complex geometry - Holomorphic vector fields acting on Dolbeault cohomology

The question.

Let $(X, J)$ be a complex manifold and $u$ a holomorphic vector field, i.e. $L_uJ = 0$. The holomorphicity of $u$ implies that the Lie derivative $L_u$ on forms preserves the (p,q) decomposition and also that it commutes with $bar{partial}$. From this it follows that $u$ acts infinitesimally on the Dolbeault cohomology groups $H^{p,q}(X)$ of $X$. My question is, does anyone know of an example in which this action is non-trivial?

Some context.

To give some context, first note that the analgous action for de Rham cohomology is always trivial: If $M$ is any smooth manifold and $v$ any vector field, then the formula $L_v = d circ i_v + i_v circ d$ shows that the infinitesimal action of $v$ on de Rham cohmology is trivial. (This is an instance of the more general fact that homotopic maps induce the same homomorphisms on singular cohomology. The field $v$ generates diffeomorphisms which are by construction isotopic to the identity map.)

Returning to Dolbeault cohomology, suppose we know that each Dolbeault class is represented by a $d$-closed form. (For example, this is true if $X$ is a compact Kähler manifold, by Hodge theory.) Then the action is necessarily trivial. The proof is as follows. Let $alpha$ be a $bar{partial}$-closed (p,q)-form which is also $d$-closed. Then we know that $L_u alpha = d(i_u alpha)$ is also of type (p,q). So,
$$
L_ualpha = bar{partial}left((i_ualpha)^{p, q-1}right) + partialleft((i_u alpha)^{p-1, q}right)
$$
and the other contributions $bar{partial}((i_ualpha)^{p-1,q}$) and $partial((i_ualpha)^{p,q-1})$ vanish. Now the fact that $barpartial((i_ualpha)^{p-1,q}) = 0$ and our hypothesis imply that there is a (p-1, q-1)-form $beta$ such that
$$
(i_ualpha)^{p-1,q}+ barpartial beta
$$
is closed. Hence
$$
partial left((i_ualpha)^{p-1,q}right) = barpartial partial beta
$$
and so
$$
L_ualpha = bar partial left( (i_u alpha)^{p,q-1} + partial betaright)
$$
which proves the action of $u$ on $H^{p,q}(X)$ is trivial.