User:Jiangpx/6321 hw1

= Problem 1 - Differentiation using Chain Rule =

= Problem 2 - Dimensional Analysis =

Given
The equation of motion of a train (either Maglev or wheel on rail) can be shown to have the following form:

The $$c_0$$ term is defined as:

Find
Perform dimensional analysis of the $$c_0$$ term and provide physical meaning of each term in $$c_0$$

Solution
According to Newton's second Law, the whole $$c_0$$ term should be the force term in the equation of motion.

For the first term,

in which, $$\left[ U_{,ss}^{2}\left( {{Y}^{1}},t \right) \right]=[1/L]$$; $$[\overline{R}]=[L]$$

so,

$$\left[ \overline{R}U_{,ss}^{2}\left( {{Y}^{1}},t \right) \right]=[1]$$

and

$$[{{F}^{1}}\left[ 1-\bar{R}U_{,ss}^{2}\left( {{Y}^{1}},t \right) \right]]=[F]$$

For the second term,

$$[{{F}^{2}}u_{,s}^{2}]=[F]\cdot [1]=[F]$$

For the third term,

$$[T/R]=[F\cdot L]/[L]=[F]$$

For the last term,

in which,

$$[M]=[M]$$

$$[1-\bar{R}u_{,ss}^{2}]=[1]$$;

$$[u_{,tt}^{1}-\bar{R}u_{,stt}^{2}]=[L/{{t}^{2}}]-[L]\cdot [L/(L\cdot {{t}^{2}})]=[L/{{t}^{2}}]$$;

$$[u_{,s}^{2}u_{,tt}^{2}]=[1]\cdot [L/{{t}^{2}}]=[L/{{t}^{2}}]$$

so,

$$[\left[ 1-\bar{R}u_{,ss}^{2} \right]\left[ u_{,tt}^{1}-\bar{R}u_{,stt}^{2} \right]]=[1]\cdot [L/{{t}^{2}}]=[L/{{t}^{2}}]$$

and

$$[M\left[ \left[ 1-\bar{R}u_{,ss}^{2} \right]\left[ u_{,tt}^{1}-\bar{R}u_{,stt}^{2} \right]+u_{,s}^{2}u_{,tt}^{2} \right]]=[M]\cdot [L/{{t}^{2}}]=[F]$$

Thus far, we have shown that each term in $${{c}_{o}}$$ has a dimension of force, $$[F]$$.

For the physical meaning of each term, in general, they are contributions of external forces $$({{F}^{1}},{{F}^{2}},T)$$ and the inertia term $$Ma$$.

= Problem 3 - Linearity Test of Maglev Train EOM Coefficient =

Problem Statement
C3 is defined as


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C_3(Y^1,t) \triangleq M\left [\left( 1-\bar Ru^2,_{ss}\left(Y^1,t\right)\right)\right]^2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (1)
 * }
 * }

Using the definition of linear operator, show that $$\displaystyle C_3$$ is nonlinear.

Solution
A linear operator is defined as follows:
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$$\displaystyle L(cx)=cL(x)$$ (1)
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$$\displaystyle L(x+y)=L(x)+L(y)$$ (2)
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 * 
 * }

where H is an arbitrary function and c is an arbitrary constant.

Consider
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$$\displaystyle {{C}_{3}}(x+y,t)=M{{[(1-\overline{R}_{,ss}^{2}(x+y,t))]}^{2}}$$ (3)
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 * 
 * }

and
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$$\displaystyle {{C}_{3}}(x,t)=M{{[(1-\overline{R}_{,ss}^{2}(x,t))]}^{2}}$$ (4)
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 * 
 * }


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$$\displaystyle {{C}_{3}}(y,t)=M{{[(1-\overline{R}_{,ss}^{2}(y,t))]}^{2}}$$ (3)
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 * 
 * }

In general, $$(3)\ne (4)+(5)$$, i.e., (2) does not hold.

Therefore, $$\displaystyle C_3(Y^1,t)$$ is nonlinear.

Jiangpx 05:13, 8 September 2010 (UTC)

=Problem 4 - Finding Coefficient from the given BVP=

= Problem 5 - Verify the Solutions to Legendre Equation (n=1) =

Problem Statement
Given

$$(1-{{x}^{2}}){{y}^{''}}-2x{{y}^{'}}+2y=0$$, and its two homogeneous solutions

$${{y}_{h1}}=x$$

$${{y}_{h2}}=\frac{x}{2}\cdot \log (\frac{1+x}{1-x})-1$$

Verify

$${{L}_{2}}({{y}_{h1}})={{L}_{2}}({{y}_{h2}})=0$$

Solution
Legendre equation with n=1:
 * {| style="width:100%" border="0"

$$\displaystyle (1-{{x}^{2}}){{y}^{''}}-2x{{y}^{'}}+2y=0$$ (1)
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 * 
 * }

1st solution is given as


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$$  \displaystyle {{y}_{h1}}=x $$  (2)
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 * 
 * }

then
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$$  \displaystyle y_{h1}^{'}=1 $$
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 * 
 * }


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$$  \displaystyle y_{h1}^{''}=0 $$
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 * 
 * }

It is obvious that eq.(1) is valid with $${{y}_{h1}}=x$$.

2nd solution is given as
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$$  \displaystyle {{y}_{h2}}=\frac{x}{2}\cdot \log (\frac{1+x}{1-x})-1 $$     (3)
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 * }

and the matlab code has developed as follows to verify the solution

Result shows that A = 0, i.e., eq.(1) holds.

Therefore yh1 and yh2 are two solutions to the homogeneous equation.

Jiangpx 05:14, 8 September 2010 (UTC)

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= Reference =