A FEniCS Tutorial - FEniCS Project

• if u 1 is to be computed by projecting I:
– define a 0 and L 0
– assemble matrix M from a 0 and vector b from L 0
– solve MU = b and store U in u 1
• else: (interpolation)
– let u 1 interpolate I
• define a and L
• assemble matrix A from a
• set some stopping time T
• t = ∆t
• while t ≤ T
– assemble vector b from L
– apply essential boundary conditions
– solve AU = b for U and store in u
– t ← t+∆t
– u 1 ← u (be ready for next step)
Before starting the coding, we shall construct a problem where it is easy to
determine if the calculations are correct. The simple backward time difference
is exact for linear functions, so we decide to have a linear variation in time.
Combining a second-degree polynomial in space with a linear term in time,
u = 1+x 2 +αy 2 +βt, (76)
yields a function whose computed values at the nodes may be exact, regardless
of the size of the elements and ∆t, as long as the mesh is uniformly partitioned.
By inserting (76) in the PDE problem (64), it follows that u 0 must be given as
(76) and that f(x,y,t) = β −2−2α and I(x,y) = 1+x 2 +αy 2 .
Anewprogrammingissueishowtodealwithfunctionsthatvaryinspaceand
time, such as the the boundary condition u 0 given by (76). A natural solution
is to apply an Expression object with time t as a parameter, in addition to the
parameters α and β (see Section 1.7 forExpression objects with parameters):
alpha = 3; beta = 1.2
u0 = Expression(’1 + x[0]*x[0] + alpha*x[1]*x[1] + beta*t’,
alpha=alpha, beta=beta, t=0)
The time parameter can later be updated by assigning values to u0.t.
Given a mesh and an associated function space V, we can specify the u 0
function as
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