User:EML5526.S11.Team3.risher/homework2/solutions

Problem 2.1
Energy balance for a control volume can be written as:

heat generated + heat flow in - heat flow out = 0

$$ s(x+\Delta x/2)\Delta x + q(x)A(x) - q(x+\Delta x)(A(x+\Delta x) $$

Balance of heat: heat flow into a control volume (H1) + heat generated (H2) = heat due to change in temperature (H3)

$$H1+H2=H3$$

Heat due to change in temperature: $$H3 = A(x) \rho(x) c\frac{\partial u}{\partial t}$$

Rearranging H1 and H2 we get:

$$ \frac{ q(x+\Delta x)(A(x+\Delta x)-q(x)A(x)}{\Delta x} + s(x+\Delta x/2) = A(x)\rho(x) c\frac{\partial u}{\partial t}$$

If we take the limit of as $$\Delta x$$ → 0, then we get:

$$ \frac{\partial (qA)}{\partial x}+s =A(x)\rho(x) c\frac{\partial u}{\partial t}$$

Using Fourier's Law,

$$ q = k \frac{\partial u}{\partial x} $$

This becomes:

$$ \frac{\partial}{\partial x}\left[ A(x)k(x) \frac{\partial u}{\partial x}\right]+f(x,t)=-A(x)\rho(x) c\frac{\partial u}{\partial t} $$

Problem 2.2
(1)

$$ det\left[ b_{jk} \right]= det\left[\begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 &6 \\ \end{matrix}\right]= -(-1*1*3)-(1*3*2)-(1*2*6)+(-1*1*6)+(1*3*3)+(1*2*2) = -8 $$

$$ det\left[\begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 &6 \\ \end{matrix}\right]=-8 $$

(2)

$$ \underline{b}_{j} = b_{jk}\underline{a}_{k} $$

$$ \underline{ \Gamma } (\underline{b}_{1},\underline{b}_{2},\underline{b}_{3}) = \underline{K} = K_{ij}=\underline{b}_{i}.\underline{b}_{j}$$

$$ \underline{b}_{i}.\underline{b}_{j}=\underline{b}_{i}.\underline{b}_{i}^T = \left[\begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & 6 \\ \end{matrix}\right]$$ $$\left[\begin{matrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ 1 & 3 & 6 \\ \end{matrix}\right]$$ $$=\left[\begin{matrix} 3 & 4 & 11 \\ 4 & 14 & 22 \\ 11 & 22 & 49 \\ \end{matrix}\right]$$

$$ \underline{ \Gamma } (\underline{b}_{1},\underline{b}_{2},\underline{b}_{3}) = \underline{K} = \left[\begin{matrix} 3 & 4 & 11 \\ 4 & 14 & 22 \\ 11 & 22 & 49 \\ \end{matrix}\right]$$

$$det \underline{ \Gamma }= 64 $$

(3)

$$\underline{F}=\underline{b}_{i}*\underline{v}$$

$$\underline{F}= \left[\begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & 6 \\ \end{matrix}\right]$$ $$ \left[\begin{matrix} 5 \\ -7  \\ -4  \\ \end{matrix}\right]$$

$$\underline{F}= \left[\begin{matrix} -6 \\ 5 \\ -23 \\ \end{matrix}\right]$$

(4)

$$ \underline{K} \underline{d} = \underline{F} $$

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

$$\underline{d}= \left[\begin{matrix} 3 & 4 & 11 \\ 4 & 14 & 22 \\ 11 & 22 & 49 \\ \end{matrix}\right]^{-1}$$ $$\left[\begin{matrix} -6 \\ 5 \\ -23 \\ \end{matrix}\right]$$ $$=\left[\begin{matrix} 8.375 \\ 5.625 \\ -4.875 \\ \end{matrix}\right]$$

$$\underline{d}=\left[\begin{matrix} 8.375 \\ 5.625 \\ -4.875 \\ \end{matrix}\right]$$

(5)

Using $$\underline{w}_{i}=\underline{a}_{i}$$ where i = 1,2,...,n and $$\underline{a}_{i}$$ is orthonormal basis:

$$\underline{a}_{i}=\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix}\right]$$

Looking at

$$ \bar{\underline{K}} \underline{d} = \bar{\underline{F}} $$

We can express this as:

$$ \left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix}\right]$$ $$\left[\begin{matrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ 1 & 3 & 6 \\ \end{matrix}\right]\underline{d}$$ $$=\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix}\right]$$ $$\left[\begin{matrix} 5 \\ -7 \\ -4 \\ \end{matrix}\right]$$

Which is simplified to:

$$\left[\begin{matrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ 1 & 3 & 6 \\ \end{matrix}\right]\underline{d}$$ $$=\left[\begin{matrix} 5 \\ -7 \\ -4 \\ \end{matrix}\right]$$

Where:

$$\bar{\underline{K}}=\left[\begin{matrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ 1 & 3 & 6 \\ \end{matrix}\right]$$

$$\bar{\underline{F}}= \left[\begin{matrix} 5 \\ -7 \\ -4 \\ \end{matrix}\right]$$

(5)

$$ \bar{\underline{K}} \underline{d} = \bar{\underline{F}} $$

$$ \underline{d} = \bar{\underline{K}}^{-1} \bar{\underline{F}} $$

$$\underline{d}=\left[\begin{matrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ 1 & 3 & 6 \\ \end{matrix}\right]^{-1}$$ $$\left[\begin{matrix} 5 \\ -7 \\ -4 \\ \end{matrix}\right]$$ $$=\left[\begin{matrix} 8.375 \\ 5.625 \\ -4.875 \\ \end{matrix}\right]$$

$$\underline{d}=\left[\begin{matrix} 8.375 \\ 5.625 \\ -4.875 \\ \end{matrix}\right]$$

Both $$\underline{d}$$ for each case are equal, which makes since.

(7)

$$\underline{K}$$ is always symmetric while $$\bar{\underline{K}}$$ is not. Due to this symmetry, the Bubnov-Galerkin method may be easier to solve, while the effort for finding $$\underline{K}$$ may be a little more labor intensive. The lack of symmetry in $$\bar{\underline{K}}$$ for the Petrov-Galerkin method may be difficult to solve, but it is very easy to find $$\bar{\underline{K}}$$.

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