0

Consider the following initial value problem $$ y' = y+ \frac12 |\sin(y^2)|,\,\,\,\,\,\, x>0,\,\,\, y(0) = -1$$ Which of the following statements are true?
1.) there exists an $\alpha \in (0,\infty)\,\,$ s.t. $\lim_{x \to \alpha^-}|y(x)| = \infty$
2.) y(x) exists on $(0,\infty)\,\,$ and its monotone.
3.) y(x) exists on $(0,\infty)\,\,$ but not bounded below.
4.) y(x) exists on $(0,\infty)\,\,$ but not bounded above.

I thought about using the following inequality

$$\because 0 \le |\sin(y^2)| \le 1\,\,,\forall x$$ Therefore the ODE can be expressed as

$$\because 0 \le y'-y \le \frac12 \,\,,\forall x$$

Post this step I'm not sure how we can proceed further. Any approach on how to solve this question would help a lot.

Thanks!

Ark
  • 359
  • multiply by $e^{-x}$ and integrate – Didier Jul 11 '24 at 18:18
  • But integration would introduce, constant integration wouldn't that interfere with the inequality? – Ark Jul 11 '24 at 18:22
  • integrate between 0 and x – Didier Jul 11 '24 at 18:32
  • This question suggest that we still need to add C https://math.stackexchange.com/questions/4348312/how-does-the-fundamental-theorem-of-calculus-help-to-simplify-difficult-exponent – Ark Jul 11 '24 at 18:41
  • 1
    There is no constant to add when you integrate over a definite interval. You already have the value of $y(0)$. – Didier Jul 11 '24 at 18:45

2 Answers2

4

You have $0 \leqslant y' - y \leqslant \frac{1}{2}$. Multiplying by $e^{-x}$ yields $0 \leqslant y'(x)e^{-x} - y(x)e^{-x} \leqslant \frac{1}{2}e^{-x}$. Integrating on $[0,x]$, after noticing that $y'(x)e^{-x} - y(x)e^{-x} = (y(x)e^{-x})'$, yields $$ 0 \leqslant y(x)e^{-x} +1 \leqslant \frac{1}{2}(1- e^{-x}). $$ Here, we have used that $y(0)=-1$. Playing around with the inequalities now gives $$ -e^x \leqslant y(x) \leqslant -\frac{1}{2}(e^x +1). $$ You should be able to conclude now regarding the boundedness of $y$. For its monotonicity, as pointed out by @HyperbolicPDEfriend, this follows from the fact that $y$ is solution to a first order autonomous ODE.

Notice that you should first justify existence of the solution (Picard-Lindelöf). Once this is done, the above thing shows that $y$ is bounded on any compact interval of the form $[0,A]$. Since $y'$ is close to $y$ (from $|y'-y| \leqslant 1/2$), then so is $y'$. It follows from classical ODE theory that the maximal interval of definition of $y$ cannot be bounded, and thus contains $[0,+\infty)$.

Didier
  • 20,820
2

You should check first that the right hand side is locally Lipschitz in $y$. Picard-Lindelöf implies existence and uniqueness.

Under these circumstances, solutions to first order autonomous ODEs (like this one) are always monotone, if the reuqirements for Picard-Lindelöf are met. Further note, that the RHS grows linearly, i.e. in this case $$ \left \lvert y + \tfrac{1}{2}\lvert \sin(y^2) \rvert \right \rvert \leq \lvert y \rvert + \tfrac{1}{2}\lvert \sin(y^2) \rvert \leq \lvert y \rvert + \tfrac{1}{2}. $$ In this case, the solution always exists on $\mathbb{R}$. The solution is bounded from above e.g. by $0$, because $0$ is a solution and cannot be crossed due to uniqueness.

You can show that it is not bounded from below using an ODE comparison principle.

I'm in a rush, but if you need details/sources, let me know.

Hyperbolic PDE friend
  • 6,919