# FEM oscillations for polynomials of degree 1

I have the following eliptic 1-D problem $$-\mu u'' + \beta u' = 1$$ $$u(0) = u'(1) = 1$$ where $\mu = 10e^{-5}$ and $\beta = 1$. For this specific problem I am using the following space steps $h=[0.1, 0.01]$

What surprises me is that this problem doesn't present oscillations. I think it is related to the Péclet number, $\mathrm{Pe}$. When the problem had a source term (right-hand side function) equals to zero I know that the Péclet number should be less than 1 to avoid this oscillations.

Is this lack of oscillations related to the local Péclet number of my problem?

I tried to use finite differences to see if I can find why this does not present oscillations but I don't know how to proceed from here

$$(\textrm{Pe}-1)u_{i+1} + 2u_i - (\textrm{Pe}+1)u_{i-1} = \frac{h^2}{\mu}$$

Any ideas on how to proceed or if I'm actually in the right direction?

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The finite difference / volume case is covered for example in Wesseling, Principles of Computational Fluid Dynamics, Springer 2001. The main idea is to take a solution of the form $u_i=A+Bz^i$, fill in and see whether it allows solutions with $z<0$ so that $z^i$ alternates in sign (wiggles). It is important to include boundary conditions as well.

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Ok. I'll check it out. I don't have a strong background in math, so If you could provide some links that have examples on what you are saying it will be perfect. –  BRabbit27 Dec 19 '12 at 10:19
Actually, does that means I am in the right way to know why there aren't any oscillations in my numerical approximation? –  BRabbit27 Dec 19 '12 at 10:54
Have you already plugged in the formula for $$u_i$$ given here into your difference formulation? What is your result? Please add your calculation in your question above! –  vanCompute Dec 19 '12 at 11:02
I don't know how should I use that formula. Where does the $u_i = A + Bz^i$ comes from? What's the name of that? –  BRabbit27 Dec 19 '12 at 11:25
I strongly suggest you first look into the textbook I mentioned, it's all very well explained in there. –  chris Dec 19 '12 at 13:17