User:Egm6321.f12.team6/report5

=Report 5= ==Problem 5.1: Matrix Exponentiation ==

Problem Statement: The Transpose Operation

 * For a square matrix of dimensions, $$ n\times n $$, Show that;

Given

 * Properties of Matrices

Nomenclature

 * The identity matrix is denoted by, $$ \mathbf I $$

Solution

 * Given equation 1.1, we will first expand the left had side of the equation using the definition of exponentiation found in equation 1.2


 * With this result in mind, we will now analyze the right hand side of equation 1.1. Inspection of the right hand side shows that we can break up the computation into two parts: 1st- Evaluate the exponentaion of, $$ \mathbf A $$, and 2nd- Take the transpose. We'll first evaluate the exponentiation of, $$ \mathbf A $$. This is done by using the definition of exponentiation found in equation 1.2, and


 * Next, we will take the transpose of equation 1.6 giving us,


 * Using equation 1.7, we can apply a property of the transpose operation found in equation 1.4 which gives us,


 * Recognizing that, $$ \mathbf I^T=\mathbf I $$, and using equation 1.8 to apply another property of the transpose operation found in equation 1.3 gives us,


 * From equation 1.5 we see that equation 1.9 is in fact, $$ \exp[\mathbf A^T] $$, and we therefore verify equation 1.1 which says;


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle \exp [\mathbf A^T]=(\exp [\mathbf A])^T $$


 * (1.10)
 * }

Part b) Find t

 * For a projectile model, we were expecting to get a parabolic curve, however it turns out that we only derive the above figure that doesn't seem to have the correct physical meaning.
 * From the above figure, one can easily find $$x=0.2299$$ when $$y$$ goes to zero.
 * And since $$z(t)$$ becomes zero when the projectile returns to ground, we can conclude that
 * {| style="width:100%" border="0"|-


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$t=0.2299$$


 * }

Problem Statement

 * For L2-ODE-VC


 * a) Verify its exactness
 * b) Show if it is in the power form of eq(3) p21-1 in the lecture notes
 * c) Verify that :$$ F(a,b;c;x)$$ is indeed a solution of eq(12.1)

Given

 * Power form for L2-ODE-VC:

Nomenclature

 * L2-ODE-VC: Linear 2nd Order ODE with varying coefficients
 * $$ F(a,b;c;x)$$: Hypergeometric Function
 * $$ \displaystyle p=y'=\frac{dy}{dx}$$
 * $$ \displaystyle y''=\frac{d^2y}{dx^2}$$
 * $$ \displaystyle \phi_x(x,y,p) =\frac{\partial\phi(x,y,p)}{\partial x}, \phi_y(x,y,p) =\frac{\partial \phi(x,y,p)}{\partial y} $$
 * $$ \displaystyle f_{xx}=\frac{\partial^2 f(x,y,p)}{\partial x^2}, f_{xy}=\frac{\partial^2 f(x,y,p)}{\partial x \partial y}, f_{xp}=\frac{\partial^2 f(x,y,p)}{\partial x \partial p} $$
 * $$ \displaystyle f_y=\frac{\partial f(x,y,p)}{\partial y}, f_{yy}=\frac{\partial^2 f(x,y,p)}{\partial y^2}, f_{yp}=\frac{\partial^2 f(x,y,p)}{\partial y \partial p} $$
 * $$ \displaystyle g_y=\frac{\partial g(x,y,p)}{\partial y}, g_{xp}=\frac{\partial^2 g(x,y,p)}{\partial x \partial p}, g_{yp}=\frac{\partial^2 g(x,y,p)}{\partial y \partial p}, g_{pp}=\frac{\partial^2 g(x,y,p)}{\partial p^2} $$

Part a) Verify Exactness

 * Re-write equation (12.1):


 * It follows that


 * Thus equation(12.1) satisfies the 1st exactness condition


 * To verify the second exactness condition, one just need to show that if equation (12.1) satisfies the following equations:


 * Calculate partial derivatives from equation (12.5) and (12.6)
 * $$ f_{xx}=-2$$
 * $$ f_{xy}=f_{yy}=g_{yp}=0$$
 * $$ g_{xp}=-(a+b+1)$$
 * $$ g_{y}=-ab$$
 * $$ f_{xp}=f_{yp}=f_{y}=g_{pp}=0$$


 * Bring them into equation (12.7) and (12.8), we have
 * $$ f_{xx}+2p\,f_{xy}+p^2\,f_{yy}=-2+0+0=-2 $$
 * $$ g_{xp}+p\,g_{yp}-g_y=-(a+b+1)+0-(-ab)=ab-(a+b+1) $$
 * Thus only when $$ a,b $$ are chosen such that $$ab-(a+b+1)=-2$$, then equation (12.7) is satisfied
 * $$ f_{xp}+p\,f_{yp}+2\,f_{y}=0+0+0=0 $$
 * $$ g_{pp}=0 $$
 * Thus equation (12.8) is satisfied


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" |


 * Therefore, only when $$ a,b $$ are chosen such that $$ab-(a+b+1)=-2$$, both the first and second exactness conditions are satisfied, and eq(12.1) is said to be exact. Otherwise, eq(12.1) is not exact.


 * }

Part b) Compare with Power Form

 * Compare terms in eq(12.1) with those of eq(12.2):
 * $$ \underbrace{x(1-x)\,y''}_{\displaystyle\color{blue}{part1}} +\underbrace{[c-(a+b+1)x]\,y'}_{\displaystyle\color{blue}{part2}} -\underbrace{ab\,y}_{\displaystyle\color{blue}{part3}}$$
 * $$ \underbrace{\alpha\,x^r y''}_{\displaystyle\color{blue}{part1}}+\underbrace{\beta\,x^s y'}_{\displaystyle\color{blue}{part2}} +\underbrace{\gamma\,x^t y }_{\displaystyle\color{blue}{part3}}$$
 * Part1: Since there are two terms containing $$x$$ in the first part of eq(12.1), while there is only one term containing $$x$$ in the first part of eq(12.2), part1 of eq(12.1) cannot be equal to that of eq(12.2)
 * Part2: If we choose $$\beta=-(a+b+1),s=1,c=0$$, then part2 of eq(12.1) has the same form as that of eq(12.2)
 * Part3: If we choose $$\gamma=-ab,t=0$$, then part3 of eq(12.1) has the same form as that of eq(12.2)


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" |


 * In conclusion, since part1 in eq(12.1) cannot be written into the corresponding part of eq(12.2), we say that eq(12.1) is not in the power form of eq(12.2)


 * }

Part c) Verify Solution

 * Eq(12.3) is equivalent to


 * Taking the first derivative of y with respect to x yields


 * Taking the second derivative of y with respect to x yields


 * Expanding eq(12.1) gives


 * For the left hand side of eq(12.12), substituting $$ y'',y',y $$ with hypergeometric functions from eq(12.9)~ eq(12.11) yields


 * Which is equivalent to


 * In order for the polynomial in (12.14) to be equal to zero, we want the sum of coefficiets of $$ x^k $$ in all five terms to vanish.
 * In other words, one just need to prove that the sum of coefficients of $$ x^k $$ is always equal to zero for any value of integar $$ k $$, where $$ k \in [0,+\infty)$$
 * Note that term1, term4 do not appear until k=1, and term2 does not appear until k=2. Thus before analyzing the coefficients in general case, we need to analyze the special cases when k=0 and k=1 in the interest of completeness.

Case 1 $$ k=0 $$

 * Polynomial (12.14) goes to

Case 2 $$ k=1 $$

 * Polynomial (12.14) goes to

Case 3 $$ k\in [2,+\infty) $$

 * Polynomial (12.14) goes to


 * $$ x^k\,[\frac{(a)_{k+1}(b)_{k+1}}{(c)_{k+1}}\frac{1}{(k-1)!} - \frac{(a)_{k}(b)_{k}}{(c)_{k}}\frac{1}{(k-2)!} + c\,\frac{(a)_{k+1}(b)_{k+1}}{(c)_{k+1}}\frac{1}{k!} -(a+b+1)\,\frac{(a)_{k}(b)_{k}}{(c)_{k}}\frac{1}{(k-1)!}-ab\,\frac{(a)_{k}(b)_{k}}{(c)_{k}}\frac{1}{k!}]$$


 * $$ = x^k\,[\frac{(a+k)(b+k)}{(c+k)(k-1)}\,\frac{(a)_{k}(b)_{k}}{(c)_{k}}\frac{1}{(k-2)!} - \frac{(a)_{k}(b)_{k}}{(c)_{k}}\frac{1}{(k-2)!} + c\,\frac{(a+k)(b+k)}{(c+k)(k-1)k}\,\frac{(a)_{k}(b)_{k}}{(c)_{k}}\frac{1}{(k-2)!} $$
 * $$ - \frac{a+b+1}{k-1}\,\frac{(a)_{k}(b)_{k}}{(c)_{k}}\frac{1}{(k-2)!} - \frac{ab}{k(k-1)}\,\frac{(a)_{k}(b)_{k}}{(c)_{k}}\frac{1}{(k-2)!}]$$


 * $$ = x^k\,\frac{(a)_{k}(b)_{k}}{(c)_{k}}\frac{1}{(k-2)!}\,[\frac{(a+k)(b+k)}{(c+k)(k-1)} - 1 + c\,\frac{(a+k)(b+k)}{(c+k)(k-1)k} - \frac{a+b+1}{k-1} - \frac{ab}{k(k-1)}] $$


 * $$ = x^k\,\frac{(a)_{k}(b)_{k}}{(c)_{k}k(c+k)(k-1)}\frac{1}{(k-2)!}\,[(a+k)(b+k)k - (c+k)(k-1)k + c(a+k)(b+k) - (a+b+1)k(c+k)-ab(c+k)] $$


 * $$ = x^k\,\frac{(a)_{k}(b)_{k}}{(c)_{k}k(k-1)}\frac{1}{(k-2)!}\,[ab+ak+bk+k^2-k^2+k-ak-bk-k-ab] $$


 * $$ = x^k\,\frac{(a)_{k}(b)_{k}}{(c)_{k}k(k-1)}\frac{1}{(k-2)!}*0 $$


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" |


 * In general, it has been shown that for each value of $$ k $$, polynomial(12.14) equals to zero, which means that eq(12.1) is satisfied when substituting hypergeometric functions into it. Therefore we can conclude that $$ F(a,b;c;x)$$ is indeed a solution of eq(12.1)


 * }

Problem Statement

 * Consider the following governing L2-ODEs-VC for some classical special functions:
 * Legendre:


 * Bessel:


 * Hermite


 * 1) Verify the exactness of the designated L2-ODEs-VC. For the 2nd exactness condition, use 2 methods:
 * 1a.(1)-(2)p.16-5
 * 1b.(1)p.22-3
 * 2) If 13.3 is not exact, check whether it is in power form, and see whether if it can be made exact using IFM with $$h(x,y)=x^my^n$$.
 * 3) The first few Hermite polynomials are


 * Verify that(1)-(3)are homogeneous solutions of the Hermite differential equation 13.3.

Given

 * The first method of second exactness condition for 2nd-ODEs


 * The second method of second exactness condition for 2nd-ODEs

Nomenclature

 * $$f_{xy}(x)=\frac{\partial f(x)}{\partial x\partial y}$$
 * $$g_n=\frac{\partial G}{\partial y^{(n)}}$$
 * where $$y^{(n)}$$means the nth derivative of $$y$$

Part 1 Verify the exactness

 * 1.1 Verify the first method of second exactness condition of equation 13.1.
 * According to equation 13.1,


 * Obviously, the first exactness condition is satisfied, and the following can be derivated


 * According to 13.7 and 13.10, if the ODE is exact, the following equation can be acquired


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" |


 * The ODE is exact only if $$n=-1$$ or $$n=0$$


 * }


 * 1.2 Verify the second method of second exactness condition of equation 13.1.


 * The ODE is exact if the following equation is satisfied. According to 13.8 and 13.12


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" |


 * The ODE is exact only if $$n=-1$$ or $$n=0$$


 * }


 * 2.1 Verify the first method of second exactness condition of equation 13.2.
 * According to equation 13.2,


 * Obviously, the first exactness condition is satisfied, and the following can be derivated


 * According to 13.7 and 13.15, if the ODE is exact, the following equation can be acquired


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" |


 * The ODE is exact only if $$x=v$$ or $$x=-v$$


 * }


 * 2.2 Verify the second method of second exactness condition of equation 13.2.


 * The ODE is exact if the following equation is satisfied. According to 13.8 and 13.17


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" |


 * The ODE is exact only if $$x=v$$ or $$x=-v$$


 * }


 * 3.1 Verify the first method of second exactness condition of equation 13.3.
 * According to equation 13.3,


 * Obviously, the first exactness condition is satisfied, and the following can be derivated


 * According to 13.7 and 13.20, if the ODE is exact, the following equation can be acquired


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" |


 * The ODE is exact only if $$n=-1$$


 * }


 * 3.2 Verify the second method of second exactness condition of equation 13.3.


 * The ODE is exact if the following equation is satisfied. According to 13.8 and 13.22


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" |


 * The ODE is exact only if $$n=-1$$


 * }

Part 2 Verify the IFM factor

 * The equation 13.3 can be written as


 * So it satisfies the power form.
 * Assume that $$h(a,b)=x^ay^b$$is the integrating factor. Multiply $$h(a,b)$$ with equation 13.3.


 * So the following can be acquired


 * According to 13.8 and 13.26, if the ODE is exact, the following equation can be derived.


 * There are two value groups of a,b,n that could make this equation valid.


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" |


 * In other words, when $$ n=-1$$, one could find $$h(x,y)=1$$ to make 13.3 exact, when $$ n=-2 $$, one could find $$ h(x,y)=x $$ to make 13.3 exact.
 * In other cases, equation can't be made exact using IFM with $$h(x,y)=x^my^n$$


 * }

Part 3 Verify the solution

 * Put the first solution $$H_0(x)=1$$ into equation 13.3,


 * The equation is satisfied.


 * Put the second solution $$H_1(x)=2x$$ into equation 13.3,


 * The equation is satisfied.


 * Put the third solution $$H_2(x)=4x^2-2$$ into equation 13.3,


 * The equation is satisfied.


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" |


 * So that the
 * $$H_0(x)=1$$
 * $$H_1(x)=2x$$
 * $$H_2(x)=4x^2-2$$
 * are homogeneous solutions for 13.3


 * }

Problem Statement
Find the expressions for$$ \displaystyle X(x)$$in terms of $$ \displaystyle \cos K x, \ \sin K x, \ \cosh K x, \ \sinh K x $$

Given

 * L4-ODE-CC
 * {| style="width:100%" border="0"

$$ \displaystyle X^{(4)}-K^4X=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * (14.1)
 * }
 * Assume
 * {| style="width:100%" border="0"

$$ \displaystyle X(x)=e^{(rx)} $$
 * style="width:95%" |
 * style="width:95%" |
 * (14.2)
 * }
 * Solution
 * {| style="width:100%" border="0"

$$ \displaystyle r_{1,2}=\pm K $$
 * style="width:95%" |
 * style="width:95%" |
 * (14.3)
 * }
 * and
 * {| style="width:100%" border="0"

$$ \displaystyle r_{3,4}=\pm i\,K$$
 * style="width:95%" |
 * style="width:95%" |
 * (14.4)
 * }

Nomenclature
$$ \displaystyle i:=\sqrt{-1} $$

Solution

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

$$ \displaystyle X{(x)}=c_1e^{Kx}+c_2e^{-Kx}+c_3e^{iKx}+c_4e^{-iKx} $$
 * style="width:95%" |
 * style="width:95%" |
 * (14.5)
 * }
 * Eqn.(14.5) can be separate for two parts
 * First,
 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} c_1e^{Kx}+c_2e^{-Kx} &= c_1(\frac{e^{Kx}+e^{-Kx}}{2}+\frac{e^{Kx}-e^{-Kx}}{2}) + c_2(\frac{e^{Kx}+e^{-Kx}}{2}-\frac{e^{Kx}-e^{-Kx}}{2}) \\&= c_1(\cosh Kx+ \sinh Kx) + c_2(\cosh Kx- \sinh Kx)\\ &=(c_1+c_2)\cosh Kx + (c_1-c_2)\sinh Kx \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * (14.6)
 * }
 * Second,
 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} c_3e^{iKx}+c_4e^{-iKx} &=c_3(\cos Kx +i\sin Kx)+c_4(\cos Kx -i\sin Kx) \\ &= (c_3+c_4)\cos Kx + (c_3-c_4)i\sin Kx \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * (14.7)
 * }
 * Combine with two parts:
 * {| style="width:100%" border="0"|-

$$ \displaystyle X{(x)}=(c_1+c_2)\cosh Kx + (c_1-c_2)\sinh Kx + (c_3+c_4)\cos Kx + (c_3-c_4)i\sin Kx $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * (14.8)
 * }

Team Contribution

 * {| class="prettytable"


 * colspan="4" | Team Contribution Table
 * colspan="4" | Compiled by: Mutian Hu
 * colspan="4" | Compiled by: Mutian Hu


 * Problems
 * Solved and Typed By
 * Solved and Inspected By
 * Skimmed By
 * Problem 5.1
 * Brandon Taylor
 * Mutian Hu
 * All Team Members
 * Problem 5.2
 * Brandon Taylor
 * Kyung-Hoon Bang
 * All Team Members
 * Problem 5.3
 * Kyung-Hoon Bang
 * CheMing Lee
 * All Team Members
 * Problem 5.4
 * Kyung-Hoon Bang
 * Brandon Taylor
 * All Team Members
 * Problem 5.5
 * Brandon Taylor
 * Mutian Hu
 * All Team Members
 * Problem 5.6
 * Tianyu Gu
 * Jiajun Han
 * All Team Members
 * Problem 5.7
 * Cheming Lee
 * Brandon Taylor
 * All Team Members
 * Problem 5.8
 * Mutian Hu
 * Brandon Taylor
 * All Team Members
 * Problem 5.9
 * Jiajun Han
 * CheMing Lee
 * All Team Members
 * Problem 5.10
 * CheMing Lee
 * Jiajun Han
 * All Team Members
 * Problem 5.11
 * Tianyu Gu
 * Kyung-Hoon Bang
 * All Team Members
 * Problem 5.12
 * Tianyu Gu
 * Jiajun Han
 * All Team Members
 * Problem 5.13
 * Jiajun Han
 * Tianyu Gu
 * All Team Members
 * Problem 5.14
 * CheMing Lee
 * Mutian Hu
 * All Team Members
 * }
 * Tianyu Gu
 * Jiajun Han
 * All Team Members
 * Problem 5.13
 * Jiajun Han
 * Tianyu Gu
 * All Team Members
 * Problem 5.14
 * CheMing Lee
 * Mutian Hu
 * All Team Members
 * }
 * All Team Members
 * }