User:Eml5526.s11.t1.ab/HW2

=Problem 2=

Given
$$\left [B _{jk} \right ]=\begin{bmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{bmatrix}$$

$$\overset{v}{\rightarrow}=5\overset{a_{1}}{\rightarrow}-7\overset{a_{2}}{\rightarrow}-4\overset{a_{3}}{\rightarrow}$$ $$\displaystyle (Eq. 2.1) $$

Find
1) $$ Det\left [B _{jk} \right ]$$.

2) $$Find \Gamma (B_{1},B_{3},B_{3})=K$$.

3) $$Find \overset{F}{\rightarrow}=\left \{ F_{i} \right \}=\left \{ b_{i}.v \right \}$$.

4) Solve $$\displaystyle (Eq. 5) $$ on P 7-2 for $$\overset{d}{\rightarrow}=\left \{v _{j} \right \}$$

5) Use 1) on P 7-4 to find $$\vec{\overset{k}{\rightarrow}}\overset{d}{\rightarrow}=\vec{\overset{F}{\rightarrow}}$$ What is $$\vec{K}$$ & $$\vec{F}$$?

6) Solve for $$\overset{d}{\rightarrow}_{i}$$ compare to $$\overset{d}{\rightarrow}$$ in 4).

7) Observe the symmetric properties of $$\overset{K}{\rightarrow}$$ & $$\overset{\vec{K}}{\rightarrow}$$. Further Discuss the Pros & Cons of Bubnov-Galerkin & Petrov-Galerkin Methods.

1)
$$ Det\left [B _{jk} \right ]$$=

$$\begin{vmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{vmatrix}=-8.$$.

2)
$$Find \Gamma (B_{1},B_{3},B_{3})=\overset{K}{\rightarrow}$$.

According to the $$\displaystyle (Eq. no 4) $$ on P 7-1, $$B_{1},B_{3},B_{3}$$ given before we have,

$$\overset{b_{j}}{\rightarrow}.d_{j}=\overset{V}{\rightarrow}$$

Successively performing multiplication by $$B_{1},B_{3},B_{3}$$  we get,

$$d_{1}\overset{b_{1}}{\rightarrow}+d_{2}\overset{b_{2}}{\rightarrow}+d_{3}\overset{b_{3}}{\rightarrow}=\overset{V}{\rightarrow}$$ $$\displaystyle (Eq. 2.2) $$

Substituting the given values we get,

$$d_{1}\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}+d_{2}\begin{bmatrix} 2\\ -1\\ 3 \end{bmatrix}+d_{3}\begin{bmatrix} 3\\ 2\\ 6 \end{bmatrix}=\begin{bmatrix} 5\\ -7\\ -4 \end{bmatrix}$$

This equation can also be written in the following format,

$$\begin{bmatrix} 1 &2 &3 \\ 1 &-1  &2 \\ 1 & 3 & 6 \end{bmatrix}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{pmatrix} 5\\ -7\\ 4 \end{pmatrix}$$

$$\displaystyle (Eq. 3) $$ on P 7-2 is obtained by multiplying $$\displaystyle (Eq. 2.2)  $$ by each $$\overset{b}{\rightarrow}$$, so we can write,

$$\overset{b_{1}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{b_{1}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{b_{1}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{b_{1}}{\rightarrow}.\overset{V}{\rightarrow}==F_{1}$$

$$\overset{b_{2}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{b_{2}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{b_{2}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{b_{2}}{\rightarrow}.\overset{V}{\rightarrow}==F_{2}$$

$$\overset{b_{3}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{b_{3}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{b_{3}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{b_{2}}{\rightarrow}.\overset{V}{\rightarrow}==F_{3}$$

These 3 equations can be further presented as below in the matrix form,

$$\begin{bmatrix} \overset{b_{1}}{\rightarrow}.\overset{b_{1}}{\rightarrow} & \overset{b_{1}}{\rightarrow}.\overset{b_{2}}{\rightarrow} & \overset{b_{1}}{\rightarrow}.\overset{b_{3}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}.\overset{b_{1}}{\rightarrow} & \overset{b_{2}}{\rightarrow}.\overset{b_{2}}{\rightarrow} & \overset{b_{2}}{\rightarrow}.\overset{b_{3}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow}.\overset{b_{1}}{\rightarrow} & \overset{b_{3}}{\rightarrow}.\overset{b_{2}}{\rightarrow} & \overset{b_{3}}{\rightarrow}.\overset{b_{3}}{\rightarrow} \end{bmatrix}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} b_{1}\\ b_{2}\\ b_{3} \end{bmatrix}\begin{pmatrix} V_{1}\\ V_{2}\\ V_{3} \end{pmatrix}$$

Which also can be written as

$$\begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix} \begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix}^{T}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix}\begin{pmatrix} V_{1}\\ V_{2}\\ V_{3} \end{pmatrix}=\overset{F}{\rightarrow}$$

Substituting the numerical values we get,

$$\begin{bmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{bmatrix}\begin{bmatrix} 1 &2 &3 \\ 1 &-1 &2\\ 1 &3  &6 \end{bmatrix}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{bmatrix}\begin{bmatrix} 5\\ -7\\ -4 \end{bmatrix}$$

Performing the product, we get:

$$\begin{bmatrix} 3 &4 &11\\ 4 &14  &22\\ 11 &22  &49 \end{bmatrix}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{bmatrix}\begin{bmatrix} 5\\ -7\\ -4 \end{bmatrix}=\overset{F}{\rightarrow}$$ $$\displaystyle (Eq. 2.3) $$

This $$\displaystyle (Eq. 2.3) $$ is equivalent to,

$$Kd=F$$, where,

$$K=\begin{bmatrix} 3 &4 &11 \\ 4 &14 &22 \\ 11&22 &49 \end{bmatrix}=Stiffness Matrix$$

Furthermore,

$$Det[K]=\begin{vmatrix} 3 &4 &11\\ 4 &14 &22 \\ 11 &22 &49 \end{vmatrix}= 64$$

3)
$$\overset{F}{\rightarrow}=\left \{ F_{i} \right \}=\left \{ b_{i}.v \right \}$$=

From $$\displaystyle (Eq. 2.3) $$

$$\overset{F}{\rightarrow}=\begin{bmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{bmatrix}\begin{bmatrix} 5\\ -7\\ -4 \end{bmatrix}=\begin{pmatrix} -6\\ 5\\ -23 \end{pmatrix}$$

4)
Solve $$\displaystyle (Eq. 5) $$ on P 7-2 for $$\overset{d}{\rightarrow}=\left \{v _{j} \right \}$$

Rearranging $$\displaystyle (Eq. 2.3) $$, We get,

$$d=K^{-1}F$$

$$d=\begin{bmatrix} 3 &4 &11 \\ 4 &14  &22 \\ 11 &22  &49 \end{bmatrix}^{-1}\begin{pmatrix} -6\\ 5\\ -23 \end{pmatrix}=\begin{pmatrix} 8.3750\\ 5.6250\\ -4.875 \end{pmatrix}$$

5)
Use 1) on P 7-4 to find $$\vec{\overset{k}{\rightarrow}}\overset{d}{\rightarrow}=\vec{\overset{F}{\rightarrow}}$$. Further, What is $$\vec{K}$$ & $$\vec{F}$$?

In this case, $$w_{i}=a_{i}$$, where $$a_{i}$$ is the orthonormal basis.

Starting with Equation no 1 on P 7-2, we get:

$$d_{1}\overset{b_{1}}{\rightarrow}+d_{2}\overset{b_{2}}{\rightarrow}+d_{3}\overset{b_{3}}{\rightarrow}=\overset{V}{\rightarrow}$$

If we multiply successively by $$a_{i}$$, we get the following equations,

$$\overset{a_{1}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{a_{1}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{a_{1}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{a_{1}}{\rightarrow}.\overset{V}{\rightarrow}=F_{1}$$

$$\overset{a_{2}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{a_{2}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{a_{2}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{a_{2}}{\rightarrow}.\overset{V}{\rightarrow}=F_{2}$$

$$\overset{a_{3}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{a_{3}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{a_{3}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{a_{3}}{\rightarrow}.\overset{V}{\rightarrow}=F_{3}$$ $$\displaystyle (Eq. 2.4) $$

Which can be further written as,

$$\begin{bmatrix} \overset{a_{1}}{\rightarrow}\\ \overset{a_{2}}{\rightarrow}\\ \overset{a_{3}}{\rightarrow} \end{bmatrix}\begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix}^{T}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} \overset{a_{1}}{\rightarrow}\\ \overset{a_{2}}{\rightarrow}\\ \overset{a_{3}}{\rightarrow} \end{bmatrix}\begin{pmatrix} V_{1}\\ V_{2}\\ V_{3} \end{pmatrix}=\overset{\vec{F}}{\rightarrow}$$

As $$ a_{i}$$ is orthonormal basis, we can write the following,

$$a_{1}=\begin{bmatrix} 1 & 0 &0 \end{bmatrix};a_{2}=\begin{bmatrix} 0&1&0 \end{bmatrix};a_{3}=\begin{bmatrix} 0& 0 &1 \end{bmatrix}$$

Substituting the respective values we have,

$$\begin{bmatrix} 1 &0 &0 \\ 0 &1  &0 \\ 0 &0  &1 \end{bmatrix}\begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix}^{T}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0& 1 \end{bmatrix}\begin{pmatrix} V_{1}\\ V_{2}\\ V_{3} \end{pmatrix}=\overset{\vec{F}}{\rightarrow}$$

Simplifying we obtain that,

$$\begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix}^{T}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{pmatrix} V_{1}\\ V_{2}\\ V_{3} \end{pmatrix}=\overset{\vec{F}}{\rightarrow}$$

Replacing the numerical values we get,

$$\begin{bmatrix} 1 & 2 & 3\\ 1 &-1 &2 \\ 1 &3  & 6 \end{bmatrix}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{pmatrix} 5\\ -7\\ -4 \end{pmatrix}$$ $$\displaystyle (Eq. 2.5) $$

This equation is nothing but the system

$$\overset{\vec{K}}{\rightarrow}\overset{d}{\rightarrow}=\overset{\vec{F}}{\rightarrow}$$

From where we can conclude that $$\overset{\vec{K}}{\rightarrow}$$ is a matrix whose columns are the vectors $$\overset{b_{j}}{\rightarrow}$$ and $$\overset{\vec{F}}{\rightarrow}= V$$.

6)
Solve for $$\overset{d}{\rightarrow}_{i}$$ compare to $$\overset{d}{\rightarrow}$$ in question 4).

$$\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} 1 & 2 & 3\\ 1 &-1 &2 \\ 1 &3  & 6 \end{bmatrix}^{-1}\begin{pmatrix} 5\\ -7\\ -4 \end{pmatrix}=\begin{pmatrix} 8.3750\\ 5.6250\\ -4.875 \end{pmatrix}$$

We can see that this result is identical to the result we obtained in 4).

7)
Observe the symmetric properties of $$\overset{K}{\rightarrow}$$ & $$\overset{\vec{K}}{\rightarrow}$$. Further Discuss the Pros & Cons of Bubnov-Galerkin & Petrov-Galerkin Methods.

We can explicitely see that $$\overset{K}{\rightarrow}$$  is always symmetric because it is obtained from the product of $$\overset{b_{j}}{\rightarrow}.\overset{b_{i}}{\rightarrow}$$; where as $$\overset{\vec{K}}{\rightarrow}$$ is not always symmetric. It will be symmetric only when the vector b is a Orthonormal basis.

The pros of Bubnov-Galerkin method is the symmetry of the stiffness matrix which is generally easy to solve, however more calculations are necessary to get the stiffness matrix. Whereas, in the Petrov-Galerking method, obtaining the stiffness matrix is easier but solving the system on other hand can be more difficult owing to its non-symmetry.

Contributing Members

 * Abhijeet Bhalerao

=Problem 3= Show that the Equation no 1 P.8-1 which states $$\overset{w}{\rightarrow}.\overset{P(v)}{\rightarrow}=0$$

$$\forall w=\in R^{n}$$ $$\displaystyle (Eq. 3.1) $$

is equivalent to

$$\overset{w}{\rightarrow}.\overset{P(v)}{\rightarrow}=0$$ where $$\overset{w}{\rightarrow}=\sum \beta _{i}.a_{i}$$

$$\forall (\beta _{1},\beta _{2}..\beta _{n})\in R^{n}$$ $$\displaystyle (Eq. 3.2) $$

Given
$$\overset{w}{\rightarrow}.\overset{P(v)}{\rightarrow}=0$$

$$\forall w\in R^{n}$$ $$\displaystyle (Eq. 3.1) $$

Find
Show that $$\displaystyle (Eq. 3.1) $$ is equivalent to $$\displaystyle (Eq. 3.2) $$

Solution
For avoiding confusion we are designating the weight function as $$\overset{W}{\rightarrow}$$ in $$\displaystyle (Eq. 3.1) $$ and the weight function as $$\overset{w}{\rightarrow}$$ in $$\displaystyle (Eq. 3.2)  $$.

Considering $$\displaystyle (Eq. 3.2) $$ and Substituting for $$\overset{W}{\rightarrow}$$

$$\sum \beta _{i}a_{i}.\overset{P(v)}{\rightarrow}=0$$

$$\forall (\beta _{1},\beta _{2}..\beta _{n})\in R^{n}$$ $$\displaystyle (Eq. 3.3) $$

As $$a_{i}$$ is an orthonormal basis, performing a successive product by $$a_{i}$$ in the $$\displaystyle (Eq. 3.3) $$, we get

$$\begin{bmatrix} 1 &0 &0  &....0 \end{bmatrix}\beta _{1}.\overset{P(v)}{\rightarrow}=0$$

In the similar way

$$\begin{bmatrix} 0 &1 &0  &....0 \end{bmatrix}\beta _{2}.\overset{P(v)}{\rightarrow}=0$$

.

.

.

$$\begin{bmatrix} 0 &0 &0  &....1 \end{bmatrix}\beta _{n}.\overset{P(v)}{\rightarrow}=0$$;

Furthermore all the equations can be combined and written as

$$ \begin{bmatrix} 1 &0 &...0 \\ 0 &1  &...0 \\ 0 &0  &...1 \end{bmatrix}\bigl(\begin{smallmatrix} \beta_{1}\\ .\\ .\\ \beta_{n} \end{smallmatrix}\bigr).\overset{P(v)}{\rightarrow}=\bigl(\begin{smallmatrix} 0\\ .\\ .\\ 0 \end{smallmatrix}\bigr) ... \forall (\beta _{1},\beta _{2}..\beta _{n})\in R^{n}$$ $$\displaystyle (Eq. 3.4) $$

We can cleraly observe that the Matrix on the left hand side is the identity matrix $$\begin{bmatrix} I \end{bmatrix}$$; Moreover we have also written $$(\beta _{1},\beta _{2}..\beta _{n})$$ as a vector which further can be called as $$\overset{\beta}{\rightarrow}$$. and $$\displaystyle (Eq. 3.4) $$ becomes

$$\overset{\beta}{\rightarrow}.\overset{P(v)}{\rightarrow}=0; \forall \overset{\beta}{\rightarrow} \in R^{n}$$

Which is Identical to $$\displaystyle (Eq. 3.4) $$.

Contributing Members

 * Abhijeet Bhalerao