# Existence and Uniqueness for a nonlinear ODE

Suppose I have the first-order nonlinear ODE: $$(y’)^2+y^2=1$$

Upon inspection, we see that $$y(t)=\pm1$$ and $$y(t)=\pm\sin(t+a)$$. ‘Another’ solution one sees upon inspection is $$y(t)=\pm\cos(t+a),$$ but this is contained in the other solution ($$\pm\sin(t+a)$$).

My question is:

Are these solutions unique? Are there any others and how do we know? Thanks in advance.

• The usual theorems on existence and uniqueness do not hold for such kind of differential equations, because this is not an ODE: this is a Differential Algebraic Equation. – Daniele Tampieri Feb 24 at 21:10
• @downvoter - I gave my thoughts, tagged it correctly, and was not convoluted. Please let me know how I can improve my question! – user573025 Feb 24 at 21:29
• I am not an expert on such objects, so I feel difficult to give advice on such matters, but this Math.SE Q&A should be useful for understanding the differences between DAEs and ODEs. For some references, I would advice you to have a look at this Scholarpedia entry – Daniele Tampieri Feb 24 at 21:40

There are also "piecewise" solutions such as $$y(t) = \cases{-1 & if t \le -\pi/2 \cr \sin(t) & if -\pi/2 \le t \le \pi/2\cr 1 & if t \ge \pi/2\cr}$$

For each initial condition $$y(t_0) = y_0$$ with $$-1 < y_0 < 1$$, there are two solutions, one increasing and one decreasing, as long as $$|y|$$ stays less than $$1$$. But once you hit $$y = \pm 1$$, there is no more uniqueness.

You need to write the ODE in the normal form $$y'=f(y)$$. Once it is written in this form, check if $$f$$ is Lipschitz. If $$f$$ is Lipschitz, the initial value problem (i.e., when you specify the value $$y(0)$$) has a unique solution.

In your example, we have 2 normal forms, or 2 branches to consider: $$f(y)=\pm\sqrt{1-y^2}.$$ Hence, in the best case scenario, we would get exactly 2 solutions for a given initial data $$y(0)$$. Sticking to reals for simplicity, we have the following.

• Exactly 2 local solutions if $$-1.
• The branches of $$f$$ are non-Lipschitz at $$y=\pm1$$, so more than 2 solutions if $$y(0)=\pm1$$.
• Different solutions can be glued together at $$t$$ whenever $$y(t)=\pm1$$.
• What happens if $f(y)$ isn’t uniquely determined in our case? $f(y)=\pm\sqrt{1-y^2}$ – user573025 Feb 24 at 20:46
• @livingtolearn-learningtolive: It just means that you have 2 cases to consider. You can see that unless the initial data is $y(0)=\pm1$, each branch has a unique solution locally in time. For $y(0)=\pm1$, each branch is non-Lipschitz, so uniqueness is not guaranteed. This is the reason why you lose uniqueness at $t$ whenever $y(t)=\pm1$. – timur Feb 24 at 20:57