How to bound a recursive Derivative definition?

Question

I need to convert derivatives of g[x,v] wrt v into derivatives wrt x using this relation:

$$\frac{dg[x,v]}{dv}=\frac{1}{2}\frac{d^{2}g}{dx^{2}}+\frac{1}{2}\left(\frac{dg}{dx}\right)^{2}$$

This code converts the 0-4th derivatives perfectly:

ord = 5;
Derivative[q_, 1][g][x, v] = D[(D[g[x, v], {x, 2}] + D[g[x, v], x]^2), {x, q}]/2;
dgdv = Table[D[g[x, v], {v, i}], {i, 0, ord - 1}];

Notice that I only defined the 1st deriv of g wrt v, but MMa was smart enough to just apply the relation repeated. However, Series[] doesn't figure that out, for example it just leaves the 2nd derivative wrt v in that form:

Series[g[x, v], {v, 0, 2}]

$g(x,0)+v g^{(0,1)}(x,0)+\frac{1}{2} v^2 g^{(0,2)}(x,0)+O\left(v^3\right)$

I attempted to fix my definition to cover higher derivatives wrt v explicitly, but then I ran into recursion problems, which I tried and failed to fix as follows:

Derivative[q_, Except[0, n_]][g][x, v] = D[(D[g[x, v], {x, 2}] + D[g[x, v], x]^2), {x, q}, {v, n - 1}]/2;

This also failed:

Derivative[q_, 0][g][x_, v_] := D[g[x, v], {x, q}];
Derivative[q_, r_][g][x_, v_] :=
  If[r > 1,
   D[D[g[x, v], {x, q}, {v, r - 1}], v],
   D[(D[g[x, v], {x, 2}] + D[g[x, v], x]^2), {x, q}]/2];

I know that's an ugly solution, but I don't know why it didn't work. Help?

Answer 1

Try this:

Derivative[m_, Except[0, n_]][g][x_, v_] := 
   1/2 Derivative[2 + m, n - 1][g][x, v] + 
   1/2 Sum[Binomial[n - 1, k] Binomial[m, j] *
      Derivative[m - j + 1, n - 1 - k][g][x, v] *
      Derivative[j + 1, k][g][x, v], {k, 0, n - 1}, {j, 0, m}]

Series[g[x, v], {v, 0, 2}] now gives the expected output.