Solutions of the Navier-Stokes equations

Question

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?

Comments

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.

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But, if we minimize a Lagrangian over curves, we can find the solution of the NS equations.

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Which Lagrangian?

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Perhaps, a Lagrangian like:

$$L(\gamma)=\int_{\gamma} ||\dot{\gamma}||^2 +\nu ||d \dot{\gamma}||^2 dx$$

with constraints $d^*\dot{\gamma}=0$.

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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.)

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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.

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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 $u$.

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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.

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It seems to be a problem of thermodynamics to give the proper Lagrangian. But I am not able to propose the right one.

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This has nothing to do with thermodynamics. Lagrangians are just mechanics. if there is no simple way to guess the Lagrangian for a particular system of PDEs, there is none.

I don't think there is a Lagrangian giving the Navier-Stokes equations. But even should there be one, it does not help at all in devising numerical methods.

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@AntoineBalan How do you propose that knowing the variational formalism will help with solutions?

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If you have a Lagrangian, you can find the solution by minimizing it. You can take the gradient of the functional, making a flow which converges toward the solution, allowing you to compute it.

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In general, the equations of motions are only stationary points, not minimizers, of the action.

Even when the action is bounded below, minimization applies only in the case of boundary conditions. But the Navier-Stokes equations are almost always solved for initial-boundary conditions. For these no action principle exists - not even for a harmonic oscillator!

Finally, minimizing a nonconvex action inherits all the difficulties of solving nonlinear PDEs.

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But, due to the fact that the energy is converted in heat because of viscosity, we may suppose that at infinity, we have $u=0$; it is a boundary condition.

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The practical interest is in the initial-boundary-value problem and finite times. Specifying an additional boundary condition at the end overdetermines the system.

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It is not an additional condition because it is due to the fact that the energy is trranformed in heat. So, it is not an overdetermined system. At the end, we have simply $u=0$, all the energy is thermodynamic.

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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. 

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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.

Answer 1

If one of the two analyzed energy types is zero, then there will be no significant  difference between a Langrangian and the underlying Hamiltonian that solves the stream-lines of the recent found N-soliton solutions of the Navier Stokes d.e. Please do see  also the following publication cited as

[1] R. Meulens , "A note on N-soliton solutions for the viscid incompressible Navier–Stokes differential equation", AIP Advances 12, 015308 (2022) https://doi.org/10.1063/5.0074083

e.g. the benchmark  driven-lid cavity fluid dynamical problem is a problem without significant effect of the potential energy. So in this case would the Langrangian be equivalent with the Hamiltonian. Below the 3d-lump (geometric sum of all solutions or a rational) solution of this problem found in [1]