User:Egm6321.f10.team3.franklin/Hwk1

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

Given
$$\displaystyle C_3$$ from the Maglev Train EOM is defined as


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

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 (Eq. 1)
 * }
 * }

Find
Using the definition of linearity of a function, show that $$\displaystyle C_3$$ is nonlinear, with respect to $$\displaystyle Y^1$$.

Solution
A linear function is a function which obeys the two functions below. $$\displaystyle H$$ is an arbitrary function while $$\displaystyle c$$ is an arbitrary constant.


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

H(cx) = cH(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * style= |
 * }


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

H(x+y) = H(x) + H(y) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * style= |
 * }

In order for $$\displaystyle C_3(Y^1,t)$$ to be linear, the following equation must be true.


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

$$
 * $$\displaystyle C_3(cx,t) = cC_3(x,t)$$
 * $$\displaystyle (Eq. 4)
 * $$\displaystyle (Eq. 4)
 * style= |
 * }

Applying the cumulative property to (Eq. 1) gives us


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

$$
 * $$\displaystyle C_3(cx,t) = M\left [\left( 1-\bar Ru^2,_{ss}\left(cx,t\right)\right)\right]^2.$$
 * $$\displaystyle (Eq. 5)
 * $$\displaystyle (Eq. 5)
 * style= |
 * }

Let us assume the function $$\displaystyle u^2,_{ss}$$ is linear, (Eq. 5) becomes


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

$$
 * $$\displaystyle C_3(cx,t) = M\left [\left( 1-c\bar Ru^2,_{ss}\left(x,t\right)\right)\right]^2.$$
 * $$\displaystyle (Eq. 6)
 * $$\displaystyle (Eq. 6)
 * style= |
 * }

Now we evaluate the right half of (Eq. 4)


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

$$
 * $$\displaystyle cC_3(x,t) = cM\left [\left( 1-\bar Ru^2,_{ss}\left(x,t\right)\right)\right]^2.$$
 * $$\displaystyle (Eq. 7)
 * $$\displaystyle (Eq. 7)
 * style= |
 * }

Since


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

$$
 * $$\displaystyle C_3(cx,t) \not= cC_3(x,t),$$
 * $$\displaystyle (Eq. 8)
 * $$\displaystyle (Eq. 8)
 * style= |
 * }

$$\displaystyle C_3(Y^1,t)$$ is nonlinear because it did not meet the linearity test of a function.

=Problem 4=

Hello, I am Hong SJ.. First of all, I am so sorry, gentlemen.. Please, be tolerant the fact that I post this on the wiki page. I am supposed to take problem 4 & 5.. My solution for #5 is almost same with Sudheesh's. However, my solution for #4 is different from the solution posted this wiki page... (I am sorry, I just found it...)

As you know, we do not have time to hold a meeting for this because the deadline for HW1 is tomorrow. That's why I should post my solution for #4 on this page in haste before the deadline. I think we should recheck this problem (#4), even if I am not sure whether my solution is correct, or not..

Here is my solution...

Given

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

$$ y \left( a\right) =  \alpha $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


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

$$ y \left( b\right) =  \beta $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Find
Find $$c,d$$ in terms of $$\alpha$$ and $$\beta$$

Solution
y(x) was defined as the following.


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

$$ y \left( x\right) =  c {y}^{1}_{H}\left( x\right) + d {y}^{2}_{H} \left( x\right) + {y}_{p} \left( x\right) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * $$\displaystyle
 * }

Apply given factors, y(a) & y(b) to y(x)


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

$$ y \left( a\right) =  c {y}^{1}_{h}\left( a\right) + d {y}^{2}_{h} \left( a\right) + {y}_{p} \left( a\right) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * $$\displaystyle
 * }


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

$$ y \left( b\right) =  c {y}^{1}_{h}\left( b\right) + d {y}^{2}_{h} \left( b\right) + {y}_{p} \left( b\right) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * $$\displaystyle
 * }

If we multiply {y}^{1}_{b}\left( b\right) to y(a), we can find the equation below.


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

$$ c{y}_{h}^1(a){y}_{h}^2(b)+d{y}_{h}^2(a){y}_{h}^2(b)+{y}_{p}(a){y}_{h}^2(b)=\alpha{y}_{h}^2(b) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle
 * }

Also, let's multiply {y}^{2}_{a}\left( b\right) to y(b).


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

$$ c{y}_{h}^1(b){y}_{h}^2(a)+d{y}_{h}^2(b){y}_{h}^2(a)+{y}_{p}(b){y}_{h}^2(a)=\beta{y}_{h}^2(a) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle
 * }

By solving simultaneous equations above, we can detemine the equation with respect to c


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

$$ c{y}_{h}^1(a){y}_{h}^2(b) - c{y}_{h}^1(b){y}_{h}^2(a) + {y}_{p}(a){y}_{h}^2(b) - {y}_{p}(b){y}_{h}^2(a) =  \alpha{y}_{h}^2(b) - \beta{y}_{h}^2(a) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle
 * }


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

$$ c = \frac{\alpha{y}_{h}^2(b)-\beta{y}_{h}^2(a) -{y}_{p}(a){y}_{h}^2(b))+{y}_{p}(b){y}_{h}^2(a)}{{y}_{h}^1(a){y}_{h}^2(b)-{y}_{h}^2(a){y}_{h}^1(b)} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\displaystyle
 * }

In the same way, we can determine the equation with respect with d.


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

$$ c{y}_{h}^1(a){y}_{h}^1(b)+d{y}_{h}^2(a){y}_{h}^1(b)+{y}_{p}(a){y}_{h}^1(b)=\alpha{y}_{h}^1(b) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle
 * }


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

$$ c{y}_{h}^1(b){y}_{h}^1(a)+d{y}_{h}^2(b){y}_{h}^1(a)+{y}_{p}(b){y}_{h}^1(a)=\beta{y}_{h}^1(a) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle
 * }


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

$$ d = \frac{\alpha{y}_{h}^1(b)-\beta{y}_{h}^1(a) -{y}_{p}(a){y}_{h}^1(b))+{y}_{p}(b){y}_{h}^1(a)}{{y}_{h}^1(a){y}_{h}^2(a)-{y}_{h}^1(a){y}_{h}^2(b)} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\displaystyle
 * }

Hong SJ Egm6321.f10.team3.Hong SJ 11:12, 14 September 2010 (UTC)

=Notes and references=