containing $t_0$, the unique solution exists on all of $I$.
Why it works
When uniqueness fails
Worked example: Two solutions from one point
- $y = 0$ works: both sides are 0.
- Try $y = t^{3}$: $y' = 3t^{2}$ and $3y^{2/3} = 3(t^{3})^{2/3} = 3t^{2}$. It works too, with $y(0) = 0$.
- In fact $y = 0$ until any time $c \ge 0$, followed by $y = (t - c)^{3}$, is also a solution: infinitely many.
- Here $\partial f/\partial y = 2y^{-1/3}$, which is undefined at $y = 0$, so the theorem never promised uniqueness.
Answer: The IVP has more than one solution.
Common mistake
How far a solution lasts
For a linear equation the answer comes from the coefficients alone: divide by the leading coefficient to reach standard form, list every $t$ where $p$ or $g$ is undefined or discontinuous, and take the gap that contains $t_0$.
Worked example: The interval for a linear IVP
- Standard form: $y' + \frac{\ln t}{t - 3}y = \frac{2t}{t - 3}$.
- $p = \frac{\ln t}{t - 3}$ needs $t > 0$ and $t \ne 3$; $g = \frac{2t}{t - 3}$ needs $t \ne 3$.
- The trouble points split the line into $(0, 3)$ and $(3, \infty)$; $t_0 = 1$ lies in $(0, 3)$.
Answer: $(0, 3)$
Worked example: A short life without any trouble point
- $f = y^{2}$ and $\partial f/\partial y = 2y$ are continuous everywhere, so there is exactly one solution near $t = 0$.
- Separating variables gives $y = \frac{1}{1 - t}$.
- It blows up as $t \to 1^{-}$, even though the equation looks perfectly smooth.
Answer: Unique, but it exists only for $t < 1$. For nonlinear equations the theorem promises just some interval.