Strong and weak form for an axial bar

(a) Strong form

Find $u(x)$ such that

$$\frac{d}{dx}\left(AE\frac{du}{dx}\right) + f = 0,\quad 0 < x < L$$

with $u(0) = 0$ (essential BC) and $AE\dfrac{du}{dx}\Big|_{x=L} = P$ (natural BC).

It requires $u$ to be twice differentiable, and the equation must hold at every point.

(b) Weak form (Galerkin)

Multiply the equation by an arbitrary weight (test) function $w(x)$ with $w(0) = 0$, and integrate over the length:

$$\int_0^L w\left[\frac{d}{dx}\left(AE\frac{du}{dx}\right) + f\right]dx = 0$$

Integrate the first term by parts:

$$\int_0^L w\,\frac{d}{dx}\left(AE\frac{du}{dx}\right)dx = \left[w\,AE\frac{du}{dx}\right]_0^L - \int_0^L \frac{dw}{dx}AE\frac{du}{dx}\,dx$$

At $x = 0$, $w = 0$. At $x = L$, $AE\,u' = P$. So the boundary term is $w(L)P$. The weak form is:

$$\int_0^L \frac{dw}{dx}\,AE\,\frac{du}{dx}\,dx = \int_0^L w f\,dx + w(L)\,P$$

for all admissible $w$.

Note that the natural BC enters through the boundary term, while the essential BC is imposed on $u$ and $w$.

(c) Why the weak form is preferred

• Only first derivatives appear, so the trial and test functions need only be $C^0$ continuous; simple linear shape functions can be used.
• Natural boundary conditions are included automatically.
• It leads to a symmetric stiffness matrix, $K_{ij} = \int AE\,\phi_i'\phi_j'\,dx$.
• It is the physical statement of the principle of virtual work, and it still works when $A$, $E$ or the load are discontinuous.