# Does this IVP have a unique solution for all $x \in \mathbb R$

Is $\displaystyle {dy\over dx}=\sin(y)$ with initial conditions $y(X)=Y$ guaranteed to have a unique solution for all $x\in\mathbb R$?

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I would suggest plotting and examining the slope field. – Ataraxia May 23 '13 at 12:37
Do you know any theorems about existence/uniqueness of solutions for IVP? Some version of : en.wikipedia.org/wiki/Picard–Lindelöf_theorem – Sam Lisi May 23 '13 at 14:29

Artem's answer is complete, though I thought I would give a few more details in case you want to learn a bit more about existence/uniqueness questions. The key part of what I have to say is the last paragraph, in which I introduce my favourite example of non-unique solutions.

First of all, the existence of solutions to an initial value problem is (perhaps surprisingly) complicated machinery. Both existence theorems I know, the Picard-Lindelöf aka Cauchy-Lipschitz existence/uniqueness theorem and the Cauchy-Peano existence theorem require some real analysis. In particular, the proofs I know set up an iteration scheme to find better and better approximations to the solution to the IVP, and show that this iteration scheme has to converge to a function. The proof of uniqueness is relatively simple, using something called Gronwall's inequality (I can give more details on this if you are interested).

In particular, a special case of Picard-Lindelöf says that if you consider an IVP of the form $y' = f(x, y)$, $y(x_0) = y_0$, and the function $f$ is continuous in $x$ and $\frac{\partial}{\partial y}$ exists and is continuous, then there exists a local solution to the initial value problem, and this local solution is unique. (A local solution is one defined for a small interval of $x$ values, $x \in (x_0 -\epsilon, x_0 + \epsilon)$ for some small value $\epsilon >0$. The theorem doesn't tell us what $\epsilon$ is, just that there is one. You can, in principle, work through the proof to get an estimate on how big $\epsilon$ is, but this is quite difficult.)

In particular, your IVP satisfies this requirement: here, $f(x,y) = \sin(y)$. This does not depend on $x$ so is obviously continuous in $x$. $\partial_y f$ exists and is continuous (even $C^\infty$ smooth as Artem points out). Thus, the theorem applies and we have a unique local solution.

Now, you are left with two questions: are these solutions defined for all $x\in \mathbb{R}$? are they globally unique?

Again, Artem gave the answer to the first question. There is a solution for all $x \in \mathbb{R}$ because, if you had a solution defined for a finite interval $x \in (a,b)$, say with $b$ finite, then any solution $y(x)$ would have to leave the domain in which $f(x,y)$ satisfies the conditions of Picard-Lindelöf as $x \to b^-$. Now, your function $f(x,y) = \sin(y)$ satisfies the conditions for all $x$ and for all $y$... thus, if $b$ were finite, $y(x)$ would have to go to $\pm \infty$ as $x \to b^-$. Artem gave two arguments to show that $y(x)$ can't actually blow up in finite time.

This is the key fact about existence of solutions: the only way a solution to an IVP can cease to exist is if you have "finite time blow-up". An example is the following IVP: $y' = y^2$, $y(0) = 1$. Note that this satisfies the requirements of Picard-Lindelöf. We can find an explicit solution by $y(x) = \frac{1}{1-x}$. This is only defined for $x < 1$. The reason the solution can't be extended to $x=1$ and beyond is that $y(x)$ does not have a limit as $x \to 1^-$.

Key example of non-uniqueness below

OK, we are now left with global uniqueness. This follows because $\sin(y)$ is actually continuously differentiable with respect to $y$ everywhere. Here is my favourite example of what can go wrong. I really think that understanding this example is key to understanding what is going on in ODE, so I highly recommend you think about it. The IVP is $$y' = 3 y^{\frac{2}{3}}, y(x_0) = y_0.$$ Note that this is of the form $y' = f(x,y)$ for $f(x,y) = 3 y^{\frac{2}{3}}$. This function $f(x,y)$ is continuous in both variables, so Cauchy-Peano's existence theorem tells us that we always have existence of a solution to the IVP. Unfortunately, Picard-Lindelöf is only verified when $y \ne 0$, so we are guaranteed local uniqueness of solutions ONLY when $y \ne 0$. When $y=0$, all bets are off.

I will now illustrate how we don't have global uniqueness of solutions.

Then, we can do like Ron did, and use separation of variables. We obtain the following: [ y(x) = (x + C)^3. ] Note that for the initial value $y = 0$, you get other solutions... notably by taking the constant solution $y(x) = 0$. You can calculate however, that you get many other solutions by taking $$y(x) = \begin{cases} (x + C)^3 &\text{ for } x < -C \\ 0 &\text{ for } -C \le x \le C' \\ (x+C')^3 &\text{ for } x > -C'. \end{cases}$$ In particular, for the initial condition $y(0) = 1$, you can find infinitely many global solutions that pass through this point BUT all these solutions that are different from each other globally are the same when you consider them in a small neighbourhood of the initial condition $x=0$, $y=1$.

A good exercise to think about is how this matches your physical intuition. (Not this specific example, but a similar one): consider a cylindrical barrel that has a small leak in the bottom. Toricelli's law says that the speed of the water coming out is proportional to $\sqrt{h}$. Thus, the height of water in the barrel satisfies an equation like $h' = - C \sqrt{h}$, where $C$ is some positive constant. Of course, this equation only makes sense for $h \ge 0$... but it exhibits similar properties to the example I just gave. What does this mean? local existence of solutions tells you that you can tell a possible future and a possible past from the initial value. Local uniqueness would say that there is only one past and only one future possible from this state. Now, if the barrel is empty at some moment in time, you can't tell the past anymore. It could have emptied just before you looked at it, or it could have emptied two months ago, or maybe there never was anything in the barrel to begin with.

OK, I probably got carried away. I hope some of this was useful.

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The equation is equivalent to

$$\frac{dy}{\sin{y}} = dx \implies \int dy \, \csc{y} = x + C$$

where $C$ is a constant of integration. Performing the integration of the left, we get

$$\log{\tan{\frac{y}{2}}} = x + C$$

$C$ may be determined from the initial condition:

$$C = \log{\tan{\frac{Y}{2}}}-X$$

Now we solve for $y$ in terms of $x$. Exponentiating, we get

$$\tan{\frac{y}{2}} = e^{x-X} \tan{\frac{Y}{2}}$$

The problem now is that the inverse tangent does not provide a one-to-one mapping, so we would need to restrict the range of $y$ to get a unique solution here.

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There are two problems with this: (1) This doesn't really answer the question asked. The OP wants to know about existence/uniqueness. This doesn't address that. (2) you seem to have a mistake in $\int \frac{1}{sin(y)}\, dy$. – Sam Lisi May 23 '13 at 14:36
@SamLisi: the integral is indeed correct. Uniqueness is addressed. Existence I guess is not, but I only saw uniqueness. But thank you for speaking up. – Ron Gordon May 23 '13 at 14:49
Indeed you are correct about the integral, sorry about my mistake there. I don't see how what you wrote addresses uniqueness fully. You are saying that if a solution exists for which $\sin(y)$ is never $0$ and for which $\tan(y/2)$ is never $0$ or undefined, it must satisfy the last displayed equation. Now, this displayed equation does not have unique solutions, so it isn't clear why the uniqueness of the solution to the IVP follows from this. One now needs an argument to say that the $y$ has to stay in the same branch as $Y$ (and deal with the edge cases of $y=0$ etc). – Sam Lisi May 23 '13 at 15:47
@SamLisi: that is pretty much what I said - you have to restrict the range of $y$ to get a complete solution - I just wasn't very specific about it because I felt that I had done enough at that point. Note that I never stated that that I proved that the IVP has a unique solution - I just stated conditions under which it would have one. Feel free to make that last argument. As for my post, I believe it addresses the OP's question. – Ron Gordon May 23 '13 at 15:59
I really appreciate both of your comments and help. My thoughts on solving this question were to look at the function dy/dx and d(dy/dx)/dy and considering the intervals that they are continuous, to be able to determine existence and uniqueness. Am I on the right lines? Thank you again. – kay May 23 '13 at 16:33

Yes, there is a unique solution to this problem on all $\mathbf R$. The local existence and uniqueness follow from the standard theorems, since the right hand side is $C^\infty(-\infty,\infty)$ in both variables. To show that the solution exists globally note that $|\sin y|\leq 1$ for any $y$, hence the solution to the problem with the initial conditions, e.g., $y(0)=0$ is between $y_{-}=-x$ and $y_{+}=x$. Since for any $x$ $y_{-}\leq y(x)\leq y_{+}$, the solution cannot blow up and can be always extended on the whole real line.

Added: However, a better argument might be that for any initial conditions $y(x_0)=y_0$ there are always two equlibria $y_{-}$ and $y_{+}$ such that $\sin y_{-}=\sin y_{+}=0$ and $y_{-}<y_0<y_{+}$, hence the solution is bounded, and hence can be extended for any interval, including $\mathbf R$.

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