For $\lambda \in \mathbb R $, the I.V.P $\frac{dy}{dx} = \lambda \sin(x+y(x))$, with $y(0) = 1$ has
- No solution in any neighborhood of $0$.
- No solution in $\mathbb R $ if $| \lambda|\ \lt 1$.
- A solution in a neighborhood of $0$.
- A solution in $\mathbb R $ only if $| \lambda |\gt 1$.
How to proceed and solve it ? Any insight .
Thanks.
According to the Picard-Lindelöf Theorem there exists a solution in a neighborhood of $0$ if the right hand side is continuous in $x$ and uniformly Lipschity in $y$. Here, the right hand side is $f(x,y) = \lambda \sin(x+y)$. Note that in the formulation of the theorem, when checking for continuity w.r.t. $x$, $y$ is viewed as a variable, and not as a function of $x$ (even though it depends on $x$ in the full ODE). The theorem requires assumptions on the r.h.s. function $f$, not on the r.h.s. of the ODE! Since we do not know anything about the solution $y(x)$ a priori, it doesn't make sense to check for continuity of $x\mapsto f(x, y(x))$, only for $x\mapsto f(x,y)$.
You mentioned that you have already shown continuity of $f$ w.r.t. $x$. It remains to show uniform Lipschitz continuity w.r.t. $y$, uniform in $x$. Recall that the global Lipschitz constant of a differentiable function is the supremum of the modulus of the first derivative, i.e. here: $$L = \sup_{y\in\mathbb R}\Big|\frac{d}{dy} f(x,y)\Big| = \sup_{y\in\mathbb R}|\lambda\cos(x+y) | = |\lambda|.$$ So $|\lambda|$ is a Lipschitz constant of $f$ w.r.t. $y$, and since it is independent of $x$, it is uniform. Hence, we have shown that the assumptions of the Picard-Lindelöf Theorem are fulfilled, and therefore we know that the ODE has a solution for $x$ in a neighborhood of $0$.
A solution exists for any choice of $\lambda$, so the other options are indeed false.
Additional remark: It is not much additional work to see that a solution is global (i.e. exists for all $x>0$). From the ODE we immediately find that $|y'|\le |\lambda|$, i.e. a solution never grows or falls at a rate greater than $|\lambda|$. Therefore a solution with IC $y(0) = 1$ will always stay between the functions $1+|\lambda|x$ and $1-|\lambda| x$ for $x>0$. This is equivalent to $y|y(x)-1|\le |\lambda |x$. Now we observe that a solution of an ODE satisfying the Picard-Lindelöf Theorem is not global iff it becomes unbounded (i.e. it blows up) in finite time (see the detailed proof for more details). But with the above trick we have seen that this cannot happen - it stays bounded (even though the bound grows; but it is enough to prevent a blow-up). Therefore any solution is global. However, this is not nedded for the solution of your problem.