Let $a,b,c \in \mathbb{R}$, with $a \neq 0$. we consider the linear homogeneous differential equation : $(E) : ay''+by'+cy=0$.
Is there an "elementary" way (i.e. without invoking "big" results like Cauchy-Lipschitz's theorem) to prove, without solving $(E)$, that if we fix an initial condition $y(t_0)=x_0$ and $y'(t_0)=z_0$, then, the solution is unique ? In other words, is there a way to prove a kind of "very weak" version of the unicity part of Cauchy Lipschitz's theorem with elementary tools in this particular case ?
The only idea that I have for the moment is the following : Let $y_1$ and $y_2$ be two solutions. Then, we can easily show "by hand" that the Wronskian determiant $W(t) :=\begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}$ is solution of the équation $W'=\frac{b}{a}W$, and, thus, that it is of the form $W(t)=\lambda e^{\frac{b}{a}t}$, for some $\lambda \in \mathbb{R}$. We deduce that it is either always equal to $0$, either never equal to $0$. In particular, if $y_1, y_2$ satisfy the same initial condition, it is always equal to $0$. Therefore, for every $t$, there exists a scalar $\alpha(t)$ such that $\begin{pmatrix}y_1(t) \\ y_1'(t) \end{pmatrix} = \alpha(t) \begin{pmatrix}y_2(t) \\ y_2'(t) \end{pmatrix}$. If we can prove that $t \mapsto \alpha(t)$ is a constant function, then we have what we want, but I don't see how we could deduce this with simple arguments... (and maybe there is a simplest way to prove that...)
Does anybody have some ideas ?
Thanks in advance.
For convenience, I'll take $a = 1$ (otherwise divide the equation by $a$). Note that if $y$ satisfies $y'' + b y' + c y = 0$, then $v = e^{bt/2} y$ satisfies $v'' + (c - b^2/4) v = 0$. It suffices to show that solutions of this equation are unique. For convenience, take $c - b^2/4 = k$, so $v$ is a solution of $v'' + k v = 0$.
Let $z(t) = v_1(t) - v_2(t)$, where $v_1$ and $v_2$ are solutions to the initial value problem $v'' + k v = 0$, $v(t_0) = v_0$, $v'(t_0) = v'_0$. Thus $z(t)$ is also a solution of the same differential equation, with initial condition $z(t_0) = z'(t_0) = 0$.
Note that $$ \dfrac{d}{dt} (k z^2 + (z')^2) = 2 k z z' + 2 z' z'' = 2 k z z' + 2 z' (- k z) = 0$$
Thus $k z^2 + (z')^2$ is constant. Since it is $0$ at $t = t_0$, it must be $0$ everywhere. If $k \ge 0$ then we immediately get $z' = 0$ everywhere, so $z$ is constant, and that constant is $0$. What about if $k < 0$? Well, then $z' = \pm \sqrt{-k} z$. On any interval where $z \ne 0$, the choice of $+$ or $-$ would be constant. And then $$ \dfrac{d}{dt} \left(e^{\mp \sqrt{-k} t} z\right) = 0$$ which implies that $e^{\mp \sqrt{-k} t} z$ is constant. But then if $z$ is nonzero at one point, it would have to be nonzero everywhere, contradicting the initial condition $z(t_0) = 0$. So again we conclude that $z = 0$ everywhere.