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In the fluid region $y>\eta (x)$, incompressibility and potential flow giveNo penetration through the exact surface gives

The condition $hk\ll 1$ says the hill slope is small; $|\nabla \varphi |\ll U$ says the disturbance velocity is small compared with the background wind. Dropping the product ${\varphi }_x{\eta }_x$ and Taylor-shifting the boundary from $y=\eta$ to $y=0$ therefore givesThe decaying solution is

Bernoulli's equation isTo first order on the surface,ThusEquivalently, crest minus trough is the negative of this. For $k{U}^2/g\ll 1$, hydrostatic elevation dominates; for $k{U}^2/g\gg 1$, the Bernoulli pressure drop caused by faster crest flow dominates.

Spatial Fourier transformation givesHenceInverting and splitting the cosine into exponentials, or applying D'Alembert formula, yieldsThus the initial Gaussian separates into two half-amplitude Gaussian pulses travelling without distortion at speeds $c$ and $-c$.

The Euler--Lagrange equation isPutting $u=\cos (\varphi /2)$ and simplifying gives

Since $1-\cos \varphi =2(1-u^2)$, $1+\cos \varphi =2u^2$, and ${\dot{u}}^2=(1-u^2){\dot{\varphi }}^2/4$, the transformed functional isIts Euler--Lagrange equation is $\ddot{u}+(g/4a)u=0$, exactly the preceding equation.

For endpoint-vanishing $\eta$, the second variation isWriting $\eta ={\sum }_{n\ge 1}{c}_n{e}_n$ and using orthonormality givesIt is positive definite when $T<\pi /\omega$ and has a negative $e_1$ direction when $T>\pi /\omega$. Therefore

A map $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is differentiable at $x$ if there is a linear map $L$ such thatthen $Df{|}_x=L$. The inverse function theorem says that if $f$ is continuously differentiable near $x$ and $Df{|}_x$ is invertible, then $f$ restricts to a $C^1$ diffeomorphism between neighborhoods of $x$ and $f(x)$.

Since $F(A)=A^{T}A$ is polynomial,Thus $\ker DF{|}_I=T$, the space of skew-symmetric matrices.

DefineThen $D\Phi {|}_I(H)=2H$, so the inverse function theorem supplies open neighborhoods $I\in U$, $0\in V$ on which $\Phi :U\to V$ is a $C^1$ diffeomorphism. Its symmetric part is $A^{T}A-I$ and its skew part is $A-A^T$. Consequentlyand therefore $\Phi (U\cap \mathcal{O})=V\cap T$.

For any $R\in \mathcal{O}$, left multiplication $A\mapsto R^{T}A$ is a homeomorphism preserving $\mathcal{O}$ and carrying $R$ to $I$. Transporting the preceding chart gives a neighborhood of $R$ in $\mathcal{O}$ homeomorphic to an open subset of $T$.