For positive wavenumber $k$, the phase velocity is $c_p=\omega /k$, the speed of a fixed phase. The group velocity is $c_g=d\omega /dk$, the leading speed of a narrow wave packet. Differentiating the supplied dispersion relation givesFor fixed $k>0$, the shallow-depth limit uses $\tanh s/s\to 1$:

Write the surface displacement and the perturbation velocity potential as small quantities about rest, and drop their quadratic products. Evaluating the remaining surface terms at the unperturbed level $y=0$ givesThe velocity components are $u={\varphi }_x$, $v={\varphi }_y$. The shift of the evaluation point from zero to $\eta$ contributes a product of perturbations, as do $u{\eta }_x$ and the kinetic-energy term, so they disappear at this order.

Condition (1b) is the kinematic free-surface boundary condition: fluid particles on the surface remain there, since the material derivative of $y-\eta (x,t)$ vanishes. Condition (1c) is the dynamic free-surface boundary condition, obtained from the unsteady Bernoulli equation with constant atmospheric pressure and a choice of potential gauge. Condition (1d) is impermeability: no normal flow through the horizontal bed.

The convective acceleration isSteady Euler equations therefore give $p_x=-\rho \sin x\cos x$ and $p_y=\rho \sin y\cos y$. Integrating these compatible equations yieldswith arbitrary additive constant $p_0$. As a check, $p/\rho +(u^2+v^2)/2=p_0/\rho +1/2-{\psi }^2$, indeed constant on each streamline. This is a Steady Euler flow with vorticity proportional to its stream function.

For this steady, inviscid, constant-density flow without a body force, Bernoulli's theorem says thatIts constant may vary from one streamline to another because the flow has vorticity.

The vorticity points normally to the plane and has scalar componentIt is nonzero in general, so the flow is rotational.

Use the stream function convention $u={\psi }_y$, $v=-{\psi }_x$. It gives $u=-\sin x\sin y$, $v=-\cos x\cos y$. Thenso the flow is incompressible.

The energy function is $E={\sum }_a{\dot{q}}_a\partial L/\partial {\dot{q}}_a-L$. Along a solution,Therefore time independence gives the conserved quantity

A coordinate absent from the Lagrangian is cyclic. Its Euler-Lagrange equation givesThus the corresponding generalized momentum $\boxed{p_1=\partial L/\partial {\dot{q}}_1}$ is conserved.

The inverse function theorem states: if $f$ is continuously differentiable on an open subset of ${\mathbb{R}}^n$ and $Df(a)$ is invertible, then some open neighbourhood of $a$ is mapped bijectively onto an open neighbourhood of $f(a)$, with continuously differentiable inverse. Its derivative at $f(a)$ is $(Df(a){)}^{-1}$.

The present map is smooth, and at $a=(\pi /2,\pi )$,Also $f(a)=(0,0)$, so the theorem applies locally and givesThe inverse is defined near the image point $(0,0)$ and takes values near $a$; no assertion of a global inverse is required.

For $f=(f_1,\dots ,f_k)$, the Jacobian matrix is the matrix of its derivative in standard bases, with entries $(Df{)}_{ij}=\partial f_i/\partial x_j$. Here

The circle is path-connected, using circular arcs, and hence connected. Its continuous image is connected by part (ii). A connected subset of the real line is an interval: if it contained values on opposite sides of a missing value $c$, intersection with $(-\infty ,c)$ and $(c,\infty )$ would separate it. Thus $f(C)$ is an interval. It is also compact, so it is a closed bounded interval, possibly reduced to one point.

If $f(A)$ were separated by two disjoint nonempty relatively open sets, their preimages under the restriction $f{|}_A$ would be disjoint nonempty relatively open sets covering $A$. Continuity makes them open in $A$, contradicting its connectedness. Therefore $f(A)$ is connected.

A subset is connected when it has no separation into two disjoint nonempty relatively open sets. It is path-connected when every pair of its points can be joined by a continuous path from $[0,1]$ lying in the subset.

Path-connectedness implies connectedness; the reverse fails in general, for example for the topologist's sine curve. They are equivalent for open subsets of every Euclidean space ${\mathbb{R}}^n$. More generally this holds for open subsets of locally path-connected spaces.

A basis for a topology is a family of open sets such that every open set is a union of members of the family. Equivalently, for every point in an open set there is a basis member containing that point and contained in the open set.

For a basis $\mathcal{B}$ on $Y$, continuity is equivalent to $f^{-1}(B)$ being open for every $B\in \mathcal{B}$. Necessity follows because basis sets are open. Conversely, taking preimages of a union of basis sets gives a union of open preimages, so every open set has an open preimage.