# Uniqueness of OdE Solution, with 0 as one of the solutions

I am totally confused, take the ODE $$u-tu'-(u')^2=0$$ Clearly $-t^2/4$ and $ct+c^2$ are both solutions, and if we take $c = 0$, then they both satisfy the initial condition $u(0) = 0$.

Why does this not contradict uniqueness?

Thank you!

## Addendum: The Existence and Uniqueness Theorem.

Consider the initial value problem $X' = F(X)$, $X(0) = X_0$ where $X_0 ∈ \Bbb R^n$. Suppose that $F : \Bbb R^n → \Bbb R^n$ is $C^1$. Then there exists a unique solution of this initial value problem. More precisely, there exists $a > 0$ and a unique solution $X : (−a, a) → \Bbb R^n$ of this differential equation satisfying the initial condition $X(0) = X_0$.

• Please add your version of the (existence and) uniqueness theorem. Perhaps the form of the statement of the IVP will already give you a hint. Dec 11, 2016 at 14:01
• The Existence and Uniqueness Theorem. Consider the initial value problem X = F(X), X(0) = X0 where X0 ∈ Rn. Suppose that F : Rn → Rn is C1. Then there exists a unique solution of this initial value problem. More precisely, there exists a > 0 and a unique solution X : (−a, a) → Rn of this differential equation satisfying the initial condition X(0) = X0. I still dont get it
– Kai
Dec 11, 2016 at 14:12
• This is a quadratic equation, one expects two solutions. Doesn't the existence and Uniqueness theorem hold only for linear equations? Dec 11, 2016 at 14:24

From your cited version of the existence and uniqueness theorem you see that it only applies to explicit differential equations $$X'(t)=F(t,X(t)).$$
The equation you consider is not explicit. You can solve it in two different ways to be explicit by the sign choices of the square root below. Even then, $$u'=f(t,u)=-\frac t2\pm\sqrt{u+\frac{t^2}4}$$ is neither differentiable nor Lipschitz at $u=-\frac{t^2}4$, so you can not expect uniqueness there.