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An integral curve of a vector field $X$ is a smooth curve $\gamma :I\to M$ satisfying $\dot{\gamma }(t)=X_{\gamma (t)}$ for every $t$ in its interval of definition. Passing through $p$ at time zero means $\gamma (0)=p$. The tangent-vector equality is intrinsic; in a coordinate chart $x:U\to {\mathbb{R}}^n$ it becomesSmoothness makes the coordinate right side locally Lipschitz continuous. The Picard-Lindelof theorem supplies a locally unique solution through each coordinate initial value.

In fact, two integral curves of $X$ with the same initial point agree throughout their common time interval containing zero. To pass from local to global uniqueness, the set of coincidence times is nonempty and closed by continuity and the Hausdorff property of the smooth manifold. At a coincidence time, put both curves in one coordinate chart; local ordinary differential equation uniqueness makes them agree on a neighborhood of that time. Thus the coincidence set is open as well as closed in the connected common interval, so it is the whole interval. This argument applies around any prescribed time, not only zero.

By joining all extensions that agree on overlaps, one obtains the unique maximal integral curve of a vector field through $p$. Its maximal interval need not be all of $\mathbb{R}$: completeness is an additional property of the vector field.

The local flow of $X$ is the smooth map $\Phi :D\to M$, where $D\subset \mathbb{R}\times M$ is the maximal open domain of initial-point/time pairs, whose curve $t\mapsto \Phi (t,q)={\varphi }_t(q)$ is the maximal integral curve of a vector field through $q$, with ${\varphi }_0(q)=q$. Smooth dependence in the ordinary differential equation theorem gives joint smoothness and openness of $D$. Each time slice is a diffeomorphism between its domain and image, with inverse ${\varphi }_{-t}$.

Here is the group law, including its domain qualification. Fix $s,q$ where the relevant trajectories exist. Both curvessolve the same autonomous ordinary differential equation and have value ${\varphi }_s(q)$ at $t=0$. The uniqueness from the first part therefore giveson their common interval. These are local identities; neither the vector field nor its local flow has been assumed complete.

For the commuting local flows ${\varphi }_t$ and ${\psi }_s$, differentiate ${\varphi }_t({\psi }_s(q))={\psi }_s({\varphi }_t(q))$ with respect to $s$ at zero. The chain rule givesThus the pushforward of a vector field $Y$ by the $X$ flow equals $Y$. Interchanging the two fields gives the analogous identity for $X$ under ${\psi }_s$.

Now let ${\mathcal{D}}_q=\mathrm{span}(X_q,Y_q)$. Pointwise linear independence makes this a rank-two smooth distribution. The identity just proved implies $[X,Y]=0$: in coordinates, differentiating $D_q{\varphi }_t{Y}_q=Y_{{\varphi }_t(q)}$ at $t=0$ gives $DX(q)Y_q=DY(q)X_q$, precisely the vanishing Lie bracket of vector fields. The bracket of two local sections is also in $\mathcal{D}$, sinceand the analogous terms for $[fX,gX]$ and $[fY,gY]$ remain in the same span. Therefore $\mathcal{D}$ is an involutive distribution.

The Frobenius theorem says that a smooth constant-rank involutive distribution has local coordinates $(u^1,\dots ,u^n)$ in which it is spanned by ${\partial }_{u^1},{\partial }_{u^2}$, and has unique maximal connected integral manifolds through its points. These are the leaves of a regular foliation. Let $S$ be the maximal leaf through $p$. The global confinement to an immersed leaf gives the answer: $S$ is a two-dimensional immersed integral manifold and every smooth $\mathcal{D}$-tangent curve starting at $p$ stays in $S$.

For the curve assertion, in each Frobenius chart the transverse coordinate functions $u^3,\dots ,u^n$ have zero derivative along such a curve. Hence its segment in that chart stays in one plaque. On any compact subinterval of the curve's parameter interval, a finite chain of overlapping such segments connects the initial plaque to the last; all belong to the same maximal leaf. Exhausting the parameter interval proves the assertion for every time.

The two commuting local flows also give the leaf parametrization directly near $p$:Its partial derivatives are ${\partial }_{s}F=X_F$ and ${\partial }_{t}F=Y_F$, by the pushforward identity. Its differential has rank two. The constant rank theorem makes a sufficiently small image an embedded local plaque with tangent plane $\mathcal{D}$. This is a concrete local construction of the integral manifold.

There is a global distinction in the word “submanifold”. If it is required to mean an embedded submanifold, the assertion for all times is not valid in general. On the three-dimensional torus ${\mathbb{R}}^3/{\mathbb{Z}}^3$, takeThe fields are independent and commute, and their dense immersed cylinder in a three-dimensional torus through zero isIt is an injectively immersed cylinder $\mathbb{R}\times (\mathbb{R}/\mathbb{Z})$ and is dense in the torus. Indeed, fixing $x\mathrm{mod}1$ and varying $s$ by integers makes $\alpha s\mathrm{mod}1$ dense by an irrational rotation of the circle; $t$ is unrestricted. Every point of this leaf is reached by a tangent curve $r\mapsto (rs,rt,\alpha rs)\mathrm{mod}{\mathbb{Z}}^3$. An embedded surface containing it would, in a submanifold chart near one of its points, be a closed coordinate plane containing a dense subset of the ambient open chart, an impossibility. Thus the global conclusion uses an immersed leaf; the embedded conclusion is local, for curve segments remaining in a suitable chart.

Use the classical risk model surplus $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, with $c>0$, and write $\phi$ for its ultimate survival probability. The first-claim decomposition for survival probability follows by conditioning on the first arrival time of the Poisson process. That time has exponential distribution of rate $\lambda$. A claim at time $t$ leaves capital $u+ct-x$ if $x\le u+ct$, after which the Markov property restarts the same risk model. ConsequentlyPut $C(v)={\int }_0^v\phi (v-x)f(x)dx$ and change variable $v=u+ct$:The convolution $C$ is continuous since $\phi$ is bounded and $f$ is integrable. Differentiating proves the survival integro-differential equation

For the specified claim law, expand its probability density function as $f(x)=\frac{1}{2}{e}^{-x}+\frac{1}{4}{e}^{-x/2}$. This is an equal-weight mixture distribution of exponential distributions with rates $1$ and $1/2$. In the convolution, set $z=u-x$ to obtainSince $\lambda /c=10/21$, the requested coefficients are

For the exponential-mixture differential equation for survival probability, define $A(u)=e^{-u}{I}_1(u)$, $B(u)=e^{-u/2}{I}_2(u)$ and $r=\lambda /c$. Differentiating the integrals yieldsApply $(D+1)(D+\frac{1}{2})$ to the last equation, where $D=d/du$. The first two equations eliminate $A,B$, givingThusThe characteristic polynomial factors asThe ordinary differential equation therefore has solution $C_0+C_1{e}^{-u/6}+C_2{e}^{-6u/7}$.

The claim expected value is $\frac{1}{2}\cdot 1+\frac{1}{2}\cdot 2=3/2$. The relative safety loading is $\rho =2.1/(3/2)-1=2/5$, and the given zero-capital survival probability is $\phi (0)=2/7$. The limit at infinity sets $C_0=1$. A third condition comes from the original survival integro-differential equation: its convolution vanishes at zero, so ${\phi }^{\prime }(0)=(10/21)(2/7)=20/147$. Thereforewhich gives $C_1=-20/29$ and $C_2=-5/203$. The final survival probability and ultimate ruin probability areBoth exponential coefficients of $\psi$ are positive. Hence $\phi$ increases from $2/7$ to one, and $\psi$ decreases from $5/7$ to zero. The initial slope is essential: the two stated boundary values alone do not determine all three constants of the eliminated differential equation.

Use $(y,w)$ to label the initial point in phase space. The characteristic equations for a transport equation are $\dot{X}=a(V)$ and $\dot{V}=0$, with $X(s)=y$ and $V(s)=w$. Since the velocity remains fixed, the ordinary differential equations integrate directly:This characteristic flow map exists for all real times, even if $a$ grows rapidly at infinity: each trajectory samples just one fixed velocity. Its inverse is $S_{t,s}$, and its Jacobian matrix is the block triangular matrixConsequently its Jacobian determinant is one. The nonzero Jacobian determinant of $a$ is unnecessary for this construction.

Parametrize the characteristics by $s$, starting from $(x,y,u)=(1,r,r)$. Their ordinary differential equations areThey give $x=e^s$, $y=e^s(r+s)$ and $u=r+s$. Thus $s=\log x$ and $r=y/x-\log x$, soIndeed, $u_x=-y/x^2$, $u_y=1/x$ give $x{u}_x+(x+y)u_y=1$, and $u(1,y)=y$. The initial line is noncharacteristic, and its characteristics cover the half-plane $x>0$. The data do not determine a solution on $x<0$; a smooth solution through the origin is impossible because the differential equation there would read $0=1$.

The complex form of the Chordal Loewner equation, driven by a continuous real Loewner driving function, isFor $z\in \mathbb{C}\setminus \{{\xi }_0\}$ this is an ordinary differential equation up to its maximal lifetime, when the solution reaches the driving singularity. The coefficients respect complex conjugation, so the lower-half-plane flow is the conjugate of the upper-half-plane flow. On surviving real points it is the real boundary flow. The hydrodynamic normalization at infinity and factor two correspond to half-plane-capacity parameterization $\mathrm{hcap}(K_t)=2t$.

Set $\tau =T-t$, $\kappa =\lambda -\rho b\theta$ and $k=\theta (\theta -1)/2$. For the Ornstein-Uhlenbeck process volatility, the transformed partial differential equation isUse the Gaussian volatility exponential-quadratic transform ansatz $V=\exp (P+Q\sigma +R{\sigma }^2)$, where the coefficients depend on $\tau$. Its derivatives satisfyMatching the constant, linear and quadratic powers of $\sigma$ givesThe terminal condition requiresThese polynomial ordinary differential equations have a unique local solution by the Picard's theorem for ordinary differential equations. Substitution then proves the desired partial differential equation solution on every horizon for which the coefficient solution remains finite. The Riccati equation for $R$ is solved first; $Q$ subsequently solves a linear equation and $P$ is an integral of known coefficients.

Unrestricted global existence needs a qualification. Take $a=0$, $\lambda =b=1$, $\rho =0$ and $\theta =2$. Then $Q=0$ and the Riccati equation becomesThis solves the initial condition but explodes at $\tau =3\pi /4$. Therefore no finite real exponential-quadratic solution with the required terminal condition exists on an entire horizon $T\ge 3\pi /4$ for these allowed parameters. The correct general claim is local existence, or existence before the Riccati moment-explosion horizon.

A useful sufficient global condition is $0\le \theta \le 1$, so $k\le 0$. If $b\ne 0$, the lower equilibriumtraps the solution in $r_{-}\le R(\tau )\le 0$: the polynomial vector field points inward at the upper endpoint and vanishes at the lower endpoint. If $b=0$, the equation for $R$ is linear. In either case there is no finite-time explosion; $Q$ is then a linear equation with coefficients bounded on compact time intervals, and $P$ is finite on those intervals. This proves the intended ansatz globally under that sufficient parameter restriction, without asserting it for every real $\theta$.

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$: its elements are smooth functions with compact support. For a fixed compact set $K$, put ${\mathcal{D}}_K=\{\phi \in C^{\infty }(\mathbb{R}):\mathrm{supp}\phi \subset K\}$ and use the seminorms $p_m(\phi )={\max }_{0\le j\le m}\Vert {\phi }^{(j)}{\Vert }_{\infty }$. The space of test functions carries the usual test-function inductive limit topology of these spaces. In particular, a sequence converges precisely when its supports eventually lie in one compact set and every derivative converges uniformly. Thus ${\phi }_j\to \phi$ means that a common $K$ contains their supports and $p_m({\phi }_j-\phi )\to 0$ for every $m$.

A distribution is a continuous linear functional on this space of test functions, and ${\mathcal{D}}^{\prime }(\mathbb{R})$ denotes their space. We use complex-linear, bilinear pairings $⟨u,\phi ⟩$. Equivalently, for every compact set $K$ there are $C_K$ and a nonnegative integer $m_K$ such thatThe usual weak convergence of distributions is $u_j\to u$ if $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every test function. This specifies the convergence used below; it does not require convergence in any norm.

The distributional derivative is defined byThe derivative map sends ${\mathcal{D}}_K$ continuously to itself, with $p_m({\phi }^{\prime })\le p_{m+1}(\phi )$. The preceding continuity estimate therefore proves that $u^{\prime }$ is again a distribution. This definition extends the ordinary derivative of a smooth function, by integration by parts.

Choose the translation of a distribution convention ${\tau }_{h}f(x)=f(x-h)$. Its action on a test function isFor fixed $h$, the translated test functions have translated compact support and unchanged derivative sup norms, so this defines a distribution. For the differentiability of distribution translations, apply the Taylor theorem in integral form:All supports lie in one slightly enlarged compact set for $|h|\le 1$, and the same identity for every derivative proves uniform convergence. The continuity of $u$ now givesThe minus sign in the translation parameter is necessary for this convention.

For (a), choose a test function $\eta$ with $\int \eta =1$. If $\int \psi =0$, then $F(x)={\int }_{-\infty }^x\psi (s)ds$ is a test function, since the zero integral makes it vanish beyond both ends of the compact support. If $u_1^{\prime }=0$, then $⟨u_1,\psi ⟩=⟨u_1,F^{\prime }⟩=0$. Decompose $\phi =(\int \phi )\eta +\psi$ to obtainConversely, these constant regular distributions have zero distributional derivative. This also proves the general fact that a distribution with zero derivative is constant.

For (b), choose a cutoff function $\rho$ equal to one near zero. Every test function decomposes asThe quotient extends as a smooth function at zero, and it has compact support. If $x{u}_2=0$, the definition of multiplication of a distribution by a smooth function gives $⟨u_2,\phi ⟩=⟨u_2,\rho ⟩\phi (0)$. Conversely, $x{\delta }_0=0$. Thus the kernel of multiplication by a coordinate is exactlyIn particular, derivatives of the Dirac delta distribution are not additional solutions: $x{\delta }_0^{\prime }=-{\delta }_0$.

Now write $D=d/dx$ and $P(z)=-z^n+{\sum }_{j=0}^{n-1}{a}_j{z}^j$, where $n\ge 1$. The characteristic roots of a constant-coefficient differential equation are the distinct roots of a polynomial ${\lambda }_1,\dots ,{\lambda }_r$ of $P$, with multiplicities $m_1,\dots ,m_r$. The distributional regularity of a constant-coefficient ordinary differential equation can be proved without assuming regularity in advance. Put $q_j(z)=(z-{\lambda }_j{)}^{m_j}$ and $Q={\prod }_j{q}_j=-P$. Since the $q_j$ are coprime polynomials, polynomial division and Bezout identity give polynomials $b_j$ such thatFor example, invert $Q/q_j$ modulo $q_j$, sum the resulting expressions, and absorb a remaining multiple of $Q$ into one coefficient. For $P(D)v=0$, the kernel decomposition for coprime polynomials therefore givesThe Leibniz rule for multiplication of a distribution by a smooth function implies $D^{m_j}(e^{-{\lambda }_{j}x}{v}_j)=0$. Repeatedly using a distribution with zero derivative is constant shows that a distribution with $m$th derivative zero is a polynomial of degree at most $m-1$: subtract the polynomial primitive of its constant $(m-1)$st derivative, and induct on $m$. Consequently the most general exponential polynomial solution of a constant-coefficient differential equation isEvery displayed term is annihilated by $P(D)$, so all coefficients are allowed. For the linear independence of these $n$ functions, on a relation, apply ${\prod }_{i\ne j}(D-{\lambda }_i{)}^{m_i}$; on $e^{{\lambda }_{j}x}$ times a polynomial of degree less than $m_j$, each remaining factor acts invertibly on that polynomial space. Hence the $j$th polynomial must vanish. Every distributional solution is an analytic function, and in particular a classical solution. For real coefficients and real-valued distributions, take complex conjugate coefficients at conjugate characteristic roots of a constant-coefficient differential equation, or equivalently use the corresponding real sine and cosine forms.

For the last equation, the operator is $xP(D)$: the coordinate multiplies the result of differentiation. The kernel of multiplication by a coordinate says precisely thatChoose the retarded fundamental solution of a constant-coefficient ordinary differential operator $E=Hw$, where $H$ is the Heaviside function and $w$ is the analytic solution of $P(D)w=0$ withSuch $w$ exists uniquely by the elementary initial value problem for a constant-coefficient ordinary differential equation; alternatively the residue construction in the next solution gives it explicitly. The distributional jump formula for a Heaviside product isIt follows by induction from $D(Hw)=H{w}^{\prime }+w(0){\delta }_0$, using integration by parts. With the chosen initial derivatives, every lower-order jump term vanishes and the leading coefficient $-1$ gives $P(D)E={\delta }_0$. Subtracting $CE$ reduces the last equation to the homogeneous one. Thus the coordinate-degenerate constant-coefficient differential equation has exactly the solutionsThere are $n+1$ independent constants. For $n\ge 2$, derivatives through order $n-2$ match at zero, while the derivative of order $n-1$ may jump; for $n=1$, the function itself may jump. An arbitrary distribution concentrated at zero cannot be added, because its image under the nonzero leading derivative would contain a nonvanishing highest derivative of the Dirac delta distribution.

Put $z=(X,V)$ and $b(t,z)=(V,F(t,X,V))$. On a small closed time interval and a closed ball around the initial point, let $M$ bound $b$ and let $K$ be its spatial Lipschitz constant. The mapsends the corresponding closed ball of continuous paths into itself when the time length times $M$ is at most the ball radius. If the time length times $K$ is less than one, it is a contraction in the supremum norm. The Banach fixed-point theorem gives a local solution and uniqueness; differentiating its integral equation gives the ordinary differential equation. Overlapping local solutions agree by this uniqueness.

For continuation, the growth assumption gives, on any bounded time interval,for forward time, with the analogous reversed-time bound. The Gronwall inequality bounds the trajectory on that interval. A finite terminal time is impossible: within the resulting compact ball the field is bounded, so the path has a limit at the endpoint, and the local construction restarts there. This proves the global characteristic flow under linear growth. No differentiability of the flow with respect to its initial point is needed for this existence and uniqueness proof.

The complete Butcher tableau in the PDF suppliesThese are the two Gauss nodes and the corresponding Gauss--Legendre Runge-Kutta method. We can verify its order directly rather than infer it from its name. Write $e=(1,1{)}^T$, $C=\mathrm{diag}(c)$ and let powers of $c$ be componentwise. The Butcher order conditions through order four areFor these coefficients $Ae=c$, $Ac=c^2/2$ and $c^2=c-e/6$. Also $b^T{c}^j=1/(j+1)$ for $j=0,1,2,3$, as follows by averaging the two values $1/2\pm s$. These identities give all eight displayed Butcher order conditions; for example $b^{T}CAc=b^T{c}^3/2=1/8$, $b^{T}A{c}^2=b^T(c^2/2-c/6)=1/12$, and $b^T{A}^{2}c=b^{T}A{c}^2/2=1/24$. They establish order at least four for smooth nonlinear ordinary differential equations.

To exclude order five, apply the method to the Dahlquist test equation. Solving its stage equations gives the stability functionThe exact solution has coefficient $1/120$ at degree five. Hence the method has order exactly four; its one-step defect is $O(h^5)$.

Assume $D_0,n_0,Q>0$. The density-dependent diffusion coefficient makes this a cubic porous medium equation, since its two-dimensional form is $n_t=[D_0/(3n_0^2)]\Delta (n^3)$. The mass-preserving similarity solution is a radial cubic-diffusion source profile. At fixed $r$, the proposed scaled density givesSeparation is possible whenwhere $\gamma$ is a constant. The resulting ordinary differential equation isRecognize the second term as $\gamma ({\xi }^{2}F{)}^{\prime }$. Regularity and zero radial flux at the origin set the integration constant to zero:In the region $F>0$, this reduces to $F{F}^{\prime }=-\gamma \xi$, giving $F^2=1-\gamma {\xi }^2$ from the central normalization. For a nonnegative profile whose front is precisely at $\xi =1$, the first zero fixes $\gamma =1$. Continuity of the density excludes truncating a positive value at the front. HenceHere integrating the scale equation gives ${\lambda }^6=6D_{0}t/r_0^2+\text{constant}$, and the point-source initial condition requires $\lambda (0)=0$.

Use the planar area element, not the one-dimensional area under the drawn profile. Mass conservation givesThus the scale and the complete density arewhere $[u{]}_{+}=\max (u,0)$. The factors of $n_0$ are necessary on dimensional grounds; the length scale is not a function of $Q$ alone.

The front and peak obeyEach radial profile starts with horizontal tangent at the origin, decreases to a square-root edge, and is identically zero beyond $R(t)$. Later profiles are wider and lower. This is finite propagation in porous-medium diffusion, because the diffusion coefficient vanishes at zero density. The sketch uses dimensionless time $\tau =6D_{0}t/r_0^2$; it preserves the radial integral $2\pi \int rndr$, not the unweighted area under each curve.

At the front $F^{\prime }$ diverges, so the profile is a weak solution, not a globally smooth classical solution. Nevertheless $F^2{F}^{\prime }=-\xi \sqrt{1-{\xi }^2}$ tends to zero there. The density and flux both match their zero exterior values, so extending the interior solution by zero creates no spurious boundary source in the conservation law. The behavior at the origin is regular because $F^{\prime }(0)=0$.

Finally, the point release is recovered as a distributional initial condition. For any continuous compactly supported test function $\phi$ on the plane, the total density remains $Q$ and all its support lies in the shrinking disk $r\le R(t)$. ThereforeThus $n(\cdot ,t)\to Q{\delta }^{(2)}$ as $t\downarrow 0$, in the sense of the Dirac delta function.

Here the singular perturbation has an exact conservative structure:The integrating factor for this first-order linear differential equation givesAt $x=1$ and $x=-1$ the integral terms have opposite signs while the exponential factors agree. Equal boundary data therefore force $C=0$, and normalization gives the unique exact solutionThis exposes the Gaussian interior layer of a conservative drift equation. Its central scale is $x=O(\sqrt{ϵ})$, where $x=\sqrt{ϵ}\xi$ gives $Y^{\prime \prime }+\xi Y^{\prime }+Y=0$. The central Gaussian function profile is $e^{-{\xi }^2/2}$ with amplitude $e^{1/(2ϵ)}$.

Away from the center, $|x|\gg \sqrt{ϵ}$, use exponential WKB approximation branches rather than an assumed bounded algebraic outer solution. The reduced ordinary differential equation $x{y}_0^{\prime }+y_0=0$ would give $C/x$; matching both equal positive endpoint values with that algebraic branch is impossible. The exact solution selects its zero coefficient and the exponentially large branch with a Gaussian function profile instead. Near either endpoint, a distance $O(ϵ)$ changes $y$ by order one relative to its endpoint value: with $r=(1-|x|)/ϵ$, $y=e^{r-ϵr^2/2}$. These endpoint scales describe normalization of the same global exponential solution.

Thus a central $O(\sqrt{ϵ})$ turning region, exponential outer regions on both sides, and $O(ϵ)$ endpoint normalization scales give the complete regional description. The peak is exponentially large, so no uniformly bounded regular outer expansion should be presumed. The exact expression already provides a uniform solution without further matching calculations.

For two Stokes flows $(u^{(1)},{\sigma }^{(1)})$ and $(u^{(2)},{\sigma }^{(2)})$ of the same dynamic viscosity in the same region, with no volume force, the Lorentz reciprocal theorem statesIndeed, the divergence of their cross-work difference isThe pressure terms vanish by incompressible flow, and the derivatives of the Cauchy stress tensors vanish by the Stokes equation. The divergence theorem proves the result.

Use the convention that $F,G$ are the force and torque exerted by the body on the fluid. By Linearity of Stokes flow, $(F,G)=\mathsf{R}(U,\Omega )$. Applying the Lorentz reciprocal theorem to two independent rigid motions gives $q_1^T\mathsf{R}{q}_2=q_2^T\mathsf{R}{q}_1$, so the hydrodynamic resistance matrix is a symmetric matrix. The boundary-power identity givesThis is strictly positive for nonzero rigid motion: equality would imply $e=0$, hence a rigid motion throughout the connected fluid, which must vanish since the fluid is at rest at infinity. The no-slip boundary condition would then force $q=0$. Therefore $\mathsf{R}$ is a positive-definite matrix. Reversing to the fluid-on-body force changes the sign of the force law, not the positive resistance coefficients.

For the two rods, take the torque about $O$ and use body axes. On the $x$-rod, $X=(s,0,0)$, $0\le s\le 2L$, and its slender-body force density isOn the $y$-rod it isIntegrate $f$ and $X\times f$ along both rods, using $\int ds=2L$, $\int sds=2L^2$ and $\int s^{2}ds=8L^3/3$. A convenient dimensionally uniform statement of the complete right-angle two-rod resistance matrix isIn physical coordinates this means translation–translation entries scale as $CL$, the two translation–rotation blocks as $C{L}^2$, and rotation–rotation entries as $C{L}^3$. The planar block has inverseThis verifies the printed hydrodynamic mobility matrix with its third velocity component $L{\Omega }_z$; the TeX aid's ${\Omega }_x$ is an OCR error.

Take laboratory vertical velocity positive upwards. The body axes are $e_1=(\cos \theta ,\sin \theta )$ and $e_2=(-\sin \theta ,\cos \theta )$. In quasistatic sedimentation the force and torque exerted on the fluid equal the gravitational resultants on the body:The out-of-plane block is unforced, so its positive hydrodynamic resistance matrix gives $U_z={\Omega }_x={\Omega }_y=0$. Substitution in the planar hydrodynamic mobility matrix givesandFor $0\le \lambda <1$, this ordinary differential equation has positive right side on $0\le \theta <\pi /4$, with a stable zero at $\pi /4$. Uniqueness prevents crossing that equilibrium in finite time. For $\lambda >1$ the initial angular velocity is negative, and the stable equilibrium reached from zero is $-3\pi /4$. The heavier-end body turns clockwise through $3\pi /4$; it does not settle at $\pi /4$. More explicitly, with $K=(1-\lambda )mg/(4C{L}^2)$,where the continuous branch has $\theta -\pi /4\in (-\pi ,0)$. At $\lambda =1$, $\theta$ remains zero.

Transforming the translational velocity back to laboratory axes givesFor $\lambda \ne 1$, eliminate time between $U_h$ and the angular velocity:At either limiting orientation $\sin \theta =\cos \theta$. Thus both cases have the same net horizontal displacement,For $\lambda <1$ the drift is monotonically rightwards. For $\lambda >1$ it first moves left, reaching $x_O-x_O(0)=(2L/3)(1-\sqrt{2})$ at $\theta =-\pi /4$, then reverses. The total horizontal path length in this second case is $(2L/3)(2\sqrt{2}-1)$, whereas its net displacement is $2L/3$ rightwards. At $\lambda =1$ it falls vertically without rotation or drift; taking the infinite-time limit before $\lambda \to 1$ is consequently singular.