The compressible laminar boundary layer equations for an axisymmetric body are
(see for example ):
\[
\frac{\partial(\rho u r_0)}{\partial x}
+ \frac{\partial(\rho v r_0)}{\partial y} = 0
\]
\[
\rho\!\left(u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y}\right)
= -\frac{dp}{dx}
+ \frac{\partial}{\partial y}\!\left(\mu\frac{\partial u}{\partial y}\right)
\]
\[
\rho c_p\!\left(u\frac{\partial T}{\partial x} + v\frac{\partial T}{\partial y}\right)
= u\frac{dp}{dx}
+ \frac{\partial}{\partial y}\!\left(k\frac{\partial T}{\partial y}\right)
+ \mu\!\left(\frac{\partial u}{\partial y}\right)^{\!2}
\]
From the y-momentum equation, pressure is uniform across the layer and imposed
by the outer inviscid flow.
Identical to the 2D case. Pressure is set by the edge conditions via the
Euler x-momentum equation:
Azimuthal symmetry eliminates any swirl contribution.
Derivation from the Navier-Stokes equations
Axisymmetric Compressible Navier-Stokes Equations
Start from the compressible NS for a Newtonian fluid with Stokes' hypothesis
(\(\lambda = -2\mu/3\)) in cylindrical coordinates \((x, r)\), retaining
azimuthal symmetry (\(\partial/\partial\phi = 0\), \(w = 0\)).
Continuity
\[
\frac{\partial\rho}{\partial t}
+ \frac{\partial(\rho u)}{\partial x}
+ \frac{1}{r}\frac{\partial(r\rho v)}{\partial r} = 0
\]
x-momentum
\[
\rho\!\left(\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x}
+ v\frac{\partial u}{\partial r}\right)
= -\frac{\partial p}{\partial x}
+ \frac{\partial}{\partial x}\!\left(\frac{4\mu}{3}\frac{\partial u}{\partial x}
- \frac{2\mu}{3}\!\left(\frac{\partial v}{\partial r} + \frac{v}{r}\right)\right)
+ \frac{1}{r}\frac{\partial}{\partial r}\!\left(r\mu\!\left(\frac{\partial u}{\partial r}
+ \frac{\partial v}{\partial x}\right)\right)
\]
r-momentum
\[
\rho\!\left(\frac{\partial v}{\partial t} + u\frac{\partial v}{\partial x}
+ v\frac{\partial v}{\partial r}\right)
= -\frac{\partial p}{\partial r}
+ \frac{\partial}{\partial x}\!\left(\mu\!\left(\frac{\partial v}{\partial x}
+ \frac{\partial u}{\partial r}\right)\right)
+ \frac{1}{r}\frac{\partial(r\tau_{rr})}{\partial r}
- \frac{\tau_{\phi\phi}}{r}
\]
where
\[
\tau_{rr} = \frac{4\mu}{3}\frac{\partial v}{\partial r}
- \frac{2\mu}{3}\!\left(\frac{\partial u}{\partial x} + \frac{v}{r}\right),
\qquad
\tau_{\phi\phi} = \frac{4\mu}{3}\frac{v}{r}
- \frac{2\mu}{3}\!\left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial r}\right)
\]
Energy (in terms of temperature)
\[
\rho c_p\!\left(\frac{\partial T}{\partial t} + u\frac{\partial T}{\partial x}
+ v\frac{\partial T}{\partial r}\right)
= \left(\frac{\partial p}{\partial t} + u\frac{\partial p}{\partial x}
+ v\frac{\partial p}{\partial r}\right)
+ \frac{\partial}{\partial x}\!\left(k\frac{\partial T}{\partial x}\right)
+ \frac{1}{r}\frac{\partial}{\partial r}\!\left(kr\frac{\partial T}{\partial r}\right)
+ \Phi
\]
where
\[
\Phi = \mu\!\left[
2\!\left(\frac{\partial u}{\partial x}\right)^{\!2}
+ 2\!\left(\frac{\partial v}{\partial r}\right)^{\!2}
+ 2\!\left(\frac{v}{r}\right)^{\!2}
+ \!\left(\frac{\partial u}{\partial r} + \frac{\partial v}{\partial x}\right)^{\!2}
- \frac{2}{3}\!\left(\frac{\partial u}{\partial x}
+ \frac{\partial v}{\partial r} + \frac{v}{r}\right)^{\!2}
\right]
\]
Steady Flow
For steady flow, all \(\partial/\partial t\) terms vanish.
The continuity, momentum, and energy equations reduce to:
\[
\frac{\partial(\rho u)}{\partial x}
+ \frac{1}{r}\frac{\partial(r\rho v)}{\partial r} = 0
\]
\[
\rho\!\left(u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial r}\right)
= -\frac{\partial p}{\partial x}
+ \frac{\partial}{\partial x}\!\left(\frac{4\mu}{3}\frac{\partial u}{\partial x}
- \frac{2\mu}{3}\!\left(\frac{\partial v}{\partial r} + \frac{v}{r}\right)\right)
+ \frac{1}{r}\frac{\partial}{\partial r}\!\left(r\mu\!\left(\frac{\partial u}{\partial r}
+ \frac{\partial v}{\partial x}\right)\right)
\]
\[
\rho\!\left(u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial r}\right)
= -\frac{\partial p}{\partial r}
+ \frac{\partial}{\partial x}\!\left(\mu\!\left(\frac{\partial v}{\partial x}
+ \frac{\partial u}{\partial r}\right)\right)
+ \frac{1}{r}\frac{\partial(r\tau_{rr})}{\partial r}
- \frac{\tau_{\phi\phi}}{r}
\]
\[
\rho c_p\!\left(u\frac{\partial T}{\partial x} + v\frac{\partial T}{\partial r}\right)
= u\frac{\partial p}{\partial x} + v\frac{\partial p}{\partial r}
+ \frac{\partial}{\partial x}\!\left(k\frac{\partial T}{\partial x}\right)
+ \frac{1}{r}\frac{\partial}{\partial r}\!\left(kr\frac{\partial T}{\partial r}\right)
+ \Phi
\]
Boundary Layer Scaling
The steady NS above simplifies when the boundary layer is thin relative to the
streamwise length scale. Introduce a reference length \(L\) and free-stream
speed \(U_\infty\), and let \(\delta \ll L\) be the BL thickness.
For BL analysis on a body of revolution, it is convenient to work in
surface-aligned coordinates: \(s\) (streamwise arc length along the body) and
\(y\) (wall-normal distance from the surface). The local body surface radius
\(r_0(s)\) can be any smooth function — no specific body geometry is assumed
at this stage. In surface-aligned coordinates the steady continuity transforms
to
\[
\frac{\partial(\rho u\, r_0)}{\partial s}
+ \frac{\partial(\rho v\, r_0)}{\partial y} = 0
\]
The \(r_0(s)\) factor arises from the azimuthal metric of the cylindrical
geometry; it is \(O(L)\) and is not removed at any stage. The curvature
corrections to the momentum and energy equations carry a factor \(y/R(s)\)
(where \(R\) is the surface radius of curvature) and are \(O(\delta/L) = O(\varepsilon)\).
They will be dropped together with the streamwise viscous terms at Step 5.
Step 1: get \(V\) scaling from continuity.
Both terms must be the same order:
\[
\frac{U_\infty}{L} \sim \frac{V}{\delta}
\qquad\Longrightarrow\qquad
V \sim \frac{\delta}{L}\,U_\infty
\]
Step 2: get \(\delta\) by balancing convection with wall-normal viscosity in x-momentum.
\[
\underbrace{\rho\,\frac{U_\infty^2}{L}}_{\text{inertia}}
\sim \underbrace{\frac{\mu\,U_\infty}{\delta^2}}_{\text{viscosity}}
\qquad\Longrightarrow\qquad
\delta \sim \frac{L}{\sqrt{Re_L}}, \qquad Re_L = \frac{\rho U_\infty L}{\mu}
\]
Step 3: define dimensionless variables.
\[
\begin{aligned}
s^* &= \frac{s}{L}, &\quad
y^* &= \frac{y}{\delta}, &\quad
u^* &= \frac{u}{U_\infty}, \\[6pt]
v^* &= \frac{v}{\varepsilon U_\infty}, &\quad
p^* &= \frac{p}{\rho_\infty U_\infty^2}, &\quad
\rho^* &= \frac{\rho}{\rho_\infty}, \\[6pt]
\mu^* &= \frac{\mu}{\mu_\infty}, &\quad
T^* &= \frac{T}{T_\infty}, &\quad
k^* &= \frac{k}{k_\infty}
\end{aligned}
\]
where \(\varepsilon = \delta/L = Re_L^{-1/2} \ll 1\).
The body radius scales as \(r_0^* = r_0/L\).
Step 4: Substitute
Using \(\partial/\partial s = (1/L)\,\partial/\partial s^*\) and
\(\partial/\partial y = (1/\delta)\,\partial/\partial y^*\), substitute into each
equation and divide out the common dimensional factor.
Continuity
\[
\cancel{\frac{\rho_\infty U_\infty}{L}}\,
\frac{\partial(\rho^* u^* r_0^*)}{\partial s^*}
+ \cancel{\frac{\rho_\infty U_\infty}{L}}\,
\frac{\partial(\rho^* v^* r_0^*)}{\partial y^*} = 0
\]
\[
\frac{\partial(\rho^* u^* r_0^*)}{\partial s^*}
+ \frac{\partial(\rho^* v^* r_0^*)}{\partial y^*} = 0
\]
Note the \(r_0^*\) factor is carried through exactly. Both terms are
\(O(1)\) — the body radius does not produce an additional small parameter.
x-momentum
The cylindrical NS introduces curvature corrections of order \(y/R(s)\)
in the inertia and viscous terms. In dimensionless form these carry a
factor \(\varepsilon^2\) (since \(y/R \sim \delta/L = \varepsilon\), and the
leading viscous term already contributes a \(1/\varepsilon^2\) factor that
cancels the \(\varepsilon^2\) of the correction — net: \(O(\varepsilon^2)\)
relative to the retained term). All other structure is identical to
the 2D case:
\[
\begin{aligned}
\rho^*\!\left(u^*\frac{\partial u^*}{\partial s^*}
+ v^*\frac{\partial u^*}{\partial y^*}\right)
&= -\frac{\partial p^*}{\partial s^*}
+ \frac{\partial}{\partial y^*}\!\left(\mu^*\frac{\partial u^*}{\partial y^*}\right) \\[6pt]
&+ \varepsilon^2\!\left[
\frac{\partial}{\partial s^*}\!\left(\frac{4\mu^*}{3}\frac{\partial u^*}{\partial s^*}
- \frac{2\mu^*}{3}\frac{\partial v^*}{\partial y^*}\right)
+ \frac{\partial}{\partial y^*}\!\left(\mu^*\frac{\partial v^*}{\partial s^*}\right)
+ \text{(curvature)}
\right]
\end{aligned}
\]
Wall-normal momentum
The same order-of-magnitude argument as the 2D y-momentum applies. All inertia
and viscous terms are \(O(\varepsilon)\) or smaller relative to \(\partial p^*/\partial y^*\):
\[
\varepsilon^2\,\rho^*\!\left(u^*\frac{\partial v^*}{\partial s^*}
+ v^*\frac{\partial v^*}{\partial y^*}\right)
= -\frac{\partial p^*}{\partial y^*}
+ \varepsilon^2[\cdots]
\]
Energy
The azimuthal terms \((v/r)^2\) in \(\Phi^*\) and the cylindrical conduction
term \(k\partial T/\partial r / r\) both carry a factor \(\varepsilon^2\) in
dimensionless form (same argument as the curvature terms above). All other
structure matches the 2D case:
\[
\begin{aligned}
\rho^*\!\left(u^*\frac{\partial T^*}{\partial s^*}
+ v^*\frac{\partial T^*}{\partial y^*}\right)
&= \mathrm{Ec}\,u^*\frac{\partial p^*}{\partial s^*}
+ \frac{1}{\mathrm{Pr}}\frac{\partial}{\partial y^*}\!\left(k^*\frac{\partial T^*}{\partial y^*}\right)
+ \frac{\mathrm{Ec}}{\mathrm{Pr}}\,\mu^*\!\left(\frac{\partial u^*}{\partial y^*}\right)^{\!2} \\[6pt]
&+ \varepsilon^2\!\left[\frac{1}{\mathrm{Pr}}\frac{\partial}{\partial s^*}\!\left(k^*\frac{\partial T^*}{\partial s^*}\right)
+ \text{(curvature \& azimuthal terms)}\right]
\end{aligned}
\]
where \(\mathrm{Pr} = \mu_\infty c_p / k_\infty\) and \(\mathrm{Ec} = U_\infty^2/(c_p T_\infty)\).
Step 5: drop \(\mathcal{O}(\varepsilon^2)\) terms.
Setting \(\varepsilon \to 0\):
\[
\frac{\partial(\rho^* u^* r_0^*)}{\partial s^*}
+ \frac{\partial(\rho^* v^* r_0^*)}{\partial y^*} = 0
\]
\[
\rho^*\!\left(u^*\frac{\partial u^*}{\partial s^*} + v^*\frac{\partial u^*}{\partial y^*}\right)
= -\frac{\partial p^*}{\partial s^*}
+ \frac{\partial}{\partial y^*}\!\left(\mu^*\frac{\partial u^*}{\partial y^*}\right)
\]
\[\frac{\partial p^*}{\partial y^*} = 0\]
\[
\rho^*\!\left(u^*\frac{\partial T^*}{\partial s^*} + v^*\frac{\partial T^*}{\partial y^*}\right)
= \mathrm{Ec}\,u^*\frac{\partial p^*}{\partial s^*}
+ \frac{1}{\mathrm{Pr}}\frac{\partial}{\partial y^*}\!\left(k^*\frac{\partial T^*}{\partial y^*}\right)
+ \frac{\mathrm{Ec}}{\mathrm{Pr}}\,\mu^*\!\left(\frac{\partial u^*}{\partial y^*}\right)^{\!2}
\]
Step 6: re-dimensionalize.
Reversing the substitutions and applying \(\partial p^*/\partial y^* = 0
\Rightarrow \partial p/\partial y = 0\) recovers the dimensional BL equations
shown at the top of this page. The x-momentum and energy equations are
identical to the 2D case; the only difference is that
the continuity equation retains the body radius \(r_0(s)\).