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Consider the nonresonant traveling forced solution, with $k>0$ and $c\ne 0$. Put $x=\bar{x}+\xi$, $y=\bar{y}+\upsilon$, where $(\bar{x},\bar{y})$ is the mean position of a fluid element. At linear order, evaluate the velocity field at this mean position: substituting $\xi ,\upsilon$ into its spatial arguments would change the already first-order velocity only at second order. With $\Theta =k(\bar{x}-ct)$, differentiation of the velocity potential givesSince $\dot{\Theta }=-kc$, integration, with zero mean displacements, yieldsIf both coefficients are nonzero, eliminating the wave phase proves that the pathline is an ellipse:Its principal axes of an ellipse are horizontal and vertical, with respective semi-axesThus the full axis lengths are $2|A|$ and $2|B|$. The signed coefficients also determine the direction of traversal. The useful checks $B(0)=b$ and $B(-d)=a$ reproduce the vertical displacements of the free surface and bottom.

To identify the major axis without comparing the two expressions separately, use ${\cosh }^{2}s-{\sinh }^{2}s=1$:Consequently the major axis is horizontal if $c^2<g/k$ and vertical if $c^2>g/k$; equality gives circles. If $A$ or $B$ vanishes at a particular depth, the first-order orbit degenerates to a line segment. These are linearized particle orbits of a bottom-forced gravity wave, not exact closed trajectories at all orders: a higher-order calculation may introduce a mean drift.

The principal axes of an ellipse are the two perpendicular symmetry axes through its center of an ellipse. For a symmetric positive-definite matrix $M$, its orthonormal eigenvectors give those directions and its positive eigenvalues ${\lambda }_1,{\lambda }_2$ give semiaxis lengths ${\lambda }_1^{-1/2},{\lambda }_2^{-1/2}$. This follows by orthogonal diagonalization of the defining quadratic form. In ${\xi }^2/A^2+{\upsilon }^2/B^2=1$, the axes are already horizontal and vertical, with semiaxes $|A|,|B|$ and full lengths $2|A|,2|B|$. Equal semiaxes give a circle, for which every perpendicular pair through the center can serve.