Blasius solution — exact laminar boundary layer solution: Question bank

Intuition What this page is for

The parent note gave you the machinery. This page attacks the ideas behind it — the places where a confident student says the wrong thing because a formula "felt right". Each item below is a one-line reveal: cover the answer, commit to a guess, then check. If you can justify all of these, you truly understand Blasius.

Before we start, a quick glossary so no symbol is unearned:

$U_{\infty}$ (say "U-infinity") = the free-stream speed far from the plate, where viscosity is no longer felt.
$u$ = fluid speed along the plate (the $x$ -direction); $v$ = fluid speed away from the plate (the $y$ -direction).
$δ$ (delta) = boundary-layer thickness, the height where $u$ has climbed to $99%$ of $U_{\infty}$ .
$η$ (eta) = the "clever height" $y U_{\infty} / (ν x)$ that makes every profile look identical.
$ψ$ (psi) = the stream function; $u = \partial ψ / \partial y$ , $v = - \partial ψ / \partial x$ (this pairing satisfies continuity automatically). For Blasius $ψ = ν x U_{\infty} f (η)$ .
$f$ = the dimensionless stream-function shape; $f^{'} = u / U_{\infty}$ is the velocity ratio (a prime means $d / d η$ ).
$ν$ (nu) = kinematic viscosity (how "sticky-and-spread-y" the fluid is); $μ = ρ ν$ = dynamic viscosity.
$R e_{x} = U_{\infty} x / ν$ = the Reynolds Number, the ratio of inertia to viscosity at position $x$ . With the plate's total length $L$ it becomes $R e_{L} = U_{\infty} L / ν$ , a single fixed number for the whole plate.
$τ_{w}$ = the wall shear stress, the tangential drag force per unit area the fluid exerts on the plate.
$C_{f} = \frac{τ _{w}}{\frac{1}{2} ρ U _{\infty}^{2}} = \frac{0.664}{R e _{x}}$ = the local skin-friction coefficient — wall shear made dimensionless by the dynamic pressure $\frac{1}{2} ρ U_{\infty}^{2}$ at position $x$ .
$C_{D} = \frac{1.328}{R e _{L}}$ = the length-averaged drag coefficient for one side of the whole plate (it turns out to be twice $C_{f}$ evaluated at $x = L$ ).
$β \approx 1.72$ = the "displacement constant": far from the wall $f (η) \to η - β$ , and this offset $β$ is the displacement thickness — how far the outer streamlines are pushed away by the slowed fluid.

Three pictures to hold in your head first

Definition Where the

$v$ -formula comes from — the $ψ \to f$ mapping We never guessed $v$ ; it follows from the stream function $ψ = ν x U_{\infty} f (η)$ by $v = - \partial ψ / \partial x$ . Since $ψ$ depends on $x$ both through the front factor $ν x U_{\infty} = (ν U_{\infty})^{1/2} x^{1/2}$ and through $η (x)$ inside $f$ , use the product rule: $v = - \frac{\partial ψ}{\partial x} = - front factor differentiated (ν U_{\infty})^{1/2} \frac{1}{2} x^{- 1/2} f + f differentiated via η (ν U_{\infty})^{1/2} x^{1/2} f^{'} \frac{\partial η}{\partial x} .$ Insert $\partial η / \partial x = - η /2 x$ into the second term: $v = - \frac{1}{2} \frac{ν U _{\infty}}{x} f + \frac{1}{2} \frac{ν U _{\infty}}{x} η f^{'} = \frac{1}{2} \frac{ν U _{\infty}}{x} (η f^{'} - f) .$ So the mysterious $(η f^{'} - f)$ is nothing but "front-factor derivative minus $η$ -chain derivative" — a direct consequence of $ψ = ν x U_{\infty} f$ , not a separate assumption. (Check: at the wall $η = 0, f = 0, f^{'} = 0$ , so $v (x, 0) = 0$ ✓.)

Intuition Why the transform kills

$x$ — the mechanism The next figure is a construction diagram. At two stations $x_{1} < x_{2}$ the physical profiles $u (y)$ look different (the right layer is taller). But rescale the height by $y / ν x / U_{\infty}$ and both collapse onto the one curve. That collapse is not luck — it is engineered by choosing $η$ so that the $x$ -factors in every derivative cancel.

Here is the cancellation in words, term by term (a prime is $d / d η$ ), using the two chain-rule facts derived just above:

$u = U_{\infty} f^{'} (η)$ — no bare $x$ survives, only through $η$ .
$\frac{\partial u}{\partial x} = U_{\infty} f^{''} \frac{\partial η}{\partial x} = U_{\infty} f^{''} (- \frac{η}{2 x})$ — brings a $1/ x$ .
$\frac{\partial u}{\partial y} = U_{\infty} f^{''} \frac{\partial η}{\partial y} = U_{\infty} f^{''} \frac{U _{\infty}}{ν x}$ , and $\frac{\partial ^{2} u}{\partial y ^{2}} = U_{\infty} f^{'''} \frac{U _{\infty}}{ν x}$ — brings a $1/ x$ .
$v = \frac{1}{2} \frac{ν U _{\infty}}{x} (η f^{'} - f)$ — brings a $1/ x$ that pairs with $\partial u / \partial y$ 's $1/ x$ . When you drop all these into $u u_{x} + v u_{y} = ν u_{y y}$ , every term carries exactly $U_{\infty}^{2} / x$ , so dividing through erases both $x$ and $U_{\infty}$ — leaving the pure ODE $f^{'''} + \frac{1}{2} f f^{''} = 0$ . That is why a PDE in two variables becomes an ODE in one.

Intuition Why

$δ \propto x$ and $τ_{w} \propto x^{- 1/2}$ — the scaling argument sketched The last figure shows the parabolic $δ (x)$ curve growing and the wall shear decaying. First, why $u \partial u / \partial x \sim U_{\infty}^{2} / x$ : the fluid speeds up from $0$ to about $U_{\infty}$ over the streamwise distance $x$ it has travelled from the leading edge, so the along-plate gradient is $\partial u / \partial x \sim U_{\infty} / x$ (change $U_{\infty}$ divided by distance $x$ ); multiplying by the typical speed $u \sim U_{\infty}$ gives the inertial term $u \partial u / \partial x \sim U_{\infty}^{2} / x$ . Now balance this inertia $\sim U_{\infty}^{2} / x$ against the cross-stream viscous force $ν \partial^{2} u / \partial y^{2} \sim ν U_{\infty} / δ^{2}$ (velocity change $U_{\infty}$ across the small height $δ$ , differentiated twice). Equating gives $δ^{2} \sim ν x / U_{\infty}$ , hence $δ \sim ν x / U_{\infty}$ . The wall slope is $u / δ \sim U_{\infty} / δ$ , so $τ_{w} \sim μ U_{\infty} / δ \propto μ U_{\infty} U_{\infty} / (ν x) \propto x^{- 1/2}$ . Both proportionalities fall straight out of one force balance.

True or false — justify

The boundary layer has a sharp top edge where the flow suddenly becomes free-stream.

False — the approach to

U_{\infty}

is smooth and asymptotic (

f^{'} (\infty) = 1

is reached only as

η \to \infty

);

δ

is just an arbitrary cutoff defined at

u = 0.99 U_{\infty}

If you double the free-stream speed

U_{\infty}

, the boundary layer at a fixed

x

gets thinner.

True —

δ \propto ν x / U_{\infty}

, so more speed means more inertia sweeps the slow zone downstream before it can thicken, shrinking

δ

1/ 2

The Blasius velocity profile changes shape as you move downstream along the plate.

False — that is the whole point of a similarity solution: plotted against

η

the profile is one universal curve; only the physical

y

-scale stretches.

The vertical velocity

v

is zero everywhere inside the boundary layer.

False — continuity

\partial u / \partial x + \partial v / \partial y = 0

forces a small upward

v = - \partial ψ / \partial x = \frac{1}{2} ν U_{\infty} / x (η f^{'} - f) \neq = 0

as the layer thickens; the combination

(η f^{'} - f)

is exactly what

- \partial ψ / \partial x

produces from

ψ = ν x U_{\infty} f

. Only

v (x, 0) = 0

(no through-flow at the wall).

The Blasius equation

f^{'''} + \frac{1}{2} f f^{''} = 0

has a neat closed-form solution.

False — it is nonlinear with no elementary closed form; the constant

f^{''} (0) = 0.332

comes from a numerical shooting method.

Wall shear stress

τ_{w}

is largest at the leading edge of the plate.

True —

τ_{w} = 0.332 μ U_{\infty} U_{\infty} / (ν x) \propto 1/ x

, which blows up as

x \to 0

(the singular leading edge) and decays downstream.

Since

τ_{w} \to \infty

x = 0

, the total drag on the plate is infinite.

False —

τ_{w} \propto x^{- 1/2}

is integrable;

\int_{0}^{L} x^{- 1/2} d x = 2 L

is finite, giving the well-behaved

C_{D} = 1.328/ R e_{L}

The Blasius result applies equally to laminar and turbulent boundary layers.

False — it assumes smooth laminar flow (

R e_{x} ≲ 5 \times 1 0^{5}

); a Turbulent Boundary Layer mixes momentum far more efficiently and grows like

x^{4/5}

, not

x^{1/2}

Doubling the plate length

L

doubles the total skin-friction drag.

False —

F_{D} \propto C_{D} \cdot L \propto L^{- 1/2} \cdot L = L^{1/2}

, so drag rises only as

L

; doubling length multiplies drag by

2 \approx 1.41

The pressure gradient

\partial p / \partial x

is assumed zero inside the Blasius layer.

True (effectively) — Prandtl's scaling shows the

y

-momentum equation collapses to

\partial p / \partial y \approx 0

, so pressure does not vary across the thin layer; it is simply impressed from the uniform outer flow, where

\partial p / \partial x = 0

for a flat plate. Hence it is zero inside too.

Spot the error

"The layer thickens downstream, so

δ

grows linearly with

x

Error: growth is

δ \propto x

, a decelerating parabolic growth; a straight line badly over-predicts thickness far downstream.

C_{D} = C_{f}

, since both are skin-friction coefficients."

Error:

C_{f} = 0.664/ R e_{x}

is local (one point);

C_{D} = 1.328/ R e_{L}

is the length-averaged value — and it happens to equal twice the local

C_{f}

evaluated at

x = L

"We can drop

\partial^{2} u / \partial y^{2}

because the layer is thin."

Error: reversed — we drop the streamwise

\partial^{2} u / \partial x^{2}

; the cross-stream

\partial^{2} u / \partial y^{2}

is the dominant viscous term because variation across the thin layer is fast.

"Since

f^{'} (\infty) = 1

, the wall condition must be

f^{'} (0) = 1

too."

Error: the wall enforces no-slip, so

f^{'} (0) = 0

(zero velocity); it is the free stream that reaches

f^{'} = 1

η = y U_{\infty} / (ν x)

, so at fixed

y

the similarity variable grows as you go downstream."

Error:

η \propto 1/ x

, so at fixed physical height

y

the value of

η

shrinks downstream — a fixed point sits lower on the universal curve as the layer swells past it.

"The Reynolds number in

R e_{x}

uses the plate length, so it is a single fixed number."

Error:

R e_{x} = U_{\infty} x / ν

uses the local distance

x

from the leading edge, so it grows along the plate;

R e_{L}

(with total length

L

) is the fixed one.

"Wall shear needs

μ

, so it depends only on the fluid, not the flow."

Error:

τ_{w} = 0.332 μ U_{\infty} U_{\infty} / (ν x)

also scales with

U_{\infty}^{3/2}

and

x^{- 1/2}

— it is very much a property of the flow state.

Why questions

Why do we use a stream function

ψ

instead of solving for

u

and

v

separately?

Because defining

u = \partial ψ / \partial y, v = - \partial ψ / \partial x

makes continuity satisfied automatically, collapsing two unknowns into one and one equation into the momentum balance alone.

Why does the similarity transform reduce a PDE (in

x, y

) to an ODE (in

η

Because when

u = U_{\infty} f^{'} (η)

is substituted, each derivative brings a power of

x

(via

\partial η / \partial x = - η /2 x

and

\partial η / \partial y = U_{\infty} / ν x

); in the momentum equation every term ends up carrying the same factor

U_{\infty}^{2} / x

, so dividing through cancels both

x

and

U_{\infty}

and leaves

f^{'''} + \frac{1}{2} f f^{''} = 0

— an ODE in

η

alone.

Why is the coefficient

5.0

δ = 5.0 ν x / U_{\infty}

not a fundamental constant?

It is tied to the arbitrary

99%

definition of "edge";

f^{'} = 0.99

happens at

η \approx 5.0

, but a

95%

99.9%

definition would give a different coefficient.

Why must the vertical velocity

v

be non-zero even though the plate is horizontal?

As the layer thickens downstream, mass conservation squeezes fluid upward;

\partial u / \partial x < 0

inside the layer forces

\partial v / \partial y > 0

, so fluid drifts gently away from the wall.

Why can Blasius ignore the full Navier–Stokes Equations terms that Prandtl dropped?

Prandtl Boundary Layer Theory shows that inside a thin layer the streamwise diffusion and the

y

-momentum equation are negligibly small compared with the retained balance of inertia against cross-stream viscous shear.

Why does higher viscosity

ν

make the boundary layer thicker, not thinner?

More viscosity means the "no-slip drag" of the wall reaches farther out into the flow, so the slow zone

δ \propto ν

spreads wider before inertia can overcome it.

Why is

f^{''} (0) = 0.332

the single most important number in the whole solution?

It is the wall slope of the velocity profile, so it directly sets the wall shear (

τ_{w} \propto f^{''} (0)

), the skin-friction coefficient

C_{f} = 0.664/ R e_{x}

, and hence the drag

C_{D} = 1.328/ R e_{L}

Edge cases

What happens to

δ

and

τ_{w}

exactly at the leading edge

x = 0

δ \to 0

(no layer has formed yet) while

τ_{w} \to \infty

— a mathematical singularity the thin-layer approximation cannot resolve, but which integrates to finite drag.

What does the Blasius profile predict in the inviscid limit

ν \to 0

δ \propto ν \to 0

: the layer collapses to the wall, recovering the slip, inviscid free-stream everywhere — consistent with viscosity being the sole cause of the layer.

Is Blasius valid at very low Reynolds number (e.g.

R e_{x} \sim 1

No — the derivation requires

δ ≪ x

, i.e. large

R e_{x}

; at

R e_{x} \sim 1

the layer is not thin and the dropped Navier–Stokes Equations terms return.

What sets the upper Reynolds limit of validity for Blasius?

Transition to a Turbulent Boundary Layer near

R e_{x} \approx 5 \times 1 0^{5}

; beyond it the smooth laminar assumption fails and one uses

δ / x \approx 0.37 R e_{x}^{- 1/5}

instead.

What is the value of

f

(not

f^{'}

) far from the wall, and what does it mean?

f (η) \to η - β

with

β \approx 1.72

for large

η

; the constant offset

β

is the displacement thickness — the effective outward shift of the streamlines caused by the slowed fluid.

At the wall (

η = 0

), which of

f, f^{'}, f^{''}

are zero and which is not?

f (0) = 0

(no through-flow) and

f^{'} (0) = 0

(no-slip) are zero, but

f^{''} (0) = 0.332 \neq = 0

— the profile leaves the wall with finite slope, giving finite skin friction.

What happens to

f^{''} (η)

and

f^{'''} (η)

η \to \infty

, and what does it mean physically?

Both vanish:

f^{'}

has flattened to the constant

1

, so its slope

f^{''} \to 0

(no shear far from the wall) and

f^{'''} \to 0

. Physically the flow is uniform out there — no velocity gradient means no viscous stress, and

v

has levelled off to its small constant entrainment value

\frac{1}{2} ν U_{\infty} / x β

What would zero pressure-gradient becoming adverse (

\partial p / \partial x > 0

) do to this solution?

Blasius no longer applies — an adverse gradient decelerates near-wall fluid and can drive separation; you'd need the more general Falkner–Skan family instead of the flat-plate case.

Recall One-line self test

If you can answer this, the page landed: Why is the Blasius profile "the same at every $x$ " yet the boundary layer visibly thickens downstream? Answer ::: The shape is identical in the stretched variable $η$ ; only the physical conversion $y = η ν x / U_{\infty}$ grows with $x$ , so the same curve is drawn on an ever-taller $y$ -axis.