# Wrong reasoning for uniqueness of solution of ODE?

Sometimes I have seen this argument to prove that a differential equation has an unique solution, but I think it's wrong.

Suppose the differential equation: $$\mathscr{D}[y(t)]=f(x)$$ where $\mathscr{D}$ is an arbitrary differential operator of order $N$. With initial conditions: $$y(0)=K_0,Dy(0)=K_1\dots,D^{N-1}y(0)=K_{N-1}$$ has two solutions $y_1,y_2$. Then $w=y_1-y_2$ is a solution of the initial value problem: $$\mathscr{D}[y(t)]=0$$ With initial conditions: $$y(0)=0, Dy(0)=0\dots,D^{N-1}y(0)=0$$ As this second problem has solution $w=0$ then $y_1=y_2$.

I think this argument is wrong because the second system could have more solutions. Then maybe it has one solution $w_1=0$, but another $w_2\neq 0$. It looks if we have just passed the issue from one initial value problem to another.

Anybody could help me to make clear this doubt?

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## 1 Answer

The reasoning is correct: there is an existence and uniquiness theorem for the homogeneous initial value problem you are considering. This is the well known Picard Lindelof theorm applied to the equivalent system of $N$ first order Cauchy's problems. Being $0$ obviously Lipschitz continuous the theorem applies, therefore there exist only one solution to the problem $$\mathcal{D}[y(t)] = 0, \qquad y(0) = 0, \dots ,\ D^{N-1}y(0) = 0.$$

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I have seen this even in PDE where we don't have Picard's theorem. Even in with fractional differential operators. So I'm meaning the use of this argument without the use of Picard's. –  Ambesh Apr 19 '13 at 19:15
It's impossible to apply this without any uniquiness result on the homogeneous probelm, that's a fact. There would be no reason to bring the issue on another problem when nothing is known. –  user01123581321345589144... Apr 19 '13 at 19:19