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 Incoming links: Explicit one-dimensional dual residual bound in L2

Past exam of the approximation theory and finite element methods course of the University of Oxford / 2007 / B / 7 / ii / Solution 2026-10-31

Put

e_{h} = u - u_{h}

and

w = z - I_{h} z

. Galerkin orthogonality gives

a (e_{h}, I_{h} z) = 0

, so the dual identity becomes

\int_{0}^{1} e_{h} ψ = a (e_{h}, z) = a (e_{h}, w) = \int_{0}^{1} f w - a (u_{h}, w) .

On

I_{i} = (x_{i - 1}, x_{i})

,

u_{h}^{'}

is constant and

w

is zero at both endpoints. Consequently

\int_{I_{i}} u_{h}^{'} w^{'} = u_{h}^{'} [w]_{x_{i - 1}}^{x_{i}} = 0

. Defining the element residual by

R (u_{h}) ∣_{I_{i}} = f - e^{x} u_{h}

gives

\int_{0}^{1} (u - u_{h}) ψ = \sum_{i = 1}^{N} \int_{I_{i}} R (u_{h}) (z - I_{h} z) .

No derivative-jump terms survive, because this interpolation error vanishes at every mesh node. This is the mechanism of the explicit one-dimensional dual residual bound in L2.

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