Tutorial : Finite Element Method Fundamentals Cont...

In the last post we had gotten as far as the weak formulation of the problem. We had also posed a few questions. Their answers will be given in this post. The discussion will be general though and applicable to all the problems amenable to finite element approximation.

Weighted Residual!!

What is meant by weighted residual? Remember from the previous post, the residual of the system were all the terms moved to the left hand side of the differential equation. The right hand side then had no terms left and it was obviously equal to zero. In other words, the residual was equal to zero. If we multiply this residual by some weighing function, the resultant is still zero (no surprises!). Now the idea of a weighted residual formulation is to multiply the residual by some weighing function and then to integrate this product over the entire domain of the problem. For a one dimensional problem, this is equivalent to integrating over the length of the domain. The integral is then put equal to zero. Of course, because the residual is zero.

$$ \int_{\Omega}r(x)w(x)dx=0$$

Here $\Omega$ is nothing but the physical domain of the problem. For a one dimensional problem and in particular for the problem of longitudinal deformation of a bar, this is the length of the bar.

A very delicate point in all this process is that since we are approximating a function, the residual is not strictly equal to zero; owing to approximation errors. To further elaborate, let us reconsider the expression of the residual function:
$$r(x)= -\frac{d}{dx}\left ( EA\frac{du}{dx} \right) - f(x)$$
Since we know that the original function $u(x)$ would be replaced by some finite element trial function $\hat{u}(x)$, which only approximates the original function as close as possible; therefore, there will be a difference in the two solutions and hence the residual is not strictly equal to zero. This brings us to another very important point.

"Since the residual is not strictly equal to zero, the whole idea of a finite element solution is to find so good an approximation that the residual nearly comes out to be zero and so that the weighted residual integral becomes valid if it is put equal to zero."

Test Function vs Weighing Function

In the previous post I wrote that the weighing function is deliberately omitted and that the test function $v(x)$ is used instead. This point is not a big deal. The weak form of the problem remains the same no matter which of the two are used. The only point worth mentioning here is that somewhere in the history we started using the following convention:

• $u$ for the trial function
• $v$ for the test function
• $w$ for the weight function

One way of obtaining finite element formulation is to derive the weak form using the weight function $w(x)$ and then substitute for it the test function $v(x)$. When this step is done, it is said that $v(x)$ is an arbitrary admissible function. One key property of this admissible function is that when the original function is supposed to be zero, the test function is also zero. This property comes in handy when applying boundary conditions. See the previous post for boundary condition application.

Why Call it Weak Form?

The next step is to apply integration by parts to the weighted residual formulation or the formulation with the test function, and render the weak form of the integral. Why call it 'weak'? It is called a weak form because in the original integral the unknown function is differentiated twice, and the weighing function or the test function is without any differential operator. However, once integration by parts is applied, both the trial and the test functions have the same number of differential operators.
$$\int_{\Omega} EA(x) \frac{du}{dx} \frac{dv}{dx} dx = \int_{\Omega}f(x)v(x)dx + \left. EA \left( \frac{du}{dx} \right) v \right |_{0}^{l}$$
This process of taking off one differential operator of the original trial function reduces the minimum smoothness criterion. In other words, for the above integral to be finite only the first derivative of the trial function $u(x)$ needs to be defined. If on the other hand we had tried solving the original integral then we would have needed the second derivative of $u(x)$ to be defined also. This is viewed as weakening of the requirement of some higher order derivative to be defined in the integral and hence, this form is called the weak form of the problem. To see the complete steps, refer to my previous post.

In the next post we will discuss the fundamental steps in realization of finite element approximation.