User:EML4500.f08.JAMAMA/FE/ex15

= HW 2.1 =

Find
By taking infinitesimal slice of the bar (shown in red in Figure 1),$$ dx $$, develop expression for elastodynamic response in Heat Problem in 1-D

Solution
Free Body Diagram:

Heat is transferred in the form of Conduction, Convection and Thermal Radiation. In this problem we will do Heat Conduction only.

Here we have consider a small infinitesimal portion of wall of thickness dx as shown in figure.

Let,The Heat Flux is represented by "q".


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$$ \alpha_2=1, \alpha_1=0, \alpha_3=...=\alpha_n=0 $$ $$
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$$ \left \{ \alpha _i \right \}=\left \{ 0,1,0,...,0 \right \} $$ $$
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 *  $$ \displaystyle (Eq. x.x)
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$$ \underline w=\sum_{i}\alpha _i\underline b_i=\sum_{i}\left \{ 0,1,0,...,0 \right \}\underline b_i=\underline b_2 $$ $$
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 *  $$ \displaystyle (Eq. x.x)
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$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
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 *  $$ \displaystyle (Eq. x.x)
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$$ \underline b_2\cdot \underline P(\textrm{\underline v})=0 $$ $$
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 *  $$ \displaystyle (Eq. x.x)
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$$ \alpha_n=1,\ \alpha_1=...=\alpha_{n-1}=0 $$ $$
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 *  $$ \displaystyle (Eq. x.x)
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$$ \left \{ \alpha_i \right \}=\left \{ 0,..., 0,1 \right \} $$ $$
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 *  $$ \displaystyle (Eq. x.x)
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$$ \underline w=\sum_{i}\alpha _i\underline b_i=\sum_{i}\left \{ 0,...,0,1 \right \}\underline b_i=\underline b_n $$ $$
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 *  $$ \displaystyle (Eq. x.x)
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$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline b_{n}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ \left \{ \alpha_1,...,\alpha_n \right \}\in \ \mathbb{R}^n $$ $$
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 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline w=\sum_{i}\alpha_i\underline b_i $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline b_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ $$
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 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \left \{\underline a_{i},i=1,...,n \right \} $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline a_{i}\cdot \underline a_{j}= \delta _{ij} $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \delta _{ij}=\left\{\begin{matrix} 1\ for\ i=j\\ 0\ for\ i\neq j \end{matrix}\right. $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \left \{\underline a_{i} \right \} $$ $$
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 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \textrm{\underline v}=\sum_{i=1}^{n}\ \textrm{v}_{i}\underline a_{i} $$ $$
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 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \textrm{\underline v}=\sum_{j=1}^{n}\ \textrm{v}_{j}\underline a_{j} $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \textrm{\underline P (\underline v)}=\sum_{j=1}^{n}\ \textrm{v}_{j}\underline a_{j} - \textrm{\underline v}=\underline 0 $$ $$
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 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline a_{i}\cdot\textrm{\underline P (\underline v)}=\underline a_{i}\cdot \sum_{j=1}^{n}\ \textrm{v}_{j}\underline a_{j} - \underline a_{i}\cdot \textrm{\underline v}=\underline 0 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline a_{i}\cdot \sum_{j=1}^{n}\ \textrm{v}_{j}\underline a_{j}=\underline a_{i}\cdot \textrm{\underline v}=\sum_{j=1}^{n}(\underline a_{i}\cdot \underline a_{j})\textrm{v}_{j}=\sum_{j=1}^{n}\ \delta_{ij}\cdot \textrm{v}_{j} $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline a_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ $$
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 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline w\cdot \underline P (\textrm{\underline v})=0 \ \forall \ \underline w\in \mathbb{R}^{n} $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline w = \sum_{i}\beta_i\underline a_i $$ $$
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$$ \underline w \cdot \underline P(\textrm{\underline v})=0 \ \forall \ \left \{ \beta_1,..., \beta_n \right \}\ \in \ \mathbb{R}^n $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline w=\sum_{i}\beta_i\underline a_i $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline a_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \beta_1=1, \beta_2=...=\beta_n=0 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \left \{\beta _1, ..., \beta _n \right \} $$ $$
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$$ \forall \ \left \{\beta _1, ..., \beta _n \right \} $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline w=\sum_{i}\beta _i\underline a_i=\sum_{i}\left \{ 1,0,...,0 \right \}\underline a_i=\underline a_1 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline a_1\cdot \underline P(\textrm{\underline v})=0 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \beta_2=1, \beta_1=0, \beta_3=...=\beta_n=0 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \left \{ \beta _i \right \}=\left \{ 0,1,0,...,0 \right \} $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline w=\sum_{i}\beta _i\underline a_i=\sum_{i}\left \{ 0,1,0,...,0 \right \}\underline a_i=\underline a_2 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline a_2\cdot \underline P(\textrm{\underline v})=0 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \beta_n=1,\ \beta_1=...=\beta_{n-1}=0 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \left \{ \beta_i \right \}=\left \{ 0,..., 0,1 \right \} $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline w=\sum_{i}\beta _i\underline a_i=\sum_{i}\left \{ 0,...,0,1 \right \}\underline a_i=\underline a_n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline a_{n}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ \left \{ \beta_1,...,\beta_n \right \}\in \ \mathbb{R}^n $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline w=\sum_{i}\beta_i\underline a_i $$ $$
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 * <p style="text-align:right"> $$ \displaystyle (Eq. x.x)
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$$ \underline a_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ $$
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= HW 2.4 =

Problem Statement
Show $$ \int_{} \frac{x^2}{1+x}dx = \frac{x^2}{2} - x +log(1+x) + k $$

Do the following problems to lead this result

1) Show $$\int{log(x) }dx = xlog|x| - x $$ 2) Show $$\int{x log(x)}dx = \frac{1}{2}x^2\left[log(x)-\frac{1}{2}\right]$$ 3) Find $$\int{\frac{x^2}{1+cx}}dx$$ 4) Find $$\int{\frac{x^2}{a+bx}}dx$$ 5) Find the exact solution of $$ \displaystyle u(x) $$ for, <P> $$ \displaystyle            \frac{d}\left[ (2+3x)\frac\right] + 5x = 0 $$ </P> 6) Plot $$ \displaystyle u(x) $$

1) Show $$\int{logx}dx = xlogx - x $$
Using Integration By Parts, (4.1) Let

(4.2)

Entering values from eqns 4.2 into eqn. 4.1 yields <P> $$xlogx-\int 1dx$$</P>

2) Show $$\int{x logx}dx = \frac{1}{2}x^2\left[logx-\frac{1}{2}\right]$$
Using Integration by parts, let

(4.3) Now entering values from eqns 4.3 into eqn. 4.1 yields <P> $$\frac{x^2}{2}logx - \int{\left(\frac{x^2}{2}\right) \left(\frac{1}{x}\right)}dx $$</P>

$$\frac{x^2}{2}logx- \frac{x^2}{4}$$

3) Find $$\int \frac{x^2}{1+cx}$$
Using Integration by parts, let <p style="text-align:right"> (4.4)

Now, entering values from eqns 4.4 into eqn 4.1 yields, <p style="text-align:right"> (4.5)

Now, integration by parts must be done again. Let (4.6)

The equation now takes the form (4.7)

Entering values from eqns 4.6 into 4.7 yields $$ \displaystyle \frac{x^{2}}{c} log(1+cx) - \left[ {\left( \frac{x^2}{c} log(1+cx)- \frac{1}{c^3}log(1+cx)- \frac{x^2 }{2c} + \frac{x}{c^2} \right)} {\left( 2x \right)} - \int {\left(2dx \right)}\right] $$

Distributing terms,

(4.8)

Solving each integral in 4.8 individually <P> $$ \displaystyle \begin{align} & {2\int{\frac{x^2}{c}log(1+cx)}\ dx} = \frac{2x^3}{3c}\log(1+cx)+\frac{2}{3c^4}\log(1+cx)-\frac{2x^3}{9c}+\frac{x^2}{3c^2} -\frac{2x}{3c^3}\\ & {-2\int{\frac{1}{c^3}log(1+cx)}\ dx} = -\frac{2x}{c^3}log(1+cx) - \frac{2}{c^4}log(1+cx) + \frac{2x}{c^3} \\ & {\int{\frac{x^2}{c}}\ dx} = \frac{x^3}{3c} \\ & {2\int{\frac{x}{c^2}}\ dx} = \frac{x^2}{c^2} \end{align} $$ <p style="text-align:right"> (4.9)

Entering values for eqns 4.9 into eqn. 4.8 yields

Which simplifies to

When c=1, the problem further simplifies to

4) Find $$\int \frac{x^2}{a+bx}$$
Using integration by parts, let

(4.10)

Now, entering values from eqns 4.10 into eqn 4.1 yields, (4.11)

Now, integration by parts must be done again. Let

(4.12)

The equation now takes the form

(4.13)

Entering the values from 4.12 into 4.13 yields,

Distributing terms,

(4.14)

Solving each integral in 4.14 individually,

$$ \displaystyle \begin{align} &\frac{2}{b}\int{xlog(a+bx)}dx = \frac{x^2}{b}\log(a+bx)-\frac{a^2}{b^3}\log(a+bx)-\frac{x^2}{2b}+\frac{ax}{b^2} \\ &\frac{2a}{b^2}\int{log(a+bx)}dx = \frac{2ax}{b^2}log(a+bx)+\frac{2a^2}{b^3}log(a+bx) - \frac{2ax}{b^2} \\ &\frac{2}{b}\int{x}dx = -\frac{x^2}{b} &\end{align} $$ <p style="text-align:right"> (4.15)

Substituting eqns 4.15 into 4.14 yields

Which simplifies to

<p style="text-align:right"> (4.16)

5) Find the exact solution for
$$ \displaystyle \frac{d}\left[ (2+3x)\frac\right] + 5x = 0 $$  ;    $$  \forall x \in ] 0,1 [ $$

With the following Boundary Conditions:

Essential: $$ u(1)=4 $$ <p style="text-align:right"> (4.17) Natural:$$ -\frac{d}{dx} u(x=0) =6 $$ <p style="text-align:right"> (4.18)

$$ \frac{d}{du} [(2+3x)\frac{du}{dx}]=-5x $$ <p style="text-align:right"> (4.19) Integrate both sides of eqn. 4.19,<P> $$ \int d [(2+3x)\frac{du}{dx}] = \int-5x dx$$</P>

$$ (2+3x)\frac{du}{dx}=\frac{-5x^2}{2}+c_1$$

$$\frac{du}{dx}=\frac{-5x^2}{2(2+3x)}+\frac{c_1}{2+3x}$$ <p style="text-align:right"> (4.20) Apply Natural B.C., eqn. 4.18, to find $$ c_1 $$,

$$-\frac{du}{dx}(x=0) =6 \implies -6= 0+ \frac{c_1}{2} $$

$$c_1=-12$$ <p style="text-align:right"> (4.21) Enter value from eqn. 4.21 into eqn 4.20

$$\frac{du}{dx}=\frac{-5x^2}{2(2+3x)}+\frac{-12}{2+3x}$$

$$du=\frac{-5x^2}{2(2+3x)}dx+\frac{-12}{2+3x}dx$$ <p style="text-align:right"> (4.22) Integrate both sides of eqn. 4.22<P> $$\int du=\int\frac{-5x^2}{2(2+3x)}dx+\int\frac{-12}{2+3x}dx$$</P>

$$u=-\frac{5}{2}\int \frac{x^2}{2+3x}dx-12\int \frac{dx}{2+3x}$$

$$u=\frac{-5}{2}[\frac{x^2}{6}-\frac{2x}{9}+\frac{4log(2+3x)}{27}]-12[\frac{log(2+3x)}{3}]+c_2$$

$$u=\frac{-5x^2}{12} +\frac{5x}{9}-\frac{98}{27}log(2+3x)+c_2$$ <p style="text-align:right"> (4.23) Apply Essential B.C., eqn 4.17, to find

$$ u(1)=4 \implies 4=\frac{-5}{12} +\frac{5}{9}-\frac{98}{27}log(2+3x)+c_2$$

$$c_2=-1.98055$$ <p style="text-align:right"> (4.24)

Enter the value from eqn 4.24 into eqn. 4.23 to obtain exact solution,

= HW 2.5 =