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# Weak form

Thu, 2009-06-04 01:25 - surajitdas

During the study of FEM, I found the term "weak form" of differential equation, which is the integral form of a differential equation along with the boundary conditions for a boundary value problem. Now my question is that why the term "**weak**" has been used.

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## Comments

## Weak or variational form of PDEs

Hello,

The term "weak form" attributes to the partial differential equations (PDE) that are to be solved. The solution of a boundary value probelm described by a particyular PDE can be achieved by setting the PDE to a weak form. In general the solution of a PDE can be performed from a "strong form" or from a "weak form". The strong form description of a PDE is the one we have already known, for example the Laplace equation. Since the finite element method is based on the discretization of the domain, the solution of the strong form of the PDEs describing the problem must be descretized. For this purpose a weak form is being developed. To develop a weak form of the equations, the integration of the product of the trial functions with the equations must be performed. The trial (or test) functions are assumed to be as smooth as posiible so as they vanish on the prescribed displacement boundary.

The term weak form is sometimes refered as variational form, because the solution of the PDEs is approximated with the use of trial functions. The Galerkin method is on of the methods that are besd on the variational form of the equations. The weak form is well defined for test and trial functions which are less smooth (Co) and indeed the test and trial functions used in FEM do not meet the smoothness requirements of C1 of the trial functions. Thus, the form is called weak because it contains only the first derivative of the test function.

## more permissive set of equations

In the book "Finite Element Method - Volume 1" from Zienkiewicz and Taylor you will find the awnser at pages 41-46. The "weak" is becouse that when you exchange your initial set of PDEs in the integral for the new ones plus the funcitions C(v) (where v is the functions that multiply the set of PDEs) your initial funcition U in Omega will need lower order of continuity, so you have more freedom to set U, you shall have a "weak" restriction.

Sorry about my english, I`m still improving it =)

## here is my 2 cents

DE is called "strong form" because the relationship MUST satisfy at every mathematical point in the domain.

a "weak form" means that the relationship (in integral form) is only satisfied in overall sense.

in another word, "it is only statisfied in an integral (sum) sense, it is not a requirement that every point in the domain MUST obey"

## I think it arises from the

I think it arises from the PDE theory of defining a "weak" solution.

For example, consider the equalibrium solution for a heterogeneous medium : div[C(x) grad u ]=0. Since C(x) is not continuous, a classic solution in C^2 doesn't exist. So we need to define what we mean by a "solution". A "weak" solution is defined by requiring u satisifes the integral equation int <grad v, C(x) grad u>=0 for all v in appropriate space. It turns out that a weak solution is the same as the strong/classic solution if a classic solution indeed exists, which justifies the name "weak solutin".

## Thank you all

Thank you all for your explanations.

What I understood from the above discussion that the strong form of the differential equation satisfied for all the infinitesimals of the domain, whereas the weak form is satisfied for finite domains (i.e the continuity is compromised). That is why the term "weak" is used.

## Still not quite it

The strong form of a differential equation is just that: the (partial) differential equation itself. Evaluating the PDE requires being able to get all the associated derivatives. It is satisfied pointwise at every point in a body, and is usually stated as D[u] = 0, where D is some partial differential operator. In this case, I am using u as the displacement. It may be more appropriate to look at, say, Cauchy stress (s) instead of displacement. Then the strong form might be something like D[s,u] = div[s] - ru,tt= 0.

The weak form is obtained by multiplying the PDE by an arbitrary function of (in most cases) the spatial variables, then integrating the result over the domain. One then requires that the result is zero for all choices of such functions. Once we start setting requirements on these functions, we "weaken" the form even more, but often provide a basis for expressing the approximate solution.

Matt Lewis

Los Alamos, New Mexico

## Correction

I have a correction to make to this post. The weak form is obtained by multiplying the PDE by an arbitrary

weighting function of (in most cases) the spatial variables, then integrating the

result over the domain. One then requires that the result is zero for

all choices of such functions. An integration by parts is performed, leading to differentiability requirements on the weighting function, but relaxing, or "weakening" the requirements on the field described by the PDE.

Matt Lewis

Los Alamos, New Mexico

## Thank you Mr. Matt Lewis

Thank you Mr. Matt Lewis for your comments. Here, my questions are:

1) why do we multiply the PDE by the weighting function.

2) How do we choose the weighting function.

Regards,

## How and why

Dear Surajit,

The "why" is to reduce differentiability requirements on our approximate solution.

The "how" is whatever works. If you know of a particular set of functions that work well in your geometry (say Bessel functions for axisymmetric problems), you can use these. We often use arbitrary combinations of nodal basis functions (compact support, C1, finite domains). If we use the same basis for our weighting functions as we use to represent our primary variable field(s), e.g. displacement, we are using a Galerkin method.

Take care,

Matt

Matt Lewis

Los Alamos, New Mexico

## Re: The Strong and Weak Forms

Hi all,

An interesting thread and very informative comments... But still...

From what I know (or am ignorant about), I think "seechew" hit the issue perfectly. In other words, the issue is the differential vs. the integral form---and not the one involving residuals as in Galerkin's or other methods of weighted residuals.

That we end up making a stronger demand on the continuity requirements to be fulfilled by the expected solution when the formulation is in the differential equation form, is a consequence and not the basic reason.

In a certain sense, even if we were not to reduce the degree of the integrand terms by performing integration by parts as an intermediate step, we would still end up being more forgiving with regards to the nature of the sought solution just by the grace of the fact that it is the integral (a sum, an average over a region/domain) that must satisfy the equality, not a bunch of differential terms individually and collectively.

To satisfy a differential form for a field phenomenon, each of the infinite points in the domain must possess the same value as the exact solution (which is being sought). To satisfy an integral form of equation for the same phenomenon you only require the final integral to come out in correspondence with a similar integral performed for the exact solution (assuming it exists).

Consider a simple case of a step function containing exactly one step within a contiguous finite interval. Suppose that this actually happens to be the exact solution which is being sought. Now, to address this matter in the differential equation terms, you will not only have to partition the domain into two halves (around the vertical jump), you would also have to ensure that each of the two solution equations come out in correspondence with the true solution. On the other hand, if you are dealing with an integral equation, apart from the step function, you could get away using a single line, even an arbitrary curve, of the same area under the curve, as does the step function. Only the area under the curve matters now---not what curve it is.

From what I think---and I could very easily be completely wrong---this difference itself tells you something about the stronger vs. the weaker nature of the constraints on the sought solution.

Unfortunately, however, the matter is not even addressed sufficiently clearly in any of the FEM related (or other) books I happened to have read. Not even in Springer's online encylopaedia: http://eom.springer.de. (I checked out "W" for "weak" and "weighted," and "R" for residual, and "G" for Galerkin, but without finding anything stated very directly about the matter under discussion---of course, granted my flimsy level of mathematical knowledge.)

Just my two cents. Hope some mathematician clarifies the matter. Thanks in advance.

## test function (weight functions)

In general how to choose the class of test functions for weal solution?

## Weak derivative

A topic that I see is not touched in the discussion is the space where the solution is looked for.

These wikipedia articles talk about it:

http://en.wikipedia.org/wiki/Weak_derivative

http://en.wikipedia.org/wiki/Weak_formulation

http://en.wikipedia.org/wiki/Weak_solution