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1 and 2 Order Boundary condition at the same place

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Old   July 9, 2005, 03:52
Default 1 and 2 Order Boundary condition at the same place
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I want to ask the following thing

in very simple geometries it is thinkable, to set a first and a second order Boundary condition at the same place, and therefor let a boundary condition at a different place variable.

e.g. Heat flux and temperature at the bottom of a "one dimensional" pot are fixed and therefor neither the Temperature nor the heat flux at the Top is defined. (No heat flux over the other walls)

But where are the limitations of these possibility, and why?

In my opinion as far as i assume a two dimensional problem with fixed but localy different boundary conditions of first and second order at one side of the geometry, it will not help to set free a boundary condition on a other side of the geometry?

Am I right? And whats the reason?

Thanks for your help

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Old   July 10, 2005, 10:05
Default Re: 1 and 2 Order Boundary condition at the same p
Jonas Holdeman
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I don't think I completely understand your statements or questions, but perhaps this will be germane.

For incompressible flow computation by the finite element method, I use divergence-free Hermite elements, where the degrees-of-freedom are the stream function and its curl (the velocity components). On no-flow non-slip boundaries, I usually assign both the stream function and velocities as boundary conditions. This seems to work very well, as the velocities are fixed or assigned at the boundary nodes and the the (constant) stream function assignment says nothing strange (no net flow) is going on between the nodes.

For this type of problem it is both permissible and useful to use two boundary conditions of different order, if that is what you meant by first and second order B.C.
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Old   July 11, 2005, 05:39
Default Re: 1 and 2 Order Boundary condition at the same p
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From what I understand/think you mean then you are mistaken - you need to look at a book on partial differential equations to see what defines a well-posed problem (incorrect use of boundary conditions for a pde usually results in an ill-posed problem).

e.g. consider Hadamard's problem

u_xx + u_yy = 0

with u = 0 on y=0,1 and the two boundary conditions u=f(y), u_x = 0 on x=0. If we write f(x) as a Fourier sine series then the solution will contain terms of the form

sum (n=1,infinity) C_n.exp(n.pi.x)sin(n.pi.y)

which, although convergent for x=0, does not converge for any x>0 (no matter how small!)
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Old   July 11, 2005, 10:38
Default Re: 1 and 2 Order Boundary condition at the same p
Jonas Holdeman
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There is a difference between continuum problems and their discrete analog. My remarks here are in the context of the finite element method in 2D using Hermite elements. I will also be talking about four-node elements, but this is not an essential restriction. For problems like the themal cavity, the DOFs of the thermal element are the temperature and temperature gradients at each node. For incompressible fluids, the DOFs are the stream function and solenoidal velocity components at each node. The scalar steam function element is cubic complete (as is the thermal element). Taking the curl of the stream function element gives a vector element that is obviously divergence-free on the element, and perhaps less obvious, has sufficient continuity at element boundaries that the divergence vanishes at element interfaces as well.

Consider the example of developed or developing flow in a channel. I will apply no BC at the outlet, the so-called "do nothing" boundary condition. No-slip velocity and constant stream function BC are applied along the sides of the channel (no pressures are involved). Start with fully-developed flow at the inlet. If one were using linear Lagrange elements for the velocity components, one might interpolate the parabolic profile at the nodes. One is then stuck with a linear behavior between the nodes. Likewise, with the Hermite element, one would have a parabolic velocity profile at the nodes, but the velocity between the nodes is undetermined by these BCs. One can assign a consistent stream function at the inlet nodes (resulting in an exact parabolic profile), or let the inlet stream function be determined by the solution, or assign stream function values inconsistent with the parabolic profile. In any case one has a consistent foumulation for developing flow that is computationally stable.

In fact, one can apply the BC above at the sides of the channel, with no inlet or outlet BC, and the problem is well-posed and converges to the parabolic profile in one step with one Newton-Raphson iteration to deal with the non=linearity.

Likewise for the thermal cavity, one assigns the temperature T=const and dT/dy=0 on the sides to get a consistent formulation.

The computations described above were done in Matlab.

In summary, one has more flexibility in forming a well-posed problem in the discrete case, but if this flexibility is abused, you may not have the specification you expect.
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Old   July 11, 2005, 11:57
Default Re: 1 and 2 Order Boundary condition at the same p
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My comment was to the original poster and is my interpretation of his question.

As for your comment about the difference between the discrete and continuous cases - if you are solving a pde you should ensure that your numerical scheme is consistent with the underlying continuous problem and is convergent to the "true" solution. You shold also understand the implication of your "auxillary" boundary conditions in the limiting process.

As another example consider the linear equation A_tt = A_x with A = exp(-x^2), A_t=0 at t = 0 and A->0 as |x|-> infty. This linear problem is well-posed on any finite time interval but, if additional diffusive are not added, is numerically ill-posed posed. This is due to the fact that the equation permits modes whose growthrate increases with increasing wavenumber. In the continuous case the boundary conditions exclude these modes but in the discrete form they need to be artificially damped since they give rise to growing grid scale noise.
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