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Energy: the energy equality, dissipation and enstrophy

The one estimate every Navier-Stokes solution obeys, what it controls, and why enstrophy saves two dimensions.

Take the dot product of the Navier-Stokes equations with \(\mathbf{u}\) and integrate over \(\mathbb{R}^3\) (with enough decay) or the torus \(\mathbb{T}^3\). Three things happen. The time derivative gives \(\frac{d}{dt}\int\tfrac12|\mathbf{u}|^2\). The nonlinear term vanishes: \[ \int \mathbf{u}\cdot(\mathbf{u}\cdot\nabla)\mathbf{u} = \int u_j\,\partial_j\!\left(\tfrac12|\mathbf{u}|^2\right) = -\int(\nabla\cdot\mathbf{u})\tfrac12|\mathbf{u}|^2 = 0. \] The pressure term vanishes too, \(\int\mathbf{u}\cdot\nabla p = -\int p\,\nabla\cdot\mathbf{u} = 0\). Integrating the viscous term by parts gives \(-\nu\int|\nabla\mathbf{u}|^2\). So every smooth solution satisfies \[ \frac{d}{dt}\,\frac12\int|\mathbf{u}|^2\,dx = -\nu\int|\nabla\mathbf{u}|^2\,dx, \] and integrating in time, the energy equality \[ \frac12\|\mathbf{u}(t)\|_{L^2}^2 + \nu\int_0^t\|\nabla\mathbf{u}(s)\|_{L^2}^2\,ds = \frac12\|\mathbf{u}_0\|_{L^2}^2. \] The nonlinearity moves energy around but never creates it; viscosity only removes it.

The rate of loss is the dissipation. For divergence-free fields, \(\int|\nabla\mathbf{u}|^2 = \int|\boldsymbol{\omega}|^2 = 2\int S:S\), so dissipation can be read as total squared vorticity or total squared strain. For the Taylor-Green field \((\sin x\cos y, -\cos x\sin y)\) on \([0,2\pi]^2\), the energy is \(\pi^2\) and \(\int|\nabla\mathbf{u}|^2 = 4\pi^2\), consistent with its exact decay like \(e^{-4\nu t}\). On the torus with zero mean, the Poincaré inequality \(\|\nabla\mathbf{u}\|_{L^2} \ge \|\mathbf{u}\|_{L^2}\) turns the energy equality into exponential decay, \(\|\mathbf{u}(t)\|^2 \le \|\mathbf{u}_0\|^2 e^{-2\nu t}\).

The energy equality gives two bounds valid for all time and all data: \(\sup_t\|\mathbf{u}(t)\|_{L^2} \le \|\mathbf{u}_0\|_{L^2}\) and \(\int_0^\infty\|\nabla\mathbf{u}\|_{L^2}^2\,dt \le \|\mathbf{u}_0\|^2_{L^2}/(2\nu)\). In three dimensions they are the basic a priori bounds for general data, and the other known global bounds are derived from them. They are strong enough to construct weak solutions, but, as the scaling lesson showed, they are weak at small scales: energy scales like \(\lambda^{-1}\).

In two dimensions there is a second conserved structure. The enstrophy \(\tfrac12\int\omega^2\) obeys, from \(D\omega/Dt = \nu\Delta\omega\), \[ \frac{d}{dt}\,\frac12\int\omega^2 = -\nu\int|\nabla\omega|^2 \le 0, \] because there is no stretching term. Since \(\|\nabla\mathbf{u}\|_{L^2} = \|\omega\|_{L^2}\), this is a uniform bound on the velocity gradient, a quantity that in 2D is subcritical (it scales like \(\lambda\), so it controls small scales), and it leads to global smooth solutions. In 3D the same calculation leaves \(\int\boldsymbol{\omega}\cdot S\boldsymbol{\omega}\) on the right, which can be positive; estimating it only gives \(\frac{d}{dt}\|\nabla\mathbf{u}\|^2 \le C\nu^{-3}\|\nabla\mathbf{u}\|^6\), an inequality that bounds the gradient for a time depending on the size of the data, and, once the dissipation term is kept, for all time if the data are small, but not for large data.

Picture it: a bank account of kinetic energy that the nonlinearity can move between scales but never top up, with viscosity charging a fee proportional to the squared gradients. Think it: the energy equality is an identity for smooth solutions, obtained by testing the equation against the solution itself; for weak solutions only an inequality survives, and whether equality holds is tied to their smoothness.

Umzekelo osebenzelayo · integrate (cos(x))^2 dx from 0 to 2*pi

Integrate cos(x)^2 from 0 to 2·pi

\int_{0}^{2 \pi} \cos^{2}{\left(x \right)}\, dx

Inyathelo ngenyathelo

  1. \int_{0}^{2 \pi} \cos^{2}{\left(x \right)}\, dx

    First find an antiderivative F, then evaluate F(b) − F(a).

  2. \cos^{2}{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}

    Rewrite the integrand into a friendlier form.

  3. \int \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\, dx = \int \frac{\cos{\left(2 x \right)}}{2}\, dx + \int \frac{1}{2}\, dx

    The integral of a sum is the sum of the integrals.

  4. \int \frac{\cos{\left(2 x \right)}}{2}\, dx = \frac{1}{2} \int \cos{\left(2 x \right)}\, dx

    Pull the constant \frac{1}{2} out of the integral.

  5. u = 2 x,\quad du = 2\, dx

    Substitute u = 2 x.

  6. \int \cos{\left(2 x \right)}\, dx = \int \frac{\cos{\left(u \right)}}{2}\, d_u

    Rewrite the integral in terms of u.

  7. \int \frac{\cos{\left(u \right)}}{2}\, d_u = \frac{1}{2} \int \cos{\left(u \right)}\, d_u

    Pull the constant \frac{1}{2} out of the integral.

  8. \int \cos{\left(u \right)}\, d_u = \sin{\left(u \right)}

    Standard trigonometric antiderivative.

  9. = \frac{\sin{\left(2 x \right)}}{2}

    Substitute back u = 2 x.

  10. \int \frac{1}{2}\, dx = \frac{x}{2}

    The integral of a constant c is c·x.

  11. F(2 \pi) - F(0) = \left(\pi\right) - \left(0\right)

    Fundamental theorem of calculus: plug in the limits.

  12. = \pi \approx 3.1416

    Simplify.

Bonisa impendulo
\pi

ngoku Khetha ingxaki, okanye ubhale okanye uzobe yakho. Inyathelo ngalinye, umfanekiso, impendulo ifihlakele de ubuze.

Ukugcina umsebenzi wakho

I akhawunti ekhululekileyo idibanisa amaphetshana kwi ncwadi nganye, irekhodi lento ogqibe ngayo, iingxaki zakho ezisoliweyo kwindawo enye, nomfundi onokuthi ubuze malunga nale phepha. IiMathematiki ngokwazo zivuliwe kubo bonke, bangeniswe okanye hayi.

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Iimpawu ezisetyenziswa apha

Nqakraza nasiphi na isibonakaliso sokuqonda okupheleleyo, umfanekiso, nokuba iileta zonke zithetha ntoni.

Indlela yokusebenza: Energy: the energy equality, dissipation and enstrophy

  1. Dot the equation with u and integrate over the whole domain.
  2. Show the advection and pressure terms vanish, using div u = 0 and integration by parts (no boundary terms on the torus or with decay).
  3. Integrate the viscous term by parts to get minus nu times the squared gradient, then integrate in time.
  4. For decay rates, combine with Poincare; for 2D, repeat the argument with the vorticity.

Imibuzo abantu bebuza

Why does the nonlinear term not change the energy?

Because it is transport by a divergence-free field, which only moves things around. Integrated over the whole domain, what leaves one region enters another.

Where does the energy go physically?

Into heat. Viscous dissipation converts kinetic energy into internal energy of the fluid; in the incompressible equations the temperature is not tracked, so the energy simply leaves the books.

What do I need before starting fluid dynamics?

Multivariable calculus (divergence, curl, the divergence and Stokes theorems), linear algebra (symmetric matrices and eigenvalues) and partial differential equations (the heat and Laplace equations). The last lessons also use Sobolev spaces, which are introduced where they are needed.

What are the Navier-Stokes equations in one sentence?

Newton's second law for each particle of a viscous incompressible fluid: acceleration equals the pressure force plus viscous diffusion of momentum, with the constraint that the velocity field is divergence-free.

Why is two-dimensional flow easier than three-dimensional flow?

In 2D the vorticity is a scalar that is only carried and diffused, so its maximum never grows; in 3D vortex lines can be stretched, which amplifies vorticity, and no known bound rules out unlimited growth.

Is this course physics or mathematics?

Both, in order. The first half derives the equations from physical principles and solves classical flows; the second half treats the equations as mathematical objects and studies which of their properties can be proved.

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