Uniqueness failure in an initial-value problem
A student studies the initial-value problem y′(t) = √|y(t)|, y(0)=0. The right-hand side f(y)=√|y| is continuous everywhere, including at y=0. The student recalls the Picard–Lindelöf theorem: if f is continuous and Lipschitz in y on a rectangle around the initial point, then a unique solution exists on some interval. They check continuity, declare the hypotheses satisfied, and conclude that y(t)=0 is the unique solution. A classmate points out that the theorem also requires a Lipschitz condition, and that the square-root function has an infinite derivative at 0. The student asks whether non-uniqueness can actually occur, how to construct a second solution, and what goes wrong in the Lipschitz check. They also wonder whether the solution can be extended for all t.
1. Show that f(y)=√|y| is not Lipschitz in any neighbourhood of y=0.
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At y=0, consider the Lipschitz quotient |f(y)−f(0)|/|y−0| = √|y|/|y| = 1/√|y|. As y→0, this tends to +∞. Hence there is no constant L with |f(y)−f(0)|≤L|y| near 0, so f is not Lipschitz on any neighbourhood of 0.
2. Construct two distinct solutions of the IVP and verify them.
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Two solutions are y₁(t)=0 for all t and y₂(t)=0 for t≤0, y₂(t)=t²/4 for t≥0. For y₂, at t>0, y₂′=t/2 and √|y₂|=√(t²/4)=t/2; at t<0, y₂′=0 and √|0|=0; at t=0, both one-sided derivatives are 0 and √|y₂(0)|=0. Thus y₂ satisfies the IVP. More generally, for any a≥0, y_a(t)=0 for t≤a and (t−a)²/4 for t≥a is a solution, so uniqueness fails badly.
3. Which hypothesis of Picard–Lindelöf fails, and does continuity alone guarantee uniqueness?
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The failing hypothesis is the Lipschitz condition in y near the initial value. Continuity of f alone guarantees existence (Peano) but not uniqueness. Picard–Lindelöf cannot be applied here because its Lipschitz hypothesis is false at y=0. The example shows that a continuous right-hand side can produce infinitely many solutions through the same initial point.