Deriving the 2D Boussinesq Vorticity Equation

Good to see you again. In the previous lesson, we wrote the forced inviscid Boussinesq system as

$${\partial }_t\theta +u\cdot \nabla \theta =f_{\theta },$$ $${\partial }_{t}u+u\cdot \nabla u+\nabla p=\theta e_2+f_u,$$ $$\mathrm{div}u=0.$$

The key structural observation was that $\theta$ exerts a vertical force, even though it is itself transported by the flow. This lesson applies the curl operator to the momentum equation. Doing so removes pressure and reveals the form used throughout the blowup construction: the horizontal temperature gradient is a direct source of vorticity.

By the end, you should be able to derive, including its sign convention,

$${\partial }_t\omega +u\cdot \nabla \omega ={\partial }_1\theta +\mathrm{curl}{f}_u.$$

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Vorticity in two dimensions

Write the planar velocity as

$$u=(u_1,u_2).$$

Embed this field in three dimensions as $(u_1,u_2,0)$. Its three-dimensional curl then points perpendicular to the plane:

$$\nabla \times u=(0,0,\omega ),$$

where the scalar vorticity is defined by

$$\omega :={\partial }_1{u}_2-{\partial }_2{u}_1.$$

For this course, we use the scalar notation

$$\mathrm{curl}v:={\partial }_1{v}_2-{\partial }_2{v}_1$$

for any planar vector field $v=(v_1,v_2)$. Thus,

$$\omega =\mathrm{curl}u.$$

This sign convention matters. It is the standard counterclockwise-positive convention: near a point, positive $\omega$ corresponds to an infinitesimal counterclockwise rotation.

For instance, consider the rigid rotation field

$$u(x)=\frac{\Omega }{2}(-x_2,x_1).$$

Then

$${\partial }_1{u}_2=\frac{\Omega }{2},{\partial }_2{u}_1=-\frac{\Omega }{2},$$

and hence

$$\omega =\Omega .$$

So a positive angular velocity produces positive vorticity.

Vorticity is not merely “rotation” in an informal sense. It is the part of the velocity gradient that records local circulation. The velocity itself remains a nonlocal quantity determined from vorticity together with incompressibility and suitable boundary or decay conditions. Later, this relation will become the Fourier-mode Biot–Savart calculation. For now, the important step is to derive the equation that governs $\omega$.

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Why take curl?

Taking curl extracts the rotational content of a force. A pressure force $\nabla p$ is a gradient field, so it is locally circulation-free:

$$\mathrm{curl}(\nabla p)={\partial }_1{\partial }_{2}p-{\partial }_2{\partial }_{1}p=0.$$

This is the main payoff: pressure is indispensable in the velocity equation because it enforces incompressibility, but it disappears from the vorticity equation.

The following short lecture excerpt gives a useful general demonstration of this term-by-term procedure. It is framed for Navier–Stokes and briefly discusses three dimensions, so keep your attention on the derivative commutations and the disappearance of pressure; our Boussinesq-specific buoyancy term comes immediately afterwards.

There is a useful physical way to anticipate the result. The buoyancy force is

$$\theta e_2=(0,\theta ).$$

If $\theta$ changes horizontally, then the upward force differs from one vertical side of a small fluid rectangle to the other. That force imbalance creates a local tendency to rotate. Mathematically, its curl is

$$\mathrm{curl}(\theta e_2)={\partial }_1\theta .$$

By contrast, if $\theta$ depends only on height $x_2$, its buoyancy force has zero curl. Indeed, in that special case it can be absorbed locally into a pressure gradient.

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Taking curl of the Boussinesq momentum equation

Start from the momentum equation:

$${\partial }_{t}u+u\cdot \nabla u+\nabla p=\theta e_2+f_u.$$

Apply $\mathrm{curl}$ to both sides:

$$\mathrm{curl}({\partial }_{t}u)+\mathrm{curl}(u\cdot \nabla u)+\mathrm{curl}(\nabla p)=\mathrm{curl}(\theta e_2)+\mathrm{curl}{f}_u.$$

We now simplify every term.

The time-derivative term

Because the spatial derivatives in curl commute with the time derivative for a smooth solution,

$$\mathrm{curl}({\partial }_{t}u)={\partial }_t(\mathrm{curl}u)={\partial }_t\omega .$$

The pressure term

As already noted,

$$\mathrm{curl}(\nabla p)=0.$$

Pressure has not ceased to matter physically or mathematically. Rather, curl filters out the irrotational part of the acceleration, and a gradient is precisely such a part.

The buoyancy term

Since

$$e_2=(0,1),$$

we have

$$\theta e_2=(0,\theta ).$$

Therefore,

$$\mathrm{curl}(\theta e_2)={\partial }_1(\theta )-{\partial }_2(0)={\partial }_1\theta .$$

The fact that the source is ${\partial }_1\theta$, not ${\partial }_2\theta$, follows directly from the geometry: curl differentiates the vertical component of a vector field in the horizontal direction.

The external-force term

Write

$$f_u=(f_{u,1},f_{u,2}).$$

Then its rotational contribution is

$$\mathrm{curl}{f}_u={\partial }_1{f}_{u,2}-{\partial }_2{f}_{u,1}.$$

A useful corollary is that a forcing of the form $f_u=\nabla \varphi$ has no effect on vorticity. It can be incorporated into a redefined pressure. Only the non-gradient, rotational component of $f_u$ enters this equation.

The only term still requiring work is the nonlinear transport term,

$$\mathrm{curl}(u\cdot \nabla u).$$

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The nonlinear term and the role of incompressibility

In three-dimensional Euler, taking curl exposes the famous vortex-stretching term. In two dimensions, there is no independent third direction in which vortex lines can be stretched. Under incompressibility, the nonlinear term reduces to pure transport:

$$\mathrm{curl}(u\cdot \nabla u)=u\cdot \nabla \omega .$$

This identity is worth deriving rather than treating as a black box.

Set

$$u=(a,b),$$

so that

$$\omega ={\partial }_{1}b-{\partial }_{2}a.$$

The advective acceleration is

$$u\cdot \nabla u=\left(a{\partial }_{1}a+b{\partial }_{2}a,a{\partial }_{1}b+b{\partial }_{2}b\right).$$

Taking curl gives

$$\mathrm{curl}(u\cdot \nabla u)={\partial }_1\left(a{\partial }_{1}b+b{\partial }_{2}b\right)-{\partial }_2\left(a{\partial }_{1}a+b{\partial }_{2}a\right).$$

Expanding with the product rule yields

$$\begin{array}{rl}\mathrm{curl}(u\cdot \nabla u)=& a{\partial }_{11}b+b{\partial }_{12}b-a{\partial }_{21}a-b{\partial }_{22}a\\ & +({\partial }_{1}a)({\partial }_{1}b)+({\partial }_{1}b)({\partial }_{2}b)\\ & -({\partial }_{2}a)({\partial }_{1}a)-({\partial }_{2}b)({\partial }_{2}a).\end{array}$$

The first line is exactly

$$a{\partial }_1\omega +b{\partial }_2\omega =u\cdot \nabla \omega .$$

The remaining terms can be collected as

$$({\partial }_{1}a+{\partial }_{2}b)({\partial }_{1}b-{\partial }_{2}a).$$

That is,

$$\mathrm{curl}(u\cdot \nabla u)=u\cdot \nabla \omega +(\mathrm{div}u)\omega .$$

Now impose incompressibility:

$$\mathrm{div}u={\partial }_{1}a+{\partial }_{2}b=0.$$

The extra term vanishes, leaving

$$\mathrm{curl}(u\cdot \nabla u)=u\cdot \nabla \omega .$$

This is a precise statement of a central two-dimensional fact: incompressible flow transports scalar vorticity without directly stretching it.

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The vorticity equation

Substituting all the simplified terms into the curled momentum equation gives

$${\partial }_t\omega +u\cdot \nabla \omega ={\partial }_1\theta +\mathrm{curl}{f}_u.$$

Equivalently, using the material derivative from the previous lesson,

$$D_t\omega ={\partial }_1\theta +\mathrm{curl}{f}_u.$$

Along a particle trajectory $X(t)$, this becomes

$$\frac{d}{dt}\omega (X(t),t)={\partial }_1\theta (X(t),t)+\mathrm{curl}{f}_u(X(t),t).$$

This equation should be read with care.

• The term $u\cdot \nabla \omega$ transports and geometrically rearranges vorticity.
• The term ${\partial }_1\theta$ produces vorticity through buoyancy.
• The term $\mathrm{curl}{f}_u$ is vorticity supplied externally.
• Pressure does not appear because its force is locally curl-free.

When both source terms vanish,

$${\partial }_1\theta =0,\mathrm{curl}{f}_u=0,$$

the vorticity of each fluid parcel is conserved:

$$\frac{d}{dt}\omega (X(t),t)=0.$$

That conservation is often contrasted with three-dimensional Euler, where vorticity may grow through vortex stretching even without external forcing or buoyancy. The Boussinesq mechanism studied here gets its derivative-level growth from a different route: it generates vorticity from a carefully arranged temperature gradient.

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Two quick sign and structure checks

Consider first a temperature anomaly depending only on vertical position:

$$\theta =\theta (x_2).$$

Then

$${\partial }_1\theta =0.$$

Although buoyancy can still accelerate the fluid vertically, it introduces no direct vorticity source. In fact, define a scalar function $Q$ satisfying

$${\partial }_{2}Q=\theta (x_2).$$

Then

$$\theta e_2=\nabla Q,$$

so this force is indistinguishable from a pressure contribution in the vorticity equation.

Now consider instead

$$\theta =x_1,f_u=0.$$

Then

$${\partial }_1\theta =1,$$

so

$$D_t\omega =1.$$

The upward force is stronger to the right than to the left. This produces a counterclockwise torque and hence a positive vorticity source, consistent with our convention.

These are local diagnostic tests worth retaining when reading formulae in the papers. They make the sign in the vorticity equation much easier to reconstruct than memorising it.

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Why this equation is the gateway to the construction

The original system has a scalar $\theta$ pushing a velocity $u$. The vorticity equation makes the derivative feedback visible:

$$D_t\theta =f_{\theta },$$ $$D_t\omega ={\partial }_1\theta +\mathrm{curl}{f}_u.$$

Thus the value of $\theta$ need not be large in order to generate large vorticity. What matters is that $\theta$ develops a large horizontal derivative. The later wave ansatz is designed precisely to exploit that distinction: a bounded-amplitude oscillation can have a temperature gradient proportional to its frequency.

This does not yet establish an instability by itself. One must also understand how vorticity determines velocity, how that velocity transports wavevectors, and how the coupled amplitudes can grow. Those are the next steps in the module.

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Takeaways and next lesson

Starting from the planar forced inviscid Boussinesq momentum equation, we defined

$$\omega ={\partial }_1{u}_2-{\partial }_2{u}_1$$

and took curl term by term. The time derivative becomes ${\partial }_t\omega$, pressure vanishes because curl of a gradient is zero, buoyancy contributes ${\partial }_1\theta$, and incompressibility reduces the curl of nonlinear advection to $u\cdot \nabla \omega$. The result is

$$\boxed{{\partial }_t\omega +u\cdot \nabla \omega ={\partial }_1\theta +\mathrm{curl}{f}_u}$$

or, along fluid trajectories,

$$D_t\omega ={\partial }_1\theta +\mathrm{curl}{f}_u.$$

Next, we will distinguish three claims that can otherwise blur together in a blowup theorem: smooth evolution at each time before the terminal time, forcing that remains smooth even at the terminal time, and the divergence of selected solution derivative norms as that time is approached.