# One-dimensional initial value problem

Consider the initial value problem

$$y'(t)=f\bigl(y(t)\bigr),\qquad y(0)=a \in \mathbf R$$

where $f\colon \mathbf R \to \mathbf R$. Which of the following statements are necessarily true?
a) There exists a continuous function $f\colon \mathbf R \to \mathbf R$ and $a \in \mathbf R$ such that the above problem does not have a solution in any neighbourhood of $0$.
b) When $f$ is twice continuously differentiable, the maximal interval of existence for the above initial value problem is $\mathbf R$.

Assuming $\mathbf{R}=\mathbb{R}$:

a.) is false, by the standard existence and uniqueness theorems for first-order ODEs. See http://www.math.uiuc.edu/~tyson/existence.pdf.

b.) is true, and to see that it is, we achieve the maximum (note that $\mathbb{R}$ itself is the largest possible maximum in $\mathbb{R}$, so this is sufficient). Let $f(y(t))=0$, and we have the solution $y(t)=a$, which exists everywhere.

• part b) is false. – Artem Dec 20 '16 at 20:58
• It says "maximal" for a reason. Unless we're interpreting the question differently; my interpretation is, "There is an $f$ that is $C^2$ such that the interval of existence is $\mathbb{R}$." I believe your interpretation is, "For any $f$ that is $C^2$, the interval of existence is $\mathbb{R}$." Since, according to conventional definition, the maximum need not be achieved everywhere in $C^2$, I believe my interpretation is correct. – probably_someone Dec 20 '16 at 21:33
• I would comment on your own answer, but I do not have sufficient reputation to do so. – probably_someone Dec 20 '16 at 21:33

a) Is FALSE. This is a consequence of Peano's Theorem.

b) Is FALSE. For example $$y'=y^2, \quad y(0)=1,$$ then the unique solution is $$y(t)=\frac{1}{1-t}.$$ Clearly, the solution is NOT globally defined, as it blows up at $t=1$, while $f=y^2$ is $C^\infty$.