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  Solutions of the Navier-Stokes equations

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The Navier-Stokes equations can be geometrized in the following form:

$$\dot{u} + \nabla_u u =\nu \Delta (u)+ df^* $$

$$d^* u=0$$

$\nabla_X Y$ is the connection $dY(X)$. If we define $u=\dot{\gamma}$, we recognize the equation of geodesics:

$$ \nabla_{\dot{\gamma}}\dot{\gamma}=0$$

Can we solve the Navier-Stokes equations with help of a lagrangian formalism?

asked Nov 10 in Mathematics by Antoine Balan (280 points) [ revision history ]
edited Nov 10 by Antoine Balan
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No. The Lagrangian formalism is just a tool to set up equations, not to solve them.

Moreover, the Navier-Stokes equations are dissipative, while the Lagrangian formalism produces conservative equations. Thus there is no useful Lagrnagian formulation of the Navier-Stokes equation.

This gives a second order equation for $\gamma$, hence with your substitution a first order equation for $u$. There is no natural way to get a dissipative equation from a Lagrangian. (One can get one - like for a damped harmonic oscillator - by doubling the fields, but then the resulting Hamiltonian has nothing to do with the energy.)

As noted, the Navier-Stokes equations contain something like the geodesic equations. With the help of a dissipation functional the NS equations can be reframed in a variational formalism, but I don't see how this gives new information about the solutions.

The NS equations are dissipative, but the energy isn't lost, simply it is transformed in heat. So, I propose to introduce a temperature parameter $T$ and to make a variational approach with a Lagrangian decomposed in a term depending on $u$ and another depending on the temperature $T$ which is a function of $u$.

You'd need to propose an actual Lagrangian that reproduces the Navier-Stokes equation. Just speculating that such a thing might exist is not enough for a discussion.

Most recent comments show all comments

Could you give more information for how this dissipation is supposed to interact with temperature (or just how temperature plays into this)? Neglecting T, the presence of a dissipation functional makes it so that the equations of motion are no longer extrema of the action functional. 

If the boundary condition at infinity is not an additional condition then the minimizer of the action is not determined by the boundary condition, as it leaves the initial conditions unspecified. Thus it cannot be used for the usual fluid flow calculations.

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