$u$ 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 $\bf c \cdot A \cdot c \geq 0 \space \forall \space c$ $\bf c \cdot A \cdot c = 0 \space |\space c = 0$ .
$\mathcal{H}^n$ 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∈H0 if ∫Ωu2dΩ<∞u∈H1 if ∫Ω(uuH0+u,iu,i)H1dΩ<∞
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⋅K⋅c=A=1∑NB=1∑NcAKABcB=A=1∑NB=1∑NcAa(NA,NB)cB=A=1∑NB=1∑Na(cANA,cBNB no free index, same)=a(A=1∑NcANA,B=1∑NcBNB)=a(wh,wh)=∫01(w,xh)2dx
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.
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→ℜ with the following properties: i) ⟨u,v⟩=⟨v,u⟩ (symmetry) ii) ⟨αu,v⟩=α⟨u,v⟩ iii) ⟨αu,v⟩=⟨u,w⟩+⟨v,w⟩ (ii) and iii) are linearity) iv) ⟨u,u⟩≥0 and ⟨u,u⟩=0 if and only if u=0 (positive definiteness)
Note
Definition: Let {A,⟨⋅⟩} be an inner produce space (i.e., a linear space A with and inner product ∈/, ’ defined on A . Then u,v∈A are said to be orthogonal (with respect to ⟨⋅⋅⟩) if ⟨u,v⟩=0
⟨u,v⟩2≤⟨u,u⟩⟨v,v⟩
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 Ω⊂ℜnμ,nsd≥1 (will be the spatial dimension −1D,2D,3D), and let u,v:Ω→ℜ The L2(Ω) (or equivalently Ho(Ω)) inner product and norm are defined by (u,v)=(u,v)0=∫ΩuvdΩ∥u∥=(u,u)1/2 The H1(Ω) inner product and norm are defined by (u,v)1=∫Ω(uv+u,iv,i)dΩ(sum1≤i≤nsd)∥u∥=(u,u)11/2
Note on Notation: follow index rules
Weighing function and trial functions have nice properties here
u∈H0 if ∫Ωu2dΩ<∞u∈H1 if ∫Ω(uu+u,iu,i)dΩ<∞
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→ℜ,κ∈ℜ,κ>0, and constants g and h, find u∈δ such that for all w∈Va(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)
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 Ω→−∫01wthis is new(uxx+f)dx=0∀w∈V
We now apply integration by parts
∫01w,xuxdx−∫01wfdx−wux∣01=0∀w∈V
apply the boundary conditions where $w(1)=0(w \in V), \text { and }-u_{, x}(0)=h$
∫01w,xu,xdx=∫01wfdx+w(0)h∀w∈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)h∀w∈V
inorder to reverse integration by parts
∫01w(u,xx+f)dx−wu,x∣01+w(0)h=0∀w∈V
and again apply boundary conditions $w(1)=0(w \in V)$
∫01w(uxx+f)dx+w(0)(ux(0)+h)=0∀w∈V∈q.A
Solution Uniqueness
Give the previous expression of functions and their spaces
a(w,u)=(w,f)+(w,h)Γ
Proof by contradiction
a(w,u1)a(w,u2)=(w,f)+(w,h)Γ=(w,f)+(w,h)Γ
difference then apply bilinearity
a(w,u1)−a(w,u2)=0→a(w,(u1−u2))=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 vh∈Vh and gh∈δh
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=1∑ndANA+B=n+1∑n+mgBNB
This is the abstract form $a(w,(v+g))=a(w, v)+a(w, g)$ so that
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
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 $g_i$
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 $i={1,2}$.
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)−f=0→D2m(ua)−f=R,ua∈δa⊂δ
Method of weighted residuals means that we are using an interval instead, so that:
∫Ωw(D2m(ua)−f)dΩ=0∀w∈V
We consolidate our search area so that weight function $w$ is also part of the finite dimensional space
∫Ωwa(D2m(ua)−f)dΩ=0∀wa∈Va⊂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
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
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 $N_A$