Exercise · ZW-EXISTENCE-AND-UNIQUENESS-THEOREM-010 · ★★★ Advanced
Does the theorem guarantee a unique solution of \(y'=y^{1/3},\ y(0)=1\)?
Choose one option.
Show answer
- A. Yes, because \(f_y=\tfrac13 y^{-2/3}\) is continuous near \((0,1)\)
Explanation
What matters is continuity near the initial point \((0,1)\), not on the whole plane. There \(f=y^{1/3}\) and \(f_y=\tfrac13 y^{-2/3}\) are continuous, so there is a unique solution on some interval around \(0\).
Review the concept: Existence and uniqueness theorem
The existence and uniqueness theorem guarantees that an initial value problem has a unique solution near the initial point, if the equation and its partial derivative with respect to the unknown are continuous there.
\[f,\ \frac{\partial f}{\partial y}\ \text{continuous near }(x_0,y_0)\ \Longrightarrow\ \text{a unique solution}\]
More practice problems: existence and uniqueness theorem
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