Page 92 - Process Modelling and Simulation With Finite Element Methods

Partial Differential Equations and the Finite Element Method 79

boundary conditions. Why is it strong? Because the field variables are required
to be continuous and have continuous partial derivatives up through the order of
the equation. That is a strong requirement. The weak form places a weaker
restriction on the functions that could satisfy the constraints - discontinuities
must be integrable.
To see the equivalence between a PDE and its weak form, consider a
stationary PDE for a single dependent variable u in three spatial dimensions in a
domain Q, e.g. the general form:

v-r(u)= F(U) (2.17)

Let's suppose that v, called a test function, is any arbitrary function defined on
the domain S2 and restricted to a class of functions v E v . Multiplying (2.17)
by v and integrating over the domain results in
jvv-r(+ =jV~(u)du (2.18)
n R
where dx is the volume element. Upon applying the divergence theorem, we
achieve
j cads -jvV.r(+ = jV~(u)h (2.19)
an Q Q

When the PDE is constrained by Neumann boundary conditions, the boundary
term on (2.19) vanishes. This is one of the reasons that FEM have Neumann
natural boundary conditions. Recall in Chapter 1 we observed that finite
difference methods have natural Dirichlet conditions. This results in the
condition on the volume integral:
J"v~(u)-vv.r(U)]du =o (2.20)
n

This must hold for every v E v . Now for the magic: finite elements and basis
functions. Let's suppose that u is decomposed onto a series of basis functions:

i
For instance, if the q$ are sines and cosines with the fundamental and progressive
harmonics, then (2.21) is a Fourier series. Instead, in FEM, the basis functions
are chosen to be functions that only have support within a single element, i.e.
they are zero in every element but one.