User:EGM6321.f09.team1.Zhichao Gong/Mtg18

Mtg 18: Thu, 1 Oct 09

page18-1

HW: Develop reduct. of order meth using diff algebraic eqs:

$${\color{blue}1)} \ y(x)=U(x) \underset{-}{+} {u}_{1}(x)$$

$${\color{blue}2)} \ y(x)=U(x)/{u}_{1}(x)$$

$${\color{blue}3)} \ y(x)={u}_{1}(x)/U(x)$$

Application: $$(1-{x}^{2}){y}^{''}-2x{y}^{'}+2y=0 \ {\color{red}(1)}$$

homog. L2-ODE-VC

Given: $${u}_{1}=x{\color{blue}={x}^{+1}} \ {\color{blue}(2)}$$

Ans: $${u}_{2}(x)=\frac{x}{2}log(\frac{1+x}{1-x})-1 \ {\color{red}(3)}$$

$${u}_{2}(x){\color{blue}\neq {x}^{-1}}$$

$${a}_{1}=\frac{-2x}{1-{x}^{2}} \ {\color{red}(4)}$$

Rem: Comp.(1) to Legendre L2-ODE-VC $$\leftarrow$$ Eq(1) p.1402

page18-2

$$ \overset{-}{{a}_{1}}(t)=\int^{}_{t}{a}_{1}(s)ds=log(1-{t}^{2})$$

Eq.(2 ) p.17-4:                            K.p.6

$${u}_{2}=x \int^{}_{x}\frac{dt}{{t}^{2}(1-{t}^{2})}$$

$$=x \int^{}_{x} \left\{\frac{1}{{t}^{2}}+\frac{1}{2(1+t)}+\frac{1}{2(1-t)} \right\}dt$$

$${\color{red}=Eq.(3)} \ {\color{blue}p.18-1}$$

HW: Find u1</SUB>(x) and u<SUB>2</SUB>(x) of Eq.(1)</P> <P>p.18-1 using 2 trial solns:</P>

$${\color{blue}1)} \ y=a{x}^{b}$$

$${\color{blue}2)} \ y={e}^{rx}$$

<P>Comp the 2 solns using b.c.'s</P> <P>y(0)=1, y(1)=2: and comp. to</P> <P>soln by reduct. of order</P> <P>method2. Plot solns with matlab</P>

page18-3

<P><U>Particular soln:</U> non-homog. L2-ODE-VC</P> <P>2 methods:</P> <P>- trial soln(undet. coeff) = not dep. on knowledge of u<SUB>1</SUB> and u<SUB>2</SUB></P> <P>- var. of parameters: (dep. on knowledge of u<SUB>1</SUB> and u<SUB>2</SUB>)</P> <P>u<SUB>1</SUB> and u<SUB>2</SUB> are homog. solns.</P> <P>Assume full soln y of non-homog L2-ODE-VC Eq(1)p.3-1</P> <P> </P>

$${\color{red}(1)} \ y(x)={c}_{1}{\color{red}(x)}{u}_{1}(x)+{c}_{2}{\color{red}(x)}{u}_{2}(x)$$

<P>Find c<SUB>1</SUB>(x) and c<SUB>2</SUB>(x) unkown</P> <P>Given u<SUB>1</SUB>(x) and u<SUB>2</SUB>(x) known</P>

$${\color{red}(2)} \ {y}^{'}=({c}^{'}_{1}{u}_{1}+{c}_{1}{u}^{'}_{1})+({c}^{'}_{2}{u}_{2}+{c}_{2}{u}^{'}_{2})$$

<P>If c<SUB>1</SUB> and c<SUB>2 </SUB>were const, then</P>

$${c}^{'}_{1}={c}^{'}_{2}=0$$

page18-4

Break thru assump:

$${\color{blue} \underset{ \overset{\neq}{0}}}{u}_{1}+{\color{blue} \underset{ \overset{\neq}{0}}}{u}_{2}=0 \ {\color{red}(1)}$$

$${\color{red}(2)} \ \underset{\Rightarrow} \ {y}^{'}={c}_{1}{u}^{'}_{1}+{c}_{2}{u}^{'}_{2} \ {\color{red}(2)}$$

$${\color{blue}Q:} \ why \ not \ {c}^{'}_{1}{u}^{'}_{1}+{c}^{'}_{2}{u}^{'}_{2}=0$$

$${y}^{}=[{c}^{'}_{1}{u}^{'}_{1}+{c}_{1}{u}^{}_{1}]+[{c}^{'}_{2}{u}^{'}_{2}+{c}_{2}{u}^{''}_{2}]$$