u = Function(W)
solve(a == L, u, solver_parameters=...)

But now I take out the line

L = (dot(u,f)+p*g)*dx

and assume that somewhere I calculate instead:

L2 = TestFunction(V2)*g*dx

L1 and L2 taken together contain the same information as L, but I can’t write L=L1+L2, because L1 and L2 were not defined using the mixed function spaces (assume that I calculated L1 and L2 somewhere else in the code, and I only use the mixed function spaces in this subroutine).

So given L1 and L2, how can I build L, which can be used in the solve? What I currently do is this:

m_u = dot(TestFunction(V1),TrialFunction(V1))*dx

m_p = TestFunction(V2)*TrialFunction(V2)*dx

f_u = Function(V1)

f_p = Function(V2)

solve(m_u=L1,f_u)

solve(m_p=L2,f_p)

L = (dot(u,f_u)+p*f_g)*dx

which I believe does the right thing, but it involves mass solves. Is there a better way? Can I copy the data over directly?

Thanks a lot,

Eike