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This question might seem silly but I'll try to make it clear. It's a question (I think) about partial differential equations systems in general, but since currently I'm studying fluid mechanics I'll ask on that context.

The equations for an incompressible flow are:

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

Where we have to find $u_1,u_2,u_3$ the components of $\mathbf{u}$, $\rho$ and $p$. The point is: these are the equations, regardless of the flow under study. So, how does this connect to the real situation I have?

For example, in Newtonian Mechanics the equation is $\mathbf{F} = m\mathbf{a}$, but here we know the connection with the problem at hand: we plug the forces there and we solve the equation.

Now, in fluid mechanics, if I consider pipe flow, or channel flow, or flow past a sphere, or inside some complicated region $D\subset \mathbb{R}^3$ the equations are just the same. Nothing changes, there's not anything connecting the equations to the problem at hand but we might expect the solutions be quite different on each case.

In that case, since there are $4$ equations and $4$ unknowns, it seems the solutions would always have to be the same. So I ask: what connects these equations to the real situation? Is it just the boundary conditions? I believe my problem is that I don't know yet how existence and uniqueness works for partial differential equations systems.

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  • $\begingroup$ I have a small remark with respect to your post. You name the equation incompressible, but still want to solve for $\rho$. Later you mention four unknowns (instead of 5). This is a contradiction. $\endgroup$ – Bernhard Oct 24 '14 at 14:57
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I didn't entirely understand your question, aside from the last bit "what makes a solution of these equations unique?"

And you are absolutely right, it is the boundary and initial conditions that entirely determine the solution (which may or may not be unique, that's a very open problem for the Navier-Stokes equations).

You say that the situation is clearer in Newtonian mechanics, you just plug in the forces and you're done. However, this is also true in fluid mechanics. The momentum equation of the Navier-Stokes is really just $F = ma$. And you really just plug in your forces -- pressure gradient force, body forces (like gravity), viscous forces. Then you plug in your other forces that arise at the boundaries of the domain -- normal forces from walls, tangential forces if you have viscosity, etc..

It's really the same situation! And the boundary conditions, along with initial conditions, do determine the solution. But existence and uniqueness of the Navier-Stokes equations is something that has yet to be proven, and there is a substantial reward if you can do it.

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Both boundary conditions and initial conditions matter equally when connecting the model to the real world.

Consider the flow around a cylinder that you mention. We know that the Reynolds number, $$ {\rm Re}=\frac{u L}{\nu}\tag{1} $$ can characterize laminar or turbulent flows, depending on the values in (1). Below are flows for different Reynolds numbers (values given in the image).

enter image description here

If we initialize the flow with different starting velocities, $u(x,t=0)=u_0$, (keeping the length scale $L$ and kinematic viscosity $\nu$ constant for each simulation) and maintain the steady flow with the boundary condition $u(x=0,t>0)=u_0$, then the two combined (IC + BC) generate different results for different IC+BC pairs.

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    $\begingroup$ Unfortunately, I disagree with this one... You say that it's due to initial conditions, and say the initial speed is the driver. But if you had all 0's for BC's, that initial speed would disappear and you lose all flow. The boundary conditions drive it entirely, the initial conditions really just determine how long it takes for the flow to become the "real" flow. For a simulation like this to work, you must impose the upstream velocity (or impose the upstream and downstream pressures to generate a gradient which drives the flow). $\endgroup$ – tpg2114 Oct 24 '14 at 14:50
  • $\begingroup$ @tpg2114 I agree with you, but I think this is more a difference in terminology, i.e., some people call the condition at x=0 initial condition as well (not me!) $\endgroup$ – Bernhard Oct 24 '14 at 15:00
  • $\begingroup$ @Bernhard I had never heard that (in the context of time-dependent equations) but then there wouldn't be a distinction made between a boundary condition and an initial condition (like the first sentence) -- they would all just be "initial" conditions. $\endgroup$ – tpg2114 Oct 24 '14 at 15:02
  • $\begingroup$ @tpg2114: I view $q(x,y,t=0)\,\forall x,y\in D$ (which includes $x,y=0$) as the initial condition. Boundary conditions are imposed at the boundaries of $D$ (at $y=0\,\forall x$ & vice versa). In this view, my statement should stand. $\endgroup$ – Kyle Kanos Oct 24 '14 at 15:10
  • $\begingroup$ That clears up the terminology but I still don't think that the initial conditions are the biggest driver. By your definition, the boundary conditions are a subset of the initial conditions -- those imposed only on the boundaries of $D$ and not the interior. I contend that if you had two simulations, the first with a set of initial conditions and the second with a set of identical boundary conditions but different initial conditions (ie. the values at $x=0$ is the same for both sets but the value in the interior of the domain is different), you will end up with the same solution eventually $\endgroup$ – tpg2114 Oct 24 '14 at 15:18
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I would like to add another point to the ones already mentioned by Kyle Kanos and tpg2114. When you deal with partial differential equations (PDE), you have to take into account more things than with ordinary differential equations. I will try to explain the some of the ideas behind them, but without entering into details.

For example in evolution equations, like the one you wrote, you are studying the evolution in time not of a point of a finite dimensional space, but of a function that can be seen as a point but of an infinite dimensional space. In those infinite dimensional spaces, there are several notions of norm and/or topology (distance); and that results in different type of spaces. Examples are the infinitely differentiable functions ($\mathscr{C}^\infty$), the Schwartz space of rapid decrease functions ($\mathscr{S}$), the $p$-integrable functions ($L^p$) ...

Each of these spaces corresponds to functions that has particular properties: the infinitely differentiable functions are very smooth; the $\mathscr{S}$ functions are smooth and decrease rapidly at infinity; the $L^p$ functions behave well w.r.t integration, and thus decay at infinity, but need not to be differentiable, or even continuous.

To investigate whether the problem you analize has a solution (in mathematical terms it is well-posed), you have to specify in which space you are investigating it: because (mathematically) you need to have a notion of proximity (i.e. convergence) for your objects, and of magnitude (i.e. norm); and because you also have physical requirements on those objects e.g. you may want your flow to be continuous (so it can not be a general $L^p$ function, that may be not-continuous).

It turns out that a problem may be well-posed on a particular space, but not on another. For example the Naiver-Stokes equation is well-posed on some spaces (that in general have non-continuous functions), but we don't know if it is well-posed on the space of smooth (infinitely differentiable) functions (and that is the millennium problem with the big money prize that was mentioned before).

Another distinction is between local and global well-posedness. As the word would suggest, local well-posedness mens that we have existence of a unique solution only for at most finite times. That is because we may stumble upon what is called a finite time blow-up, i.e. the norm of the solution diverges in a finite time. Global well-posedness on the contrary means that for any finite time we have a finite solution, and its norm could eventually diverge only at infinite times. Again there are equations that admit only local well-posedness on some spaces (there is a finite time blow-up), but global well-posedness on others.

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