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Suppose we consider the flow of a fluid through a channel, considered to happen in a plane. We want to consider the region $D = [0,L]\times [-a,a]\subset \mathbb{R}^2$ representing the piece of the chanel we are looking at. Furthermore, considering constant density $\rho_0$ this gives incompressible flow with equations

$$\begin{cases}\nabla\cdot \mathbf{u} &= 0 & (1) \\ \dfrac{D\mathbf{u}}{Dt} &= -\dfrac{1}{\rho_0} \nabla p + \nu \nabla^2\mathbf{u} & (2) \end{cases}$$

Where $\nu = \mu/\rho_0$. We have then the no-slip boundary condition $\mathbf{u}(x,-a)=\mathbf{u}(x,a)=0$. Books usually give this example, but they impose much more conditions to solve it:

  • The pressure satisfies $p_1 = p(0, y) > p(L,y) = p_2$.
  • The velocity field is of the form $\mathbf{u}(x,y,t) = (u_1(x,y), 0)$

Using these two properties we can solve the equations and then after some manipulations we get

$$\begin{align}p(x,y) &= p_1 + \dfrac{p_2-p_1}{L} x \\ u_1(x,y) &= \dfrac{p_2-p_1}{2\mu L}(y^2-a^2)\end{align}$$

In that case we've found a solution, but for that we needed

  • To specify a boundary condition on the pressure.
  • To consider $u_2 = 0$ reducing the number of unknowns.
  • To consider $u_1$ independent of $t$.

My question is: would it be possible to determine the solution without this? I mean, using the general form $\mathbf{u}(x,y,t) = (u_1(x,y,t), u_2(x,y,t))$ would it be possible to obtain the solution?

The reason for this question is: I thought that the equation was capable of encoding all the infromation about the flow. If without that kind of imposed condition we can't predict the flow, it seems that the equations "lack information". Is that true? Aren't the Navier-Stokes equation together with continuity equation enough to predict the flow behavior?

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In your problem you have set up, you have 3 unknowns -- $u_1$, $u_2$, and $P$. You also have 3 equations, so all the information you have to solve it is there. To be more technically correct, you still have a $u_3$ but the problem is considered infinite in that direction so you assume there is no variation in that direction.

Let's look at the "extra" things imposed. You are looking for the steady-state solution here so all of your time derivatives go away. You have a channel with top and bottom walls, and flow is not allowed to go through the walls. At least that's what is being imposed here -- you could actually solve this same problem with an imposed mass-flow through the walls. At any rate, you know that $u_2 = 0$ at $y = -a$ and $y = a$. So you go to your handy equation for $u_2$:

$$ u_1 \frac{\partial u_2}{\partial x} + u_2 \frac{\partial u_2}{\partial y} = - \frac{1}{\rho_0}\frac{\partial P}{\partial y} + \nu \left(\frac{\partial^2 u_2}{\partial x^2} + \frac{\partial^2 u_2}{\partial y^2}\right)$$

This equation has to hold at both boundaries. The pressure gradient normal to a wall is $0$. Because these are no-slip walls, $u_1(x,-a) = u_1(x,a) = 0$ as well. Typically one will also impose boundary conditions at $x = 0$ (the upstream side) which will say that the flow enters parallel, ie. ($u_2 = 0$). So the result of the imposition that $u_2 = 0$ is determined completely by the boundary conditions! It's not an extra imposition on the problem -- it is a result of the problem setup. You could assume the flow is not parallel entering the channel, that's perfectly valid, but makes the analysis more complicated and would not give you a nice, analytical solution at the end of it.

The problem lies with your book (or your interpretation of the book). It likely didn't say "we enforce the no-slip condition, initial density and set $u_2 = 0$ everywhere". It more likely said something like "based on the boundary conditions, $u_2 = 0$" which is a very bad way to say "based on the boundary conditions, it is obvious that $u_2 = 0$" which is also a bad way of saying "Let's solve for $u_2$ based on the boundary conditions....... And we see $u_2 = 0$".

Now the reason they give some information about pressure. These are the upstream and downstream boundary conditions. You could impose a BC on $u_1$ at upstream and downstream, but frankly that's not physical. Pressure gradients determine velocities, not the other way around. So the physical boundary conditions that you can set are the pressures. In real life, you might have a 2 atmosphere pressure chamber upstream of a pipe open to the atmosphere. So you would know $P(0,y)$ and $P(L,y)$. And that's what is imposed.

It seems as if you're having a hard time wrapping your head around how differential equations work and what is needed (or not needed) to solve them. That's natural, and the Navier-Stokes equations are not really all that easy to dive into if you're unfamiliar with differential equations. Here's some things to help:

  • You must have the same number of equations as unknowns. Here you do (because the density is specified), however in compressible flows you have 5 equations but 6 unknowns. You need the equation of state to link two variables together, giving you the required 6 equations.
  • When it is viscous, the equations are a 2nd order PDE. This means you must have 2 boundary conditions on each variable on each boundary in order to solve the equations.
  • When it is inviscid, the equations are a 1st order PDE. This means you must have 1 BC on each variable on each boundary (which is why the no-slip BC no longer is applied and the velocity is not zero at walls)
  • The boundary conditions you apply can be 2 types. You can impose a specific value, say $P(0,y) = 101325$ or you can specify the gradients at the boundary $\partial P/\partial x = 10$. Same holds at walls, but typically for velocity we impose the value (no-slip, no flow through the surface). But temperature and pressure gradients at walls are commonly chosen BCs.

That's all the information that is required. And that's all that you have in this case as well, the result that there is no flow in the $y$ direction is a consequence of the equations and not an imposition on them as you seem to think.

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  • $\begingroup$ Thanks for your answer @tpg2114, it helped me to get a better understanding. I'm really having a hard time with PDE's, in fact I've never studied them, I've just studied ODE's and the method of separation of variables for PDE's (which worked well for quantum mechanics until now). Can you recommend a book about PDE's for Physicists? Something like Arfken's book which goes straight to the point. I tried his book, but he doesn't cover systems of PDE's. $\endgroup$ – user1620696 Oct 25 '14 at 23:11
  • $\begingroup$ @user1620696 Unfortunately I cannot, all of my study has been in fluid dynamics courses and the majority of it is in grad school. I've taken many numerical PDE classes, but that won't help all that much here. You can learn what you need from fluid dynamics books if you get good ones. The viscous flow book I used in grad school, which covers this problem and similar, is Viscous Fluid Flow and I found it pretty easy to follow. Just make sure you repeat all derivations yourself to see what really happens! $\endgroup$ – tpg2114 Oct 25 '14 at 23:14
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About the $t$ independence part of your question, the answer is yes, you can find a lot more solutions which do depend on the time. Once you get to the PDE for $u_1(t,y)$, you will find it is like the forced ODE's you studied, where you can add a solution of the homogeneous equation. Specifically here the PDE is $$\frac{\partial u_1}{\partial t} = \frac{p_2-p_1}{2\mu L}+\frac{\partial^2 u_1}{\partial y^2}$$ and the homogeneous equation is the heat equation $$\frac{\partial u_1}{\partial t} = \frac{\partial^2 u_1}{\partial y^2}$$ so you can add to your solution any heat solution which satisfies the BC, say any number of terms of the form $$ ce^{-\nu d^2 t}\cos(dy)$$ or $\sin(dy)$ where the constants $c$ are arbitrary and the $d$ is chosen to keep $u_1$ zero when $y=\pm a$.

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I am glad you asked this question because I have worked on this problem in my post grad years, as well as the 3D channel. There are two answers, the short one and the long one, I am going to give you the long one...

First of all there has to be a pressure gradient because it is equivalent to a body force exerted on the fluid and it is this what drives the flow. If there where no pressure gradient there are still solutions but they consist of vortices that decay exponentially in time and the flow eventually stops. Not an interesting situation.

What you have described is the "velocity profile" which is the steady-state of the system. Let me call it $\mathbf{U}$, ie

$$U_i=U(x_2) \delta_{i1}= \dfrac{F}{2\mu}(x_2^2-a^2) \delta_{i1}$$

where $F= (p_2-p_1)/L$ is the pressure gradient or body force. An easy calculation shows that

$$\mathbf{U}. \nabla \mathbf{U}=0$$

and

$$ -\dfrac{1}{\rho_0} \mathbf{F} + \frac{\mu}{\rho_0} \nabla^2\mathbf{U}=0$$

So it's a solution of the Navier Stokes and also $\nabla. \mathbf{U}=0$ which we need for incompressible flow.

So far so good, now we ask the question are there any other flows that satisfy incompressibility and N-S equations. Let me introduce a small deviation from the steady state, by substituting $\mathbf{U}+\mathbf{u}$ instead of $\mathbf{u}$ into the N-S equation (which is actually the momentum conservation equation):

$$\frac{\partial \mathbf{u}}{\partial t}+\mathbf{u}.\nabla \mathbf{U}+\mathbf{U}.\nabla \mathbf{u}+\mathbf{u}.\nabla \mathbf{u}=\nu \nabla^2 \mathbf{u}$$

Happily the force dropped out of the problem because it balanced out viscosity, $\nu\nabla^2\mathbf{U}$. But we have introduced two more convective terms:

$$\mathbf{u}.\nabla \mathbf{U}=u_2U' \delta_{i1}$$ $$\mathbf{U}.\nabla{u}=U \frac{\partial u_i}{ \partial x_1}$$

Both of these terms are linear with respect to $\mathbf{u}$ but we cannot get rid of the term $\mathbf{u}.\nabla \mathbf{u}$ which is nonlinear. The viscosity term is also linear in $\mathbf{u}$ so we can write the equation finally as

$$\frac{\partial \mathbf{u}}{\partial t}=-\mathbf{u}.\nabla \mathbf{u}+\mathbf{L} \mathbf{u}$$

where $\mathbf{L}$ is a linear operator acting on $\mathbf{u}$. If $\mathbf{u}$ is really tiny the quadratic term is one order of magnitude smaller than the linear terms, so we might as well solve

$$\frac{\partial \mathbf{u}}{\partial t}=\mathbf{L} \mathbf{u}$$

together with our usual incompressibility constraint

$$\nabla . \mathbf{u}=0$$

This is typically solved by the method of eigenvalues and eigenfunctions of the linear operator:

$$\mathbf{u}( \mathbf{x},t)=\sum c_n e^{\alpha_n t} \mathbf{\phi_n(\mathbf{x})}$$

The time dependence is the essential thing here. For small Reynolds numbers all the eigenvalues $\alpha_n$ have negative real part and so the flow soon enough returns to the steady state $\mathbf{U}$. But there is a threshold above which some eigenvalues have positive real part and so we have exponential growth! And soon enough the nonlinear term cannot be neglected any more. In this case we have chaotic turbulent flow. It's not like genuine 3D turbulence, but it's still a kind of turbulence. You cannot have that in the linear case.

Now, all this was the framework, I still haven't answered your question! And the answer applies to the majority of nonlinear systems which are chaotic: Suppose we Fourier expand the solution in terms of $y$:

$$\mathbf{u}(x,y,t)=\sum_{n=1}^N \mathbf{v}_n(x,t) \sin{( \frac{n \pi y}{a})}$$

respecting the boundary conditions at $y=\pm a$ and keeping a finite number of terms. Never mind about the $\cos$ for the moment, let's assume it's odd in $y$ and keeping only a finite number of modes $N$. In our equation

$$\frac{\partial \mathbf{u}}{\partial t}=\sum_{n=1}^N \frac{\partial \mathbf{v}_n}{\partial t} \sin{( \frac{n \pi y}{a})}$$

i.e. is also expressible as a sum of the same modes $n=1,2, \dots,N$. The same applies with the linear term $\mathbf{L} \mathbf{u}$. Now we come to the "juice", the nonlinear term $\mathbf{u}.\nabla \mathbf{u}$. That is going to produce modes greater than $N$ that cannot be balanced out since all the other terms in the equation involve modes up to $N$. For example if we had the first mode only say $f(y)=\sin(\frac{\pi y}{a})$, the quadratic term

$$f(y) f'(y)=\frac{\pi}{a} \sin(\frac{\pi y}{a}) \cos(\frac{\pi y}{a})=\frac{\pi}{2a}\sin(\frac{2 \pi y}{a}) $$

which involves mode 2. Similarly with the $\cos$. And in general any finite sum of modes will give you modes more than you had considered. This is the so called "energy cascade", the mechanism through which energy is transfered from larger scales to smaller scales. Big eddies break up and transfer their energy to smaller and smaller eddies until viscosity is the major factor, not the nonlinear terms, and then friction takes over and their energy becomes heat. You are doing work pumping the fluid down the channel, where is that work going, it becomes heat. The energy cascade is essential to energy balance. So there is an energy leak in any finite solution to the N-S equation. We introduce more energy than we can handle.

And this is another reason why the N-S equations are incomplete: you need to introduce the heat equation as well to get it all balanced properly, which is coupled with the momentum equation but that's another story... In theory you could consider the infinite sum, i.e. all modes to infinity but then how are you going to simulate an infinite system on the computer?

Hope this helps, I 've just summed up two year's hard work!

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