Test 1 Study Guide
use maniatty nomenclature for vector and tensor spaces (lowercase bold for vector, capital bold for 2+)
$\textcolor{red}{\texttt{this shit is fucked up}}$
Notation
Comma notation
Index Notation
Bracket Notation
Inner Product Spaces
energy inner product and standard inner product which we use to define it
a ( w , u ) = ā« 0 1 w , x u , x d x = ( w , d ) + w ( 0 ) h ( w , l ) = ā« 0 1 w ā d x \begin{aligned} a(w, u) &=\int_{0}^{1} w_{, x} u_{, x} d x = (w, d)+w(0) h \\(w, l) &=\int_{0}^{1} w \ell d x \\
\end{aligned} a ( w , u ) ( w , l ) ā = ā« 0 1 ā w , x ā u , x ā d x = ( w , d ) + w ( 0 ) h = ā« 0 1 ā w ā d x ā symmetric and bilinear in each slot
symmetric $\begin{aligned} a(u, v) &=a(v, u) \(u, v) &=(v, u) \end{aligned}$
bilinear $\left(c{1} u+c {2} v, w\right)=c{1}(u, w)+c {2}(v, w)$
[TOC]
how the fuck am I going to study for this test?
practice test
Hey, do you have the old exams for Shephardās FEM? Iād like to take them as practice tests - to rui
Chapter 1
Finite elements is a solution to a boundary value problem usually a PDE
i. variational weak form
ii. approximate solution to new weakened pde using finite element functions
Start with an ODE $u_{, x x}+f=0$ mapped onto the unit interval $f : [0,1] \rightarrow \mathbb{R}$ where [0,1] is the domain
solution to strong form $u(x)=q+(1-x) h+\int{x}^{1}\left{\int {0}^{y} \ell(z) d z\right} d y$
dummy variables not really represent directional stuff
we can weaken this with method of weighted residuals
let u be a trial function we need n derivatives so the nth derivative has a nice quality
square integrable :$\int{0}^{1}\left(u {, x}\right)^{2} d x<\infty$ or $\mathcal{H}^1$
all those that work make the collection $\delta = \left{u | u \in H^1, u(1)=g\right}$ where g is a is a nonzero essential boundary condition
the other is the weighing function space $\mathcal{V} = \left{w | w \in H^{1}, w(1)=0\right}$
this is called virtual work/displacement/principals where the $w,_x$ is the virtual part and the generalized part
Galerkin
$u^{h}=v^{h}+a^{h}$
Descretize function space
Ī“ h ā Ī“ Ā (i.e.,Ā ifĀ u h ā S h , Ā thenĀ u h ā E ) v Ī¼ ā U ( Ā i.e.,Ā ifĀ w h ā U h , Ā thenĀ w h ā U ) \begin{array}{ll}{\delta^{h} \subset \delta} & { \text { (i.e., if }\left.u^{h} \in \mathcal{S}^{h}, \text { then } u^{h} \in \mathcal{E}\right)} \\ {v^{\mu} \subset \mathcal{U}} & {\left(\text { i.e., if } w^{h} \in \mathcal{U}^{h}, \text { then } w^{h} \in \mathcal{U}\right)}\end{array} Ī“ h ā Ī“ v Ī¼ ā U ā Ā (i.e.,Ā ifĀ u h ā S h , Ā thenĀ u h ā E ) ( Ā i.e.,Ā ifĀ w h ā U h , Ā thenĀ w h ā U ) ā write variational weak form
$a\left(w^{h}, u^{h}\right)=\left(w^{h}, \ell\right)+w^{h}(0) h$
$a\left(w^{h}, v^{h}\right)=(w^{h},f)+w^{h}(0) k-a\left(w^{h}, q^{h}\right)$
Bubnov Galerkin
pose the weak form in the finite space
Petrov galerkin is when the weigh function is not of the homogenous variety?
Intro-to-1D-second-order-pde.pdf
1D ODE $\left(\kappa u{, x}\right) {, x}+f=0 \text { in } \Omega$
x ā [ a , b ] Ā f ā R , Īŗ ā R + x \in [a,b]\space f\in \mathbb{R}, \kappa \in \mathbb{R}^+ x ā [ a , b ] Ā f ā R , Īŗ ā R + $u$ dependent variable $f$ forcing function $\kappa$ material parameter $\Omega$ domain. Material is dependent but set to 1 in textbook for constants $\left(\kappa u{x}\right) {, x}=\kappa u_{x x}=\kappa \frac{d^{2} u}{d x^{2}}$
no boundary: $\Omega=] a, b[, a<x<b$
closure includes boundary: $\bar{\Omega}=[a, b], a \leq x \leq b$
Positive Definite
definition: matrix $\bf A$ is positive definite if $\bf c \cdot A \cdot c \geq 0 \space \forall \space c$ $\bf c \cdot A \cdot c = 0 \space |\space c = 0$ .
properties:
eigenvalues $\mathbb{R}^+$
Vector Space Definitions
Weight space
$\delta^h \in \delta \subset {w| w\in \mathcal{H}^n, w|_{\Gamma_g} =0} $
Trial space
$\mathcal{V}^h \in \delta \subset {w| w\in \mathcal{H}^n, w|_{\Gamma_g} =0} $
whatchamacallit space
$\mathcal{H}^n$ is the collection of square integral spaces, which measures how many inner products on derivative you can take before one of them blows up or has a singularity.
u ā H 0 Ā ifĀ ā« Ī© u 2 d Ī© < ā u ā H 1 Ā ifĀ ā« Ī© ( u u ā H 0 + u , i u , i ) ā H 1 d Ī© < ā u \in H^{0} \text { if } \int_{\Omega} u^{2} d \Omega<\infty
\\
u \in H^{1} \text { if } \int_{\Omega}
\overbrace{
\left(
\overbrace{u u}^{H_0}
+u_{, i} u_{, i}\right)
}^{H_1}
d \Omega<\infty u ā H 0 Ā ifĀ ā« Ī© ā u 2 d Ī© < ā u ā H 1 Ā ifĀ ā« Ī© ā ( uu H 0 ā + u , i ā u , i ā ) ā H 1 ā ā d Ī© < ā remember that our stiffness matrix is found by the energy inner product, so that $\int{0}^{1} w {, x} \kappa u_{, x} d x$. This is well behaved if the function is square integrable on $H$.
Stiffness Matrix Positive Definite
c ā
K ā
c = ā A = 1 N ā B = 1 N c A Ā K A B Ā c B = ā A = 1 N ā B = 1 N c A Ā a ( N A , N B ) Ā c B = ā A = 1 N ā B = 1 N a ( c A N A , c B N B ā Ā noĀ freeĀ index,Ā same ) = a ( ā A = 1 N c A N A , ā B = 1 N c B N B ) = a ( w h , w h ) = ā« 0 1 ( w , x h ) 2 d x {\bf c} \cdot {\bf K} \cdot {\bf c} =
\sum_{A=1}^{N}
\sum_{B=1}^{N}
c_A
\space
K_{AB}
\space
c_B
=
\sum_{A=1}^{N}
\sum_{B=1}^{N}
c_A
\space
a(N_A,N_B)
\space
c_B
=
\sum_{A=1}^{N}
\sum_{B=1}^{N}
a(
\overbrace{c_A N_A,c_BN_B}^\text{ no free index, same}
)
\\
=
a(
\sum_{A=1}^{N}c_A N_A,\sum_{B=1}^{N}c_BN_B
)
= a(w^h, w^h) = \int_0^1(w^h_{,x})^2dx c ā
K ā
c = A = 1 ā N ā B = 1 ā N ā c A ā Ā K A B ā Ā c B ā = A = 1 ā N ā B = 1 ā N ā c A ā Ā a ( N A ā , N B ā ) Ā c B ā = A = 1 ā N ā B = 1 ā N ā a ( c A ā N A ā , c B ā N B ā ā Ā noĀ freeĀ index,Ā same ā ) = a ( A = 1 ā N ā c A ā N A ā , B = 1 ā N ā c B ā N B ā ) = a ( w h , w h ) = ā« 0 1 ā ( w , x h ā ) 2 d x Interpolating Shape Functions ${\bf N}_a \text{a}$
${\bf N}_a(x)$ in the 1D case, but $\bf x = = $ are the more general versions
Decomposition of Trial Space : ${\bf u}^h = {\bf v}^h + {\bf g}^h$
Appendix4.I.1.pdf
Linear space: a collection of objects that satisfy the following: If u and v are members of a linear space and Ī± and Ī² are scalars, then Ī±u + Ī²v is also a member of that linear space.
addition is component wise
u ā + v ā = ( u 1 , u 2 , u 3 , ā¦ u n ) + ( v 1 , v 2 , v 3 , ā¦ v n ) = ( u 1 + v 1 , u 2 + v 2 , u 3 + v 3 , ā¦ u n + v n ) \begin{array}{l}{\vec{u}+\vec{v}=\left(u_{1}, u_{2}, u_{3}, \dots u_{n}\right)+\left(v_{1}, v_{2}, v_{3}, \dots v_{n}\right)=} \\ {\left(u_{1}+v_{1}, u_{2}+v_{2}, u_{3}+v_{3}, \dots u_{n}+v_{n}\right)}\end{array} u + v = ( u 1 ā , u 2 ā , u 3 ā , ā¦ u n ā ) + ( v 1 ā , v 2 ā , v 3 ā , ā¦ v n ā ) = ( u 1 ā + v 1 ā , u 2 ā + v 2 ā , u 3 ā + v 3 ā , ā¦ u n ā + v n ā ) ā scalar multiplication is distributed
Ī± u ā = ( Ī± u 1 , Ī± u 2 , Ī± u 3 , ā¦ Ī± u n ) \alpha \vec{u}=\left(\alpha u_{1}, \alpha u_{2}, \alpha u_{3}, \dots \alpha u_{n}\right) Ī± u = ( Ī± u 1 ā , Ī± u 2 ā , Ī± u 3 ā , ā¦ Ī± u n ā ) Ī± u ā + Ī² v ā = ( Ī± u 1 + Ī² v 1 , Ī± u 2 + Ī² v 2 , Ī± u 3 + Ī² v 3 , ā¦ Ī± u n + Ī² v n ) \alpha \vec{u}+\beta \vec{v}=\left(\alpha u_{1}+\beta v_{1}, \alpha u_{2}+\beta v_{2}, \alpha u_{3}+\beta v_{3}, \dots \alpha u_{n}+\beta v_{n}\right) Ī± u + Ī² v = ( Ī± u 1 ā + Ī² v 1 ā , Ī± u 2 ā + Ī² v 2 ā , Ī± u 3 ā + Ī² v 3 ā , ā¦ Ī± u n ā + Ī² v n ā ) Linear spaces have very nice properties that make it easy for us to āproveā things will behave the way we would like. Thus we want to be sure to know when the contributions to our FE weak forms are members of linear spaces. (For this class they will be, as you get to more complex problems they may not be, then things you have to figure out what you can use. Key linear space properties we like to employ are inner products (like our integrals to be inner products) and norms (which will represent a measure of size).
Inner product
Ā Definition:Ā AnĀ innerĀ productĀ āØ ā
, ā ā© Ā onĀ aĀ realĀ linearĀ Ā spaceĀ A Ā isĀ aĀ mapĀ āØ ā
; A Ć A ā ā Ā withĀ theĀ followingĀ Ā properties:Ā Ā i)Ā āØ u , v ā© = āØ v , u ā© Ā (symmetry)Ā Ā ii)Ā āØ Ī± u , v ā© = Ī± āØ u , v ā© Ā iii)Ā āØ Ī± u , v ā© = āØ u , w ā© + āØ v , w ā© Ā (ii)Ā andĀ iii)Ā areĀ linearity)Ā Ā iv)Ā āØ u , u ā© ā„ 0 Ā andĀ āØ u , u ā© = 0 Ā ifĀ andĀ onlyĀ ifĀ u = 0 Ā (positiveĀ definiteness)Ā \begin{array}{l}{\text { Definition: An inner product }\langle\cdot, \bullet\rangle \text { on a real linear }} \\ {\text { space } A \text { is a map }\langle\cdot ; A \times A \rightarrow \Re \text { with the following }} \\ {\text { properties: }} \\ {\text { i) }\langle u, v\rangle=\langle v, u\rangle \text { (symmetry) }} \\ {\text { ii) }\langle\alpha u, v\rangle=\alpha\langle u, v\rangle} \\ {\text { iii) }\langle\alpha u, v\rangle=\langle u, w\rangle+\langle v, w\rangle \quad \text { (ii) and iii) are linearity) }} \\ {\text { iv) }\langle u, u\rangle \geq 0 \text { and }\langle u, u\rangle= 0 \text { if and only if } u=0} \\ {\text { (positive definiteness) }}\end{array} Ā Definition:Ā AnĀ innerĀ productĀ āØ ā
, ā ā© Ā onĀ aĀ realĀ linearĀ Ā spaceĀ A Ā isĀ aĀ mapĀ āØ ā
; A Ć A ā ā Ā withĀ theĀ followingĀ Ā properties:Ā Ā i)Ā āØ u , v ā© = āØ v , u ā© Ā (symmetry)Ā Ā ii)Ā āØ Ī±u , v ā© = Ī± āØ u , v ā© Ā iii)Ā āØ Ī±u , v ā© = āØ u , w ā© + āØ v , w ā© Ā (ii)Ā andĀ iii)Ā areĀ linearity)Ā Ā iv)Ā āØ u , u ā© ā„ 0 Ā andĀ āØ u , u ā© = 0 Ā ifĀ andĀ onlyĀ ifĀ u = 0 Ā (positiveĀ definiteness)Ā ā Ā Definition:Ā LetĀ { A , āØ ā
ā© } Ā beĀ anĀ innerĀ produceĀ spaceĀ Ā (i.e.,Ā aĀ linearĀ spaceĀ A Ā withĀ andĀ innerĀ productĀ ā , Ā āĀ Ā definedĀ onĀ A Ā .Ā ThenĀ u , v ā A Ā areĀ saidĀ toĀ beĀ orthogonalĀ Ā (withĀ respectĀ toĀ āØ ā
ā
ā© ) Ā ifĀ āØ u , v ā© = 0 \begin{array}{l}{\text { Definition: Let }\{A,\langle\cdot\rangle\} \text { be an inner produce space }} \\ {\text { (i.e., a linear space } A \text { with and inner product } \notin, \text { ' }} \\ {\text { defined on } A \text { . Then } u, v \in A \text { are said to be orthogonal }} \\ {\text { (with respect to }\langle\cdot \cdot\rangle) \text { if }\langle u, v\rangle= 0}\end{array} Ā Definition:Ā LetĀ { A , āØ ā
ā©} Ā beĀ anĀ innerĀ produceĀ spaceĀ Ā (i.e.,Ā aĀ linearĀ spaceĀ A Ā withĀ andĀ innerĀ productĀ ā / , Ā āĀ Ā definedĀ onĀ A Ā .Ā ThenĀ u , v ā A Ā areĀ saidĀ toĀ beĀ orthogonalĀ Ā (withĀ respectĀ toĀ āØ ā
ā
ā©) Ā ifĀ āØ u , v ā© = 0 ā āØ u , v ā© 2 ā¤ āØ u , u ā© āØ v , v ā© \langle u, v\rangle^{2} \leq\langle u, u\rangle\langle v, v\rangle āØ u , v ā© 2 ā¤ āØ u , u ā© āØ v , v ā© Norm on linear space is an operator with properties
SemiNorm is positive semidefinite: where the inner product with itself returns 0
Natural norm or a true norm $|u|=\langle u, u\rangle^{1 / 2}$
Sobolev Inner Product and Norm
Ā ConsiderĀ aĀ domainĀ Ī© ā ā n Ī¼ , n s d ā„ 1 Ā (willĀ beĀ theĀ spatialĀ Ā dimensionĀ ā 1 D , 2 D , 3 D ) , Ā andĀ letĀ u , v : Ī© ā ā Ā TheĀ L 2 ( Ī© ) Ā (orĀ equivalentlyĀ H o ( Ī© ) ) Ā innerĀ productĀ andĀ Ā normĀ areĀ definedĀ byĀ ( u , v ) = ( u , v ) 0 = ā« Ī© u v d Ī© ā„ u ā„ = ( u , u ) 1 / 2 Ā TheĀ H 1 ( Ī© ) Ā innerĀ productĀ andĀ normĀ areĀ definedĀ byĀ ( u , v ) 1 = ā« Ī© ( u v + u , i v , i ) d Ī© ( sum ā” 1 ā¤ i ā¤ n s d ) ā„ u ā„ = ( u , u ) 1 1 / 2 \begin{array}{l}{\text { Consider a domain } \Omega \subset \Re^{n_{\mu}}, n_{s_{d}} \geq 1 \text { (will be the spatial }} \\ {\text { dimension }-1 D, 2 D, 3 D), \text { and let } u, v : \Omega \rightarrow \Re}\end{array}
\\
\begin{array}{l}{\left.\text { The } L_{2}(\Omega) \text { (or equivalently } H^{o}(\Omega)\right) \text { inner product and }} \\ {\text { norm are defined by }} \\ {(u, v)=(u, v)_{0}=\int_{\Omega} u v d \Omega} \\ {\|u\|=(u, u)^{1 / 2}}\end{array}
\\
\begin{array}{l}{\text { The } H^{1}(\Omega) \text { inner product and norm are defined by }} \\ {(u, v)_{1}=\int_{\Omega}\left(u v+u_{, i} v_{, i}\right) d \Omega\left(\operatorname{sum} 1 \leq i \leq n_{s d}\right)} \\ {\|u\|=(u, u)_{1}^{1 / 2}}\end{array} Ā ConsiderĀ aĀ domainĀ Ī© ā ā n Ī¼ ā , n s d ā ā ā„ 1 Ā (willĀ beĀ theĀ spatialĀ Ā dimensionĀ ā 1 D , 2 D , 3 D ) , Ā andĀ letĀ u , v : Ī© ā ā ā Ā TheĀ L 2 ā ( Ī© ) Ā (orĀ equivalentlyĀ H o ( Ī© ) ) Ā innerĀ productĀ andĀ Ā normĀ areĀ definedĀ byĀ ( u , v ) = ( u , v ) 0 ā = ā« Ī© ā uv d Ī© ā„ u ā„ = ( u , u ) 1/2 ā Ā TheĀ H 1 ( Ī© ) Ā innerĀ productĀ andĀ normĀ areĀ definedĀ byĀ ( u , v ) 1 ā = ā« Ī© ā ( uv + u , i ā v , i ā ) d Ī© ( sum 1 ā¤ i ā¤ n s d ā ) ā„ u ā„ = ( u , u ) 1 1/2 ā ā Note on Notation: follow index rules
Weighing function and trial functions have nice properties here
u ā H 0 Ā ifĀ ā« Ī© u 2 d Ī© < ā u ā H 1 Ā ifĀ ā« Ī© ( u u + u , i u , i ) d Ī© < ā \begin{array}{c}{u \in H^{0} \text { if } \int_{\Omega} u^{2} d \Omega<\infty} \\ {u \in H^{1} \text { if } \int_{\Omega}\left(u u+u_{, i} u_{, i}\right) d \Omega<\infty}\end{array} u ā H 0 Ā ifĀ ā« Ī© ā u 2 d Ī© < ā u ā H 1 Ā ifĀ ā« Ī© ā ( uu + u , i ā u , i ā ) d Ī© < ā ā Recall that $f : \Omega \rightarrow \mathfrak{R}, \kappa \in \mathfrak{R}$$$
$\int{0}^{1} w {, x} \kappa u_{, x} d x$it is clear that it will be well behaved for u and w in H1
we want weighting function to be in the space $V=\left{w\left|w \in H^{1}, w\right|{\Gamma {g}}=0\right}$
which is the set of functions where the weight on the closure of the set is 0
trial space is similar but not homogenous bc $\boldsymbol{\delta}=\left{u\left|u \in H^{1}, u\right|{\Gamma {s}}=g\right}$
Ā GivenĀ hadĀ f : Ī© ā¾ ā ā , Īŗ ā ā , Īŗ > 0 , Ā andĀ constantsĀ g Ā andĀ h , Ā findĀ u ā Ī“ Ā suchĀ thatĀ forĀ allĀ w ā V a ( w , u ) = ( w , f ) + ( w , h ) Ī Ā forĀ theĀ problemĀ weĀ haveĀ thusĀ farĀ weĀ have:Ā a ( w , u ) = ā« 0 1 w x Īŗ u x d x ( w , f ) = ā« 0 1 w f d x ( w , h ) Ī = w ( 0 ) h Ā WeĀ canĀ checkĀ theĀ symmetryĀ andĀ bilinearityĀ ofĀ theĀ a ( w , u ) Ā andĀ ( w , f ) \begin{array}{l}{\text { Given had } f : \overline{\Omega} \rightarrow \Re, \kappa \in \Re, \kappa>0, \text { and constants } g} \\ {\text { and } h, \text { find } u \in \delta \text { such that for all } w \in V} \\ {\qquad a(w, u)=(w, f)+(w, h)_{\Gamma}} \\ {\text { for the problem we have thus far we have: }} \\ {a(w, u)=\int_{0}^{1} w_{x} \kappa u_{x} d x} \\ {(w, f)=\int_{0}^{1} w f d x} \\ {(w, h)_{\Gamma}=w(0) h} \\ {\text { We can check the symmetry and bilinearity of the }} \\ {a(w, u) \text { and }(w, f)}\end{array} Ā GivenĀ hadĀ f : Ī© ā ā , Īŗ ā ā , Īŗ > 0 , Ā andĀ constantsĀ g Ā andĀ h , Ā findĀ u ā Ī“ Ā suchĀ thatĀ forĀ allĀ w ā V a ( w , u ) = ( w , f ) + ( w , h ) Ī ā Ā forĀ theĀ problemĀ weĀ haveĀ thusĀ farĀ weĀ have:Ā a ( w , u ) = ā« 0 1 ā w x ā Īŗ u x ā d x ( w , f ) = ā« 0 1 ā w fd x ( w , h ) Ī ā = w ( 0 ) h Ā WeĀ canĀ checkĀ theĀ symmetryĀ andĀ bilinearityĀ ofĀ theĀ a ( w , u ) Ā andĀ ( w , f ) ā Equivalence of S and W.pdf
This shows how that the strong and weak forms are the same thing: all that separates the two is the application of $\textcolor{red}{\texttt{INTEGRATION BY PARTS}}$.
how did we get here
??
Strong solution satisfies the weak
We have a solution to the strong form $u{, x x}+f=0 \text { in } \Omega$, where $u(1) = g$ and $-u (0) = h$ and we want to show that this also holds over an interval
u x x + f = 0 Ā inĀ Ī© ā ā ā« 0 1 w ā thisĀ isĀ new ( u x x + f ) d x = 0 ā Ā w ā V u_{x x}+f=0 \text { in } \Omega
\rightarrow
\overbrace{
-\int_0^1 w
}^\text{this is new}(u_{x x}+f) dx = 0 \forall \space w \in \mathcal{V} u xx ā + f = 0 Ā inĀ Ī© ā ā ā« 0 1 ā w ā thisĀ isĀ new ā ( u xx ā + f ) d x = 0ā Ā w ā V We now apply integration by parts
ā« 0 1 w , x u x d x ā ā« 0 1 w f d x ā w u x ā£ 0 1 = 0 ā w ā V \int_{0}^{1} w_{, x} u_{x} d x-\int_{0}^{1} w f d x-\left.w u_{x}\right|_{0} ^{1}=0 \forall w \in V ā« 0 1 ā w , x ā u x ā d x ā ā« 0 1 ā w fd x ā w u x ā ā£ 0 1 ā = 0ā w ā V apply the boundary conditions where $w(1)=0(w \in V), \text { and }-u_{, x}(0)=h$
ā« 0 1 w , x u , x d x = ā« 0 1 w f d x + w ( 0 ) h ā w ā V \int_{0}^{1} w_{, x} u_{, x} d x=\int_{0}^{1} w f d x+w(0) h \quad \forall w \in V ā« 0 1 ā w , x ā u , x ā d x = ā« 0 1 ā w fd x + w ( 0 ) h ā w ā V weak solution satisfies the strong
we donāt need to do this again, but needs to be done to show that they are equivalent
weakĀ form:Ā ā« 0 1 w x u x d x = ā« 0 1 w f d x + w ( 0 ) h ā w ā V \text{weak form: } \int_{0}^{1} w_{x} u_{x} d x=\int_{0}^{1} w f d x+w(0) h \forall w \in \mathcal{V} weakĀ form:Ā ā« 0 1 ā w x ā u x ā d x = ā« 0 1 ā w fd x + w ( 0 ) h ā w ā V inorder to reverse integration by parts
ā« 0 1 w ( u , x x + f ) d x ā w u , x ā£ 0 1 + w ( 0 ) h = 0 ā w ā V \int_{0}^{1} w\left(u_{, x x}+f\right) d x-\left.w u_{, x}\right|_{0} ^{1}+w(0) h=0 \quad \forall w \in V ā« 0 1 ā w ( u , xx ā + f ) d x ā w u , x ā ā£ 0 1 ā + w ( 0 ) h = 0 ā w ā V and again apply boundary conditions $w(1)=0(w \in V)$
ā« 0 1 w ( u x x + f ) d x + w ( 0 ) ( u x ( 0 ) + h ) = 0 ā w ā V ā q . A \int_{0}^{1} w\left(u_{x x}+f\right) d x+w(0)\left(u_{x}(0)+h\right)=0 \forall w \in V \in q . A ā« 0 1 ā w ( u xx ā + f ) d x + w ( 0 ) ( u x ā ( 0 ) + h ) = 0ā w ā V ā q . A Solution Uniqueness
Give the previous expression of functions and their spaces
a ( w , u ) = ( w , f ) + ( w , h ) Ī a(w, u)=(w, f)+(w, h)_{\Gamma} a ( w , u ) = ( w , f ) + ( w , h ) Ī ā Proof by contradiction
a ( w , u 1 ) = ( w , f ) + ( w , h ) Ī a ( w , u 2 ) = ( w , f ) + ( w , h ) Ī \begin{aligned} a(w, u 1) &=(w, f)+(w, h)_{\Gamma} \\ a(w, u 2) &=(w, f)+(w, h)_{\Gamma} \end{aligned} a ( w , u 1 ) a ( w , u 2 ) ā = ( w , f ) + ( w , h ) Ī ā = ( w , f ) + ( w , h ) Ī ā ā difference then apply bilinearity
a ( w , u 1 ) ā a ( w , u 2 ) = 0 ā a ( w , ( u 1 ā u 2 ) ) = 0 a(w, u 1)-a(w, u 2)=0
\rightarrow
a(w,(u 1-u 2))=0 a ( w , u 1 ) ā a ( w , u 2 ) = 0 ā a ( w , ( u 1 ā u 2 )) = 0 by positive definiteness, this is only possible if $u 1-u 2=0 \text { or } u 1=u 2$.
this is all inservice to show that this is as good as it is going to get and the finite dimensional denoted with superscript $(\cdot)^h$.
Finite Dimensional Subspace
If the exact solution is in there, what if we use the finite dimensional subspace
$V^{h} \subset V \text { and } \delta^{h} \subset \delta$ and get corresponding weight and trial functions $w^{h} \in V^{h} \text { and } u^{h} \in \delta^{h}$.
deal with essential boundary conditions by decomposing linear
u h = v h + g h Ā whereĀ v h ā V h Ā andĀ g h ā Ī“ h u^{h}=v^{h}+g^{h} \text { where } v^{h} \in V^{h} \text { and } g^{h} \in \delta^{h} u h = v h + g h Ā whereĀ v h ā V h Ā andĀ g h ā Ī“ h Use interpolating shape functions so that $w^{h}=C{A} N {A}=\sum{A=1}^{n} C {A} N{A}=C {1} N{1}+C {2} N{2}+C {3} N{3}+\ldots+C {n} N_{n}$
so in terms of essential BC in nonzero and homogenous parts
u h = v h + g h = ā A = 1 n d A N A + ā B = n + 1 n + m g B N B u^{h}=v^{h}+g^{h}=\sum_{A=1}^{n} d_{A} N_{A}+\sum_{B=n+1}^{n+m} g_{B} N_{B} u h = v h + g h = A = 1 ā n ā d A ā N A ā + B = n + 1 ā n + m ā g B ā N B ā This is the abstract form $a(w,(v+g))=a(w, v)+a(w, g)$ so that
a ( w , v ) = ( w , f ) + ( w , h ) Ī ā a ( w , g ) a(w, v)=(w, f)+(w, h)_{\Gamma}-a(w, g) a ( w , v ) = ( w , f ) + ( w , h ) Ī ā ā a ( w , g ) apply summations
a ( ā A = 1 n C A N A , ā B = 1 n d B N B ) = ( ā 1 n C A N A , f ) + ( ā 1 n C A N A h ) Ī ā a ( ā A = 1 n C A N A , ā B = n + 1 n + m g B N B ) a\left(\sum_{A=1}^{n} C_{A} N_{A}, \sum_{B=1}^{n} d_{B} N_{B}\right)=\left(\sum_{1}^{n} C_{A} N_{A}, f\right)+\left(\sum_{1}^{n} C_{A} N_{A} h\right)_{\Gamma}
-a\left(\sum_{A=1}^{n} C_{A} N_{A}, \sum_{B=n+1}^{n+m} g_{B} N_{B}\right) a ( A = 1 ā n ā C A ā N A ā , B = 1 ā n ā d B ā N B ā ) = ( 1 ā n ā C A ā N A ā , f ) + ( 1 ā n ā C A ā N A ā h ) Ī ā ā a ( A = 1 ā n ā C A ā N A ā , B = n + 1 ā n + m ā g B ā N B ā ) where the summation can be pulled out $a\left(\sum{A=1}^{n} w {A}, v\right)=\sum{A=1}^{n} a\left(w {A}, v\right)$ and $a\left(\sum{A=1}^{n} w {A}, \sum{B=1}^{m} v {B}\right)=\sum{A=1}^{n} \sum {B=1}^{m} a\left(w{A}, v {B}\right)$ so that
ā A = 1 n ā B = 1 n a ( C A N A , d B N B ) = ā A = 1 n ( C A N A , f ) + ā A = 1 n ( C A N A , h ) Ī ā ā A = 1 n ā B = n + 1 n + m a ( C A N A , g B N B ) \sum_{A=1}^{n} \sum_{B=1}^{n} a\left(C_{A} N_{A}, d_{B} N_{B}\right)=\sum_{A=1}^{n}\left(C_{A} N_{A}, f\right)+\sum_{A=1}^{n}\left(C_{A} N_{A}, h\right)_{\Gamma}
-\sum_{A=1}^{n} \sum_{B=n+1}^{n+m} a\left(C_{A} N_{A}, g_{B} N_{B}\right) A = 1 ā n ā B = 1 ā n ā a ( C A ā N A ā , d B ā N B ā ) = A = 1 ā n ā ( C A ā N A ā , f ) + A = 1 ā n ā ( C A ā N A ā , h ) Ī ā ā A = 1 ā n ā B = n + 1 ā n + m ā a ( C A ā N A ā , g B ā N B ā ) Factor summation on arbitrary constant
ā A = 1 n C A [ ā B = 1 n a ( N A , N B ) d B ā ( N A , f ) ā ( N A , h ) Ī + ā B = n + 1 n + m a ( N A , N B ) g B ] ā G A = 0 \sum_{A=1}^{n} C_{A}
\overbrace{
\left[\sum_{B=1}^{n} a\left(N_{A}, N_{B}\right) d_{B}-\left(N_{A}, f\right)-\left(N_{A}, h\right)_{\Gamma}+\sum_{B=n+1}^{n+m} a\left(N_{A}, N_{B}\right) g_{B}\right]}^{G_A}
=0 A = 1 ā n ā C A ā [ B = 1 ā n ā a ( N A ā , N B ā ) d B ā ā ( N A ā , f ) ā ( N A ā , h ) Ī ā + B = n + 1 ā n + m ā a ( N A ā , N B ā ) g B ā ] ā G A ā ā = 0 For all integer values A, this must be zero. This is only true by our finite element system
ā B = 1 n a ( N A , N B ) d B = ( N A , f ) + ( N A , h ) Ī ā ā B = n + 1 n + m a ( N A , N B ) g B \sum_{B=1}^{n} a\left(N_{A}, N_{B}\right) d_{B}=\left(N_{A}, f\right)+\left(N_{A}, h\right)_{\Gamma}-\sum_{B=n+1}^{n+m} a\left(N_{A}, N_{B}\right) g_{B} B = 1 ā n ā a ( N A ā , N B ā ) d B ā = ( N A ā , f ) + ( N A ā , h ) Ī ā ā B = n + 1 ā n + m ā a ( N A ā , N B ā ) g B ā This defines the stiffness matrix $K{A B}=a\left(N {A}, N{B}\right)$ where the RHS is the forcing function $F {A}=\left(N{A}, f\right)+\left(N {A}, h\right){\Gamma}-\sum {B=n+1}^{n+m} a\left(N{A}, N {B}\right) g{B}$. Where n square matrix equation is $[K] {n x n}{d}{n x 1}={F} {n x 1}$.
1-2DOF-example.pdf
MWR.pdf
Lets state the generic form of the problem $\textcolor{red}{\texttt{probs need to move this up to match overage order}}$
Given: $f: \Omega \rightarrow \mathbb{R}$ and known $g_i$
Find: $u:\overline{\Omega} \rightarrow \mathbb{R}$
approach
Let $D^j$ and $B^j$ be differential operators of order $m$, and $\Gammai$ are appropriate portions of the boundary $\Gamma$. At every point of the boundary, then there $m$ boundary conditions, corresponding to the $m$ directions, or $n {sd},$ spatial dimensions. In the 2D case of beam bending.
Given: $f: \Omega \rightarrow \mathbb{R}$ with constants $g_i,h_i$, where $i={1,2}$.
Find: $u:\overline{\Omega}\rightarrow \mathbb{R} $
such that $E I u_{x x x x}-f=0 \text { on } \Omega$
$\left.u\right|{\Gamma {u}}=g_{1}$ displacement BC
$\left.u{,x}\right| {\Gamma{\theta}}=g {1}$ rotation BC
$\left.EIu{,xx}\right| {\Gamma{M}}=h {1}$ moment BC
$\left.EIu{,xxx}\right| {\Gamma{Q}}=h {2}$ shear BC
We cannot find the strong form, the only equation which will but an approximation $u^a = u^h$
D 2 m ( u a ) ā f ā 0 ā D 2 m ( u a ) ā f = R , u a ā Ī“ a ā Ī“ D^{2 m}\left(u^{a}\right)-f \neq 0
\rightarrow
D^{2 m}\left(u^{a}\right)-f=R, u^{a} \in \delta^{a} \subset \delta D 2 m ( u a ) ā f ī = 0 ā D 2 m ( u a ) ā f = R , u a ā Ī“ a ā Ī“ Method of weighted residuals means that we are using an interval instead, so that:
ā« Ī© w ( D 2 m ( u a ) ā f ) d Ī© = 0 ā w ā V \int_{\Omega} w\left(D^{2 m}\left(u^{a}\right)-f\right) d \Omega=0 \forall w \in V ā« Ī© ā w ( D 2 m ( u a ) ā f ) d Ī© = 0ā w ā V We consolidate our search area so that weight function $w$ is also part of the finite dimensional space
ā« Ī© w a ( D 2 m ( u a ) ā f ) d Ī© = 0 ā w a ā V a ā V \int_{\Omega} w^{a}\left(D^{2 m}\left(u^{a}\right)-f\right) d \Omega=0 \quad \forall w^{a} \in \mathcal{V}^{a} \subset \mathcal{V} ā« Ī© ā w a ( D 2 m ( u a ) ā f ) d Ī© = 0 ā w a ā V a ā V When applying the method of weighted residuals, we need to use an even function of order $2m$
, so that we move half of them over onto a weight function. Method of weighted residuals reduces the order $\textcolor{red}{\texttt{by half}}$. Many ways to do this
Collocation
Force the residual to be zero at node points
ā« Ī© Ī“ ( x ā x A ) ā unappealingĀ part ( D 2 m ( u a ) ā f ) d Ī© = 0 ā D 2 m ( u a ( x A ) ā f ) = 0 , A = 1 ( 1 ) n \int_{\Omega}
\overbrace{\delta\left(x-x_{A}\right)}^\text{unappealing part}
\left(D^{2 m}\left(u^{a}\right)-f\right) d \Omega=0
\rightarrow
D^{2 m}\left(u^{a}\left(x_{A}\right)-f\right)=0, \quad A=1(1) n ā« Ī© ā Ī“ ( x ā x A ā ) ā unappealingĀ part ā ( D 2 m ( u a ) ā f ) d Ī© = 0 ā D 2 m ( u a ( x A ā ) ā f ) = 0 , A = 1 ( 1 ) n Least Squares
Minimize the squared residual $\operatorname{Min} .\left(\int{\Omega}\left(D^{2 m}\left(u\left(x, d {A}\right)\right)-f\right)^{2} d \Omega\right)$, where $d_A$ are unknown parameters
ā ā d A ( ā« Ī© ( D 2 m ( u ( x , d A ) ) ā f ) 2 d Ī© ) = 0 , A = 1 ( 1 ) n ā ā« Ī© ā ( D 2 m ( u ( x , d A ) ) ā f ) ā d A ( D 2 m ( u ( x , d A ) ) ā f ) d Ī© = 0 , A = 1 ( 1 ) n \frac{\partial}{\partial d_{A}}\left(\int_{\Omega}\left(D^{2 m}\left(u\left(x, d_{A}\right)\right)-f\right)^{2} d \Omega\right)=0, A=1(1) n
\rightarrow
\int_{\Omega} \frac{\partial\left(D^{2 m}\left(u\left(x, d_{A}\right)\right)-f\right)}{\partial d_{A}}\left(D^{2 m}\left(u\left(x, d_{A}\right)\right)-f\right) d \Omega=0, A=1(1) n ā d A ā ā ā ( ā« Ī© ā ( D 2 m ( u ( x , d A ā ) ) ā f ) 2 d Ī© ) = 0 , A = 1 ( 1 ) n ā ā« Ī© ā ā d A ā ā ( D 2 m ( u ( x , d A ā ) ) ā f ) ā ( D 2 m ( u ( x , d A ā ) ) ā f ) d Ī© = 0 , A = 1 ( 1 ) n Galerkin Methods
These are all the other choices for weighting functions. The bubnov sub method is just where the weight function is an interpolating shape function and the non essential components use them as well $v_a$.
Step 1: weight and trail functions in terms of $N_A$
u h = v h + g h = d A N A + g B N B = ā A = 1 n d A N A + ā B = n + 1 n + m g B N B u^{h}=v^{h}+g^{h}=d_{A} N_{A}+g_{B} N_{B}=\sum_{A=1}^{n} d_{A} N_{A}+\sum_{B=n+1}^{n+m} g_{B} N_{B} u h = v h + g h = d A ā N A ā + g B ā N B ā = A = 1 ā n ā d A ā N A ā + B = n + 1 ā n + m ā g B ā N B ā MWR-example.pdf
global-local.pdf
Chapter 2
Chapt-II-heat-transfer.pdf
Notation and Variables
$n_{SD}$ - number of spatial dimensions
$\Omega \subset \mathbb{R}^{n_{sd}}$ is the domain
$\Gamma$ boundary or closure of the domain
$\bar{\Omega} = \Omega \cup \Gamma$
Heat conduction equation $q_{i,i} = f$
Fourierās Law $\kappa{ij} u {,j} + qi =0$ where $\kappa$ constant isotropic such that $\kappa\delta {ij}$
Strong Form
Given $f: \Omega \rightarrow \mathbb{R}$, $h: \Gamma_h \rightarrow \mathbb{R}$
Find: $u:\bar{\Omega} \rightarrow \mathbb{R}$
Such that: $q_{i,i} = f$, $u = g \text{ on } \Gamma$, $-q_i n_i \text{ on } \Gamma_h$
Application of method of weighted residuals on the interval
ā« Ī© Ļ ( q i , i ā f ) d Ī© ā Ļ ā V \int_\Omega \omega (q_{i,i} - f)d\Omega \forall \omega \in \mathcal{V} ā« Ī© ā Ļ ( q i , i ā ā f ) d Ī©ā Ļ ā V Chapt-II-elastostatics.pdf
FE-Analysis-pseudo-code.pdf
Chapter 3
Continuity Requirements
Intraelement
Interelement
Continuity