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User:Egm6321.f10.team2.zou/HW3

= Problem 1: Derive EOM for the particle =

From the lecture slide Mtg 13-1.

Given
The picture was given in the MT13-1 for the moving particle :

For the dynamic objects Newton's Second law cites that: The total force applied on a body is equal to the time derivative of linear momentum of the body.See Newton's Law of Motion According to this law :
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$$  \displaystyle \vec{F}=\frac{\mathrm{d} (m\vec{v})}{\mathrm{d} t} $$ (Eq 1.1)
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If we manipulate this equation for the x and y axis;
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$$  \displaystyle \sum F_{x}=ma_{x} $$     (Eq 1.2)
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$$  \displaystyle \sum F_{y}=ma_{y} $$     (Eq 1.3)
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m: mass of the particle

k:air resistence constant

g:acceleration of gravity

Find
Case 1:Derive the EOM's for the moving particle with air resistance.

Case 2:k=0 no air resistance.

Case 3:Exist air resistance.

Case 1
For the x axis:
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$$  \displaystyle a_{x}=\frac{\mathrm{d} v_{x}}{\mathrm{d} t} $$ (Eq 1.4)
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$$  \displaystyle F_{air_x}=kv^{n}cos(\alpha ) $$     (Eq 1.5) Then now we can say that according to Newton's Law: Total force acting on the particle=Particle momentum change. Since there is only force that acting on the particle on x axis is just the x component of air resistance.By tis way we can concluded that:
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$$  \displaystyle m\frac{\mathrm{d} v_{x}}{\mathrm{d} t}=-kv^{n}cos(\alpha ) $$     (Eq 1.6)
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Similarly if we write EOM for the y axis
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$$  \displaystyle F_{air_y}=kv^{n}sin(\alpha ) $$     (Eq 1.7)
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$$  \displaystyle a_{y}=\frac{\mathrm{d} v_{y}}{\mathrm{d} t} $$ (Eq 1.8)
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And total forces acting on the particle are y companent of air resistance and gravity force.

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$$  \displaystyle m\frac{\mathrm{d} v_{y}}{\mathrm{d} t}=-kv^{n}sin(\alpha )-mg $$     (Eq 1.9)
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Case 2
For k=0;
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$$  \displaystyle m\frac{\mathrm{d} v_{x}}{\mathrm{d} t}=0 $$     (Eq 1.10)
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$$  \displaystyle m\frac{\mathrm{d} v_{y}}{\mathrm{d} t}=-mg $$     (Eq 1.11)
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$$  \displaystyle \int dv_{x}=\int dt $$ (Eq 1.12)
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$$  \displaystyle v=V_{xo} $$     (Eq 1.13)
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$$  \displaystyle x(t)=\int V_{x0}dt+x(0) $$     (Eq 1.14)
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$$  \displaystyle x(t)=V_{x0}t+x(0) $$     (Eq 1.15)
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$$  \displaystyle \int dv_{y}=-\int gdt $$     (Eq 1.16)
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$$  \displaystyle v_{y}=-gt+V_{y0} $$     (Eq 1.17)
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$$  \displaystyle x(t)=\int (-gt+V_{y0})+x(0) $$     (Eq 1.18)
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$$  \displaystyle x(t)=-\frac{gt^{2}}{2}+V_{y0}t+y(0) $$     (Eq 1.19)
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Case 3
To alleviate notion, we define,

 * $$\displaystyle z\equiv {{v}_{y}}$$

Then the original equation can be written as,

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$$ \displaystyle m{z}'=-k{{z}^{n}}-mg$$ (Eq 1.21)
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This nonlinear first order ODE can be solved by the separation of variables. By some manipulation, we can write Eq 1.21 as,

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$$ \displaystyle -\frac{m}{mg+k{{z}^{n}}}dz=dt$$ (Eq 1.22)
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Then we integrate Eq 1.22, we have,

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$$ \displaystyle \int{\frac{m}{mg+k{{z}^{n}}}dz=C-t}$$ (Eq 1.23)
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where $$C$$ is some constant.

If you want to use Euler interation factor method, we can still get the same answer as Eq 1.23.

The original equation can be written as,

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$$ \displaystyle k{{z}^{n}}+mg+m{z}'=0$$ (Eq 1.24)
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It's easy to see that the Eq 1.24 is not exact. So we expect we can multiple an integrating factor $$\displaystyle h$$ to make it exact.

Define:

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& M\equiv k{{z}^{n}}+mg \\ & N\equiv m \\ \end{align} \right.$$

The prerequisite that Eq 1.24 being exact is,

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$$ \displaystyle \frac{\partial (hM)}{\partial z}=\frac{\partial (hN)}{\partial t}$$ (Eq 1.25)
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Thus,

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$$ \displaystyle {{h}_{t}}N-{{h}_{z}}M+h({{N}_{t}}-{{M}_{z}})=0$$ (Eq 1.26)
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If we assume $$\displaystyle h$$ is a function only of $$\displaystyle z$$, meaning $$\displaystyle {{h}_{t}}=0$$, then substituting into Eq 1.26 $$\displaystyle M$$ and $$\displaystyle N$$, we have

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$$ \displaystyle \frac{h}=\frac{1}{M}({{N}_{t}}-{{M}_{z}})=\frac{-kn{{z}^{n-1}}}{k{{z}^{n}}+mg}$$ (Eq 1.27)
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By integrating Eq 1.27, we have,

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$$ \displaystyle h=\frac{1}{k{{z}^{n}}+mg}$$ (Eq 1.28)
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We can see the assumption "$$\displaystyle h$$ is only a function of $$\displaystyle z$$" is satisfied.

Then we mutiple $$\displaystyle h$$ to both sides of Eq 1.24, we have,

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$$ \displaystyle 1+\frac{m}{k{{z}^{n}}+mg}{z}'=0$$ (Eq 1.29)
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We can easily see that Eq 1.29 is indeed exact. Because it's exact we can rewrite Eq 1.29 as,

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$$ \displaystyle \frac{d}{dt}(t+\int{\frac{m}{mg+k{{z}^{n}}}dz})=0$$ (Eq 1.30)
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That means,

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$$ \displaystyle t+\int{\frac{m}{mg+k{{z}^{n}}}dz}=C$$ (Eq 1.31)
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or

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$$ \displaystyle \int{\frac{m}{mg+k{{z}^{n}}}dz=C-t}$$ (Eq 1.32)
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where $$C$$ is some constant.

= Problem 2 =

From the lecture slide Mtg 13-3

Given
The pendulum problem was shown in MT13-3 as

Angular momentum of a particle is gives as :
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$$  \displaystyle \vec{L}=\vec{r}\times \vec{p} $$     (Eq 2.1) Where r=the position of the particle from the origin
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p=The linear momentum of a particle.

Find
Derive the EOM.

Solution
For deriving EOM's we think in angular basis. Since we are interested in angular momentum we can enlarge Newton's Law for linear momentum into angular momentum.According to Newton's Law Total moment acting on the particle=Total angular momentum chance of particle.So

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$$  \displaystyle \vec{r}\times \frac{\mathrm{d}(m\vec{v)}}{\mathrm{d} t}=\sum \vec{r}\times \vec{F} $$     (Eq 2.2)
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$$  \displaystyle \sum \vec{M}=\sum \vec{r}\times \vec{F} $$     (Eq 2.3)
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Total moment acting on the particle mass of m1=Spring moment+gravity force moment+control force moment.

Spring moment

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$$  \displaystyle \vec{M}_{spring}=\vec{a}\times k\vec{a}(sin\theta _{1}-sin\theta _{2}) $$     (Eq 2.4)
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Gravity force moment

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$$  \displaystyle \vec{M}_{gravity}=\vec{l}sin\theta _{1}\times m\vec{g} $$     (Eq 2.5) For small angular displacements
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$$  \displaystyle sin\theta _{1}\approx \theta _{1} $$     (Eq 2.6)
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Control Force moment
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$$  \displaystyle \vec{M}_{control}=\vec{l}\times \vec{u_{1}} $$     (Eq 2.7)
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If we subtitute all term in angular momentum equation we concluded

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$$  \displaystyle m_{1}l^{2}\ddot{\theta _{1}}=-ka^{2}(\theta _{1}-\theta _{2})-m_{1}gl\theta_{1}+u_{1}l $$     (Eq 2.8)
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Similarly for the second pendulum

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$$  \displaystyle m_{2}l^{2}\ddot{\theta _{2}}=-ka^{2}(\theta _{2}-\theta _{1})-m_{2}gl\theta_{2}+u_{2}l $$     (Eq 2.9)
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We can write this equations in matrix form for the SC_L1_ODEs.The form can be shown as basically:

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$$  \displaystyle \dot{X}_{4\times 1}=A(t)_{4\times 4}.X_{4\times 1}+B(t)_{4\times 2}.U_{2\times 1} $$     (Eq 2.10)
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$$  \displaystyle X=\begin{bmatrix} \theta _{1}\\ \dot{\theta _{1}}\\ \theta _{2}\\ \dot{\theta _{2}}\\
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\end{bmatrix} $$     (Eq 2.11)
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$$  \displaystyle \left[ \begin{matrix} \overset{.}{\mathop}\, \\ \overset{..}{\mathop}\, \\ \overset{.}{\mathop}\, \\ \overset{..}{\mathop}\, \\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\   -\frac{k{{a}^{2}}}-\frac{g}{l}  \\ 0 \\   \frac{k{{a}^{2}}}  \\ \end{matrix}\begin{matrix} 1 \\   0  \\   0  \\   0  \\ \end{matrix}\begin{matrix} 0 \\   \frac{k{{a}^{2}}}  \\ 0 \\   -\frac{k{{a}^{2}}}-\frac{g}{l}  \\ \end{matrix}\begin{matrix} 0 \\   0  \\   1  \\   0  \\ \end{matrix} \right]\left[ \begin{matrix} {{\theta }_{1}} \\ \overset{.}{\mathop}\, \\ {{\theta }_{2}} \\ \overset{.}{\mathop}\, \\ \end{matrix} \right]+\left[ \begin{matrix} 0 \\   \frac{1}{{{m}_{1}}l}  \\ 0 \\   0  \\ \end{matrix}\begin{matrix} 0 \\   0  \\   0  \\   \frac{1}{{{m}_{2}}l}  \\ \end{matrix} \right]\left[ \begin{matrix} {{u}_{1}} \\ {{u}_{2}} \\ \end{matrix} \right] $$     (Eq 2.11)
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= Problem 3 =

From the lecture slide Mtg 14-1

Given
Linear first order ODE:

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$$ \displaystyle \dot{x}(t)=a(t)x(t)+b(t)u(t)$$ (Eq 3.1)
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Initial Condition:

 * $$\displaystyle {{t}_{0}}$$ and $$\displaystyle x({{t}_{0}})$$.

Find
$$\displaystyle x(t)$$ when $$\displaystyle a(t)$$ and $$\displaystyle b(t)$$ are constants and varying coefficients respetively.

Case of Varying Coefficients
First assume $$\displaystyle a(t)$$ and $$\displaystyle b(t)$$ are varying coefficients since constants are only a particular case.

The Eq 3.1 can be written as:

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$$ \displaystyle \dot{x}(t)-a(t)x(t)=b(t)u(t)$$ (Eq 3.2)
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Thus we have the Euler integrating factor:

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$$ \displaystyle h(t)=\exp \int_ – ^{t}{-a(s)ds}$$ (Eq 3.3)
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Or:

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$$ \displaystyle h(t)=\exp \int_^{t}{-a(s)ds}+{{k}_{1}}$$ (Eq 3.4)
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where $$\displaystyle {{t}_{M}}$$ is the lower limit of integral with respect to integration constant $$\displaystyle {{k}_{1}}$$.

Then we have $$\displaystyle x(t)$$:

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$$ \displaystyle \begin{align} x(t) & =\exp \int_ – ^{t}{a(s)ds*\int_ – ^{t}{\left[ \exp \int_ – ^{\tau }{-a(s)ds} \right]}}*bu(\tau )d\tau \\ & =\exp \int_^{t}{a(s)ds}*\left\{ \int_^{t}{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau +{{k}_{2}} \right\} \end{align}$$ (Eq 3.5)
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where $$\displaystyle {{t}_{N}}$$ is the lower limit of integral with respect to integration constant $$\displaystyle {{k}_{2}}$$. Note that $$\displaystyle {{k}_{1}}$$ disappeared because Patterson and Rasul have proved that it is unnecessary. In fact $$\displaystyle {{t}_{M}}$$ will also be canceled in the following procedure.

To derive $$\displaystyle {{t}_{N}}$$ and $$\displaystyle {{k}_{2}}$$, we should substituting the initial condition into Eq 3.5:

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$$ \displaystyle x({{t}_{o}})=\exp \int_^{a(s)ds}*\left\{ \int_^{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau +{{k}_{2}} \right\}$$ (Eq 3.6)
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then we get $$\displaystyle {{k}_{2}}$$:

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$$ \displaystyle {{k}_{2}}=\frac{x({{t}_{0}})}{\exp \int_^{a(s)ds}}-\int_^{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau $$ (Eq 3.7)
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Substituting $$\displaystyle {{k}_{2}}$$ back into Eq 3.5, we derive:

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$$ \displaystyle \begin{align} x(t) & =\exp \int_^{t}{a(s)ds}*\left\{ \int_^{t}{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau +\frac{x({{t}_{0}})}{\exp \int_^{a(s)ds}}-\int_^{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau \right\} \\ & =x({{t}_{0}})\exp \left[ \int_^{t}{a(s)ds}-\exp \int_^{a(s)ds} \right]+\exp \int_^{t}{a(s)ds}*\left\{ \int_^{t}{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau -\int_^{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau \right\} \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\exp \int_^{t}{a(s)ds}*\int_^{t}{\left[ \exp \int_{\tau }^{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\exp \int_^{t}{a(s)ds}*\int_^{t}{\left[ \exp (\int_{\tau }^{t}{a(s)ds+\int_{t}^{a(s)ds)}} \right]}*b(\tau )u(\tau )d\tau \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\exp \int_^{t}{a(s)ds}*\exp \int_{t}^{a(s)ds*}\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \end{align}$$ (Eq 3.8)
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So we have got the $$\displaystyle x(t)$$ when $$\displaystyle a(t)$$ and $$\displaystyle b(t)$$ are varying coefficients:

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$$ \displaystyle x(t)=x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau $$ (Eq 3.9)
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Verifying of result
Note that the derivation rule of $$\displaystyle \frac{d}{dt}\int_^{t}{f(\tau )d\tau }$$ between that of $$\displaystyle \frac{d}{dt}\int_^{t}{f(\tau ,t)d\tau }$$ are totaly different:

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$$ \displaystyle \frac{d}{dt}\int_^{t}{f(\tau )d\tau =}f(t)$$, (Eq 3.10)
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However,

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$$ \displaystyle \frac{d}{dt}\int_^{t}{f(\tau ,t)d\tau =}f(t,t)+\int_^{t}{\frac{\delta f(\tau ,t)}{\delta t}d\tau }$$ (Eq 3.11)
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More generally,
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$$ \displaystyle \frac{d}{dt}\int_{b(t)}^{a(t)}{f(\tau ,t)d\tau =}\frac{da(t)}{dt}f\left( a(t),t \right)-\frac{db(t)}{dt}f\left( b(t),t \right)+\int_{b(t)}^{a(t)}{\frac{\delta f(\tau ,t)}{\delta t}d\tau }$$ (Eq 3.12)
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Thus by Eq 3.11,

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$$ \displaystyle
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\begin{align} \dot{x}(t) & ={{\left[ x({{t}_{0}})\exp \int_^{t}{a(s)ds} \right]}^{\prime }}+{{\left\{ \int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \right\}}^{\prime }} \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}*a(t)+\exp \int_{t}^{t}{a(s)ds}*b(t)u(t)+\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*a(t)b(\tau )u(\tau )d\tau \\ & =a(t)\left\{ x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \right\}+b(t)u(t) \\ & =a(t)x(t)+b(t)u(t) \end{align}$$ (Eq 3.13)
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Thus the result is veryfied. See also Team2 discussion on Mtg 14.

Case of Constant Coefficients
As constant coefficients is only a particular case of varying coefficients, we can just replace $$\displaystyle a(s)$$ by $$\displaystyle a$$ in the Eq 3.9,

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$$ \displaystyle \begin{align} x(t) & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \\ & =x({{t}_{0}})[\exp a(t-{{t}_{0}})]+\int_^{t}{\left[ \exp a(t-\tau ) \right]}*b(\tau )u(\tau )d\tau \end{align}$$ (Eq 3.14)
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Then this is the answer to the case of constant coefficent.

Another way to solve the Case of Constant Coefficients(by Team2. Yaakov)
The problem is time independent; therefore, we can rewrite the initial equation as:

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$$ \displaystyle \dot{x}(t)-ax(t)=bu(t)$$ (Eq 3.15)
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To calculate the integrating factor we find

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$$ \displaystyle \exp (\int{-adt)=\exp (-at})$$ (Eq 3.16)
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Now we can multiply the entire equation 3.16 with the integrating factor

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$$ \displaystyle \exp (-at)\dot{x}(t)-\exp (-at)ax(t)=\exp (-at)bu(t)$$ (Eq 3.17)
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It is noteworthy to realize that the LHS of the equation is equal to $$\displaystyle [\exp (-at)x(t)]'$$

Integrating both sides of the new equation from $$\displaystyle {{t}_{0}}$$ to $$\displaystyle t$$ and changing the variables from $$\displaystyle t$$ to $$\displaystyle \tau $$ for integration will yield us the following equation:

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$$ \displaystyle \int\limits_^{t}{[\exp (-a\tau )x(\tau )]'=}\int\limits_^{t}{\exp (-a\tau )bu(\tau )dt}$$ (Eq 3.18)
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$$ \displaystyle \exp (-at)x(t)-\exp (-a{{t}_{0}})x({{t}_{0}})=\int\limits_^{t}{\exp (-a\tau )bu(\tau )dt}$$ (Eq 3.19)
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Solving the equation for $$x(t)$$ will yield us to the proof:

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$$ \displaystyle x(t)=[\exp a(t-{{t}_{0}})]x({{t}_{0}})+\int\limits_^{t}{[\exp a(t-\tau )]bu(\tau )d\tau }$$ (Eq 3.20)
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= Problem 4 =

From the lecture slide Mtg 14-2

Given
Solution of system of coupled linear first order ODE of constant coefficient:

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$$ \displaystyle \underline{x}(t)=[\exp (t-{{t}_{0}})\underline{A}]\underline{x}({{t}_{0}})+\int_^{t}{\left[ \exp (t-\tau )\underline{A} \right]\underline{B}\underline{u}(\tau )}d\tau $$ (Eq 4.1)
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Find
Solution of System of coupled linear first order ODE of varying coefficient

Solution
Just turn constant cofficient matrices $$\underline{A}$$ and $$\underline{B}$$ in Eq 4.1 into varying coefficient matrices case, and via similarity to Eq 3.9, we have,

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$$ \displaystyle \underline{x}(t)=[\exp \int_^{t}{\underline{A}(s)ds}]\underline{x}({{t}_{0}})+\int_^{t}{\left[ \exp \int_{\tau }^{t}{\underline{A}(s)ds} \right]\underline{B}(\tau )\underline{u}(\tau )}d\tau $$ (Eq 4.2)
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= Problem 5 =

From the lecture slide Mtg 15-1

Given
System of Linear first order ODE:

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$$ \displaystyle \underline(t)=\underline{A}\underline{x}(t)+\underline{B}\underline{u}(t)$$ (Eq 5.1)
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where $$\displaystyle \underline{A}$$ and $$\displaystyle \underline{B}$$ are constant matrices.

Initial condition:

 * $$\displaystyle {{t}_{0}}$$ and $$\displaystyle \underline{x}({{t}_{0}})$$.

Find
$$\displaystyle \underline{x}(t)$$

Solution
From Eq 5.1 we derive Euler integrating factor,

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$$ \displaystyle \underline{h}(t)=\exp \int_ – ^{t}{-\underline{A}ds}=\exp (\underline{M}-t\underline{A})$$ (Eq 5.2)
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where $$\displaystyle \underline{M}$$ is a matrix consists of constants.

Thus we get $$\displaystyle \underline{x}(t)$$,

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$$ \displaystyle \underline{x}(t)=\exp (t\underline{A}-\underline{M})*\left\{ \int_^{t}{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )d\tau +\underline{N}} \right\}$$ (Eq 5.3)
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where $$\displaystyle {{t}_{N}}$$ is lower limit of integral with respect to $$\displaystyle \underline{N}$$ which is also a matrix consists of constants.

To derive $$\displaystyle \underline{N}$$ we should use the initial condition,

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$$ \displaystyle \underline{x}({{t}_{0}})=\exp (t\underline{A}-\underline{M})*\left\{ \int_^{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )d\tau +\underline{N}} \right\}$$ (Eq 5.4)
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Since,

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$$ \displaystyle \exp (t\underline{A}-\underline{M})*\exp (M-t\underline{A})=1$$
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Then $$\displaystyle \exp (M-t\underline{A})$$ are the inverse matrix of $$\displaystyle \exp (t\underline{A}-\underline{M})$$.

Then, by manipulation of Eq 5.4, we have,

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$$ \displaystyle \underline{N}=\exp (M-{{t}_{0}}\underline{A})\underline{x}({{t}_{0}})-\int_^{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )}d\tau $$ (Eq 5.5)
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Substituting back to Eq 5.4, we have

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$$ \displaystyle \begin{align} \underline{x}(t) & =\exp (t\underline{A}-\underline{M})*\left\{ \int_^{t}{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )d\tau +\exp (M-{{t}_{0}}\underline{A})\underline{x}({{t}_{0}})-\int_^{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )}d\tau } \right\} \\ & =\exp (t\underline{A}-\underline{M}+M-{{t}_{0}}\underline{A})\underline{x}({{t}_{0}})+\exp (t\underline{A}-\underline{M})*\int_^{t}{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )}d\tau \\ & =[\exp (t-{{t}_{0}})\underline{A}]\underline{x}({{t}_{0}})+\int_^{t}{\left[ \exp (t-\tau )\underline{A} \right]\underline{B}\underline{u}(\tau )}d\tau \end{align}$$ (Eq 5.6)
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Thus we have derive $$\displaystyle \underline{x}(t)$$:

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$$ \displaystyle \underline{x}(t)=[\exp (t-{{t}_{0}})\underline{A}]\underline{x}({{t}_{0}})+\int_^{t}{\left[ \exp (t-\tau )\underline{A} \right]\underline{B}\underline{u}(\tau )}d\tau $$ (Eq 5.7)
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 *

 * }

= References =

title: User:Egm6321.f10.team2.zou/HW3


text:
Towards a better India

This learning resource analyzes core problems that India faces and looks for solutions. Though mainly focused on India, the intent is to find feasible solutions for Third World countries by using India as a model.

Contents

 * /Introduction/
 * /Past and Present/
 * /Objectives of a Modern India/
 * /Major Problems/
 * /Possible Solutions/
 * /Our Future/
 * Further Reading

Participants

 * Balajiviswanathan
 * Goldenburg111

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Towards a better India


text:
Towards a better India/Introduction

India - Towards a better India
India is a land of paradoxes and extremes. A country with the hottest of lands (around 50C in Rajasthan) and coldest of mountains (around -50C in Dras, Kashmir), the highest and lowest of terrain, the driest and wettest of surfaces, the richest and poorest of people, a nation with a hyper techno-savvy upper middle class co-existing with world's largest illiterate group. The nation has every mix of races (Negroids, Caucasian and Mongoloid strains) and every religion (has the highest Hindu population, second highest Muslim population, one of the top 10 highest populations for Christianity and Buddhism, besides being birth for Sikhism, Jainism, one of the first acceptors of Judaism, Zoroastrianism, Christianity and Baha'i).

Irrespective of the good public opinion generated by its savvy graduates both in India and overseas (particularly in US & Singapore) there are mountains of problems underneath. Though the change in world opinion towards India, in the last 10 years, was good to a great extent, it has the real danger of masking the great realities of India. Let us have a small snapshot of them.

 * 1) India is the home to the world's highest number of illiterates (430 million, in 2000)
 * 2) 6 children out of 100 die before they reach the age of 1 (2003, stats)
 * 3) World's second highest AIDS population - 5.3 million (2003)
 * 4) World's highest concentration of Polio, Tuberculosis and Typhoid

title: Towards a better India/Introduction


text:
Towards a better India/Past and Present

Bronze Age
Historically, India had one of the oldest civilizations that dated back to 3500 BC. The Indus Valley Civilization had an extremely advanced town planning, architecture and pottery design for its age and it reached its zenith at around 1500 BC. Roughly around the same time, there was a parallel civilization in the south called the Dravidian Civilization.

Colonial Times
At the time Mughals, were establishing themselves in India, the other notable development was the arrival of the Europeans. Vasco da Gama, of Portugal, landed on the Western Indian coast in 1498. Initially the Europeans were interested in trade with India but subsequently started establishing settlements and cornering political power mainly by cleverly dividing the Indian Princes and turning them against each other and superior armed forces.

However, as European history has it, the major European powers like the UK, France, the Netherlands and Portugal were engaged in a major struggle to gain upper hand in political power as well as trade, they engaged in similar struggle in India too.

The UK displaced almost completely all others and the British Queen took over the Indian Government from the administration of the East India Company after 1857, following a revolt by Indians of assorted interests, including the last Mughal king, Bahadur Shah Jaffer. The British rule continued until 1947, when India gained independence after a prolonged struggle for independence lead by Mahatma Gandhi.

Free India Post-Liberalization
--Chirax (talk) 12:09, 15 July 2008 (UTC)

title: Towards a better India/Past and Present


text:
Towards a better India/Objectives of a Modern India

Requirements of a modern state varies from culture to culture. However, by commonly accepted world standards and principles of the Human Development Index, we would like to state the minimum achievement required for a Better India to be:
 * 1) Complete literacy (everyone should be able to read and write) and every adult should have completed high school
 * 2) Poverty Eradication - every Indian should be able to earn above the bare minimum of $3/day
 * 3) Eradication of major communicable diseases including but not limited to Polio, Tuberculosis, Malaria, and Typhoid
 * 4) Eradication of child labor (no child should be forced to work, at the cost of his/her education)
 * 5) Removal of social bias and untouchability (no person should be judged based on his/her birth)
 * 6) Eradication of gender bias and crimes against women (rape, sexual harassment, dowry)
 * 7) Promotion of communal harmony to eradicate religion-based violence (inter-faith acrimony should stop)
 * 8) Corruption at all levels of governance should get closer to nil
 * 9) Cross-border terrorism, sectarianism, separationism, and naxalitism should come to an end
 * 10) Promotion of harmony with all neighbors including Pakistan
 * 11) Increasing the use of alternative fuels such as ethanol and biodiesel


text:
Towards a better India/Major Problems

The major problems that India faces are:

 * Corruption
 * Poverty
 * Hunger
 * Crime
 * Pollution
 * Communalism
 * Illiteracy
 * Population Explosion
 * Rural undevelopment
 * Terrorism
 * Crime against Women
 * Water Scarcity
 * Unemployment
 * Brain Drain

title: Towards a better India/Major Problems


text:
Towards a better India/Major Problems/Corruption

India suffers from corruption at all levels. To get your work done, you have to offer bribe or else you will have to wait for a few months for even small things like getting a power connection for a newly constructed home. This hampers the efficiency with which things are done.

corruption had spread like a cancer in India,people are giving money mainly for the work to be done in a faster rate. corruption had spread from east to west and north to south.corruption rate has been increasing in these decades at a rapid rate

The root cause of corruption is the Nehruvian economic setup of the country and the civil services. Through the economic structure is changing, Civil services are still the same. A recent study by reputed Hong Kong based institute, found the Indian Civil Services as most corrupt in Asia. Among all Civil Services, IRS (Indian Revenue Services) is the most corrupt. There is an estimate that if IRS is abolished, the GDP growth will be increased by 2% automatically.


text:
Towards a better India/Major Problems/Poverty

Poverty is a significant problem in India considering the fact that India is the home of nearly 400 million poor people. Poverty is a root cause for many other problems including population explosion, child labor and crime proliferation. For India to prosper, the most important goal is to achieve poverty alleviation for millions of its citizens and poverty reduction would have a significant impact on reaching the other objectives for a modern state. However, poverty removal would require the partial solution to the other major problems, including illiteracy and gender bias.

Factors Responsible for Poverty
Incidence of poverty is strongly dependent of various social, economic and natural factors. Being a nation of 1 billion people compacted in a region one-third the size of United States and heavy concentration of population in lands far from the access to sea, complicates the problem. The nation has a large variety of tribals, who do not have access to any of the gifts of modern world but suffer the negative aspects of modernization, including pollution and global warming.

Natural Factors
The natural factors responsible for poverty include:
 * 1) Heavy concentration of population over a large area (the country houses more people than Europe, Africa, North America & South America).
 * 2) Predominant dependence on seasonal rainfall for agriculture (agriculture employs nearly 70% of the people and majority of the cultivation depend solely upon rain and the rainfall pattern can vary significantly from year-to-year).
 * 3) Majority of dense populated regions (Bihar, Uttar Pradesh, Punjab) are far from the access to the sea (oceans and seas are very critical to the economic development of nations).

Social Factors
The social factors include:
 * 1) Around 430 million people are illiterate (majority of them are poor) and illiteracy  forces people to stick to ancestral jobs and prevents people from having job flexibility
 * 2) Caste System prohibited access to lucrative jobs to a vast majority of the people. Though, such institutions have been dismantled in the last 50 years, its presence can still be seen in rural areas.


text:
Towards a better India/Major Problems/Hunger

Hunger is a major problem in South Asia and Africa, where millions die to lack of proper nutrition and growth of millions of children stunted due to malnutrition.


text:
Towards a better India/Major Problems/Gangs

Crime is our enemy and as one problem leads to another crime is also followed by one or another problem.


text:
Towards a better India/Major Problems/Communalism

We can categorise the communalism into two
 * 1) Majority communalism
 * 2) Minority communalism

 * 1) Majority communalism is not a threat to india as Majority of indians are not communal -simple reason being that, they never voted to forces with communal background to power.
 * 2) Minority communalism is definitely a threat to India, examples being Kashmir,North east states & other parts of the country.  As minorities are good vote banks for politicians, politicians are ready to do anything from allowing illegal immigrant to get their names added into voterlist to getting the legal documents of stay.


text:
Towards a better India/Major Problems/Terrorism

Since its independence, India has faced the problem of terrorism and militancy in different areas within the country.

When talking of terrorism and India, the main terrorism threat that comes to everyone's mind is that of Kashmiri terrorists fighting against the Indian government in Jammu and Kashmir. While this has been going on for a very long time it saw a resurgence in 1989.

In the late 70's and early 80's India was engulfed by the terrorist threats of Sikhs demanding a separate state. The major casualities of this threat was the assasination of Indian Prime Minister Indira Gandhi by her own bodyguards and the crashing of Kanishka, an Air India flight flying between India and Canada.

Other terrorist activities that continue to haunt India is from the Mao rebels in West Bengal, Infiltrators from Bangladesh, Tamil Rebel Tigers from Sri Lanka (who killed Indian Prime Minister Rajiv Gandhi), ultras in North-East India, Peoples War Group in Andhra Pradesh and threats in Mumbai from the underworld and Dawood Ibrahim who is also on the most wanted men of FBI.

Clearly, India has a very big problem of terrorism which has engulfed it since its independence and across the country.


text:
Towards a better India/Major Problems/Brain Drain

Brain Drain is one of the big problem india facing since last few years. Most of the educated and highly qualified indians like to work in developed countries. 38% of doctors in US are indians. 36% of NASA scientists are indians. 28% IBM employees are indians. 34% Microsoft employees are indians.

These data shows that india facing very big problem of highly quailified brain drains to developed countries.