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

\begin{align} q_{i, i}&=f \quad \text { in } \Omega &\quad \text { (Heat Equation) } \label{eqnHeat} \\ u &= g \quad \text { on } \Gamma_{g} \quad &\text { (Essential Boundaries) } \\ -q_{i} n_{i} &= h \quad \text { on } \Gamma_{h} \quad &\text { (Natural Boundaries) } \end{align}

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

$$\begin{align} \delta &= \{u\big|u \in H^1(\Omega), u(\Gamma) = g \}\\ \mathcal{V} &=\{w\big|w= H^1(\Omega), w(\Gamma) = 0\} \end{align}$$

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

\begin{align} 0 &= \int_{\Omega} \overbrace{w\left(q_{i, i}-f\right) }^\text{distribute} d \Omega = \overbrace{\int_{\Omega} wq_{i, i} d \Omega }^\text{weaken w/ IBP} - \int_{\Omega} w f d \Omega \\ &= - \overbrace{\int_{\Omega} w_{, i} q_{i} d \Omega + \int_{\Gamma_h} w \underbrace{q_{i} n_{i}}_\text{BC} d \Gamma }^\text{from IBP}- \int_{\Omega} w f d \Omega \label{eqnLambda}\\ &=-\int_{\Omega} w_{, i} q_{i} d \Omega-\int_{\Gamma_{h}} w h d \Gamma-\int_{\Omega} w f d \Omega \end{align}

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

\boxed{ -\int_{\Omega} w_{, i} q_{i} d \Omega = \int_{\Gamma_{h}} w h d \Gamma +\int_{\Omega} w f d \Omega } \label{eqn2BC}

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

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

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:

\left[q_{n}\right] =\left(q_{i}^{+}-q_{i}^{-}\right) n_{i}^{+} =q_{i}^{+} n_{i}^{+}+q_{i}^{-} n_{i}^{-} ;\quad n_i^+ = -n_i^- \label{eqnJump}

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.

$$-\int_{\Omega} w_{, i} q_{i} d \Omega = \int_{\Gamma_{h}} w h d \Gamma +\int_{\Omega} w f d \Omega \\ 0 = \int_{\Omega} w_{, i} q_{i} d \Omega + \int_{\Gamma_{h}} w h d \Gamma +\int_{\Omega} w f d \Omega \\ 0 = \overbrace{ \underbrace{\int_{\Omega} w_{, i} q_{i} d \Omega}_\text{IBP} + \int_{\Omega} w f d \Omega }^\text{collect same domain} + \int_{\Gamma_{h}} w h d \Gamma \\$$

where IBP was

$$\int_\Omega w\left(q_{i,i}\right)d\Omega+\int_\Omega w_{,i}q_i d\Omega = \int_\Gamma wh d\Gamma$$

so substitution into

$$0 = - \int_\Omega wq_{i,i}d\Omega + \int_\Omega wf d\Omega -\int_\Gamma whd\Gamma + \int_{\Gamma_h} whd\Gamma \\ 0 = \int_\Omega wq_{i,i}d\Omega - \int_\Omega wf d\Omega +\int_\Gamma whd\Gamma - \int_{\Gamma_h} whd\Gamma$$

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

0 = \sum_{e=1}^{n_{el}} \left\{ \int_{\Omega^e} \left( \overbrace{w q_{i,i} - w f}^{\text{factor the }w} \right)d\Omega +\int_\Gamma whd\Gamma - \int_{\Gamma_h} whd\Gamma \right\} \\ = \underbrace{\sum_{e=1}^{n_{el}} \int_{\Omega^e} w\left( q_{i,i} - f \right)d\Omega}_\text{keep this term} + \underbrace{\sum_{e=1}^{n_{el}} \int_\Gamma}_\text{union} whd\Gamma - \underbrace{\sum_{e=1}^{n_{el}}\int_{\Gamma_h}}_\text{fixed # BC} whd\Gamma \\

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

$$\boxed{ 0=\sum_{e=1}^{n_{e l}} \int_{\Omega^{e}} w\left(q_{i, i}-f\right) d \Omega-\int_{\Gamma_{h}} w\left(q_{n}+h\right) d \Gamma+\int_{\Gamma_{i n t}} w\left[q_{n}\right] d \Gamma }$$

Problem 3

Exercise 1 on page 71 of the text book

Let

$$\underset{n_{e n} \times 1}{\underbrace{{\bf d}^{e}}}=\left\{d_{a}^{e}\right\}=\left\{\begin{array}{c}{d_{1}^{e}} \\ {d_{2}^{e}} \\ {\vdots} \\ {d_{n_{e n}}^{e}}\end{array}\right\}$$

where

$$d_{a}^{e}=u^{h}\left(x_{a}^{e}\right)$$

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

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

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

$$\lambda u-q_{i} n_{i}=h \quad \text { on } \Gamma_{h}$$

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

$$- \int_{\Omega} w_{, i} q_{i} d \Omega + \int_{\Gamma_h} w \overbrace{q_{i} n_{i}}^{\text{new }h} d \Gamma - \int_{\Omega} w f d \Omega = -\int_{\Omega} w_{, i} q_{i} d \Omega -\int_{\Gamma_{h}} w \left(h - \lambda u\right) d \Gamma -\int_{\Omega} w f d \Omega \\ = -\int_{\Omega} w_{, i} q_{i} d \Omega -\int_{\Gamma_{h}} w h d \Gamma +\int_{\Gamma_{h}} w \lambda u d \Gamma -\int_{\Omega} w f d \Omega$$

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 $.