## You are here

# Problems with MLPG and collocation

I fight with my models... Maybe can somebody of us help me...

I model the three dimensional body, those two sizes length and width are much larger then the thickness. For solving PDE i use MLPG Method. After defining my hexahedral support domain with MLS and hexahedral test doman with Heaviside function, i make the nodal integration. For that i divide the hole problem in hexahedrals, in the centers of those the particles are placed. Then i compute the surface of the hexahedrals in test domain through gauss quadratur to assemble the stiffness matrix

Follows problems occure:

- the excursion depends on the amount of nodes that i use. Why?

- if i change the material parameters also the form of excursion changes. What could be wrong?

- The size of the support domain hast also large influence.

Do i make something wrong?

If i use the collocation method for the solving of PDE i have also the problem that by changing of support domain the form and amount of execution changes. Is there any possibility to get the both meshless methods stable?

Thank you very much in advance.

- Mike W. Long's blog
- Log in or register to post comments
- 5943 reads

## Comments

## MLPG

Mike

Changing the support domain will change the results just as changing a FE mesh will change the results there. I don't understand what you mean by "excursion"? Can you clarify?

Charles

## MLPG

Sorry, there were technical problems to post the comment. I hope that it goes now.

Thanks a lot for your reply.

Excursion stays for the displacement or change of the displacement of the body under external forces. This result i'll get from my model. May be it's not the right word for that.

I suppose that by changing of the support domain the size of displacement changes because of the smoothing of weight function. By changing of the form of displacement the model does not converge. Is there any reason why it could be unstable for every support size?

I noted that the amount of gauss integration points also defines the size of the displacement. Is there any explanation?

## Re: MLPG

Dear Mike,

The issue of the support of the basis function in meshfree methods is not new. Here at Imechanica you can find several discussions on the same (see links provided below). The problems in meshfree methods which use Gaussian integration on a background cell arise from two specific points:

1) Basis functions are rational functions (non-polynomial) and hence they are not accurately integrated

2) The region used to compute numerical integration (background cell) does not coincide with the intersection of supports of two basis functions.

Point 2) is the most critical since it is present in the computation of the stiffness matrix. The latter involves the product of two basis functions derivatives which is much more complex to integrate than the basis function itself. An excellent article that discusses this issue is Numerical integration of Galerkin weak form in meshfree methods.

Now, moving to your problem. In the light of the above two issues, you can expect that several Gauss points (typically much more than in FE) are needed to get convergence of the solution and minimize the error in the numerical integration. This is the reason why you obtain different results for the displacement solution.

The instability you obtain for different basis function support size is due to a combination of number of Gauss points used and number of nodes that are covered by the meshfree basis function. Possibly, the second is more important. Let us assume that you always use the same number of Gauss points. This will leave us only with the issue of the support size of the basis function. Let us also assume that the basis functions are constructed so that they reproduce an arbitrary linear displacement field. Note that in 2D you need at least 3 points to reproduce a linear field. So, if for one Guass point evaluation there are less than three basis functions contributing at that Gauss point, there is no possibility of reproducing a linear displacement field in 2D. Thus, if the support size of the meshfree basis function is too small so that this minimum requirement is not met, non-convergence might arise or your numerical results might be totally damaged.

You can get much more information in the following posts:

http://www.imechanica.org/node/490

http://www.imechanica.org/node/1085

http://www.imechanica.org/node/402

http://www.imechanica.org/node/502

Hope this helps.

Alejandro

## Stability

Dear Alejandro,

thank you very much for the information. I've read these topics more time and find them very usefull. I tried to consider all comments by constructing of my model. But i wonder how i can get it stable?

I use the MLPG where the shape function and test function are defined over different domains. The support domains of shape function for approximation of displacement overlap. As the test function the heaviside function is used, so that i get rid of the product of two basis function derivatives. The quadratur domains are the same as test domains. They don't overlap and don't include any nodes. There is also gap between test domains. Only for this setup of domain the model function. Is it right? Can they overlap and include orther nodes?

The quadratur domains are hexahedrons and for integration over every side of hexahedron i use about 30 points. After integration over every hexahedron the stiffness matrix is filled for every node. I think it must be enough for reproduction of linear field. The support domains are so large that at least the minimal amount of nodes is included. As you wrote it is very important for stability of model.

As you said the region used

to compute numerical integration (background cell) does not coincide

with the intersection of supports of two basis functions. What does it really mean for MLPG? The quadratur or in this case also the test domain should be large enough that the integration points contains minimal number of shape function in stiffness matrix. It force the matrix A of MLS to stay non-singular.

Ther are also cases that the deformation in my model changes in the opposite direction if i change the support size. The form of the deformation can remain. If i give the force on the lower edge of the body to deform it from bottom to top, it deforms in the opposite direction depending on the size of support or test domain. May be there is a mistake in the model?

Mike

## MLPG

Mike

With MLPG the weak form is satisfied only over the test regions, so if there are gaps there are areas in your domain where the weak form is not satisfied/enforced (whatever you wish to call it). Quite what this means in practice is variable as those regions may or may not be important. You say you are using a Heaviside function for the test function, which I think has been called MLPG5 by Aturi & Shen () and in the excellent survey by Fries and Matthies, is not something I have tried so I can't tell if that is a source of difficulties. How do you enforce essential boundary conditions?

Charles

## MLPG

Hello Charles,

You are right it is MLPG5 method. This method don't need the volume integration. Unfortunaly it does not matter how large are the test regions, the model is unstable for small and large regions.

I use colocation method for enforcing of essential boundary condition.

Mike

## Re: Stability

Dear Mike,

First I have to say that I wanted to say that you need at least three nodes to reproduce a linear field in 2D. I guess you are thinking in Guass points instead. Second, I am not familiar with MLPG but I have read something and still there are some similarities, for example, with EFG. So, most of the problems of Galerkin-based methods should remain in MLPG (which is not strictly Galerkin-based).

I think your problem might be when gaps are present. I could be completely wrong with what I am going to say (maybe other experienced user should clarify). If you have gaps between local domains used for the test functions, you will not cover all the domain of analysis. This way, I am not sure that equilibrium equations along with boundary conditions will be satisifed.

Alejandro.