Stationary increments give $N_5-N_3\sim \mathrm{Poisson}(2\lambda )$. ThereforeThere is zero probability of an arrival at either deterministic endpoint, so endpoint inclusion does not alter the result.

Since $N_3\sim \mathrm{Poisson}(3\lambda )$,

Assume convergence in probability and fix $\epsilon >0$. On the event $|Y_n-Y|\le \epsilon$, $Y\le x-\epsilon$ implies $Y_n\le x$, while $Y_n\le x$ implies $Y\le x+\epsilon$. HenceTaking lower and upper limits in $n$ removes the error probabilities. If $x$ is a continuity point of $F_Y$, letting $\epsilon \downarrow 0$ squeezes both bounds to $F_Y(x)$. Thus

In the analytic continuation inside the circle, only the image location $c=a^2/b$ is singular; $b$ lies outside. Squaring the velocity givesThe last term has no residue, and the middle term has residuebecause $AB=m^2{a}^2/b^2$ and $c-b=-(b^2-a^2)/b$. The residue theorem and part (a) therefore yieldThus $F_y=0$ and $F_x>0$. The force points towards the planar potential doublet and is independent of its orientation, since the factors $e^{i\alpha }$ and $e^{-i\alpha }$ cancel in $AB$.

The complex velocity obtained from part (b) isFor a nonzero planar potential doublet, $B/A=(a^2/b^2)e^{-2i\alpha }$. Its zeros therefore satisfyand henceThe denominators are nonzero because $a/b<1$. For either sign,so $|z_{\pm }|=a$. They are distinct because $t_{+}\ne t_{-}$ and $t\mapsto (c-bt)/(1-t)$ is injective, its determinant being $c-b\ne 0$. Clearing the velocity denominators gives a quadratic with these two roots, so there are no further finite stagnation points. Thus both stagnation points lie on the obstacle boundary.

The Milne-Thomson circle theorem states that if $f$ is an analytic complex potential with no singularity on or inside $|z|=a$, thenmakes $|z|=a$ a streamline while retaining the original exterior singularities and adding reflected singularities inside the circle. On the circle the two terms are conjugates, so the resulting streamfunction is zero there.

Put $A=m{e}^{i\alpha }$ and $c=a^2/b$. Applying the formula to the exterior planar potential doublet givesEquivalently, up to an irrelevant constant,The image planar potential doublet is inside the obstacle at $c<a$; the only physical fluid singularity is the prescribed one at $b$.

Orient the body boundary counterclockwise and write its tangent displacement as $(dx,dy)$. The outward normal from the body has displacement-weighted components $(dy,-dx)$. Pressure therefore exerts the force components $F_x=-\oint pdy$ and $F_y=\oint pdx$, orFor steady irrotational incompressible flow at constant density, Bernoulli gives $p=p_0-\rho |w^{\prime }(z){|}^2/2$, with any constant far-field kinetic term absorbed into $p_0$. Its constant contribution integrates to zero. The boundary is a streamline, so $\mathrm{Im}w$ is constant there and $w^{\prime }(z)dz$ is real. Thus $w^{\prime }dz=\overline{w^{\prime }}d\bar{z}$, giving $|w^{\prime }{|}^{2}d\bar{z}=(w^{\prime }{)}^{2}dz$. This identity element remains valid at a stagnation point as both sides vanish. Substitution yields Blasius force theorem:

The Born probability is the integral of $|{\psi }_n{|}^2$ over the indicated interval. Changing variables to $\xi$ cancels the dimensional normalization factors and givesSince $\frac{d}{d\xi }(\xi e^{-{\xi }^2})=e^{-{\xi }^2}-2{\xi }^2{e}^{-{\xi }^2}$, integration givesAs $\ell \to \infty$, the correction term tends to zero and $P_0(\ell )\to 1$ by normalization. Hence $\boxed{P_1(\ell )\to 1}$. This is expected because the expanding interval exhausts the whole position space and the state is normalized: the particle is somewhere on the line with total probability one.

Choose the $z$-axis along the cylinder and the origin at its midpoint, so $x^2+y^2\le b^2$ and $-a\le z\le a$. Reflection symmetry puts the centroid at the origin and makes all products of inertia zero. The mass isCylindrical integration gives ${\int }_A\rho x^{2}dV={\int }_A\rho y^{2}dV=m{b}^2/4$, while the uniform height coordinate has ${\int }_A\rho z^{2}dV=m{a}^2/3$. ConsequentlyThe two equal transverse entries reflect the rotational symmetry about the cylinder axis.

Write $r=(x_1,x_2,x_3{)}^T$ for position relative to the centroid. The inertia tensor there isIn particular its diagonal entries integrate the squares of distances from the corresponding axes, and its off-diagonal entries are the negative products of coordinates.

Insert ${\mathbf{v}}_Q={\mathbf{v}}_G+\mathbf{\omega }\times r$ into $T=\frac{1}{2}{\int }_A\rho |{\mathbf{v}}_Q{|}^{2}dV$. The cross term vanishes because ${\int }_A\rho rdV=0$. Using $|\mathbf{\omega }\times r{|}^2=|\mathbf{\omega }{|}^2|r{|}^2-(\mathbf{\omega }\cdot r{)}^2$ givesIf $a$ is the vector from $G$ to $O$, positions relative to $O$ are $r-a$. Expanding the defining tensor integral with this substitution again removes the terms linear in $r$. Thus the tensor parallel axis theorem isSince $O$ is fixed, ${\mathbf{v}}_G=\mathbf{\omega }\times (-a)$, so $m|{\mathbf{v}}_G{|}^2=m{\mathbf{\omega }}^T(|a{|}^{2}I-a{a}^T)\mathbf{\omega }$. Substitution into the energy decomposition yields

For uniform density, the centroid is the volume average of position and coincides with the centre of mass:A point fixed in the body has ${\mathbf{r}}_{OQ}={\sum }_i{q}_i{\mathbf{e}}_i$ with time-independent body coordinates $q_i$. Since $O$ is fixed, part (a) givesThe centroid is fixed relative to the body as well, so ${\mathbf{v}}_G=\mathbf{\omega }\times {\mathbf{r}}_{OG}$. Using ${\mathbf{r}}_{OQ}={\mathbf{r}}_{OG}+{\mathbf{r}}_{GQ}$ therefore gives