Solution of $x'(t)=x(x(t))$

135 Views Asked by At

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?

1

There are 1 best solutions below

4
On

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.