User:EML5526.S11.Team3.vnarayanan/Homework2

Problem 2.7
Consider F = {$$1,x,x^{2},x^{3},x^{4}$$} in the interval $$ \Omega=[0,1]$$ (2.7.1) Construct $$\underline{\Gamma }$$(F) (2.7.2) Find the determinant of $$\underline{\Gamma }$$(F) (2.7.3) Is F Orthogonal Family?

Solution(2.7.1)
F = {$$1,x,x^2,x^3,x^4$$} = {$$b_{1}(x),b_{2}(x),b_{3}(x),b_{4}(x),b_{5}(x)$$} $$\Gamma = \begin{bmatrix} \Gamma_{11}&\Gamma_{12}&\Gamma_{13} &\Gamma_{14}  &\Gamma_{15} \\ \Gamma_{21}&\Gamma_{22}&\Gamma_{23} &\Gamma_{24}  &\Gamma_{25} \\ \Gamma_{31}&\Gamma_{32}&\Gamma_{33} &\Gamma_{34}  &\Gamma_{35} \\ \Gamma_{41}&\Gamma_{42}&\Gamma_{43} &\Gamma_{44}  &\Gamma_{45} \\ \Gamma_{51}&\Gamma_{52}&\Gamma_{53} &\Gamma_{54}  &\Gamma_{55} \end{bmatrix}$$

$$\;\Gamma_{ij}=\int_{\Omega}b_{i}(x)b_{j}(x)dx$$ $$\Gamma_{11} =  = \int_{0}^{1}1 dx = 1$$ $$\Gamma_{12} =  = \int_{0}^{1}1.x dx =\frac{1}{2} $$ $$\Gamma_{13} =  = \int_{0}^{1}1.x^2 dx =\frac{1}{3} $$ $$\Gamma_{14} =  = \int_{0}^{1}1.x^3 dx =\frac{1}{4} $$ $$\Gamma_{15} =  = \int_{0}^{1}1.x^4 dx =\frac{1}{5} $$ $$\Gamma_{21} =  = \int_{0}^{1}x.1 dx =\frac{1}{2} $$ $$\Gamma_{22} =  = \int_{0}^{1}x.x dx =\frac{1}{3} $$ $$\Gamma_{23} =  = \int_{0}^{1}x.x^2 dx =\frac{1}{4}$$ $$\Gamma_{24} =  = \int_{0}^{1}x.x^3dx =\frac{1}{5} $$ $$\Gamma_{25} =  = \int_{0}^{1}x.x^4dx =\frac{1}{6} $$ $$\Gamma_{31} =  = \int_{0}^{1}x^2.1 dx =\frac{1}{3}$$ $$\Gamma_{32} =  = \int_{0}^{1}x^2.x dx =\frac{1}{4}$$ $$\Gamma_{33} =  = \int_{0}^{1}x^2.x^2 dx =\frac{1}{5} $$ $$\Gamma_{34} =  = \int_{0}^{1}x^2.x^3 dx =\frac{1}{6} $$ $$\Gamma_{35} =  = \int_{0}^{1}x^2.x^4 dx =\frac{1}{7} $$ $$\Gamma_{41} =  = \int_{0}^{1}x^3.1 dx =\frac{1}{4} $$ $$\Gamma_{42} = <b_{4}(x),b_{2}(x)> = \int_{0}^{1}x^3.x dx =\frac{1}{5} $$ $$\Gamma_{43} = <b_{4}(x),b_{3}(x)> = \int_{0}^{1}x^3.x^2 dx =\frac{1}{6}$$ $$\Gamma_{44} = <b_{4}(x),b_{4}(x)> = \int_{0}^{1}x^3.x^3 dx =\frac{1}{7} $$ $$\Gamma_{45} = <b_{4}(x),b_{5}(x)> = \int_{0}^{1}x^3.x^4 dx =\frac{1}{8} $$ $$\Gamma_{51} = <b_{5}(x),b_{1}(x)> = \int_{0}^{1}x^4.1 dx =\frac{1}{5} $$ $$\Gamma_{52} = <b_{5}(x),b_{2}(x)> = \int_{0}^{1}x^4.x dx =\frac{1}{6} $$ $$\Gamma_{53} = <b_{5}(x),b_{3}(x)> = \int_{0}^{1}x^4.x^2 dx =\frac{1}{7} $$ $$\Gamma_{54} = <b_{5}(x),b_{4}(x)> = \int_{0}^{1}x^4.x^3 dx =\frac{1}{8} $$ $$\Gamma_{55} = <b_{5}(x),b_{5}(x)> = \int_{0}^{1}x^4.x^4 dx =\frac{1}{9} $$

$$ \underline{\Gamma}(F) = \begin{bmatrix}

1 &\frac{1}{2} & \frac{1}{3}  & \frac{1}{4}  &\frac{1}{5} \\

\frac{1}{2} & \frac{1}{3} & \frac{1}{4}  & \frac{1}{5}  & \frac{1}{6} \\

\frac{1}{3} &\frac{1}{4} &\frac{1}{5}  &\frac{1}{6}  &\frac{1}{7} \\

\frac{1}{4} & \frac{1}{5} &\frac{1}{6} &\frac{1}{7}  &\frac{1}{8} \\

\frac{1}{5} &\frac{1}{6} &\frac{1}{7}  &\frac{1}{8}  &\frac{1}{9} \end{bmatrix}$$

Solution(2.7.2)
The determinant of $$\underline{\Gamma}(F)\;\;$$ is

$$\left | \underline{\Gamma}(F) \right | = 3.74\times 10^{-12}$$

Solution(2.7.3)
Consider any two basis functions. Say, $$ b_{i}$$ and $$b_{j}$$. If $$\Gamma_{ij} = <b_{i},b_{j}> = \delta_{ij} $$, then F is a family of orthogonal basis of functions. In other words, if $$\;\;\;\delta_{ij} = 1 \;for\; i=j\;$$  or if $$\;\;\; \delta_{ij} = 0\; for\; i\neq j$$, then F is a family of orthogonal basis of functions.

But in the $$\underline{\Gamma}(F)$$ matrix of Solution (2.7.1) , we find none of the elements satisfying the above condition. Hence, F is not a family of orthogonal basis of functions.

Problem 2.8

 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \int_{\Omega}w^{h}(x)P(u^{h}(x))dx = 0 \;\; \forall\; w^{h}(x)\;\;$$ where $$\;w^{h}(x) = \sum_{i=1}^{n}c_{i}b_{i}(x)\;$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)


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 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \int_{\Omega}b_{i}(x)P(u^{h}(x))dx = 0 \;\; for\;\; i = 1,2,...,n$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)


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Show that Eq.1 $$\;\Leftrightarrow\; $$ Eq.2

Part 1
Eq.1 $$\;\Rightarrow\;\;\int_{\Omega}w^{h}(x)P(u^{h}(x))dx = 0 \;\; \forall\;\; w^{h}(x)$$ where $$\;\;w^{h}(x) = \sum_{i=1}^{n}c_{i}b_{i}(x)$$

Case (1) : $$\;\; c_{1} = 1, c_{2}=c_{3}=...=c_{n} = 0$$

$$\therefore\;\; w^{h}(x) = b_{1}(x) $$

Hence, Eq.1 becomes
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \;\;\int_{\Omega}b_{1}(x)P(u^{h}(x))dx = 0$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)


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Case (2) : $$\;\; c_{2}=1,c_{1} = c_{3}=...=c_{n} = 0$$

$$\therefore\;\; w^{h}(x) = b_{2}(x) $$

Hence, Eq.1 becomes
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \;\;\int_{\Omega}b_{2}(x)P(u^{h}(x))dx = 0$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)


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Case (3) : $$\;\; c_{1} =c_{2}= c_{3}=...=c_{n-1} = 0,c_{n}=1,$$

$$\therefore\;\; w^{h}(x) = b_{n}(x) $$

Hence, Eq.1 becomes
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \;\;\int_{\Omega}b_{n}(x)P(u^{h}(x))dx = 0$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)


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 * }

Hence, from Eq.(3), Eq.(4) and Eq.(5) it can be concluded that


 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \int_{\Omega}b_{i}(x)P(u^{h}(x))dx = 0 \;\; for\;\; i = 1,2,...,n$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)


 * }
 * }

Part 2
Eq.(2) $$\; \Rightarrow\;\int_{\Omega}b_{i}(x)P(u^{h}(x))dx = 0 \;\; for\;\; i = 1,2,...,n $$

Multiply $$\;\;c_{i}\;\;$$ on both sides of the Eq.(2). We get, $$\;\int_{\Omega}c_{i}b_{i}(x)P(u^{h}(x))dx = 0 \;\; for\;\; i = 1,2,...,n$$

For, $$\; i = 1\;$$ $$\;\int_{\Omega}c_{1}b_{1}(x)P(u^{h}(x))dx = 0 $$

For, $$\; i = 2\;$$ $$\;\int_{\Omega}c_{n}b_{n}(x)P(u^{h}(x))dx = 0 $$

For, $$\; i = n\;$$ $$\;\int_{\Omega}c_{n}b_{n}(x)P(u^{h}(x))dx = 0 $$

Since, $$\;\int_{\Omega}c_{i}b_{i}(x)P(u^{h}(x))dx = 0 \;\; for\;\; i = 1,2,...,n\;\;$$ individually, the sum of all terms should also be equal to 0 $$\therefore\;\;\int_{\Omega}c_{1}b_{1}(x)P(u^{h}(x))dx + \int_{\Omega}c_{2}b_{2}(x)P(u^{h}(x))dx + .\; .\;. + \int_{\Omega}c_{n}b_{n}(x)P(u^{h}(x))dx = 0$$

$$\Rightarrow\;\;\int_{\Omega}\left [ (c_{1}b_{1}(x)P(u^{h}(x))) + (c_{2}b_{2}(x)P(u^{h}(x)))+ .\; .\;. + (c_{n}b_{n}(x)P(u^{h}(x)))\right ]dx = 0$$

$$\Rightarrow\;\;\int_{\Omega}\left [ c_{1}b_{1}(x) + c_{2}b_{2}(x)+ .\; .\;. + c_{n}b_{n}(x)\right ]P(u^{h}(x))dx = 0$$
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \Rightarrow\;\;\int_{\Omega}\sum_{i=1}^{n}c_{i}b_{i}(x)P(u^{h}(x))dx = 0$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)


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We know that, $$\;w^{h}(x) = \sum_{i=1}^{n}c_{i}b_{i}(x)\;$$

$$\therefore\; Eq.(7) $$ becomes,


 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \int_{\Omega}w^{h}(x)P(u^{h}(x))dx = 0$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)


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 * }