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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.

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    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 '19 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! –  Feb 24 '19 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 '19 at 21:40

2 Answers2

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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.

Robert Israel
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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 $-1<y(0)<1$.
  • 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 $y(t)=\pm1$.
timur
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  • What happens if $f(y)$ isn’t uniquely determined in our case? $f(y)=\pm\sqrt{1-y^2}$ –  Feb 24 '19 at 20:46
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    @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 '19 at 20:57