Finite Element Method (Weak form)

In order to apply Finite Element Method to solve PDEs, we need to convert the PDEs into an equivalent “weak form”.

Example: Axially loaded Elastic Bar

Strong Form: \frac{d}{dx}\big(AE\frac{du}{dx}\big)+b=0

with boundary conditions:

at x=0, traction/force per unit area \bar{t}, is prescribed: \sigma(x=0) = \big(E\frac{du}{dx}\big)_{x=0}=\frac{p(0)}{A(0)}=-\bar{t}

at x=l, displacement \bar{u} is prescribed: u(x=l) = \bar{u}

 

Weak Form: w(x) is an arbitrary weight function, multiplied by the governing equation and boundary conditions and integrated over the domain.

Governing equation: \int^l_0 w\big[\frac{d}{dx}\big(AE\frac{du}{dx}\big)+b\big]=0

Boundary condition (traction force): \big(wA\big(E\frac{du}{dx}+\bar{t}\big)\big)_{x=0}=0

Boundary condition (end-point displacement): w(l) = 0

 

Reminder:

\frac{d}{dx}(wf) = w\frac{df}{dx}+f\frac{dw}{dx} \Rightarrow w\frac{df}{dx}=\frac{d}{dx}(wf)-f\frac{dw}{dx}

Therefore, when we integrate the equation over [0,l]:

\int^l_0 w\frac{df}{dx} dx = \int^l_0 \frac{d}{dx}(wf) dx - \int^l_0 f\frac{dw}{dx} dx

\Rightarrow \int^l_0 w\frac{df}{dx} dx = wf\Big|^l_0 - \int^l_0 f\frac{dw}{dx} dx

Substitute in f = AE\frac{du}{dx}:

\int^l_0 w\frac{d}{dx}(AE\frac{du}{dx}) dx = (wAE\frac{du}{dx})\Big|^l_0 - \int^l_0 (AE\frac{du}{dx})\frac{dw}{dx} dx

  1. From the governing equation, the first term becomes: -\int^l_0 wb dx
  2. From the relation \sigma = E\frac{du}{dx}, the middle term (first term on the right hand side) becomes: (wA\sigma)_{x=l}-(wA\sigma)_{x=0}

And the equation becomes:

(wA\sigma)_{x=l}-(wA\sigma)_{x=0} - \int^l_0 (AE\frac{du}{dx})\frac{dw}{dx} dx + \int^l_0 wb dx = 0

Consider the boundary conditions:

  1. Boundary condition (traction force): \big(wA\big(E\frac{du}{dx}=-\bar{t}\big)\big)_{x=0}
  2. Boundary condition (end-point displacement): w(l) = 0

Now, let’s say hello to the slimmed and toned equation:

\int^l_0 (AE\frac{du}{dx})\frac{dw}{dx} dx = \bar{t}\big)\big)_{x=0} + \int^l_0 wb dx

where we only need to recover u(x) such that it satisfies u(l) = \bar{u} .

 

This form has a weaker continuity requirements than the strong form, hence the name “weak form”.

 

Sources:

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