 Solution

ID: past-exam-of-the-numerical-solutions-of-partial-differential-equations-course-of-the-university-of-oxford/2023/1/c/solution

Use

u_{i} = u (x_{i})

for the sampled exact solution, and let

e_{i} = U_{i} - u_{i}

e_{0} = e_{N} = 0

φ

gives the exact discrete equation

- D_{x}^{+} D_{x}^{-} u_{i} + u_{i} + u_{i}^{3} = f_{i} - φ_{i} .

Subtracting it from the numerical equation therefore gives

- D_{x}^{+} D_{x}^{-} e_{i} + e_{i} + (U_{i}^{3} - u_{i}^{3}) = φ_{i} .

The sign on the right is positive because

φ_{i}

was defined as the difference approximation minus the exact second derivative. Multiply by

h e_{i}

, sum over the interior nodes and apply discrete summation by parts:

∥ D_{x}^{-} e] ∥_{h}^{2} + ∥ e ∥_{h}^{2} + h \sum_{i = 1}^{N - 1} (U_{i}^{3} - u_{i}^{3}) e_{i} = (φ, e)_{h} .

Again

(U_{i}^{3} - u_{i}^{3}) e_{i} = e_{i}^{2} (U_{i}^{2} + U_{i} u_{i} + u_{i}^{2}) \geq 0

(φ, e)_{h} \leq \frac{1}{2} ∥ φ ∥_{h}^{2} + \frac{1}{2} ∥ e ∥_{h}^{2}

. Hence

∥ D_{x}^{-} U - D_{x}^{-} u] ∥_{h}^{2} + \frac{1}{2} ∥ U - u ∥_{h}^{2} \leq \frac{1}{2} ∥ φ ∥_{h}^{2} .

No fourth derivative is needed for this step: the assumed twice continuous differentiability suffices to define

φ

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