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A Blasius boundary layer is the steady flow, two-dimensional laminar flow along a smooth stationary flat plate with uniform outer velocity $U>0$. Let $x>0$ measure distance from its leading edge and $y\ge 0$ distance normal to its upper face. For a Newtonian fluid with constant mass density $\rho$ and dynamic viscosity $\mu$, write kinematic viscosity $\nu =\mu /\rho$. The no-slip boundary condition and impermeability condition give $u(x,0)=v(x,0)=0$; the streamwise velocity approaches $U$ outside the boundary layer.

For example, choose $U=10\mathrm{m}{\mathrm{s}}^{-1}$ and $\nu =1.5\times 10^{-5}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. At $x=0.30\mathrm{m}$, the local Reynolds number is ${\mathrm{Re}}_x=Ux/\nu =2.0\times 10^5$ and the boundary layer thickness is about $3.29\mathrm{mm}$. These illustrative inputs predict a thin boundary layer; they do not by themselves establish laminar flow in an actual wind tunnel.

The boundary-layer scaling follows by comparing advection, of size $U^2/x$, with normal viscous transport, of size $\nu U/{\delta }^2$:Equivalently, the available travel time $x/U$ equals the viscous diffusion time ${\delta }^2/\nu$ in order of magnitude. Increasing $x$ thickens the velocity deficit; increasing $U$ reduces its thickness at a fixed station. This boundary-layer scaling requires ${\mathrm{Re}}_x\gg 1$. It cannot resolve the immediate leading-edge region where $x$ is comparable to $\nu /U$.^[ref]

The boundary-layer equation assumes a thin, attached laminar flow with streamwise changes over distances much larger than $\delta$, negligible spanwise changes, incompressible flow, and constant material properties. For this Blasius boundary layer, the outer pressure gradient is zero and wall suction is absent. Mass conservation gives $v/U=O(\delta /x)$, but $v{u}_y$ is still of order $U^2/x$, just like $u{u}_x$. Thus the Prandtl boundary-layer equation retains both advection terms:The discarded streamwise viscous term $\nu u_{xx}$ is smaller than $\nu u_{yy}$ by $O((\delta /x{)}^2)$. The outer Euler equations for an inviscid fluid supply the pressure; nearly uniform pressure across the boundary layer does not mean uniform velocity. Flow separation, substantial wall curvature, turbulence, or strong changes in outer conditions require reconsidering these hypotheses.

Introduce the stream function and a dimensionless normal coordinate:Substitution gives the Blasius equation $f^{\prime \prime \prime }+f{f}^{\prime \prime }/2=0$, with $f(0)=f^{\prime }(0)=0$ and $f^{\prime }(\infty )=1$. Its numerical boundary value problem gives $f^{\prime \prime }(0)\simeq 0.332057$. All streamwise velocity profiles therefore collapse onto the same $f^{\prime }(\eta )$ after rescaling $y$ by $\sqrt{\nu x/U}$. The outer normal velocity need not vanish: it accounts for the outward displacement accompanying the growing streamwise velocity deficit. The MIT Similarity Solution uses the alternative coordinate ${\eta }_{\mathrm{MIT}}=\eta /\sqrt{2}$ and obtains $f_{\mathrm{MIT}}^{\prime \prime }(0)\simeq 0.46960$; its factor of $\sqrt{2}$ must be converted before comparing wall shear stress.

The historical Motion of fluids with very little viscosity by L. Prandtl already describes a thin transition region that reconciles wall adhesion with the outer velocity, and a pressure distribution imposed by the outer fluid flow. Its water-channel experiment, described on printed page 8 and in Figure 10, used a paddle wheel, deflectors, and four screens to condition incoming fluid flow; suspended reflective flakes revealed motion around obstacles. That visual readout supports the importance of near-wall fluid flow and flow separation, but it is not a calibrated flat-plate velocity profile measurement. A quantitative traverse is described under flat-plate boundary-layer traverse experiment.

For a steady incompressible viscous boundary layer over a flat plate with constant outer speed $U$, the stream function $\psi =\sqrt{\nu Ux}f(y\sqrt{U/(\nu x)})$ reduces the Prandtl boundary-layer equation to the displayed ordinary differential equation. The wall and outer conditions are $f(0)=f^{\prime }(0)=0$ and $f^{\prime }(\infty )=1$. The wall shear is $\rho U\sqrt{\nu U/x}{f}^{\prime \prime }(0)$, so its integral on one plate face of length $L$ is $2\rho U^{2}L{f}^{\prime \prime }(0)/\sqrt{UL/\nu }$. Both equally exposed faces contribute twice this value.

In a unit square with prescribed conductive boundary temperatures and impermeable boundaries, the stream function convention $\mathbf{u}=(-{\psi }_z,0,{\psi }_x)$ gives linear equations $\Delta \psi =R{\theta }_x$, ${\theta }_t={\psi }_x+\Delta \theta$, with $\psi =\theta =0$ on all sides. The velocity equation follows by taking the vertical-plane curl of Darcy law, and the temperature equation by linearizing the heat equation about $1-z$. Unlike a laterally unbounded Darcy layer, the conducting side boundaries produce a two-dimensional real marginal eigenspace at $R_c=8{\pi }^2$.

A convection roll is a circulating cell pattern whose ideal velocity and temperature fields are periodic across a chosen horizontal direction and independent of the coordinate along the convection roll axis. In a two-dimensional stream function representation, its velocity field is $(-{\psi }_z,0,{\psi }_x)$. Rayleigh-Bénard convection can select stationary or oscillatory convection rolls; a two-dimensional convection roll may be unstable to three-dimensional disturbances, a different convection roll orientation, or slow amplitude and phase modulations. Existence of a saturated Landau amplitude equation branch does not alone establish its stability.

For a stationary cylinder of radius $a$ in the far-field shear $(\Gamma y,0)$, the stream function is $\psi =(\Gamma /4)[r^2-2a^2\log (r/a)-a^2-(r^2-2a^2+a^4/r^2)\cos 2\phi ]$. It solves the biharmonic equation and the no-slip boundary condition. The fluid exerts torque $-2\pi \mu \Gamma a^2$ per unit length, because the angular part of the surface traction averages to zero and the mean tangential traction is $-\mu \Gamma$.

For a sphere of radius $a$ rotating with angular velocity $\Omega$, the primary Stokes flow has $v_{\varphi }=K\sin \theta /r^2$, $K=\Omega a^3$. In the weak, inertialess second-order fluid correction, put $c={\psi }_1+2{\psi }_2$. The stream function in the title hasBoth meridional components vanish at $r=a$ and decay at infinity. Indeed, $E_{r\varphi }=-3K\sin \theta /(2r^3)$ gives $[\nabla \times \nabla \cdot (2c{E}^2){]}_{\varphi }=72c{K}^2\sin \theta \cos \theta /r^8$, and the curl of the momentum equation reduces to $(d^2/d{r}^2-6/r^2{)}^{2}H=72c{K}^2/(\mu r^7)$. For $c>0$ the secondary flow enters towards the equator and leaves near the poles; $c<0$ reverses it and $c=0$ removes this correction. The meridional circulation is even in the sign of sphere rotation. Inertial secondary flow is excluded from this calculation.

A Galerkin method retaining one stream function mode, two temperature modes and two transverse-velocity modes produces five coupled amplitudes. A useful normalization gives $a^{\prime }=-\sigma a+\sigma rb-{\sigma }^{2}qd$, $b^{\prime }=a-b-ac$, $c^{\prime }=h(-c+ab)$, $d^{\prime }=a-\sigma d-ae$, $e^{\prime }=-(4-h)\sigma e+had$. Here $\sigma >0$ is the Prandtl number, $0<h<4$, and $q$ measures squared rotation. The trivial equilibrium point has characteristic polynomial $(s+\sigma {)}^2(s+1)-\sigma r(s+\sigma )+{\sigma }^{2}q(s+1)$ in its three coupled linear modes; the remaining two modes decay. Stationary bifurcations and oscillatory onset meet at $q=(1+\sigma )/(1-\sigma )$, $r=2/(1-\sigma )$ when $\sigma <1$.

A free streamline is a streamline that forms an unknown boundary of a fluid region, typically against a region of prescribed pressure. In steady inviscid potential flow with gravity and surface tension neglected, the Bernoulli equation makes its speed constant when its pressure is constant. It is simultaneously a level set of the stream function. These two boundary conditions take particularly simple forms in the complex potential of a two-dimensional potential flow and the hodograph plane.

The method of images enforces impermeability at a plane wall by reflecting a line vortex across it and reversing its circulation. The two logarithmic stream functions cancel on the wall, which is consequently a streamline. For a vortex of circulation $\Gamma$ at $(0,b)$ above $y=0$, the image of circulation $-\Gamma$ at $(0,-b)$ induces velocity $(\Gamma /(4\pi b),0)$ at the real vortex. The signed drift parallel to the wall is therefore $\Gamma /(4\pi b)$.

The incompressible planar field $(\alpha x-\beta y,\beta x-\alpha y)$ has the displayed stream function under $(u,v)=({\psi }_y,-{\psi }_x)$ and vorticity $2\beta$. In rotated coordinates $s=(x+y)/\sqrt{2}$, $t=(x-y)/\sqrt{2}$, $\psi =((\alpha -\beta )s^2-(\alpha +\beta )t^2)/2$. For positive parameters its nonstationary streamlines are closed ellipses if $\alpha <\beta$, hyperbolas and straight separatrices if $\alpha >\beta$, and parallel straight lines with a stationary diagonal if $\alpha =\beta$.

An ideal straight line vortex in a two-dimensional incompressible flow has circulation $\Gamma$ and induces velocityIts stream function, with $\mathbf{u}=({\psi }_y,-{\psi }_x)$, is $\psi =-\Gamma \log |\mathbf{r}-{\mathbf{r}}_0|/(2\pi )$. Positive $\Gamma$ means counterclockwise circulation. In an ideal point-vortex description, the vortex position is advected by the regular velocity produced by the other vortices and the boundaries; its own singular velocity is excluded.

Perturb the sinusoidal cellular flow by $ϵ\cos x\cos y\sin (\omega t)$ in its stream function. On $y=0$, parameterizing the unperturbed orbit with $q_0(0)=(X_0,0)$ gives $x(s)=2\arctan e^{s+a}$, $a=\log \tan (X_0/2)$. The heteroclinic Melnikov function for a periodic planar flow has integrand ${\mathrm{sech}}^2(t-t_0+a)\sin (\omega t)$. Its odd part vanishes and the Fourier transform of ${\mathrm{sech}}^2$ gives the displayed result. For each fixed nonzero frequency its simple zeros produce transverse intersections at sufficiently small perturbation, hence exchange of fluid lobes across the former cell boundary. The leading splitting is exponentially small at high frequency; that estimate is not uniform in frequency and perturbation size.

The Milne-Thomson circle theorem constructs a circular impermeable boundary in a two-dimensional potential flow. Suppose an incident complex potential of a two-dimensional potential flow $f$ is analytic on and inside $|z|=a$, with its physical singularities outside that disk. Then the displayed $W$ has constant stream function on $|z|=a$. The proof is the identity $a^2/\bar{z}=z$ on the circle: there $W=f(z)+\overline{f(z)}$ is real. Hence the circle is a streamline and satisfies the no-penetration boundary condition. The reflected term has its image singularities inside the obstacle. Additional imposed circulation must be specified separately. The theorem is named after Louis Melville Milne-Thomson.

For indentation $F=b{x}^{1/3}$, the common layer displacement is $H=b{x}^{1/3}/2$. The boundary layer stream function $\psi =x^{2/3}G(z{x}^{-1/3})$ gives velocity $U=x^{1/3}{G}^{\prime }$ and pressure $P=\bar{p}{x}^{2/3}$. Substitution into the Prandtl boundary-layer equation yields the displayed equation, with $G(0)=G^{\prime }(0)=0$ and $G^{\prime }\sim 6\eta +3b$. This is a conditional similarity construction: the constant $\bar{p}$ must permit a solution of the boundary value problem.

Let $L_{\perp }=k_{\perp }^{-1}$ and let $U$ be the common size of horizontal and vertical velocity perturbations, as found in part b. The linear inertial wave time and nonlinear advection time areThe critical balance in a rapidly rotating fluid is thereforeThe Rossby number is of order $ϵ$. Horizontal geostrophic flow balances the leading pressure gradient against Coriolis acceleration, soThese are dimensionally comparable statements, rather than comparisons of a dimensional frequency with a dimensional velocity. Equivalently, use $L_{\perp }$ and $(2\Omega {)}^{-1}$ as units. ThenThus rotation stays $O(1)$ in these units, while the motions evolve on the slower $O({ϵ}^{-1})$ time. Vertical advection $u_{\parallel }{\partial }_z$ is smaller than horizontal advection by $ϵ$.

Write ${\mathbf{u}}_{\perp }={\mathbf{u}}_{\perp }^{(0)}+{\mathbf{u}}_{\perp }^{(1)}+\cdots$, with ${\mathbf{u}}_{\perp }^{(1)}=O(ϵU)$, and $p=p_0+p_1+\cdots$. At leading horizontal order,Define the stream function $\Phi =p_0/(2\Omega )$, including its horizontally uniform part in this convention. ThenThe next-order incompressibility condition is not zero horizontal divergence:This term has size $ϵk_{\perp }U$ and is essential: its product with the large rotation rate is a leading contribution to the vertical vorticity equation.

To see that contribution explicitly, write $D_{\perp }={\partial }_t+{\mathbf{u}}_{\perp }^{(0)}\cdot {\nabla }_{\perp }$. At the first horizontal inertial order,Taking its vertical curl eliminates $p_1$. The two-dimensional advection identity and the two divergence orders giveEquivalently the ageostrophic flow satisfieswhose divergence is $-D_{\perp }\zeta /(2\Omega )$. This explicitly verifies that the next-order divergence has the required sign.

The vertical momentum equation, at the same slow order, isFinally ${\mathbf{u}}_{\perp }^{(0)}\cdot {\nabla }_{\perp }f=\{\Phi ,f\}$, where the canonical two-dimensional Poisson bracket is $\{f,g\}=f_x{g}_y-f_y{g}_x$. The reduced equations for an anisotropic rotating fluid areThe omitted vertical-advection, ageostrophic-advection and relative-vorticity stretching terms are smaller by $O(ϵ)$ in the equations retained here. Linearizing these equations recovers ${\omega }^2=4{\Omega }^2{k}_{\parallel }^2/k_{\perp }^2$, the anisotropic limit of part a.

The imaginary part of the complex potential of a two-dimensional potential flow is the stream function. Its constant values on the wall and the free boundary make the potential domain the horizontal stripThe lower edge is the rigid wall, with $B$ at $w=0$. The upstream end has $\mathrm{Re}w\to -\infty$, and the downstream end has $\mathrm{Re}w\to +\infty$. The upper edge is the free streamline. Thus $AB$ and $BC$ are the negative and positive portions of the real $w$ axis, respectively.

Write $q=w^{\prime }=u-iv$ for the complex velocity. Downstream $q\to 1$. Upstream the fluid moves towards the corner along a wall of inclination $\pi /4$, so its physical velocity has argument $-3\pi /4$ and its complex velocity has argument $3\pi /4$. Along each straight wall the velocity direction is fixed. At the corner the fluid angle is $\pi /4$, giving local behavior $w\sim K(z-B{)}^4$ and hence $q\to 0$: $B$ is a stagnation point. On the free streamline, the speed is one, so $|q|=1$.

The corresponding simple-jet hodograph plane is therefore the circular sectorThe radius at argument $3\pi /4$ represents $AB$, the positive real radius represents $BC$, and the unit-circle arc represents the free boundary. The upstream and downstream ends are $e^{3\pi i/4}$ and one, respectively; the corner is the sector's origin.

Use the torsion stress function for a simply connected bar. The only nonzero strain components are $e_{xz}=(w_x-\Omega y)/2$ and $e_{yz}=(w_y+\Omega x)/2$. ThusThe axial equilibrium equation is ${\partial }_x{\tau }_{xz}+{\partial }_y{\tau }_{yz}=0$. Since the section is simply connected, the planar divergence-free stress field has a stream function:Equality of the mixed warping derivatives gives $\boxed{\Delta \varphi =-2\mu \Omega }$. On the lateral boundary the axial traction is ${\varphi }_y{n}_x-{\varphi }_x{n}_y$, the tangential derivative of $\varphi$. Traction-free walls therefore make $\varphi$ constant on the connected boundary, and its arbitrary additive constant can be chosen so $\boxed{\varphi =0\text{ on }\partial D}$.

The axial torque isThe divergence theorem removes the boundary term because $\varphi =0$.

Tresca's yield criterion requires the maximum shear stress, half the largest principal-stress difference, not to exceed the material's shear yield stress ${\tau }_Y$. The stress eigenvalues here are $0,\pm \sqrt{{\tau }_{xz}^2+{\tau }_{yz}^2}$, so the maximum shear is $|\nabla \varphi |$ andIf yield is instead quoted as a uniaxial tensile stress, that parameter is $2{\tau }_Y$; the present convention is the source's shear parameter.

For the ellipse, the proposed quadratic vanishes on its boundary. Its Laplacian fixesThe corresponding warping is $w=\Omega (b^2-a^2)xy/(a^2+b^2)+\text{constant}$. Elliptic polar coordinates $x=ar\cos \vartheta$, $y=br\sin \vartheta$ give ${\int }_D\varphi =C\pi ab/2$, henceTo locate first yield, maximize $|\nabla \varphi |=2|C|\sqrt{x^2/a^4+y^2/b^4}$. It increases radially to the boundary. For $a>b$, the maximum there occurs at $(0,\pm b)$ and equals $2|C|/b$. ThereforeThe positive-twist value is the requested expression. The ends of the minor axis yield first, not the ends of the longer axis. This is elliptical-bar torsion and first yield.

Use $W=\varphi +i\psi$ for the complex potential of a two-dimensional potential flow, so $W^{\prime }(z)=u-iv$. The Milne-Thomson circle theorem says that if $f(z)$ is an incident complex potential of a two-dimensional potential flow whose singularities lie outside $|z|=a$, thenmakes $|z|=a$ a streamline. Indeed, on that circle $a^2/\bar{z}=z$, so $W=f+\bar{f}$ is real and the stream function is constant. This imposes the no-penetration boundary condition.

For the two-dimensional potential doublet, put $M=m{e}^{i\alpha }$ and $f=M/(z-b)$, where $b>a$ is real. The theorem givesUp to an additive constant, this is equivalentlyThe method of images has placed the image two-dimensional potential doublet at $a^2/b$, inside the obstacle. We use the solution with zero added circulation and no imposed far-field flow; the original doublet is the only singularity in the liquid.

Choose the stream function convention $u=-{\psi }_y$, $v={\psi }_x$, so that this is an incoming stagnation point flow rather than the reversed flow. The velocity field isAt the plate $v(x,0)=0$, satisfying the no-penetration boundary condition. The nonzero tangential velocity $u(x,0)=Ax$ is permitted for an inviscid flow; the no-slip boundary condition is not imposed on the Euler equations for an inviscid fluid.

The steady convective acceleration is $(A^{2}x,A^{2}y)$. Thus the Euler equations for an inviscid fluid requireIntegrating givesThe printed formula omits $\rho$ from its quadratic term. The source uses physical pressure, not pressure divided by mass density, as is clear from its $\rho gy$ term and its Navier-Stokes equation. The omitted factor is also required by dimensional analysis. For a concrete substitution check, the printed expression gives $-p_x/\rho =A^{2}x/\rho$ rather than the required $A^{2}x$ when $\rho \ne 1$ in the chosen numerical units.

Use the same stream function convention and denote differentiation with respect to the dimensionless variable $\eta =y/\delta$ by a prime. The proposed similarity solution givesIts continuity equation holds identically. Substitution in the tangential Prandtl boundary-layer equation givesChoose the constant scale $\delta =\sqrt{\nu /A}$. The Hiemenz flow equation and its boundary conditions areThe first two conditions impose the no-penetration boundary condition and no-slip boundary condition. The last matches the tangential outer velocity. One also has $f^{\prime \prime }\to 0$ and $f(\eta )/\eta \to 1$. It is too strong to impose $f-\eta \to 0$: the displacement thickness generally permits a constant offset, $f\sim \eta -B$.

The normal Navier-Stokes equation makes the approximation in part (c) explicit. With this similarity solution, it givesAt the wall this differs from $(p_{\infty }{)}_y=-\rho g$ by $-\rho A^2\delta f^{\prime \prime }(0)$. This is a concrete next-order counterexample to interpreting the printed normal pressure gradient equality exactly; the tangential Hiemenz flow equation remains valid at its stated order.