Visual walkthrough — Blasius solution — exact laminar boundary layer solution

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Step 1 — The picture we are trying to explain

WHAT. A flat, thin plate sits edge-on in a wind. Far away the air all moves at one steady speed we call $U_{\infty }$ ("U-infinity" — the free-stream speed, the speed the air has when the plate is nowhere near it). Right at the surface, air molecules stick to the plate and cannot move — this is the no-slip rule.

WHY. Between "stuck at the wall" and "zooming at $U_{\infty }$", the air speed must climb smoothly. That thin climbing region is the boundary layer. Our whole job is to find its shape (how speed rises as we move away from the wall) and its thickness.

PICTURE. Below: the free stream (top arrows all equal length), the wall (bottom, arrows zero), and the layer where the arrows grow from nothing to full length.

Two coordinates are used everywhere:

• $x$ ::: distance along the plate, measured from its leading (front) edge.
• $y$ ::: distance away from the plate, straight out into the air.

The speed of the air in the $x$-direction is written $u(x,y)$ — it depends on where you stand. The much smaller sideways speed (in the $y$-direction) is $v(x,y)$. We will meet $v$ properly in Step 5.

We also fix the name of the fluid's density now, because it returns at the drag stage:

• $\rho$ (Greek "rho") ::: the fluid's density — its mass per unit volume, in ${\text{kg/m}}^3$ (for air, about $1.2{\text{kg/m}}^3$).

See Prandtl Boundary Layer Theory for the physical origin of this picture.

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Step 2 — The two laws the air must obey

WHAT. Two physical statements govern the flow inside the layer.

Law A — mass is conserved (continuity): $\underset{\text{how }u\text{ changes along plate}}{\underbrace{\frac{\partial u}{\partial x}}}+\underset{\text{how }v\text{ changes away from wall}}{\underbrace{\frac{\partial v}{\partial y}}}=0$

The symbol $\partial u/\partial x$ (read "partial-u-partial-x") just means "how fast $u$ changes as you take a tiny step in $x$, holding $y$ fixed." The equation says: if the along-flow $u$ speeds up in one direction, the sideways flow $v$ must adjust so no fluid is created or destroyed.

Law B — force = mass × acceleration for a fluid blob (the Navier–Stokes $x$-momentum equation): $\underset{\text{inertia: the blob’s acceleration}}{\underbrace{u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}}}=\underset{\text{viscous friction across the layer}}{\underbrace{\nu \frac{{\partial }^{2}u}{\partial y^2}}}$

WHY. The left side is inertia — a moving blob carrying its own momentum. The right side is viscosity — neighbouring layers of air rubbing and dragging each other. Here $\nu$ (Greek "nu") is the kinematic viscosity: how strongly the fluid resists shearing, measured in ${\text{m}}^2/\text{s}$. The double symbol ${\partial }^{2}u/\partial y^2$ is the curvature of the speed profile — how bent the velocity curve is.

PICTURE. A single fluid blob with the inertia arrows (carrying it forward) fighting the viscous arrows (the sticky drag from the slower layer below).

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Step 3 — The scaling guess: how thick should the layer be?

WHAT. Before solving anything, we estimate the thickness $\delta$ (Greek "delta") by balancing the two sides of Law B in size only.

WHY. In the layer, inertia and viscosity are comparable (that is the definition of the layer). So set their rough sizes equal: $\underset{\text{inertia size}}{\underbrace{\frac{U_{\infty }^2}{x}}}\sim \underset{\text{viscous size}}{\underbrace{\frac{\nu U_{\infty }}{{\delta }^2}}}$ Here $U_{\infty }^2/x$ is "speed² per length travelled" and $\nu U_{\infty }/{\delta }^2$ is "viscosity × speed ÷ thickness²". Solving for $\delta$: $\boxed{\delta \sim \sqrt{\frac{\nu x}{U_{\infty }}}}$

PICTURE. The layer edge traced as a curve growing like $\sqrt{x}$ — steep at first, then flattening. Contrast with a (wrong) straight line.

This single estimate is the seed of the whole solution: it tells us the natural yardstick for measuring height inside the layer. Notice the connection to Reynolds Number: $\delta /x\sim 1/\sqrt{U_{\infty }x/\nu }$, and $U_{\infty }x/\nu$ is exactly $R{e}_x$.

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Step 4 — Stretch the ruler: the similarity variable $\eta$

WHAT. Instead of raw height $y$, measure height in units of the natural thickness from Step 3. Call this stretched height $\eta$ (Greek "eta"): $\boxed{\eta =y\sqrt{\frac{U_{\infty }}{\nu x}}}$

• $y$ ::: raw distance from wall.
• $\sqrt{U_{\infty }/(\nu x)}$ ::: the stretch factor — big near the leading edge (thin layer), small far downstream (thick layer).

WHY. In Step 3 we found the layer thickens as $\sqrt{x}$. If we divide height by that same $\sqrt{x}$, the profiles at every $x$ line up. This is a similarity solution: one universal curve, no matter where along the plate you look. (See Similarity Solutions in PDEs.)

PICTURE. Left panel: raw profiles $u(y)$ at three values of $x$ — different widths, a fan. Right panel: the same data plotted against $\eta$ — they collapse onto a single curve.

To handle both Law A and Law B at once, we introduce the stream function $\psi$ (Greek "psi"). By definition $u=\partial \psi /\partial y$ and $v=-\partial \psi /\partial x$; writing velocities this way makes Law A (continuity) automatically true — see Stream Function and Vorticity. We write it as $\psi =\underset{\text{amplitude}}{\underbrace{\sqrt{\nu x{U}_{\infty }}}}f(\eta )$

where $f(\eta )$ is a dimensionless unknown function — the shape we must find. Everything now hangs on this one function $f$.

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Step 5 — Reading the velocities off $f$

WHAT. Turn the stream function into the actual velocities.

For the along-flow speed: $u=\frac{\partial \psi }{\partial y}=\sqrt{\nu x{U}_{\infty }}{f}^{\prime }(\eta )\cdot \underset{=\sqrt{U_{\infty }/(\nu x)}}{\underbrace{\frac{\partial \eta }{\partial y}}}=U_{\infty }{f}^{\prime }(\eta )$ so $\boxed{u=U_{\infty }{f}^{\prime }(\eta )}$

WHY this is beautiful. The prime means derivative: $f^{\prime }(\eta )=\mathrm{d}f/\mathrm{d}\eta$. So $f^{\prime }$ is literally the velocity ratio $u/U_{\infty }$ — a number between $0$ (at the wall) and $1$ (in the free stream). The shape of the boundary layer is the graph of $f^{\prime }$.

The sideways speed, step by step. By definition $v=-\partial \psi /\partial x$. The tricky part: $\psi =\sqrt{\nu x{U}_{\infty }}f(\eta )$ depends on $x$ in two places — through the amplitude $\sqrt{\nu x{U}_{\infty }}$ and through $\eta$ (because $\eta =y\sqrt{U_{\infty }/(\nu x)}$ itself contains $x$). So we need the product rule:

= -\underbrace{\frac{\partial\sqrt{\nu x U_\infty}}{\partial x}}_{\text{amplitude changes}}f \; -\; \sqrt{\nu x U_\infty}\,f'\underbrace{\frac{\partial\eta}{\partial x}}_{\eta\text{ changes}}$$ Now the two pieces: - $\dfrac{\partial}{\partial x}\sqrt{\nu x U_\infty} = \dfrac12\sqrt{\dfrac{\nu U_\infty}{x}}$ (since $\sqrt{x}$ differentiates to $\tfrac{1}{2\sqrt{x}}$ — **this is where the $\tfrac12$ is born**). - $\dfrac{\partial\eta}{\partial x} = y\sqrt{\dfrac{U_\infty}{\nu}}\cdot\big(-\tfrac12\big)x^{-3/2} = -\dfrac{\eta}{2x}$ (again the $\sqrt{x}$ in the denominator gives the $-\tfrac12$). Substituting both and simplifying (the second term becomes $+\tfrac12\sqrt{\nu U_\infty/x}\,\eta f'$): $$\boxed{\;v = \frac{1}{2}\sqrt{\frac{\nu U_\infty}{x}}\,\big(\eta f' - f\big)\;}$$ - The bracket $(\eta f' - f)$ is generally **not zero** — so $v \neq 0$ inside the layer! As the layer thickens, mass conservation *forces* a gentle upward drift. **PICTURE.** The layer with big horizontal $u$-arrows (from $f'$) and small upward $v$-arrows (from the bracket), showing fluid being nudged outward as the layer grows. ![[deepdives/dd-physics-2.2.22-d2-s05.png]] --- ## Step 6 — The miracle: everything collapses to one ODE **WHAT.** Substitute $u = U_\infty f'$ and $v = \tfrac12\sqrt{\nu U_\infty/x}(\eta f' - f)$, together with their $x$- and $y$-derivatives, back into Law B (the momentum equation). **Building the inertia side, term by term.** We need four derivatives. Using $\partial\eta/\partial x = -\eta/(2x)$ and $\partial\eta/\partial y = \sqrt{U_\infty/(\nu x)}$: $$\frac{\partial u}{\partial x} = U_\infty f''\frac{\partial\eta}{\partial x} = -\frac{U_\infty}{2x}\,\eta f'',\qquad \frac{\partial u}{\partial y} = U_\infty f''\sqrt{\frac{U_\infty}{\nu x}}$$ Now form the two inertia pieces: $$u\frac{\partial u}{\partial x} = U_\infty f'\cdot\Big(-\frac{U_\infty}{2x}\eta f''\Big) = -\frac{U_\infty^2}{2x}\,\eta f' f''$$ $$v\frac{\partial u}{\partial y} = \tfrac12\sqrt{\tfrac{\nu U_\infty}{x}}(\eta f'-f)\cdot U_\infty f''\sqrt{\tfrac{U_\infty}{\nu x}} = \frac{U_\infty^2}{2x}f''(\eta f'-f)$$ **Add them** — watch the $\eta f' f''$ terms cancel exactly: $$u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y} = \frac{U_\infty^2}{2x}\big(-\eta f' f'' + \eta f' f'' - f f''\big) = -\frac{U_\infty^2}{2x}\,f f''$$ That surviving $-\tfrac12 f f''$ is **exactly where the $\tfrac12$ coefficient in the Blasius equation comes from** — it is the leftover after the $\eta f' f''$ pieces annihilate. **The viscous side:** $$\nu\frac{\partial^2 u}{\partial y^2} = \nu U_\infty f'''\Big(\frac{U_\infty}{\nu x}\Big) = \frac{U_\infty^2}{x}f'''$$ **Set inertia = viscous** and multiply through by $x/U_\infty^2$: $$-\tfrac12 f f'' = f''' \;\Longrightarrow\; f''' + \tfrac12 f f'' = 0$$ **WHY it works.** Every appearance of $x$, of $U_\infty$, and of $\nu$ **cancels** — because the amplitude $\sqrt{\nu x U_\infty}$ (Step 4) was *chosen* to make the common factor $U_\infty^2/x$ appear on **both** sides. What survives is a pure equation in $f$ and $\eta$ alone: > [!formula] The Blasius equation > $$\boxed{\;f''' + \tfrac{1}{2}\,f\,f'' = 0\;}$$ > - $f'''$ ::: third derivative — comes from the viscous term $\nu\,\partial^2 u/\partial y^2$. > - $\tfrac12 f f''$ ::: the leftover of the inertia terms after $\eta f' f''$ cancels. > > **Boundary conditions** (the physics that pins down $f$): > $$f(0)=0,\qquad f'(0)=0,\qquad f'(\infty)=1$$ > - $f(0)=0$ ::: no fluid passes *through* the wall ($v=0$ at $y=0$). > - $f'(0)=0$ ::: no-slip — speed is zero at the wall. > - $f'(\infty)=1$ ::: far out, $u \to U_\infty$, so $u/U_\infty \to 1$. **PICTURE.** A "funnel" diagram: two messy PDEs (Laws A + B) plus the similarity transform pour in the top and one clean ODE comes out the bottom. ![[deepdives/dd-physics-2.2.22-d2-s06.png]] > [!mistake] Expecting a tidy formula for $f$ > **Why it feels right:** most physics problems end in a neat expression. > **The fix:** $f'''+\tfrac12 f f''=0$ is **nonlinear** and has **no closed form**. It is solved numerically (the *shooting method*: guess $f''(0)$, integrate, adjust until $f'(\infty)=1$). The magic guess that works is $f''(0)=0.332$. --- ## Step 7 — The universal shape and its one magic number **WHAT.** Numerically integrating gives the curve $f'(\eta)$ — the velocity profile — and the crucial slope at the wall, $f''(0)=0.332$. **WHY $f''(0)$ matters.** The slope of the velocity curve *at the wall* controls how hard the air drags on the plate (Step 8). And the profile reaches $f'=0.99$ at $\eta \approx 5.0$ — that is where we say the layer "ends". **PICTURE.** The Blasius curve: $\eta$ up the side, $f'=u/U_\infty$ across. Mark $f'(0)=0$ (no-slip), the initial slope $0.332$, and the point $\eta=5.0$ where $f'=0.99$. ![[deepdives/dd-physics-2.2.22-d2-s07.png]] From $\eta \approx 5.0$ at the layer edge: $$\boxed{\;\delta \approx 5.0\sqrt{\frac{\nu x}{U_\infty}} = \frac{5.0\,x}{\sqrt{Re_x}}\;},\qquad Re_x = \frac{U_\infty x}{\nu}$$ Exactly the $\sqrt{x}$ growth guessed in Step 3, now with the precise number $5.0$ attached. --- ## Step 8 — From slope to drag **WHAT.** The wall shear stress $\tau_w$ ("tau-w", the frictional force per unit area on the plate) is proportional to the velocity slope at the wall: $$\tau_w = \mu\left.\frac{\partial u}{\partial y}\right|_{y=0} = \mu U_\infty\sqrt{\frac{U_\infty}{\nu x}}\;\underbrace{f''(0)}_{=\,0.332} = 0.332\,\mu U_\infty\sqrt{\frac{U_\infty}{\nu x}}$$ Here $\mu = \rho\nu$ is the *dynamic* viscosity — the density $\rho$ (defined in Step 1) times the kinematic viscosity $\nu$. **WHY.** Friction from a fluid is proportional to how sharply the speed changes at the surface — a steeper profile means neighbouring layers slide past faster, dragging harder. That steepness *is* $f''(0)$. Dividing by the dynamic pressure $\tfrac12\rho U_\infty^2$ (the natural "flow-energy per volume", built from density $\rho$ and speed $U_\infty$) gives the dimensionless local [[Skin Friction Drag|skin-friction coefficient]]: $$\boxed{\;C_f = \frac{0.664}{\sqrt{Re_x}}\;}\qquad(0.664 = 2\times 0.332)$$ and integrating $\tau_w$ over a plate of total length $L$ gives the total-drag coefficient. Here: - $L$ ::: the **full length of the plate** from leading to trailing edge (in metres). - $Re_L$ ::: the Reynolds number built on that full length, $Re_L = \dfrac{U_\infty L}{\nu}$ — the same $Re_x$ formula but evaluated at $x=L$. $$\boxed{\;C_D = \frac{1.328}{\sqrt{Re_L}}\;}\qquad(1.328 = 2\times 0.664)$$ **PICTURE.** The wall with the tangent line to the profile at $y=0$; its slope (marked $0.332$) feeds the drag arrow on the plate. Downstream, the slope softens and the drag arrow shrinks. ![[deepdives/dd-physics-2.2.22-d2-s08.png]] --- ## Step 9 — The edge cases (never leave the reader stranded) **WHAT & WHY.** - **Leading edge, $x\to 0$.** The formula gives $\delta\to 0$ and $\tau_w\to\infty$. This is the *leading-edge singularity*: the thin-layer assumption ($\delta \ll x$) breaks right at the tip, so Blasius is invalid in a tiny nose region. Everywhere downstream it is excellent. - **Far downstream / high $Re_x$.** Once $Re_x \gtrsim 5\times10^5$ the flow trips **turbulent** (see [[Turbulent Boundary Layer]]); then $\delta$ grows like $x^{4/5}$, not $\sqrt{x}$, and Blasius no longer applies. - **Zero viscosity, $\nu\to 0$.** The layer vanishes ($\delta\to 0$) — an ideal *inviscid* flow slips freely past the plate. Viscosity is precisely what creates the layer. **PICTURE.** A single strip along the plate: a shaded *invalid* nose region ($x\to0$), the wide *valid laminar Blasius* band, and the *turbulent* zone past the transition line. ![[deepdives/dd-physics-2.2.22-d2-s09.png]] --- ## The one-picture summary ![[deepdives/dd-physics-2.2.22-d2-s10.png]] The whole derivation on one canvas: raw profiles at several $x$ (top-left) → collapsed by the stretch $\eta = y\sqrt{U_\infty/\nu x}$ into **one** curve $f'$ (centre), governed by the single ODE $f'''+\tfrac12 f f''=0$, delivering the three headline numbers $5.0,\;0.332,\;0.664$ and the drag law $C_D=1.328/\sqrt{Re_L}$. > [!recall]- Feynman retelling of the whole walkthrough > Air blows past a flat plate. Right at the plate the air sticks (Step 1). Two rules apply: nothing is created (mass is conserved) and pushes equal accelerations (Newton) — Step 2. Because the sticky "slow zone" is thin, only the sharpest bending of the speed matters, which lets us throw away one hard term. We *guessed* the zone's thickness by matching how strongly the air's momentum shoves against how strongly stickiness drags — it grows like the square root of distance (Step 3). So we measured height in *stretched* units $\eta$ that grow the same way, and — miracle — the profile looks identical at every spot along the plate (Step 4). Writing the flow through a clever helper $f$ whose amplitude we *chose* so the $x$'s cancel, the horizontal speed became simply $U_\infty$ times the slope of $f$, and there was a small but real upward drift $v$ born from the product rule (Step 5). Plugging everything back in, the extra $\eta f' f''$ pieces cancelled and left one clean equation, $f'''+\tfrac12 f f''=0$, with three common-sense conditions (Step 6). A computer solves it: the profile hits $99\%$ speed at stretched height $5.0$, and its steepness at the wall is $0.332$ (Step 7). That steepness is the drag: friction coefficient $0.664/\sqrt{Re}$, total drag $1.328/\sqrt{Re}$ (Step 8). It fails only at the very tip and once the flow turns turbulent (Step 9). One curve, three numbers, the whole flat-plate boundary layer. > [!mnemonic] The numbers double > **Five thick · three-thirty steep · six-six-four friction · one-three-two-eight total.** > $0.664 = 2\times0.332$, and $1.328 = 2\times0.664$ — each step doubles. --- ## #flashcards/physics Why keep only $\partial^2 u/\partial y^2$ in the momentum equation? ::: The layer is thin, so speed bends sharply across it (in $y$) but gently along it (in $x$); the $y$-curvature dominates. What natural thickness comes from balancing inertia against viscosity? ::: $\delta \sim \sqrt{\nu x/U_\infty}$, i.e. it grows like $\sqrt{x}$. What does $f'(\eta)$ represent physically? ::: The velocity ratio $u/U_\infty$ — the shape of the boundary-layer profile. Why is the stream-function amplitude chosen as $\sqrt{\nu x U_\infty}$? ::: So that $u=\partial\psi/\partial y$ comes out as exactly $U_\infty f'$ with no leftover $x$; the amplitude is forced, not guessed. Where does the $\tfrac12$ in $f'''+\tfrac12 f f''=0$ come from? ::: From differentiating $\sqrt{x}$ (and $x^{-1/2}$) in the product rule; after adding the inertia terms the $\eta f' f''$ parts cancel, leaving $-\tfrac12 f f''$. Where does the drag number $0.664$ come from? ::: From the wall slope $f''(0)=0.332$, doubled when dividing $\tau_w$ by the dynamic pressure $\tfrac12\rho U_\infty^2$.