Past exam of the numerical solution of differential equations course of the University of Oxford / 2010 / B / 5 / c

The displayed estimate requires a nonnegative barrier. As printed, a counterexample is obtained with two interior nodes, each having row

L_{h} v_{p} = v_{a} - v_{p}

to one boundary node. Set

e_{a} = 0

e_{p} = 1

, and

ϕ_{a} = - 1

ϕ_{p} = - 2

. Both interior rows have

∣ T_{p} ∣ = 1

and

L_{h} ϕ_{p} = 1

. Put one node in each of

J_{1}, J_{2}

and take

T = S = d_{1} = d_{2} = 1

. Then the proposed right-hand side is

- 1

while

E = 1

For the intended use, suppose

ϕ_{p} \geq 0

and the stencil has the maximum principle established in part (a). Set

C = max {T / d_{1}, S / d_{2}}

. On either subregion,

L_{h} (\pm e + Cϕ) \geq - ∣ T_{p} ∣ + C d_{i} \geq 0.

At boundary nodes these functions equal

C ϕ_{a}

. Hence

\pm e_{p} \leq C (max_{a} ϕ_{a} - ϕ_{p}) \leq C max_{a} ϕ_{a},

which proves the intended discrete elliptic barrier estimate

E \leq a \in J_{\partial Ω} max ϕ_{a} max {T / d_{1}, S / d_{2}} .

For a normalized stencil with

L_{h} 1 = 0

and a sign-changing barrier, shifting it by its minimum gives the valid general replacement

E \leq C (max_{a} ϕ_{a} - min_{p} ϕ_{p})

. The supplied barrier in part (d) is positive, so the printed estimate applies there without a sign problem.

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