Existence and Uniqueness
Ch 2 — Part I — Explicit & Qualitative Methods
When does \(y' = f(t,y),\; y(t_0) = y_0\) have a solution, and when is it unique? The Picard–Lindelöf theorem gives a clean sufficient condition.
Picard's theorem
If \(f(t,y)\) is continuous and \(\partial f/\partial y\) is continuous on a rectangle around \((t_0, y_0)\), then the IVP has a unique solution on some open interval containing \(t_0\).
Things that can go wrong
- Non-uniqueness: \(y' = \sqrt{|y|},\; y(0)=0\) admits both \(y\equiv 0\) and \(y(t) = t^2/4\) (for \(t\ge 0\)). Here \(\partial f/\partial y\) blows up at \(y=0\).
- Finite-time blow-up: \(y' = y^2,\; y(0)=1\) gives \(y(t) = 1/(1-t)\) — only defined for \(t < 1\). The solution exists but does not extend globally.
Linear case is nicer
For linear ODEs \(y' + p(t) y = g(t)\) with \(p,g\) continuous on an interval \(I\), solutions exist and are unique on all of \(I\). No blow-up surprises.
Practice Problems
Hint
Solution
\(\partial_y(y^{1/3}) = \tfrac{1}{3} y^{-2/3}\) is undefined at \(y=0\). Uniqueness fails: both \(y\equiv 0\) and \(y(t) = (2t/3)^{3/2}\) are solutions.
Answer: No — non-unique.
Hint
Solution
\(\int dy/(1+y^2) = \int dt\Rightarrow \arctan y = t + C\). \(C=0\), so \(y = \tan t\). Blows up at \(t = \pi/2\).
Answer: \(t \in (-\pi/2, \pi/2)\).
Hint
Solution
We need \(p, g\) continuous on \(I\). The solution exists and is unique on all of \(I\).
Answer: \(p, g\) continuous on \(I\).