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In his book on ordinary differential equations, Arnold says that, in general, equations which contain unknown functions and their derivatives are not differential equations. For example, Arnold says,

$$ \frac{dx}{dt}=x(x(t)) $$

is not a differential equation.

How can one solve this non differential equation?

zar
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1 Answers1

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Playing around.

If $x'(t) = x(x(t))$, $x''(t) = x'(t)x'(x(t)) = x(x(t))x(x(x(t))) $. By induction, $x^{(k)}(t) $ involves terms with $x(x(...x(t)...)$ nested up to $k+1$ deep.

If $x(t) = \sum_{n=0}^{\infty} a_n t^n $, $x'(t) = \sum_{n=1}^{\infty} na_n t^{n-1} = \sum_{n=0}^{\infty} (n+1)a_{n+1} t^{n} $.

We need $x(0) = 0$ for the composition to be defined.

Therefore $x'(0) =x(0) = 0 $, $x''(0) = 0 $.

By induction on $k$, $x^{(k)}(0) =0 $ for all $k$.

Therefore $x(t)$ can not be a infinite power series.

If $x(t) =b t^a $, $x'(t) =abt^{a-1} $ and $x(x(t)) =b(bt^a)^a =b^{a+1}t^{a^2} $, so we need $ab=b^{a+1}$ (so $a = b^a$ or $b = a^{1/a})$ and $a^2=a-1$, so that

$a =\frac12(1\pm\sqrt{1-4}) =\frac12(1\pm i\sqrt{3}) $.

For this,

$\begin{array}\\ t^a &=t^{\frac12(1\pm i\sqrt{3})}\\ &=t^{1/2}e^{\pm \ln(t)\frac12( i\sqrt{3})}\\ &=t^{1/2}\left(\cos(\ln(t)\frac12( \sqrt{3}))\pm i\sin(\ln(t)\frac12( \sqrt{3}))\right)\\ \end{array} $

From $b = a^{1/a}$ we get $b$, but I'm not going to work that out.

That is all I can come up with for now.

marty cohen
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    Shouldn't you have that $ab=b^{a+1}$, so that $a=b^a$ instead of $b^a=1$ – Mark Schultz-Wu Aug 03 '16 at 21:18
  • You are right. Thanks. I'll fix that. Upvoted. – marty cohen Aug 03 '16 at 23:40
  • Didn't you already prove that $x(t) \neq bt^a$? Such an equation qualifies as a power series. You just proved it wasn't. Rather, what I believed you proved is that $x(t)$ has no meaningful Taylor series representation at 0. The Taylor series expansion of $x(t)$ is by your induction, equivalent to $0$, but the function itself is not necessarily. – Axoren Aug 04 '16 at 00:04
  • No. I proved that it could not be an infinite power series - that is what requires that x(x(t)) has x(0) = 0. As you can see, if x(t) is a single term, or a finite number of terms, x(x(t)) is quite well defined. – marty cohen Aug 04 '16 at 05:05