-1
$\begingroup$

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 $\mathbf R$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ part b) is false. $\endgroup$ – Artem Dec 20 '16 at 20:58
  • $\begingroup$ 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. $\endgroup$ – probably_someone Dec 20 '16 at 21:33
  • $\begingroup$ I would comment on your own answer, but I do not have sufficient reputation to do so. $\endgroup$ – probably_someone Dec 20 '16 at 21:33
2
$\begingroup$

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 $C^\infty$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.