No slip on the two fixed radial walls gives $g(\pm \alpha )=0$. The outward flux through a circular arc is ${\int }_{-\alpha }^{\alpha }{u}_{r}rd\theta ={\int }_{-\alpha }^{\alpha }gd\theta$, so source strength $Q$ requires that integral to equal $Q$. Set $g=QG$ and $ϵ=Q/\nu$. ThenAt leading order $G_0^{(3)}+4G_0^{\prime }=0$, whose general solution is a constant plus sine and cosine of $2\theta$. The symmetric source-driven branch has no sine term; for generic $\alpha$ this also follows directly by subtracting its two wall conditions. ThusIndeed its integral is $C(\sin 2\alpha -2\alpha \cos 2\alpha )=1$. At isolated angles where an odd zero-flux homogeneous mode satisfies both wall conditions, symmetry selects the stated branch.

Set $u_{\theta }=0$. Continuity gives ${\partial }_r(r{u}_r)=0$, hence $\boxed{u_r=g(\theta )/r}$. Substituting into radial and angular momentum givesFor example ${\nabla }^2{u}_r-u_r/r^2=g^{\prime \prime }/r^3$, and the radial acceleration is $-g^2/r^3$. Equality of the mixed pressure derivatives now requiresThus the equation for radial viscous flow in a wedge isIt can also be integrated once as $\nu (g^{\prime \prime }+4g)+g^2=\text{constant}$.

Use $u={\psi }_Y$, $V=-{\psi }_x$, and put $\eta =Y/x^{\gamma }$, $\psi =x^{\beta }f(\eta )$. ThenMatching $u\sim x^m$ requires $\beta -\gamma =m$. Equating the powers in streamwise inertia $x^{2m-1}$ and transverse diffusion $x^{m-2\gamma }$ givesThe two convective terms combine to $x^{2m-1}[m(f^{\prime }{)}^2-\beta f{f}^{\prime \prime }]$, while $U{U}^{\prime }=m{x}^{2m-1}$. Hence the Falkner-Skan equation in this scaling isThe additive stream-function constant is fixed by $f(0)=0$; when $\beta \ne 0$ this also encodes no penetration. For $\beta =0$ it is a harmless choice of additive constant choice. Existence of an attached physical branch for a given $m$ is a separate issue from deriving the similarity reduction.

On an attached branch with $w=f^{\prime }>0$ away from the wall, use $f$ itself as the independent variable. The chain rule gives $f^{\prime \prime }=w{w}_f$ and $f^{(3)}=w(w_f^2+w{w}_{ff})$. ThereforeIf the wall shear $s=f^{\prime \prime }(0)>0$, then $f\sim s{\eta }^2/2$, so $w(f)\sim \sqrt{2sf}$. Thus $w_f$ is singular at zero while $w{w}_f\to s$ is finite. Imposing a finite arbitrary $w_f(0)$ would remove the usual attached branch. This is the velocity as dependent variable in the Falkner-Skan equation reduction.

Let $\delta =R{e}^{-1/2}$, stretch $Y=y/\delta$ and write the original transverse velocity as $v=\delta V$. Continuity then gives $u_x+V_Y=0$. In the streamwise momentum equation $u_{yy}/Re=u_{YY}$ is order one, while $u_{xx}/Re$ is negligible. The transverse equation gives $p_Y=0$ to leading order, so pressure is supplied by the outer Euler flow. Its streamwise balance is $p_x=-U{U}^{\prime }$. ConsequentlyFor a stationary impermeable plate, $\boxed{u(x,0)=V(x,0)=0}$, with matching $\boxed{u(x,Y)\to U(x)}$ as $Y\to \infty$. The transverse velocity must match the outer displacement flow; it is generally wrong to impose $V\to 0$ independently at infinity.

At the leading edge the wall no-slip data meet the external stream abruptly, and the boundary-layer approximation is nonuniform. For a constant external speed, the thickness is $O(\sqrt{x/Re})$, whose ratio to $x$ ceases to be small when the local Reynolds number is order one. A leading-edge region retaining neglected derivatives is needed there. An adverse pressure gradient can also cause separation farther downstream; a large global Reynolds number alone does not prevent it.

Take $\mathbf{u}=(u(y),0)$ and no imposed streamwise pressure gradient. The normal momentum balance gives $p=p_{\ast }-\rho gy\cos \alpha$. The streamwise equation is $\mu u^{\prime \prime }+\rho g\sin \alpha =0$, since the convective acceleration vanishes. With no slip at the lower wall and signed upper-wall traction $\tau$ in the positive $x$ direction, the conditions areIntegration givesThe required upper-wall velocity is its endpoint value $u(h)=\rho g{h}^2\sin \alpha /(2\mu )+\tau h/\mu$. The unfactored formula remains meaningful when $\sin \alpha =0$.

The fluid exerts shear $\boxed{\mu u^{\prime }(0)=\rho gh\sin \alpha +\tau }$ per unit length on the lower plane in the positive $x$ direction. The force of that plane on the fluid has the opposite sign. The total wall force on the fluid balances its downslope weight, providing an independent sign check.

Use conservation of mass and Newton's momentum balance, in addition to the supplied constitutive law. For a material volume,The Reynolds transport theorem and ${\rho }_t+\nabla \cdot (\rho \mathbf{u})=0$ reduce the left side to $\int \rho D\mathbf{u}/Dt$. The divergence theorem then gives $\rho D{u}_i/Dt={\partial }_j{\sigma }_{ij}+\rho g_i$. For constant $\mu$,Incompressibility removes the last term. ThusThe stress formula and zero divergence alone are not a dynamical law: the force balance supplies the required physical input.

Label fluid particles by $\mathbf{a}$, with flow map $\mathbf{x}=\chi (\mathbf{a},t)$ satisfying ${\chi }_t=\mathbf{u}(\chi ,t)$. If $J=\det D_{\mathbf{a}}\chi$, differentiation of the determinant gives $J_t=J\nabla \cdot \mathbf{u}$. Pull the material-volume integral back to a fixed initial volume:The chain rule and the Jacobian identity give the integrand $J(F_t+\mathbf{u}\cdot \nabla F+F\nabla \cdot \mathbf{u})$. Transforming back yields the Reynolds transport theoremEquivalently the additional surface flux is ${\int }_{\partial V(t)}F\mathbf{u}\cdot \mathbf{n}dS$. The material boundary moves with the same normal velocity as the fluid, which is the key distinction from a fixed control volume.

The long-time Spitzer winding theorem states that, for the continuous unwrapped angle ${\Theta }_t$ of planar Brownian motion started at $1$,Here $C$ has the standard symmetric Cauchy distribution. The notation means convergence in distribution: for every real $x$,Equivalently, expectations converge for every bounded continuous test function. This is a statement about the laws at large deterministic times, without asserting almost-sure convergence of sample paths. The angle is unwrapped; a principal argument is bounded and would instead give zero after division by $\log t$.

The second coordinate ${\zeta }^2$ is one-dimensional Brownian motion, whose zero set is unbounded almost surely by recurrence. The clock $\tau$ and its inverse are continuous, increasing, and unbounded, so the unwrapped angle ${\Theta }_t={\zeta }_{\tau (t)}^2$ has zeros at arbitrarily large original times.

To ensure the times are stopping times for the original Brownian filtration, define directlyThe logarithmic lift makes $\Theta$ a continuous adapted process. Hitting a closed set after a stopping time is a stopping time under the usual filtration conditions, and the unbounded zero set proves inductively that each $T_n$ is finite almost surely. Since $T_n\ge T_{n-1}+1$, the sequence tends to infinity, and continuity givesGeometrically, the Brownian path returns to the positive real ray at arbitrarily late times with zero net unwrapped winding angle about the origin. The assertion is stronger than principal argument zero, which also allows any integer number of completed turns.

Assume $0<r<1<R$, as required for the annulus and logarithms. The logarithmic radius of planar Brownian motion is a local martingale. Stopping at $T=T_{r,R}$ keeps it between the constants $\log r$ and $\log R$, so the stopped process is a bounded martingale. The exit time is finite almost surely because the annulus is bounded and Brownian motion eventually exits its containing disk.

The optional stopping theorem followed by dominated convergence gives $\mathbb{E}\log |Z_T|=\log |Z_0|=0$. Continuity puts the exit radius at either $r$ or $R$. If $p$ is the inner-exit probability, then $p\log r+(1-p)\log R=0$. HenceStopping before taking expectations matters: the unstopped logarithmic-radius process is a strict local martingale, so one cannot assume its unrestricted expectation is constant.

Use first that planar Brownian motion avoids a fixed point, so the path starting at $1$ never reaches zero. Its continuous logarithmic lift, initialized at zero, has real part $U_t=\log |Z_t|$ and imaginary part the continuous unwrapped angle ${\Theta }_t$. Local branches of the complex logarithm, patched along the path, and the Itô formula show that both coordinates are continuous local martingales withThe clock is finite on every compact time interval, since a continuous nonvanishing path stays a positive distance from zero there. It is continuous and strictly increasing. Moreover $\tau (\infty )=\infty$: by recurrence, planar Brownian motion spends infinite total time in a fixed disk, and inside the unit disk the clock integrand is at least one. This is the divergence of a positive planar Brownian occupation integral.

The common-clock time change of orthogonal local martingales, using the Lévy characterization of multidimensional Brownian motion, supplies independent Brownian coordinates ${\zeta }^1,{\zeta }^2$ such that $U_t+i{\Theta }_t={\zeta }_{\tau (t)}$. Exponentiating givesTo identify the inverse clock, write $A={\tau }^{-1}$. The derivative of an inverse function and the representation just obtained giveConsequentlyThe strict inequality in this generalized inverse has the same infimum as the unique inverse value, since $A$ is continuous and strictly increasing onto the half-line.

The disk exit time is finite almost surely; for example, either coordinate of the planar Brownian motion eventually leaves $(-1,1)$. By path continuity, the exit lies on the unit circle. Starting at the origin, rotational invariance of planar Brownian motion makes the exit law invariant under every circle rotation. The unique rotation-invariant probability measure is normalized arc length. Consequently $Z_{T_D}$ is uniform on the unit circle, or equivalently its angle has density $1/(2\pi )$ on $[0,2\pi )$.

There is a printed domain error: $\mathrm{Re}z>0$ defines the right half-plane. Taken literally, continuity forces $\mathrm{Re}{Z}_{T_H}=0$ almost surely, so its law is a point mass at zero. The displayed density belongs to the intended upper half-plane $\mathrm{Im}z>0$, with $b>0$. We prove that intended assertion explicitly.

Write $Z_t=a+X_t+i(b+Y_t)$ for independent standard Brownian motions. The exit time $T$ is the first time $Y$ hits $-b$, and depends only on $Y$. Its density, by the supplied Brownian first-passage time law, is $b(2\pi t^3{)}^{-1/2}{e}^{-b^2/(2t)}$. Conditional on $T=t$, the real exit coordinate is normally distributed with mean $a$ and variance $t$. Independence therefore givesWith $u=[b^2+(x-a{)}^2]/(2t)$, the integral is $2/[b^2+(x-a{)}^2]$. Thus the Cauchy exit law from a Brownian half-plane isFor the literal right half-plane, the corresponding nondegenerate coordinate is $\mathrm{Im}{Z}_{T_H}$, whose density is $a/\{\pi [a^2+(y-b{)}^2]\}$ for $a>0$.

The Itô formula for $f(x)=x^4$ givesThe stochastic integral has expectation zero: its square integrability on a finite horizon follows from the finite sixth moments of a normal distribution. Since $\mathbb{E}{B}_s^2=s$, integration and expectation yield