2
$\begingroup$

Consider the function $h:[0,1]\to \mathbb{R}$ $$h(\theta):=\sum_{k\geq 1}\frac{a_{k}}{\sqrt{k}}\cos(2\pi k \theta)+\frac{b_{k}}{\sqrt{k}}\sin(2\pi k \theta),$$ where $a_{k},b_{k}\in\mathbb{R}$. For simplicity let's take $\big(\frac{a_{k}}{\sqrt{k}}\big),\big(\frac{b_{k}}{\sqrt{k}}\big)\in \ell^{1}$, so that $h\in H^{1}(S^{1})$.

The ODE we have is

$$y'=\exp(-h(y))\;\Leftrightarrow \;y(x)=\int_{0}^{x}\exp(-h(y(t)))dt.$$

Does this fit anywhere into the ODE theory (e.g. fixed points of integral equations)? Maybe the simpler problem with a truncated series

$$y'=\exp(-h^{N}(y)),$$

where $h^{N}(\theta)=\sum_{k=1}^{N}\frac{a_{k}}{\sqrt{k}}\cos(2\pi k \theta)+\frac{b_{k}}{\sqrt{k}}\sin(2\pi k \theta)$. Or even simpler with a single term:

$$y'=\exp\Big(-\big(\frac{a_{k}}{\sqrt{k}}\cos(2\pi k y)+\frac{b_{k}}{\sqrt{k}}\sin(2\pi k y)\big)\Big).$$

Or even more simply

$$y'=\exp(-(\cos(2\pi y)).$$

Our main question is to obtain a (series) approximation solution to $y$ (for any of the above ODEs)?

$\endgroup$
2
  • $\begingroup$ I am not an expert in ODEs, but I thought that any equation not involving $x$ of the form $y'=f(y)$ has the trivial solution $x$ in terms of $y$ of the form $x=\int^y dt/f(t)$, of which you then take the inverse function to get $y$ in terms of $x$. $\endgroup$ Jan 31, 2019 at 22:04
  • $\begingroup$ indeed, we know x and we are trying to get anything possible on its inverse. $\endgroup$ Feb 1, 2019 at 0:33

2 Answers 2

1
$\begingroup$

If $f$ is a smooth enough function, then higher-order derivatives of a solution $y$ of the ODE \begin{equation} y'=f(y) \end{equation} can be found by successive differentiation of both sides of the ODE, giving $y''=f'(y)y'$ and, more generally, a recursion of the form \begin{equation} y^{(n)}=f_n(y,y',\dots,y^{(n-1)}) \end{equation} for natural $n$, where $f_n$ is a certain function, depending on $f$, which can be given an explicit, albeit complicated, expression -- writing \begin{equation} f_n(y,y',\dots,y^{(n-1)})=(f\circ y)^{(n-1)} \end{equation} and then using the Faà di Bruno formula.

So, given an initial value $y_0=y(t_0)$, one obtains a Taylor approximation to $y$ in a neighborhood of $t_0$. E.g., rescaling the argument of $y$, rewrite your last display as \begin{equation} y'=e^{-\cos y}. \end{equation} Then, doing as described above and assuming $y(0)=0$, we get \begin{equation} y(t)=\frac1{1!e}\,t+\frac1{3!e^3}\,t^3+\frac6{5!e^5}\,t^5+\frac{87}{7!e^7}\,t^7+\cdots \end{equation} for $t$ near $0$. (The sequence of the numerators here does not match any of the sequences in The On-Line Encyclopedia of Integer Sequences, which strongly suggests that none of your ODE's has an explicit series approximation to its solutions.)

$\endgroup$
4
  • $\begingroup$ Yes, hence my comment that it is only a distribution over that space (cf. GFF in the literature) and to assume $h\in H^1$. I might remove it because it confuses people. $\endgroup$ Feb 1, 2019 at 0:31
  • $\begingroup$ Because originally $a_k$ and $b_k$ were random variables, I thought you meant the term "distribution" in the sense "probability distribution". So, I thought you actually confused the random function with its distribution. Now I see you meant "distribution" in the sense of generalized functions. $\endgroup$ Feb 1, 2019 at 0:54
  • $\begingroup$ No problem. If you don't mind, please remove this answer to avoid confusing others. $\endgroup$ Feb 1, 2019 at 1:38
  • $\begingroup$ I have deleted my previous answer and replaced it by a new one, with an updated understanding of your problem. $\endgroup$ Feb 1, 2019 at 13:25
1
$\begingroup$

It can fall under " nonlinear Volterra integral equations":

$$u(x)=\int_{0}^{x} K(x,t)F(u(t))dt,$$

and so many approaches are possible:

  • successive approximation
  • series approximation as suggested from the other answer.
  • Adomian Decomposition Method

See that and many other methods in the wonderful reference: "Linear and nonlinear integral equations"

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.