2
$\begingroup$

I've heard that an $n$-th order differential equation will always have exactly $n$ arbitrary constants in its solution; that is, if $y$ satisfies the differential equation $f(y(t),y^{(1)}(t),\cdots, y^{(n)}(t), t) = 0$ then it will always be of the form $y(t) = g(t; C_1, \cdots, C_n)$ where $C_1, \cdots C_n \in \mathbb{C}$. This makes intuitive sense, as one might have to integrate $n$ times to get to the solution, which would mean $n$ constants of integration. It can even be proven quite easily in the constant-coefficient case; given that $\sum _{i = 1} ^{n} a_i \hat{D}^i y(t) = f_0(t)$, one can factor the differential operator $\sum _{i = 1} ^{n} a_i \hat{D}^i$ that acts on $y$ into $n$ first order derivatives according to solutions to the characteristic equation, which are all integrated to create $n$ arbitrary constants.

However, I am not satisfied with this argument. Consider, for example, the following second-order differential equation: \begin{align} \sin(y'') + (y')^2 = y\cos(t^2y''). \end{align} I cannot imagine how one would be able to apply either of the previous arguments (factorisation of the differential operator or $n$ integrations) to show that this has exactly two arbitrary constants. And even if one could show that there are two integrations in this case, the general case is still uncertain: Are there always going to be $n$ arbitrary constants? Is there perhaps a rigorous proof of this?

$\endgroup$
1
  • $\begingroup$ Interesting question. $\endgroup$ Commented Aug 14, 2020 at 13:24

1 Answer 1

3
$\begingroup$

This is, under some extra conditions on the functions defining the ODE, a consequence of the Picard-Lindelof Theorem for vector functions. Any n-th order ODE $y^{(n)} = f(t,y,y',...,y^{(n-1)})$ can be reduced to an ODE of the form $v'=F(v,t)$ where $v: \mathbb{R} \rightarrow \mathbb{R}^{n}$ is given by $ v_{1} = y, v_{2} = y',...,v_{n} = y^{(n-1)} $ and so on. To see this take $(F(v,t))_{i} = v_{i+1}$ and $(F(v,t))_{n}$ to be the function defining the ODE, with $y^{(k)}$ replaced appropriately. If we take $F$ to be continuous in t and lipschitz in $v$ we get that any solution of the ODE is uniquely determined by the initial condition of $v$, which is n initial conditions on y and its' derivatives up to $n-1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .