User:Eml5526.s11.team7.jb/work51

For Polynomial Basis
The corresponding family $${{\bar{F}}_{P}}=\left\{ {{{\bar{b}}}_{j}} \right\}$$ of polynomial basis function satisfying the Constraint Breaking Solution (CBS) for $$\Gamma_g = \left \{ 0 \right \}$$ is:

$${{F}_{P}}=\left\{ {{\left( x \right)}^{j}},\text{ }j=0,1,2,\ldots \right\}$$

Now that the family of basis functions are determined, we recall that the discrete weak form is defined by

$$ \mathbf{\tilde{d}} \cdot \mathbf{b}( \beta ) = g \qquad \text{and} \qquad \mathbf{\tilde{c}} \cdot \left[ \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} + \mathbf{\tilde{K}} \mathbf{\tilde{d}} - \mathbf{\tilde{F}} \right] = 0 $$

where the expression $$\displaystyle \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} = 0 $$ for the static case. Then the discrete weak form is simplified to:

\mathbf{K} \mathbf{d} = \mathbf{F} \qquad \Rightarrow \qquad \mathbf{K} \mathbf{d} = \mathbf{F}_F - \mathbf{K}_{FE} g. $$

The first constant, $$\displaystyle d_0$$, is determined from the formulation of the CBS. It is known that:


 * $$\displaystyle u^h(\beta) = d_0 b_0(\beta)=g$$


 * $$\displaystyle d_0= \frac{g}{b_0(\beta)}=4/1=4$$

Now the matrices used to determine the rest of the coefficients are determined as follows. Note that the number of terms needed in the approximate solution, $$\displaystyle n$$, is determined by the convergence interval given. The first few terms of each matrix computation is shown. A Matlab script was written to iterate these computation for increasing values of $$\displaystyle n$$ until the desired accuracy was reached.



\displaystyle

{F_F} = \left\{ {{F_i},i = 1,2,...,n} \right\} = \left\{ {\tilde f\left( \right),i = 1,2,...,n} \right\}

$$



\displaystyle

\tilde f({b_i}) = {b_i}(\alpha )h + \int\limits_\alpha ^\beta {{b_i}fdx}

$$



\displaystyle

h = {a_2}(\alpha )\frac = (2+3\alpha)\frac = 5*12/5=12

$$



\displaystyle

\tilde f({b_1}) = {b_1}(1)h + \int\limits_0^1 {{b_1}(5x)dx}=1*12+\int\limits_0^1  {{(x)}(5x)dx}=41/3

$$



\displaystyle

\tilde f({b_2}) = {b_2}(1)h + \int\limits_0^1 {{b_2}(5x)dx}=1*12 + \int\limits_0^1  {{(x)^2}(5x)dx}=53/4

$$

$$ \displaystyle

\therefore {F_F} = \left[ {\begin{array}{ccccccccccccccc} {41/3} \\    {53/4} \\   {.} \\   {.} \\  {.} \end{array}} \right]

$$



\displaystyle

{K_{FE}} = \left\{ {{K_{0j}};j = 1,...,n} \right\} = \left\{ {\tilde k\left( {{b_0},{b_j}} \right),j = 1,...,n} \right\}

$$



\displaystyle

\tilde k({b_i},{b_j}) = \int\limits_0^1 {b{'_i}{a_2}b{'_j}dx}

$$



\displaystyle

\tilde k({b_0},{b_1}) = \int\limits_0^1 {b{'_0}{a_2}b{'_1}dx} = \int\limits_0^1 {b{'_0}(2+3x)b{'_1}dx}  = 0 = \tilde k({b_0},{b_2}) = \tilde k({b_0},{b_3}) = 0

$$



\displaystyle

\therefore {K_{EF}} = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   . \\ .\\ . \end{array}} \right]

$$

And：

\displaystyle

\because {K_{FE}} = {K_{EF}}^T = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   . \\ .\\ . \end{array}} \right]

$$



\displaystyle

{K_{FF}} = \left[ {{K_{ij}};i,j = 1,2,...} \right] = \tilde k({b_i},{b_j})

$$



\displaystyle

\tilde k({b_1},{b_1}) = \int\limits_0^1 {b{'_1}{2+3x}b{'_1}dx} = \int\limits_0^1 {1*(2+3x)*1dx}=7/2

$$



\displaystyle

\tilde k({b_1},{b_2}) = \int\limits_0^1 {b{'_1}{2+3x}b{'_2}dx} = \int\limits_0^1 {1*(2+3x)*2(x)dx}=4

$$



\displaystyle

\tilde k({b_2},{b_1}) = \int\limits_0^1 {b{'_2}{2+3x}b{'_1}dx} = \int\limits_0^1 {2(x)*(2+3x)*1dx}=4

$$



\displaystyle

\tilde k({b_2},{b_2}) = \int\limits_0^1 {b{'_2}{2+3x}b{'_2}dx} = \int\limits_0^1 {2(x)*(2+3x)*2(x)dx}=17/3

$$

\displaystyle \therefore {K_{FF}} = \begin{bmatrix} 7/2&4 &.&. \\  4&17/3  & .&.\\ . &.  & .&.\\ . &.  & .&. \end{bmatrix} $$

The matrices are then solved for the coefficients:

\displaystyle

Kd = F $$



\displaystyle

F = {F_F} - K_{FE}g = {F_F} $$



\displaystyle

K_{FF}d_F=F_F $$



\displaystyle

d_F=K_{FF}^{-1} F_F $$

The approximate solution, $$\displaystyle u^h(x)$$ is defined as,


 * $$\displaystyle u^h(x)= d_0b_0+\sum_{j=1}^{n}d_jb_j(x) \qquad \text{where} \quad d_0=4 \ \text{and} \ b_0=1$$.

As stated above a Matlab script was written to iterate $$\displaystyle n$$ until the desired accuracy was reached. At $$\displaystyle n=6$$ the error was $$\displaystyle  O(3.5840*10^{-6})$$. The following plots were generated: