# University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg37

EGM6321 - Principles of Engineering Analysis 1, Fall 2009

Mtg 37: Thurs, 17Nov09

### Page 37-1

P.36-4 continued

$P_{n}(x)\$ $P_{1}(x)=x\$ $Q_{n}(x)\$ From P.18-1 :

 \displaystyle {\begin{aligned}Q_{1}(x)={\frac {1}{2}}xlog\left({\frac {1+x}{1-x}}\right)-1=x\tanh ^{-1}x-1\end{aligned}} (1)

HW show $Q_{1}(x)={\frac {1}{2}}xlog\left({\frac {1+x}{1-x}}\right)-1=x\tanh ^{-1}x-1\$ ref K p33 for $Q_{2},Q_{3}...\$ \displaystyle {\begin{aligned}Q_{n}(x)=P_{n}(x)\tanh ^{-1}x-2\sum _{j=1,3,5}^{J}{\frac {2n-2j+1}{(2n-j+1)j}}P_{n-j}(x)\end{aligned}} (2)

$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

### Page 37-2

 \displaystyle {\begin{aligned}\left\langle L_{n},L_{m}\right\rangle =0\end{aligned}} (1)

for $n\neq 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

$\left\langle L_{n},L_{m}\right\rangle =\left\langle P_{n},P_{m}\right\rangle \ =$ Eq.(3) P.33-1

$\left\langle L_{n},L_{m}\right\rangle =\left\langle P_{n},Q_{m}\right\rangle \ =0$ for $n\neq m\$ Proof: Legendre equation, Eq.(1) P.14-2 : 2) $\left[(1-x^{2})y'\right]'+n(n+1)y=0\$ Where

 \displaystyle {\begin{aligned}\left[(1-x^{2})y'\right]'=(1-x^{2})y''-2xy'\end{aligned}} (2)

$\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_{m}L_{n}\,dx=0\$ Where $L_{m}\left[(1-x^{2})L_{n}'\right]'=\alpha \ \$ ### Page 37-3

Integrate $\alpha \ \$ by parts:

 \displaystyle {\begin{aligned}-\int _{-1}^{1}(1-x^{2})L_{n}'L_{m}'\,dx+(n)(n+1)\left\langle L_{n},L_{m}\right\rangle =0\end{aligned}} (1)

Interchange n and m:

 \displaystyle {\begin{aligned}-\int _{-1}^{1}(1-x^{2})L_{m}'L_{n}'\,dx+(m)(m+1)\left\langle L_{m},L_{n}\right\rangle =0\end{aligned}} (2)

Eq(1)-Eq(2):

$\left[n(n+1)-m(m+1)\right]\left\langle L_{n},L_{m}\right\rangle =0$ Where $\left[n(n+1)-m(m+1)\right]\neq 0$ since $n\neq m\$ $\Rightarrow \ \left\langle L_{n},L_{m}\right\rangle =0\$ when $n\neq m\$ cf.K.p41