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vincent April 13, 1999 08:26

Boundary conditions in a Poisson's equation?
In order to obtain the pressure field one can use div operator to Navier-Stokes equation so we can obtains a poisson equation for the pressure. In order to solve this equation we need boundary conditions on the pressure. My question is: Is the problem well-posed with Dirichlet's conditions on p?? Is the same problems still well-posed with Neuman's conditions on p?? Could some one help me to remind the criteria of convergence in this two cases. Thanks a lot. Vincent

andy April 13, 1999 10:45

Re: Boundary conditions in a Poisson's equation?
It depends on what you want to do with this pressure equation and what meaning you wish to attach to its solution.

(1) Consider it to have no physical meaning. Dirichlet conditions will lead to a unique answer and Neumann conditions to a set of answers all differing by a constant (fix a value somewhere and then you will have a unique solution).

(2) Consider it to be the pressure field for the specified velocity field. In order to determine this you need to know how the momentum equation used the pressure field. This is (usually) relatively straightforward for the internal field but more difficult on the boundaries and will almost certainly involve the boundary conditions for the velocity. You cannot consider the pressure conditions in isolation from the velocity boundary conditions or (usually) the numerical scheme.

(3) Consider it to be one equation of a set used to solve the Navier-Stokes equations. If you also solve for the component momentum equations (very likely) then it is redundant since it was derived directly from these equations. (Hint: the information in the continuity equation is missing).

Anil Shenoy April 13, 1999 16:36

Re: Boundary conditions in a Poisson's equation?
Hi Vincent,

This is not an answer to your question but may serve as a pointer.

Let's consider incompressible, steady flows to begin with. In this case the equations are elliptic which means pressure travels at infinite speed and the presssure in the interior depends on all the boundary conditions.

As you might have seen the poissons equation for pressure has the following form:

d2p/dx2 + d2p/dy2 + d2p/dz2 = momentum + viscous terms.

we can call all the terms on the RHS as source terms.

To get a better understanding of the above equation, consider a simpler equation. Lets consider a one-dimensional heat transfer equation with a constant source in the domain. The equation can be given as

d2T/dx2 = S

integrating the equation we have

T = (S/2)x2 + ( B )x + C , where varies from 0 to 1

Lets consider the different boundary conditions :

(1) Two drichlet boundary conditions T = T0 at x = 0 T = T1 at x = 1 this gives

T = (S/2)x(x-1) + (T1-T0)x + T0

so the boundary condition uniquely determines the temperature field. But we have no control on the temperature derivatives. The temperature derivatives must meet only the following criterion:

dT/dx|x=1 - dT/dx|x=0 = S Note: Temperature derivatives can be compared with the flow rates in a fluid flow problem

So drawing an analogy to pressure poisson equation we can expect that if we use only drichlet boundary condition we may not get correct flow rates. Furthermore if the source S is non-linear (as in the pressure poissons equation) and use of an iterative solution can lead to wrong results. A good guess of the flow becomes important.

(2) One drichlet and one Nuemann condition T=T0 at x =0 dT/dx = k at x =1

this gives

T = Sx(x/2-1) + (k )x + T0

in this case we can determine the gradient at x =0 accurately. The temperature at the x = 1 would depend on the source term S, gradient k and T0. For a non-linear problem this can lead to inaccuarate pressure results at the boundary with gradient specification.

(3) Two Nuemann boundary condition

dT/dx = k1 at x = 0 dT/dx = k2 at x = 1

as we have seen before this not possible unless

k2 = k1 + S

extending this to the pressure poissons equation you cannot arbitrarly specify the Nuemann boundary conditions, they must specify a global criterion (as given in the above equation) obtained by integrating the poissons equation over the domain. Furthermore, assuming k2 =k1 + S we have

T = (S/2)x2 + (k1) x + C

so we see that we cannot uniquely determine the temperature (i.e. we are yet to determine C)

extending it to pressure poissons equation we can say that we cannot get an unique solution for pressure, infinite solutions differing by a constant can be obtained.

Things get a little messy with compressibility and unsteady behaviour. But I am sure you can figure it out. Try to build a simple model and extend it..... It works most of the time...........

Good luck !!!!!!!!!


John C. Chien April 15, 1999 14:01

Re: Boundary conditions in a Poisson's equation?
(1). There are two cases where different types of pressure boundary conditions are used. For example,... (2). Flow over a flat plate case, where the pressure can be specified as a constant value. In this case, the flow field is obtained from the continuity equation, momentum equation without the pressure term contributions. You get the flow over the flat plate boundary layer solution. ( How to keep the pressure field everywhere equal to a constant value is unknown. But it can be used as an assumption. assuming that it can be done.) (3). The second case is the fully developed flow in a tube, where the axial pressure gradient is a constant value, and the radial pressure gradient normal to the wall is a zero constant value. In this case, the pressure gradients can be specified at the boundary. Physically, the fully developed flow actually exists. In this case, the flow field is obtained from the continuity equation, the momentum equations without the convection term contributions. (4). Other than these two simple cases, the pressure at two points are related to the integration of the momentum equations between these two points, and the pressure gradient at a point is related to the momentum equations. That is, the pressure gradient boundary condition is derived from the momentum equations. In other words, the derived higher-order equation requires lower-order terms as boundary conditions. Therefore, the derived higher order pressure equation still needs momentum equations as boundary conditions.

Gary Dantinne April 16, 1999 02:19

Re: Boundary conditions in a Poisson's equation?

in incompressible flows pressure is not a thermodynamic variable, it is something like a Lagrange multiplier to enforce the divergence-free condition on the velocity field. For the time-dependent incompressible N-S equations there are no (a priori) boundary conditions on the pressure. Only the hydrostatic pressure level is needed.

If you write a Poisson equation for the pressure the 'best' boundary conditions to apply are Neumann conditions. These conditions can be derived by taking the normal component of the momentum equations on the boundary. Dirichlet conditions can lead to some problems.

If you want to impose an external pressure gradient, playing the role of a driving force, you can do it by adding a 'body force' to the momentum equation.

You can find a lot of interesting development in Gresho and Sani 's article 'On pressure boundary conditions for the incompressible Navier-Stokes equations' Int. J. Num. Meth. in Fluids vol 7, 1111-1145 (1987)

Hope it helps


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