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Fix for Cofunction self-assignment via interpolation #3939

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7 changes: 6 additions & 1 deletion firedrake/interpolation.py
Original file line number Diff line number Diff line change
Expand Up @@ -859,7 +859,12 @@ def _interpolate(self, *function, output=None, transpose=False, **kwargs):
V = self.V
result = output or firedrake.Function(V)
with function.dat.vec_ro as x, result.dat.vec_wo as out:
mul(x, out)
if x is not out:
mul(x, out)
else:
out_ = out.duplicate()
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Shouldn't a no-op be the right thing?

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I don't think so, e.g.

u = fd.Function(space, name="u")
v = fd.Function(space, name="v")
w = fd.Cofunction(space.dual(), name="w")
...
fd.assemble(fd.action(
    fd.adjoint(fd.derivative(interpolate(v * u, space), u)),
    w), tensor=w)

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I see, it only is a no-op in the MFE from the issue: assemble(L(w), tensor=w) with L being the identity.

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Maybe it'd be good to add a more complicated test then

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Done. I think there are also be issues with complex here (transpose should be Hermitian for the adjoint actions?).

mul(x, out_)
out_.copy(result=out)
return result

else:
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7 changes: 7 additions & 0 deletions tests/firedrake/regression/test_interp_dual.py
Original file line number Diff line number Diff line change
Expand Up @@ -34,6 +34,13 @@ def f1(mesh, V1):
return Function(V1).interpolate(expr)


def test_interp_self(V1):
a = assemble(TestFunction(V1) * dx)
b = assemble(TestFunction(V1) * dx)
a.interpolate(a)
assert (a.dat.data_ro == b.dat.data_ro).all()
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Not sure if this should be a np.allclose test

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I think you're right as written. No flops should be performed so that should be an equality.

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I don't think it's actually no flops, it's interpolation which happens to be the identify except for roundoff.

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Changed to np.allclose, I'm pretty sure now that's correct



def test_assemble_interp_operator(V2, f1):
# Check type
If1 = Interpolate(f1, V2)
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