= Solution
Differentiating the <velocity potential> gives $u=a\omega e^{kz}\cos(kx-\omega t)$ and $w=a\omega e^{kz}\sin(kx-\omega t)$. Let $A=a e^{kz}$ and $\vartheta=kx-\omega t$. To first order in the <wave steepness> $ka$, evaluate these velocities at the initial parcel position and integrate from time zero:
$$
\boxed{X-x=A[\sin(kx)-\sin\vartheta]+O(a^2k),\qquad
Z-z=A[\cos\vartheta-\cos(kx)]+O(a^2k).}
$$
These expressions satisfy both initial conditions and distinguish <initial and mean parcel labels in a surface wave>. Their actual short-time limits are
$$
\boxed{X-x=a\omega e^{kz}\cos(kx)t+O(t^2),\qquad
Z-z=a\omega e^{kz}\sin(kx)t+O(t^2).}
$$
The printed expressions omit the integration constants and therefore cannot be the small-time displacement from the prescribed initial point. For example, at $x=0$, they give $Z(0)-z=a e^{kz}\ne0$. This is a genuine inconsistency, rather than a missing step in the calculation.
The intended oscillatory orbit is recovered by using mean parcel coordinates instead. Set $x_c=x+A\sin(kx)$ and $z_c=z-A\cos(kx)$ to this order. Then
$$
X-x_c\simeq-ae^{kz_c}\sin(kx_c-\omega t),\qquad
Z-z_c\simeq ae^{kz_c}\cos(kx_c-\omega t).
$$
Thus a <deep-water gravity wave> produces approximately circular parcel orbits with radius $ae^{kz_c}$, decaying exponentially with depth. The approximation is one of small amplitude over a wave cycle, not an assertion that these oscillatory coordinates vanish at time zero.
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