Our aim is to extend the ideas for approximating \( f \) by \( u \), or solving
$$ u = f $$
to real, spatial differential equations like
$$ -u'' + bu = f,\quad u(0)=C,\ u'(L)=D $$
$$
\begin{equation*}
\mathcal{L}(u) = 0,\quad x\in\Omega \end{equation*}
$$
Examples (1D problems):
$$
\begin{align*}
\mathcal{L}(u) &= \frac{d^2u}{dx^2} - f(x),\\
\mathcal{L}(u) &= \frac{d}{dx}\left(\dfc(x)\frac{du}{dx}\right) + f(x),\\
\mathcal{L}(u) &= \frac{d}{dx}\left(\dfc(u)\frac{du}{dx}\right) - au + f(x),\\
\mathcal{L}(u) &= \frac{d}{dx}\left(\dfc(u)\frac{du}{dx}\right) + f(u,x)
\end{align*}
$$
$$
\begin{equation*}
\mathcal{B}_0(u)=0,\ x=0,\quad \mathcal{B}_1(u)=0,\ x=L
\end{equation*}
$$
Examples:
$$
\begin{align*}
\mathcal{B}_i(u) &= u - g,\quad &\hbox{Dirichlet condition}\\
\mathcal{B}_i(u) &= -\dfc \frac{du}{dx} - g,\quad &\hbox{Neumann condition}\\
\mathcal{B}_i(u) &= -\dfc \frac{du}{dx} - h(u-g),\quad &\hbox{Robin condition}
\end{align*}
$$
Much is similar to approximating a function (solving \( u=f \)), but two new topics are needed:
Inserting \( u=\sum_jc_j\baspsi_j \) in \( \mathcal{L}=0 \) gives a residual \( R \)
$$
\begin{equation*}
\mathcal{L}(u) = \mathcal{L}(\sum_j c_j \baspsi_j) = R \neq 0
\end{equation*}
$$
Goal: minimize \( R \) with respect to \( \sequencei{c} \) (and hope it makes a small \( e \) too)
$$ R=R(c_0,\ldots,c_N; x)$$
Idea: minimize
$$
\begin{equation*}
E = ||R||^2 = (R,R) = \int_{\Omega} R^2 dx
\end{equation*}
$$
Minimization wrt \( \sequencei{c} \) implies
$$
\frac{\partial E}{\partial c_i} =
\int_{\Omega} 2R\frac{\partial R}{\partial c_i} dx = 0\quad
\Leftrightarrow\quad (R,\frac{\partial R}{\partial c_i})=0,\quad
i\in\If
$$
\( N+1 \) equations for \( N+1 \) unknowns \( \sequencei{c} \)
Idea: make \( R \) orthogonal to \( V \),
$$
(R,v)=0,\quad \forall v\in V
$$
This implies
$$
(R,\baspsi_i)=0,\quad i\in\If
$$
\( N+1 \) equations for \( N+1 \) unknowns \( \sequencei{c} \)
Generalization of the Galerkin method: demand \( R \) orthogonal to some space \( W \), possibly \( W\neq V \):
$$
(R,v)=0,\quad \forall v\in W
$$
If \( \{w_0,\ldots,w_N\} \) is a basis for \( W \):
$$
(R,w_i)=0,\quad i\in\If
$$
Idea: demand \( R=0 \) at \( N+1 \) points in space
$$ R(\xno{i}; c_0,\ldots,c_N)=0,\quad i\in\If$$
The collocation method is a weighted residual method with delta functions as weights
$$ 0 = \int_\Omega R(x;c_0,\ldots,c_N)
\delta(x-\xno{i})\dx = R(\xno{i}; c_0,\ldots,c_N)$$
$$
\hbox{property of } \delta(x):\quad
\int_{\Omega} f(x)\delta (x-\xno{i}) dx = f(\xno{i}),\quad \xno{i}\in\Omega
$$
Exemplify the least squares, Galerkin, and collocation methods in a simple 1D problem with global basis functions.
$$ -u''(x) = f(x),\quad x\in\Omega=[0,L],\quad u(0)=0,\ u(L)=0$$
Basis functions:
$$ \baspsi_i(x) = \sinL{i},\quad i\in\If$$
Residual:
$$
\begin{align*}
R(x;c_0,\ldots,c_N) &= u''(x) + f(x),\nonumber\\
&= \frac{d^2}{dx^2}\left(\sum_{j\in\If} c_j\baspsi_j(x)\right)
+ f(x),\nonumber\\
&= \sum_{j\in\If} c_j\baspsi_j''(x) + f(x)
\end{align*}
$$
Since \( u(0)=u(L)=0 \) we must ensure that all \( \baspsi_i(0)=\baspsi_i(L)=0 \), because then
$$ u(0) = \sum_jc_j{\color{red}\baspsi_j(0)} = 0,\quad
u(L) = \sum_jc_j{\color{red}\baspsi_j(L)} =0 $$
$$
(R,\frac{\partial R}{\partial c_i}) = 0,\quad i\in\If
$$
$$
\begin{equation*}
\frac{\partial R}{\partial c_i} =
\frac{\partial}{\partial c_i}
\left(\sum_{j\in\If} c_j\baspsi_j''(x) + f(x)\right)
= \baspsi_i''(x)
\end{equation*}
$$
Because:
$$
\frac{\partial}{\partial c_i}\left(c_0\baspsi_0'' + c_1\baspsi_1'' + \cdots +
c_{i-1}\baspsi_{i-1}'' + {\color{red}c_i\baspsi_{i}''} + c_{i+1}\baspsi_{i+1}''
+ \cdots + c_N\baspsi_N'' \right) = \baspsi_{i}''
$$
$$
\begin{equation*}
(\sum_j c_j \baspsi_j'' + f,\baspsi_i'')=0,\quad i\in\If
\end{equation*}
$$
Rearrangement:
$$
\begin{equation*}
\sum_{j\in\If}(\baspsi_i'',\baspsi_j'')c_j = -(f,\baspsi_i''),\quad i\in\If \end{equation*}
$$
This is a linear system
$$
\begin{equation*} \sum_{j\in\If}A_{i,j}c_j = b_i,\quad i\in\If
\end{equation*}
$$
$$
\begin{align*}
A_{i,j} &= (\baspsi_i'',\baspsi_j'')\nonumber\\
& = \pi^4(i+1)^2(j+1)^2L^{-4}\int_0^L \sinL{i}\sinL{j}\, dx\nonumber\\
&= \left\lbrace
\begin{array}{ll} {1\over2}L^{-3}\pi^4(i+1)^4 & i=j \\ 0, & i\neq j
\end{array}\right.
\\
b_i &= -(f,\baspsi_i'') = (i+1)^2\pi^2L^{-2}\int_0^Lf(x)\sinL{i}\, dx
\end{align*}
$$
Useful property of the chosen basis functions:
$$
\begin{equation*}
\int\limits_0^L \sinL{i}\sinL{j}\, dx = \delta_{ij},\quad
\quad\delta_{ij} = \left\lbrace
\begin{array}{ll} \half L & i=j \\ 0, & i\neq j
\end{array}\right.
\end{equation*}
$$
\( \Rightarrow\ (\baspsi_i'',\baspsi_j'') = \delta_{ij} \), i.e., diagonal \( A_{i,j} \), and we can easily solve for \( c_i \):
$$
\begin{equation*}
c_i = \frac{2L}{\pi^2(i+1)^2}\int_0^Lf(x)\sinL{i}\, dx
\end{equation*}
$$
Let sympy do the work (\( f(x)=2 \)):
from sympy import *
import sys
i, j = symbols('i j', integer=True)
x, L = symbols('x L')
f = 2
a = 2*L/(pi**2*(i+1)**2)
c_i = a*integrate(f*sin((i+1)*pi*x/L), (x, 0, L))
c_i = simplify(c_i)
print c_i
$$
\begin{equation*}
c_i = 4 \frac{L^{2} \left(\left(-1\right)^{i} + 1\right)}{\pi^{3}
\left(i^{3} + 3 i^{2} + 3 i + 1\right)},\quad
u(x) = \sum_{k=0}^{N/2} \frac{8L^2}{\pi^3(2k+1)^3}\sinL{2k}
\end{equation*}
$$
Fast decay: \( c_2 = c_0/27 \), \( c_4=c_0/125 \) - only one term might be good enough:
$$
\begin{equation*} u(x) \approx \frac{8L^2}{\pi^3}\sin\left(\pi\frac{x}{L}\right) \end{equation*}
$$
\( R=u''+f \):
$$
\begin{equation*}
(u''+f,v)=0,\quad \forall v\in V,
\end{equation*}
$$
or rearranged,
$$
\begin{equation*}
(u'',v) = -(f,v),\quad\forall v\in V \end{equation*}
$$
This is a variational formulation of the differential equation problem.
\( \forall v\in V \) is equivalent with \( \forall v\in\baspsi_i \), \( i\in\If \), resulting in
$$
\begin{equation*}
(\sum_{j\in\If} c_j\baspsi_j'', \baspsi_i)=-(f,\baspsi_i),\quad i\in\If \end{equation*}
$$
$$
\begin{equation*}
\sum_{j\in\If}(\baspsi_j'', \baspsi_i) c_j=-(f,\baspsi_i),\quad i\in\If \end{equation*}
$$
Since \( \baspsi_i''\propto -\baspsi_i \), Galerkin's method gives the same linear system and the same solution as the least squares method (in this particular example).
\( R=0 \) (i.e.,the differential equation) must be satisfied at \( N+1 \) points:
$$
\begin{equation*}
-\sum_{j\in\If} c_j\baspsi_j''(\xno{i}) = f(\xno{i}),\quad i\in\If
\end{equation*}
$$
This is a linear system \( \sum_j A_{i,j}=b_i \) with entries
$$
\begin{equation*} A_{i,j}=-\baspsi_j''(\xno{i})=
(j+1)^2\pi^2L^{-2}\sin\left((j+1)\pi \frac{x_i}{L}\right),
\quad b_i=2
\end{equation*}
$$
Choose: \( N=0 \), \( x_0=L/2 \)
$$ c_0=2L^2/\pi^2 $$
>>> import sympy as sym
>>> # Computing with Dirichlet conditions: -u''=2 and sines
>>> x, L = sym.symbols('x L')
>>> e_Galerkin = x*(L-x) - 8*L**2*sym.pi**(-3)*sym.sin(sym.pi*x/L)
>>> e_colloc = x*(L-x) - 2*L**2*sym.pi**(-2)*sym.sin(sym.pi*x/L)
>>> # Verify max error for x=L/2
>>> dedx_Galerkin = sym.diff(e_Galerkin, x)
>>> dedx_Galerkin.subs(x, L/2)
0
>>> dedx_colloc = sym.diff(e_colloc, x)
>>> dedx_colloc.subs(x, L/2)
0
# Compute max error: x=L/2, evaluate numerical, and simplify
>>> sym.simplify(e_Galerkin.subs(x, L/2).evalf(n=3))
-0.00812*L**2
>>> sym.simplify(e_colloc.subs(x, L/2).evalf(n=3))
0.0473*L**2
Second-order derivatives will hereafter be integrated by parts
$$
\begin{align*}
\int_0^L u''(x)v(x) dx &= - \int_0^Lu'(x)v'(x)dx
+ [vu']_0^L\nonumber\\
&= - \int_0^Lu'(x)v'(x) dx
+ u'(L)v(L) - u'(0)v(0)
\end{align*}
$$
Motivation:
Dirichlet conditions: \( u(0)=C \) and \( u(L)=D \). Choose for example
$$ B(x) = \frac{1}{L}(C(L-x) + Dx):\qquad B(0)=C,\ B(L)=D $$
$$
\begin{equation*}
u(x) = B(x) + \sum_{j\in\If} c_j\baspsi_j(x),
\end{equation*}
$$
$$ u(0) = B(0)= C,\quad u(L) = B(L) = D $$
Dirichlet condition: \( u(L)=D \). Choose for example
$$ B(x) = D:\qquad B(L)=D $$
$$
\begin{equation*}
u(x) = B(x) + \sum_{j\in\If} c_j\baspsi_j(x),
\end{equation*}
$$
$$ u(L) = B(L) = D $$
The finite element literature (and much FEniCS documentation) applies an abstract notation for the variational formulation:
Find \( (u-B)\in V \) such that
$$ a(u,v) = L(v)\quad \forall v\in V $$
$$ -u''=f, \quad u'(0)=C,\ u(L)=D,\quad u=D + \sum_jc_j\baspsi_j$$
Variational formulation:
$$
\int_{\Omega} u' v'dx = \int_{\Omega} fvdx - v(0)C
\quad\hbox{or}\quad (u',v') = (f,v) - v(0)C
\quad\forall v\in V
$$
Abstract formulation: find \( (u-B)\in V \) such that
$$ a(u,v) = L(v)\quad \forall v\in V$$
We identify
$$ a(u,v) = (u',v'),\quad L(v) = (f,v) -v(0)C $$
Linear form means
$$ L(\alpha_1 v_1 + \alpha_2 v_2)
=\alpha_1 L(v_1) + \alpha_2 L(v_2),
$$
Bilinear form means
$$
\begin{align*}
a(\alpha_1 u_1 + \alpha_2 u_2, v) &= \alpha_1 a(u_1,v) + \alpha_2 a(u_2, v),
\\
a(u, \alpha_1 v_1 + \alpha_2 v_2) &= \alpha_1 a(u,v_1) + \alpha_2 a(u, v_2)
\end{align*}
$$
In nonlinear problems: Find \( (u-B)\in V \) such that \( F(u;v)=0\ \forall v\in V \)
$$ a(u,v) = L(v)\quad \forall v\in V\quad\Leftrightarrow\quad
a(u,\baspsi_i) = L(\baspsi_i)\quad i\in\If$$
We can now derive the corresponding linear system once and for all by inserting \( u = B + \sum_jc_j\baspsi_j \):
$$ a(B + \sum_{j\in\If} c_j \baspsi_j,\baspsi_i) = L(\baspsi_i)\quad i\in\If$$
Because of linearity,
$$ \sum_{j\in\If} \underbrace{a(\baspsi_j,\baspsi_i)}_{A_{i,j}}c_j =
\underbrace{L(\baspsi_i) - a(B,\baspsi_i)}_{b_i}\quad i\in\If$$
If \( a \) is symmetric: \( a(u,v)=a(v,u) \),
$$ a(u,v)=L(v)\quad\forall v\in V$$
is equivalent to minimizing the functional
$$ F(v) = {\half}a(v,v) - L(v) $$
over all functions \( v\in V \). That is,
$$ F(u)\leq F(v)\quad \forall v\in V $$
Derive variational formulations for some prototype differential equations in 1D that include
$$
\begin{equation*}
-\frac{d}{dx}\left( \dfc(x)\frac{du}{dx}\right) = f(x),\quad x\in\Omega =[0,L],\
u(0)=C,\ u(L)=D
\end{equation*}
$$
$$
u(x) = B(x) + \sum_{j\in\If} c_j\baspsi_i(x),\quad
$$
$$ R = -\frac{d}{dx}\left( a\frac{du}{dx}\right) -f $$
Galerkin's method:
$$
(R, v) = 0,\quad \forall v\in V
$$
or with integrals:
$$
\int_{\Omega} \left(-\frac{d}{dx}\left( \dfc\frac{du}{dx}\right) -f\right)v \dx = 0,\quad \forall v\in V
$$
$$ -\int_{\Omega} \frac{d}{dx}\left( \dfc(x)\frac{du}{dx}\right) v \dx
= \int_{\Omega} \dfc(x)\frac{du}{dx}\frac{dv}{dx}\dx -
\left[\dfc\frac{du}{dx}v\right]_0^L
$$
Boundary terms vanish since \( v(0)=v(L)=0 \)
Find \( (u-B)\in V \) such that
$$
\int_{\Omega} \dfc(x)\frac{du}{dx}\frac{dv}{dx}dx = \int_{\Omega} f(x)vdx,\quad
\forall v\in V
$$
Compact notation:
$$ \underbrace{(\dfc u',v')}_{a(u,v)} = \underbrace{(f,v)}_{L(v)},
\quad \forall v\in V $$
With
$$ a(u,v) = (\dfc u', v'),\quad L(v) = (f,v) $$
we can just use the formula for the linear system:
$$
\begin{align*}
A_{i,j} &= a(\baspsi_j,\baspsi_i) = (\dfc \baspsi_j', \baspsi_i')
= \int_\Omega \dfc \baspsi_j' \baspsi_i'\dx =
\int_\Omega \baspsi_i' \dfc \baspsi_j'\dx \quad (= a(\baspsi_i,\baspsi_j) = A_{j,i})\\
b_i &= (f,\baspsi_i) - (\dfc B',\baspsi_i') = \int_\Omega (f\baspsi_i -
\dfc L^{-1}(D-C)\baspsi_i')\dx
\end{align*}
$$
\( v=\baspsi_i \) and \( u=B + \sum_jc_j\baspsi_j \):
$$
(\dfc B' + \dfc \sum_{j\in\If} c_j \baspsi_j', \baspsi_i') =
(f,\baspsi_i), \quad i\in\If
$$
Reorder to form linear system:
$$ \sum_{j\in\If} (\dfc\baspsi_j', \baspsi_i')c_j =
(f,\baspsi_i) - (aL^{-1}(D-C), \baspsi_i'), \quad i\in\If
$$
This is \( \sum_j A_{i,j}c_j=b_i \) with
$$
\begin{align*}
A_{i,j} &= (a\baspsi_j', \baspsi_i') = \int_{\Omega} \dfc(x)\baspsi_j'(x)
\baspsi_i'(x)\dx\\
b_i &= (f,\baspsi_i) - (aL^{-1}(D-C),\baspsi_i')=
\int_{\Omega} \left(f\baspsi_i - \dfc\frac{D-C}{L}\baspsi_i'\right) \dx
\end{align*}
$$
$$
-u''(x) + bu'(x) = f(x),\quad x\in\Omega =[0,L],\
u(0)=C,\ u'(L)=E
$$
New features:
Initial steps:
$$ u = C + \sum_{j\in\If} c_j \baspsi_i(x)$$
Galerkin's method: multiply by \( v \), integrate over \( \Omega \), integrate by parts.
$$ (-u'' + bu' - f, v) = 0,\quad\forall v\in V$$
$$ (u',v') + (bu',v) = (f,v) + [u' v]_0^L, \quad\forall v\in V$$
\( [u' v]_0^L = u'(L)v(L) - u'(0)v(0)= E v(L) \) since \( v(0)=0 \) and \( u'(L)=E \)
$$ (u',v') + (bu',v) = (f,v) + Ev(L), \quad\forall v\in V$$
$$ (u',v') + (bu',v) = (f,v) + Ev(L), \quad\forall v\in V$$
Important observations:
Abstract notation:
$$ a(u,v)=L(v)\quad\forall v\in V$$
With
$$ (u',v') + (bu',v) = (f,v) + Ev(L), \quad\forall v\in V$$
we have
$$
\begin{align*}
a(u,v)&=(u',v') + (bu',v)\\
L(v)&= (f,v) + E v(L)
\end{align*}
$$
Insert \( u=C+\sum_jc_j\baspsi_j \) and \( v=\baspsi_i \) in
$$ (u',v') + (bu',v) = (f,v) + Ev(L), \quad\forall v\in V$$
and manipulate to get
$$
\sum_{j\in\If}
\underbrace{((\baspsi_j',\baspsi_i') + (b\baspsi_j',\baspsi_i))}_{A_{i,j}}
c_j =
\underbrace{(f,\baspsi_i) + E \baspsi_i(L)}_{b_i},\quad i\in\If
$$
Observation: \( A_{i,j} \) is not symmetric because of the term
$$
(b\baspsi_j',\baspsi_i)=\int_{\Omega} b\baspsi_j'\baspsi_i dx
\neq \int_{\Omega} b \baspsi_i' \baspsi_jdx = (\baspsi_i',b\baspsi_j)
$$
$$ (u',v') + (bu',v) = (f,v) + u'(L)v(L) - u'(0)v(0)$$
It is easy to forget the boundary term when integrating by parts. That mistake may prescribe a condition on \( u' \)!
Problem:
$$
\begin{equation*}
-(\dfc(u)u')' = f(u),\quad x\in [0,L],\ u(0)=0,\ u'(L)=E
\end{equation*}
$$
Galerkin: multiply by \( v \), integrate, integrate by parts
$$ \int_0^L \dfc(u)\frac{du}{dx}\frac{dv}{dx}\dx =
\int_0^L f(u)v\dx + [\dfc(u)vu']_0^L\quad\forall v\in V
$$
$$ \int_0^L \dfc(u)\frac{du}{dx}\frac{dv}{dx}v\dx =
\int_0^L f(u)v\dx + \dfc(u(L))v(L)E\quad\forall v\in V
$$
or
$$ (\dfc(u)u', v') = (f(u),v) + \dfc(u(L))v(L)E\quad\forall v\in V
$$
$$
\begin{equation*}
-u''(x)=f(x),\quad x\in \Omega=[0,1],\quad u'(0)=C,\ u(1)=D
\end{equation*}
$$
Variational formulation: find \( (u-B)\in V \) such that
$$
(u',\baspsi_i') = (f,\baspsi_i) - C\baspsi_i(0),\ i\in\If
$$
Insert \( u(x) = B(x) + \sum_{j\in\If}c_j\baspsi_j \) and derive
$$ \sum_{j\in\If} A_{i,j}c_j = b_i,\quad i\in\If$$
with
$$ A_{i,j} = (\baspsi_j',\baspsi_i')
$$
$$ b_i = (f,\baspsi_i) - (D,\baspsi_i') -C\baspsi_i(0) $$
$$ A_{i,j} = (\baspsi_j',\baspsi_i') = \int_{0}^1 \baspsi_i'(x)\baspsi_j'(x)dx
= \int_0^1 (i+1)(j+1)(1-x)^{i+j} dx
$$
Choose \( f(x)=2 \):
$$
\begin{align*}
b_i &= (2,\baspsi_i) - (D,\baspsi_i') -C\baspsi_i(0)\\
&= \int_0^1 \left( 2(1-x)^{i+1} - D(i+1)(1-x)^i\right)dx -C\baspsi_i(0)
\end{align*}
$$
Can easily do the integrals with sympy. \( N=1 \) and \( \If = \{0,1\} \):
$$
\begin{equation*}
\left(\begin{array}{cc}
1 & 1\\
1 & 4/3
\end{array}\right)
\left(\begin{array}{c}
c_0\\
c_1
\end{array}\right)
=
\left(\begin{array}{c}
-C+D+1\\
2/3 -C + D
\end{array}\right)
\end{equation*}
$$
$$ c_0=-C+D+2, \quad c_1=-1,$$
$$ u(x) = 1 -x^2 + D + C(x-1)\quad\hbox{(exact solution)} $$
Assume that apart from boundary conditions, \( \uex \) lies in the same space \( V \) as where we seek \( u \):
$$
\begin{align*}
u &= B + {\color{red}F},\quad F\in V\\
a(B+F, v) &= L(v),\quad\forall v\in V\\
\uex & = B + {\color{red}E},\quad E\in V\\
a(B+E, v) &= L(v),\quad\forall v\in V
\end{align*}
$$
Subtract: \( a(F-E,v)=0\ \Rightarrow\ E=F \) and \( u = \uex \)
Tasks:
$$ -u''(x) = 2,\quad x\in (0,L),\ u(0)=u(L)=0,$$
Variational formulation:
$$ (u',v') = (2,v)\quad\forall v\in V $$
Since \( u(0)=0 \) and \( u(L)=0 \), we must force
$$ v(0)=v(L)=0,\quad \baspsi_i(0)=\baspsi_i(L)=0$$
Let's choose the obvious finite element basis: \( \baspsi_i=\basphi_i \), \( i=0,\ldots,N_n-1 \)
Problem: \( \basphi_0(0)\neq 0 \) and \( \basphi_{N_n-1}(L)\neq 0 \)
Solution: we just exclude \( \basphi_0 \) and \( \basphi_{N_n-1} \) from the basis and work with
$$ \baspsi_i=\basphi_{i+1},\quad i=0,\ldots,N=N_n-3$$
Introduce index mapping \( \nu(i) \): \( \baspsi_i = \basphi_{\nu(i)} \)
$$ u = \sum_{j\in\If}c_j\basphi_{\nu(j)},\quad i=0,\ldots,N,\quad \nu(j) = j+1$$
Irregular numbering: more complicated \( \nu(j) \) table
$$
\begin{equation*}
A_{i,j}=\int_0^L\basphi_{i+1}'(x)\basphi_{j+1}'(x) dx,\quad
b_i=\int_0^L2\basphi_{i+1}(x) dx
\end{equation*}
$$
Many will prefer to change indices to obtain a \( \basphi_i'\basphi_j' \) product: \( i+1\rightarrow i \), \( j+1\rightarrow j \)
$$
\begin{equation*}
A_{i-1,j-1}=\int_0^L\basphi_{i}'(x)\basphi_{j}'(x) \dx,\quad
b_{i-1}=\int_0^L2\basphi_{i}(x) \dx
\end{equation*}
$$
$$ \basphi_i' \sim \pm h^{-1} $$
$$ A_{i-1,i-1} = h^{-2}2h = 2h^{-1},\quad
A_{i-1,i-2} = h^{-1}(-h^{-1})h = -h^{-1}$$
and \( A_{i-1,i}=A_{i-1,i-2} \)
$$ b_{i-1} = 2({\half}h + {\half}h) = 2h$$
$$
\begin{equation*}
{
\frac{1}{h}\left(
\begin{array}{ccccccccc}
2 & -1 & 0
&\cdots &
\cdots & \cdots & \cdots &
\cdots & 0 \\
-1 & 2 & -1 & \ddots & & & & & \vdots \\
0 & -1 & 2 & -1 &
\ddots & & & & \vdots \\
\vdots & \ddots & & \ddots & \ddots & 0 & & & \vdots \\
\vdots & & \ddots & \ddots & \ddots & \ddots & \ddots & & \vdots \\
\vdots & & & 0 & -1 & 2 & -1 & \ddots & \vdots \\
\vdots & & & & \ddots & \ddots & \ddots &\ddots & 0 \\
\vdots & & & & &\ddots & \ddots &\ddots & -1 \\
0 &\cdots & \cdots &\cdots & \cdots & \cdots & 0 & -1 & 2
\end{array}
\right)
\left(
\begin{array}{c}
c_0 \\
\vdots\\
\vdots\\
\vdots \\
\vdots \\
\vdots \\
\vdots \\
\vdots\\
c_{N}
\end{array}
\right)
=
\left(
\begin{array}{c}
2h \\
\vdots\\
\vdots\\
\vdots \\
\vdots \\
\vdots \\
\vdots \\
\vdots\\
2h
\end{array}
\right)
}
\end{equation*}
$$
General equation at node \( i \):
$$
-\frac{1}{h}c_{i-1} + \frac{2}{h}c_{i} - \frac{1}{h}c_{i+1} = 2h
$$
Now, \( c_i = u(\xno{i+1})\equiv u_{i+1} \). Writing out the equation at node \( i-1 \),
$$
-\frac{1}{h}c_{i-2} + \frac{2}{h}c_{i-1} - \frac{1}{h}c_{i} = 2h
$$
translates directly to
$$
-\frac{1}{h}u_{i-1} + \frac{2}{h}u_{i} - \frac{1}{h}u_{i+1} = 2h
$$
The standard finite difference method for \( -u''=2 \) is
$$ -\frac{1}{h^2}u_{i-1} + \frac{2}{h^2}u_{i} - \frac{1}{h^2}u_{i+1} = 2 $$
Multiply by \( h \)!
The finite element method and the finite difference method are identical in this example.
(Remains to study the equations at the end points, which involve boundary values - but these are also the same for the two methods)
$$
\begin{equation*}
A_{i-1,j-1}^{(e)}=\int_{\Omega^{(e)}} \basphi_i'(x)\basphi_j'(x) \dx
= \int_{-1}^1 \frac{d}{dx}\refphi_r(X)\frac{d}{dx}\refphi_s(X)
\frac{h}{2} \dX,
\end{equation*}
$$
$$ \refphi_0(X)=\half(1-X),\quad\refphi_1(X)=\half(1+X)$$
$$ \frac{d\refphi_0}{dX} = -\half,\quad \frac{d\refphi_1}{dX} = \half $$
From the chain rule
$$ \frac{d\refphi_r}{dx} = \frac{d\refphi_r}{dX}\frac{dX}{dx}
= \frac{2}{h}\frac{d\refphi_r}{dX}$$
$$
\begin{equation*}
A_{i-1,j-1}^{(e)}=\int_{\Omega^{(e)}} \basphi_i'(x)\basphi_j'(x) \dx
= \int_{-1}^1 \frac{2}{h}\frac{d\refphi_r}{dX}\frac{2}{h}\frac{d\refphi_s}{dX}
\frac{h}{2} \dX = \tilde A_{r,s}^{(e)}
\end{equation*}
$$
$$
\begin{equation*}
b_{i-1}^{(e)} = \int_{\Omega^{(e)}} 2\basphi_i(x) \dx =
\int_{-1}^12\refphi_r(X)\frac{h}{2} \dX = \tilde b_{r}^{(e)},
\quad i=q(e,r),\ r=0,1
\end{equation*}
$$
Must run through all \( r,s=0,1 \) and \( r=0,1 \) and compute each entry in the element matrix and vector:
$$
\begin{equation*}
\tilde A^{(e)} =\frac{1}{h}\left(\begin{array}{rr}
1 & -1\\
-1 & 1
\end{array}\right),\quad
\tilde b^{(e)} = h\left(\begin{array}{c}
1\\
1
\end{array}\right)
\end{equation*}
$$
Example:
$$ \tilde A^{(e)}_{0,1} =
\int_{-1}^1 \frac{2}{h}\frac{d\refphi_0}{dX}\frac{2}{h}\frac{d\refphi_1}{dX}
\frac{h}{2} \dX
= \frac{2}{h}(-\half)\frac{2}{h}\half\frac{h}{2} \int_{-1}^1\dX
= -\frac{1}{h}
$$
For \( e=0 \) and \( =N_e \):
$$
\tilde A^{(e)} =\frac{1}{h}\left(\begin{array}{r}
1
\end{array}\right),\quad
\tilde b^{(e)} = h\left(\begin{array}{c}
1
\end{array}\right)
$$
Only one degree of freedom ("node") in these cells (\( r=0 \) counts the only dof)
4 P1 elements:
vertices = [0, 0.5, 1, 1.5, 2]
cells = [[0, 1], [1, 2], [2, 3], [3, 4]]
dof_map = [[0], [0, 1], [1, 2], [2]] # only 1 dof in elm 0, 3
Python code for the assembly algorithm:
# Ae[e][r,s]: element matrix, be[e][r]: element vector
# A[i,j]: coefficient matrix, b[i]: right-hand side
for e in range(len(Ae)):
for r in range(Ae[e].shape[0]):
for s in range(Ae[e].shape[1]):
A[dof_map[e,r],dof_map[e,s]] += Ae[e][i,j]
b[dof_map[e,r]] += be[e][i,j]
Result: same linear system as arose from computations in the physical domain
Define
The general formula for \( B \) is now
$$
\begin{equation*}
B(x) = \sum_{j\in\Ifb} U_j\basphi_j(x)
\end{equation*}
$$
Suppose we have a Dirichlet condition \( u(\xno{k})=U_k \), \( k\in\Ifb \):
$$
u(\xno{k}) = \sum_{j\in\Ifb} U_j\underbrace{\basphi_j(x_k)}_{\neq 0
\hbox{ only for }j=k} +
\sum_{j\in\If} c_j\underbrace{\basphi_{\nu(j)}(\xno{k})}_{=0,\ k\not\in\If}
= U_k $$
$$ -u''=2, \quad u(0)=C,\ u(L)=D $$
$$ \int_0^L u'v'\dx = \int_0^L2v\dx\quad\forall v\in V$$
$$ (u',v') = (2,v)\quad\forall v\in V$$
$$
\begin{equation*}
B(x) = \sum_{j\in\Ifb} U_j\basphi_j(x)
\end{equation*}
$$
Here \( \Ifb = \{0,N_n-1\} \), \( U_0=C \), \( U_{N_n-1}=D \); \( \baspsi_i \) are the internal \( \basphi_i \) functions:
$$ \baspsi_i = \basphi_{\nu(i)}, \quad \nu(i)=i+1,\quad i\in\If =
\{0,\ldots,N=N_n-3\} $$
$$
\begin{align*}
u(x) &= \underbrace{C\basphi_0 + D\basphi_{N_n-1}}_{B(x)}
+ \sum_{j\in\If} c_j\basphi_{j+1}\\
&= C\basphi_0 + D\basphi_{N_n-1} + c_0\basphi_1 + c_1\basphi_2 +\cdots
+ c_N\basphi_{N_n-2}
\end{align*}
$$
Insert \( u = B + \sum_j c_j\baspsi_j \) in variational formulation:
$$ (u',v') = (2,v)\quad\Rightarrow\quad (\sum_jc_j\baspsi_j',\baspsi_i')
= (2,\baspsi_i)-(B',\baspsi_i')\quad \forall v\in V$$
$$
\begin{align*}
A_{i-1,j-1} &= \int_0^L \basphi_i'(x)\basphi_j'(x) \dx\\
b_{i-1} &= \int_0^L (f(x)\basphi_i(x) -
B'(x)\basphi_i'(x))\dx,\quad B'(x)=C\basphi_{0}'(x) + D\basphi_{N_n-1}'(x)
\end{align*}
$$
for \( i,j = 1,\ldots,N+1=N_n-2 \).
New boundary terms from \( -\int B'\basphi_i'\dx \): add \( C/h \) to \( b_0 \) and \( D/h \) to \( b_N \)
From the first cell:
$$
\tilde b_0^{(1)} = \int_{-1}^1 \left(f\refphi_1 -
C\frac{2}{h}\frac{d\refphi_0}{dX}\frac{2}{h}\frac{d\refphi_1}{dX}\right)
\frac{h}{2} \dX = \frac{h}{2} 2\int_{-1}^1 \refphi_1 \dX
- C\frac{2}{h}(-\frac{1}{2})\frac{2}{h}\frac{1}{2}\frac{h}{2}\cdot 2
= h + C\frac{1}{h}\tp
$$
From the last cell:
$$
\tilde b_0^{(N_e)} = \int_{-1}^1 \left(f\refphi_0 -
D\frac{2}{h}\frac{d\refphi_1}{dX}\frac{2}{h}\frac{d\refphi_0}{dX}\right)
\frac{h}{2} \dX = \frac{h}{2} 2\int_{-1}^1 \refphi_0 \dX
- D\frac{2}{h}\frac{1}{2}\frac{2}{h}(-\frac{1}{2})\frac{h}{2}\cdot 2
= h + D\frac{1}{h}\tp
$$
Method 2: always choose \( \baspsi_i = \basphi_i \) for all \( i\in\If \) and set
$$
\begin{equation*}
u(x) = \sum_{j\in\If}c_j\basphi_j(x),\quad \If=\{0,\ldots,N=N_n-1\}
\end{equation*}
$$
\( u \) is treated as unknown at all boundaries when computing entries in the linear system
$$ -u''=2,\quad u(0)=0,\ u(L)=D$$
Assemble as if there were no Dirichlet conditions:
$$
\begin{equation*}
{
\frac{1}{h}\left(
\begin{array}{ccccccccc}
1 & -1 & 0
&\cdots &
\cdots & \cdots & \cdots &
\cdots & 0 \\
-1 & 2 & -1 & \ddots & & & & & \vdots \\
0 & -1 & 2 & -1 &
\ddots & & & & \vdots \\
\vdots & \ddots & & \ddots & \ddots & 0 & & & \vdots \\
\vdots & & \ddots & \ddots & \ddots & \ddots & \ddots & & \vdots \\
\vdots & & & 0 & -1 & 2 & -1 & \ddots & \vdots \\
\vdots & & & & \ddots & \ddots & \ddots &\ddots & 0 \\
\vdots & & & & &\ddots & \ddots &\ddots & -1 \\
0 &\cdots & \cdots &\cdots & \cdots & \cdots & 0 & -1 & 1
\end{array}
\right)
\left(
\begin{array}{c}
c_0 \\
\vdots\\
\vdots\\
\vdots \\
\vdots \\
\vdots \\
\vdots \\
\vdots\\
c_{N}
\end{array}
\right)
=
\left(
\begin{array}{c}
h \\
2h\\
\vdots\\
\vdots \\
\vdots \\
\vdots \\
\vdots \\
2h\\
h
\end{array}
\right)
}
\end{equation*}
$$
$$
\begin{equation*}
{
\frac{1}{h}\left(
\begin{array}{ccccccccc}
h & 0 & 0
&\cdots &
\cdots & \cdots & \cdots &
\cdots & 0 \\
-1 & 2 & -1 & \ddots & & & & & \vdots \\
0 & -1 & 2 & -1 &
\ddots & & & & \vdots \\
\vdots & \ddots & & \ddots & \ddots & 0 & & & \vdots \\
\vdots & & \ddots & \ddots & \ddots & \ddots & \ddots & & \vdots \\
\vdots & & & 0 & -1 & 2 & -1 & \ddots & \vdots \\
\vdots & & & & \ddots & \ddots & \ddots &\ddots & 0 \\
\vdots & & & & &\ddots & \ddots &\ddots & -1 \\
0 &\cdots & \cdots &\cdots & \cdots & \cdots & 0 & 0 & h
\end{array}
\right)
\left(
\begin{array}{c}
c_0 \\
\vdots\\
\vdots\\
\vdots \\
\vdots \\
\vdots \\
\vdots \\
\vdots\\
c_{N}
\end{array}
\right)
=
\left(
\begin{array}{c}
0 \\
2h\\
\vdots\\
\vdots \\
\vdots \\
\vdots \\
\vdots \\
2h\\
D
\end{array}
\right)
}
\end{equation*}
$$
In cell 0 we know \( u \) for local node (degree of freedom) \( r=0 \). Replace the first cell equation by \( \tilde c_0 = 0 \):
$$
\begin{equation*}
\tilde A^{(0)} =
A = \frac{1}{h}\left(\begin{array}{rr}
h & 0\\
-1 & 1
\end{array}\right),\quad
\tilde b^{(0)} = \left(\begin{array}{c}
0\\
h
\end{array}\right)
\end{equation*}
$$
In cell \( N_e \) we know \( u \) for local node \( r=1 \). Replace the last equation in the cell system by \( \tilde c_1=D \):
$$
\begin{equation*}
\tilde A^{(N_e)} =
A = \frac{1}{h}\left(\begin{array}{rr}
1 & -1\\
0 & h
\end{array}\right),\quad
\tilde b^{(N_e)} = \left(\begin{array}{c}
h\\
D
\end{array}\right)
\end{equation*}
$$
Algorithm for incorporating \( c_i=U_i \) in a symmetric way:
$$
\begin{equation*}
{
\frac{1}{h}\left(
\begin{array}{ccccccccc}
h & 0 & 0
&\cdots &
\cdots & \cdots & \cdots &
\cdots & 0 \\
0 & 2 & -1 & \ddots & & & & & \vdots \\
0 & -1 & 2 & -1 &
\ddots & & & & \vdots \\
\vdots & \ddots & & \ddots & \ddots & 0 & & & \vdots \\
\vdots & & \ddots & \ddots & \ddots & \ddots & \ddots & & \vdots \\
\vdots & & & 0 & -1 & 2 & -1 & \ddots & \vdots \\
\vdots & & & & \ddots & \ddots & \ddots &\ddots & 0 \\
\vdots & & & & &\ddots & \ddots &\ddots & 0 \\
0 &\cdots & \cdots &\cdots & \cdots & \cdots & 0 & 0 & h
\end{array}
\right)
\left(
\begin{array}{c}
c_0 \\
\vdots\\
\vdots\\
\vdots \\
\vdots \\
\vdots \\
\vdots \\
\vdots\\
c_{N}
\end{array}
\right)
=
\left(
\begin{array}{c}
0 \\
2h\\
\vdots\\
\vdots \\
\vdots \\
\vdots \\
\vdots \\
2h +\frac{D}{h}\\
D
\end{array}
\right)
}
\end{equation*}
$$
Symmetric modification applied to \( \tilde A^{(N_e)} \):
$$
\begin{equation*}
\tilde A^{(N_e)} =
A = \frac{1}{h}\left(\begin{array}{rr}
1 & 0\\
0 & h
\end{array}\right),\quad
\tilde b^{(N_e)} = \left(\begin{array}{c}
h + D/h\\
D
\end{array}\right)
\end{equation*}
$$
How can we incorporate \( u'(0)=C \) with finite elements?
$$ -u''=f,\quad u'(0)=C,\ u(L)=D$$
Galerkin's method:
$$
\begin{equation*}
\int_0^L(u''(x)+f(x))\baspsi_i(x) dx = 0,\quad i\in\If
\end{equation*}
$$
Integration of \( u''\baspsi_i \) by parts:
$$
\begin{equation*}
\int_0^Lu'(x)\baspsi_i'(x) \dx -(u'(L)\baspsi_i(L) - u'(0)\baspsi_i(0)) -
\int_0^L f(x)\baspsi_i(x) \dx =0
\end{equation*}
$$
$$
\begin{equation*}
\int_0^Lu'(x)\basphi_i'(x) dx =
\int_0^L f(x)\basphi_i(x) dx - C\basphi_i(0),\quad i\in\If
\end{equation*}
$$
$$
\begin{equation*}
\sum_{j=0}^{N}\left(
\int_0^L \basphi_i'\basphi_j' dx \right)c_j =
\int_0^L\left(f\basphi_i -D\basphi_N'\basphi_i\right) dx
- C\basphi_i(0)
\end{equation*}
$$
for \( i=0,\ldots,N=N_n-2 \).
We can therefore forget about the term \( u'(L)\basphi_i(L) \)!
Boundary terms \( u'\basphi_i \) at points \( \xno{i} \) where Dirichlet values apply can always be forgotten.
$$
\begin{equation*}
u(x) = \sum_{j=0}^{N=N_n-1} c_j\basphi_j(x)
\end{equation*}
$$
$$
\begin{equation*}
\sum_{j=0}^{N=N_n-1}\left(
\int_0^L \basphi_i'(x)\basphi_j'(x) dx \right)c_j =
\int_0^L f(x)\basphi_i(x) dx - C\basphi_i(0)
\end{equation*}
$$
Assemble entries for \( i=0,\ldots,N=N_n-1 \) and then modify the last equation to \( c_N=D \)
The extra term \( C\basphi_0(0) \) affects only the element vector from the first cell since \( \basphi_0=0 \) on all other cells.
$$
\begin{equation*}
\tilde A^{(0)} =
A = \frac{1}{h}\left(\begin{array}{rr}
1 & 1\\
-1 & 1
\end{array}\right),\quad
\tilde b^{(0)} = \left(\begin{array}{c}
h - C\\
h
\end{array}\right)
\end{equation*}
$$
The differential equation problem defines the integrals in the variational formulation.
Request these functions from the user:
integrand_lhs(phi, r, s, x)
boundary_lhs(phi, r, s, x)
integrand_rhs(phi, r, x)
boundary_rhs(phi, r, x)
Must also have a mesh with vertices, cells, and dof_map
<Declare global matrix, global rhs: A, b>
# Loop over all cells
for e in range(len(cells)):
# Compute element matrix and vector
n = len(dof_map[e]) # no of dofs in this element
h = vertices[cells[e][1]] - vertices[cells[e][0]]
<Declare element matrix, element vector: A_e, b_e>
# Integrate over the reference cell
points, weights = <numerical integration rule>
for X, w in zip(points, weights):
phi = <basis functions + derivatives at X>
detJ = h/2
x = <affine mapping from X>
for r in range(n):
for s in range(n):
A_e[r,s] += integrand_lhs(phi, r, s, x)*detJ*w
b_e[r] += integrand_rhs(phi, r, x)*detJ*w
# Add boundary terms
for r in range(n):
for s in range(n):
A_e[r,s] += boundary_lhs(phi, r, s, x)*detJ*w
b_e[r] += boundary_rhs(phi, r, x)*detJ*w
for e in range(len(cells)):
...
# Incorporate essential boundary conditions
for r in range(n):
global_dof = dof_map[e][r]
if global_dof in essbc_dofs:
# dof r is subject to an essential condition
value = essbc_docs[global_dof]
# Symmetric modification
b_e -= value*A_e[:,r]
A_e[r,:] = 0
A_e[:,r] = 0
A_e[r,r] = 1
b_e[r] = value
# Assemble
for r in range(n):
for s in range(n):
A[dof_map[e][r], dof_map[e][r]] += A_e[r,s]
b[dof_map[e][r] += b_e[r]
<solve linear system>
$$
\begin{equation*}
-\int_{\Omega} \nabla\cdot (\dfc(\x)\nabla u) v\dx =
\int_{\Omega} \dfc(\x)\nabla u\cdot\nabla v \dx -
\int_{\partial\Omega} a\frac{\partial u}{\partial n} v \ds
\end{equation*}
$$
$$
\begin{align*}
\v\cdot\nabla u + \beta u &= \nabla\cdot\left( \dfc\nabla u\right) + f,
\quad & \x\in\Omega\\
u &= u_0,\quad &\x\in\partial\Omega_D\\
-\dfc\frac{\partial u}{\partial n} &= g,\quad &\x\in\partial\Omega_N
\end{align*}
$$
Method 1 with boundary function and \( \baspsi_i=0 \) on \( \partial\Omega_D \) (ensures \( u=u_0 \) condition):
$$ u(\x) = B(\x) + \sum_{j\in\If} c_j\baspsi_j(\x),\quad B(\x)=u_0(\x) $$
Galerkin's method: multiply by \( v\in V \) and integrate over \( \Omega \),
$$
\int_{\Omega} (\v\cdot\nabla u + \beta u)v\dx =
\int_{\Omega} \nabla\cdot\left( \dfc\nabla u\right)v\dx + \int_{\Omega}fv \dx
$$
Integrate the second-order term by parts according to the formula:
$$
\int_{\Omega} \nabla\cdot\left( \dfc\nabla u\right) v \dx =
-\int_{\Omega} \dfc\nabla u\cdot\nabla v\dx
+ \int_{\partial\Omega} \dfc\frac{\partial u}{\partial n} v\ds,
$$
Galerkin's method then gives
$$
\int_{\Omega} (\v\cdot\nabla u + \beta u)v\dx =
-\int_{\Omega} \dfc\nabla u\cdot\nabla v\dx
+ \int_{\partial\Omega} \dfc\frac{\partial u}{\partial n} v\ds
+ \int_{\Omega} fv \dx
$$
Note: \( v\neq 0 \) only on \( \partial\Omega_N \) (since \( v=0 \) on \( \partial\Omega_D \)):
$$ \int_{\partial\Omega} \dfc\frac{\partial u}{\partial n} v\ds
= \int_{\partial\Omega_N} \underbrace{\dfc\frac{\partial u}{\partial n}}_{-g} v\ds
= -\int_{\partial\Omega_N} gv\ds
$$
The final variational form:
$$
\int_{\Omega} (\v\cdot\nabla u + \beta u)v\dx =
-\int_{\Omega} \dfc\nabla u\cdot\nabla v \dx
- \int_{\partial\Omega_N} g v\ds
+ \int_{\Omega} fv \dx
$$
Or with inner product notation:
$$
(\v\cdot\nabla u, v) + (\beta u,v) =
- (\dfc\nabla u,\nabla v) - (g,v)_{N} + (f,v)
$$
\( (g,v)_{N} \): line or surface integral over \( \partial\Omega_N \).
$$
A_{i,j} = (\v\cdot\nabla \baspsi_j, \baspsi_i) +
(\beta \baspsi_j ,\baspsi_i) + (\dfc\nabla \baspsi_j,\nabla \baspsi_i)
$$
$$
b_i = (g,\baspsi_i)_{N} + (f,\baspsi_i) -
(\v\cdot\nabla u_0, \baspsi_i) + (\beta u_0 ,\baspsi_i) +
(\dfc\nabla u_0,\nabla \baspsi_i)
$$
We want to compute an integral in the physical domain by integrating over the reference cell.
$$
\begin{equation*}
\int_{{\Omega}^{(e)}} \dfc(\x)\nabla\basphi_i\cdot\nabla\basphi_j\dx
\end{equation*}
$$
Mapping from reference to physical coordinates:
$$ \x(\X) $$
with Jacobian \( J \),
$$ J_{i,j}=\frac{\partial x_j}{\partial X_i} $$
Can derive
$$
\begin{align*}
\nabla_{\X}\refphi_r &= J\cdot\nabla_{\x}\basphi_i\\
\nabla_{\x}\basphi_i &= \nabla_{\x}\refphi_r(\X)
= J^{-1}\cdot\nabla_{\X}\refphi_r(\X)
\end{align*}
$$
Integral transformation from physical to reference coordinates:
$$
\begin{equation*}
\int_{\Omega^{(e)}} \dfc(\x)\nabla_{\x}\basphi_i\cdot\nabla_{\x}\basphi_j\dx =
\int_{\tilde\Omega^r} \dfc(\x(\X))(J^{-1}\cdot\nabla_{\X}\refphi_r)\cdot
(J^{-1}\cdot\nabla\refphi_s)\det J\dX
\end{equation*}
$$
Numerical integration over reference cell triangles and tetrahedra:
$$ \int_{\tilde\Omega^r} g\dX = \sum_{j=0}^{n-1} w_j g(\bar\X_j)$$
Module numint.py contains different rules:
>>> import numint
>>> x, w = numint.quadrature_for_triangles(num_points=3)
>>> x
[(0.16666666666666666, 0.16666666666666666),
(0.66666666666666666, 0.16666666666666666),
(0.16666666666666666, 0.66666666666666666)]
>>> w
[0.16666666666666666, 0.16666666666666666, 0.16666666666666666]