Picard-Lindelöf and First-Order Ordinary Differential Equations: Examples and Counterexamples

The cited construction gives $u_m(t)={\sum }_{k=0}^m{t}^k\mathrm{/}k!$, beginning with $u_0=1$, and [L1] gives compact-uniform convergence to $\exp$.

The limit satisfies the Picard equation, and [L2] identifies it with the unique solution of $y^{\mathrm{\prime }}=y$, $y(0)=1$.

Differentiation gives $((1-t{)}^{-1}{)}^{\mathrm{\prime }}=(1-t{)}^{-2}=y(t{)}^2$ and $y(0)=1$. On the connected component of a solution's nonzero set containing $0$, [L2] gives $(1\mathrm{/}y{)}^{\mathrm{\prime }}=-1$, so integration forces $1\mathrm{/}y=1-t$. If that component had a finite boundary $c$ inside the solution interval, continuity would give $y(t)\to 0$ there and hence $\mathrm{\mid }1\mathrm{/}y(t)\mathrm{\mid }\to +\mathrm{\infty }$, while the identity gives $1\mathrm{/}y(t)\to 1-c\in \mathbb{R}$, a contradiction. Thus the component is the whole solution interval.

On every compact state interval $\mathrm{\mid }y\mathrm{\mid },\mathrm{\mid }z\mathrm{\mid }\le R$, one has $\mathrm{\mid }{y}^2-z^2\mathrm{\mid }\le 2R\mathrm{\mid }y-z\mathrm{\mid }$, so the polynomial field satisfies the local state-Lipschitz hypothesis of [L3]. The formula is defined on $(-\mathrm{\infty },1)$ and tends to $+\mathrm{\infty }$ as $t\uparrow 1$; no finite value permits continuation through $1$, while the formula continues indefinitely to the left. Thus it is the unique maximal solution from [L3], consistently with the compact-escape conclusion [L1].

On the first piece $y_c^{\mathrm{\prime }}=0=2\sqrt{\mathrm{\mid }{y}_c\mathrm{\mid }}$ by [L1], on the second $y_c^{\mathrm{\prime }}=2(t-c)=2\sqrt{(t-c{)}^2}$, and at $t=c$ both one-sided derivatives are $0$, so every $y_c$ is a solution through $(0,0)$.

Distinct delays give distinct solutions, while $\mathrm{\mid }f(y)-f(0)\mathrm{\mid }\mathrm{/}\mathrm{\mid }y\mathrm{\mid }=2\mathrm{/}\sqrt{y}$ for $y>0$, so [L2] rules out every finite local Lipschitz constant at zero.

By [L2] and [L3], $F$ takes only $0$ and $1$, and takes both values on every nondegenerate interval.

Suppose a differentiable solution existed; its derivative would equal $F$ but omit every value strictly between $0$ and $1$, contrary to [L1], so no local solution exists.

The quotient $\mathrm{\mid }f(x)-f(0)\mathrm{\mid }\mathrm{/}\mathrm{\mid }x\mathrm{\mid }=1+\mathrm{\mid }\log \mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }$ is unbounded as $x\to 0$ by [L2], so $f$ is not locally Lipschitz at zero.

Define $\varphi (r)=r(1-\log r)$ for $0<r\le e^{-1}$ with $\varphi (0)=0$, and put $\rho =2\varphi$ on $[0,e^{-1}]$. Differentiation using [L2] shows that $\varphi$ is increasing and concave on this interval. On one side of zero, concavity with $\varphi (0)=0$ gives $\mathrm{\mid }\varphi (a)-\varphi (b)\mathrm{\mid }\le \varphi (\mathrm{\mid }a-b\mathrm{\mid })$; on opposite sides, $\mathrm{\mid }a-b\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }$ and monotonicity gives $\varphi (\mathrm{\mid }a\mathrm{\mid })+\varphi (\mathrm{\mid }b\mathrm{\mid })\le 2\varphi (\mathrm{\mid }a-b\mathrm{\mid })$. Since $f$ is the odd extension of $\varphi$ near zero, $\rho$ is therefore a state modulus there. It has the properties in [L3], and the substitution $u=1-\log r$ gives its reciprocal divergence, so [L1] makes the zero solution unique through the origin.

Suppose the false claim were true; [L1] gives a continuous right-hand side with distinct solutions through the same initial value.

This contradicts the asserted uniqueness, so the claim is false.

Suppose the false claim were true; the vector field in [L1] is defined for all real times, but its solution has finite maximal endpoint $1$.

The asserted extension would continue this maximal solution past $1$, a contradiction.

Suppose the false claim were true; [L1] gives uniqueness at zero while its vector field has no local Lipschitz constant there.

This contradicts the asserted necessity, so the claim is false.