Homework 4

Chris Nkinthorn

Finite Element Methods

Problem 1

Prompt: In section 2.3 of the text (pg. 60-64) the strong form and weak form of the equations for linear heat conduction are given. Using a Galerkin based method of weighted residuals, show the construction of the given weak form starting from the strong form, stated as:

Given $f : \Omega \rightarrow \mathbb{R}, q : \Gamma{o} \rightarrow \mathbb{R} \text { and } h : \Gamma{h} \rightarrow \mathbb{R}, \text { find } u : \overline{\Omega} \rightarrow \mathbb{R}$ such that

With the functions from candidate spaces: $u \in \delta$ and $w \in \mathcal{V}$.

δ={u∣u∈H1(Ω),u(Γ)=g}V={w∣w=H1(Ω),w(Γ)=0}\begin{align} \delta &= \{u\big|u \in H^1(\Omega), u(\Gamma) = g \}\\ \mathcal{V} &=\{w\big|w= H^1(\Omega), w(\Gamma) = 0\} \end{align}

Beginning with the Heat Equation $[\ref{eqnHeat}]$, we cannot solve in the strong form presented. Instead, we minimize a weighting function $w$ in the domain $\Omega$, where the closure of $\Omega$ is $\Gamma$. Distributing the weight function to produce two integral terms. Solving the first term using integration by parts (IBP) produces

Algebraic rearrangement with knowns on the right and the unknown on the left:

Problem 2

$\textcolor{red}{\texttt{To Be Graded }}$ Exercise 1 on page 68 of the text book. This exercise is a multidimensional analog of the the one contained in Sec. 1.8. Let

Γint=(⋃e=1nelΓe)−Γ (interior element boundaries) \Gamma_{\mathrm{int}} =\left( \bigcup_{e = 1}^{n_{el}}\Gamma^e\right) -\Gamma \quad \quad \text { (interior element boundaries) }

One side of $\Gamma{int}$ is arbitrarily designated to be the “+ side” and the other is the “- side”. Let $n^+$ and $n^-$ be unit normals to $\Gamma{int}$, with the relationship:

Consider the weak formulation$[\ref{eqn2BC}]$ and assume that $w$ and $u$ are smooth on the element interiors but may experience discontinuities in gradient across element boundaries. Restating $[\ref{eqn2BC}]$, the equation is made homogenous by moving the left hand side (LHS) term to the right hand side (RHS) of the equation.

−∫Ωw,iqidΩ=∫ΓhwhdΓ+∫ΩwfdΩ0=∫Ωw,iqidΩ+∫ΓhwhdΓ+∫ΩwfdΩ0=∫Ωw,iqidΩ⏟IBP+∫ΩwfdΩ⏞collect same domain+∫ΓhwhdΓ-\int_{\Omega} w_{, i} q_{i} d \Omega = \int_{\Gamma_{h}} w h d \Gamma +\int_{\Omega} w f d \Omega \\ 0 = \int_{\Omega} w_{, i} q_{i} d \Omega + \int_{\Gamma_{h}} w h d \Gamma +\int_{\Omega} w f d \Omega \\ 0 = \overbrace{ \underbrace{\int_{\Omega} w_{, i} q_{i} d \Omega}_\text{IBP} + \int_{\Omega} w f d \Omega }^\text{collect same domain} + \int_{\Gamma_{h}} w h d \Gamma \\

where IBP was

∫Ωw(qi,i)dΩ+∫Ωw,iqidΩ=∫ΓwhdΓ\int_\Omega w\left(q_{i,i}\right)d\Omega+\int_\Omega w_{,i}q_i d\Omega = \int_\Gamma wh d\Gamma

so substitution into

0=−∫Ωwqi,idΩ+∫ΩwfdΩ−∫ΓwhdΓ+∫ΓhwhdΓ0=∫Ωwqi,idΩ−∫ΩwfdΩ+∫ΓwhdΓ−∫ΓhwhdΓ0 = - \int_\Omega wq_{i,i}d\Omega + \int_\Omega wf d\Omega -\int_\Gamma whd\Gamma + \int_{\Gamma_h} whd\Gamma \\ 0 = \int_\Omega wq_{i,i}d\Omega - \int_\Omega wf d\Omega +\int_\Gamma whd\Gamma - \int_{\Gamma_h} whd\Gamma

This statement is local, so on the global mesh is the sum of all elements from $1$ to $n_{el}$:

The normal flux relationship of Eqn $[\ref{eqnJump}]$ is applied into the middle term. Decomposing the $\Gamma = \cancelto{0}{\Gamma_g} + \Gamma_h$, produces

0=∑e=1nel∫Ωew(qi,i−f)dΩ−∫Γhw(qn+h)dΓ+∫Γintw[qn]dΓ\boxed{ 0=\sum_{e=1}^{n_{e l}} \int_{\Omega^{e}} w\left(q_{i, i}-f\right) d \Omega-\int_{\Gamma_{h}} w\left(q_{n}+h\right) d \Gamma+\int_{\Gamma_{i n t}} w\left[q_{n}\right] d \Gamma }

Problem 3

Exercise 1 on page 71 of the text book

Let

de⏟nen×1={dae}={d1ed2e⋮dnene}\underset{n_{e n} \times 1}{\underbrace{{\bf d}^{e}}}=\left\{d_{a}^{e}\right\}=\left\{\begin{array}{c}{d_{1}^{e}} \\ {d_{2}^{e}} \\ {\vdots} \\ {d_{n_{e n}}^{e}}\end{array}\right\}

where

dae=uh(xae)d_{a}^{e}=u^{h}\left(x_{a}^{e}\right)

${\bf d}^e$ is the element temperature vector. Show that the heat flux vector at point $x \in \Omega^e$ can be calculated from the formulation

q(x)=−D(x)B(x)de=−D(x)∑a=1nenBadaeq(x)=-D(x) B(x) d^{e}=-D(x) \sum_{a=1}^{n_{en}} B_{a} d_{a}^{e}

Not sure how this expression was constructed, to move the $B$ matrix as a summation of element nodes. It looks like collocation because of evaluation at specific locations but I’m not sure how to prove it.

Problem 4

Exercise 2 on page 71 of the textbook, Consider the strong statement where the replacement on $\Gamma_h$ is

λu−qini=h on Γh\lambda u-q_{i} n_{i}=h \quad \text { on } \Gamma_{h}

where $\lambda\geq0$ is a function of $x \in\Gamma_h$. To generalize, continue from $[\ref{eqnLambda}]$, where the additional term creates:

−∫Ωw,iqidΩ+∫Γhwqini⏞new hdΓ−∫ΩwfdΩ=−∫Ωw,iqidΩ−∫Γhw(h−λu)dΓ−∫ΩwfdΩ=−∫Ωw,iqidΩ−∫ΓhwhdΓ+∫ΓhwλudΓ−∫ΩwfdΩ- \int_{\Omega} w_{, i} q_{i} d \Omega + \int_{\Gamma_h} w \overbrace{q_{i} n_{i}}^{\text{new }h} d \Gamma - \int_{\Omega} w f d \Omega = -\int_{\Omega} w_{, i} q_{i} d \Omega -\int_{\Gamma_{h}} w \left(h - \lambda u\right) d \Gamma -\int_{\Omega} w f d \Omega \\ = -\int_{\Omega} w_{, i} q_{i} d \Omega -\int_{\Gamma_{h}} w h d \Gamma +\int_{\Gamma_{h}} w \lambda u d \Gamma -\int_{\Omega} w f d \Omega

Assuming that the original weak form expression was also positive definite, then the additional contribution to $k_{ab}^e$ is based purely on $\lambda$. To prove that $\bf K$ is positive definite, then we would need to show that $\bf c^T\cdot K \cdot c \geq 0$ and $0$ only when $\bf c$ is the 0 vector, $\bf 0 $.

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