Question

Let $j$ be the Klein $j$-invariant (from the theory of modular functions).
Does $j$ satisfy a differential equation of the form $j^\prime (z) = f(j(z),z)$ for any rational function $f$?

Answer 1

No. Conceptually, the reason is that $j'(z)$ is a weakly holomorphic (= holomorphic except at the cusp at infinity, where it has a pole) modular form of weight $2$, so it cannot be expressed in terms of $j$ (weakly holomorphic modular form of weight $0$) and $z$ (not anywhere near being a modular form).

For a rigorous proof:

Note that $j(z+1) = j(z)$, so $j'(z+1) = j'(z)$.

Suppose that the $j$ invariant did satisfy a differential equation of your form. Then we'd have $f(j(z), z) = f(j(z+1), z+1) = f(j(z), z+1)$. Note that the functions $z$ and $j(z)$ are algebraically independent (this is just saying that $j(z)$ is a transcendental function). Hence the underlying two-variable rational function $f(x, y)$ satisfies $f(x, y) = f(x, y+1)$. This then easily implies that $f(x, y)$ must be independent of $y$, e.g. $f(x, y) = g(x)$ for some rational function $g$.

So our original differential equation must actually take the form $j'(z) = f(j(z))$. But the left hand side is a nonzero (weakly holomorphic) modular form of weight $2$ while the right hand side has weight 0, and a nonzero modular form has a unique weight, so this is impossible.

Answer 2

The j-function satisfies a third-order differential equation. I've learned this from an old paper of Daniel Bertrand which I am having trouble locating right now. Maybe is this one:

MR0550281 (81i:10042) Bertrand, Daniel Propriétés arithmétiques des dérivées de la fonction modulaire $j(\tau )$. (French) Séminaire de Théorie des Nombres 1977–1978, Exp. No. 22, 4 pp., CNRS, Talence, 1978, 10F37 (10F35)

Answer 3

(this is too long for a comment)

Here is the explicit equation of order three for the $q$-expansion of $j$ multiplied by $q$. Keep in mind that this does not prove that there is no order one differential equation, so it is not an answer to the question.

       n
     [x ]f(x):
                3    4        4    3           5    2  ,        2    5
             (2x f(x)  - 6912x f(x)  + 5971968x f(x) )f (x) - 2x f(x)

           + 
                  3    4           4    3
             6912x f(x)  - 5971968x f(x)
        *
            ,,,
           f   (x)

       + 
              3    4         4    3           5    2  ,,   2
         (- 3x f(x)  + 10368x f(x)  - 8957952x f(x) )f  (x)

       + 
                2    4         3    3            4    2  ,             5
             (6x f(x)  - 20736x f(x)  + 17915904x f(x) )f (x) - 6x f(x)

           + 
                   2    4            3    3
             20736x f(x)  - 17915904x f(x)
        *
            ,,
           f  (x)

       + 
           3    2        4               5  ,   4
         (x f(x)  - 1968x f(x) + 2654208x )f (x)

       + 
              2    3        3    2            4      ,   3
         (- 4x f(x)  + 7872x f(x)  - 10616832x f(x))f (x)

       + 
                 4        2    3            3    2  ,   2
         (5x f(x)  - 8352x f(x)  + 12939264x f(x) )f (x)

       + 
               5            4           2    3  ,              5               4
       (- 2f(x)  + 960x f(x)  - 4644864x f(x) )f (x) + 1488f(x)  - 331776x f(x)

         =
         0
     ,
                            2            3      4
    f(x)= 1 + 744x + 196884x  + 21493760x  + O(x )]

Answer 4

A third-order differential equation for $j(\tau)$ is gotten via the Schwarzian derivative. The result is (1.13) of the paper by Harnad: http://arxiv.org/abs/solv-int/9902013