# Alexandre Chorin steady state solution of Navier Stokes

I'm trying to learn fluid dynamics and numerical methods for solving the NS equations by reading research papers. But a lot of them are complicated so I kept searching for older references that look easier to understand. I found papers I like by Alexndre Chorin. I've been trying to work through one of his papers on a steady state finite difference method for the NS equations that uses artificial compressibility. Before the numerical part of the paper he introduces

the auxiliary system of equations $$\partial_t u_i+R\partial_i (u_i u_i) = - \partial_i p + \Delta u_i + F_i \hskip{1em}(1)$$

$$\partial_t \rho+\partial_j u_j = 0 \hskip{1em},\hskip{1em} p=\rho / \delta \hskip{1em}(2)$$

then writes "It is not indispensable that the solution of the differential system $(1)$ and $(2)$ tend to a steady limit, as long as the solution to the difference system does. It is however believed that the solution of $(1)$ and $(2)$ does tend to a steady limit, at least in the absence of external forces, under quite general conditions. This can be proved in the limiting case $R=0$, for problems in which the velocities are prescribed at the boundary. By linearity it is sufficient to consider the case of zero velocities at the boundary. From $(1)$ and $(2)$ the following equality can be obtained"

$$\frac{1}{2}\partial_t \int_D \left(\frac{1}{2} u_i u_i + \frac{\rho^2}{\delta} \right) \, dV = - \int_D \sum_{i,j} (\partial_i u_j)^2 \, dV \hskip{1em}(3)$$

Alexandre then finishes the section by stating "The integrands on both sides are positive; hence the $u_i$ tend to the limit $u_i=0$, and $p$ to a limit independent of $t$. From $(1)$ and $(2)$ one sees that this limit is independent of $x_i$ and therefore is a constant."

This reminds me of Bernoulli's equation. I've seen derivation of Bernoulli's equation from Navier Stokes online but they used a lot of vector calculus and in this paper Alexandre doesn't make use of vector calc methods at all so that leads me to think that Alexandre found a simple method to go from the auxiliary NS equations to the equation that looks similar to Bernoulli's equation but not exactly like Bernoulli's equation.

My best attempt was to work backwards from the left hand side by moving the derivative inside the integral sign since it is in time and the integral is in space $$\frac{1}{2} \int_D \left(\frac{1}{2} \partial_t(u_i^2) + \partial_t (\frac{\rho^2}{\delta}) \right) \, dV \hskip{1em}(4)$$ Then taking the derivative $$\frac{1}{2} \int_D \left(\frac{2}{2} u_i \partial_t(u_i) + \frac{2}{\delta} \rho \, \partial_t\rho \right) \, dV \hskip{1em}(5)$$

Then I wrote the auxiliary equations in my preferred notation. I set $R=0$ because I think that's what Alexandre meant when he wrote "the case of zero velocities at the boundary", and because the Reynold's number is not in the integral equation we are trying to derive. The first NS equation becomes

$$\frac{\partial u_i}{\partial t} + \frac{\partial p}{\partial x} = \Delta u \hskip{1em}(6)$$

Then I tried to relate equations $(5)$ and $(6)$ using the continuity equation and multiplying by $u_i$ but it didn't work. I don't see how to go from $(6)$ to $(5)$.

My question has three parts

1. How to go from the Navier Stokes equations to equation (3) that Alexandre derived?
2. Have I interpreted Alexandre correctly by setting R=0 which led me to equations (5) and (6)?
3. What does Alexandre mean when he states, "From $(1)$ and $(2)$ one sees that this limit is independent of the x_i and therefore a constant"?


## 1 Answer

The derivation of equation (3) requires two independent assumptions: (1) The Reynolds number $R$ is set to $0$; (2) the fluid velocity at the boundary is taken to be zero (justified by the linearity argument in the reference). This along with the following derivation should answer your first and second questions.

To derive (3), first take the dot product of the velocity vector with both sides of equation (1). We can ignore the body force or absorb it if conservative into the pressure field. Setting $R = 0$ and using the Einstein summation convention we obtain

$$u_i \partial_t u_i = - u_i \partial_i p - u_i \Delta u_i,$$

This reduces to

$$\tag{a} \frac{1}{2} \partial_t (u_i u_i) = -\partial_i(pu_i) + \frac{\rho}{\delta}\partial_i u_i + u_i \Delta u_i .$$

Using equation (2) we find

$$\frac{\rho}{\delta}\partial_i u_i = \frac{\rho}{\delta}\partial_j u_j = -\frac{\rho}{\delta}\partial_t\rho = - \frac{1}{2}\partial_t\frac{\rho^2}{\delta}.$$

Expanding the term $u_i \Delta u_i$ we obtain

\begin{align}u_i \Delta u_i &= u_i \partial_j^2u_i \\ &=\partial_j (u_i \partial_j u_i)- (\partial_ju_i)(\partial_j u_i) \\ &= \partial_j (u_i \partial_j u_i)- \sum_{i,j}(\partial_ju_i)^2 \end{align},

where the summation is shown explicitly for one term (without the Einstein convention) to be consistent with the reference.

Substituting into (a) and rearranging we obtain

$$\tag{b} \frac{1}{2} \partial_t \left( u_i u_i + \frac{\rho^2}{\delta}\right) = -\partial_i(pu_i) + \partial_j (u_i \partial_j u_i)- \sum_{i,j}(\partial_ju_i)^2.$$

Integrating (b) over the domain $D$ we get

$$\tag{c} \int_D \frac{1}{2} \partial_t \left( u_i u_i + \frac{\rho^2}{\delta}\right) \, dV = - \int_D \partial_i(pu_i) \, dV + \int_D \partial_j (u_i \partial_j u_i) \, dV- \int_D \sum_{i,j}(\partial_ju_i)^2 \, dV.$$

The first and second terms on the right-hand side of (c) vanish as a consequence of the divergence theorem and zero velocity on the boundary. Note that

$$\int_D\partial_i(pu_i) \, dV = \int_{\partial D} pu_i n_i \, dS = 0, \\ \int_D \partial_j (u_i \partial_j u_i) \, dV = \int_{\partial D} u_i\partial_j u_i n_j \, dS = 0.$$

Assuming that the velocity and pressure fields are sufficiently smooth (the usual assumption for fluid flow), the time derivative can be brought outside of the integral on the left-hand side of (c) to obtain the final result

$$\tag{d} \frac{1}{2} \partial_t \int_D \left( u_i u_i + \frac{\rho^2}{\delta}\right) \, dV = - \int_D \sum_{i,j}(\partial_ju_i)^2 \, dV.$$

Equation (d) differs from equation (3) only by a factor of $1/2$ inside the integral, and I believe this is a typographical error in the original reference. Needless to say the presence or absence of this factor is irrelevant in the subsequent argument.

Regarding your third question, the integral on the left-hand side is positive, hence bounded below, and a decreasing function time. Therefore it converges to a non-negative constant in the limit as $t \to \infty$. The time-derivative and the integral on the right-hand side must therefore decay to zero as $t \to \infty$, whence, velocity and pressure gradients vanish at steady state.

• I see. But why do we only set those two terms to zero for the zero velocity at the boundary? Is it because those are the only terms we can relate to the boundary using the divergence theorem... We wouldn't want to set all the u_i's to zero or there would be no equation to solve. But how do we justify not setting all the u_i's to zero when the boundary velocity is zero, such as $\partial_t u_i = 0$ and $(\partial_i u_i)^2=0$? Feb 23, 2017 at 6:43
• @betafractal: You answered the first question yourself. I only applied the divergence theorem where the integrand takes the form of a divergence. The resulting surface integrals vanish because velocity components appear and are zero on the boundary. The velocity is in general not zero in the interior of $D$ so we can't just declare the remaining integrals to be zero at all times. Furthermore, even if velocity components are zero on the boundary that does not mean velocity gradients are zero there.
– RRL
Feb 23, 2017 at 6:55