ETH_NumericalMethodsForPDE

Info

HPK D 3 (13:45, Thu) https://polybox.ethz.ch/index.php/s/dM5NqE3wqeWYsNp
CAB G 61 (14:00, Thu) https://polybox.ethz.ch/index.php/s/Xen2eXQaoQd6Css

Übungen

#timestamp 2026-02-26 (CAB G 61)

\begin{aligned} min_{u} J (u); J (u) & = \frac{1}{2} a (u, u) - l (u) + C & (QMP) \\ \leftrightarrow a (\underset{weak solution}{\underset{⏟}{u}}, v) & = l (v) & (LVP) & (weak form) \end{aligned}

as opposed to

\begin{aligned} - Δ \underset{s t r o n g s o l u t i o n}{\underset{⏟}{u}} & = f & (strong form) \end{aligned}

#todo learn Green's formula

Green's formula

$\begin{aligned} \int_{Ω} j \cdot grad (v) d x & = - \int_{Ω} div (j) v d x + \int_{\partial Ω} (j \cdot n) v d S \\ \int_{Ω} j \cdot \nabla v d x & = - \int_{Ω} (\nabla \cdot j) v d x + \int_{\partial Ω} (j \cdot n) v d S \end{aligned}$

fundamental lemma of variational calculus

$\int g \cdot v = 0 \forall v \in V ⟹ g \equiv 0$

strong->weak form:
$\cdot v$ -> $\int_{Ω} \cdot$ -> apply Green's formula -> bring in form $a (u, v) = l (v)$

(if the b.c. are not Dirichlet, the boyndary term drops:)

\begin{aligned} a (u, v) + k (u, v) & = l (v) \forall v \in H^{1} \\ a (u, v) & = l (v) \forall v \in H_{0}^{1} \end{aligned}

weak->strong form:
apply Green's formula backwards -> bring everything onto one side -> apply fundamental lemma of variational calculus

\begin{aligned} ∥ u ∥_{L^{2} (Ω)} & = ∥ u ∥_{0} = {(\int_{Ω} (u \cdot u)^{2} d x)}^{1 / 2} \\ | u |_{H^{1} (Ω)} & = {(\int_{Ω} grad (u)^{2} d x)}^{1 / 2} \\ ∥ u ∥_{H^{1} (Ω)} & = ∥ u ∥_{L^{2} (Ω)} + | u |_{H^{1} (Ω)} \end{aligned}

$L^{2}$ norm
$H^{1}$ -seminorm, how much function changes (?)
$H^{1}$ norm

some shortcuts:

\begin{aligned} ∥ u ∥_{L^{2} (Ω)} & \leq C \cdot diam (Ω) \cdot ∥ grad (u) ∥_{L^{2} (Ω)} & C > 0, \forall u \in H_{*}^{1} (Ω) & 2nd Poincaré (1.8.0.20) \\ ∥ u ∥_{L^{2} (Ω)} & \leq diam (Ω) \cdot ∥ grad (u) ∥_{L^{2} (Ω)} & \forall u \in H_{0}^{1} (Ω) \\ ∥ u ∥_{L^{2} (\partial Ω)}^{2} & \leq C \cdot ∥ u ∥_{L^{2} (Ω)} \cdot ∥ u ∥_{H^{1} (Ω)} & multipl. trace ineq. \\ | l (u) | & \leq C ∥ u ∥_{L^{2} (Ω)} & \forall u \in L^{2} (Ω) & bounded lin. operator \end{aligned}

for mutipl. trace. ineq, note the ^2 and the boundary $\partial Ω$
the bounded lin. operator is important for existence or something?

boundary conditions

\begin{aligned} - Δ u & = f on Ω \\ u & = g on Ω & dirichlet \\ j (u) \cdot \vec{n} & = h on Ω & neumann \\ j (u) \cdot \vec{n} & = Ψ (u) on \partial Ω & radiation \end{aligned}

to look up until next week:

Gauss
general product rule

#timestamp 2026-03-05

Global assembly
Pasted image 20260305150408.png
red edge:

for the lower triagle, value goes at index 1,3 (edge between point 1, 3)
for the upper tirangle, value goes at index 1,2 (edge between point 1, 2)

combined:

A_{i j} = A_{K 1}^{(1, 3)} + A_{K 2}^{(1, 2)}

#timestamp 2026-05-21

If two elements linearly dependent of integral form, $= 0$ :

\begin{aligned} ω^{2} (v, v, k) & \underset{arguments swapped}{\underset{⏟}{=}} - ω^{2} (v, v, k) \\ ⟹ = 0 \end{aligned}

usage:

\begin{aligned} ω_{p} \land μ_{q} (v_{1}, \dots, v_{p + q}) \\ p + q < n, \underset{basis functions vectorspace}{\underset{⏟}{(β_{1}, \dots, β_{n})}} \end{aligned}

If $p + q > n$ , of the two integral forms $ω_{p}, μ_{q}$ then some elements are (since n basis vectors) linearly dependent -> $= 0$