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Taking the vertical component of the curl of the momentum equation givesThe continuity equation gives ${\eta }_t+h_0\nabla \cdot \mathbf{u}=0$, and henceSince initially $\mathbf{u}=0$ and $\eta ={\eta }_0$,Taking the divergence of the momentum equation and usinggivesDifferentiate continuity in time and substitute the last two identities to obtain

Let the Rossby deformation radius beThe even, decaying steady solution ofwith continuous value and derivative at $x=\pm a$ isThe steady momentum balance is a geostrophic balance:Thus the flow is parallel to the two edges of the raised strip, in opposite $y$-directions on the two sides; for $f>0$, it points toward $+y$ on the left and $-y$ on the right. Its magnitude is concentrated within a few deformation radii of the edges.

WriteVariation of $\varphi$ gives $\nabla \cdot \mathbf{u}=0$. Variation of $\alpha$ givesand variation of $\beta$ givesUsing incompressibility, the middle equation becomesThese are the three required Euler-Lagrange equations.

Let $D_t={\partial }_t+\mathbf{u}\cdot \nabla$. Since $D_t\alpha =D_t\beta =0$, differentiating $u_i={\partial }_i\varphi +\beta {\partial }_i\alpha$ givesNowbecause $\mathbf{u}\cdot \nabla \alpha =-{\alpha }_t$. HencewhereThis is the Clebsch-potential variational derivation of incompressible Euler flow.

Differentiability at $x\in {\mathbb{R}}^2$ means that there is a linear map $Df{|}_x:{\mathbb{R}}^2\to \mathbb{R}$ such thatThe partial derivatives are $D_{i}f(x)=Df{|}_x(e_i)$ when the derivative exists; independently, they may be defined by the corresponding one-variable difference quotients.

Suppose the partial derivatives exist near $x=(x_1,x_2)$ and are continuous at $x$. Apply the one-dimensional mean value theorem along the two coordinate segments from $x$ to $x+h$. It gives intermediate points ${\xi }_h,{\eta }_h\to x$ such thatContinuity of the partials makes the remainder after subtractingequal to $o(\Vert h\Vert )$. Hence $f$ is differentiable and this is its derivative.

For the given function, writing $r=\sqrt{x^2+y^2}$, at $r>0$ we haveAt the origin both partial derivatives are zero, since $f(h,0)/h$ and $f(0,h)/h$ tend to zero. Neither partial is continuous there: along its corresponding coordinate axis the cosine term oscillates without a limit. Neverthelessso $f$ is differentiable at the origin with derivative zero. This illustrates that existence of partial derivatives does not imply their continuity.

The final assertion is false. DefineAgain $|g|\le r^2$, so $g$ is differentiable at the origin and it is smooth elsewhere. But along the $x$-axis,is unbounded near zero, and similarly $D_{2}g(0,y)$ is unbounded. Thus both partial derivatives can be unbounded in every neighbourhood of a point even when the function is differentiable everywhere.