Euler's Equations as the Inviscid Limit

Derivation

Euler's Equations as the Inviscid Limit

D-319 Home PU-307 Threads force · energy · fields Depends on Navier-Stokes from the Newtonian Constitutive Law

Statement

Starting from the incompressible Navier–Stokes momentum equation with the Newtonian constitutive closure, we recover the inviscid Euler equations \(\rho\left(\partial_t\vec{u}+(\vec{u}\cdot\nabla)\vec{u}\right)=-\nabla p+\rho\vec{f}\) by letting the kinematic viscosity \(\nu=\mu/\rho\to 0\) (equivalently the Reynolds number \(\mathrm{Re}=UL/\nu\to\infty\)). We show that this limit is singular: it deletes the highest-order spatial derivative, drops the differential order from two to one, forces the no-slip condition to be abandoned in favour of no-penetration, and thereby leaves an unresolved boundary layer of thickness \(\delta\sim L\,\mathrm{Re}^{-1/2}\) that never disappears no matter how small \(\nu\) becomes.

Why it matters

Almost every analytic result in classical fluid dynamics — Bernoulli's theorem, Kelvin's circulation theorem, potential flow, aerofoil lift by the Kutta–Joukowski relation, sound and gravity waves — lives inside the Euler equations, not Navier–Stokes. Understanding how Euler emerges tells you precisely where those clean results are trustworthy (the irrotational core of a high-Reynolds-number flow) and where they silently break (any thin region adjacent to a wall or wake).

The limit is also the archetypal singular perturbation in mathematical physics. The small parameter \(1/\mathrm{Re}\) multiplies the highest derivative, so uniform convergence fails and a matched-asymptotic (inner/outer) treatment is unavoidable. The same structure recurs in the WKB limit \(\hbar\to0\), the geometrical-optics limit of wave equations, and vanishing-diffusivity limits throughout transport theory.

Assumptions

Incompressible, constant-density flow (\(\nabla\cdot\vec{u}=0\), \(\rho=\text{const}\)).If dropped, the viscous stress carries an extra \(\tfrac{1}{3}\mu\nabla(\nabla\cdot\vec{u})\) dilatational term and the inviscid limit becomes the compressible Euler system with its own energy equation; the boundary-layer scaling still holds but acoustics enter.

Newtonian constitutive law \(\boldsymbol{\tau}=\mu\left(\nabla\vec{u}+\nabla\vec{u}^{\mathsf T}\right)\) with constant \(\mu\).If dropped (shear-thinning, viscoelastic, or turbulent eddy-viscosity fluids) the term removed in the limit is no longer \(\mu\nabla^2\vec{u}\), so the boundary-layer thickness scaling and even the order-reduction argument change.

Smooth solution on the interior away from walls (bounded velocity gradients as \(\nu\to0\)).If dropped — e.g. the Euler solution itself develops singularities or sustains an energy cascade to arbitrarily small scales — the viscous term \(\nu\nabla^2\vec u\) need not vanish even as \(\nu\to0\) (anomalous dissipation, the Onsager conjecture), and the naive limit is false in the bulk, not merely at boundaries.

The relevant length and velocity scales \(L,U\) are those of the imposed geometry, used to build \(\mathrm{Re}\).If dropped, a single global \(\mathrm{Re}\) is meaningless; near a wall the correct inner length is \(\delta\ll L\), which is exactly why one number cannot describe the whole field.

Derivation

\[\rho\left(\frac{\partial\vec{u}}{\partial t}+(\vec{u}\cdot\nabla)\vec{u}\right)=-\nabla p+\mu\nabla^2\vec{u}+\rho\vec{f},\qquad \nabla\cdot\vec{u}=0.\]

Incompressible Navier–Stokes: substitute the Newtonian stress into Cauchy's momentum balance; \(\nabla\cdot\boldsymbol{\tau}=\mu\nabla^2\vec u\) because \(\nabla\cdot\vec u=0\) kills the second stress term. A

\[\vec{x}=L\,\vec{x}^{*},\quad \vec{u}=U\,\vec{u}^{*},\quad t=\frac{L}{U}\,t^{*},\quad p=\rho U^2 p^{*}.\]

Introduce dimensionless variables built from the geometric scale \(L\) and flow scale \(U\); the inertial pressure scale \(\rho U^2\) is chosen so the pressure gradient balances advection. A

\[\frac{\rho U^2}{L}\left(\frac{\partial\vec{u}^{*}}{\partial t^{*}}+(\vec{u}^{*}\cdot\nabla^{*})\vec{u}^{*}\right)=-\frac{\rho U^2}{L}\nabla^{*}p^{*}+\frac{\mu U}{L^2}\nabla^{*2}\vec{u}^{*}+\rho\vec{f}.\]

Apply the chain rule term by term: each \(\partial_x\to L^{-1}\partial_{x^*}\), giving the prefactors shown. A

\[\frac{\partial\vec{u}^{*}}{\partial t^{*}}+(\vec{u}^{*}\cdot\nabla^{*})\vec{u}^{*}=-\nabla^{*}p^{*}+\frac{\mu}{\rho U L}\,\nabla^{*2}\vec{u}^{*}+\frac{L}{U^2}\vec{f}.\]

Divide through by \(\rho U^2/L\). The viscous coefficient collapses to \(\mu/(\rho U L)=\nu/(UL)\). A

\[\frac{\partial\vec{u}^{*}}{\partial t^{*}}+(\vec{u}^{*}\cdot\nabla^{*})\vec{u}^{*}=-\nabla^{*}p^{*}+\frac{1}{\mathrm{Re}}\,\nabla^{*2}\vec{u}^{*}+\frac{1}{\mathrm{Fr}^2}\hat{g},\qquad \mathrm{Re}\equiv\frac{\rho U L}{\mu}=\frac{U L}{\nu}.\]

Define the Reynolds number as the sole coefficient of the viscous term (and, for gravity, the Froude number \(\mathrm{Fr}=U/\sqrt{gL}\)). Only one dimensionless group governs the viscous–inertial competition. B

\[\lim_{\mathrm{Re}\to\infty}\frac{1}{\mathrm{Re}}\,\nabla^{*2}\vec{u}^{*}=0\quad\text{pointwise, provided }\ \nabla^{*2}\vec u^{*}=O(1).\]

Take \(\nu\to0\) at fixed \(U,L\). If the interior gradients stay bounded (interior-smoothness assumption), the viscous term is uniformly small and drops out. B

\[\frac{\partial\vec{u}^{*}}{\partial t^{*}}+(\vec{u}^{*}\cdot\nabla^{*})\vec{u}^{*}=-\nabla^{*}p^{*}+\frac{1}{\mathrm{Fr}^2}\hat g\;\;\Longleftrightarrow\;\; \rho\!\left(\frac{\partial\vec u}{\partial t}+(\vec u\cdot\nabla)\vec u\right)=-\nabla p+\rho\vec f.\]

Restore dimensions: this is the incompressible Euler equation. The limit reduced the spatial order from 2 (\(\nabla^2\)) to 1 (\(\nabla\)); the equation can no longer accept two vector boundary conditions per wall. B

\[\text{No-slip }\vec u|_{\text{wall}}=0\ \ (\text{2 conditions})\ \longrightarrow\ \text{no-penetration }\vec u\cdot\hat n|_{\text{wall}}=0\ \ (\text{1 condition}).\]

Order reduction forces us to discard the tangential (no-slip) condition; Euler retains only impermeability. The discarded condition cannot in general be met by the outer solution — the tell-tale of a singular perturbation. C

\[\text{Inner rescale near wall: } \eta=\frac{y}{\delta}.\quad \frac{1}{\mathrm{Re}}\,\partial_y^2\sim\frac{1}{\mathrm{Re}}\frac{1}{\delta^2}\ \overset{!}{\sim}\ \partial_x\sim O(1)\ \Rightarrow\ \boxed{\ \delta\sim L\,\mathrm{Re}^{-1/2}.\ }\]

Restore the viscous term in a thin inner layer of unknown thickness \(\delta\) and demand a dominant balance (Prandtl): viscous diffusion must re-enter to satisfy no-slip. The distinguished scaling \(\delta\sim L\,\mathrm{Re}^{-1/2}\) is the layer that survives the limit. C

Result

\[\rho\left(\frac{\partial\vec{u}}{\partial t}+(\vec{u}\cdot\nabla)\vec{u}\right)=-\nabla p+\rho\vec{f},\qquad \nabla\cdot\vec u=0,\qquad \vec u\cdot\hat n\big|_{\text{wall}}=0,\]

with the layer that the limit cannot resolve set by \(\displaystyle \frac{\delta}{L}\sim \mathrm{Re}^{-1/2}=\left(\frac{\nu}{UL}\right)^{1/2}\).

Reading. As inertia overwhelms viscosity (\(\mathrm{Re}\to\infty\)) the bulk flow obeys Euler and is governed by inertia and pressure alone. Viscosity does not vanish everywhere: it is squeezed into a wall layer whose fractional thickness scales as \(\mathrm{Re}^{-1/2}\). The limit is singular precisely because the small parameter \(1/\mathrm{Re}\) multiplies the highest derivative, so one boundary condition must be shed in the outer (Euler) problem and recovered only inside the inner (boundary-layer) problem.

Units check. Every term in the momentum equation has units \([\rho][u]^2/[L]=(\mathrm{kg\,m^{-3}})(\mathrm{m^2\,s^{-2}})(\mathrm{m^{-1}})=\mathrm{kg\,m^{-2}\,s^{-2}}=\mathrm{Pa\,m^{-1}}\), matching \(\nabla p\). The Reynolds number \(\mathrm{Re}=UL/\nu\) has units \((\mathrm{m\,s^{-1}})(\mathrm m)/(\mathrm{m^2\,s^{-1}})=1\), dimensionless, and \(\delta\sim L\,\mathrm{Re}^{-1/2}\) carries units of length.

Limiting cases

\(\mathrm{Re}\to 0\): the opposite limit — drop inertia instead, giving the linear Stokes equations \(\nabla p=\mu\nabla^2\vec u\); creeping flow, no boundary layer, time-reversible.
\(\mathrm{Re}\to\infty\), irrotational, steady: Euler integrates to Bernoulli, \(\tfrac12\rho u^2+p+\rho g z=\text{const}\) along (indeed across) streamlines.
Finite but large \(\mathrm{Re}\): outer Euler core matched to an inner Prandtl boundary layer of thickness \(\delta\approx 5L/\sqrt{\mathrm{Re}}\) (Blasius, laminar flat plate).
\(\nu\to0\) with singular Euler solution: anomalous dissipation — \(\nu\|\nabla\vec u\|^2\) may stay \(O(1)\); the inviscid limit dissipates energy (Onsager).
Free-slip / periodic domains: no walls, no no-slip to shed — the limit becomes regular and Euler is uniformly valid (subject to interior smoothness).

Breaks when

Near any solid boundary. Euler permits fluid to slip along the wall; the real fluid does not. In the layer \(y\lesssim\delta\sim L\,\mathrm{Re}^{-1/2}\) the neglected \(\nu\nabla^2\vec u\) is restored to \(O(1)\), and the outer inviscid solution is simply wrong there, however large \(\mathrm{Re}\).
Wherever the boundary layer separates. Under an adverse pressure gradient the layer detaches, throwing vorticity into the bulk; the flow becomes globally rotational and the potential-flow reading of Euler (and d'Alembert's zero-drag prediction) fails badly — this is the physical origin of pressure drag and stall.
When interior gradients blow up (turbulence / Euler singularities). If the cascade drives \(\|\nabla\vec u\|\) to grow without bound as \(\nu\to0\), the product \(\nu\nabla^2\vec u\) need not vanish; anomalous dissipation means the inviscid limit is not the naive Euler solution even in the core.
At low Reynolds number. When \(\mathrm{Re}\lesssim O(1)\) the viscous term is comparable to or larger than inertia; deleting it is quantitatively indefensible and Stokes flow, not Euler, is the correct reduction.

Failure modes

Imposing no-slip on the Euler solution. Students try to force \(\vec u_\parallel=0\) at a wall in the inviscid problem; the first-order equation is over-determined and has no solution. Euler takes only \(\vec u\cdot\hat n=0\).
"The limit is regular because the term is small." Treating \(1/\mathrm{Re}\,\nabla^2\vec u\) as a uniformly small regular perturbation. It multiplies the highest derivative, so convergence is non-uniform and a boundary layer is mandatory.
Claiming zero drag is physical. Reading d'Alembert's paradox as "no drag at high Re" instead of "the inviscid limit misses the boundary-layer/separation physics that produces drag."
Confusing \(\mu\to0\) with \(\rho\to\infty\). Only the combination \(\nu=\mu/\rho\) (via \(\mathrm{Re}\)) matters for the momentum balance; students vary the wrong parameter.
Using the wrong length in \(\mathrm{Re}\). Plugging the domain scale \(L\) into a near-wall estimate instead of the inner scale \(\delta\), then concluding viscosity is negligible in the layer.
Assuming Bernoulli holds across the boundary layer. Total head is not conserved there because viscosity dissipates it; Bernoulli is an Euler (outer) result only.

Discussion

The single most important physical statement is that viscosity is never truly negligible; it is merely confined. At \(\mathrm{Re}=10^6\) the boundary layer on a metre-scale body is about a millimetre thick, yet that millimetre carries all the shear stress the wall exerts on the fluid and, through separation, controls the entire pressure field and hence the drag. The Euler equations describe the 99.9% of the volume where the fluid behaves as if ideal, but the physically decisive events happen in the 0.1% they cannot see. This is why "solve Euler everywhere" is a good outer approximation and a catastrophic global one.

Mathematically, the order reduction from second to first spatial derivative is the whole story. A second-order elliptic/parabolic operator wants two conditions per boundary (no-slip: both velocity components); a first-order operator can accept only one (no-penetration). The mismatch is resolved by matched asymptotics: an outer Euler solution valid for \(y\gg\delta\), an inner Prandtl solution valid for \(y\lesssim\delta\), stitched together in an overlap region. The distinguished limit \(\delta\sim L\,\mathrm{Re}^{-1/2}\) is exactly the thickness at which the retained viscous term balances advection again — it is selected, not assumed.

The connection to circulation and lift is the payoff. In the inviscid outer flow Kelvin's theorem conserves circulation, so an aerofoil started from rest would generate no lift — yet aircraft fly. The resolution is that the boundary layer, and specifically its separation at a sharp trailing edge (the Kutta condition), sheds a starting vortex and sets the circulation that Euler alone leaves undetermined. Viscosity, banished from the bulk, returns through the boundary condition it silently fixes. The clean inviscid theory of lift is therefore parasitic on the viscous layer it pretends to ignore.

At the deepest level the limit \(\nu\to0\) is not guaranteed to commute with the flow's own dynamics. Onsager conjectured, and modern work has largely confirmed, that below Hölder regularity \(1/3\) Euler solutions can dissipate kinetic energy without any viscosity at all: the term \(\varepsilon=\nu\langle|\nabla\vec u|^2\rangle\) can tend to a strictly positive constant as \(\nu\to0\) because \(\langle|\nabla\vec u|^2\rangle\sim\varepsilon/\nu\) diverges in step. This anomalous dissipation underlies the Kolmogorov cascade and means the inviscid limit of Navier–Stokes is, in the turbulent regime, a weak (dissipative) Euler solution rather than the smooth one this derivation assumed — a live research frontier, not a settled reduction.

Common misconceptions. "High Reynolds number means viscosity can be ignored" — only in the bulk, and only if you separately supply the boundary-layer/separation physics; the wall stress and the drag both come from the viscosity you discarded. And "the Euler limit is the \(\hbar\to0\) of fluids" is apt precisely because both are singular: the small parameter multiplies the highest derivative in each case.

Worked examples

Boundary-layer thickness for air over a wing panel. Given \(U=50\ \mathrm{m\,s^{-1}}\), \(L=2\ \mathrm{m}\), \(\nu_{\text{air}}=1.5\times10^{-5}\ \mathrm{m^2\,s^{-1}}\).

Set up the Reynolds number; symbols first. A

\[\mathrm{Re}=\frac{UL}{\nu}=\frac{(50)(2)}{1.5\times10^{-5}}=6.7\times10^{6}.\]

Insert numbers; large \(\mathrm{Re}\) confirms the inviscid core is a good outer model. A

\[\delta\approx\frac{5L}{\sqrt{\mathrm{Re}}}=\frac{5(2)}{\sqrt{6.7\times10^{6}}}=\frac{10}{2.58\times10^{3}}=3.9\times10^{-3}\ \mathrm{m}.\]

Blasius laminar estimate; \(\delta\sim L\,\mathrm{Re}^{-1/2}\) with prefactor \(5\). B

\[\boxed{\ \delta\approx 3.9\ \mathrm{mm}\ }\qquad \frac{\delta}{L}\approx1.9\times10^{-3}.\]

Reading. Over a 2 m panel the layer is under 4 mm: Euler is excellent everywhere except a paper-thin sheet at the surface — which nonetheless carries all the skin friction.

Units check. \(\mathrm{Re}\) dimensionless; \(\delta=[\mathrm m]/\sqrt{1}=[\mathrm m]\). Consistent.

Is the inviscid core valid? Compare viscous to inertial term magnitudes for water in a 5 cm pipe, \(U=1\ \mathrm{m\,s^{-1}}\), \(\nu_{\text{water}}=1.0\times10^{-6}\ \mathrm{m^2\,s^{-1}}\). Estimate the fractional pressure drop across a smooth contraction that halves the area.

Order-of-magnitude ratio then a Bernoulli application; symbols first. A

\[\frac{|\nu\nabla^2\vec u|}{|(\vec u\cdot\nabla)\vec u|}\sim\frac{\nu U/L^2}{U^2/L}=\frac{\nu}{UL}=\frac{1}{\mathrm{Re}},\qquad \mathrm{Re}=\frac{(1)(0.05)}{1.0\times10^{-6}}=5\times10^{4}.\]

The viscous-to-inertial ratio is \(1/\mathrm{Re}\) — here \(2\times10^{-5}\), so the bulk is inviscid to five significant figures. B

\[A_2=\tfrac12 A_1\ \Rightarrow\ U_2=2U_1=2\ \mathrm{m\,s^{-1}};\qquad \Delta p=\tfrac12\rho\left(U_2^2-U_1^2\right).\]

Mass conservation \(A_1U_1=A_2U_2\), then Euler integrated along a streamline (Bernoulli), legitimate because the core is inviscid. B

\[\Delta p=\tfrac12(1000)\left(2^2-1^2\right)=\tfrac12(1000)(3)=1.5\times10^{3}\ \mathrm{Pa}.\]

Insert \(\rho=1000\ \mathrm{kg\,m^{-3}}\); pressure falls where the flow speeds up. A

\[\boxed{\ \mathrm{Re}=5\times10^{4},\quad \Delta p=p_1-p_2=1.5\ \mathrm{kPa}\ }\]

Reading. At \(\mathrm{Re}=5\times10^4\) the viscous term is 20 000 times smaller than inertia in the core, so the Euler/Bernoulli pressure estimate is reliable there — while a \(\sim\!0.2\ \mathrm{mm}\) wall layer still governs the friction loss the calculation omits.

Units check. \(\tfrac12\rho U^2=(\mathrm{kg\,m^{-3}})(\mathrm{m^2\,s^{-2}})=\mathrm{kg\,m^{-1}\,s^{-2}}=\mathrm{Pa}\). Correct.

Problems

Glycerine (\(\nu=1.2\times10^{-3}\ \mathrm{m^2\,s^{-1}}\)) flows at \(U=0.1\ \mathrm{m\,s^{-1}}\) through an \(L=0.02\ \mathrm{m}\) gap. Compute \(\mathrm{Re}\) and state whether the Euler or the Stokes reduction is appropriate.
Solution
\(\mathrm{Re}=UL/\nu=(0.1)(0.02)/(1.2\times10^{-3})=1.67\). Since \(\mathrm{Re}=O(1)\lesssim 1\), viscous and inertial terms are comparable; neither limit is clean, but the low-Re Stokes reduction (\(\nabla p=\mu\nabla^2\vec u\)) is far closer to correct than Euler. Deleting viscosity here is indefensible.
A model ship hull of length \(L=3\ \mathrm{m}\) moves at \(U=4\ \mathrm{m\,s^{-1}}\) in water (\(\nu=1.0\times10^{-6}\ \mathrm{m^2\,s^{-1}}\)). Estimate the laminar boundary-layer thickness at the stern using \(\delta\approx 5L\,\mathrm{Re}^{-1/2}\).
Solution
\(\mathrm{Re}=UL/\nu=(4)(3)/10^{-6}=1.2\times10^{7}\). \(\sqrt{\mathrm{Re}}=3.46\times10^{3}\). \(\delta\approx 5(3)/3.46\times10^{3}=15/3460=4.3\times10^{-3}\ \mathrm{m}\approx 4.3\ \mathrm{mm}\). (In reality this \(\mathrm{Re}\) exceeds transition, \(\sim5\times10^{5}\), so the real layer is turbulent and thicker, \(\delta\sim 0.37L\,\mathrm{Re}^{-1/5}\approx 4.3\ \mathrm{cm}\); the laminar formula gives an order-of-magnitude floor.)
Water discharges from a large open tank through a small side hole a depth \(h=2.0\ \mathrm{m}\) below the free surface. Using Euler integrated to Bernoulli, find the exit speed (Torricelli). State the assumption that makes viscosity ignorable.
Solution
Bernoulli from surface (speed \(\approx0\), gauge pressure \(0\)) to jet (pressure \(0\)): \(\rho g h=\tfrac12\rho v^2\Rightarrow v=\sqrt{2gh}=\sqrt{2(9.81)(2.0)}=\sqrt{39.24}=6.26\ \mathrm{m\,s^{-1}}\). Valid because the flow accelerates through a short, high-\(\mathrm{Re}\) contraction where the boundary layer is thin relative to the orifice; the real efflux is \(\sim2\%\) lower (discharge coefficient) from that neglected layer.
Show, by non-dimensionalising, that if two flows share the same geometry and the same Reynolds number they satisfy identical dimensionless equations (dynamic similarity). Then find the air-tunnel speed that matches \(\mathrm{Re}\) of a car at \(U_{\text{car}}=30\ \mathrm{m\,s^{-1}}\), \(L_{\text{car}}=4\ \mathrm{m}\) using a \(1{:}4\) model.
Solution
Step 5 of the derivation shows the only parameter multiplying the viscous term is \(1/\mathrm{Re}\); with identical geometry and boundary conditions, equal \(\mathrm{Re}\) gives identical \(\vec u^*(\vec x^*,t^*)\). Matching: \(U_mL_m=U_cL_c\) (same fluid, same \(\nu\)). \(L_m=L_c/4=1\ \mathrm m\), so \(U_m=U_cL_c/L_m=(30)(4)/1=120\ \mathrm{m\,s^{-1}}\). (At that speed compressibility, \(\mathrm{Ma}\approx0.35\), starts to matter — a real caveat of similarity testing.)
Explain d'Alembert's paradox quantitatively for steady inviscid irrotational flow past a body: state the predicted drag and identify precisely which step of the inviscid limit is responsible for the wrong answer.
Solution
For steady, incompressible, irrotational Euler flow past a closed body the pressure field is fore–aft symmetric, so the net pressure force (drag) integrates to exactly zero: \(D_{\text{inviscid}}=0\). Real bodies experience substantial drag. The culprit is Step 8: shedding the no-slip condition. Without no-slip there is no boundary layer, hence no separation, hence no fore–aft pressure asymmetry and no wake. Restoring viscosity in the thin layer (Step 9) produces separation and the low-pressure wake that generates pressure drag. The paradox is thus a signature that the inviscid limit is singular, not a claim that real drag vanishes.