Neglecting terms of order ${\Omega }^2$ and linearizing in $\theta$ givesThe second equation and the initial data implyIts contribution to the first equation is of order ${\Omega }^2$, so the retained equation isThusIntegrating the associated $y$ velocity and using $y(0)=0$ gives

The circular constraint isIts tangent in the $xz$-plane is $(\cos \theta ,0,-\sin \theta )$, while the normal force is radial. Projecting the equations from part (a) onto this tangent eliminates the normal force and givesBecause the ramp is translation-invariant along $y$, $N_y=0$; using $\dot{x}=R\dot{\theta }\cos \theta$ gives

For rest in the rotating frame, $\dot{\theta }=\dot{y}=\ddot{\theta }=\ddot{y}=0$. Assuming $\Omega \ne 0$, the second equation gives $y=0$, and the first givesOn the semicircular ramp the rest points are thereforeand, when ${\Omega }^{2}R\ge g$,

The term $-2\mathbf{\Omega }\times \dot{\mathbf{x}}$ is the Coriolis acceleration, and $-\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{x})$ is the centrifugal acceleration. With $\mathbf{\Omega }=-\Omega \hat{\mathbf{z}}$, the Cartesian components are

Part (c) givesThe half-space $x_1>0$ is a simply connected domain, so the Poincaré lemma makes this curl-free field a conservative vector field: it is the gradient of a single-valued scalar potential.

Using $\tan u+\cot u=2/\sin (2u)$,The spherical-coordinate formula from part (b) shows that a potential must satisfy $f_{\varphi }=2$ and may be independent of $r,\theta$. On $x_1>0$ the azimuthal angle has the single-valued branchHence one solution, up to an additive constant, is

Each field has only an azimuthal component and is independent of $r$ and $\varphi$. The curl in spherical coordinates therefore reduces toand its $\varphi$ component vanishes. The half-angle identitiesshow in both cases that

Apply part (a) with $g(r)=r^{-3}$. Thenaway from the origin, explicitly confirming that each resulting curl has zero divergence.

Differentiating the spherical coordinate systemgives the orthogonal decompositionwhereThus the scale factors of orthogonal coordinates are

Since the total differential satisfiescomparison of coefficients gives the gradient

Since $r=(x_j{x}_j{)}^{1/2}$, the chain rule givesFor the radial vector field $g(r)\mathbf{x}$, the product rule and Einstein summation convention give

Reflection symmetry of the cube in each coordinate plane makes every off-diagonal integralvanish because its integrand is odd in one coordinate. Permuting the three coordinate axes leaves the cube unchanged, so its three diagonal moments are equal. Thus its inertia tensor at the centre has the isotropic form

Forpart (a) and $\lambda =M{\ell }^2/6$ giveThe displayed symmetric matrix has orthogonal eigenvectorswith corresponding eigenvaluesFor a unit angular velocity, $|\mathbf{L}|=|I\mathbf{\omega }|$ is minimized and maximized along eigenvectors for the smallest and largest eigenvalues. Their ratio is

Let a change of orthonormal coordinates be represented by a rotation matrix $R$. Since both angular momentum and angular velocity are vectors,Using $L_i=I_{ij}{\omega }_j$ in both frames givesfor every vector $\mathbf{\omega }$. Multiplication by $R^{-1}=R^T$ yieldswhich is exactly the transformation law for a rank-two tensor. Thus $I_{ij}$ is a rank-two tensor.

Put $\mathbf{r}=\mathbf{x}-\mathbf{a}$. The vector triple product givesThereforeAt the centre of mass,Expanding the first formula, the terms linear in $x_i$ vanish because the centre of mass is the origin, while $\rho {\int }_{B}dV=M$. Hence the parallel axis theorem in tensor form is

Put $\eta =\psi -\varphi$. The common Dirichlet boundary condition gives $\eta =0$ on $S$. Expanding the energy functional,Integration by parts and the field equation make the cross term zero:Consequently

For $m>0$, equality holds exactly when $\eta =0$. For $m=0$, equality holds exactly when $\nabla \eta =0$, so $\eta$ is constant on the connected region; its zero boundary value then again forces $\eta =0$. The minimizing property therefore proves uniqueness for the Dirichlet problem in both cases:

Let ${\varphi }_1,{\varphi }_2$ be two solutions with the same Neumann boundary condition, and set $u={\varphi }_1-{\varphi }_2$. ThenMultiplying by $u$ and applying Green's first identity givesThe integrand is nonnegative. If $m>0$, both terms can vanish only when $u=0$, so the solution of the modified Helmholtz equation is unique.

If $m=0$, the identity only forces $\nabla u=0$. On a connected region, $u$ may be any constant. Thus solutions of the Laplace equation with prescribed normal derivative are unique only up to an additive constant, so

No. A standard counterexample comes from the log-normal distribution. Letand, for a fixed $0<|\epsilon |\le 1$, letThis is a nonnegative density distinct from $f$. For every nonnegative integer $n$, substituting $z=\log x$ makes the difference of the $n$th moments proportional toIt is the imaginary part ofwhich vanishes because its phase is $2\pi n$. The case $n=0$ also proves that $f_{\epsilon }$ is normalized. Thus the two distributions have every finite moment equal but are different. Unbounded random variables need not be determined by their moments.

The normalizing constant satisfiesFor an integer $k$,Translation by the integer $k$ merely permutes the summation indices, so the final sum is $B^{-1}$. ConsequentlyThe last equality uses the moment-generating function of the standard normal variable from part (b).

Normalization of the Gaussian distribution givesCompleting the square,Thereforeand hence

Because $U$ and $V$ are bounded, their exponential series may be integrated term by term for every real $t$. Thus their moment-generating functions satisfyand similarly for $V$. Equality of every moment gives equality term by term, so

The only positive equilibrium solves $1-y=0$ and $x-1=0$, hence isAway from a nullcline,This is separable:Integration gives the first integralThus one convenient conserved quantity is

Its gradient isso $(1,1)$ is its only stationary point in the positive quadrant. Its Hessian matrix iswhich is positive definite everywhere. Hence $(1,1)$ is a strict, indeed global, minimum.

The nearby level curves of $E$ are closed curves surrounding this minimum. Since $E$ is constant along every trajectory, a solution starting on a sufficiently small nearby level set cannot leave the region bounded by a slightly larger level set. This proves Lyapunov stability of the equilibrium: solutions initially close to $(1,1)$ remain close for all time.

Write the vector field asThe equilibrium points of a dynamical system in the closed first quadrant areThe Jacobian matrix is

At $(0,0)$ its eigenvalues are $8,-1$, so this is a saddle equilibrium. Its unstable direction is the positive $x$-axis and its stable direction is the positive $y$-axis; to first order, nearby trajectories satisfy $\dot{x}=8x$, $\dot{y}=-y$.

At $(4,0)$ the eigenvalues are $-8,3$, so it is also a saddle. The $x$-axis is the stable direction, while trajectories entering the quadrant in the $y$ direction move away.

At $(1,\sqrt{3})$,whose eigenvalues areIt is therefore a stable spiral. A point immediately to its right moves upward, so nearby trajectories spiral counterclockwise into the equilibrium. These eigendirections and the inward spiral give the requested local phase-portrait sketches.