# 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”.

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”.

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