Test 1 Study Guide

use maniatty nomenclature for vector and tensor spaces (lowercase bold for vector, capital bold for 2+)

$\textcolor{red}{\texttt{this shit is fucked up}}$


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)=01w,xu,xdx=(w,d)+w(0)h(w,l)=01wdx\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}

symmetric and bilinear in each slot

symmetric $\begin{aligned} a(u, v) &=a(v, u) \(u, v) &=(v, u) \end{aligned}$

bilinear $\left(c{1} u+c{2} v, w\right)=c{1}(u, w)+c{2}(v, w)$


how the fuck am I going to study for this test?

  • study guide

    • notes

    • handouts

    • textbook

  • practice test

    • Hey, do you have the old exams for Shephard’s FEM? I’d like to take them as practice tests - to rui

    • 90 minute exam

  • test prep

    • aaaaayyyyyyyyy

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

Strong Form

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

Weak Variational Form

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 $\mathcal{V} = \left{w | w \in H^{1}, w(1)=0\right}$

\text{weak form} = \left\{ \begin{array} & \text{given before }\\ \int_{0}^{1} w_{, x} u_{, x} d x=\int_{0}^{1} w f d x+w(0) h \end{array} \right.

this is called virtual work/displacement/principals where the $w,_x$ is the virtual part and the generalized part



Descretize function space

δhδ (i.e., if uhSh, then uhE)vμU( i.e., if whUh, then whU)\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}

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?


1D ODE $\left(\kappa u{, x}\right){, x}+f=0 \text { in } \Omega$

x[a,b] fR,κR+x \in [a,b]\space f\in \mathbb{R}, \kappa \in \mathbb{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 $\bf c \cdot A \cdot c \geq 0 \space \forall \space c$ $\bf c \cdot A \cdot c = 0 \space |\space c = 0$ .


  • unique inverse

  • 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.

uH0 if Ωu2dΩ<uH1 if Ω(uuH0+u,iu,i)H1dΩ<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

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

cKc=A=1NB=1NcA KAB cB=A=1NB=1NcA a(NA,NB) cB=A=1NB=1Na(cANA,cBNB no free index, same)=a(A=1NcANA,B=1NcBNB)=a(wh,wh)=01(w,xh)2dx{\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

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: ${\bf u}^h = {\bf v}^h + {\bf g}^h$


  • Linear space: a collection of objects that satisfy the following: If u and v are members of a linear space and α and β are scalars, then αu + βv is also a member of that linear space.

  • addition is component wise

u+v=(u1,u2,u3,un)+(v1,v2,v3,vn)=(u1+v1,u2+v2,u3+v3,un+vn)\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}
  • scalar multiplication is distributed

    αu=(αu1,αu2,αu3,αun)\alpha \vec{u}=\left(\alpha u_{1}, \alpha u_{2}, \alpha u_{3}, \dots \alpha u_{n}\right)
αu+βv=(αu1+βv1,αu2+βv2,αu3+βv3,αun+βvn)\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)

Linear spaces have very nice properties that make it easy for us to “prove” 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

     Definition: An inner product , on a real linear  space A is a map ;A×A with the following  properties:  i) u,v=v,u (symmetry)  ii) αu,v=αu,v iii) αu,v=u,w+v,w (ii) and iii) are linearity)  iv) u,u0 and u,u=0 if and only if u=0 (positive definiteness) \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}
    • Note

    •  Definition: Let {A,} be an inner produce space  (i.e., a linear space A with and inner product , ’  defined on A . Then u,vA are said to be orthogonal  (with respect to ) if u,v=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}
u,v2u,uv,v\langle u, v\rangle^{2} \leq\langle u, u\rangle\langle v, v\rangle
  • 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

     Consider a domain Ωnμ,nsd1 (will be the spatial  dimension 1D,2D,3D), and let u,v:Ω The L2(Ω) (or equivalently Ho(Ω)) inner product and  norm are defined by (u,v)=(u,v)0=ΩuvdΩu=(u,u)1/2 The H1(Ω) inner product and norm are defined by (u,v)1=Ω(uv+u,iv,i)dΩ(sum1insd)u=(u,u)11/2\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}

Note on Notation: follow index rules

Weighing function and trial functions have nice properties here

uH0 if Ωu2dΩ<uH1 if Ω(uu+u,iu,i)dΩ<\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}

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 $\boldsymbol{\delta}=\left{u\left|u \in H^{1}, u\right|{\Gamma{s}}=g\right}$

 Given had f:Ω,κ,κ>0, and constants g and h, find uδ such that for all wVa(w,u)=(w,f)+(w,h)Γ for the problem we have thus far we have: a(w,u)=01wxκuxdx(w,f)=01wfdx(w,h)Γ=w(0)h We can check the symmetry and bilinearity of the a(w,u) and (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}

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

uxx+f=0 in Ω01wthis is new(uxx+f)dx=0 wVu_{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}

We now apply integration by parts

01w,xuxdx01wfdxwux01=0wV\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

apply the boundary conditions where $w(1)=0(w \in V), \text { and }-u_{, x}(0)=h$

01w,xu,xdx=01wfdx+w(0)hwV\int_{0}^{1} w_{, x} u_{, x} d x=\int_{0}^{1} w f d x+w(0) h \quad \forall w \in V

weak solution satisfies the strong

we don’t need to do this again, but needs to be done to show that they are equivalent

weak form: 01wxuxdx=01wfdx+w(0)hwV\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}

inorder to reverse integration by parts

01w(u,xx+f)dxwu,x01+w(0)h=0wV\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

and again apply boundary conditions $w(1)=0(w \in V)$

01w(uxx+f)dx+w(0)(ux(0)+h)=0wVq.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

Solution Uniqueness

Give the previous expression of functions and their spaces

a(w,u)=(w,f)+(w,h)Γa(w, u)=(w, f)+(w, h)_{\Gamma}

Proof by contradiction

a(w,u1)=(w,f)+(w,h)Γa(w,u2)=(w,f)+(w,h)Γ\begin{aligned} a(w, u 1) &=(w, f)+(w, h)_{\Gamma} \\ a(w, u 2) &=(w, f)+(w, h)_{\Gamma} \end{aligned}

difference then apply bilinearity

a(w,u1)a(w,u2)=0a(w,(u1u2))=0a(w, u 1)-a(w, u 2)=0 \rightarrow a(w,(u 1-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

uh=vh+gh where vhVh and ghδhu^{h}=v^{h}+g^{h} \text { where } v^{h} \in V^{h} \text { and } g^{h} \in \delta^{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

uh=vh+gh=A=1ndANA+B=n+1n+mgBNBu^{h}=v^{h}+g^{h}=\sum_{A=1}^{n} d_{A} N_{A}+\sum_{B=n+1}^{n+m} g_{B} N_{B}

This is the abstract form $a(w,(v+g))=a(w, v)+a(w, g)$ so that

a(w,v)=(w,f)+(w,h)Γa(w,g)a(w, v)=(w, f)+(w, h)_{\Gamma}-a(w, g)

apply summations

a(A=1nCANA,B=1ndBNB)=(1nCANA,f)+(1nCANAh)Γa(A=1nCANA,B=n+1n+mgBNB)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)

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

A=1nB=1na(CANA,dBNB)=A=1n(CANA,f)+A=1n(CANA,h)ΓA=1nB=n+1n+ma(CANA,gBNB)\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)

Factor summation on arbitrary constant

A=1nCA[B=1na(NA,NB)dB(NA,f)(NA,h)Γ+B=n+1n+ma(NA,NB)gB]GA=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

For all integer values A, this must be zero. This is only true by our finite element system

B=1na(NA,NB)dB=(NA,f)+(NA,h)ΓB=n+1n+ma(NA,NB)gB\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}

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 $[K]{n x n}{d}{n x 1}={F}{n x 1}$.



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: $u:\overline{\Omega} \rightarrow \mathbb{R}$


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 $u^a = u^h$

D2m(ua)f0D2m(ua)f=R,uaδaδD^{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

Method of weighted residuals means that we are using an interval instead, so that:

Ωw(D2m(ua)f)dΩ=0wV\int_{\Omega} w\left(D^{2 m}\left(u^{a}\right)-f\right) d \Omega=0 \forall w \in V

We consolidate our search area so that weight function $w$ is also part of the finite dimensional space

Ωwa(D2m(ua)f)dΩ=0waVaV\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}

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


Force the residual to be zero at node points

Ωδ(xxA)unappealing part(D2m(ua)f)dΩ=0D2m(ua(xA)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

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

dA(Ω(D2m(u(x,dA))f)2dΩ)=0,A=1(1)nΩ(D2m(u(x,dA))f)dA(D2m(u(x,dA))f)dΩ=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

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 $N_A$

uh=vh+gh=dANA+gBNB=A=1ndANA+B=n+1n+mgBNBu^{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}
  • Step 2

  • Step 3

  • Step 4



Chapter 2


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 $q_{i,i} = f$

  • Fourier’s 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

Ωω(qi,if)dΩωV\int_\Omega \omega (q_{i,i} - f)d\Omega \forall \omega \in \mathcal{V}



Chapter 3

Continuity Requirements




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