How to solve the initial value problem $y'(x)=\lambda \sin(x+y(x))$, $y(0)=1$.

Question

For $\lambda \in \mathbb{R}$, consider the initial value problem $y'(x)=\lambda \sin(x+y(x))$, $y(0)=1$. Then this initial value problem has

1. no solution in any neighbourhood of $0$.
2. a solution in $\mathbb{R}$ if $|\lambda|<1$
3. a solution in a neighbourhood of $0$.
4. a solution in $\mathbb{R}$ only if $|\lambda|>1$.

This is the first time I have encountered this kind of IVP, and have no idea to proceed. The entanglement of $x$ and $y(x)$ in $\sin(x+y(x))$ is what causing me the trouble to make any headway. So please help me to solve this. Thanks.

Answer 1

Hint: Observe we have \begin{align} y' = F(x, y) \end{align} where $F$ is Lipschitz in $y$ variable since \begin{align} |F(x, y_1) -F(x, y_2)| = |\lambda| |\sin(x+y_1)-\sin(x+y_2)| \leq |\lambda||y_1-y_2|. \end{align} Note we have used the fact $|\sin u-\sin v| \leq |u-v|$.

Now, by Picard-Lindelof theorem, one can guarantee local existence, i.e. (3) holds if we do not know anything about $\lambda$.

Moreover, one can use a Banach fixed point argument to show that the ode has a global solution, i.e. solution on all of $\mathbb{R}$ if $|\lambda|<1$, which means (2) holds.

Answer 2

You first make a string of numbers for the range of $x$. Let's say $x\in[0,1]$. You could select $x=0, 0.1 , 0.2, ..., 1.0$.

Next, substitute $x=0$ to the equation to get: $y^\prime(0) = \lambda \text{sin}(0+y(0)) = 0$.

Then use:

$y(0.1) = y(0) + 0.1 y^\prime(0) = 0$.

Next, substitute $x=0.1$ in the equation:

$y^\prime(0.1) = \lambda \text{sin}(0.1+y(0.1)) = \lambda \text{sin}(0.1)$

You continue the iteration up to $x=1.0$. The finer the meshing, the more precise the answer.