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ifmath.isclose(u.offset[dim] %1, 0.5): # cell center
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# make the linearly interpolated value equal to the boundary by setting
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# the padded values to the negative symmetric values
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# on the left side if u.offset is either 0.5 or 1.5, width = -1
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# then (u - bc_val)/u.offset = (u_padded - bc_val)/(u.offset + width)
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# here the implemenation assumes that the left pad and the right pad widths are the same when u.offset == 0.5 and only left needs to be padded when u.offset == 1.5
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