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The equivalence between the weak and strong formulation of PDEs for FEA


#1

What is the equivalence between the weak and strong forms of PDEs

During the process of deriving the weak formulation of the FEA, we get a form that looks like the formula given below.

\int_0^l \frac{dw}{dx}AE\frac{du}{dx} dx = \left(wAE\frac{du}{dx}\right) \Biggr|^l_0 - \int_0^l w \frac{d}{dx}\left(AE\frac{du}{dx}\right) dx \tag{1}

Substituting the formula above in the weak formulation

\int_0^l\frac{dw}{dx}AE\frac{du}{dx}dx=(wA\overline{t})_{x=0}+\int _0^lwbdx~~~~~~~~~~~~~~\forall{w} ~~with ~~w(l)=0 \tag{2}

and placing the integral terms on the left-hand side and the boundary terms on the right-hand side gives:

\int_0^l w \left[\frac{d}{dx}\left(AE\frac{du}{dx}\right) + b \right] dx + wA(\overline{t} + \sigma)_{x=0} = 0 \tag{3}

The quintessence of making the proof possible is the arbitrariness of w(x). First we let

w = \psi(x) \left[\frac{d}{dx}\left(AE\frac{du}{dx}\right) + b \right] = 0 \tag{4}

where \psi(x) is smooth \psi(x) > 0 on 0 < x < l and \psi(x) vanishes on the boundaries. An example of a function satisfying the requirements above is \psi(x) = x(l-x). Because of how \psi(x) is constructed, it follows that w(l) = 0, so that w=0 on the prescribed displacement boundary, the essential boundary is met.

Combining the formulas above gives us:

w = \psi(x) \left[\frac{d}{dx}\left(AE\frac{du}{dx}\right) + b \right]^2 dx = 0 \tag{5}

The boundary term vanishes because we have constructed the weight function so that w(0)=0. As the integrand is the product of a positive function and the square of a function, it must be positive at every point in the problem domain. The only way equality is met in this equation is if the integrand is zero at every point! It follows that:

\frac{d}{dx}\left(AE\frac{du}{dx}\right) +b , \qquad 0<x<l \tag{6}

which is exactly the strong form.


Please go to our related article to find out more about the weak and strong formulation in our SimWiki article about the Finite Element Analysis.