text:
University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg37

Mtg 37: Thurs, 17Nov09

[[media: Egm6321.f09.mtg37.djvu | Page 37-1]]
[[media: Egm6321.f09.mtg36.djvu | P.36-4]] continued $$ P_n(x) \ $$ $$ P_1(x)=x \ $$ $$ Q_n(x) \ $$ From [[media: Egm6321.f09.mtg18.djvu  |  P.18-1 ]]: HW show $$ Q_1(x)=\frac{1}{2}xlog \left ( \frac{1+x}{1-x} \right ) -1 = x \tanh ^{-1}x-1 \ $$ ref [[media: http://books.google.com/books?id=9Cg3HWCnCjAC&printsec=frontcover&dq=differential+equations+billingham&ei=pGR4SpPVLojSMpb07Qw#v=onepage&q=&f=false | K p33 ]] for $$ Q_2, Q_3...\ $$ $$ Q_0(x)= \tanh ^{-1}(x) \ \ $$ is odd HW: Use Eq(2) to show when $$ Q_n \ $$ is even or odd, depending on "n" HW: Plot $$ \left \{ P_0, P_1,...,P_4 \right \} \ $$ and $$ \left \{ Q_0, Q_1,...,Q_4 \right \} \ $$ Legendre function $$ L_n(x)=P_n(x) \ $$ or $$ Q_n(x) \ $$ solution of Legendre solution

[[media: Egm6321.f09.mtg37.djvu | Page 37-2]]
for $$ n \ne m \ $$ for $$ L_n =P_n, \left \langle L_n,L_n \right \rangle = \left \langle P_n,P_n \right \rangle = \frac{2}{2n+1}  \ $$ for $$ \left \langle L_n,L_n \right \rangle = \left \langle P_n,Q_n \right \rangle \ =  $$ HW $$ \left \langle L_n,L_m \right \rangle =  \left \langle P_n,P_m \right \rangle \ =  $$ [[media: Egm6321.f09.mtg33.djvu  | Eq.(3) P.33-1]] $$ \left \langle L_n,L_m \right \rangle = \left \langle P_n,Q_m \right \rangle \ = 0 $$ for $$ n \ne m \ $$ Proof: Legendre equation, [[media: Egm6321.f09.mtg14.djvu | Eq.(1) P.14-2 ]]: 2) $$ \left [ (1-x^2)y' \right ]'+n(n+1)y=0 \ $$   Where $$ \Rightarrow \ \left [ (1-x^2)L_n' \right ]'+n(n+1)L_n=0 \ $$  Multiply by $$ L_m \ $$ and integrate from  -1 to +1: $$ \int_{-1}^{1} L_m \left [ (1-x^2)L_n' \right ]' \, dx + n(n+1) \int_{-1}^{1} L_mL_n \, dx=0 \ $$ Where  $$ L_m \left [ (1-x^2)L_n' \right ]' = \alpha\ \ $$

[[media: Egm6321.f09.mtg37.djvu | Page 37-3]]
Integrate $$ \alpha\ \ $$ by parts: Interchange n and m: Eq(1)-Eq(2): $$ \left [ n(n+1)-m(m+1) \right ] \left \langle L_n,L_m \right \rangle =0 $$ Where $$ \left [ n(n+1)-m(m+1) \right ] \ne 0 $$ since $$ n \ne m \ $$ $$ \Rightarrow \ \left \langle L_n,L_m \right \rangle =0 \ $$ when $$ n \ne m \ $$ [[media: http://books.google.com/books?id=9Cg3HWCnCjAC&printsec=frontcover&dq=differential+equations+billingham&ei=pGR4SpPVLojSMpb07Qw#v=onepage&q=&f=false | cf.K.p41]]


text:
University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg39

EGM6321 - Principles of Engineering Analysis 1, Fall 2009
Mtg 39: Thurs, 19Nov09

[[media: Egm6321.f09.mtg39.djvu | Page 39-1]]
Note: Other Application: Quantum Mechanics discrete variables <-- coding theory Ref: [[media: http://books.google.com/books?id=4Q0ZAQAAIAAJ&q=nikiforov+functions&dq=nikiforov+functions&ei=EAsAS8yGKZbQNO-9vYkP | Nikiforov, et.al (1991) ]] Probability (Queing theory, birth and death) Generating functions: -Legendre Polynomial $$ P_n \ $$: [[media: Egm6321.f09.mtg38.djvu | Eq.(5) P.38-3]] - $$ {r \choose k} \ = $$ "r choose k" : $$ (1+x)^r = \sum_{k=0}^\infty {r \choose k} x^k \ $$ for $$ \left | x \right \vert \le 1 \ $$ $$ A( \mu\, \rho\ ):=1-2 \mu\ \rho\ + \rho\ ^2) \ $$ From [[media: Egm6321.f09.mtg38.djvu | Eq.(6) P.38-3]] and [[media: Egm6321.f09.mtg38.djvu | Eq.(7) P.38-3]] : $$ \frac{1}{\sqrt{A( \mu\ , \rho\ )}} = \alpha\ _0 + \alpha\ _1(2 \mu\ \rho\ - \rho\ ^2)+ \alpha\ _2(2 \mu\ \rho\ - \rho\ ^2)^2+...= \alpha\ _0+ (2 \mu\ \alpha\ _1) \rho\ + (-\alpha\ _1+4 \mu\ ^2 \alpha\ _2) \rho\ ^2 +...  \ $$ Where $$ (2 \mu\ \rho\ - \rho\ ^2) = 4 \mu\ ^2 \rho\ ^2-4 \mu\ \rho\ ^3 + \rho\ ^4 \ $$ and $$ \alpha\ _0 = P_0( \mu\ ) \ $$ and $$ 2 \mu\ \alpha\ _1 = P_1( \mu\ ) \ $$ and $$ (-\alpha\ _1+4 \mu\ ^2 \alpha\ _2) = P+2( \mu\ ) \ $$

[[media: Egm6321.f09.mtg39.djvu | Page 39-2]]
HW: Continue the power series development to find $$ P_3, P_4, P_5 \ $$ and complete result to that obatined by [[media: Egm6321.f09.mtg31.djvu | Eq.(6) P.31-3]] or [[media: Egm6321.f09.mtg31.djvu  | Eq.(7) P.31-3]] 2 recurrence formulas Plan: Find $$ \frac{d}{d \mu\ } \frac{1}{ \sqrt{A}} \ $$ and $$ \frac{d}{d \rho\ } \frac{1}{ \sqrt{A}} \ $$ 1) $$\frac{d}{d \rho\ } \frac{1}{ \sqrt{A}} = -\frac{1}{2}A^{\frac{-3}{2}}\frac{dA}{d \mu\ } = \frac{ \rho\ }{A^{\frac{-3}{2}}} \ $$ Recall [[media: Egm6321.f09.mtg38.djvu | Eq.(5) P.38-3]], now  $$ \frac{d}{d \mu\ } \frac{1}{ \sqrt{A}} = \frac{ \rho\ }{A^{\frac{3}{2}}} =  \ $$ Where $$ P_0( \mu\ )=1 \Rightarrow \ P_0'=0 \ $$ and $$ P_n'( \mu\ ) = \frac{d}{d  \mu\ } P_n( \mu\ ) \ $$

[[media: Egm6321.f09.mtg39.djvu | Page 39-3]]
Compare Eq(4) and Eq(5) $$ \frac{ \rho\ ( \mu\ - \rho\ )}{A^{\frac{3}{2}}} = ( \mu\ - \rho\ ) \sum_{n=1}^\infty P_n' ( \mu\ ) \rho\ ^{n} = \rho\ \sum_{n=0}^\infty P_n ( \mu\ ) n \rho\ ^{n-1} = \rho\ \sum_{n=1}^\infty P_n { \mu\ } n \rho\ ^{n-1} $$ Where n=1 in summation due to factor u being introduced $$ \Rightarrow \ \mu\  \sum_{n=1} P_n' \rho\  - \sum_{n=1} P_n' \rho\ ^{n-1} = \sum_{n=1} P_n \mu\ \rho\  \ $$ $$ \Rightarrow \ \mu\ P_1' \rho\ - P_1 1 \rho\ + \sum_{n=2} \left [ -P_{n-1}' + \mu\ P_n'-nP_n \right ] \rho\ ^n = 0 \ $$

[[media: Egm6321.f09.mtg39.djvu | Page 39-4]]
$$ \Rightarrow \ -P_{n-1}' + \mu\ P_n' - nP_n = 0 \ $$

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


text:
University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg41

EGM6321 - Principles of Engineering Analysis 1, Fall 2009
Mtg 41: Tues, 1Dec09

[[media: Egm6321.f09.mtg41.djvu | Page 41-1]]
Review for exam 2 - Historical development - Legendre functions Question: How to obtain $$ P_n \ $$ based on known $$ P_{n-1}, P_{n-2},... \ $$ ? - 2 recurring relationships. Same technique in power series. Solution: Frobenius method Question: Find a differential equation governing all $$ \left \{ P_n \right \} \ $$ ? - Legendre differential equations 2 families of homogeneous solutions: - Legendre functions= $$ \left \{ P_n \right \} \ $$ +  $$ \left \{ Q_n \right \} \ $$ $$ L_n = P_n \ $$ or $$ L_n = Q_n \ $$ Newtonian potential is solution of Laplace equation i.e., $$ \Delta\ \left ( \frac{1}{r} \right ) = 0 \ $$ $$ \frac{1}{r} = \frac{1}{r_{PQ}} = \frac{1}{r_Q} \left ( 1-2 \mu\ \rho\ + \rho\ ^2 \right )^{- \frac{1}{2}} \ $$

[[media: Egm6321.f09.mtg41.djvu | Page 41-2]]
$$ = \frac{1}{r_Q} \sum_{n} P_n ( \mu\ ) \rho\ ^n \ $$, where $$ = \rho\ := \frac{r_P}{r_Q} \ $$ $$ \Rightarrow \ \frac{1}{r} = \sum_{n} P_n ( \mu\ ) \frac{r_P^n}{r_Q^{n+1}} = \sum_{n} H_n \left ( \mu\, r_P, r_Q \right ) \ $$ , where $$ H_n \left ( \mu\ , r_P, r_Q \right ) = H_n \left ( x,y,z \right ) \ $$ $$ \Delta\ \frac{1}{r} = 0 = \sum_{n} \Delta\ H_n(x,y,z) \ \Rightarrow \ \Delta\ H_n = 0$$ Where this argument is based on the power series Laplace equations in a sphere axisymmetrical case [[media: Egm6321.f09.mtg29.djvu |  P.29-1 ]] separation of variables [[media: Egm6321.f09.mtg30.djvu | P.30-1]] General solution of axisymmetrical Laplace equations in a sphere $$ \psi\ (r, \theta\ ) = \sum_{n} (A_nr^n+B_nr^{n+1})(C_nP_n+D_nQ_n) \ $$ Where $$ A_n r^n + B_n r^{n+1} \ $$ can be found on [[media: Egm6321.f09.mtg31.djvu | P.31-2]] and $$ C_n P_n + D_n Q_n \ $$ can be found on [[media: Egm6321.f09.mtg32.djvu | P.32-1]] and $$ \mu\ = \sin \theta\ \ $$


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg1

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009=

Mtg 1: Tue, 25 Aug 09

[[media: Egm6321.f09.mtg1.djvu | Page 1-1]]
1-1 defined as (Meeting Number) - (Slide Number) German Transrapid Emsland 500 km/h, youtube, Uploaded by TransrapidSupporter on Feb 14, 2007 Vu-Quoc & Olson (1989) CMAME The magnet (i.e vehicle) and structure (i.e. guideway) interaction is shown below. Where: $$ y^1(t)= \ $$ nominal motion $$ u^1(s,t)= \ $$ axial displacement of guideway $$ u^2(s,t)= \ $$ transversal displacement of guideway

[[media: Egm6321.f09.mtg1.djvu | Page 1-2]]
$$ f(y^1(t),t) \ $$ Total time derivative of f is given as: Where $$ \dot y^1(t)=\frac{d}{dt}y^1(t) \ $$ $$ \frac{d}{dt}f=f_{,s}(y^1,t) \dot y^1+f_{,t}(y^1,t) $$ Where $$ \dot y^'= \dot y^2 $$ Coriolis forces - Dynamics

Material time derivative - Continuum Mechanics

Reynolds transport theorem - Continuum Mechanics is an example of a nonlinear ordinary differential equation (ODE)

title: User:EGM6321.F10.TEAM1.WILKS/Mtg1


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg3

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009=

Mtg 3: Thur, 27 Aug 09

[[media: Egm6321.f09.mtg3.djvu | Page 3-1]]
To find class homepage, go to Wikiversity: Main Page (http://en.wikiversity.org/wiki/Wikiversity:Main_Page) and search for--> user:egm6321.f09

My Wiki address is: http://clesm.mae.ufl.edu/wiki.vq/index.php/Main_Page

From [[media: Egm6321.f09.mtg2.djvu | Eq.(1)p.2-3]]: If $$ P(x)\ne \ 0 \ \forall x$$, divide throughout by $$ P(x) \ $$ to get:

Where:

$$\forall \ $$ is defined as "for all"

$$ a_2(x)=1 \ $$,

$$ a_1(x)=\frac{Q}{P} $$,

$$ a_0(x)=\frac{R}{P} $$, and

$$ f(x)=\frac{F}{P} $$

$$ \forall x_0 \ $$ such that $$ P(x_0) \ne\ 0 \ $$ then $$ x_0 \ $$ is a regular point

Any $$ x_0 \ $$ such that $$ P(x_0)=0 \ $$ is a regular point

[[media: Egm6321.f09.mtg3.djvu | Page 3-2]]
2nd order--> need 2 conditions to solve for 2 constraints

Boundary Value Problem (BVP)

Prescribe:

where $$ \alpha\ \ $$ and $$ \beta\ \ $$ are known values

Initial Value Problem (IVP)

Prescribe:

where $$ \alpha\ \ $$ and $$ \beta\ \ $$ are known values

Solve IVP by ODE from p3-1 Eq(1) or initial condition p3-2 Eq(2)

Two points:

1) Existence and uniqueness of solution

[[media: Egm6321.f09.mtg3.djvu | Page 3-3]]
2) Superposition based on linearity of differential operation L(.)

Where the 2 in $$ L_2(y) \ $$ is defined as 2nd order

$$ \forall u,v \ $$ in a function of x

and $$ \forall \alpha\, \beta\ \ $$ belonging to $$ \mathbb R \ $$ (scalars, real numbers);

$$ L( \alpha\ u+ \beta\ v)= \alpha\ L(u)+ \beta\ L(v) \ $$

Where $$ \mathbb R \ $$ is defined as a set of real numbers Example: Matrix Algebra

$$ \mathbf{A} \in \mathbb R \ ^{nxm} $$ matrix with n rows and m columns of real numbers

$$\forall \mathbf{u}, \mathbf{v} \in \mathbb R \ ^{mx1} \ $$ is a column matrix

$$ \forall \alpha\, \beta\, \in \mathbb R \ \ $$

[[media: Egm6321.f09.mtg3.djvu | Page 3-4]]
Clearly: $$ \mathbf{A}(\alpha\ \mathbf{u} + \beta\ \mathbf{v}) = \alpha\ \mathbf{A}\mathbf{u} + \beta\ \mathbf{A} \mathbf{v} $$

Example:

$$ \frac{d}{dx}(.) \ $$ is a linear operation

$$ (\alpha\ u+ \beta\ v)'= \alpha\ u'+\beta\ v' \ $$

linearity allows the use of superposition

$$ y=y_H+y_P \ $$

$$ L(y) = L(y_H) + L(y_P) \ $$, where the subscripts H and P stand for homogeneous and particular in respective order.

title: User:EGM6321.F10.TEAM1.WILKS/Mtg3


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg5

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 5: Thur, 03 Sept 09

[[media: Egm6321.f09.mtg5.djvu | Page 5-1]]
Note: [[media: Egm6321.f09.mtg4.djvu | Eq.(2)p.4-2 ]] and [[media: Egm6321.f09.mtg4.djvu | Eq.(4)p.4-2 ]]

$$ y'(x)=p(x) \ $$

Integrate from a to x:

$$ \int_{s=a}^{s=x} y'(s)\, ds = \int_{s=a}^{s=x} p(s)\, ds $$

$$ \left [ y(s) \right ]_{s=a}^{s=x} = y(x)-y(a) $$, where $$ y(a) = constant \ $$,

another way: $$ y(x)=\int_{}^{} p(x)\, dx +k $$

Where $$ \int_{}^{} p(x)\, dx = F(x) $$ and $$ k=constant \  $$

$$ \Rightarrow \ y(a)=F(a)+k \ $$

$$ \Rightarrow \ k=y(a)-F(a) \ $$

$$ \Rightarrow \ y(x)=F(x)-F(a)+y(a) \ $$

[[media: Egm6321.f09.mtg5.djvu | Page 5-2]]
But $$ F(x)-F(a)=\int_{s=a}^{s=x} p(s)\, ds \ $$

[[media: Egm6321.f09.mtg4.djvu | Eq.(3)p.4-2 ]]: Why this form of nonlinear 1st order ODE?

Most general form:

Application:

Where $$ x^2y^5+6(y')^2 \ $$ is defined as $$ F(x,y,y') \ $$

HW Show that $$ F(x,y,y')=0 \ $$ in Eq(3) is a nonlinear 1st order ODE.

Hint: Define the differential operator $$ D(.) \ $$ associated with Eq(3). END HW

[[media: Egm6321.f09.mtg5.djvu | Page 5-3]]
Form for exact nonlinear 1st order ODE:

$$ F(x,y,y')=0 \ $$ is exact if $$ \exists \ $$ a function $$ \phi\ (x,y) $$ such that

where:

$$ \exists \ $$ is defined as "there exists"

$$ \phi_x = M(x,y) $$

$$ \phi_y = N(x,y) $$

Multiply Eq1) thru by $$ dx \ $$ to get:

[[media: Egm6321.f09.mtg4.djvu | Eq.(3)p.4-2 ]] : $$ M(x,y)dx + N(x,y)dy=0 \ $$

NOTE: If $$ F(x,y,y')=0 \ $$ does not have the form:

Then $$ F(.) \ $$ cannot be exact.

Application: [[media: Egm6321.f09.mtg5.djvu | Eq.(3)p.5-2 ]] is not exact because of the nonlinear term $$ (y')^2 \ $$

[[media: Egm6321.f09.mtg5.djvu | Page 5-4]]
Exactness test (continued)

[[media: Egm6321.f09.mtg5.djvu | Eq.(1)p.5-3 ]]:

$$ M(x,y)= \phi_x(x,y) \ $$

$$ N(x,y)= \phi_y(x,y) \ $$

Since $$ \phi_{xy} = \phi_{yx} \ $$

and $$ \phi_{xy} = \frac{\partial ^2 \phi\ }{\partial x \partial y} \ $$

and $$ \phi_{yx} = \frac{\partial ^2 \phi\ }{\partial y \partial x} \ $$

and $$ \frac{\partial ^2 \phi\ }{{\partial x} {\partial y} } = (\phi_{x})_y \ $$

and $$ \frac{\partial ^2 \phi\ }{{\partial y} {\partial x} } = (\phi_{y})_x \ $$

Application: [[media: Egm6321.f09.mtg4.djvu | Eq.(1)p.4-3 ]] Not Exact

$$ M(x,y)=2x^2 + \sqrt{y} \Rightarrow \ M_y= \frac{1}{2 \sqrt{y}} \ $$

$$ N(x,y)=x^5y^3 \Rightarrow \ N_x= 5x^4y^3 \ $$

$$ M_y \ne \ N_x \ $$

title: User:EGM6321.F10.TEAM1.WILKS/Mtg5


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg7

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 7: Tues, 08Sept 09

[[media: Egm6321.f09.mtg7.djvu | Page 7-1]]
Mediawiki Hints: Alt+Shift+E = Edit Alt+Shift+P = Preview Alt+Shift+S = Save [[media: Egm6321.f09.mtg6.djvu | p.6-3]]: Case2: $$ h_xN=0 \ \Rightarrow \ -h_yM+h(N_x-M-y)=0 \Rightarrow \ \frac{h_y}{h}=\frac{1}{M} \left ( N_x-M_y \right )=-g(y) \ $$ HW  Complete details to obtain  $$ h_y \ $$ END HW

title: User:EGM6321.F10.TEAM1.WILKS/Mtg7


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg9

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 9: Thur, 10Sept09

[[media: Egm6321.f09.mtg9.djvu | Page 9-1]]
Note: Symbol notations: $$ \Delta\ $$ defined as equal by definition Non-symmetric notation $$ := $$ means equal by definition as well, better notation than $$ \Delta\ $$ Goal: Derive mathematical structure of a class of exact N1_ODEs Exact L1_ODE_VC: Application: Just invent any $$ a(x),b(x) \ $$ Let $$ a(x)=x^4 \ $$ $$ b(x)=x \Rightarrow \ \bar b \ (x)=\frac{1}{2}x^2 \ $$ $$ k=10 \ $$

[[media: Egm6321.f09.mtg9.djvu | Page 9-2]]
HW Show that Eq(3) L1_ODE_VC is "exact". See Note [[media: Egm6321.f09.mtg10.djvu | P.10-1 ]] END HW Question: How about N1_ODEs? Eq.(2)P.6-4 N1 means Nonlinear, 1st Order Where $$ \int_{}^{x} b(s) = \bar b \ (x) \ $$ [[media: Egm6321.f09.mtg6.djvu | Eq.(2) P.6-4]]: $$ \frac{1}{N}(N_x-M_y)=\frac{1}{ \bar b \ (x) c(y)} \left [ b(x)c(y)-a(x)c(y) \right ]=-f(x) \ $$

[[media: Egm6321.f09.mtg9.djvu | Page 9-3]]
[[media: Egm6321.f09.mtg9.djvu | Eq.(4) P.9-2 ]] $$ M(x,y)=a(x)\int_{}^{y} c(s) \ $$ Where $$ \int_{}^{y} c(s)= \bar c \ (y) \ $$ From [[media: Egm6321.f09.mtg9.djvu | Eq.(6) P.9-3 ]], [[media: Egm6321.f09.mtg9.djvu  | Eq.(4) P.9-2 ]] and [[media: Egm6321.f09.mtg4.djvu  | Eq.(3) P.4-2 ]] obtain: Application: Consider the following $$ a(x)=5x^3+2 \ $$ $$ b(x)=x^2 \Rightarrow \ \bar b \ (x)=\frac{1}{3}x^3 \ $$ $$ c(y)=y^4 \Rightarrow \ \bar c \ (y)=\frac{1}{5}y^5 \ $$ HW Show Eq(8) is "exact" N1_ODE. Note [[media: Egm6321.f09.mtg10.djvu | P.10-1 ]] END HW

[[media: Egm6321.f09.mtg9.djvu | Page 9-4]]
L2_ODE_VC with missing dependant variable $$ P(x)y''+Q(x)y'+R(x)y=S(x) \ $$ Where $$ R(x) \rightarrow \ 0 \ $$ due to missing dependant variable y. [[media: Egm6321.f09.mtg2.djvu | Eq.(1)P.2-3]] $$ p:=y' \rightarrow \ P(x)p'+Q(x)p=S(x) \ $$ is a L1_ODE_VC Solution: [[media: Egm6321.f09.mtg8.djvu | Eq.(4)P.8-2]] Exact N2_ODEs: General N2_ODEs: $$ F(x,y,y',y'')=0 \ $$ Application: $$ (x^3+2x^5y^2)y''+(x^{\frac{3}{2}}+10) \sqrt{y} y'+y^{100}=0 \ $$ $$ F=0 \ $$ is exact means $$ \exists \phi\ (x,y,y') \ $$ such that $$ F(x,y,y',y'')=\frac{d}{dx} \phi\ (x,y,y') \ $$

title: User:EGM6321.F10.TEAM1.WILKS/Mtg9


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg11

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 11: Thur, 17Sept09

[[media: Egm6321.f09.mtg11.djvu | Page 11-1]]
Note: $$ f(x)= \alpha\ x^a+k $$ Let $$ \alpha\ = 0 $$ and $$ k=1 \Rightarrow \ f(x)=1 $$ $$ f(x=2)=f(x=11)=f(x=10^6)=1 \ $$ Thus $$ f(x)=y \ $$ and $$ f(x)=y' \ $$ is not possible Three arguments: $$ f(x,y,p)= \alpha\ x^a+ \beta\ y^b+ \gamma\ p^c+k $$ and k=5 $$ \Rightarrow \ f(x,y,p)=5 \ $$ $$ a = b =1 \ $$ $$ \gamma\ =0 \ $$ and $$ k=0 \ $$ $$ f(x,y,p)=xy \ $$ Thus $$ f(x,y,p)=y'' \ $$ is not possible

[[media: Egm6321.f09.mtg11.djvu | Page 11-2]]
Derive [[media: Egm6321.f09.mtg10.djvu | Eq.(4) P.10-2 ]]  and [[media: Egm6321.f09.mtg10.djvu  | Eq.(5) P.10-2 ]], second condition of exactness for N2_ODE Recall N1_ODE: $$ F(x,y,y')=0 \ $$ is exact if $$ \exists \ \phi\ (x,y) \ $$ such that $$ F=\frac{d}{dx} \phi\ (x,y)= \phi_x+\phi_y y' \ $$ Where $$ \phi_x = M(x,y) :=g(x,y) \ $$ and $$ \phi_y = N(x,y) :=f(x,y) \ $$ $$ F=f(x,y)y'+g(x,y)=0 \ $$ Exactness condition 2: $$ f_x=g_y \ $$ = [[media: Egm6321.f09.mtg5.djvu | Eq.(1) P.5-4 ]] $$ N_x=M_y \ $$ Now N2_ODE: HW [[media: Egm6321.f09.mtg10.djvu | Eq.(5) P.10-2 ]]Derive this relationship by differentiating [[media: Egm6321.f09.mtg10.djvu  | Eq.(3) P.10-1 ]] with respect to $$ p=y' \ $$ i.e., $$ g_{pp} \ $$ (2nd relationship in exactness condition 2.) END HW

[[media: Egm6321.f09.mtg11.djvu | Page 11-3]]
[[media: Egm6321.f09.mtg10.djvu | Eq.(4) P.10-2 ]] 1st relationship in exactness condition 2 Recall $$ \phi\ (x,y,p) \ $$ Recall from [[media: Egm6321.f09.mtg10.djvu | Eq.(2) P.10-1 ]]) and [[media: Egm6321.f09.mtg10.djvu  | Eq.(3) P.10-2 ]]: $$ f= \phi\ _p \ $$ and $$ g= \phi_x + \phi_y p \ $$ $$ \phi_{xp} = \phi_{px} \Rightarrow \ ( \phi\ )_p = ( \phi_p)_x  \ $$ Where $$( \phi_x )_p =  \frac{\partial}{\partial p} \left ( \frac{\partial \phi\ }{\partial x} \right )  \ $$ and $$( \phi_p )_x =  \frac{\partial}{\partial x} \left ( \frac{\partial \phi\ }{\partial p} \right )  \ $$ and $$\frac{\partial}{\partial p} \left ( \frac{\partial \phi\ }{\partial x} \right ) = (g- \phi_y p)_p  \ $$ and $$\frac{\partial}{\partial x} \left ( \frac{\partial \phi\ }{\partial p} \right ) = f_x  \ $$ and $$ (g- \phi_y p)_p = f_x  \ $$ Where $$ ( \phi_{py} = \phi_{yp} = ( \phi_p)_y = f_y $$

[[media: Egm6321.f09.mtg11.djvu | Page 11-4]]
$$ \phi_{yp} = \phi_{py} \Rightarrow \ ( \phi_y)_p = ( \phi_p)_y \ $$ Where $$ ( \phi_y)_p = \left [ \frac{1}{p} (g- \phi_x) \right ]_p \ $$ and $$ ( \phi_p)_y = f_y $$ and $$ \left [ \frac{1}{p} (g- \phi_x) \right ]_p = f_y \ $$ $$ \Rightarrow \ -\frac{1}{p^2} (g- \phi_x) + \frac{1}{p}(g_p- \phi_{xp}) = f_y \ $$

title: User:EGM6321.F10.TEAM1.WILKS/Mtg11


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg13

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 13: Tues, 22Sept09

[[media: Egm6321.f09.mtg13.djvu | Page 13-1]]
Euler integrating factor method [[media: Egm6321.f09.mtg12.djvu | Eq.(1) P.12-3 ]] Where: $$ (x^wy^n)=h(x,y) \ $$ HW Find (m,n) such that Eq(1) is exact. END HW Result: A first integration is: Where $$ p=y' \ $$ and $$ k_1 \ $$ and $$ k_2 \ $$ are constants HW Solve Eq(2) for $$ y(x) \ $$ HINT: L1_ODE_VC (integrating factor) END HW A class of exact L2_ODE_VC (how to invent more exact L2_ODE_VC) HW Find mathematical structure of $$ \phi\ \ $$ that yields the above class END HW

[[media: Egm6321.f09.mtg13.djvu | Page 13-2]]
$$ F=\frac{d \phi\ }{dx}= \phi_x(x,y,p)+ \phi_yp+ \phi_pp'=P(x)y''+Q(x)y'+R(x)y \ $$ Where $$ p=y' \ $$ and $$ p'=y'' \ $$ and $$ P(x)= \phi_p \ $$ and $$ Q(x)= \phi_y \ $$ and $$ R(x)y= \phi_x \ $$ Answer: [[media: Egm6321.f09.mtg13.djvu | Eq.(2) P.13-1 ]] : $$ P(x)=x \ $$ and $$ T(x)=2x^{\frac{3}{2}}-1 \ $$ and $$ k=k_1 \ $$ Exact Nn_ODE's, where N means nonlinear and n means nth order Where $$y^{(0)}=y \ $$ and $$y^{(1)}=y' \ $$ and $$y^{(n)}= \frac{d^ny}{dx^n} \ $$

[[media: Egm6321.f09.mtg13.djvu | Page 13-3]]
Condition 1 for exactness: Where $$ \phi_x+ \phi_{y(0)}y^{(1)}+...+ \phi_{y(n-1)}y^{(n)} \ $$ is related term by term to $$ x,y^{(0)},...,y^{n-1} \ $$ Condition 2 of exactness: $$ f_i:= \frac{\partial F}{\partial y^{(i)}} \ $$, where $$ i=1,...,n \ $$ HW Case $$ n=1 \ $$ (N1_ODE) END HW $$ F(x,y,y')=0=\frac{d}{dx} \phi\ (x,y) \ $$ $$ f_0-\frac{df_1}{dx}=0 \Leftrightarrow \phi_{xy}= \phi_{yx} \ $$ HINT: $$ f_1= \phi_y \ $$

[[media: Egm6321.f09.mtg13.djvu | Page 13-4]]
Case $$ n=2 \ $$ (N2_ODE) $$ f(x,y,y',y'')=0=\frac{d \phi\ }{dx}(x,y,y') \ $$ $$ f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0 \ $$

title: User:EGM6321.F10.TEAM1.WILKS/Mtg13


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg15

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 15: Thur, 24Sept09

[[media: Egm6321.f09.mtg15.djvu | Page 15-1]]
[[media: Egm6321.f09.mtg14.djvu | Ex.1 P.14-4 ]] : $$ N_J(x)= \ $$ finite element basis function associated with node J = "hat" function. $$ N_J \in C^0 \ $$, but $$ N_J \notin C^1 \ $$ [[media: Egm6321.f09.mtg14.djvu | Ex.1 P.14-4 ]]: Cubic Spline (Bexler, cubic Hermitian)

[[media: Egm6321.f09.mtg15.djvu | Page 15-2]]
Cubic $$ \Rightarrow \ \ $$ 4 Coefficients $$ \Rightarrow \ \ $$ 4 degrees of freedom per element $$ \Rightarrow \ \ $$ 2 degrees of freedom per node (each element has 2 nodes) HW $$ N_J^{ \alpha\ }(x) \in C^1 \ $$, but $$ N_J^{ \alpha\ } \notin C^2 \ $$, for $$ \alpha\ =1,2 \ $$ END HW

[[media: Egm6321.f09.mtg15.djvu | Page 15-3]]
$$ L_2(y_H^{ \alpha\ })=0 \ $$, $$ \alpha\ =1,2 \ $$ $$ L_2(y_P)=f(x) \ $$ $$ y=Ay_H^1+By_H^2+y_P \ $$, where A and B are constants Where this can be rewritten for x as: $$ x=Ax_H^1+Bx_H^2+x_P \ $$, where A and B are constants $$ L_2(y)=f \Leftrightarrow y=L_2(f) \ $$

[[media: Egm6321.f09.mtg15.djvu | Page 15-4]]
Euler Equations: Special Homogeneous Ln_ODE_VC $$ a_nx^ny^{(n)})+a_{n-1}x^{n-1}y^{n-1}+...+a_1xy'+a_0y=0 \Leftrightarrow \sum_{i=0}^n a_ix^iy^{(i)}=0 \ $$ Where $$ y'=y^{(1)} \ $$ and $$ y=y^{(0)} \ $$ Two methods of solution: Method 1: Transfer of variables $$ x=e^t \ $$ Method 2: Method of undetermined coefficients $$ y=x^r \ $$ (Or Trial Solution) [[media: http://books.google.com/books?id=9Cg3HWCnCjAC&printsec=frontcover&dq=differential+equations+billingham&ei=pGR4SpPVLojSMpb07Qw#v=onepage&q=&f=false | K.etal(2003) ]]

title: User:EGM6321.F10.TEAM1.WILKS/Mtg15


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg17

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 17: Thur, 10Oct09

[[media: Egm6321.f09.mtg17.djvu | Page 17-1]]
Linearity $$ \Rightarrow \ \ $$ superposition $$ y=y_H+y_P \ $$ Homogeneous Solution $$ y_H \ $$ - Euler Equations - Trial solution (undefined coefficient) - Reduction of order method 2: Undetermined factor Homogeneous L2_ODE_VC: cf. [[media: Egm6321.f09.mtg3.djvu | Eq.(1) P.3-1 ]] Where $$ a_0(x) \ $$ can be substituted for $$ a_1(x) \ $$ or $$ a_0y \ $$ in Eq(1) Given one homogeneous solution $$ u_1(x) \ $$ known Find second homogeneous solution $$ u_2(x) \ $$ such that Where$$ k_1,k_2 \ $$ are constants Assume full homogeneous solution Where$$ U(x) \ $$ is an unknown to be determined Where$$ u_1(x) \ $$ is known

[[media: Egm6321.f09.mtg17.djvu | Page 17-2]]
"Full" = includes $$ u_2(x) \ $$ Add the following: $$ a_0(x) \left [ y=Uu_1 \right ] \ $$ and $$ a_1(x) \left [ y'=Uu_1'+U'u_1 \right ] $$ and $$ \left [ y=Uu_1+2U'u_1'+U''u_1 \right ] $$ To get $$ a_0y+a_1y'+y=U \left [ a_0u_1+a_1u_1'+u_1 \right ] + U'\left [ a_1u_1+2u_1' \right ]+U''u_1=0 \ $$ by Eq(1) p17-1 Reduce to $$ u_1''+a_1u_1'+a_0u_1=0 \ $$ Since $$ u_1 \ $$ is a homogeneous solution $$ \Rightarrow \ 0=U'(a_1u_1+2u_1')+U''u_1 \ $$, NOTE missing dependent variable U in front of $$ U' \ $$ term Let $$ Z:=U' \Rightarrow \ \ $$ homogeneous L1_ODE_VC for Z

[[media: Egm6321.f09.mtg17.djvu | Page 17-3]]
Solve for Z, HW - integration factorial method END HW - Direct integration (because Eq(1) is homogeneous) $$ \Rightarrow \ \frac{Z'}{Z}+(a_1+\frac{2u_1'}{u_1})=0 \ $$ Where $$ a_1, \frac{2u_1'}{u_1} \ $$ are known Integrate $$ \log \left | Z \right \vert +2 \log \left | u_1 \right \vert+ \int_{}^{x} a_1(s)\, ds=k $$, where k is a constant where $$ c=e^k \ $$ $$ \Rightarrow \ U(x)=\int_{}^{x} \frac{c}{(u_1(t))^2}e^ \, dt+\tilde{c} \ $$ where $$ \bar a \ _1(t)=\int_{}^{t} a_1(s)\, ds \ $$

[[media: Egm6321.f09.mtg17.djvu | Page 17-4]]
Homogeneous solution where $$ \tilde{c}=k_1 \ $$ and $$ c=k_2 \ $$ HW obtain [[media: Egm6321.f09.mtg17.djvu | Eq.(2) P.17-3 ]] $$ Z(x) \ $$ using the integrating factor method END HW

title: User:EGM6321.F10.TEAM1.WILKS/Mtg17


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg19

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 19: Tues, 5Oct09

[[media: Egm6321.f09.mtg19.djvu | Page 19-1]]
HW Legendre differential [[media: Egm6321.f09.mtg14.djvu | Eq.(1) P.14-2 ]] with $$ n=0 \ $$, such that homogeneous solution $$ u_1(x)=1 \ $$. Use reduction of order method 2 (undetermined factor) to find $$ u_2(x) \ $$, second homgenous solution END HW HW [[media: http://books.google.com/books?id=9Cg3HWCnCjAC&printsec=frontcover&dq=differential+equations+billingham&ei=pGR4SpPVLojSMpb07Qw#v=onepage&q=&f=false | K. p28, pb. 1.1.b. ]] END HW Variation of parameters (continued)   P.18-4

Use expression for $$ y' \ $$  Eq.(2) P.18-4  and $$ y'' \ $$   Eq.(3) P.18-4  in non-homogeneous L2_ODE_VC [[media: Egm6321.f09.mtg3.djvu  | Eq.(1) P.3-1 ]]

Where $$ u_1''+a_1u_1'+a_0u_1 \Rightarrow 0 \ $$, because $$ u_1 \ $$ is a homogeneous solution Where $$ u_2''+a_1u_2'+a_0u_2 \Rightarrow 0 \ $$, because $$ u_2 \ $$ is a homogeneous solution

[[media: Egm6321.f09.mtg19.djvu | Page 19-2]]
2 equations Eq.(1) P.18-4  and  Eq.(1) P.19-1  for two unknowns $$ \begin{Bmatrix} c_1' \\ c_2' \end{Bmatrix} \ $$ In matrix form: $$ \begin{bmatrix} u_1 & u_2 \\ u_1' & u_2' \end{bmatrix}  \begin{Bmatrix} c_1' \\ c_2'\end{Bmatrix} = \begin{Bmatrix} 0 \\ f \end{Bmatrix} \ $$ Where $$ \begin{bmatrix} u_1 & u_2 \\ u_1' & u_2' \end{bmatrix} \ $$ is the Wronskian matrix designated as $$ \underline{W} \ $$ The Wronskian, W, is the determinant of $$ \underline{W} \ $$ $$ W= det \underline{W} \ $$ If $$ W \ne \ 0 \ $$, then $$ \underline{W}^{-1} \ $$ exists and $$ \begin{Bmatrix} c_1' \\ c_2'\end{Bmatrix} =  \underline{W} ^{-1}  \begin{Bmatrix} 0 \\ f \end{Bmatrix} \ $$ Theorem: $$\forall \!\,$$ $$ u_1, u_2 \ $$ (function of x) are linearly independant if $$ \underline{W} \ne \ 0  \ $$, where $$ 0= \ $$ zero function.

[[media: Egm6321.f09.mtg19.djvu | Page 19-3]]
Where $$ \begin{Bmatrix} -u_2f \\ u_1f \end{Bmatrix} \ $$ are known Where $$ \int_{}^{x} \frac{u_2(s)f(s)}{W_s}\, ds = d_1(x) \ $$ Where $$ \int_{}^{x} \frac{u_1(s)f(s)}{W_s}\, ds= d_2(x) \ $$

title: User:EGM6321.F10.TEAM1.WILKS/Mtg19


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg21

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 21: Thurs, 8Oct09

[[media: Egm6321.f09.mtg21.djvu | Page 21-1]]
P.20-4 (continued) $$ \Rightarrow ax^{-1} \ $$ is a homgenous solution $$ \forall \!\, a\ $$ $$ u_1(x)=c_1x^{-1} \ $$ 2) $$ b=2 \Rightarrow ax^2 \ $$ is another homogeneous solution since $$ b^2-b-2=0 \ $$ $$ u_2(x)=c_2x^2 \ $$ (Verify $$ u_1 \ $$ and $$ u_2 \ $$ are linearly independant components of $$ W \ $$ 3) $$ b=6, a=\frac{1}{4} \Rightarrow \ $$ left hand side of Eq(1) p20-4 $$ =7x^4 \ $$, where $$ =7x^4 \ $$  is the 1st term on the right hand side for $$ b=6, a=\frac{1}{4} \Rightarrow  \frac{1}{4}x^6 \ $$  4) $$ b=5, a=\frac{1}{6} \Rightarrow \ $$ left hand side of Eq(1) p20-4  $$ =3x^3 \ $$ , where $$ =3x^3 \ $$  is the 2nd term on the right hand side for $$ b=5, a=\frac{1}{6} \Rightarrow \frac{1}{6}x^5 \ $$

[[media: Egm6321.f09.mtg21.djvu | Page 21-2]]
Llinearity of ordinary differential equation $$ \Rightarrow \ $$ superposition $$ y_P(x)=\frac{1}{4}x^6+\frac{1}{6}x^5 \ $$ $$ y(x)=c_1x^{-1}+c_2x^2+y_P(x) \ $$, where  $$  c_1x^{-1}+c_2x^2=y_H(x) \ $$ Alternative method to obtain full solution for non-homogeneous L2_ODE_VC knowing only one homogeneous solution (e.g. obtained by trial solution) (bypassing reduction of order method2-undertermined factor for $$ u_2 \ $$ and variation of parameter method) [[media: Egm6321.f09.mtg3.djvu | Eq.(1) P.3-1 ]] =    $$ y''+a_1(x)y'+a_0(x)y=f(x) \ $$ Assume having found $$ u_1(x) \ $$, a homogeneous solution: $$ u_1''+a_1(x)u_1'+a_0(x)u_1=0 \ $$ Consider: $$ y(x)=U(x)u_1(x) \ $$, where $$ U(x) \ $$ is an undetermined factor

[[media: Egm6321.f09.mtg21.djvu | Page 21-3]]
Follow the same argument as on [[media: Egm6321.f09.mtg17.djvu | P.17-2 ]] to obtain: NOTE: this equation is missing the dependant variable $$ U \ $$ in front of $$ U' \ $$ term due to reduction of order method $$ \phi\ \ $$ where $$ u_1(x) \ $$ and $$ \left [ a_1(x)u_1(x)+2u_1'(x) \right ] \ $$ are known Non-homogeneous L1_ODE_VC solution for $$ Z(x) \ $$ : [[media: Egm6321.f09.mtg8.djvu |  Eq.(4) P.8-2 ]] ref: [[media: http://books.google.com/books?id=9Cg3HWCnCjAC&printsec=frontcover&dq=differential+equations+billingham&ei=pGR4SpPVLojSMpb07Qw#v=onepage&q=&f=false | K p.28, problem 1.1ab ]] a) $$ u_1(x)=e^x \ $$, $$ (x-1)y''-xy'+y=0 \ $$ Trial solution $$ y(x)=e^{rx} \ $$ , where $$ r= \ $$ constant

[[media: Egm6321.f09.mtg21.djvu | Page 21-4]]
Find $$ r_1,r_2 \ $$ How many valid homogeneous solutions to $$ u_1=e^{r_1x} \ $$, find $$ u_2 \ $$ using undetermined factor method

title: User:EGM6321.F10.TEAM1.WILKS/Mtg21


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg23

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 23: Thurs, 15Octt09

[[media: Egm6321.f09.mtg23.djvu | Page 23-1]]
HW Non-homogeneous L2_ODE_VC [[media: http://books.google.com/books?id=9Cg3HWCnCjAC&printsec=frontcover&dq=differential+equations+billingham&ei=pGR4SpPVLojSMpb07Qw#v=onepage&q=&f=false | (cf. K. p28, problem 1.1.ab) ]] a) $$ (x-1)y-xy'+y=x \ $$ b) $$ xy+2y'+xy=x \ $$ using: 1) Variation of parameters (after knowing $$ u_1(x), u_2(x)) \ $$ 2) Alternative method: [[media: Egm6321.f09.mtg21.djvu |  P.21-2 ]] and [[media: Egm6321.f09.mtg21.djvu  |  P.21-3 ]] END HW Engineering application: Motivate L2_ODE_VC Ware equation: Drums=vibrating membrane Rectangular membrane Circular membrane (Bessel functions) Bessel differential equation Chladni: Experimental method for vibrating modes of plates

[[media: Egm6321.f09.mtg23.djvu | Page 23-2]]
Wave eq: (PDE) solution by separation of variables

title: User:EGM6321.F10.TEAM1.WILKS/Mtg23


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg25

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= [tp://upload.wikimedia.org/wikiversity/en/e/e5/Egm6321.f09.mtg25.djvu Mtg 25:] Tues, 20Oct09

[[media: Egm6321.f09.mtg25.djvu | Page 25-1]]
Pre-recorded Thursday, 15 Oct09 [[media: Egm6321.f09.mtg23.djvu |  P.23-1 ]] : Motivate L2_ODE_VC Review: Class exact L2_ODE_VC [[media: Egm6321.f09.mtg13.djvu | P.13-2 ]] : $$ \phi\ (x,y,p)=P(x)p+T(x)y+k \ $$, where  $$ p=y' \ $$ Let: $$ P(x)=x^2 \ $$ and $$ T(x)=x^4 \ $$, where $$ k=0 \ $$ $$ \phi\ = x^2p+x^4y=C \ $$ where $$ C= \ $$ constant $$ \phi_x =2xp+4x^3y \ $$ $$ \phi_y =x^4 \ $$ $$ \phi_p =x^2 \ $$ $$ \frac{d \phi\ }{dx}= \left [ 2xp+4x^3y \right ] + \left [ x^4 \right ]p + \left [ x^2 \right ]p'=0 \ $$ Where $$ p'=y'' \ $$ Exact L2_ODE_VC $$ x^2y''+x(2+x^3)y'+4x^3y=0 \ $$

[[media: Egm6321.f09.mtg25.djvu | Page 25-2]]
Generate non-exact L2_ODE_VC that can be made exact by intergral factor method: $$ x^a \left [ x^{2-a}y''+x^{1-a}(2+x^3)y'+4x^{3-a}y \right ]=0 \ $$ Where $$ x^a \ $$ is the integrating factor $$ h(x,y)=x^my^n\ $$ and $$ \left [ x^{2-a}y''+x^{1-a}(2+x^3)y'+4x^{3-a}y \right ]=0 \ $$ is L2_ODE_VC, not necessarily exact "$$ a \ $$" can be any number, $$ a=1, \frac{3}{2},5... \ $$

title: User:EGM6321.F10.TEAM1.WILKS/Mtg25


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg29

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 29: Thurs, 29Oct09

[[media: Egm6321.f09.mtg29.djvu | Page 29-1]]
Eq.(3) P.28-3 : L2_ODE_VC $$ g_i ( \xi_i) \Rightarrow \ $$ linear $$ \phi_{ij}( \xi_i) \Rightarrow \ $$ linear Simplify: $$ \xi_i \Rightarrow x \ $$ $$ x_i ( \xi_i) \Rightarrow y(x) \ $$ $$ g_i ( \xi_i) \Rightarrow g(x) \ $$ $$ \left [ \sum \phi_{ij}(k_j)^2 \right ]  \Rightarrow a_0(x) \ $$ Eq.(3) P.28-3 : $$ \frac{1}{g(x)} \frac{d}{dx} \left [ g(x) \frac{dy}{dx} \right ] +a_0(x)y=0  \Rightarrow y''+ \frac{g'}{g}y' + a_0y=0 \ $$ Where $$ \frac{g'}{g}y' = a_1 \ $$ L2_ODE_VC, cf. [[media: Egm6321.f09.mtg3.djvu | Eq.(1) P.3-1 ]] Ellipsoidal coordinates: $$ a, b, c=0 \ $$

[[media: Egm6321.f09.mtg29.djvu | Page 29-2]]
$$ g_i ( \xi_i)= \left [ \frac{ \xi_i^2-a^2}{ \xi_i^2-b^2} \right ]^{\frac{1}{2}} \ $$ $$ \phi_{i1}( \xi_i) = 1  \ $$ $$ \phi_{i2}( \xi_i) = \frac{1}{ \xi_i^2-a^2 } \ $$ Heat condition on a sphere Steady State Heat Eq(1): $$ ( \underline{k} \ $$ gradient $$  \psi\ )+f=0 \ $$ Where $$ \underline{k} = \ $$ conductivity and $$ f= \ $$ tensor matrix Homogeneous isotropic material where $$ k= \ $$ condition and $$  \underline{I} = \ $$ Identity matrix If no heat source, $$ f=0 \ $$ Laplace equation Laplace differential operation on orthogonal curviture coordinates (e.g. spherical coordinates, ellipsoidal, etc.)

[[media: Egm6321.f09.mtg29.djvu | Page 29-3]]
$$ \Delta\ \psi\ = \frac{1}{h_1h_2h_3} \sum_{i=1}^3 \frac{\partial}{\partial \xi_i} \left [ \frac{h_1h_2h_3}{(h_i)^2} \frac{\partial \psi\ }{\partial \xi_i} \right ] \ $$ $$ \bar \theta\ \ $$ is " $$ \theta\ \ $$" in $$ K \ $$ Here $$ \theta\ \ $$ is defined such that equator is at $$ 0^o \ $$ latitude $$ x=r \cos \theta\ \cos \varphi \ $$ $$ y=r \cos \theta\ \sin \varphi \ $$ $$ x=r \sin \theta\ \ $$ Eq.(4) P.29-2 $$ ds^2=1dr^2+r^2 \cos ^2 \theta\ d \varphi ^2 + r^2d \theta\ ^2 \ $$ Where $$ 1= (h_1)^2 \ $$, $$ dr^2=(d \xi_1)^2 \ $$ , $$ r^2 \cos ^2 \theta\ = (h_2)^2 \ $$ , $$ d \varphi ^2 = (d \xi_2)^2 \ $$ , $$ r^2 = (h_3)^2 \ $$ and $$ d \theta\ ^2 = (d \xi_3)^2 \ $$ Then $$ h_1 ( \xi\ )=1 \ $$ $$ \xi\ = ( \xi_1, \xi_2, \xi_3) \ $$ $$ h_2 ( \xi\ )= r \cos \theta\ = \xi_1 \cos \xi_2 \ $$ $$ h_3 ( \xi\ )= r = \xi_1 \ $$

[[media: Egm6321.f09.mtg29.djvu | Page 29-4]]
HW END HW HW END HW For axisymmetric problem: no dependancy on $$ \varphi \ $$ cf. [[media: http://books.google.com/books?id=9Cg3HWCnCjAC&printsec=frontcover&dq=differential+equations+billingham&ei=pGR4SpPVLojSMpb07Qw#v=onepage&q=&f=false | Kp.45) ]]

[[media: Egm6321.f09.mtg29.djvu | Page 29-5]]
Where $$ \psi\ (1, \theta\ ) = f( \theta\ ) = T_0 \sin ^4 \bar \theta\ \  \ $$ Where $$ T_0= \ $$ constant and $$ \bar \theta\ = K \ $$ Then $$ \bar \theta\ :=\frac{ \pi\ }{2}- \theta\ \ $$

title: User:EGM6321.F10.TEAM1.WILKS/Mtg29


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg31

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 31: Tues, 3Nov09

[[media: Egm6321.f09.mtg31.djvu | Page 31-1]]
HW Circular cylindrical coordinates $$ x=r \cos \theta\ = \xi_1 \cos \xi_2 \ $$ $$ y=r \sin \theta\ = \xi_1 \sin \xi_2 \ $$ $$ z= \xi_3 \ $$ 1) Find $$ \left \{ dx_i \right \} = \left \{ dx_1, dx_2, dx_3 \right \} \ $$ in terms of $$ \left \{ \xi_j \right \} = \left \{ \xi_1, \xi_2, \xi_3 \right \}  \ $$ and $$ \left \{ d \xi_k \right \}  \ $$ 2) Find $$ ds^2 = \sum_{i} (dx_i)^2 = \sum_{k} (h_k)^2(d \xi_k)^2 \ $$. Identify $$ \left \{ h_i \right \} \ $$ in terms of $$ \left \{ \xi_j \right \}  \ $$ 3) Find Laplacian in this curviture coordinate END HW Note: Prelude to Bessel functions; Bessel Differential Equation HW Spherical coordinates. Repeat 1), 2) and 3) above steps. See [[media: Egm6321.f09.mtg29.djvu |  P.29-3 ]], use astronautical convention END HW

[[media: Egm6321.f09.mtg31.djvu | Page 31-2]]
Conventions: Astronomy: $$ \left ( r, \varphi, \theta\ \right ) \ $$ [[media: Egm6321.f09.mtg29.djvu  |  P.29-3 ]] Math/Physics: $$ \left ( r, \varphi, \bar \theta\ \right ) \ $$, where $$ \bar \theta\ = \frac{ \pi\ }{2} - \theta\ \ $$, [[media: http://books.google.com/books?id=9Cg3HWCnCjAC&printsec=frontcover&dq=differential+equations+billingham&ei=pGR4SpPVLojSMpb07Qw#v=onepage&q=&f=false  | K p39 ]] HW Find Laplacian in Math/Physics convention. $$ \left ( r, \varphi, \bar \theta\ \right ) \ $$ END HW Continued from [[media: Egm6321.f09.mtg30.djvu  |  P.30-4]] : Heat problem in sphere From [[media: Egm6321.f09.mtg30.djvu | Eq.(2) P.30-4]] and [[media: Egm6321.f09.mtg30.djvu  | Eq.(5) P.30-2]] : Where: and HW Solve Eq(1) (quadratic equation) END HW [[media: Egm6321.f09.mtg30.djvu | Eq.(3) P.30-2]] : $$ R(r)=A_nr^n+B_nr^{-(n+1)}  \ $$ One of two homogeneous solutions of Legendre differential equations, [[media: Egm6321.f09.mtg30.djvu | Eq.(1) P.30-4]] is the Legendre ppolynomial $$ P_n( \mu\ ) \ $$. How...given later

[[media: Egm6321.f09.mtg31.djvu | Page 31-3]]
HW show that Eq(6) = Eq(7) END HW $$ \left [ \frac{n}{2} \right ]  = \ $$ integer part of  $$ \frac{n}{2} \ $$ HW Verify that Eq(1) thru Eq(5) can be written as Eq(6) or Eq(7) END HW

title: User:EGM6321.F10.TEAM1.WILKS/Mtg31


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg33

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 33: Thurs, 5Nov09

[[media: Egm6321.f09.mtg33.djvu | Page 33-1]]
Where: $$ \sin \theta\ = \mu\ \ $$ and $$ d \theta\ \ $$ becomes $$ d \mu\ \ $$

Similarly for $$ \left \langle P_n, P_m \right \rangle $$

Orthogonality of Legendre polynomial

Where $$ \delta_{mn} = \ $$ kronecker delta

[[media: Egm6321.f09.mtg33.djvu | Page 33-2]]
$$ \Rightarrow \ \underline{ \Gamma\ } (F) \ $$ is diagonal with diagonal coefficient:

F is complete, i.e. any continuous function, f, can be expressed as an infinite series of function in F:

Eq(4) is an equality due to the completeness of F

p29-5: $$ f( \theta\ )= T_0 \cos ^4 \theta\ = T_0(1- \mu\ ^2)^2\sum_{u=0}^\infty A_nP_n( \mu\ ) \ $$

Where $$ \mu\ = \sin |theta\ \ $$

Where n=0,1,2...n

[[media: Egm6321.f09.mtg33.djvu | Page 33-3]]
HW $$ f= \sum_{i} g_i \ $$

Show that if $$ \left \{ g_i \right \} \ $$ is odd, then f is odd

Show that if $$ \left \{ g_i \right \} \ $$ is even, then f is even END HW

HW Show that $$ P_{2k} \ $$ is even for k=0,1,2... and $$ P_{2k+1} \ $$ is odd END HW

[[media: Egm6321.f09.mtg33.djvu | Eq.(5) P.33-2]] $$ A_n=\frac{ \left \langle f, P_n \right \rangle }{ \left \langle P_n, P_n \right \rangle } \ $$, f even

$$ \Rightarrow \ A_n=0 \ $$ for  $$  n=2k+1 \ $$, since $$  P_{2k+1}(x) \ $$ is odd

$$ A_1=A_3=A_5=...=0 \ $$

It turns out that $$ A_n=0 \ $$ for all $$ n \ge 5 \ $$ due to linear independance of $$ F= \left \{ P_n \right \} \ $$ and the orthogonality of $$ F \ $$

[[media: Egm6321.f09.mtg33.djvu | Page 33-4]]
Linear independance of $$ F \ $$

$$ P_n(x) \ $$ is a polynomial of order n

$$ P_n \in \Rho\ _n \ $$ set of all polynomials of degree (order)  $$ \le \ n \ $$

HW $$ q \in \Rho\ _4 \ $$

$$ q(x) = \sum_{i=0}^4 c_ix^i \ $$

Given $$ c_0=3, c_1=10, c_2=15, c_3=-1, c_4=5 \ $$

Find $$ \left \{ a_i \right \} \ $$ such that $$ q(x) = \sum_{i=0}^4 a_iP_i \ $$

Plot $$ q = \sum_{i} c_ix^i = \sum_{i} a_iP_i \ $$

Where $$ \sum_{i} c_ix^i = \ $$ figure 1 and $$ \sum_{i} a_iP_i= \ $$ figure 2 END HW

Othogonality of $$ F= \left \{ P_k \right \} \ $$ [[media: Egm6321.f09.mtg33.djvu | Eq.(3) P.33-1]]

title: User:EGM6321.F10.TEAM1.WILKS/Mtg33


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg35

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 35: Thurs, 12Nov09

[[media: Egm6321.f09.mtg35.djvu | Page 35-1]]
Gauss Legendre quadrature (numerical integration} Quadrature; QUAD-->quadrilateral-->Greek: measuring areas Area $$ = \sum $$ Quadrilaterals Cubature; CUBE; Volume $$ = \sum  $$ cubes $$ I(f):=\int_{-1}^{1} f(x)\, dx \ $$ $$ I_n(f):=\sum_{j=1}^n w_jf(x_j)\, dx \ $$ with $$ \left \{ x_j \right \} \ $$ the roots for $$ P_n(x)=0 \ $$, where n is the degree of $$ P_n(x) \ $$ and $$ w_j \ $$ being the weight $$ -1[[media: Egm6321.f09.mtg35.djvu | Page 35-2]]
where j=1,2,...,n for $$ \eta\ \in \left [ -1,+1 \right ] \ $$ Ex: $$ n=2 \ $$ (2 point interpolation) [[media: Egm6321.f09.mtg31.djvu | Eq.(3) P.31-3]]  $$ P_2(x)=\frac{1}{2}(3x^2-1) \ $$ $$ \Rightarrow \ x_{1,2}= \pm \ \frac{1}{ \sqrt{3} } \ $$ [[media: Egm6321.f09.mtg31.djvu | Eq.(4) P.31-3]] $$ P_2'(x)=3x, P_3(x)=\frac{1}{2}(5x^3-3x) \ $$ $$ W_1=\frac{2}{(2+1)(3)(\frac{-1}{ \sqrt{3} }) \frac{1}{2} \left [ 5 (\frac{-1}{\sqrt{3}}) ^3-3(\frac{-1}{\sqrt{3}}) \right ] }=1 \ $$

[[media: Egm6321.f09.mtg35.djvu | Page 35-3]]
$$ W_2=1 \ $$ HW verify table for Gauss Legendre quadrature in wikipedia, analytical expression of $$ \left \{ x_j \right \} \ $$ and $$ \left \{ w_j \right \}, j=1,...,n \ $$ and  $$ n=1,...,5 \ $$ (n=integration points) after verifying the expression for  $$ P_n(x) \ $$ with $$ n=1,...,6 \ $$; (see [[media: Egm6321.f09.mtg31.djvu  | HW p31-3]] ) Evaluate numerically $$ \left \{ x_j \right \} \ $$ and $$ \left \{ w_j \right \} \ $$ and compute results with Abram & Stegum (see lecture plan) Question: How does Gauss Legendre quadrature compare to other quadrature methods, e.g. trapezoidal rule? Answer: Look at $$ E_n(f) \ $$, [[media:Egm6321.f09.mtg35.djvu | Eq.(3) P.35-2]]. Consider  $$ f \in \Rho\ _{2n-1} \ $$... END HW

title: User:EGM6321.F10.TEAM1.WILKS/Mtg35


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg37

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 37: Thurs, 17Nov09

[[media: Egm6321.f09.mtg37.djvu | Page 37-1]]
[[media: Egm6321.f09.mtg36.djvu | P.36-4]] continued $$ P_n(x) \ $$ $$ P_1(x)=x \ $$ $$ Q_n(x) \ $$ From [[media: Egm6321.f09.mtg18.djvu  |  P.18-1 ]]: HW show $$ Q_1(x)=\frac{1}{2}xlog \left ( \frac{1+x}{1-x} \right ) -1 = x \tanh ^{-1}x-1 \ $$ END HW ref [[media: http://books.google.com/books?id=9Cg3HWCnCjAC&printsec=frontcover&dq=differential+equations+billingham&ei=pGR4SpPVLojSMpb07Qw#v=onepage&q=&f=false  | K p33 ]] for $$ Q_2, Q_3...\ $$ $$ Q_0(x)= \tanh ^{-1}(x) \ \ $$ is odd HW Use Eq(2) to show when $$ Q_n \ $$ is even or odd, depending on "n" END HW HW Plot $$ \left \{ P_0, P_1,...,P_4 \right \} \ $$ and $$ \left \{ Q_0, Q_1,...,Q_4 \right \} \ $$ END HW Legendre function $$ L_n(x)=P_n(x) \ $$ or $$ Q_n(x) \ $$ solution of Legendre solution

[[media: Egm6321.f09.mtg37.djvu | Page 37-2]]
for $$ n \ne m \ $$ for $$ L_n =P_n, \left \langle L_n,L_n \right \rangle = \left \langle P_n,P_n \right \rangle = \frac{2}{2n+1}  \ $$ HW for $$ \left \langle L_n,L_n \right \rangle = \left \langle P_n,Q_n \right \rangle \ =  $$ END HW $$ \left \langle L_n,L_m \right \rangle =  \left \langle P_n,P_m \right \rangle \ =  $$ [[media: Egm6321.f09.mtg33.djvu  | Eq.(3) P.33-1]] $$ \left \langle L_n,L_m \right \rangle = \left \langle P_n,Q_m \right \rangle \ = 0 $$ for $$ n \ne m \ $$ Proof: Legendre equation, [[media: Egm6321.f09.mtg14.djvu | Eq.(1) P.14-2 ]]: 2) $$ \left [ (1-x^2)y' \right ]'+n(n+1)y=0 \ $$   Where $$ \Rightarrow \ \left [ (1-x^2)L_n' \right ]'+n(n+1)L_n=0 \ $$  Multiply by $$ L_m \ $$ and integrate from  -1 to +1: $$ \int_{-1}^{1} L_m \left [ (1-x^2)L_n' \right ]' \, dx + n(n+1) \int_{-1}^{1} L_mL_n \, dx=0 \ $$ Where  $$ L_m \left [ (1-x^2)L_n' \right ]' = \alpha\ \ $$

[[media: Egm6321.f09.mtg37.djvu | Page 37-3]]
Integrate $$ \alpha\ \ $$ by parts: Interchange n and m: Eq(1)-Eq(2): $$ \left [ n(n+1)-m(m+1) \right ] \left \langle L_n,L_m \right \rangle =0 $$ Where $$ \left [ n(n+1)-m(m+1) \right ] \ne 0 $$ since $$ n \ne m \ $$ $$ \Rightarrow \ \left \langle L_n,L_m \right \rangle =0 \ $$ when $$ n \ne m \ $$ [[media: http://books.google.com/books?id=9Cg3HWCnCjAC&printsec=frontcover&dq=differential+equations+billingham&ei=pGR4SpPVLojSMpb07Qw#v=onepage&q=&f=false | cf.K.p41]]

title: User:EGM6321.F10.TEAM1.WILKS/Mtg37


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg39

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 39: Thurs, 19Nov09

[[media: Egm6321.f09.mtg39.djvu | Page 39-1]]
Note: Other Application: Quantum Mechanics

discrete variables <-- coding theory

Ref: [[media: http://books.google.com/books?id=4Q0ZAQAAIAAJ&q=nikiforov+functions&dq=nikiforov+functions&ei=EAsAS8yGKZbQNO-9vYkP | Nikiforov, et.al (1991) ]]

Probability (Queing theory, birth and death)

Generating functions:

-Legendre Polynomial $$ P_n \ $$: [[media: Egm6321.f09.mtg38.djvu | Eq.(5) P.38-3]]

- $$ {r \choose k} \ = $$ "r choose k" :

$$ (1+x)^r = \sum_{k=0}^\infty {r \choose k} x^k \ $$ for $$ \left | x \right \vert \le 1 \ $$

$$ A( \mu\, \rho\ ):=1-2 \mu\ \rho\ + \rho\ ^2) \ $$

From [[media: Egm6321.f09.mtg38.djvu | Eq.(6) P.38-3]] and [[media: Egm6321.f09.mtg38.djvu | Eq.(7) P.38-3]] : $$ \frac{1}{\sqrt{A( \mu\, \rho\ )}} = \alpha_0 + \alpha_1(2 \mu\ \rho\ - \rho\ ^2)+ \alpha_2(2 \mu\ \rho\ - \rho\ ^2)^2+...= \alpha_0+ (2 \mu\ \alpha_1) \rho\ + (-\alpha_1+4 \mu\ ^2 \alpha_2) \rho\ ^2 +... \ $$

Where $$ (2 \mu\ \rho\ - \rho\ ^2) = 4 \mu\ ^2 \rho\ ^2-4 \mu\ \rho\ ^3 + \rho\ ^4 \ $$

and $$ \alpha_0 = P_0( \mu\ ) \ $$

and $$ 2 \mu\ \alpha_1 = P_1( \mu\ ) \ $$

and $$ (-\alpha_1+4 \mu\ ^2 \alpha_2) = P+2( \mu\ ) \ $$

[[media: Egm6321.f09.mtg39.djvu | Page 39-2]]
HW Continue the power series development to find $$ P_3, P_4, P_5 \ $$ and complete result to that obatined by [[media: Egm6321.f09.mtg31.djvu | Eq.(6) P.31-3]] or [[media: Egm6321.f09.mtg31.djvu  | Eq.(7) P.31-3]] END HW

2 recurrence formulas

Plan: Find $$ \frac{d}{d \mu\ } \frac{1}{ \sqrt{A}} \ $$ and $$ \frac{d}{d \rho\ } \frac{1}{ \sqrt{A}} \ $$

1) $$\frac{d}{d \rho\ } \frac{1}{ \sqrt{A}} = -\frac{1}{2}A^{\frac{-3}{2}}\frac{dA}{d \mu\ } = \frac{ \rho\ }{A^{\frac{-3}{2}}} \ $$

Recall [[media: Egm6321.f09.mtg38.djvu | Eq.(5) P.38-3]], now  $$ \frac{d}{d \mu\ } \frac{1}{ \sqrt{A}} = \frac{ \rho\ }{A^{\frac{3}{2}}} =  \ $$

Where $$ P_0( \mu\ )=1 \Rightarrow \ P_0'=0 \ $$ and $$ P_n'( \mu\ ) = \frac{d}{d \mu\ } P_n( \mu\ ) \ $$

[[media: Egm6321.f09.mtg39.djvu | Page 39-3]]
Compare Eq(4) and Eq(5)

$$ \frac{ \rho\ ( \mu\ - \rho\ )}{A^{\frac{3}{2}}} = ( \mu\ - \rho\ ) \sum_{n=1}^\infty P_n' ( \mu\ ) \rho\ ^{n} = \rho\ \sum_{n=0}^\infty P_n ( \mu\ ) n \rho\ ^{n-1} = \rho\ \sum_{n=1}^\infty P_n { \mu\ } n \rho\ ^{n-1} $$

Where n=1 in summation due to factor u being introduced

$$ \Rightarrow \ \mu\  \sum_{n=1} P_n' \rho\  - \sum_{n=1} P_n' \rho\ ^{n-1} = \sum_{n=1} P_n \mu\ \rho\  \ $$

$$ \Rightarrow \ \mu\ P_1' \rho\ - P_1 1 \rho\ + \sum_{n=2} \left [ -P_{n-1}' + \mu\ P_n'-nP_n \right ] \rho\ ^n = 0 \ $$

[[media: Egm6321.f09.mtg39.djvu | Page 39-4]]
$$ \Rightarrow \ -P_{n-1}' + \mu\ P_n' - nP_n = 0 \ $$

title: User:EGM6321.F10.TEAM1.WILKS/Mtg39


text:
User:EGM6321.F10.TEAM1.WILKS/Mtg41

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 41: Tues, 1Dec09

[[media: Egm6321.f09.mtg41.djvu | Page 41-1]]
Review for exam 2 - Historical development - Legendre functions Question: How to obtain $$ P_n \ $$ based on known $$ P_{n-1}, P_{n-2},... \ $$ ? - 2 recurring relationships. Same technique in power series. Solution: Frobenius method Question: Find a differential equation governing all $$ \left \{ P_n \right \} \ $$ ? - Legendre differential equations 2 families of homogeneous solutions: - Legendre functions= $$ \left \{ P_n \right \} \ $$ +  $$ \left \{ Q_n \right \} \ $$ $$ L_n = P_n \ $$ or $$ L_n = Q_n \ $$ Newtonian potential is solution of Laplace equation i.e., $$ \Delta\ \left ( \frac{1}{r} \right ) = 0 \ $$ $$ \frac{1}{r} = \frac{1}{r_{PQ}} = \frac{1}{r_Q} \left ( 1-2 \mu\ \rho\ + \rho\ ^2 \right )^{- \frac{1}{2}} \ $$

[[media: Egm6321.f09.mtg41.djvu | Page 41-2]]
$$ = \frac{1}{r_Q} \sum_{n} P_n ( \mu\ ) \rho\ ^n \ $$, where $$ = \rho\ := \frac{r_P}{r_Q} \ $$ $$ \Rightarrow \ \frac{1}{r} = \sum_{n} P_n ( \mu\ ) \frac{r_P^n}{r_Q^{n+1}} = \sum_{n} H_n \left ( \mu\, r_P, r_Q \right ) \ $$ , where $$ H_n \left ( \mu\ , r_P, r_Q \right ) = H_n \left ( x,y,z \right ) \ $$ $$ \Delta\ \frac{1}{r} = 0 = \sum_{n} \Delta\ H_n(x,y,z) \ \Rightarrow \ \Delta\ H_n = 0$$ Where this argument is based on the power series Laplace equations in a sphere axisymmetrical case [[media: Egm6321.f09.mtg29.djvu |  P.29-1 ]] separation of variables [[media: Egm6321.f09.mtg30.djvu | P.30-1]] General solution of axisymmetrical Laplace equations in a sphere $$ \psi\ (r, \theta\ ) = \sum_{n} (A_nr^n+B_nr^{n+1})(C_nP_n+D_nQ_n) \ $$ Where $$ A_n r^n + B_n r^{n+1} \ $$ can be found on [[media: Egm6321.f09.mtg31.djvu | P.31-2]] and $$ C_n P_n + D_n Q_n \ $$ can be found on [[media: Egm6321.f09.mtg32.djvu | P.32-1]] and $$ \mu\ = \sin \theta\ \ $$

title: User:EGM6321.F10.TEAM1.WILKS/Mtg41


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg15

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010=

Mtg 15: Thu, 23 Sep 10

[[media: 2010_09_23_14_52_54.djvu | Page 15-1]]
SC-L1-ODE-CC:

HW Use [[media:2010_09_23_13_57_17.djvu  | Eq.(1) and Eq.(2) p.14-3]], together with Eq.(1) to obtain [[media:2010_09_23_13_57_17.djvu | Eq.(4)p.14-1, p.14-2]] END HW

= Example: Roll control of rocket - prevent rolling by activating ailerons =

Bryson & Ho 1975 p.169; F10 Lecture plan $$ \delta\ = \ $$ aileron angle (deflection) $$ \phi\ = \ $$ roll angle $$ \omega\ = \ $$ roll angle velocity $$ Q = \ $$ aileron effectiveness $$ \tau\ = \ $$ roll time constant

[[media: 2010_09_23_14_52_54.djvu | Page 15-2]]
HW  Put Eq.(2) thru Eq.(4) p.15-1 in form of Eq(1)p.15-1 END HW

= 2nd order ODEs (cont'd from [[media:2010_09_16_14_59_27.djvu | Page 12-4]]) =

General N2-ODE: $$ F(x,y,y',y'')=0 \ $$

$$ F(.)=0 \ $$ is exact means $$ \exists \ \phi\ (x,y,y') =k \ $$ such that $$ F(x,y,y',y'') = \frac{d}{dx} \phi\ (x,y,y') \ $$

Define $$ p := y' \ $$ and thus $$ y'' = p' \ $$.

where Eq.(2) is equivalent to the first condition of exactness.

[[media: 2010_09_23_14_52_54.djvu | Page 15-3]]
2nd condition of exactness:

If both exactness conditions are satisfied, then from Eq.(4)p.15-2, we have where $$ h(x,y) \ $$ is a function of integration determined using Eq.(3) p.15-2.

Application:

Consider $$ x(y')^2+yy'+(xy)y''=0 \ $$ Where $$ x(y')^2+yy' = g(x,y,y') \ $$ and $$ (xy) = f(x,y,y') \ $$ and $$ y'=p \ $$

1st exact condition (1) is satisfied.

2nd exact condition (2) is also satisfied (F09 HW; see [[media: egm6321.f09.mtg10.djvu | F09 Mtg 10, p10-3]]).

NOTE:

For lectures on the exact conditions for N2-ODE, see: F09 Lecture plan, [[media: egm6321.f09.mtg10.djvu | Mtg 10]], [[media: egm6321.f09.mtg11.djvu | Mtg 11]], [[media: egm6321.f09.mtg12.djvu | Mtg 12]], [[media: egm6321.f09.mtg13.djvu | Mtg 13]], [[media: egm6321.f09.mtg14.djvu | Mtg 14]].

ENDNOTE

$$ f_x = y \Rightarrow \ f_{xx}=0 \ $$

[[media: 2010_09_23_14_52_54.djvu | Page 15-4]]
$$ g_y=p \Rightarrow \ g_{yp}=1 \ $$ Eq.(3)p15-3 $$ \Rightarrow \ \ $$ where $$ \int_{}^{} (xy)\, dp=xyp \ $$ using definition of g in Eq(3)p.15-2:

title: User:EGM6341.s11.TEAM1.WILKS/Mtg15


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg13

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010= Mtg 13: Tue, 21 Sep 10

[[media: 2010_09_21_14_56_48.djvu | Page 13-1]]
= Application (engineering): Motion of particle (e.g.,rocket) in air =

where $$ v := \left \Vert \boldsymbol v \right \| \ $$, norm = magnitude $$ k,n \in \mathbb R \ $$ $$ m= \ $$ mass of the particle $$ g= \ $$ acceleration of gravity HW  1) Derive equation of motion Where Eq.(1) and Eq.(2) are a system of coupled N1_ODEs = SC_N1_ODE (numerical methods) 2) Particular case $$ k=0 \ $$ : verify $$ y(x) \ $$ is parabola

[[media: 2010_09_21_14_56_48.djvu | Page 13-2]]
3) Consider $$ k \ne 0 \ $$, $$ v_{x0}=0 \ $$ 3.1) Find $$ v_y(t), y(t) \ $$ for $$ m=constant \ $$ 3.2) $$ m=m(t) \ $$

Is Eq.(1) "exact" (= either exact or can be made exact by integrating factor method) (IFM) END HW

= Application: Control Engineering - Linear Systems =

1) Time invariant 2) Time variant: Where $$ \mathbf{\dot{x}}(t) \ $$ is a nx1 matrix, $$ \mathbf{A} (t) \ $$ is a nxn matrix, $$ \mathbf{x}(t) \ $$ is a nx1 matrix, $$ \mathbf{B}(t) \ $$ is a nxm matrix, $$ \mathbf{u} (t) \ $$ is a mx1 matrix

[[media: 2010_09_21_14_56_48.djvu | Page 13-3]]
Time invariant $$ \mathbf{A}, \mathbf{B} \ $$ constant [[media: 2010_09_21_14_56_48.djvu | Eq.(2)P.13-2]] : SC-L1-ODEs EXAMPLE Application (Engineering)

END EXAMPLE HW 1) Derive 2) Write Eq.(1) and Eq.(2) in form of [[media: 2010_09_21_14_56_48.djvu | Eq.(2)P.13-2]] $$ \mathbf{x}:= \left \lfloor \theta_1 \ \dot{ \theta_1 } \ \theta_2 \ \dot{ \theta_2 }    \right \rfloor \ ^T $$ Where $$ \mathbf{x} $$ is a 4x1 matrix Find $$ \mathbf{A}, \mathbf{B} , \mathbf{u}:= \begin{Bmatrix} u_1l \\ u_2l \end{Bmatrix} \ $$ Where $$ \mathbf{A} $$ is a 4x4 matrix $$ \mathbf{B} $$ is a 4x2 matrix $$ \mathbf{u} $$ is a 2x1 matrix END HW

title: User:EGM6341.s11.TEAM1.WILKS/Mtg13


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg11

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010= Mtg 11: Thu, 24 Aug 10

[[media: 2010_09_16_13_50_21.djvu | Page 11-1]]
[[media: 2010_09_14_15_00_52.djvu | HW P.10-4 (continued)]] 2) Assume $$ a_1(x) \ne 0 \forall \ x \ $$, [[media: 2010_09_14_15_00_52.djvu | Eq(8) P.10-3 ]] becomes $$ y'+\frac{a_0(x)}{a_1(x)}y=\frac{b(x)}{a_1(x)} \ $$ Where $$ \frac{a_0(x)}{a_1(x)}= P(x) \ $$ and $$ \frac{b(x)}{a_1(x)} = Q(x) \ $$ from [[media: http://uf.catalog.fcla.edu/uf.jsp?Ntt=king+differential+equations&I=1&Submit=Find&N=20&S=0781249404236215&Ntk=Keyword&V=D&Nty=1#top | K.p.512]] Find expression for $$ y(x) \ $$ in terms of $$ a_0, a_1, b \ $$. 3) $$ a_1(x)=x^2+1 \ $$ $$ b(x)=2x \ $$ $$ a_0(x)=x \ $$ END HW NOTE: [[media: http://uf.catalog.fcla.edu/uf.jsp?Ntt=king+differential+equations&I=1&Submit=Find&N=20&S=0781249404236215&Ntk=Keyword&V=D&Nty=1#top | cf. to K.p.512]] 1) [[media: http://uf.catalog.fcla.edu/uf.jsp?Ntt=king+differential+equations&I=1&Submit=Find&N=20&S=0781249404236215&Ntk=Keyword&V=D&Nty=1#top | K. etal.]] did not derive expression [[media: 2010_09_14_15_00_52.djvu | Eq.(1)p.10-3]] $$ h(x)=e^{\left [ \int_{}^{x} a_0(s)\, ds \right ]} \ $$ "pulling rabbit out of hat" 2) $$ \oint_{}^{x} f(s)\, ds:= \int_{}^{x} f(s)\, ds \ $$ in [[media: http://uf.catalog.fcla.edu/uf.jsp?Ntt=king+differential+equations&I=1&Submit=Find&N=20&S=0781249404236215&Ntk=Keyword&V=D&Nty=1#top | K.2003]] $$ \equiv \int f(x)\, dx$$ without constant

[[media: 2010_09_16_13_50_21.djvu | Page 11-2]]
Lecture: $$ \int_{}^{x} f(s)\, ds = \int f(x)\, dx+k= \oint_{}^{x} f(s)\, ds+k \ $$ [[media: 2010_09_14_15_00_52.djvu | Eq.(6)p.10-3]] :2 constants $$ k_1 \ $$ and $$ k_2 \ $$ [[media: 2010_09_14_15_00_52.djvu | Eq.(1)p.10-3]] : $$ h(x) \rightarrow \ k_1 \ $$ [[media: 2010_09_14_15_00_52.djvu | Eq.(6)p.10-3]] : $$ \int_{}^{x} h(s)b(s) \, ds \rightarrow \ k_2  \ $$ But [[media: 2010_09_14_15_00_52.djvu | Eq.(5)p.10-2]] is L1_ODE_VC HW  $$ \alpha\ \ $$ Show that $$ k_1  \ $$ is not necessary. END HW HW  $$ \beta\ \ $$ Show  [[media: 2010_09_14_15_00_52.djvu | Eq.(6)p.10-3]] agrees with [[media: http://uf.catalog.fcla.edu/uf.jsp?Ntt=king+differential+equations&I=1&Submit=Find&N=20&S=0781249404236215&Ntk=Keyword&V=D&Nty=1#top | K.p.512]], i.e. $$ y(x)=Ay_H(x)+y_P(x) \ $$ END HW HW  $$ \gamma\ \ $$ Find $$ y_H(x) \ $$ independantly, i.e. solve $$ y'+a_0y=0 \ $$ END HW $$ \delta\ \ $$ How about $$ y_P(x) \ $$ ? $$ \Rightarrow \ \ $$ Variation fo parameters (later) END NOTE

[[media: 2010_09_16_13_50_21.djvu | Page 11-3]]
= A class of exact N1-ODE =

Recall [[media: 2010_09_14_15_00_52.djvu | Eq.(7)p.10-1]] (Case 1) One possibility to satisfy this condition: Consider:

Where Eq(6) is a L1-ODE-VC (not necessarily exact, but can be made exact: integrating factor method) APPLICATION Consider $$ a(x)= x^4 \ne \ b(x)= x \Rightarrow \ \bar b \ (x) = \frac{1}{2}x^2 \ $$ $$ k=10 \ $$

F09: Find $$ h(x) \ $$ such that Eq.(7) is exact END APPLICATION

[[media: 2010_09_16_13_50_21.djvu | Page 11-4]]
Question: But [[media: 2010_09_16_13_50_21.djvu | Eq.(6)p.11-3]] is linear! Find N1-ODEs that are exact or can be made exact by integrating factor method.

title: User:EGM6341.s11.TEAM1.WILKS/Mtg11


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg9

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010= Mtg 9: Thu, 9 Sep 10

[[media: 2010_09_09_14_48_41.djvu | Page 9-1]]
NOTE: Smooth functions are functions that are infinitely differentiable, e.g., $$ \cos x, e^x, \log x,x,x^2 \ $$ Non-Smooth:

The derivative is a generalized function: $$ \frac{dH (x - \hat x)}{dx} = \delta\ (x-\hat{x}) \ $$

Continuity of $$ f(x) \ $$ at $$ \hat{x} \ $$ : $$ \lim_{x \to \hat{x} ^-}f(x) = \lim_{x \to \hat{x} ^+}f(x) \ $$

Derivative in the generalized sense.

[[media: 2010_09_09_14_48_41.djvu | Page 9-2]]
Hat function not differentiable in classical sense at $$ x_1, x_2, x_3 \ $$ since $$ \lim_{x \to x_i ^-} \frac{df(x)}{dx} \ne \lim_{x \to x_i ^+} \frac{df(x)}{dx} \ $$ for $$ i=1,2,3 \ $$ limit on left of $$ x_i \ $$: $$ \lim_{x \to x_i ^-} \frac{df(x)}{dx} \ $$ limit on right of $$ x_i \ $$: $$ \lim_{x \to x_i ^+} \frac{df(x)}{dx} \ $$ smooth $$ \equiv \ $$ no kinks END NOTE HW  $$ M(x,y) + N(x,y) f(y')=0 \ $$, Find $$ f(y') \ $$ such that there is no analytical solution to $$ f(y')=-\frac{M}{N} \ $$ (i.e., such N1_ODE cannot be exact) END HW SUMMARY: 2 conditions of exactness 1) [[media: 2010_09_09_13_54_49.djvu | Eq.(3)P.8-2]] 2) [[media: 2010_09_09_13_54_49.djvu | Eq.(2)P.8-4]] END SUMMARY [[media: 2010_09_07_14_59_22.djvu | Ex. P.7-3]], [[media: 2010_09_09_13_54_49.djvu | P.8-2]]: [[media: 2010_09_09_13_54_49.djvu | Eq.(3)P.8-2]]

[[media: 2010_09_09_14_48_41.djvu | Page 9-3]]
Verify 2 conditions of exactness: 1) Condition 1: [[media: 2010_09_09_13_54_49.djvu | Eq.(3)P.8-2]] : where $$ M(x,y)=75x^4 \ $$ and $$ N(x,y)= \cos y \ $$ 2) Condition 2: $$ \Rightarrow \ \ $$ [[media: 2010_09_09_13_54_49.djvu | Eq.(3)P.8-2]] is exact. Suppose $$ \phi\ \ $$ can be found to be [[media: 2010_09_07_14_59_22.djvu | Eq.(4)P.7-3]], i.e., a reduced order equation $$ \phi\ (x,y)=k \ $$. If $$ y \ $$ can be solved in terms of $$ x \Rightarrow \ \ $$ explicit solution (otherwise get implicit solution) Here: Where $$ k \ $$ is the integration constant HW  verify Eq(4) satisfies Eq(1) END HW Question: Suppose condition 1 of exactness (continued on p9-4)

[[media: 2010_09_09_14_48_41.djvu | Page 9-4]]
(continued from 9-3) is satisfied, i.e. [[media: 2010_09_09_13_54_49.djvu | Eq.(3)P.8-2]] but not condition 2 of exactness, i.e., Can we make the original N1_ODE exact? $$ \Rightarrow \ \ $$ Euler integration factor method Question: Find $$ h(x,y) \ $$ such that Eq(2) is exact.

title: User:EGM6341.s11.TEAM1.WILKS/Mtg9


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg1

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010= Mtg 1: Tue, 24 Aug 10

[[media: 2010_08_24_15_22_30.djvu | Page 1-1]]
- course website, wiki - high-speed trains German Transrapid Emsland 500 km/h, youtube, Uploaded by TransrapidSupporter on Feb 14, 2007 German transrapid (electromagentic attraction) Japanese Maglev (electrodynamic repulsion) French TGV (wheel on rail) Vu Quoc and Olsson 1989 CMAME vehicle/structure interaction, where vehicle is the high speed maglev and the structure is the the flexible guideway

[[media: 2010_08_24_15_22_30.djvu | Page 1-2]]
$$ Y^1(t)= \ $$ nominal position of wheel (w/o guideway deformation) $$ S=x^1 \ $$, horizontal coordinate $$u^1 (S,t)= \ $$ axial deformation (displacement) of guideway, where $$t \ $$ is the time parameter $$u^2 (S,t)= \ $$ transverse deformation (displacement) of guideway $$u^2_{,s}:=\frac{\partial u^2(S,t)}{\partial S} \ $$, where := means equal by definition (non symmetric) NOTE: $$ \Delta\ $$ and def are symbols (no direction) $$ A:=B \ $$ means A is defined by B

[[media: 2010_08_24_15_22_30.djvu | Page 1-3]]
$$ A=:B \ $$ means B is defined by A

Axial displacement under moving wheel/magnet

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

$$  \displaystyle \left. u^1(S,t) \right|_{S=Y^1(t)} =  u^1(Y^1(t),t) $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

where $$\displaystyle \left. u^1(S,t) \right|_{S=Y^1(t)}$$ is $$\displaystyle u^1(S,t)$$ evaluated at $$\displaystyle S=Y^1(t)$$.

General setting:

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

$$  \displaystyle \left. f(S,t) \right|_{S=Y^1(t)} =  f(Y^1(t),t) $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \frac{d}{dt}f(Y^1(t),t)= \frac{\partial f(Y^1(t),t)}{\partial S} \dot Y ^1 + \frac{\partial f(Y^1(t),t)}{\partial t} $$ (3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

where $$ \dot Y ^1 = \frac{dY^1 (t)}{dt} $$

Where $$ f_{,S}(Y^1,t)=\frac{\partial f(Y^1,t)}{\partial S} \ $$ and $$ f_{,SS}(Y^1,t)=\frac{\partial ^2f(Y^1,t)}{\partial S^2} \ $$.

title: User:EGM6341.s11.TEAM1.WILKS/Mtg1


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg3

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010= Mtg 3: Thur, 26 Aug 10

[[media: 2010_08_26_14_53_13.djvu | Page 3-1]]
NOTE: - page numbering 3-1 defined as meeting number 3, page 1 - T = torque [[media: 2010_08_24_15_22_30.djvu | Fig.p.1-1]] - HW* END NOTE [[media: 2010_08_26_13_57_13.djvu | Eq.(3)P.2-1]] : "Ordinary" Differential Equation (ODE) order = highest order of derivative Nonlinearity = What is linearity? ; use intuition for now, formal definition soon. System has 3 unknowns: $$ Y^1(t) \ $$ $$ u^1(S,t) \ $$ $$ u^2(S,t) \ $$ 1 nonlinear ODE (Ordinary Differential Equation) for $$ Y^1(t) \ $$, and 2 PDEs (Partial Differential Equations) for $$ u^1(S,t) \ $$ and $$ u^2(S,t) \ $$. 3 equations are coupled $$ \Rightarrow \ \ $$ Numerical Methods; Ref: VQ&O 1989

= Simplify for analytical solution =

2nd Order $$ \rightarrow \ \ $$ 2nd Order nonlinear $$ \rightarrow \ \ $$ linear unknown varying coefficient $$ \rightarrow \ \ $$ known varying coefficient

[[media: 2010_08_26_14_53_13.djvu | Page 3-2]]
NOTE: Math structure of coefficient  $$ c_i(Y^1,t) \ $$ for $$ i=0,1,...3 \ $$ is known, but not their values until $$ u^1 \ $$ and $$ u^2 \ $$ are known (solved for) END NOTE General structure of Linear 2nd order ODEs with varying coefficients (L2-ODE-VC) where $$ y''=\frac{d^2y}{dx^2} \ $$ $$ x= \ $$ independent variable $$ y(x)= \ $$ dependent variable (unknown function to solve for) Many linear PDEs in engineering applications can be solved by separation of variables. Some examples include, but are not limited to: Heat, solids, fluids, acoustics and electromagnetics. Examples of these types equations are: the Helmholz equation: $$ \Delta X+k^2X=0 \ $$ and the Laplace equation: $$ \Delta X=0 \ $$ [[media: Egm6321.f09.mtg28.djvu| Ref F09 Mtg.28]], [[media: Egm6321.f09.mtg29.djvu| Ref F09 Mtg.29]] , [[media: Egm6321.f09.mtg30.djvu| Ref F09 Mtg.30]]

[[media: 2010_08_26_14_53_13.djvu | Page 3-3]]
In 3-D,, a point $$\displaystyle x$$ has coordinates $$\displaystyle (x_1,x_2,x_3)$$ and is written as $$\displaystyle x=(x_1,x_2,x_3)$$.

where the lowercase $$ x \ $$ in the left-hand-side term $$ X(x) \ $$ is defined as $$ x=(x_1,x_2,x_3) \ $$ and $$ X_1(x_1)X_2(x_2)X_3(x_3) \ $$ is the separation of variables

[[media:EGM6341.s11.TEAM1.WILKS EC3.vql.svg|figure svg file]] Where $$ \xi\ \ $$ in the first term $$ X( \xi\ ) \ $$ is defined as $$ \xi\ =( \xi_1, \xi_2, \xi_3) \ $$ and $$ X_1( \xi_1)X_2( \xi_2)X_3( \xi_3) \ $$ is the separation of variables Separated equations for $$ i=1,2,3 \ $$

[[media: 2010_08_26_14_53_13.djvu | Page 3-4]]
Simplify: $$ \xi_i \rightarrow \ x \ $$ $$ X_i ( \xi_i) \rightarrow \ y(x) \ $$ $$ g_i ( \xi_i) \rightarrow \ g(x) \ $$ $$ f_i ( \xi_i) \rightarrow \ a_0(x) \ $$ [[media: 2010_08_26_14_53_13.djvu | Eq.(3)p.3-3]]: Where $$ \frac{g'(x)}{g(x)} = a_1(x) \ $$ Particular case of [[media: 2010_08_26_14_53_13.djvu | Eq.(1)p.3-2]]

= Definition: Linearity =

Let $$ F(.) \ $$ be an operator. $$ u \ $$ and $$ v \ $$ are 2 possible arguments (could be functions) of $$ F(.) \ $$ $$ F( \alpha\ u+ \beta\ v = \alpha\ F(u) + \beta\ F(v) \ $$ Where $$ \alpha\ \ $$ and $$ \beta\ \ $$ are any arbitrary number.

END Definition

title: User:EGM6341.s11.TEAM1.WILKS/Mtg3


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg5

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010= Mtg 5: Thu, 2 Sep 10

[[media: 2010_09_02_13_55_50.djvu | Page 5-1]]
L2-ODE-VC: linear 2nd order $$ \Rightarrow \ \ $$ need 2 conditions to solve for 2 constants Boundary Value Problem (BVP) Prescribe known values: Initial Value Problem (IVP) Prescribe known values: Solve IVP: 1) ODE [[media: 2010_08_31_15_01_03.djvu | Eq.(3)p.4-3]] for $$ x \in \left [ a, \infty \right [ \ $$, where left [ is defined as closed and right [ defined as open 2) Initial Condition Eq.2 Two Points: A) solution exists and is unique, existence and uniqueness emphasized. Math Proof not done here. Accept as facts.

[[media: 2010_09_02_13_55_50.djvu | Page 5-2]]
B) Superposition of solution B1) Homogeneous Solution $$ y_H(x) \ $$ B2) Particular Solution $$ y_P(x) \ $$ Homogeneous Solution: Particular Solution: Add Eq.(3) and Eq.(4) to get: Linearity of $$ L_2 \Rightarrow L_2(y_H + y_P) \ $$

[[media: 2010_09_02_13_55_50.djvu | Page 5-3]]
Existence and uniqueness of solution $$ y \ $$ in [[media: 2010_09_02_13_55_50.djvu | Eq.(2)p.5-2]], then by [[media: 2010_09_02_13_55_50.djvu | Eq.(5)p.5-2]], we have: $$ y(x)=y_H(x)+y_P(x) \ $$ Question: $$ L_2 \ $$ is second order differential operator; there are 2 constants of integration, thus we need 2 boundary conditions or 2 initial conditions to solve for these 2 constants; where are those constants ? Answer: There are actually 2 homogeneous solutons $$ y_H^1(x) \ $$ and $$ y_H^2(x) \ $$, such that $$ L_2(y_H^1)=L_2(y_H^2)=0 \ $$; in addition, $$ y_H^1 \ $$ and $$ y_H^2 \ $$ are linearly independent. EXAMPLE : $$ \sin x \ $$ and $$ \cos x \ $$ are examples of 2 linearly independent functions. END EXAMPLE $$ y_H \ $$ is a linear combination of $$ y_H^1 \ $$ and $$ y_H^2 \ $$: $$ y_H=cy_H^1+dy_H^2 \ $$, where $$ c,d \in \mathbb R \ $$ 2 parameters to fit 2 boundary conditions or 2 initial conditions $$ y(x)=cy_H^1+dy_H^2+y_P(x) \ $$, where $$ y_H(x)=cy_H^1+dy_H^2 \ $$

[[media: 2010_09_02_13_55_50.djvu | Page 5-4]]
HW Boundary Value Problem: $$ y(a)= \alpha\ \ $$, $$ y(b)= \beta\ \ $$ , find $$ c,d \ $$ in terms of $$ \alpha\ , \beta\ \ $$ END HW EXAMPLE Legendre Differential Equation [[media: http://uf.catalog.fcla.edu/uf.jsp?Ntt=king+differential+equations&I=1&Submit=Find&N=20&S=0781249404236215&Ntk=Keyword&V=D&Nty=1#top | (K.p.31)]]

title: User:EGM6341.s11.TEAM1.WILKS/Mtg5


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg7

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010= Mtg 7: Tue, 7 Sept10

[[media: 2010_09_07_14_59_22.djvu | Page 7-1]]
NOTE Where $$ z \ $$ is a dummy integration variable. Where $$ s \ $$ is a dummy integration variable and $$ y(a) \ $$ is an integration constant. Where $$ k \ $$ is an integration constant. END NOTE [[media: Egm6321.f09.mtg5.djvu| Ref F09 Mtg.5]] for details of Eq.(3) and Eq.(4) Euler integration factor method (continued) From [[media: 2010_09_02_14_58_46.djvu | Eq.(5)p.6-3]]: In general N1_ODE HW  Verify the above statement and give example for L1_ODE END HW Application:

HW Verify Eq.(7) is N1_ODE END HW

[[media: 2010_09_07_14_59_22.djvu | Page 7-2]]
NOTE [[media: 2010_09_02_14_58_46.djvu | Eq.(5)p.6-3]] is in general a N1_ODE, but is not the most general N1_ODE (i.e. it is still a particular N1_ODE) END NOTE EXAMPLE [[media: 2010_09_02_14_58_46.djvu | Eq.(5)p.6-3]] Where $$ \bar M \ (x,y) = M(x,y)+k \ $$ and Eq.(2) has the same structure as [[media: 2010_09_02_14_58_46.djvu| Eq.(5)p.6-3]]. [[media: 2010_09_02_14_58_46.djvu | Eq.(5)p.6-3]] Where $$ \bar M \ (x,y) := M(x,y)-P(x,y) \ $$ and Eq.(3) has the same structure as [[media: 2010_09_02_14_58_46.djvu | Eq.(5)p.6-3]]. see [[media: 2010_09_09_13_54_49.djvu | p.8-2]] and [[media: 2010_09_09_13_54_49.djvu | p.8-3]] END EXAMPLE  Most general N1_ODE: DEFINITION: Exact N1_ODE - Eq(5) is an exact N1_ODE if $$ \exists \ \ $$ a function $$ \phi\ (x,y) \ $$ such that $$ F=\frac{d \phi\ }{dx} \ $$, where $$ \phi\ (x,y) = \ $$ constant k. (continued on p7-3)

[[media: 2010_09_07_14_59_22.djvu | Page 7-3]]
(continued from p7-2) i.e., END DEFINITION From Eq(1) Where $$ \frac{\partial \phi\ (x,y) }{ \partial x} = M(x,y) \ $$ and $$ \frac{\partial \phi\ (x,y) }{\partial y}=N(x,y) \ $$

NOTE If Eq(2) can be found $$ \Rightarrow \ $$ reduce order ($$ y' \ $$ no longer present) in principle, $$ y(x) \ $$ can be obtained by solving Eq(2).$$ k= \ $$ integration constant END NOTE EXAMPLE: Find $$ F(x,y,y') = 0 = \frac{d \phi\ (x,y)}{dx} \ $$ END EXAMPLE:

title: User:EGM6341.s11.TEAM1.WILKS/Mtg7


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg29

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010=

[[media: 2010_10_26_15_07_13.djvu | Mtg 29:]] Tue, 26 Oct 10

= Application: Nonhomogeneous L2-ODE-CC cont'd =

[[media: 2010_10_26_15_07_13.djvu | Page 29-1]]
[[media: 2010_10_21_14_48_38.djvu | HW5.5 p.28-3]] : continued 2.8) Now solve the nonhomogeneous L2-ODE-CC with $$ f(t)=e^{(-t^2)} \ $$, i.e., Gaussian distribution. Coefficients: $$ a_2, a_1, a_0 \ $$ : Find these coefficients such that the L2-ODE-CC [[media: 2010_10_21_14_05_59.djvu | Eq.(2)p.27-2]] accepts as characteristic equation: 2.8.1) $$ (r+1)(r-2)=0 \ $$ 2.8.2) $$ (r-4)^2=0 \ $$

END HW5.5

= Variation of parameters: Reduction-of-order method 2 (cont'd) =

Instead of doing the homogeneous case in [[media: 2010_10_21_14_48_38.djvu | Eq.(1)p.28-4]] (see F09 Mtgs [[media: Egm6321.f09.mtg17.djvu | 17]], [[media: Egm6321.f09.mtg18.djvu | 18]], [[media: Egm6321.f09.mtg19.djvu | 19]], and King ), let's consider directly the non-homogeneous case, then recover the homogeneous as a particular case.

So consider: $$ y''+a_1(x)y'+a_0(x)y=f(x) \ $$ Given: $$ u_1(x) \ $$ homogeneous solution Find: $$ u_2(x) \ $$ 2nd homogeneous solution

[[media: 2010_10_26_15_07_13.djvu | Page 29-2]]
$$ y_P(x) \ $$ particular solution

Method
where $$ U(x) \ $$ is unknown and $$ u_1(x) \ $$ is known Add the following equations: $$ \left [ y=Uu_1 \right ] \times a_0(x) \ $$ $$ \left [ y'=Uu_1'+U'u_1 \right ] \times a_1(x) \ $$ $$ \left [ y=Uu_1+2U'u_1'+U''u_1 \right ] \times a_2(x) \ $$, where $$ a_2(x)=1 \ $$ After summing: $$ a_0y+a_1y'+y= U \left [ a_0 u_1 + a_1 u'_1 + u_1 \right ] + U' \left [ a_1u_1+2u_1' \right ] +U''u_1 = f(x) \Rightarrow \ \ $$ L2-ODE-VC with missing dependent variable $$ U \ $$, because $$ \left [ a_0u_1+a_1u'_1+u_1'' \right ] = 0 \ $$ since $$ u_1 \ $$ is a homogeneous solution. Order-reduction method 0: $$ Z:=U' \ $$ $$ \tilde{a}_1(x) Z'+\tilde{a}_0(x) Z = \frac{f(x)}{u_1(x)} \ $$ where $$ \tilde{a}_1(x)=1 \ $$ and $$ \tilde{a}_0(x)= \frac{ \left [ a_1(x)u_1(x)+2u'_1(x) \right ] }{u_1(x)} \ $$.

[[media: 2010_10_26_15_07_13.djvu | Page 29-3]]
Solve for $$ Z(x) \ $$ given by [[media: 2010_09_14_15_00_52.djvu | Eq.(6)p.10-3]] with $$ h(x) \ $$ given by [[media: 2010_09_14_15_00_52.djvu | Eq.(1)p.10-3]]. [[media: 2010_09_14_15_00_52.djvu | Eq.(1)p.10-3]] : $$ h(x) = \exp \left[ \int \tilde{a}_0(x) \, dx \right] = \exp \left[ \int \left\{ a_1(x) + \frac{2u_1'(x)}{u_1(x)} \right\} \, dx \right] \ $$

and

$$ \int \frac{2u_1'(x)}{u_1(x)}\,dx = 2 \log {u_1(x)}  = \log {u_1^2(x)} \ $$

Thus

From [[media: 2010_09_14_15_00_52.djvu | Eq.(6)p.10-3]]:

$$ Z(x) = \frac{1}{h(x)} \left[ k_2 + \int h(x) \frac{f(x)}{u_1(x)} \, dx \right] \ $$ ,

Since

$$ Z(x)=U'(x) \ $$

we have

$$ U(x) = k_1+ \int_{}^{} Z(x)\, dx \ $$

From [[media: 2010_10_26_15_07_13.djvu | Eq.(1)p.29-2]]: $$ y(x) = U(x)u_1(x) \ $$

where $$ u_2(x) = u_1(x) \int \frac{1}{h(x)} \, dx \ $$.

title: User:EGM6341.s11.TEAM1.WILKS/Mtg29


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg27

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010=

[[media: 2010_10_21_14_05_59.djvu | Mtg 27:]] Thu, 21 Oct 10

= Application: Euler L2-ODE-VC cont'd =

[[media: 2010_10_21_14_05_59.djvu | Page 27-1]]
[[media: 2010_10_19_14_56_03.djvu | Eq.(7)p.26-4]]:

[[media: 2010_10_19_14_56_03.djvu | Eq.(8)p.26-4]]:

Define

and apply the Euler's formula [[media: 2010_10_19_14_56_03.djvu | Eq.(9)p.26-4]].

= Application: Euler L2-ODEs with multiple roots, method of variation of parameters =

HW 5.3

Consider characteristic equation:

where $$ \lambda\ \ $$ is a given number, e.g., $$ \lambda\ =5 \ $$ 1) Euler L2-ODE-VC 1.1) Find $$ a_2, a_1, a_0 \ $$ such that Eq.(4) is characteristic equation of Eq.(5) 1.2) 1st homogeneous solution: $$ y_1=x^{ \lambda\ } \ $$ 1.3) 2nd homogeneous solution: $$ y=c(x)y_1(x) \ $$, where $$ c(x)=u(x) \ $$ which is an unknown.

This is called the Euler method of variation of constants or variation of parameters.

Find the differential equation governing $$ u(x) \ $$.

[[media: 2010_10_21_14_05_59.djvu | Page 27-2]]
Solve for $$ u \ $$ then $$ y_2 \ $$ 1.4) General homogeneous solution 2) Euler L2-ODE-CC Repeat above steps for Eq.(1)

END HW 5.3 HW 5.4

Recall [[media: 2010_09_16_13_50_21.djvu | Page 11-2]], use same idea of variation of constants (parameters) to find the particular solution $$ y_P \ $$ after knowing the homogeneous solution $$ y_H \ $$, i.e., let $$ y(x)=A(x) y_H(x) \ $$.

END HW 5.4

= Application: Nonhomogeneous Euler L2-ODE-CC =

HW 5.5 Non-homogeneous L2-ODE-CC

Equation of motion: 1) Find the partial differential equation(s) governing the integrating factor $$ h(t,y) \ $$.

Reference [[media: 2010_09_23_14_52_54.djvu | Eq.(1)p.15-3]] and [[media: 2010_09_23_14_52_54.djvu | Eq.(2)p.15-3]] (condition 2 of exactness: 2 relationships)

to be continued

title: User:EGM6341.s11.TEAM1.WILKS/Mtg27


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg25

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010=

[[media: 2010_10_14_14_56_10.djvu | Mtg 25:]] Thu, 14 Oct 10

= Euler-Bernoulli beam (cont'd) =

[[media: 2010_10_14_14_56_10.djvu | Page 25-1]]
NOTE: Finite Element (FE) discretization (continued)

[[media: 2010_10_14_14_02_27.djvu | Eq.(2)p.24-2]] : $$ \mathbf{M} \mathbf{\ddot{d}} + \mathbf{K} \mathbf{d} = \mathbf{F} \ $$

the origin of each term is given below:

$$ \mathbf{M} \mathbf{\ddot{d}} \leftrightarrow m \frac{\partial ^2u}{\partial t^2} \ $$

$$ \mathbf{K} \mathbf{d} \leftrightarrow +EI \frac{\partial ^4u}{\partial x^4} \ $$ where the positive sign in front of $$ EI \ $$ is changed twice due to 2 integration by parts

$$ \mathbf{F} \leftrightarrow f \ $$

END NOTE

Free vibration
Back to free vibration: [[media: 2010_10_14_14_02_27.djvu | Eq.(1)p.24-3]]

Separation of variables; consider the trial solution of the form

Assume:

where $$ i^2=-1 \ $$.

is an L4-ODE-CC.

Definition:

Leads to

Method of trial solution
Method of trial solution (undetermined coefficient) [http://uf.catalog.fcla.edu/uf.jsp?Ntt=king+differential+equations&I=1&Submit=Find&N=20&S=0781249404236215&Ntk=Keyword&V=D&Nty=1#top Ref:K. p513 Sec A5.3]

Assume

where r is an undetermined coefficient; r=root of a characteristic equation.

Use Eq.(1) in Eq.(6) p.25-1:

$$ K^4r^4=1 \Rightarrow \ r^4=\frac{1}{K^4} \ $$

$$ r_{1,2} = \pm \ \frac{1}{K} \ $$

$$ r_{3,4} = \pm \ \frac{i}{K} \ $$, where $$ i= \sqrt{-1} \ $$

HW  (4.3): Find $$ X(x) \ $$ in terms $$ \sin , \cos  ,\sinh  , \cosh   \ $$ where $$ e^{i \theta\ } = \cos \theta\ + i \sin \theta\ \ $$ similar relationship for $$ \sinh \theta\ \ $$ and $$ \cosh \theta\ \ $$ END HW

= Generalization: Euler equation, special Ln-ODE-VC =

Generalization: Euler equation; special Homogeneous Ln-ODE-VC and $$ \equiv \ \ $$ Euler Ln-ODE-VC

[[media: 2010_10_14_14_56_10.djvu | Page 25-3]]
where $$ y'=y^{(1)} \ $$ and $$ y=y^{(0)} \ $$, and the ODE is homogeneous because its r.h.s. is equal to 0.

Euler equation with constant coefficient

is an Euler Ln-ODE-CC (which includes Eq.(6)p.25-1 as particular case).

Two methods for solving Eq.(2).

Method 1
Method 1:

Stage 1: Transformation of variables

$$ x=e^t \Rightarrow \ \ $$ Ln-ODE-CC

Stage 2: Trial Solution $$ e^{rt} \ $$

Method 2
Method 2: Trial solution $$ x^r \ $$

Application: Homogeneous L2-ODE-VC (Euler L2-ODE-VC)
Method 1:

Stage 1: Let

[[media: 2010_10_14_14_56_10.djvu | Page 25-4]]
current variable $$ x \rightarrow \ \ $$ new variable $$ t  \ $$

$$ y(x(t)) \Rightarrow \ \frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt} \ $$

Using

$$ \frac{dy}{dt} =: y_t \ $$

$$ \frac{dy}{dx} =: y_x \ $$

and

$$ \frac{dx}{dt} = e^t \ $$

we obtain

Another way:

$$ y_x= \frac{dy}{dx}=\frac{dy}{dt} \frac{dt}{dx} = \left ( \frac{dt}{dx} \frac{d}{dt} \right ) y \ $$

where $$ \left ( \frac{dt}{dx} \frac{d}{dt} \right ) = \frac{d}{dx} \ $$

$$ \frac{dy}{dx} =: y_x \ $$

$$ \frac{dy}{dt} =: y_t \ $$

$$ \frac{dt}{dx} = \left ( \frac{dx}{dt} \right )^{-1} \ $$

Then $$ y_x = y_t \underbrace{\left ( \frac{dx}{dt} \right )^{-1}}_{\displaystyle =e^{-t}} \ $$

Hence:

$$ y_{xx}= \frac{d^2y}{dx^2} = \frac{d}{dx} \left ( \frac{d}{dx}y \right ) = \frac{dt}{dx} \frac{d}{dt} \left ( \frac{dt}{dx}\frac{d}{dt}y \right ) \ $$

with

$$ \frac{dt}{dx} = e^{-t} \ $$

and

$$ \frac{d}{dt}y =: y_t \ $$

and

$$ \frac{d}{dt} ( e^{-t} y_t ) = - e^{-t}y_t+e^{-t}y_{tt} \ $$

we obtain:

title: User:EGM6341.s11.TEAM1.WILKS/Mtg25


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg23

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010=

[[media: 2010_10_12_15_02_11.djvu | Mtg 23:]] Tue, 12 Oct 10

= Lecture plan, contents, servers =

[[media: 2010_10_12_15_02_11.djvu | Page 23-1]]
-Lecture plan, table of contents: Course organization; References specific to sections F09 Apollo server: http://clesm.mae.ufl.edu/~vql/courses/pea1/2009.fall/ clesm.ippd: http://clesm.ippd.ufl.edu/~vql/courses/pea1/2009.fall/

= Soccer physics =

NOTE: Soccer physics: sport projectiles $$ \Rightarrow \ \ $$ must account for air resistance: [[media: 2010_09_21_14_56_48.djvu | see Mtg 13]]

Drag force: $$ F_D= \rho\ \frac{U^2}{2}AC_D \ $$ $$ \rho\ = \ $$ air mass density $$ A = \ $$ area of cross section $$ C_D = \ $$ drag coefficient

[[media: 2010_10_12_15_02_11.djvu | Page 23-2]]
$$ D = \ $$ diameter of ball $$ \nu\ = \ $$ kinematic viscosity $$ \mu\ = \ $$ dynamic/kinetic viscosity

END NOTE

= Exact Nn-ODEs [[media: 2010_10_08_17_07_22.djvu | p.22.5]] cont'd =

Case $$ n = 2 \ $$ : N2_ODE

[[media: 2010_10_12_15_02_11.djvu | Page 23-3]]
Eq.(1)p.23-2 is equivalent to 2 relationships for exactness of condition 2 for N2-ODEs: See F09 [[media: Egm6321.f09.mtg13.djvu | Mtg 13]] [[media: Egm6321.f09.mtg14.djvu | Mtg 14]] for the details on the homework problem statement. Below is a detailed explanation.

where $$ p=y' \ $$ $$ q=y'' \ $$ $$ \phi_x + \phi_y p= g(x,y,p) \ $$ $$ \phi_p = f(x,y,p) \ $$

where $$ q=y'' \ $$

where $$ q=y'' \ $$

[[media: 2010_10_12_15_02_11.djvu | Eq.(1)p.23-2]] $$ \Rightarrow \ \ $$

Define

$$\bar g := f_{xx}+2pf_{xy}+p^2f_{yy}-g_{px}-pg_{py}+g_y \ $$

$$\bar f := (f_{xp}+pf_{yp}+2f_y-g_{pp})\ $$

[[media: 2010_10_12_15_02_11.djvu | Page 23-4]]
[[media: 2010_10_12_15_02_11.djvu | Eq.(6)p.23-3]] can be rewritten as

$$ \bar g \cdot 1 + \bar f \cdot q = 0 \ $$

For [[media: 2010_10_12_15_02_11.djvu | Eq.(6)p.23-3]] to hold, the independence of $$ 1 $$ and $$ q $$ (since $$ q = y'' $$ is not a constant in general) implies that $$ \bar g =0 \ $$ and $$ \bar f =0 \ $$.

Recall $$ \phi\ (x,y,p) \ $$

title: User:EGM6341.s11.TEAM1.WILKS/Mtg23


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg17

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010=

[[media: 2010_09_29_07_17_06.djvu | Mtg 17:]] Wed, 29 Sep 10

[[media: 2010_09_29_07_17_06.djvu | Page 17-1]]
Recall: [[media: 2010_09_28_15_08_25.djvu | Eq.(1)p.16-1]] :

Find $$\displaystyle h$$.

[[media: 2010_09_28_15_08_25.djvu | p.16-1]]: If $$ h=k_1 \ $$ (constant) $$ \Rightarrow \ \ $$ Eq.(1) is satisfied, i.e., h = constant is one solution  of Eq.(1). HW Question: Without assuming a priority that h=constant, discuss the search for the solution of Eq.(1). END HW

= Application: Generate exact N2-ODEs (reverse engineering) =

Method:

1) Select:

where $$ p=y' \ $$.

Then

2) Differentiate $$\displaystyle \phi(x,y,p) = k$$ obtain an exact N2-ODE:

End Method

Examples
Example 1) $$ \phi\ (x,y,p) = 4x^5p^2+2x^2y \ $$, see [[media: Egm6321.f09/Lecture_plan | F09]] for details Example 2) $$ \phi\ (x,y,p) = 3p^5 \cos x^2+2xy^3=k \ $$ $$ \phi_x = 3p^5 (-\sin x^2) (2x)+2y^3 \ $$ $$ \phi_y = 6xy^2 \ $$

[[media: 2010_09_29_07_17_06.djvu | Page 17-2]]
N2-ODE (exact): Eq.(3)p.17-1 $$ \Rightarrow \ \ $$ [[media: 2010_09_23_14_52_54.djvu | Eq.(2)p.15-2]], [[media: 2010_09_23_14_52_54.djvu| Eq.(3)p.15-2]] , [[media: 2010_09_23_14_52_54.djvu | Eq.(4)p.15-2]]

where $$ \phi_p = f \ $$ and $$ \phi_yy'+ \phi_x =g \ $$

where $$ (15p^4 \cos x^2) = \phi_p \ $$ and $$ (6xy^2) = \phi_y \ $$ and $$ \left [ -6xp^5 \sin x^2+2y^3 \right ] = \phi_x \ $$

Show that the above N2-ODE (Eq.(3)) is exact.

Exact condition 1:

$$ f= \phi_p=15p^4 \cos x^2 \ $$

$$ g= \phi_y y'+ \phi\ _x = (6xy^2)y'+ \left [ -6xp^5 \sin x^2 + 2y^3 \right ] \ $$

HW Exact condition 2: END HW

[[media: 2010_09_29_07_17_06.djvu | Page 17-3]]
Find first integral $$ \phi\ \ $$ :

since $$ f(x,y,p) = 15p^4 \cos x^2\ $$, and $$ h(x,y) \ $$ can be found using the definition of $$ g \ $$. $$ g= \phi_yy'+ \phi_x \ $$, with $$ y'=p \ $$

2 choices

title: User:EGM6341.s11.TEAM1.WILKS/Mtg17


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg19

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010=

[[media: 2010_09_30_15_51_32.djvu | Mtg 19:]] Thu, 30 Sep 10

[[media: 2010_09_30_15_51_32.djvu | Page 19-1]]
= Review =

= Non-exact L2-ODE-VC =

APPLICATION: [[media: Egm6321.f09/Lecture_plan | F09]]: Verify Exactness of Eq.(1). END APPLICATION

Recipe
For some special power forms (not $$ \sin x, \cos x ... \ $$) N1-ODE: $$ x^ay^b(py+qxy')+x^cy^d(ry+sxy') \ $$ Where $$ (a,b,c,d) \ $$ and $$ (p,q,r,s) \in \mathbb R \ $$ L2-ODE-VC: $$ \alpha\ x^ry''+ \beta\ x^sy'+ \gamma\ x^ty \ $$ Where $$ ( \alpha\, \beta\ , \gamma\ ) \ $$ and $$ (r,s,t) \in \mathbb R \ $$ Consider integrated factor $$ h(x,y)=x^my^n \ $$

title: User:EGM6341.s11.TEAM1.WILKS/Mtg19


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg31

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010=

[[media: 2010_10_28_14_54_20.djvu | Mtg 31:]] Thu, 28 Oct 10

= Application: King 2003 p.28 (Pb.1.1 ab) cont'd =

[[media: 2010_10_28_14_54_20.djvu | Page 31-1]]
Wolfram Alpha $$ \Rightarrow \ \ $$ $$ r_1=1 \ $$ $$ r_2=\frac{1}{x-1} \ $$ See F09 [[media: Egm6321.f09.mtg21.djvu |p.21-4]] for the problem statement on the method of finding $$ u_2(x) \ $$.

= Application: Obtaining L2-ODE-VC from characteristic equation =

Reverse engineering.

Consider two roots: $$ r_1=2 \ $$ and $$ r_2=\frac{1}{x+1} \ $$. Thus the characteristic function is

$$ \ (r-r_1)(r-r_2)=0 \ $$ $$ \Rightarrow \ r^2 - 2r - \frac{r}{x+1} + \frac{2}{x+1}=0 \ $$

So even though $$\displaystyle r_2(x) = \frac{1}{x+1}$$ is not a valid root, and since $$\displaystyle r_1 = 2 = \text{constant}$$ is a valid root, we still have $$\displaystyle y(x) = e^{r_1 x}$$ as a valid trial solution, which leads to the above L2-ODE-VC. On the other hand, the above L2-ODE-VC for sure will lead to the above characteristic equation.

The point is there must be at least one valid root, i.e., constant root, for the above procedure of generating L2-ODE-VC to be valid. In other words, the following two roots $$ r_3 = 2x \ $$ and $$ r_4 = \frac{1}{x+1} \ $$ are not valid, and cannot be used to construct an L2-ODE-VC with the characteristic function being

$$ \ (r-r_3)(r-r_4)=0 \ $$

HW5.7

Find $$ u_1, u_2 \ $$ (2 homogeneous solutions) of Eq.(1) using trial solution $$ y=e^{rx} \ $$

END HW5.7

= Application: Trial solution with singularity =

Consider: (see also F09 [[media: Egm6321.f09.mtg22.djvu | p.22-2]])

Given:

[[media: 2010_10_28_14_54_20.djvu | Page 31-2]]
1) Verify exactness of [[media: 2010_10_28_14_54_20.djvu | Eq.(2)p.31-1]] 2) Can you use the Integrating Factor Method ? 3) Trial Solution 3.1) $$ y=e^{rx} \ $$, with r=constant 3.2) $$ y=xe^{rx} \ $$, with r=constant 3.3) $$ y=\frac{e^{rx}}{x} \ $$ , with r=constant Find $$ u_1 \ $$ in [[media: 2010_10_28_14_54_20.djvu | Eq.(3)p.31-1]] 4) Find $$ u_2 \ $$ using variation of parameters. Wolfram Alpha for $$ y=\frac{e^{rx}}{x} \Rightarrow \ r_1=i \ $$ and $$ r_2=-i \ $$, where $$ r_1, r_2 \ $$ are constant

$$ \Rightarrow \ u_1=\frac{e^{ix}}{x} \ $$ $$ \Rightarrow \ u_2=\frac{e^{-ix}}{x} \ $$ $$ \Rightarrow \ \bar u \ _1 =\frac{\cos x}{x} \ $$ $$ \Rightarrow \ \bar u \ _2 =\frac{\sin x}{x} \ $$

Obtain L2-ODE-VC from a known trial solution and a characteristic equation
Reverse engineering

Add the following equations: $$ \left [ \frac{e^{rx}}{x} \right ] \times a_0(x) \ $$ $$ \left [ \frac{d}{dx} \left ( \frac{e^{rx}}{x} \right ) = \left ( r-\frac{1}{x} \right ) \frac{e^{rx}}{x} \right ] \times a_1(x) \ $$ $$ \left [ \frac{d^2}{dx^2} \left ( \frac{e^{rx}}{x} \right ) = \left ( r^2+\frac{2}{x^2}-\frac{2r}{x} \right ) \frac{e^{rx}}{x} \right ] \times a_2(x) \ $$ Obtain the following: $$ \frac{e^{rx}}{x} \left [ a_2 \left ( r^2+\frac{2}{x^2}-\frac{2r}{x} \right ) +a_1 \left ( r-\frac{1}{x} \right ) +a_0 \right ]=\frac{e^{rx}}{x}(r^2+1) =0 \ $$, where $$ r=\pm i \ $$ for the last term $$ (r^2+1) \ $$ 3 equations for $$ a_2, a_1, a_0 \ $$ $$ r^2(a_2)+r \left ( \frac{-2}{x}a_2+a_1 \right ) +1 \left ( \frac{2}{x^2}a_2-\frac{a_1}{x}+a_0 \right )=r^2+0r+1 \ $$ $$ \Rightarrow \ a_2=1 \ $$ $$ \Rightarrow \ a_1= \frac{2}{x} \ $$ $$ \Rightarrow \ a_0=1 \ $$ The previous three equations lead to $$ \Rightarrow \ y''+\frac{2}{x}y'+y=0 \ $$ which is the same as [[media: 2010_10_28_14_54_20.djvu | Eq.(2)p.31-1]]

[[media: 2010_10_28_14_54_20.djvu | Page 31-4]]
HW5.8 Find L2-ODE-VC (homogeneous) such that trial solution $$ \frac{e^{rx}}{x^2} \ $$ and such that $$ r^2+3=0 \ $$ is the characteristic equation. END HW5.8

title: User:EGM6341.s11.TEAM1.WILKS/Mtg31


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg33

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010=

[[media: 2010_11_04_13_57_39.djvu | Mtg 33:]] Thu, 4 Nov 10

= Examples: Non-homogeneous L2-ODE-VC (cont'd) =

[[media: 2010_11_04_13_57_39.djvu | Page 33-1]]
HW6.2 continued F09 F10 END HW6.2

HW6.3 Non-homogeneous Legendre Equation $$ n=1 \ $$ [[media: 2010_09_02_13_55_50.djvu | Eq.(1)p.5-4]]: [[media: 2010_09_02_14_58_46.djvu | Eq.(1)p.6-1]]

See K.p.34 for the example. Note that there was a misprint in King 2003 p.34: The excitation function should be

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

$$  \displaystyle f(x) = 1 $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

and NOT

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

$$  \displaystyle f(x) = \frac{1}{1-x^2} $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Solve Eq.(3) using direct method ([[media: 2010_10_26_15_07_13.djvu | Mtg29]] and [[media: 2010_10_28_14_00_19.djvu | Mtg30]] ; See [[media: 2010_10_26_15_07_13.djvu | Eq.(2)p.29-3]])

END HW6.3 [[media: 2010_10_28_14_00_19.djvu | Eq.(2)p.30-2]] $$ u_1(x) = x \equiv P_1(x) \ $$.

= Heat conduction on a sphere =

Transient heat equation:

where $$ \mathbf{k} \ $$ is the conductivity tensor (think matrix) For a homogeneous isotropic material: $$ \mathbf{k} = k \mathbf{I} \ $$, where $$ \mathbf{I} \ $$ is an identity matrix For steady state: $$ \frac{\partial \psi}{\partial t} =0 \ $$ For no heat source: $$ f =0 \ $$ From Eq.(4) $$ \Rightarrow \ k\ {\rm div} ( {\rm grad} \, \psi ) = 0 \ $$ ,

where

$$ {\rm div} ( {\rm grad} \, \psi ) = \nabla \cdot \nabla \psi = \nabla^2 \psi = \Delta \psi \ $$

is the Laplace operator acting on $$\displaystyle \psi$$.

[[media: 2010_11_04_13_57_39.djvu | Page 33-2]]
Repeated index of $$ x_i \ $$ in Eq.(1) is a summation convention $$ m= \ $$ space dimension (1, 2, or 3) Consider an infinitesimal segment $$\displaystyle \mathbf{ds}$$ (vector quantity)

$$ \mathbf{ds} = dx_i \mathbf{e}_i \Rightarrow \ ds^2= \mathbf{ds} \cdot \mathbf{ds} = (dx_i \mathbf{e}_i) \cdot (dx_j \mathbf{e}_j) = dx_i \, dx_j (\mathbf{e}_i \cdot \mathbf{e}_j) \ $$ where $$ \mathbf{e}_i \cdot \mathbf{e}_j = \delta_{ij} \ $$, which is the Kronecker Delta defined as follows

$$ \Rightarrow \ ds^2= dx_i dx_i = \sum_{i=1}^3 (dx_i)^2 \ $$

Spherical coordinates
Orthogonal curvilinear coordinates: Spherical coordinates $$ ( \xi_1, \xi_2, \xi_3 ) \ $$ (General curvilinear coordinates: $$ ( \xi_1, \xi_2, \xi_3 ) \ $$)

[[media: 2010_11_04_13_57_39.djvu | Page 33-3]]

Astronomy convention: $$ ( r, \theta, \varphi ) \ $$ Mathematical-physics convention: $$ ( r, \bar \theta, \varphi ) \ $$ where $$ \bar \theta := \frac{\pi}{2} - \theta \ $$ From F09: $$ \xi_1 =r \ $$ $$ \xi_2 = \varphi\ \ $$ $$ \xi_3 = \theta\ \ $$

title: User:EGM6341.s11.TEAM1.WILKS/Mtg33


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg35

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010=

[[media: 2010_11_05_07_46_02.djvu | Mtg 35:]] Fri, 5 Nov 10

= Legendre functions =

[[media: 2010_11_05_07_46_02.djvu | Page 35-1]]
NOTE:

1. Legendre polynomials $$ \{ P_i(x), i=0,1,2,... \} \ $$ : Finite series, where $$ \{ P_i(x) \} \ $$ is the family of 1st homogeneous solutions of the Legendre equation. 2. Legendre functions: $$ \begin{cases} {P_i(x)} & \mbox{ }\mbox{ polynomial} \\ {Q_i(x)} & \mbox{ } \mbox{ functions} \end{cases} \ $$ 3. Solving Laplace equation: Use the two families $$ \{ P_i \} \ $$ and $$ \{ Q_i \} \ $$ (infinite series); similar to the use of Fourier series (see also [[media: 2010_11_02_14_58_42.djvu | p.32-2]] and [[media: 2010_11_02_14_58_42.djvu | p.32-3]]).

END NOTE

= Heat conduction on a sphere (cont'd) =

[[media: 2010_11_04_14_51_31.djvu | Eq.(2)p.34-4]] : Euler L2-ODE-VC Trial solution: where $$ \lambda\ \ $$ is a constant to be determined, and $$ r \ $$ is an independent variable.

$$ R'=\frac{dR}{dr} = \lambda\ r^{ \lambda\ -1} \ $$ $$ R''= \lambda\ ( \lambda\ -1) r^{ \lambda\ -2} \ $$ [[media: 2010_11_04_14_51_31.djvu | Eq.(2)p.34-4]] $$ \Rightarrow \ r^2R''+2rR'-kR=0 \ $$

[[media: 2010_11_05_07_46_02.djvu | Page 35-2]]
Hence the characteristic equation:

 Goal:  Transform 2nd separation [[media: 2010_11_04_14_51_31.djvu | Eq.(3)p.34-4]] into Legendre equation of order n, i.e. [[media: 2010_09_02_13_55_50.djvu | Eq.(1)p.5-4]].

With $$ \cos^2 \theta = 1 - \sin^2 \theta \ $$, define $$ \mu^2 := \sin ^2 \theta \ $$, thus

Think of $$ \mu\ \ $$ as "$$\displaystyle x$$" (independent variable) for the Legendre polynomial (function).

Transform the independent variable from $$ \theta\ \ $$ to $$ \mu\ \ $$ $$ \frac{d}{d \theta\ } = \frac{d}{d \mu\ }\frac{d \mu\ }{d \theta\ } \ $$ , where $$ \frac{d \mu\ }{d \theta\ } =\cos \theta\ \ $$ from Eq.(1) $$ \Rightarrow \ \frac{1}{\cos \theta\ }\frac{d(.)}{d \theta\ } = \frac{d(.)}{d \mu\ } \ $$ , where $$ C := \cos \theta\ \ $$ [[media: 2010_11_04_14_51_31.djvu | Eq.(3)p.34-4]] : $$ \left ( \frac{1}{C}\frac{d}{d \theta\ } \right ) \left [ C^2 \left ( \frac{1}{C}\frac{d}{d \theta\ } \right ) \Theta\ \right ] = k \Theta\ \ $$

[[media: 2010_11_05_07_46_02.djvu | Page 35-3]]
where $$ (1- \mu^2) = C^2 \ $$.

is the Legendre equation [[media: 2010_09_02_13_55_50.djvu | Eq.(1)p.5-4]] if

Next step: solve for $$ \lambda, \ R(r), \ \Theta\ (\theta) \ $$.

= Heat conduction on a cylinder =

HW6.5  Circular Cylinder Coordinates (cylindrical)

Given: $$ x=r\cos \theta\ = \xi_1 \cos \xi_2 \ $$ $$ y=r\sin \theta\ = \xi_1 \sin \xi_2 \ $$ $$ z= \xi_3 \ $$ 1) Find $$ \left \{ dx_i \right \} = \left \{ dx_1, dx_2, dx_3  \right \}  \ $$

[[media: 2010_11_05_07_46_02.djvu | Page 35-4]]
In terms of $$ \left \{ \xi_j \right \} = \left \{ \xi_1, \xi_2, \xi_3  \right \}  \ $$ and  $$ \left \{ d \xi_k  \right \} \ $$. 2) Find $$ ds^2= \sum_{i} (dx_i)^2 = \sum_{k}  (h_k)^2 (d \xi_k)^2 \ $$.

Identify $$ \{ h_i \} \ $$ in terms of $$ \{ \xi_j \}  \ $$. 3) Find $$ \Delta \Psi \ $$ in cylindrical coordinates ($$ \Rightarrow \ \ $$ Bessel equation [[media: 2010_10_14_14_02_27.djvu | Eq.(2)p.24-1]])

END HW6.5 HW6.6

Find $$ \Delta \Psi\ $$ in spherical coordinates using math/physics convention of $$ (r, \bar \theta , \varphi ) = ( \xi_1, \xi_2, \xi_3 ) \ $$

END HW6.6

title: User:EGM6341.s11.TEAM1.WILKS/Mtg35


text:
User:EGM6341.s11.TEAM1.WILKS/Mtg37

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010=

[[media: 2010_11_10_14_40_12.djvu | Mtg 37:]] Wed, 10 Nov 10

= Heat conduction on a sphere (cont'd) =

[[media: 2010_11_10_14_40_12.djvu | Page 37-1]]
Heat Problem continued [[media: 2010_11_09_15_00_14.djvu | P.36-3]]

Physically meaningful boundary conditions
[[media: 2010_11_09_15_00_14.djvu | Eq.(1)P.36-2]] : $$ R_n(r) \ $$

$$ r \rightarrow \ 0, r^{-(n+1)} \rightarrow \ + \infty \ $$

$$ n \ge 0 \Rightarrow n+1 > 0 \ $$

$$ \Rightarrow \ $$ set:

[[media: 2010_11_09_15_00_14.djvu | Eq.(1)P.36-3]] : $$ \Theta_n ( \theta\ ) \ $$

$$ \theta \rightarrow \pm \frac{ \pi\ }{2} \Rightarrow \mu = \sin \theta \rightarrow \pm 1 \ $$

$$ \left | Q_n( \mu ) \right| \rightarrow + \infty \ $$

$$ \Rightarrow \ \ $$ set:

HW7.1

Plot $$ \left \{ P_0, P_1, P_2, P_3 \right \} \ $$ in Figure 1

and $$ \left \{ Q_0, Q_1, Q_2, Q_3 \right \} \ $$ in Figure 2.

Observe $$ P_n( \mu ) \ $$ and $$ Q_n( \mu ) \ $$ as $$ \mu \rightarrow \pm 1 \ $$.

END HW7.1

Series solution for homogeneous heat equation
Combine Eq.(1), Eq.(2) and [[media: 2010_11_09_15_00_14.djvu | Eq.(2)P.36-3]] :

$$ \Psi (r, \theta ) = \sum_{n=0}^\infty (A_n r^n) [ C_n P_n( \sin \theta )] =\sum_{n=0}^\infty \bar A_n r^n P_n ( \sin \theta ) \ $$ ,

where $$\bar A_n = A_n C_n \ $$ and $$ \sin \theta = \mu \ $$.

[[media: 2010_11_10_14_40_12.djvu | Page 37-2]]
which is similar to the Fourier-Legendre series (see below), since the basis functions involve the Legendre polynomials $$\displaystyle P_n(\mu)$$, with $$\displaystyle \mu = \sin \theta$$.

Fourier series: Projection on Fourier basis
NOTE: Fourier series $$ f( \theta) = 1 \cdot a_0 + \sum_{n=1}^\infty a_n \cos n \omega \theta + \sum_{n=1}^\infty b_n \sin n \omega \theta \ $$ Basis functions $$ = \left \{ 1, \cos n \omega \theta, \sin n \omega \theta \right \} \ $$ , where $$ n=1,2,3,... \ $$ linearly independent. END NOTE NOTE: Vector space $$\displaystyle \mathbb R^n$$

$$ \left \{ \mathbf b _1, \mathbf b _2 , ... , \mathbf b _n \right \} \in \mathbb R^n \ $$ basis vectors.

$$ \left \{ \mathbf{b}_1, \mathbf{b}_2 , ... , \mathbf{b}_n \right \} \ $$ are linearly independent. Consider: $$ \mathbf{v} \in \mathbb R^n \Rightarrow \ \mathbf v = \sum_{i=1}^n a_i \mathbf b  _i = a_i b_i  \ $$ Find $$ a_i \ $$ for $$ i=1,2,...,n \ $$ $$ \mathbf{v} \cdot \mathbf{b} _j = \sum_{i} a_i \mathbf{b} _i \cdot \mathbf{b} _j \ $$, for $$ j=1,2,...,n \ $$

[[media: 2010_11_10_14_40_12.djvu | Page 37-3]]
$$ \Rightarrow \left[ (\mathbf b _i \cdot \mathbf b  _j) \right] \left\{ a_i \right\} = \left\{ (\mathbf v  \cdot \mathbf b  _j) \right\} \ $$ where $$ [ (\mathbf b _i \cdot \mathbf b _j)] \in \mathbb R^{n \times n} \ $$ is the Gram Matrix $$ \mathbf \Gamma \ $$ and is n rows by n columns, $$ \left \{ a_i \right \} \ $$ is a n row by 1 column matrix and $$ \left\{ (\mathbf v \cdot \mathbf b _j) \right\} \ $$ is a n row by 1 column matrix. Theorem: $$ \left \{ (\mathbf{b}_1, \cdots, \mathbf{b}_n) \right \} \ $$ is linearly independent iff the determinant of $$ \mathbf { \Gamma\ } \ne 0 \ $$ , where iff is defined as "if and only if" $$ ( \iff ) \ $$ END NOTE Generalize to functions: Inner (scalar) product of two functions: $$ g: \left [ a,b \right ] \rightarrow \ \mathbb R \ $$ $$ h: \left [ a,b \right ] \rightarrow \ \mathbb R \ $$ where $$ \left [ a,b \right ] \ $$ is the domain and $$ \mathbb R \ $$ is the range. $$ \left \langle g,h \right \rangle = \int_{a}^{b} g(x) h(x) \, dx \ $$

Fourier-Legendre series: Projection on Legendre polynomial basis
Consider: $$ f( \theta ) = \sum_{n=0}^\infty A_n P_n(\sin \theta ) \ $$ Question: Find $$ A_n, n=0,1,...,\infty \ $$

[[media: 2010_11_10_14_40_12.djvu | Page 37-4]]
where $$ \left [ \left \langle P_n,P_m  \right \rangle \right ] \ $$ is the Gram Matrix $$ \mathbf { \Gamma\ } (P_0, P_1, P_2, \cdots ) \ $$ that has infinite rows and columns.

$$ \left \{ A_n \right \} \ $$ is an infinite row by 1 column matrix and $$ \left \{ \left \langle f, P_m \ \right \rangle \right \} \ $$ is an infinite row by 1 column matrix.

where $$ \sin \theta\ = \mu\ \ $$ Similarly for $$ \left \langle P_n,P_m \right \rangle \ $$

Orthogonality of Legendre Polynomials
where $$ \delta_{mn} \ $$ is the Kronecker delta [[media: 2010_11_04_13_57_39.djvu| Ref Eq.(2)P.33-2]] $$ \Rightarrow \ \mathbf{ \Gamma\ }(P_0, P_1,...) \ $$ is diagonal and $$ \Gamma\ ^{-1} \ $$ can be found easily

Computation of projection coefficients
Use $$ \mu\ = \sin \theta\ \ $$ as integrating variable instead fo $$ \theta\ \ $$ due to [[media:2010_11_04_14_51_31.djvu | Eq.(4)P.34-4]].

Rewrite [[media:2010_11_04_14_51_31.djvu | Eq.(3)P.34-4]] as:

$$\displaystyle \mathcal F$$ is complete, i.e., any continuous function $$\displaystyle f$$ can be expressed as an infinite series of functions in $$\displaystyle \mathcal F$$:

Similar to Fourier series, Eq.(3) is an equality, not an approximation, due to competeness of $$\displaystyle \mathcal F$$.

Example:

title: User:EGM6341.s11.TEAM1.WILKS/Mtg37