Update time-dependent expression in PDE

variational formulation
1

Hi everyone, I am new to fenics and I wonder if you can kindly help me on the problem.

I am trying to solve a time-dependent PDE where in the rhs there are some time-dependent coefficients. For simplicity, lets say the equation is:

\frac{\partial u}{\partial t} = f(t)\frac{\partial u}{\partial x}

where f(t) is a time dependent scalar function. I start this project by using a constant first, f(t) = const and everything is fine. Now I am considering the actual time-dependent f(t), I wonder how to do this. In my opinion, I am thinking to change the expression at each iteration in solving the PDE (assuming f(t) = sin(\omega t)):

# here just ignore the actual variational formulation of the PDE, I just wonder how to update with t
f_t = Expression('sin(omega t) * x[0]' , omega=1, t=0))
...
for n in range(1, max_step):
   t += dt
   u_D.t = t
    f_t.user_paremeters['t'] = t
    # a and L are lhs and rhs of the PDE
    solve(a == L, u)

I doubt whether this will update the f_t expression actually. I read from this link:

possibly I can use a self-defined UserExpression, but this is extremely slow on my machine. I wonder what is the simplest way to do this, since my expression is constant on the whole mesh but only dependent on time.

Thanks in advance for any help!

2

If the expression is constant on the whole mesh, you can use a Constant and update it using assign, e.g.

f_t = Constant(0)
omega = 1
...
for n in range(1, max_step):
    t += dt
    u_D.t = t
    f_t.assign(np.sin(omega*t))
    # a and L are lhs and rhs of the PDE
    solve(a == L, u)
1 Like
3

Hi @conpierce8 , thanks for your reply. However in my problem the whole expression is not independent over the mesh, but some coefficients in the expression only depends on time, e.g.

f_t = Expression('sin(omega * t) * x[0]', omega=1, t=0)

where the term sin(\omega t) only depends on time but the whole expression is different through the mesh. I wonder if there is a update method to do so, something like

for for n in range(1, max_step):
    t += dt
    u_D.t = t
    f_t.update(t) # does this work?
    # a and L are lhs and rhs of the PDE
    solve(a == L, u)

Thanks for your help again and look forward to hearing from you.

4

Hi @Zhuonan_Lin,

The Expression has instance variables (e.g. f_t.omega, f_t.t) which you can use to update the values of the parameters, i.e.:

from fenics import *

f_t = Expression("sin(omega*t) * x[0]", degree=1, omega=1, t=0)

t = 0
dt = 0.1
for i in range(10):
    print(f_t([0.5, 0.5]))
    
    t += dt
    f_t.t = t
1 Like