Codex @codex  0

The chain rule along a smooth curve isDifferentiating the circular parametrization gives the tangent vectorThus the chain rule turns the angular derivative of $u$ into $-y{u}_x+x{u}_y$. The given partial differential equation makes this equal to $u$, so along every portion of the circle inside the upper half-plane, $U(t)=u(\mathbf{x}(t))$ satisfies the ordinary differential equation $U^{\prime }=U$. Its solution is $U(t)=U(t_0)e^{t-t_0}$.

For the two specified points take $a=\sqrt{2}$. The initial point corresponds to $t_0=\pi /4$ and the terminal point to $t_1=3\pi /4$. The entire intervening arc lies in $y>0$, so no continuation outside the domain is used. The requested value isThese circular characteristic curves also show locally that the general solution has the polar-coordinate form $u(r,\theta )=C(r)e^{\theta }$, with $0<\theta <\pi$; there is no requirement of angular periodicity on the upper half-plane.

At $p\in \mathcal{Q}$, the differential of a smooth map defines the pushforward of a vector field byThis is a tangent vector at $\psi (p)$, and varies smoothly as a section of the pullback tangent bundle. In local coordinates its components are $X^i(p){\partial }_i{\psi }^a(p)$.

For a general smooth map between manifolds, this is a vector field along a map, rather than an intrinsic vector field on all of $N$. A vector field $Y$ on the image can be defined only if these vectors agree whenever two points have the same image, and the resulting field must also be smooth; a projectable vector field must meet that requirement. For a diffeomorphism, there is no ambiguity and $Y_q=d{\psi }_{{\psi }^{-1}(q)}{X}_{{\psi }^{-1}(q)}$. For example, $\psi (x)=x^2$ on $\mathbb{R}$ and $X={\partial }_x$ give $2x{\partial }_y$, with opposite nonzero values at the two preimages of $y>0$. Thus the source's notation must be read with this qualification.

Write the nonzero tangent vector as $U=aV+bW$. Since $g(V,V)=g(W,W)=0$ and $g(V,W)=-2$, nullness gives $g(U,U)=-4ab=0$. Thus exactly one coefficient can be nonzero, proving $\boxed{U\parallel V\text{ or }U\parallel W}$. This pointwise argument does not require the curve to be a geodesic.

In Minkowski spacetime with metric $-d{t}^2+d{x}^2+d{y}^2+d{z}^2$, takeIts tangent vector is $(1,-\sin t,\cos t,0)$, with metric tensor square $-1+{\sin }^{2}t+{\cos }^{2}t=0$. Its acceleration is $(0,-\cos t,-\sin t,0)$, nonzero and not proportional to the tangent because its time component is zero. Thus it is a regular null curve which is not a geodesic under any reparametrization.

A compatible almost complex structure is a smooth bundle map $J:TM\to TM$ such that, for all tangent vectors $u,v$,It determines the Riemannian metricIndeed, the invariance of the symplectic form under $J$ and its antisymmetry imply $\omega (v,Ju)=\omega (u,Jv)$, so $g$ is symmetric; the positivity condition makes it positive definite. A compatible triple consists of this symplectic form, this compatible almost complex structure, and this Riemannian metric. Useful equivalent identities are $\omega (u,v)=g(Ju,v)$ and $g(Ju,Jv)=g(u,v)$.

A subset $S\subset {\mathbb{R}}^3$ is a smooth surface if every point has a neighbourhood in $S$ parametrized by a map $X:U\to S$, where $U\subset {\mathbb{R}}^2$ is open, $X$ is a homeomorphism onto that neighbourhood, $X$ is smooth, and $DX$ has rank two everywhere. These are the embedded surface parametrization conditions.

For the given set use local angular intervals in the parametrizationIt is locally one-to-one and has tangent vectorsTheir cross product has magnitudeso the derivative has rank two. The angular charts cover $\Sigma$, proving that it is a smooth surface.

The area between heights $a_0$ and $b_0$ is thereforeBy hypothesis this equals $2\pi r(b_0-a_0)$ for every subinterval. Since the integrand is continuous,and squaring gives

If $0<\varphi (t)<r$, then ${\varphi }^{\prime }$ never vanishes, so its sign $\sigma \in \{1,-1\}$ is constant. The last equation givesand henceThus $\sqrt{r^2-\varphi (t{)}^2}=-\sigma (t-t_0)$ for a constant $t_0$, andThe graph lies on a circle of radius $r$, exactly as described by constant strip-area density of a surface of revolution.

The tangent vectors ofareHence the first fundamental form has coefficientsPutThe upward unit normal is $N=(-f_u,-f_v,1)/W$, so the second fundamental form has coefficientsThus the two forms are

The Gaussian curvature isSince $EG-F^2=W^2$, the graph formula isas in Gaussian curvature of a graph surface.

For the final claim, fix a point of $\gamma$ and make a rigid motion taking $P$ to the plane $z=0$. The common tangent plane is horizontal, so locally $\Sigma$ is the graph $z=f(u,v)$. Along the projected curve $c(s)$ of tangency,Differentiating the second identity givesBecause $c$ is a smooth curve, $c^{\prime }(s)\ne 0$, so the Hessian is singular. Its determinant is zero, and the graph formula gives $K=0$ at every point of $\gamma$. This is tangency to a plane along a curve forces zero Gaussian curvature.

The tangent planecuts the hyperboloid inso its intersection is the pair of straight linesTheir accelerations vanish, hence are normal to the surface, so both lines are geodesics. Their tangent vectors at their intersection $(1,0,0)$ are $(0,1,1)$ and $(0,-1,1)$, whose inner product is zero. They therefore intersect at a right angle, giving the second case of plane-section geodesics of the unit one-sheet hyperboloid.

There are also geodesics entirely contained in $z>0$. In the coordinatesthe metric isFor a unit-speed geodesic, the Clairaut first integral for a surface of revolution is $c=(1+z^2)\dot{\varphi }$, and the constant-speed equation becomesChoose $c>1$ and initial height $z_0=\sqrt{c^2-1}$. The geodesic initially tangent to the parallel has $\dot{z}=0$, and the displayed identity prevents it from entering $0<z<z_0$. Since the parallel at $z_0>0$ is not itself a geodesic, the curve turns there and otherwise has $z>z_0$. Hence it remains entirely in $z>0$; this is a geodesic trapped in one half of the unit one-sheet hyperboloid.

Finally, the waistis a geodesic and is preserved setwise by every isometry of the hyperboloid. Indeed, its Gaussian curvature isThe value $K=-1$ occurs exactly at $z=0$. Since Gaussian curvature is intrinsic and therefore preserved by isometries, every isometry preserves the waist. Thus the answer to the final question is also yes, with the isometry-invariant waist geodesic of the unit one-sheet hyperboloid as an example.

Write $q^1=u$, $q^2=v$, ${\sigma }_i={\partial }_i\sigma$, and let $g_{ij}={\sigma }_i\cdot {\sigma }_j$ be the first fundamental form. Sincedifferentiating ${\sigma }_u\cdot \dot{\gamma }$ and ${\sigma }_v\cdot \dot{\gamma }$, or equivalently taking the two inner products of the second formula with ${\sigma }_k$, givesThe matrix $(g_{k\ell })$ is invertible because $\sigma$ is a regular embedded surface parametrization. Consequently the two geodesic equations hold exactly when $\ddot{\gamma }$ is orthogonal to both tangent vectors ${\sigma }_u$ and ${\sigma }_v$, which says precisely that $\ddot{\gamma }$ is a normal vector. This is the ambient acceleration criterion for a surface geodesic.

Because $\dot{\gamma }$ is tangent and $\ddot{\gamma }$ is normal,Thus an affinely parametrized geodesic has constant speed, as recorded by constant speed of an affinely parametrized geodesic.

For a surface of revolution, use profile arc length $s$ and azimuth $\varphi$. Its Riemannian metric isThe azimuth is an ignorable coordinate, so the corresponding geodesic equation has the first integralIf $v=|\dot{\gamma }|$ and $\theta$ is the oriented angle with the parallel, then the component of velocity along the parallel isSince $v$ is constant,This is the Clairaut first integral for a surface of revolution.

Use the global parametrizationIts tangent vectors areandThus $X$ is a regular parametrization and the graph is a smooth surface.

The first fundamental form isChoosing the upward orientation, the Gauss map is

Put $W=\sqrt{1+h_x^2+h_y^2}$. An upward-pointing Gauss map isFor tangent vectors $v,w$, differentiation ofgivesThe Weingarten map is self-adjoint and $S=-dN$, so the right side is $-2I{I}^S(v,w)$. HenceAlternatively differentiating $⟨N,{\varphi }_i⟩=0$ givesThereforeThis is the fundamental forms of a graph surface.

For every tangent vector $v\perp x_0$, stationarity of $Q$ on the sphere gives $2v^{T}A{x}_0=0$. Hence $A{x}_0$ is parallel to $x_0$, say $A{x}_0=E{x}_0$. Taking the inner product with $x_0$ gives $E=Q(x_0)$. The spectral theorem shows that this is the largest eigenvalue.

For $A=\left(\begin{array}{cc}1& t\\ t& 1\end{array}\right)$ the eigenvalues are $1\pm t$, soIts graph is a V translated upward and is convex.

WriteDifferentiating the prescribed identity $u(\psi (r,z),r,z)=0$ in its two tangential directions givesThe unit normal is $(1,-{\psi }_r,-{\psi }_z)/\sqrt{D}$, so the second item of Cauchy data becomesConsequentlySubstitution into the principal symbol from part a shows that the graph is non-characteristic exactly where

For the surface $F(x,z,t)=z-\zeta (x,t)=0$, an outward normal vector and positively oriented tangent vector areWith $\zeta =ϵe^{ikx}$ and $|kϵ|\ll 1$, linearization gives

A tangential boundary derivative differentiates the boundary restriction of a function along a tangent vector. Prescribing it fixes the boundary value only up to a constant on each connected boundary component. This differs from a Neumann boundary condition, which specifies a normal derivative.